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\ifsolutions
\title[Math 618 (Spring 2024): Solutions~3]{Math 618 (Spring 2024,
        Phillips): Solutions to Homework~3}
\date{24~April 2024}
\else
\title[Math 618 (Spring 2024): Homework~3]{Math 618 (Spring 2024,
        Phillips): Homework~3}
\date{17~April 2024}
\fi

\begin{document}

%  \ifsolutions
%  \else
%  \vspace*{-1ex}
%  \fi

\maketitle

% \setcounter{section}{-1}

\indent

\ifsolutions
This assignment is due on Canvas on Wednesday 24~April 2024 at 9:00 pm.

Problems and all other items use two independent numbering sequences.
This is annoying, but necessary to preserve
the problem numbers in the solutions files.

Little proofreading has been done.

Some parts of problems have several different solutions.
\fi


\begin{pbm}[Problem~3
  in Chapter~10 of Rudin's book]\label{Pb_2024_3_10_3}
Suppose that $f$ and $g$ are entire functions,
and that $| f (z) | \leq | g (z) |$ for all $z \in \C$.
What conclusion can you draw?
\end{pbm}

\ifsolutions

\begin{proof}[Solution]
The conclusion is that there is a constant $c$ such that $| c | \leq 1$
and $f = c g$.

No stronger conclusion is possible,
since, if $g$ is an entire function and $c \in \C$ satisfies
$| c | \leq 1$,
then $f = c g$ is an entire function
such that $| f (z) | \leq | g (z) |$ for all $z \in \C$.

We prove the claimed conclusion.
Set $Z = \big\{ z \in \C \colon g (z) = 0 \big\}$.
There are two cases.
First, assume that $Z$ has a limit point in~$\C$.
Then $g = 0$.
Therefore also $f = 0$,
and $c = 1$ (or $c = 0$, or $c = \frac{1}{\pi}$)
will do.

Suppose now that $Z$ has no limit point in~$\C$.
Define a holomorphic function $h_0$ on $\C \SM Z$
by $h_0 (z) = f (z) / g (z)$ for all $z \in \C \SM Z$.
Then $| h_0 (z) | \leq 1$ for all $z \in \C \SM Z$.
Since the points of $Z$ are isolated,
it follows that they are all removable singularities of~$h_0$.
Therefore there exists an entire function $h$
such that $h |_{\C \SM Z} = h_0$.
By continuity,
$| h (z) | \leq 1$ for all $z \in \C$.
Therefore Liouville's Theorem
implies that there is a constant $c \in \C$
such that $h (z) = c$ for all $z \in \C$.
Clearly $| c | \leq 1$.
It follows that $f (z) = c g (z)$ for all $z \in \C \SM Z$.
By continuity,
we must have $f (z) = c g (z)$ for all $z \in \C$.
\end{proof}
\fi

The following problem counts as $1{.}5$ ordinary problems.
Most of it is in Problem~14 in Chapter~10 of Rudin's book.

\begin{pbm}[Expansion of Problem~14
  in Chapter~10 of Rudin's book]\label{Pb_2024_3_R10_14}
Let $\Om_1, \Om_2 \S \C$ be connected open sets.
Let $f \colon \Om_1 \to \C$ and $g \colon \Om_2 \to \C$ be
nonconstant functions.
Suppose that $f (\Om_1) \S \Om_2$ and that $g \circ f$
is holomorphic.
%
\begin{enumerate}
%
\item\label{I_2404_Pb_10_14_a}
If $f$ is holomorphic, can we conclude anything about~$g$?
%
\item\label{I_2404_Pb_10_14_b}
If $g$ is holomorphic, can we conclude anything about~$f$?
\end{enumerate}
%
Can you improve the situation by adding mild extra hypotheses?
\end{pbm}

\ifsolutions
\begin{proof}[Outcome]
In neither case can we conclude that the other function is holomorphic.
However, in both cases,
under reasonable additional assumptions,
we can conclude that the other function is holomorphic.
In the first case, the positive result can be interpreted
as something about $g$ in the general case.
Specifically, $f (\Om_1)$ is open (by the Open Mapping Theorem),
and we can conclude that $g |_{f (\Om_1)}$ is holomorphic.
In the second case, $f$ need not be \ct,
but, if $f$ is \ct, then $f$ is holomorphic.
\end{proof}

\begin{proof}[Counterexample for Part~(\ref{I_2404_Pb_10_14_a})]
Take $\Om_1 = \C \SM \{ 0 \}$, $\Om_2 = \C$,
$f (z) = z$ for $z \in \Om_1$, and
\[
g (z) = \left\{ \begin{array}{ll}
     z          & \hspace{3em}  z \neq 0 \\
     17 \pi^3   & \hspace{3em}  z = 0.
    \end{array} \right.
\]
Then $g (f (z)) = z$ for all $z \in \Om_1$,
but $g$ is not \ct.
\end{proof}

Requiring that $g$ be \ct\  doesn't help,
as the following example shows.

