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\begin{document}
\title{Topics in Fourier and Complex Analysis
\\Part III, Autumn 2009}
\author{T.~W.~K\"{o}rner}
\maketitle
\begin{footnotesize}
\noindent
{\bf Small print}
This is just a first draft of the
first part of the course. I suspect these
notes will cover the
first 16 hours but I will not be unduly
surprised if it
takes the entire course to cover the material.
The content of the course will be what I say, not what
these notes say.
Experience shows that skeleton notes (at least
when I write them) are very error prone so
use these notes with care.
I should {\bf very much} appreciate being told
of any corrections or possible improvements
and might even part with a small reward to the
first finder of particular errors.

This course definitely requires a first course in
complex variable and enough analysis to be
happy with terms like norm, complete metric
space and compact. I am happy to give classes on any 
topics that people request. At at least one point, the course
requires measure theory, but you need only quote
the required results in examination.
\end{footnotesize}
\tableofcontents
\section{Non-existence of functions of several variables}
\begin{theorem}\label{T;Kolmogorov} Let $\lambda$ be irrational
We can find increasing continuous functions
$\phi_{j}:[0,1]\rightarrow{\mathbb R}$ $[1\leq j\leq 5]$
with the following property. Given any continuous
function
$f:[0,1]^{2}\rightarrow{\mathbb R}$ we can find a
function $g:{\mathbb R}\rightarrow{\mathbb R}$ such that
\[f(x,y)=\sum_{j=1}^{5}g(\phi_{j}(x)+\lambda\phi_{j}(y)).\]
\end{theorem}
The main point of Theorem~\ref{T;Kolmogorov}
may be expressed as follows.
\begin{theorem}\label{T;Gilbert}
Any continuous function of two variables
can be written in terms of 
continuous functions of one variable
and addition.
\end{theorem}
That is, there are no true functions of two variables!

For the moment we merely observe that the result is due
in successively more exact forms to Kolmogorov, Arnol'd
and a succession of mathematicians ending with Kahane 
whose proof we use here. It is, of course, much easier to
prove a specific result like Theorem~\ref{T;Kolmogorov}
than one like Theorem~\ref{T;Gilbert}.

Our first step is to observe that Theorem~\ref{T;Kolmogorov}
follows from the apparently simpler result that follows.
\begin{lemma}\label{L;Hilbert A} Let $\lambda$ be irrational.
We can find increasing continuous functions
$\phi_{j}:[0,1]\rightarrow{\mathbb R}$ $[1\leq j\leq 5]$
with the following property. Given any continuous
function
$F:[0,1]^{2}\rightarrow{\mathbb R}$ we can find a
function $G:{\mathbb R}\rightarrow{\mathbb R}$ such that
$\|G\|_{\infty}\leq\|F\|_{\infty}$ and 
\[\sup_{(x,y)\in[0,1]^{2}}
\left|F(x,y)-\sum_{j=1}^{5}G(\phi_{j}(x)+\lambda\phi_{j}(y))
\right|\leq \frac{999}{1000}\|F\|_{\infty}.\]
\end{lemma}
(The choice of the constant $999/1000$ is, of course,
pretty arbitrary.)


Next we make the following observation.
\begin{lemma}\label{L;Hilbert B} We can find
a sequence of functions $f_{n}:[0,1]^{2}\rightarrow{\mathbb R}$
which are uniformly dense in $C([0,1])^{2}$.
\end{lemma}
This enables us to obtain Lemma~\ref{L;Hilbert A}
from a much more specific result.
\begin{lemma}\label{L;Hilbert C} Let $\lambda$ be irrational
and let the $f_{n}$ be as in Lemma~\ref{L;Hilbert B}.
We can find increasing continuous functions
$\phi_{j}:[0,1]\rightarrow{\mathbb R}$ $[1\leq j\leq 5]$
with the following property. We can find 
functions $g_{n}:{\mathbb R}\rightarrow{\mathbb R}$ such that
$\|g_{n}\|_{\infty}\leq\|f_{n}\|_{\infty}$ and 
\[\sup_{(x,y)\in[0,1]^{2}}
\left|f_{n}(x,y)-\sum_{j=1}^{5}g_{n}(\phi_{j}(x)+\lambda\phi_{j}(y))
\right|\leq \frac{998}{1000}\|f_{n}\|_{\infty}.\]
\end{lemma}


One of Kahane's contributions is the observation
that the proof Theorem~\ref{L;Hilbert C} is made easier
by the use of Baire category. Although most
of the audience is familiar with Baire's category
theorem, we shall reprove it here.

\begin{theorem}{\bf [Baire's category theorem]}
If $(X,d)$ is a complete metric space then
$X$ can not be written as the union of a countable collection
of closed sets with empty interior.
\end{theorem}

One way of thinking of a closed set $E$ with empty interior
is the following. The property of belonging to $E$ is \emph{unstable}
since arbitrarily small changes take one outside $E$ but the the property
of not belonging to $E$ is \emph{stable} since, if we are at a
point outside $E$ all sufficiently small changes keep us outside $E$.

