% To change:
% PPP Prob #, CCC chapter #.

% This is in AMSLaTeX.
% On NeXT and darkwing, use LaTeX.
\documentclass[10pt]{amsart}
% \renewcommand{\baselinestretch}{1.27}
% \usepackage{showkeys}
% \usepackage{pdfsync}
% \usepackage{showlabels}
\usepackage{amssymb}
\usepackage{amscd}
\usepackage[all]{xy}

% \numberwithin{equation}{section}

\def\today{\number\day\space\ifcase\month\or   January\or February\or
   March\or April\or May\or June\or   July\or August\or September\or
   October\or November\or December\fi\   \number\year}

\newif\ifsolutions
% \solutionstrue
\solutionsfalse

\theoremstyle{definition}
% \newtheorem{thm}{Theorem}[section]
\newtheorem{thm}{Theorem}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prp}[thm]{Proposition}
\newtheorem{dfn}[thm]{Definition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{cnj}[thm]{Conjecture}
\newtheorem{cnv}[thm]{Convention}
\newtheorem{rmk}[thm]{Remark}
\newtheorem{ntn}[thm]{Notation}
\newtheorem{exa}[thm]{Example}
% \newtheorem{pbm}[thm]{Problem}
\newtheorem{wrn}[thm]{Warning}
\newtheorem{qst}[thm]{Question}
\newtheorem{pbm}{Problem}

\newcommand{\beq}{\begin{equation}}
\newcommand{\eeq}{\end{equation}}
\newcommand{\beqr}{\begin{eqnarray*}}
\newcommand{\eeqr}{\end{eqnarray*}}
\newcommand{\bal}{\begin{align*}}
\newcommand{\eal}{\end{align*}}
\newcommand{\bei}{\begin{itemize}}
\newcommand{\eei}{\end{itemize}}
\newcommand{\limi}[1]{\lim_{{#1} \to \infty}}

\newcommand{\af}{\alpha}
\newcommand{\bt}{\beta}
\newcommand{\gm}{\gamma}
\newcommand{\dt}{\delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\zt}{\zeta}
\newcommand{\et}{\eta}
\newcommand{\ch}{\chi}
\newcommand{\io}{\iota}
\newcommand{\te}{\theta}
\newcommand{\ld}{\lambda}
\newcommand{\sm}{\sigma}
\newcommand{\kp}{\kappa}
\newcommand{\ph}{\varphi}
\newcommand{\ps}{\psi}
\newcommand{\rh}{\rho}
\newcommand{\om}{\omega}
\newcommand{\ta}{\tau}

\newcommand{\Gm}{\Gamma}
\newcommand{\Dt}{\Delta}
\newcommand{\Et}{\Eta}
\newcommand{\Th}{\Theta}
\newcommand{\Ld}{\Lambda}
\newcommand{\Sm}{\Sigma}
\newcommand{\Ph}{\Phi}
\newcommand{\Ps}{\Psi}
\newcommand{\Om}{\Omega}

\newcommand{\Q}{{\mathbb{Q}}}
\newcommand{\Z}{{\mathbb{Z}}}
\newcommand{\R}{{\mathbb{R}}}
\newcommand{\C}{{\mathbb{C}}}
% \newcommand{\N}{{\mathbb{N}}}
\newcommand{\N}{{\mathbb{Z}}_{> 0}}
\newcommand{\Nz}{{\mathbb{Z}}_{\geq 0}}