\begin{proof}[Second counterexample for Part~(\ref{I_2404_Pb_10_14_a})]
Take $\Om_1 = \{ z \in \C \colon {\operatorname{Im}} (z) > 0 \}$,
$\Om_2 = \C$, $f (z) = z$ for $z \in \Om_1$, and
\[
g (z) = \left\{ \begin{array}{ll}
     z            & \hspace{3em}  {\operatorname{Im}} (z) \geq 0 \\
 {\overline{z}}   & \hspace{3em}  {\operatorname{Im}} (z) \leq 0.
    \end{array} \right.
\]
Then $g (f (z)) = z$ for all $z \in \Om_1$,
but $g' (z)$ does not exist for $z \in \R$.
\end{proof}

However, the following proposition
implies, in particular, that $g$ is holomorphic on $f (\Om_1)$.
(The set $f (\Om_1)$ is a region:
it is open by the Open Mapping Theorem,
and is connected because it is the image of a connected
set under a \cfn.)

\begin{prp}\label{Prop:10-14-a}
Assume the hypotheses for Part~(\ref{I_2404_Pb_10_14_a}) above,
and assume in addition that $f (\Om_1) = \Om_2$.
Then $g$ is holomorphic.
\end{prp}

We do not need to assume that $g$ is \ct.

\begin{proof}[Proof of Proposition~\ref{Prop:10-14-a}]
Define $Z \S \Om_2$ by
\[
Z = \big\{ z \in \Om_2 \colon
   {\mbox{$f' (w) = 0$ for every $w \in \Om_1$ such that $f (w) = z$}}
    \big\}.
\]

We first claim that $g' (a)$ exists for $a \in \Om_2 \SM Z$.
To see this,
let $a \in \Om_2 \SM Z$.
Choose $b \in \Om_1$ such that $f (b) = a$.
Then $f' (b) \neq 0$.
Therefore there are open sets $U \S \Om_1$ and $V \S \Om_2$
with $b \in U$
such that $f |_{U}$ is a bijection from $U$ to $V$ with a
holomorphic inverse $h \colon V \to U$.
For $z \in V$,
we then have $g (z) = (g \circ f) (h (z))$.
Since $h$ and $g \circ f$ are holomorphic,
so is $g |_V$.
In particular, $g' (a)$ exists.
This proves the claim.

We next claim that every point of $Z$ is isolated in~$\Om_2$.
So let $a \in Z$.
Choose $b \in \Om_1$ such that $f (b) = a$.
Then $f' (b) = 0$.
Since $f$ is not constant and $\Om_1$ is connected,
the zeros of $f'$ are isolated,
so there is an open set $U \S \Om_1$ such that $b \in U$
and $f' (w) \neq 0$ for $w \in U \SM \{ b \}$.
The Open Mapping Theorem implies that $f (U)$ is open.
Since $f (U)$ contains no points of $Z$ except for~$a$,
the claim follows.

We finish the proof
by showing that $g$ is \ct\  at every point $a \in Z$.
This will imply that $g' (a)$ exists,
by the theorem on removability of singularities.
So let $\ep > 0$.
Choose $b \in \Om_1$ such that $f (b) = a$.
Choose $\rh > 0$ such that
$| w - b | < \rh$ implies $| g (f (w)) - g (f (b)) | < \ep$.
The Open Mapping Theorem implies that there is $\dt > 0$
such that $B_{\dt} (a) \S f (B_{\rh} (b))$.
If $| z - a | < \dt$,
then there is $w \in B_{\rh} (b)$ such that $f (w) = z$,
and then
\[
| g (z) - g (a) |
  = | g (f (w)) - g (f (b)) | < \ep.
\]
This proves continuity at~$a$.
\end{proof}

The following alternate proof of continuity of~$g$ is taken from
a student solution.
Once one has this, the rest of the problem is fairly easy.

\begin{proof}[Alternate proof of continuity of~$g$]
Let $w \in \Om_2$.
Choose $z \in \Om_1$ such that $f (z) = w$.
Following Theorem 10.32 of Rudin, choose a neighborhood $V$ of $z$,
a number $r > 0$, $m \in \N$,
and a bijective \hfn{} $k \colon V \to B_r (0)$,
such that for all $y \in V$ we have $f (y) = w + k (y)^m$.


We prove sequential continuity of $g$ at~$w$.
Since $f (V)$ is open, it is enough to let $(w_n)_{n \in \N}$
be a sequence in $f (V)$ such that $w_n \to w$,
and prove that $g (w_n) \to g (w)$.
For $n \in \N$ choose $z_n \in V$ such that $f (z_n) = w_n$.
We claim that $z_n \to z$.
To prove this, use $f (z_n) \to w$ and $f (z_n) = w + k (z_n)^m$
to deduce that $k (z_n)^m \to 0$,
from which it follows that $k (z_n) \to 0$.
Continuity of $k^{-1}$ now implies $z_n \to z$, proving the claim.