We shall prove a slightly stronger version of Baire's theorem.
\begin{theorem}\label{T;Baire 2} Let $(X,d)$ be a complete metric space.
If $E_{1}$, $E_{2}$, \dots are closed sets with empty interiors
then $X\setminus \bigcup_{j=1}^{\infty}E_{j}$ is dense in $X$.
\end{theorem}
\begin{exercise} (If you are happy with general topology.)
Show that a result along the same lines holds true
for compact Hausdorff spaces.
\end{exercise}
For historical reasons Baire's category theorem is associated
with some rather peculiar nomenclature.
\begin{definition} Let $(X,d)$ be a metric space. We say that a
a subset $A$ of $X$ is \emph{of the first category} if
it is a subset of the union of a countable collection
of closed sets with empty interior\footnote{This usage is not
universal. Some authors use the older definition
which says that
a subset $A$ of $X$ is of the first category if
it is the union of a countable collection
of closed sets with empty interior.However, so far as
I know, all authors who use `quasi-all' use it in he same way.}
We say that \emph{quasi-all}
points of $X$ belong to the complement $X\setminus A$ of $X$.
\end{definition}
The following observations
are trivial but useful.
\begin{lemma} (i) The countable union of first
category sets is itself of first category.

(ii) If $(X,d)$ is a complete metric
space, then Baire's theorem asserts that $X$
is not of first category.
\end{lemma}

Since Lemma~\ref{L;Hilbert B} only involve a \emph{countable}
set of conditions we can use a Baire category argument.
provided that we can find the correct metric space.
\begin{lemma}\label{L;Kahane space}
The space $Y$ of continuous functions
${\boldsymbol \phi}:[0,1]\rightarrow{\mathbb R}^{5}$
with norm
\[\|{\boldsymbol \phi}\|_{\infty}=
\sup_{t\in[0,1]}\|{\boldsymbol \phi}(t)\|\]
is complete. The subset $X$ of $Y$ consisting of
those ${\boldsymbol \phi}$ such that each $\phi_{j}$
is increasing is a closed
subset of $Y$. Thus if $d$ is the metric on $X$
obtained by restricting the metric on $Y$ derived from
$\|\ \|_{\infty}$  we have $(X,d)$ complete.
\end{lemma}
\begin{exercise} Prove Lemma~\ref{L;Kahane space}
\end{exercise}
\begin{lemma}\label{L;Hilbert D}  Let
$f:[0,1]^{2}\rightarrow{\mathbb R}$ be continuous
and let $\lambda$ be irrational. Consider the
set $E$ of ${\boldsymbol \phi}\in X$ such that
there exists a 
$g:{\mathbb R}\rightarrow{\mathbb R}$ such that
$\|g\|_{\infty}\leq\|f\|_{\infty}$
\[\sup_{(x,y)\in[0,1]^{2}}
\left|f(x,y)-\sum_{j=1}^{5}g(\phi_{j}(x)+\lambda\phi_{j}(y))
\right|<\frac{998}{1000}\|f\|_{\infty}.\]
Then $X\setminus E$ is a closed set with dense complement
in $(X,d)$.
\end{lemma}
(Notice that it is important to take `$<$' rather than
`$\leq$' in the displayed formula of Lemma~\ref{L;Hilbert D}.)
Lemma~\ref{L;Hilbert D} is the heart of the proof
and once it is proved we can easily retrace our steps
and obtain Theorem~\ref{T;Kolmogorov}.

By using appropriate notions of information 
Vistu{\v s}kin\footnote{For those who wish to
dispense with accents  Vitushkin.}
was able to show that we can not replace
continuous by continuously differentiable
in Theorem~\ref{T;Gilbert}. Thus
Theorem~\ref{T;Kolmogorov} is an `exotic'
rather than a `central' result.
The next two sections are devoted  to
the proof of Vistu{\v s}kin's simplest result.



We conclude the section with some exercises
intended to help the reader
understand the proof of Theorem~\ref{T;Kolmogorov}.
\begin{exercise} Let $(X,d)$ be as in
Exercise~\ref{L;Kahane space}. Show that
quasi-all  ${\mathbf \phi}\in X$ have the property that
$\phi_{j}$ is strictly increasing (that is to say $\phi_{j}(s)<\phi_{j}(t)$
for $0\leq s<t\leq 1$) for each $1\leq j\leq 5$. Why does this
immediately tell us that we can replace the word `increasing'
by the words `strictly increasing' in Theorem~\ref{T;Kolmogorov}.
\end{exercise}

\begin{exercise} We say that real numbers $\lambda_{1}$, $\lambda_{2}$,
\ldots, $\lambda_{n}$ are independent over ${\mathbb Q}$ if the equation
\[\sum_{j=1}^{n}q_{j}\lambda_{j}=0\]
has no solution with $q_{j}\in{\mathbb Q}$ $[1\leq j\leq n]$ apart
from the trivial solution with all $q_{j}=0$.