\pagenumbering{arabic}

\newcommand{\inv}{{\operatorname{inv}}}
\newcommand{\tsr}{{\operatorname{tsr}}}
\newcommand{\RR}{{\operatorname{RR}}}
\newcommand{\Tr}{{\operatorname{Tr}}}
\newcommand{\id}{{\operatorname{id}}}
\newcommand{\ev}{{\operatorname{ev}}}
\newcommand{\sint}{{\operatorname{int}}}
\newcommand{\diam}{{\operatorname{diam}}}
\newcommand{\dist}{{\operatorname{dist}}}
\newcommand{\sa}{{\operatorname{sa}}}
\newcommand{\spec}{{\operatorname{sp}}}
\newcommand{\Prim}{{\operatorname{Prim}}}
\newcommand{\diag}{{\operatorname{diag}}}
\newcommand{\supp}{{\operatorname{supp}}}
\newcommand{\rank}{{\operatorname{rank}}}
\newcommand{\sgn}{{\operatorname{sgn}}}
\newcommand{\spn}{{\operatorname{span}}}
\newcommand{\card}{{\operatorname{card}}}
\newcommand{\Aut}{{\operatorname{Aut}}}
\newcommand{\Ad}{{\operatorname{Ad}}}
\newcommand{\Aff}{{\operatorname{Aff}}}

\newcommand{\Ind}{{\operatorname{Ind}}}
\newcommand{\Ran}{{\operatorname{Ran}}}
\newcommand{\Res}{{\operatorname{Res}}}

\newcommand{\dirlim}{\varinjlim}
% \newcommand{\dirlim}{\displaystyle \lim_{\longrightarrow}}
\newcommand{\invlim}{\varprojlim}
\newcommand{\Mi}{M_{\infty}}

\newcommand{\andeqn}{\qquad {\mbox{and}} \qquad}
\newcommand{\QED}{\rule{0.4em}{2ex}}
\newcommand{\ts}[1]{{\textstyle{#1}}}
\newcommand{\ds}[1]{{\displaystyle{#1}}}
\newcommand{\ssum}[1]{{\ts{ {\ds{\sum}}_{#1} }}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\wolog}{without loss of generality}
\newcommand{\Wolog}{Without loss of generality}
\newcommand{\Tfae}{The following are equivalent}
\newcommand{\tfae}{the following are equivalent}
\newcommand{\ifo}{if and only if}
\newcommand{\wrt}{with respect to}

\newcommand{\ca}{C*-algebra}
\newcommand{\uca}{unital C*-algebra}
\newcommand{\usa}{unital subalgebra}
\newcommand{\hm}{homomorphism}
\newcommand{\uhm}{unital homomorphism}
\newcommand{\am}{automorphism}
\newcommand{\fd}{finite dimensional}
\newcommand{\tst}{tracial state}
\newcommand{\hsa}{hereditary subalgebra}

\newcommand{\pj}{projection}
\newcommand{\mops}{mutually orthogonal \pj s}
\newcommand{\nzp}{nonzero projection}
\newcommand{\mvnt}{Murray-von Neumann equivalent}
\newcommand{\mvnc}{Murray-von Neumann equivalence}

\newcommand{\ct}{continuous}
\newcommand{\cfn}{continuous function}
\newcommand{\nbhd}{neighborhood}
\newcommand{\cms}{compact metric space}
\newcommand{\cpt}{compact Hausdorff}
\newcommand{\chs}{compact Hausdorff space}
\newcommand{\hme}{homeomorphism}
\newcommand{\mh}{minimal homeomorphism}
\newcommand{\mf}{manifold}
\newcommand{\cmf}{compact smooth \mf}
\newcommand{\hl}{holomorphic}
\newcommand{\mrc}{meromorphic}
\newcommand{\mb}{measurable}
\newcommand{\mfn}{measurable function}
\newcommand{\hfn}{holomorphic function}

\newcommand{\cp}{crossed product}
\newcommand{\tgca}{transformation group \ca}
\newcommand{\tggp}{transformation group groupoid}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newcommand{\rsz}[1]{\raisebox{0ex}[0.8ex][0.8ex]{$#1$}}
\newcommand{\subrsz}[1]{\raisebox{0ex}[0.8ex][0.8ex]{${\scriptstyle{#1}}$}}

% Temporary abbreviations:

\renewcommand{\S}{\subset}
\newcommand{\ov}{\overline}
\newcommand{\SM}{\setminus}
\newcommand{\I}{\infty}
\newcommand{\E}{\varnothing}