Using continuity of $g \circ f$, we now see that
$g (w_n) = (g \circ f) (z_n) \to (g \circ f) (z) = g (w)$,
as desired.
\end{proof}

\begin{proof}[Counterexample for Part~(\ref{I_2404_Pb_10_14_b})]
Take $\Om_1 = \Om_2 = \C$,
take
\[
f (z) = \left\{ \begin{array}{ll}
     1          & \hspace{3em}  z \neq 0 \\
     - 1        & \hspace{3em}  z = 0,
    \end{array} \right.
\]
and take $g (z) = z^2$.
Then $g \circ f$ is the constant function~$1$,
but $f$ is not \ct.
\end{proof}

The following example shows that it does not help to assume
that $f$ is injective.

\begin{proof}[Second counterexample for Part~(\ref{I_2404_Pb_10_14_b})]
Take $\Om_1 = \Om_2 = \C$,
for $z = r e^{i \te}$ with $r \geq 0$ and $\te \in [0, 2 \pi)$
take
$f (z) =  r^{1/2} e^{i \te / 2}$,
and take $g (z) = z^2$.
Then $(g \circ f) (z) = z$ for all $z \in \C$,
but $f$ is not \ct.
\end{proof}

We now give an additional condition under which
the answer to the second part is yes.
(This is not required for a solution to the problem
as stated in Rudin's book,
but is certainly worth doing.
It is required with the extra request,
``Can you improve the situation by adding mild extra hypotheses?'')

\begin{prp}\label{Prop:10-14-b}
Assume the hypotheses for Part~(\ref{I_2404_Pb_10_14_b}) above,
and assume in addition that $f$ is \ct.
Then $f$ is holomorphic.
\end{prp}

\begin{proof}
We first claim that $g \circ f$ is not constant.
If the claim is false,
then there is $c \in \C$
such that $g (f (z)) = c$ for all $z \in \Om_1$.
Since $\Om_2$ is connected and $g$ is not constant,
the set $S = \{ z \in \Om_2 \colon g (z) = c \}$ is discrete.
Since $\Om_1$ is connected, $f$ is \ct, and $f (\Om_1) \S S$,
it follows that $f$ is constant,
contradicting the hypotheses.
This proves the claim.

Now set $Z = \{ z \in \Om_1 \colon g' (f (a)) = 0 \}$.
We claim that $Z$ is discrete.
Suppose not, and let $a \in \Om_1$ be a limit point of~$Z$.
Then there is a sequence $(z_n)_{n \in \N}$
in $Z \SM \{ a \}$ such that $\limi{n} z_n = a$.
Since $g$ is holomorphic and not constant,
$f (a)$ is an isolated point of
$\{ w \in \Om_2 \colon g' (w) = 0 \}$.
Since $f$ is \ct,
it follows that $f (z_n) = f (a)$ for all sufficiently large~$n$.
Then $g (f (z_n)) = g (f (a))$ for all sufficiently large~$n$.
It follows that $a$ is a limit point of
$\{ z \in \Om_1 \colon g (f (z)) = g (f (a)) \}$.
Since $g \circ f$ is not constant,
this contradicts the assumption that $g \circ f$ is holomorphic.
The claim is proved.

We next claim that $f |_{\Om_2 \SM Z}$ is holomorphic.
Let $a \in \Om_2 \SM Z$.
Since $g' (f (a)) \neq 0$,
there are open sets $U \S \Om_2$ and $V \S \C$ with $f (a) \in U$
such that $g |_{U}$ is a bijection from $U$ to $V$ with a
holomorphic inverse $h \colon V \to U$.
For $z \in f^{-1} (U)$,
we then have $f (z) = h ((g \circ f) (z))$.
Since $h$ and $g \circ f$ are holomorphic, so is $f |_{f^{-1} (U)}$.
In particular, $f' (a)$ exists.
This proves the claim.

We have shown that $f$ is holomorphic on the complement of
a discrete set.
Since $f$ is \ct, the theorem on removability of singularities
implies that $f$ is holomorphic.
\end{proof}
\fi

The following problem counts as $2{.}5$ ordinary problems.
It is all but the last part of Problem~25 in Chapter~10 of Rudin.
You will want the Global Cauchy Formula,
part of Theorem 10.35 of Rudin,
but I don't think anything from later is needed.
Even the Global Cauchy Formula can be avoided by a slight trick.

\begin{pbm}[Most of Problem~25
  in Chapter~10 of Rudin's book]\label{Pb_2024_3_}
Let $r_1, r_2 \in \R$ satisfy $0 < r_1 < r_2$.
Let $A$ be the annulus
\[
A = \big\{ z \in \C \colon r_1 < | z | < r_2 \big\}.
\]
%
\begin{enumerate}
%
\item\label{Sec_10_25_a}
Let $\ep > 0$ satisfy $r_1 + \ep < r_2 - \ep$.
Define closed curves $\gm_1, \gm_2 \colon [0, 2 \pi] \to A$
by
\[
\gm_1 (t) = (r_1 + \ep) e^{- i t}
\andeqn
\gm_2 (t) = (r_2 - \ep) e^{i t}
\]
for $t \in [0, 2 \pi]$.
For a holomorphic function $f$ on~$A$, prove that
if $z \in \C$ satisfies $r_1 + \ep < | z | < r_2 - \ep$, then
\[
f (z) = \frac{1}{2 \pi i}
        \left( \int_{\gm_1} \frac{ f (\zt)}{\zt - z} \, d \zt
        + \int_{\gm_2} \frac{ f (\zt)}{\zt - z} \, d \zt \right).
\]
%
\item\label{Sec_10_25_b}
Let $f$ be a \hfn{} on~$A$.
Use part~(\ref{Sec_10_25_a}) to prove that there are a \hfn{} $f_1$
on $\C \SM {\overline{B_{r_1} (0)}}$
and a \hfn{} $f_2$ on $B_{r_2} (0)$ such that $f = f_1 |_A + f_2 |_A$.
Further prove that it is possible to require
that $f_1 (z) \to 0$ as $| z | \to \I$,
and that then $f_1$ and $f_2$ are unique.
%
\item\label{Sec_10_25_c}
Use the decomposition of part~(\ref{Sec_10_25_b})
to associate to each \hfn{} $f$ on $A$ its {\emph{Laurent series}}
\[
f (z) = \sum_{n = - \I}^{\I} c_n z^n
\]
with coefficients $c_n \in \C$ for $n \in \Z$.
Prove that the coefficients $c_n$ are uniquely determined by~$f$,
and that the series converges uniformly to $f$
on compact subsets of~$A$.
%
\item\label{Sec_10_25_d}
Let $f$ be a bounded \hfn{} on~$A$.
Prove that the functions $f_1$ and $f_2$ of part~(\ref{Sec_10_25_b})
(satisfying $f_1 (z) \to 0$ as $| z | \to \I$) are bounded.
%
\item\label{Sec_10_25_e}
How much of parts (\ref{Sec_10_25_a}), (\ref{Sec_10_25_b}),
(\ref{Sec_10_25_c}), and~(\ref{Sec_10_25_d})
can be extended to the cases $r_1 = 0$, $r_2 = \I$, or both?
\end{enumerate}
%
\end{pbm}