By using the fact that the real numbers are uncountable, or otherwise,
show that, if $\lambda_{1}$, $\lambda_{2}$,
\ldots, $\lambda_{n}$ are independent over ${\mathbb Q}$
we can find $\lambda_{n+1}$ such that
$\lambda_{1}$, $\lambda_{2}$,
\ldots, $\lambda_{n+1}$ are independent over ${\mathbb Q}$.
Deduce that we can find $\lambda_{1}$, $\lambda_{2}$,
\ldots such that $\lambda_{1}$, $\lambda_{2}$,
\ldots, $\lambda_{n}$ are independent over ${\mathbb Q}$
for all $n$.
\end{exercise}
\begin{exercise} Prove that,
if $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$.
are independent over ${\mathbb Q}$
we can find increasing continuous functions
$\phi_{j}:[0,1]\rightarrow{\mathbb R}$ $[0\leq j\leq 6]$,
with the following property. Given any continuous
function
$f:[0,1]^{3}\rightarrow{\mathbb R}$ we can find a
function $g:{\mathbb R}\rightarrow{\mathbb R}$ such that
\[f(x_{1},x_{2},x_{3})
=\sum_{j=0}^{6}g\big(\lambda_{1}\phi_{j}(x_{1})+
\lambda_{2}\phi_{j}(x_{2})
+\lambda_{3}\phi(x_{3})\big).\]

\noindent[You may prefer
to identify the only part of the proof which might fail
and concentrate on that.]
\end{exercise}
\begin{exercise} Investigate to what extent the proof
would be simplified if we replaced Theorem~\ref{T;Kolmogorov}
by the following slightly less demanding result.

Given any continuous
function
$f:[0,1]^{2}\rightarrow{\mathbb R}$
we can find continuous functions
$\phi_{r}:[0,1]\rightarrow{\mathbb R}$ $[0\leq r\leq 9]$
and continuous
functions $g_{j}:{\mathbb R}\rightarrow{\mathbb R}$
$[0\leq j\leq 4]$
such that
\[f(x,y)=\sum_{j=0}^{4}g_{j}\big(\phi_{2j}(x)+\phi_{2j+1}(y)\big).\]
\end{exercise}
\begin{exercise} Show that
the result corresponding to Theorem~\ref{T;Kolmogorov}
is false if $\lambda=1$ or $\lambda=-1$.
Show that it is true for all $\lambda\neq 1,\,-1$.
\end{exercise}
\section{Fourier series on the circle} We work on the circle
${\mathbb T}={\mathbb R}/2\pi{\mathbb Z}$ (that is on
the interval $[0,2\pi]$ with the two ends $0$ and $2\pi$
identified). If $f:{\mathbb T}\rightarrow{\mathbb C}$
is integrable\footnote{That is
to say Lebesgue integrable
or Riemann integrable according to the reader's
background.} we write
\[\hat{f}(n)=\frac{1}{2\pi}\int_{\mathbb T}f(t)\exp -int\,dt.\]
We shall see (Lemma~\ref{Unique}) that $f$ is uniquely
determined by its Fourier coefficients $\hat{f}(n)$.
Indeed it is clear that there is a `natural identification'
(where natural is deliberately used in a vague sense)
\[f(t)\sim\sum_{r=-\infty}^{\infty}\hat{f}(r)\exp irt.\]
However, we shall also see that, even when $f$ is continuous,
$\sum_{r=-\infty}^{\infty}\hat{f}(r)\exp irt$ may
fail to converge at some points $t$.

Fej\'{e}r discovered that, although
\[S_{n}(f,t)=\sum_{r=-n}^{n}\hat{f}(r)\exp irt\]
may behave badly as $n\rightarrow\infty$, the
average
\[\sigma_{n}(f,t)=(n+1)^{-1}\sum_{m=0}^{n}S_{m}(f,t)
=\sum_{r=-n}^{n}\frac{n+1-|r|}{n+1}\hat{f}(r)\exp irt\]
behaves much better. (We call $\sigma_{n}(f,t)$ the
Fej\'{e}r sum. We also write $S_{n}(f,t)=S_{n}(f)(t)$
and $\sigma_{n}(f,t)=\sigma_{n}(f)(t)$.)

\begin{exercise} Let $a_{1}$, $a_{2}$, \dots be a sequence
of complex numbers.

(i) Show that, if $a_{n}\rightarrow a$, then
\[\frac{a_{1}+a_{2}+\dots+a_{n}}{n}\rightarrow a\]
as $n\rightarrow\infty$.

(ii) By taking an appropriate sequence of $0$s and $1$s
or otherwise find a sequence $a_{n}$ such that
$a_{n}$ does not tend to a limit as $n\rightarrow\infty$
but $(a_{1}+a_{2}+\dots+a_{n})/n$ does.