\newcommand{\Lem}[1]{Lemma~\ref{#1}}
\newcommand{\Def}[1]{Definition~\ref{#1}}
\newcommand{\Thm}[1]{Theorem~\ref{#1}}
\newcommand{\Prp}[1]{Proposition~\ref{#1}}
\newcommand{\Exa}[1]{Example~\ref{#1}}

\ifsolutions
\title[Math 618 (Spring 2024): Solutions~8]{Math 618 (Spring 2024,
        Phillips): Solutions to Homework~8}
\date{3~June 2024}
\else
\title[Math 618 (Spring 2024): Homework~8]{Math 618 (Spring 2024,
        Phillips): Homework~8}
\date{22~May 2024}
\fi

\begin{document}

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

\maketitle

% \setcounter{section}{-1}

\indent

This assignment is due on Canvas on Monday 3~June 2024 at 9:00 pm.

Problems from Chapters 12 and~13 of Rudin do not, I hope,
depend on material from those chapters not discussed in the course.

\ifsolutions
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~18
  in Chapter~14 of Rudin's book]\label{Pb_2024_7_18}
Let $D = \{ z \in \C \colon | z | < 1 \}$
be the open unit disk.
Let $\Om$ be a simply connected region with $\Om \neq \C$,
and let $z_0 \in \Om$.
Suppose $f, g \colon \Om \to D$ are \hl{} and bijective,
and satisfy $f (z_0) = g (z_0) = 0$.
How are $f$ and $g$ related?
Now let $a \in D$ be arbitrary.
If $f, g \colon \Om \to D$ are \hl{} and bijective,
and satisfy $f (z_0) = g (z_0) = a$,
how are $f$ and $g$ related?
\end{pbm}


\ifsolutions
\bigskip


\begin{lem}\label{L_2024_6_Lem_}
\end{lem}

\begin{proof}
\end{proof}

\begin{proof}[Solution]
\end{proof}

\begin{proof}[Alternate solution (sketch)]
\end{proof}
\fi


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{pbm}[Problem~3
  in Chapter~12 of Rudin's book]\label{Pb_2024_8_12_3}
Let $\Om \S \C$ be a region.
Determine exactly when a \hfn~$f$ has the property that $| f |$
has a local {\emph{minimum}} on~$\Om$.
\end{pbm}


\ifsolutions
% \bigskip

\begin{proof}[Solution]
We claim that this happens \ifo{} $f$ is constant
or there exists $z \in \Om$ such that $f (z) = 0$.

First, if one or other of these conditions is satisfied, then
obviously $| f |$ has a local minimum on~$\Om$.

For the reverse, assume that $f$ vanishes nowhere on~$\Om$
and that $| f |$ has a local minimum on~$\Om$:
there is $a \in \Om$ such that $| f (z) | \geq | f (a) |$
for all $z \in \Om$..
Since $f$ vanishes nowhere on~$\Om$, the function
$z \mapsto g (z) = f (z)^{-1}$ is \hl{} on~$\Om$.
Clearly $g$ has a local maximum at~$a$.
Therefore $g$ is constant by the Maximum Modulus Theorem.
So $f$ is constant.
\end{proof}
\fi


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{pbm}[Problem~4
  in Chapter~12 of Rudin's book]\label{Pb_2024_8_12_4}
\mbox{}
%
\begin{enumerate}
% [label=$\mathrm{(\arabic*)}$]
%
\item\label{I_Pb_2024_8_12_4_Disk}
Let $\Om \S \C$ be a region, let $a \in \Om$, and let $r > 0$
satisfy ${\ov{B_r (a)}} \S \Om$.
Let $f$ be a nonconstant \hfn{} on~$\Om$
such that $| f |$ is constant on $\partial B_r (a)$.
Prove that $f$ has a zero in $B_r (a)$.
%
\item\label{I_Pb_2024_8_12_4_Abs}
Find all entire functions $f$ such that $| f (z) | = 1$
whenever $| z | = 1$.
%
\end{enumerate}
%
\end{pbm}