To keep the amount of writing down, I suggest writing appropriate
steps as lemmas which can be used in part~~(\ref{Sec_10_25_e})
as well as in the earlier parts.

\bigskip

The last part of Problem~25 in Chapter~10 of Rudin
asks how much of parts (\ref{Sec_10_25_a}), (\ref{Sec_10_25_b}),
(\ref{Sec_10_25_c}), and~(\ref{Sec_10_25_d})
can be extended to regions bounded by finitely many
(more than two) circles.
I could not figure out what was intended here.
If the circles are concentric,
you don't get a region.
If even two circles are not concentric,
the situation is much more complicated
than in the rest of the problem.


\ifsolutions
\bigskip

The following convenient notation will be used
throughout the solution.

\begin{ntn}\label{N_3528_Circle}
For $r \in (0, \I)$,
we define a closed curve
$\sm_r \colon [0, 2 \pi] \to \C$
by $\sm_r = r e^{i t}$ for $t \in [0, 2 \pi]$.
Also, for $r, s \in [0, \I]$ with $r < s$,
we set
\[
A_{r, s} = \big\{ z \in \C \colon r < | z | < s \big\}.
\]
\end{ntn}

Thus, the set $A$ of part~(\ref{Sec_10_25_a})
is equal to $A_{r_1, r_2}$,
but the notation also covers the cases
of part~(\ref{Sec_10_25_e}).

\begin{rmk}\label{R_3528_Identify}
Using the notation of part~(\ref{Sec_10_25_a}),
we then have $\gm_2 = \sm_{r_2 - \ep}$.
Moreover,
$\Ran (\sm_{r_1 + \ep}) = \Ran (\gm_1)$ and
\[
\int_{\sm_{r_1 + \ep}} g ( \zt) \, d \zt
  = - \int_{\gm_1} g ( \zt) \, d \zt
\]
for every \cfn{} $g$ on $\Ran (\sm_{r_1 + \ep})$.
\end{rmk}

In view of Remark~\ref{R_3528_Identify},
the following lemma gives part~(\ref{Sec_10_25_a})
and also the corresponding part
of part~(\ref{Sec_10_25_e}).

\begin{lem}\label{L_3528_CF}
Let $r_1, r_2 \in [0, \I]$ satisfy $r_1 < r_2$.
Let $s_1, s_2 \in (0, \I)$ satisfy $r_1 < s_1 < s_2 < r_2$.
Let $f$ be a \hfn{} on $A_{r_1, r_2}$,
and let $z \in A_{s_1, s_2}$.
Then
\[
f (z) = \frac{1}{2 \pi i}
        \left( \int_{\sm_{s_2}} \frac{ f (\zt)}{\zt - z} \, d \zt
        - \int_{\sm_{s_2}} \frac{ f (\zt)}{\zt - z} \, d \zt \right).
\]
\end{lem}

\begin{proof}
By Theorem 10.11 of Rudin,
for $r \in (0, \I)$,
we have $\Ind_{\sm_r} (z) = 0$ when $| z | > r$
and $\Ind_{\sm_r} (z) = 1$ when $| z | < r$.
Therefore $\Ind_{ \sm_{s_2} - \sm_{s_1} } (z) = 0$
when $z \in \C \SM A_{r_1, r_2}$
and $\Ind_{ \sm_{s_2} - \sm_{s_1} } (z) = 1$
when $z \in \C \SM A_{s_1, s_2}$.
The result now follows from the global Cauchy formula,
Theorem 10.35 of Rudin.
\end{proof}

\begin{lem}\label{L_3601_Inner}
Let $r \in (0, \I)$,
set $S_r = \{ z \in \C \colon | z | = r \}$,
and let $h \colon S_r \to \C$ be \ct.
Then the function $g \colon B_r (0) \to \C$,
given by
\[
g (z) = \frac{1}{2 \pi i} \int_{\sm_r} \frac{ h (\zt)}{\zt - z} \, d \zt
\]
for $z \in B_r (0)$,
is \hl{} on $B_r (0)$
and has a unique representation as a power series
%
\begin{equation}\label{Eq_3601_PSR}
g (z) = \sum_{n = 0}^{\I} c_n z^n
\end{equation}
%
for $z \in B_r (0)$.
Moreover,
this series converges uniformly on compact subsets of $B_r (0)$.
\end{lem}