(iii) By taking an appropriate sequence of $0$s and $1$s
or otherwise find a bounded sequence $a_{n}$ such that
$(a_{1}+a_{2}+\dots+a_{n})/n$ does not tend to a limit
as $n\rightarrow\infty$.
\end{exercise}

In what follows we define
\[f*g(t)=\frac{1}{2\pi}\int_{\mathbb T}f(t-s)g(s)\,ds\]
(for appropriate $f$ and $g$).
\begin{lemma} If $f$ is integrable we have
\begin{align*}
S_{n}(f)&=f*D_{n}\\
\sigma_{n}(f)&=f*K_{n}.
\end{align*}
where
\begin{align*}
D_{n}(t)&=\frac{\sin((n+\tfrac{1}{2})t)}{\sin(\tfrac{1}{2}t)}\\
K_{n}(t)&=\frac{1}{n+1}\left(\frac{\sin(\tfrac{n+1}{2}t)}
{\sin(\tfrac{1}{2}t)}\right)^{2}
\end{align*}
for $t\neq 0$.
\end{lemma}

The key differences between the Dirichlet kernel $D_{n}$
and the Fej\'{e}r kernel $K_{n}$ are illustrated
by the next two lemmas.

\begin{lemma} (i)
${\displaystyle \frac{1}{2\pi}\int_{\mathbb T}D_{n}(t)\, dt=1}$.

(ii) If $t\neq \pi$, then $D_{n}(t)$ does not tend to
a limit as $n\rightarrow\infty$.

(iii) There is a constant $A>0$ such that
\[\frac{1}{2\pi}\int_{\mathbb T}|D_{n}(t)|\, dt\geq A\log n\]
for $n\geq 1$.
\end{lemma}

\begin{lemma}\label{Fejer positive}
(i) $\displaystyle{\frac{1}{2\pi}\int_{\mathbb T}K_{n}(t)\, dt=1}$.

(ii) If $\eta>0$, then $K_{n}\rightarrow 0$ uniformly
for $|t|\geq \eta$
as $n\rightarrow\infty$.

(iii) $K_{n}(t)\geq 0$ for all $t$.
\end{lemma}

The properties set out in Lemma~\ref{Fejer positive}
show why Fej\'{e}r sums work so well.

\begin{theorem}~\label{Fejer convergence}
(i) If $f:{\mathbb T}\rightarrow{\mathbb C}$
is integrable and $f$ is continuous at $t$, then
\[\sigma_{n}(f,t)\rightarrow f(t)\]
as $n\rightarrow\infty$.

(ii) If $f:{\mathbb T}\rightarrow{\mathbb C}$
is continuous, then
\[\sigma_{n}(f)\rightarrow f\]
uniformly as $n\rightarrow\infty$.
\end{theorem}
\begin{exercise} Suppose that
$L_{n}:{\mathbb T}\rightarrow{\mathbb R}$
is continuous (if you know Lebesgue theory you
merely need integrable) and

(A) ${\displaystyle \frac{1}{2\pi}\int_{\mathbb T}L_{n}(t)\, dt=1}$,

(B) If $\eta>0$, then $L_{n}\rightarrow 0$ uniformly
for $|t|\geq \eta$
as $n\rightarrow\infty$,

(C) $L_{n}(t)\geq 0$ for all $t$.

(i) Show that, if $f:{\mathbb T}\rightarrow{\mathbb C}$
is integrable and $f$ is continuous at $t$, then
\[L_{n}*f(t)\rightarrow f(t)\]
as $n\rightarrow\infty$.

(ii) Show that, if $f:{\mathbb T}\rightarrow{\mathbb C}$
is continuous, then
\[L_{n}*f\rightarrow f\]
uniformly as $n\rightarrow\infty$.

(iii) Show that condition (C) can be replaced by

(C') There exists a constant $A>0$ such that
\[\frac{1}{2\pi}\int_{\mathbb T}|L_{n}(t)|\, dt\leq A\]
in parts (i) and (ii). [You need only give the proof in
one case and say that the other is `similar'.]
\end{exercise}
\begin{exercise} Suppose that
$L_{n}:{\mathbb T}\rightarrow{\mathbb R}$
is continuous but that
\[\sup_{n}\frac{1}{2\pi}
\int_{\mathbb T}|L_{n}(t)|\, dt=\infty.\]
Show that we can find a sequence of continuous functions
$g_{n}:{\mathbb T}\rightarrow{\mathbb R}$ with
$|g_{n}(t)|\leq 1$ for all $t$, $L_{n}*g_{n}(0)\geq 0$
for all $n$ and
\[\sup_{n}L_{n}*g_{n}(0)=\infty.\]

(i) If you know some functional analysis deduce the
existence of a continuous function $f$ such that
\[\sup_{n}L_{n}*f(0)=\infty.\]

(ii) Even if you can obtain the result of (i) by
slick functional analysis there is some point
in obtaining the result directly.