\ifsolutions
% \bigskip

\begin{proof}[Solution for part~(\ref{I_Pb_2024_8_12_4_Disk})]
Suppose $| f |$ is constant on $\partial B_r (a)$ and
$f (z) \neq 0$ for all $z \in B_r (a)$.
Let $M$ be the constant value of $| f |$ on $\partial B_r (a)$.
By compactness, $| f |$ has a global maximum on
${\ov{B_r (a)}}$, say at $z \in {\ov{B_r (a)}}$.
If $| f (z) | > M$, then $z \not\in \partial B_r (a)$,
so $| f |$ has a local maximum at $z \in B_r (a)$.
The function $f$ is not constant because $| f (z) | > M$,
so this contradicts the Maximum Modulus Theorem.
Therefore $| f | \leq M$ on $B_r (a)$.
Since $f (z) \neq 0$ for all $z \in B_r (a)$, it follows that $M > 0$.

Similarly, and using $M > 0$, the function
$z \mapsto g (z) = f (z)^{-1}$ is \hl{} on an open subset $U \S \Om$
and satisfies $| g | \leq M^{-1}$ on $B_r (a)$.
Combining this with the previous inequality,
we get $| f | = M$ on $B_r (a)$.
So $f$ is constant by the Maximum Modulus Theorem.
\end{proof}

\begin{proof}[Solution for part~(\ref{I_Pb_2024_8_12_4_Abs}) (sketch)]
Set $D = \{ z \in \C \colon | z | < 1 \}$.
Recall the functions
\[
\ph_a (z) = \frac{z - a}{1 - {\ov{a}} z}
\]
for $a \in D$, each defined on a \nbhd{} of~${\ov{D}}$.

Let $a_1, a_2, \ldots, a_n \in D$ be the zeros of $f$ in~$D$,
with multiplicities $m_1, m_2, \ldots, m_n \in \N$.
There is a \hfn{} $g$ defined on some \nbhd{} $U$ of~${\ov{D}}$
such that
\[
f (z) = \ph_{a_1} (z)^{m_1} \cdot \ph_{a_2} (z)^{m_2} \cdot
   \cdots \cdot \ph_{a_n} (z)^{m_n},
\]
and $g$ is never zero on $D$ and satisfies $| g (z) | = 1$
for all $z \in \partial D$
(because this is true of $f$
and $\ph_{a_1}, \ph_{a_2}, \ldots, \ph_{a_n}$).
Therefore  $g$ is constant by part~(\ref{I_Pb_2024_8_12_4_Disk}),
and the constant $\ld$ must satisfy $| \ld | = 1$.
Now check which of the functions
\[
z \mapsto \ld \ph_{a_1} (z)^{m_1} \cdot \ph_{a_2} (z)^{m_2} \cdot
   \cdots \cdot \ph_{a_n} (z)^{m_n}
\]
extends to an entire function.
\end{proof}
\fi


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{pbm}[Problem~5
  in Chapter~12 of Rudin's book]\label{Pb_2024_8_}
Let $\Om \S \C$ be a bounded region,
let $(f_n)_{n \in \N}$ be a sequence in $C \bigl( {\ov{\Om}} \bigr)$
such that $f_n |_{\Om}$ is \hl{} for all $n \in \N$,
and suppose that there is a function $g$ on $\partial \Om$
such that $(f_n |_{\partial \Om})_{n \in \N}$ converges
uniformly to~$g$.
Prove that there is a function $f$ on $\Om$
such that $(f_n)_{n \in \N}$ converges uniformly to~$f$.
\end{pbm}


\ifsolutions
% \bigskip

\begin{proof}[Solution]
It is enough to prove that $(f_n)_{n \in \N}$ is uniformly Cauchy.
So let $\ep > 0$.