\begin{proof}
It follows from Theorem~10.7 of Rudin that
$f$ is representable by power series on $B_r (0)$,
which implies that $f$ is \hl{}
and implies the existence of the series~(\ref{Eq_3601_PSR}).
Uniqueness follows from uniqueness of the power series
representation of a function,
and uniform convergence on compact sets follow from
the fact that power series uniformly on compact subsets of
the open disk of convergence.
(I didn't find the last statement in Rudin,
but it was proved in the lectures.)
\end{proof}

\begin{lem}\label{L_3601_Outer}
Let $r \in (0, \I)$,
set $S_r = \{ z \in \C \colon | z | = r \}$,
and let $h \colon S_r \to \C$ be \ct.
Then the function $g \colon \C \SM {\overline{B_r (0)}} \to \C$,
given by
\[
g (z) = \frac{1}{2 \pi i} \int_{\sm_r} \frac{ h (\zt)}{\zt - z} \, d \zt
\]
for $z \in \C \SM {\overline{B_r (0)}}$,
is \hl{} on $\C \SM {\overline{B_r (0)}}$,
has a unique representation as a series
%
\begin{equation}\label{Eq_3601_LauSR}
g (z) = \sum_{n = 1}^{\I} c_n z^{-n}
\end{equation}
%
for $z \in \C \SM {\overline{B_r (0)}}$,
and this series converges uniformly
on compact subsets of $\C \SM {\overline{B_r (0)}}$.
Moreover, $g (z) \to 0$ as $| z | \to \I$.
\end{lem}

One can prove this lemma by the same method as used for
Theorem~10.7 of Rudin.
Specifically,
for $| \zt | = r$ and $| z | > r$,
write
\[
\frac{1}{\zt - z}
  = - \left( \frac{1}{\zt z} \right)
    \left( \frac{1}{\frac{1}{\zt} - \frac{1}{z}} \right).
\]
Since $\big| \tfrac{1}{z} \big| < \big| \tfrac{1}{\zt} \big|$,
the right hand side can be expanded as a power series in~$\frac{1}{z}$.
However, by some manipulation,
one can reduce to Lemma~\ref{L_3601_Inner}.
(The resulting argument isn't really shorter.)
However, the alternate solution below is shorter.

\begin{proof}[Proof of Lemma~\ref{L_3601_Outer}]
Define a \cfn{} $k \colon S_{1/r} \to \C$
by $k (\zt) = - \tfrac{1}{\zt} h \big( \tfrac{1}{\zt} \big)$
for $\zt \in S_{1/r}$.
Apply Lemma~\ref{L_3601_Inner} with $k$ in place of $h$
and with $\tfrac{1}{r}$ in place of~$r$,
getting a \hfn{} $f \colon B_{1/r} (0) \to \C$
with a power series representation
%
\begin{equation}\label{Eq_3601_FromPSR}
f (z) = \sum_{n = 0}^{\I} b_n z^n
\end{equation}
%
which converges uniformly on compact subsets of $B_{1/r} (0)$.

We claim that for all $z \in \C \SM {\overline{B_r (0)}}$,
we have
\[
\frac{1}{2 \pi i} \int_{\sm_{1/r}} \frac{ k (\zt)}{\zt - z} \, d \zt
  = \left( \frac{1}{z} \right) f \left( \frac{1}{z} \right).
\]
Given the claim, it follows that $g$ is \hl{} on $\C \SM {\overline{B_r (0)}}$
and that the series expansion~(\ref{Eq_3601_LauSR})
is valid if we take $c_n = b_{n - 1}$
for $n \in \N$.
Moreover,
as $| z | \to \I$,
we have $f \big( \tfrac{1}{z} \big) \to f (0)$
and $\tfrac{1}{z} \to 0$,
so $g (z) \to 0$.
Finally, if $K \S \C \SM {\overline{B_r (0)}}$ is compact,
then $L = \{ z^{-1} \colon z \in K \}$
is a compact subset of $B_{1/r} (0)$.
Since the series~(\ref{Eq_3601_FromPSR})
converges uniformly on~$L$,
it follows that the series~(\ref{Eq_3601_LauSR})
converges uniformly on~$K$ to
$g (z) = \tfrac{1}{z} f \big( \tfrac{1}{z} \big)$.
Uniqueness of the series for $g$ follows from
uniqueness of the series for~$f$.