(a) Suppose that we have defined positive integers
$n(1)<n(2)<\dots<n(k)$, a continuous function
$g_{k}$ and a real number $\epsilon(k)$ with
$2^{-k}>\epsilon(k)>0$.
Show that there is an $\epsilon(k+1)$ with
$\epsilon(k)/2>\epsilon(k+1)>0$ such that
whenever $g$ is a continuous function with
$\|g-g_{k}\|_{\infty}<2\epsilon(k+1)$
we have $|L_{n(j)}*g(0)-L_{n(j)}*g_{k}(0)|\leq 1$.
for $1\leq j\leq k$.

(b) Continuing with the notation of (a), show that
there exists an $n(k+1)>n(k)$ and a continuous
function $g_{k+1}$ with $\|g_{k+1}-g_{k}\|_{\infty}\leq\epsilon(k+1)$
such that $|L_{n(k+1)}*g_{k+1}(0)|>2^{k+1}$.

(c) By carrying out the appropriate induction
and considering the uniform limit of $g_{k}$
obtain (i).


(iii) Show that there exists a continuous function
$f$ such that $S_{n}(f,0)$ fails to converge as
$n\rightarrow\infty$.
\end{exercise}

Theorem~\ref{Fejer convergence} has several very useful
consequences.
\begin{theorem}[Density of trigonometric polynomials]%
\label{density} The trigonometric polynomials are
uniformly dense in the continuous functions on ${\mathbb T}$.
\end{theorem}
\begin{lemma}[Riemann-Lebesgue lemma] If $f$ is an
integrable function on ${\mathbb T}$, then
$\hat{f}(n)\rightarrow 0$ as $|n|\rightarrow\infty$.
\end{lemma}
\begin{theorem}[Uniqueness]\label{Unique}
If $f$ and $g$ are
integrable functions on ${\mathbb T}$ with
$\hat{f}(n)=\hat{g}(n)$ for all $n$, then $f=g$.
\end{theorem}
\begin{lemma} If $f$ is an
integrable function on ${\mathbb T}$ and
$\sum_{j}|\hat{f}(j)|$ converges, then $f$ is continuous
and $f(t)=\sum_{j}\hat{f}(j)\exp ijt$.
\end{lemma}
As a preliminary to the next couple of results we
need the following temporary lemma (which will be immediately
superseded by Theorem~\ref{Parseval}).
\begin{lemma}[Bessel's inequality]
If $f$ is
a continuous function on ${\mathbb T}$, then
\[\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^{2}
\leq\frac{1}{2\pi}\int_{\mathbb T}|f(t)|^{2}\,dt.\]
\end{lemma}
\begin{theorem} [Mean square convergence]%
\label{mean square} If $f$ is
a continuous function on ${\mathbb T}$, then
\[
\frac{1}{2\pi}\int_{\mathbb T}
| f(t)-S_{n}(f,t) |^{2}\,dt\rightarrow 0
\]
as $n\rightarrow\infty$.
\end{theorem}
\begin{theorem} [Parseval's Theorem]\label{Parseval}
If $f$ is
a continuous function on ${\mathbb T}$, then
\[\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^{2}
=\frac{1}{2\pi}\int_{\mathbb T}|f(t)|^{2}\,dt.\]
More generally, if $f$ and $g$ are continuous
\[\sum_{n=-\infty}^{\infty}\hat{f}(n)\hat{g}(n)^{*}
=\frac{1}{2\pi}\int_{\mathbb T}f(t)g(t)^{*}\,dt.\]
\end{theorem}


(The extension to all $L^{2}$ functions of Theorems~\ref{mean square}
and~\ref{Parseval} uses easy measure
theory.)
\begin{exercise}
If you use Lebesgue integration,
state and prove Theorems~\ref{mean square}
and~\ref{Parseval} for $(L^{2}({\mathbb T}),\|\ \|_{2})$.

If you use Riemann integration, extend and prove
Theorems~\ref{mean square} and~\ref{Parseval}
for all Riemann integrable function.
\end{exercise}
Note the following complement to the Riemann-Lebesgue lemma.
\begin{lemma} If
$\kappa(n)\rightarrow \infty$ as $n\rightarrow\infty$,
then we can find a continuous function $f$
such that $\limsup_{n\rightarrow\infty}\kappa(n)\hat{f}(n)=\infty$.
\end{lemma}

The proof of the next result is perhaps more interesting
than the result itself.
\begin{lemma}\label{pointwise}
Suppose that $f$ is an
integrable function on ${\mathbb T}$ such that there
exists an $A$ with $|\hat{f}(n)|\leq A|n|^{-1}$ for
all $n\neq 0$. If $f$ is continuous at $t$, then
$S_{n}(f,t)\rightarrow f(t)$ as $n\rightarrow\infty$.
\end{lemma}
\begin{exercise} Suppose that $a_{n}\in{\mathbb C}$
and there
exists an $A$ with $|a_{n}|\leq A|n|^{-1}$ for
all $n\geq 1$. Write
\[s_{n}=\sum_{r=0}^{n}a_{r}.\]
Show that, if
\[\frac{s_{0}+s_{1}+\dots+s_{n}}{n+1}\rightarrow s\]
as $n\rightarrow\infty$, then $s_{n}\rightarrow s$
as $n\rightarrow\infty$. (Results like this are
called Tauberian theorems.)