Since ${\ov{\Om}}$ is compact, for every $m, n \in \N$
there is $a_{m, n} \in {\ov{\Om}}$ such that
$| f_m (z) - f_n (z) | \leq | f_m (a_{m, n}) - f_n (a_{m, n})|$
for all $z \in {\ov{\Om}}$.
By the Maximum Modulus Theorem,
we may choose $a_{m, n} \in \partial \Om$.
Since the sequence $(f_n |_{\partial \Om})_{n \in \N}$ converges
uniformly, it is uniformly Cauchy.
Therefore there is $N \in \N$ such that whenever $m, n \geq N$ then
\[
\sup_{z \in \partial \Om} | f_m (z) - f_n (z) | < \ep.
\]
In particular, $| f_m (a_{m, n}) - f_n (a_{m, n})| < \ep$.
By the choice of $a_{m, n}$, we have
\[
\sup_{z \in \partial \Om} | f_m (z) - f_n (z) |
  \leq | f_m (a_{m, n}) - f_n (a_{m, n})|
  < \ep.
\]
So $(f_n)_{n \in \N}$ is uniformly Cauchy, as desired.
\end{proof}
\fi


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{pbm}[Problem~10
  in Chapter~13 of Rudin's book]\label{Pb_2024_8_}
Let $\Om \S \C$ be a region, and let $f$ be a \hfn{} on~$\Om$
which is not the constant function zero.
Suppose that for every $n \in \N$ there is a \hfn{} $g$ on~$\Om$
such that $g (z)^n = f (z)$ for every $z \in \Om$.
Prove that there is a \hfn{} $h$ on~$\Om$
such that $\exp (h (z)) = f (z)$ for every $z \in \Om$.
\end{pbm}


\ifsolutions
\bigskip


\begin{proof}[Solution (sketch)]
We first claim that $f (z) \neq 0$ for all $z \in \Om$.
Otherwise, a zero of $f$ must be isolated.
If $g^n = f$, then $g$ also has an isolated zero at~$z$.
The theorem on local behavior of $g$ near a zero can be used
to show that the order of the zero of $f$ at $z$ must be divisible
by~$n$.
This can't be true for all $n \in \N$, proving the claim.

Next, we claim that if $\gm \colon [\af, \bt] \to \Om$
is any piecewise~$C^1$ closed curve,
then
\[
\int_{\gm} \frac{f' (\zt)}{f (\zt)} \, d \zt = 0.
\]
For any \hfn{} $g$ on~$\Om$ which does not vanish on $\Ran (\gm)$,
consider the function
\[
\ph (t) = \int_{\af}^t
   \left( \frac{g' (\gm (t))}{g' (\gm (t))} \right) \gm' (t) \, d t.
\]
The function
\[
\ps (t) = \frac{\exp (\ph (t))}{g (\gm (t))}
\]
has $\ps' (t) = 0$ (except at finitely many points), so is constant.
This shows that
\[
\int_{\gm} \frac{g' (\zt)}{g (\zt)} \, d \zt \in 2 \pi i \Z.
\]
Now let $n \in \N$ be arbitrary, and choose $g$ such that $g^n = f$.
Then
\[
\frac{f' (\zt)}{f (\zt)} = \frac{n g' (\zt)}{g (\zt)}
\andeqn
\int_{\gm} \frac{g' (\zt)}{g (\zt)} \, d \zt \in 2 \pi i \Z,
\]
so
\[
\int_{\gm} \frac{f' (\zt)}{f (\zt)} \, d \zt \in 2 \pi i n \Z.
\]
Since $n$ is arbitrary, the claim follows.

Using the claim, show by methods we have already seen
that there is a function $h_0$ on~$\Om$ such that
\[
h_0' (z) = \frac{f' (z)}{f (z)}
\]
for all $z \in \Om$.
Use a suitable constant multiple of $h_0$ to solve the problem.
\end{proof}
\fi

\end{document}