It remains to prove the claim.
We have,
changing variables from $t$ to $- t$ at the fourth step,
\[
\begin{split}
\frac{1}{2 \pi i} \int_{\sm_r} \frac{ k (\zt)}{\zt - z} \, d \zt
& = \frac{1}{2 \pi i} \int_{0}^{2 \pi}
           \frac{ k \left( \frac{1}{r} e^{i t} \right)}{
                 \frac{1}{r} e^{i t} - z}
                  \left( \frac{1}{r} \right) i e^{i t} \, d t
   \\
& = - \frac{1}{2 \pi i} \int_{0}^{2 \pi}
           \frac{ r e^{- i t} h \left( r e^{-i t} \right)}{
                 \left( \frac{1}{r e^{-i t}} - z \right) r e^{-i t}}
                   i \, d t
   \\
& = - \frac{1}{2 \pi i} \int_{0}^{2 \pi}
           \frac{h \left( r e^{-i t} \right)}{
                 z \left( \frac{1}{z} - r e^{- i t} \right)}
                  r i e^{- i t}  \, d t
   \\
& = - \frac{1}{2 \pi i} \int_{0}^{2 \pi}
           \frac{h \left( r e^{i t} \right)}{
                 z \left( \frac{1}{z} - r e^{i t} \right)}
                   r i e^{i t} \, d t
   \\
& = \frac{1}{2 \pi i} \left( \frac{1}{z} \right) \int_{0}^{2 \pi}
           \frac{h \left( r e^{i t} \right)}{
                 r e^{i t} - \frac{1}{z}}
                   r i e^{i t} \, d t
   \\
& = \frac{1}{2 \pi i} \left( \frac{1}{z} \right) \int_{\sm_r}
           \frac{h (\zt)}{\zt - \frac{1}{z}}
                  \, d \zt
  = \frac{1}{z} f \left( \frac{1}{z} \right).
\end{split}
\]
The claim is proved.
\end{proof}

\begin{proof}[Alternate proof of Lemma~\ref{L_3601_Outer}]
It follows from Theorem~10.7 of Rudin that
$g$ is representable by power series on $\C \SM {\overline{B_r (0)}}$,
which implies that $g$ is \hl{} on $\C \SM {\overline{B_r (0)}}$.
Using the definition of the path integral, we get
\[
g (z)
 = \frac{1}{2 \pi i} \int_0^{2 \pi}
        \frac{h (r e^{i \te}) i r e^{i \te}}{r e^{i \te} - z} \, d \te.
\]
As $| z | \to \I$, the integrand converges uniformly to zero.
Therefore $\lim_{| z | \to \I} g (z) = 0$.
By the theorem on removable singularities, the formula
\[
k (z) = \begin{cases}
   0 & \hspace*{1em} z = 0
        \\
   g \left( \frac{1}{z} \right) & \hspace*{1em} 0 < | z | < \frac{1}{r}
\end{cases}
\]
defines a \hfn{} on $B_{1 / r} (0)$.
Therefore it has a power series representation
$k (z) = \sum_{n = 0}^{\I} c_n z^n$,
which converges uniformly on compact subsets of $B_{1 / r} (0)$.
Since $k (0) = 0$, we have $c_0 = 0$.
Therefore the series $\sum_{n = 1}^{\I} c_n z^{- n}$
converges uniformly to $g (z)$
on compact subsets of $\C \SM {\overline{B_r (0)}}$.
Uniqueness of the series for $g$ follows from
uniqueness of the series for~$k$.
\end{proof}

\begin{lem}\label{L_3602_OutFcn}
Let $r_1, r_2 \in [0, \I]$ satisfy $r_1 < r_2$.
Let $f$ be a \hfn{} on $A_{r_1, r_2}$.
Then there exists a \hfn{} $g$ on $A_{r_1, \I}$
such that, whenever $r \in (0, \I)$ and $z \in \C$
satisfy $r_1 < r < \min (r_2, | z |)$,
we have
\[
g (z)
 = \frac{1}{2 \pi i} \int_{\sm_r} \frac{ f (\zt)}{\zt - z} \, d \zt.
\]
\end{lem}

\begin{proof}
For $r \in (r_1, r_2)$,
Lemma~\ref{L_3601_Outer} provides
a \hfn{} $g_r \colon \C \SM {\overline{B_r (0)}} \to \C$
such that
\[
g_r (z)
 = \frac{1}{2 \pi i} \int_{\sm_r} \frac{ f (\zt)}{\zt - z} \, d \zt
\]
for $z \in \C \SM {\overline{B_r (0)}}$.

Suppose $r, s \in (r_1, r_2)$ satisfy $r < s$.
We claim that $g_r |_{\C \SM {\overline{B_s (0)}} } = g_s$.
So let $z \in \C \SM {\overline{B_s (0)}}$.
Choose $s_0 \in \R$
with $s < s_0 < \min (r_2, | z |)$.
As in the proof of Lemma~\ref{L_3528_CF},
we have $\Ind_{\sm_s - \sm_r} (w) = 0$
for all $w \not\in A_{r_1, s_0}$.
Since the function
\[
\zt \mapsto \frac{ f (\zt)}{\zt - z}
\]
is \hl{} on $A_{r_1, s_0}$,
we get
\[
g_s (z) - g_r (z)
 = \frac{1}{2 \pi i} \int_{\sm_s} \frac{ f (\zt)}{\zt - z} \, d \zt
   - \frac{1}{2 \pi i} \int_{\sm_r} \frac{ f (\zt)}{\zt - z} \, d \zt
 = \frac{1}{2 \pi i}
    \int_{\sm_r - \sm_s} \frac{ f (\zt)}{\zt - z} \, d \zt
 = 0.
\]
The claim is proved.