\end{exercise}
\begin{exercise} (i)
Suppose that $f:[-\pi,\pi)\rightarrow{\mathbb R}$
is increasing and bounded. Write $f(\pi)=\lim_{t\rightarrow 0}
f(\pi -t)$.
Show that
\[\int_{-\pi}^{\pi}f(t)\exp it\, dt=
\int_{0}^{\pi}(f(t)-f(t-\pi))\exp it\, dt\]
and deduce that $|\hat{f}(1)|\leq 
(f(\pi)-f(-\pi))/2\leq(f(\pi)-f(-\pi))$.

(ii) Under the assumptions of (i) show that
\[|\hat{f}(n)|\leq (f(\pi)-f(-\pi))/|n|\]
for all $n\neq 0$.

(iii) (Dirichlet's theorem) Suppose that $g=f_{1}-f_{2}$
where $f_{k}:[-\pi,\pi)\rightarrow{\mathbb R}$
is increasing and bounded $[k=1,2]$. (It can be
shown that functions $g$ of this form are the,
so called, functions of bounded variation.)
Show that if $g$ is continuous at $t$, then
$S_{n}(g,t)\rightarrow f(t)$ as $n\rightarrow\infty$.
\end{exercise}


Most readers will already be aware of the next fact.
\begin{lemma} If $f:{\mathbb T}\rightarrow{\mathbb C}$
is continuously differentiable, then
\[(f')\hat{\ }(n)=in\hat{f}(n).\]
\end{lemma}
This means that Lemma~\ref{pointwise} applies,
but we can do better.
\begin{lemma}\label{once is}
If $f:{\mathbb T}\rightarrow{\mathbb C}$
is continuously differentiable, then
\[\sum_{n=-\infty}^{\infty}|\hat{f}(n)|<\infty.\]
\end{lemma}

Here is a beautiful application due to Weyl
of Theorem~\ref{density}. If $x$ is real,
let us write $\langle x\rangle$ for the fractional part
of $x$, that is, let us write
\[\langle x\rangle=x-[x].\]
\begin{theorem}\label{Weyl} If $\alpha$ is an irrational
number and $0\leq a\leq b\leq 1$, then
\[\frac{\card\{1\leq n\leq N \mid \langle n\alpha\rangle
\in [a,b]\}}{N}
\rightarrow b-a\]
as $N\rightarrow\infty$. The result is false
if $\alpha$ is rational.
\end{theorem}
(Of course this result may be deduced from the ergodic
theorem and Theorem~\ref{density} itself can be deduced
from the Stone-Weierstrass theorem but the techniques
used can be extended in directions not covered by
the more general theorems.)
\section{Jackson's theorems} 
Once we have the idea of 
using different kernels such as Dirichlet's kernel
and  F{\'e}jer's kernel we can try our hand at designing
kernels for a particular purpose. The proof 
of the next theorem provides an excellent example.
\begin{theorem}{\bf [Jackson's first theorem]}\label{T;Jackson one}
There exists a constant $C$ with the following property.
If $f:{\mathbb T}\rightarrow{\mathbb R}$ is once continuously
differentiable then given $n\geq 1$ we can find
a real trigonometric polynomial $P_{n}$ of degree at most $n$
such that 
\[\|P_{n}-f\|_{\infty}\leq Cn^{-1}\|f'\|_{\infty}\]
\end{theorem}
Jackson's theorem provides a quantitative statement
of the idea that
well behaved functions are easier to approximate.
\begin{exercise}\label{E;Easy Jackson} It is easy to obtain a weak
quantitative statement of the idea
well behaved functions are easier to approximate.
Show by integrating by parts that if
$f:{\mathbb T}\rightarrow{\mathbb R}$ is
$k$ times continuously differentiable
\[|\hat{f}(r)|\leq A_{k}|r|^{-k}\|f^{(r)}\|_{\infty}\]
for all $r\neq 0$ and some constant $A_{k}$ independent of $f$.