The claim implies that there is a well defined function
$g \colon A_{r_1, \I} \to \C$
such that $g (z) = g_r (z)$ whenever $z \in A_{r_1, \I}$
and $r_1 < r < \min (r_2, | z |)$.
It remains only to show that
$g$ is \hl.
Let $z \in A_{r_1, \I}$.
Choose $r \in \R$ such that $r_1 < r < \min (r_2, | z |)$.
Then $g$ agrees with $g_r$ on an open set containing~$z$,
and $g_r ' (z)$ exists,
so $g' (z)$ exists.
\end{proof}

\begin{lem}\label{L_3602_InnFcn}
Let $r_1, r_2 \in [0, \I]$ satisfy $r_1 < r_2$.
Let $f$ be a \hfn{} on $A_{r_1, r_2}$.
Then there exists a \hfn{} $h$ on $B_{r_2} (0)$
such that, whenever $r \in (0, \I)$ and $z \in \C$
satisfy $\max (r_1, | z |) < r < r_2$,
we have
\[
h (z)
 = \frac{1}{2 \pi i} \int_{\sm_r} \frac{ f (\zt)}{\zt - z} \, d \zt.
\]
\end{lem}

\begin{proof}
The proof is essentially the same as
the proof of Lemma~\ref{L_3602_OutFcn},
but uses Lemma~\ref{L_3601_Inner} in place of Lemma~\ref{L_3601_Outer}.
\end{proof}

The following lemma gives existence in part~(\ref{Sec_10_25_b})
and existence of the series representation
in part~(\ref{Sec_10_25_c}),
as well as the corresponding parts
of part~(\ref{Sec_10_25_e}).

\begin{lem}\label{L_3602_Decomp}
Let $r_1, r_2 \in [0, \I]$ satisfy $r_1 < r_2$.
Let $f$ be a \hfn{} on $A_{r_1, r_2}$.
Then there exist a \hfn{} $f_1$ on
$\C \SM {\overline{B_{r_1} (0)}}$ (on $\C \SM \{ 0 \}$ if $r_1 = 0$)
and a \hfn{} $f_2$ on $B_{r_2} (0)$
such that $f = f_1 |_A + f_2 |_A$,
such that $f_1 (z) \to 0$ as $| z | \to \I$,
and such that there are series representations
%
\begin{equation}\label{Eq_3602_SerReps}
f_1 (z) = \sum_{n = - \I}^{- 1} c_n z^n
\andeqn
f_2 (z) = \sum_{n = 0}^{\I} c_n z^n,
\end{equation}
%
the series for $f_1$ converging uniformly on
compact sets in $\C \SM {\overline{B_{r_1} (0)}}$
(in $\C \SM \{ 0 \}$ if $r_1 = 0$)
and the series for $f_1$ converging uniformly on
compact sets in $B_{r_2} (0)$.
\end{lem}

\begin{proof}
Let $g$ be the \hfn{} of Lemma~\ref{L_3602_OutFcn},
set $f_1 = - g$,
and let $f_2$ be the \hfn{} $h$ of Lemma~\ref{L_3602_InnFcn}.
It follows from Lemma~\ref{L_3528_CF}
that for $z \in A_{r_1, r_2}$ we have
\[
0 = h (z) - g (z) = f_1 (z) + f_2 (z).
\]

It remains to prove the existence of the series
in~(\ref{Eq_3602_SerReps}).
Existence of the series for $f_2$ follows from the fact that
$f_2$ is \hl{} on $B_{r_2} (0)$.
We prove the existence of the series for~$f_1$.
For every $r \in (r_1, r_2)$,
Lemma~\ref{L_3601_Outer} provides
unique
$c_{-1, r}, \, c_{-2, r}, \, \ldots \in \C$
such that the series $\sum_{n = 1}^{\I} c_{-n, r} z^{-n}$
converges to $f_1 (z)$ for all $z \in A_{r, r_2}$.
Moreover, the convergence is uniform on compact subsets of $A_{r, r_2}$.
Fix $r_0 \in (r_1, r_2)$.
Let $K \S A_{r_1, r_2}$ be compact.
We claim that
$\sum_{n = 1}^{\I} c_{-n, r_0} z^{-n}$
converges uniformly to $f_1 (z)$ on~$K$.
Since $K$ is compact and contained in $A_{r_1, r_2}$,
we have
$\inf_{z \in K} | z | > r_1$.
Choose $r$ such that $r_1 < r < \inf_{z \in K} | z |$ and $r < r_0$.
Then $\sum_{n = 1}^{\I} c_{-n, r} z^{-n}$
converges uniformly to $f_1 (z)$
on compact subsets of $A_{r, r_2}$.
In particular,
we have uniform convergence on $K$
and on compact subsets of $A_{r_0, r_2}$.
Uniqueness of $c_{-1, r_0}, \, c_{-2, r_0}, \, \ldots$
therefore implies that $c_{- n, r} = c_{- n, r_0}$
for all $n \in \N$.
So $\sum_{n = 1}^{\I} c_{-n, r_0} z^{-n}$
converges uniformly to $f_1 (z)$ on~$K$.
\end{proof}

Without the condition on the limit of $f_1$ at~$\I$,
the functions $f_1$ and $f_2$ are certainly not unique.
For any entire function~$g$,
one may replace $f_1$ by $f_1 + g$ and $f_2$ by $f_2 - g$.