Deduce that, if $k\geq 2$
\[\|S_{n}(f)-f\|_{\infty}\leq B_{k}|n|^{2-k}\]
for all $n\neq 0$ and some constant $B_{k}$ independent of $f$.
\end{exercise}
Our proof of Theorem~\ref{T;Jackson one} depends
on properties of the Jackson kernel $J_{n}$
defined  by
\[J_{n}(t)=\gamma_{n}^{-1}K_{n}(t)^{2}=
\lambda_{n}^{-1}\left(\frac{\sin\big(nt/2\big)}{\sin t/2}\right)^{4}\]
for $t\neq 0$, $J_{n}(0)=\lambda_{n}^{-1}n^{2}$
where $\gamma_{n}$ and $\lambda_{n}$ are chosen so
that
\[\frac{1}{2\pi}\int_{\mathbb T}J_{n}(t)\,dt=1.\]
\begin{exercise}\label{E;sine estimates} 
By using convexity, or otherwise, show that
\[t\geq \sin t\geq \frac{2t}{\pi}\]
for all $t\in[0,\pi/2]$  
\end{exercise}
\begin{lemma} There exist strictly
positive constants $A$, $A'$ and $B$
such that
\[An^{3}\geq\lambda_{n}\geq A'n^{3}\]
and 
\[\frac{1}{2\pi}\int_{\mathbb T}|t|J_{n}(t)\,dt\leq Bn^{-1}.\]
\end{lemma}
We can now prove a version of Theorem~\ref{T;Jackson one}.
\begin{theorem}\label{T;Jackson double}
There exists a constant $C'$ with the following property.
If $f:{\mathbb T}\rightarrow{\mathbb R}$ is once continuously
differentiable then given $n\geq 1$ we can find
a real trigonometric polynomial $Q_{n}$ of degree at most $2(n-1)$
such that 
\[\|Q_{n}-f\|_{\infty}\leq C'n^{-1}\]
\end{theorem}

\begin{exercise} Deduce Theorem~\ref{T;Jackson one}
from Theorem~\ref{T;Jackson double}.
\end{exercise}

It is easy to guess the generalisation to higher derivatives.
\begin{theorem}{\bf [Jackson's second theorem]}\label{T;Jackson two}
There exists a constant $C_{k}$ with the following property.
If $f:{\mathbb T}\rightarrow{\mathbb R}$ is 
$k$ times continuously
differentiable then given $n\geq 1$ we can find
a real trigonometric polynomial $P_{n}$ of degree at most $n$
such that 
\[\|P_{n}-f\|_{\infty}\leq C_{k}n^{-k}\|f^{(k)}\|_{\infty}\]
\end{theorem}
It is also easy to guess one of the tools used.
\begin{exercise} (i) Suppose that we set
\[J_{n,r}(t)=\gamma_{n}^{-1}K_{n}(t)^{2r}=
\lambda_{n,r}^{-1}\left(\frac{\sin\big(nt/2\big)}{\sin t/2}\right)^{2r}\]
for $t\neq 0$, $J_{n}(0)=\lambda_{n}^{-1}n^{2}$
where $\gamma_{n,r}$ and $\lambda_{n,r}$ are chosen so
that
\[\frac{1}{2\pi}\int_{\mathbb T}J_{n,r}(t)\,dt=1.\]
Show that there exist
constants $B_{n,r,j}$
such that 
\[\frac{1}{2\pi}\int_{\mathbb T}|t|^{j}J_{n}(t)\,dt\leq 
B_{n,r,j}n^{-j}\]
for all $0\leq j\leq 2r-2$.

(ii) It is instructive (though not necessary) to see that
our particular choice of kernel is not unique. 
Construct another trigonometric polynomial 
$\tilde{J}_{n,r}$ of degree at most $Arn$ (for some suitable
constant A) which is everywhere real and positive 
and satisfies
\[\frac{1}{2\pi}\int_{\mathbb T}|t|^{j}\tilde{J}_{n,r}(t)\,dt\leq 
\tilde{B}_{n,r,j}n^{-j}\]
for all $0\leq j\leq 2r-2$.
\end{exercise}
Our proof of Jackson's first theorem depended
on the mean value inequality
\[|f(s)-f(t)|\leq\|f'\|_{\infty}|t-s|.\]
We extend this to higher derivatives by using the
difference operator $\triangle_{h}$ defined by
\[(\triangle_{h}f)(x)=f(x+h)-f(x).\]
We write $\triangle_{h}^{1}f=\triangle_{h} f$
and $\triangle_{h}^{n}f=\triangle_{h}(\triangle^{n-1} f)$
\begin{exercise}\label{E;difference operator} 
Let $f\in C_{\mathbb R}({\mathbb T})$. 

(i) Using induction, or otherwise,
show that, if $f$ is $k$ times continuously differentiable,
then
\[\|\triangle_{h}^{k}f\|_{\infty}\leq k!|h|^{k}\|f^{(k)}\|_{\infty}.\]

(ii) Using induction, or otherwise, show that
\[\triangle_{h}^{k}f(x)=\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}f(x+jh).\]
\end{exercise}
As before we prove a slight variant of the theorem as stated.
\begin{theorem}\label{T;Jackson second}
There exist constants $C_{k}'$ with the following property.
If $f:{\mathbb T}\rightarrow{\mathbb R}$ is 
$k$ times continuously
differentiable then given $n\geq 1$ we can find
a real trigonometric polynomial $Q_{n}$ of degree at most $2(n-1)k$
such that 
\[\|Q_{n}-f\|_{\infty}\leq C_{k}'n^{-k}\|f^{(k)}\|_{\infty}\]
\end{theorem}
\begin{exercise} Here is a another proof of Jackson's second
theorem. We did not use it because we want to extend the proof to $n$
dimensions.