The following lemma gives uniqueness of the series representation
in part~(\ref{Sec_10_25_c}),
as well as the corresponding part
of part~(\ref{Sec_10_25_e}).

\begin{lem}\label{L_3611_Uniq}
Let $r_1, r_2 \in \R$ satisfy $0 < r_1 < r_2$.
Let $c_n, d_n \in \C$ for $n \in \Z$.
Suppose that the series
$\sum_{n = - \I}^{\I} c_n z^n$
and $\sum_{n = - \I}^{\I} d_n z^n$
both converge on $A_{r_1, r_2}$ to the same function~$f$.
Then $c_n = d_n$ for all $n \in \Z$.
\end{lem}

\begin{proof}
Since the series $\sum_{n = 0}^{\I} c_n z^n$ converges on $A_{r_1, r_2}$,
the general properties of power series imply
that this series converges uniformly on compact subsets of $B_{r_2} (0)$.

The series $\sum_{n = - 1}^{- \I} c_n \left( \frac{1}{z} \right)^n$
converges whenever $\frac{1}{z} \in A_{r_1, r_2}$.
This series is a power series,
so the general properties of power series imply
that it converges uniformly on compact subsets of $B_{1 / r_1} (0)$.
Therefore $\sum_{n = - 1}^{- \I} c_n z^n$
converges uniformly on compact subsets
of $\C \SM {\overline{B_{r_1} (0)}}$ (of $\C \SM \{ 0 \}$ if $r_1 = 0$).

We conclude that $\sum_{n = 0}^{\I} c_n z^n$
converges uniformly on compact subsets of $A_{r_1, r_2}$.
Similarly $\sum_{n = 0}^{\I} d_n z^n$
converges uniformly on compact subsets of $A_{r_1, r_2}$.

Choose $r \in (r_1, r_2)$.
Define a closed curve $\gm$ by $\gm (t) = r e^{- i t}$.
Recall that $\int_{\gm} z^m \, d z = 2 \pi i$ when $m = - 1$
and $\int_{\gm} z^m \, d z = 0$ otherwise.
In the following computation,
the interchanges of summation and integration
are justified by uniform convergence of the series on $\Ran (\gm)$.
For $n \in \N$, we have
\[
\begin{split}
c_n
& = \frac{1}{2 \pi i} \sum_{m = - \I}^{\I}
      \int_{\gm} c_m z^{m - n - 1} \, d z
      \\
& = \frac{1}{2 \pi i} \int_{\gm} z^{- n - 1} f (z) \, d z
  = \frac{1}{2 \pi i} \sum_{m = - \I}^{\I}
      \int_{\gm} d_m z^{m - n - 1} \, d z
  = d_n.
\end{split}
\]
This completes the proof.
\end{proof}

The following lemma proves part~(\ref{Sec_10_25_d})
and the corresponding parts
of part~(\ref{Sec_10_25_e}).

\begin{lem}\label{L_3602_Bdd}
Let $r_1, r_2 \in [0, \I]$ satisfy $r_1 < r_2$.
Let $f$ be a bounded \hfn{} on $A_{r_1, r_2}$.
Let $f_1$ and $f_2$ be as in Lemma~\ref{L_3602_Decomp}.
Then $f_1$ and $f_2$ are bounded.
\end{lem}

The conclusion of the lemma
(and the requested statement in part~(\ref{Sec_10_25_d}))
is that $f_1$ is bounded on
$\C \SM {\overline{B_{r_1} (0)}}$ (on $\C \SM \{ 0 \}$ if $r_1 = 0$)
and that $f_2$ is bounded on $B_{r_2} (0)$,
not merely that they are bounded on $A_{r_1, r_2}$.

If $r_1 = 0$,
it follows that $f_1$ has a removable singularity at~$0$.
This isn't really new,
since it already follows that $f$ has a removable singularity at~$0$.

\begin{proof}[Proof of Lemma~\ref{L_3602_Bdd}]
Fix $s_1, s_2 \in \R$
with $r_1 < s_1 < s_2 < r_2$.
We first observe that $f_2$ is bounded on ${\overline{B_{s_2} (0)}}$
because ${\overline{B_{s_2} (0)}}$ is compact.
Also,
there is $M$ such that
$| f_1 (z) | < 1$ for all $z \in \C \SM B_M (0)$,
and $f_1$ is bounded on
${\overline{B_{M} (0)}} \SM B_{s_1} (0)$ by compactness,
so $f_1$ is bounded on $\C \SM B_{s_1} (0)$.

Now assume that $f$ is bounded.
On $B_{r_2} (0) \SM {\overline{B_{s_2} (0)}}$,
we have $f_2 = f - f_1$.
The function $f_1$ is bounded there by the previous paragraph.
The function $f$ is bounded there by hypothesis.
So $f_2$ is bounded on $B_{r_2} (0) \SM {\overline{B_{s_2} (0)}}$,
and hence on
${\overline{B_{s_2} (0)}}
  \cup \big[ B_{r_2} (0) \SM {\overline{B_{s_2} (0)}} \big]
 = B_{r_2} (0)$.

The proof that $f_1$ is bounded on $\C \SM {\overline{B_{r_1} (0)}}$
(on $\C \SM \{ 0 \}$ if $r_1 = 0$)
is essentially the same.
\end{proof}
\fi

\end{document}