(i) Suppose that $f:{\mathbb T}\rightarrow{\mathbb R}$
is once continuously differentiable and in addition
$\int_{\mathbb T}f(t)\,dt=0$. If $P$ is a real trigonometric
polynomial of such that $\|f-P\|_{\infty}\leq\epsilon$ show that
$|\hat{Q}(0)|\leq\epsilon$.

(ii) Suppose that $f:{\mathbb T}\rightarrow{\mathbb R}$
is once continuously differentiable and in addition
$\int_{\mathbb T}f(t)\,dt=0$.
Show that there exists a real trigonometric polynomial
$Q_{n}$ of degree at most $n$ with $\hat{Q}_{n}(0)=0$
and
\[\|Q_{n}-f\|_{\infty}\leq 2C_{1}n^{-1}\|f'\|_{\infty}\]
where $C_{1}$ is the constant that occurs in Theorem~\ref{T;Jackson one}.

(iii) Suppose that $f:{\mathbb T}\rightarrow{\mathbb R}$
is twice continuously differentiable. By using~(ii), show
that
there exists a real trigonometric polynomial
$Q_{n}$ of degree at most $n$ with $\hat{Q}_{n}(0)=0$
and
\[\|Q_{n}-f'\|_{\infty}\leq 2C_{1}n^{-1}\|f''\|_{\infty}.\]
Hence show that there is real trigonometric polynomial
$R_{n}$ of degree at most $n$ with
\[\|R_{n}-f\|_{\infty}\leq 2C_{1}^{2}n^{-2}\|f''\|_{\infty}.\]

(iv) Prove Theorem~\ref{T;Jackson two}
\end{exercise}

\begin{exercise}\label{E;Best Jackson} (i) If $f\in C({\mathbb T}$
and $P$ is trigonometric polynomial of degree at most $n$
show that
\[\|f-P\|_{\infty}\geq\left(\frac{1}{2\pi}\int_{\mathbb T}
|f(t)-P(t)|^{2}\,dt\right)^{1/2}\geq |\hat{f}(m)|\]
for any $|m|>n$.

(ii) Let $0<n(1)<n(2)<\ldots$. If
\[f(t)=\sum_{j=0}^{\infty} 2^{-j}n(j)^{-1}\cos n(j)t\]
show that $f$ is a well defined once continuously differentiable function
with $\|f'\|_{\infty}\leq 1$ and
\[\inf\{\|f-P\|_{\infty}\,:\, \text{$P$ a trigonometric
polynomial of degree at most $n(j)-1$}\}\geq 2^{-j}n(j)^{-1}.\]
Deduce that, if $\kappa_{n}$ is a sequence of strictly positive numbers
with $\kappa_{n}n\rightarrow \infty$
there exists a continuously differentiable function
$g$ with
\[\limsup \kappa_{n}\inf\{\|f-P\|_{\infty}\,:\, \text{$P$ a trigonometric
polynomial of degree at most $n$}\}=\infty.\]
In this sense, Jackson's first theorem is best possible.

(iii) Show that Jackson's second theorem is best possible
in a similar sense.
\end{exercise}
We shall need a multidimensional version of
Jackson's second theorem. The proof is a natural induction
but notationally a bit tedious.

 
We use the norm $\|\ \|_{p}$ on $C^{p}({\mathbb T}^{n})$
(the space of real $p$ times continuously differentiable
functions on $C^{p}({\mathbb T}^{n})$
given by
\[\|f\|_{p}=\|f\|+\frac{1}{n}
\sum_{1\leq p_{1}+p_{2}+\ldots+p_{n}\leq p}
\sup_{{\mathbf t}\in{\mathbb T}^{n}}\left|
\frac{\partial^{p_{1}+p_{2}+\ldots+p_{n}}f}
{\partial^{p_{1}}x_{1} \partial^{p_{2}}x_{2}
\ldots  \partial^{p_{n}}x_{n}}({\mathbf t})
\right|.\]
\begin{theorem}\label{T;Jackson many}
There exist constants $A_{p,n}$ with the following property.
If $f\in C^{p}({\mathbb T}^{n})$,
then, given $N\geq 1$ we can find
a real trigonometric polynomial
\[P_{N}({\mathbf t})=\sum_{|j(u)|\leq N}a_{j(1)j(2)\ldots j(n)}
\exp\left(i\sum_{v=1}^{n}j(v)t_{v}\right)\]
such that
\[\|P_{N}-f\|_{\infty}\leq A_{p,n}N^{-p}\|f\|_{p}.\]
\end{theorem}
\section{Vistu{\v s}kin's theorem} We now prove
a theorem of Vistu{\v s}kin
which shows that, if we demand continuous differentiability,
there are genuine functions of several variables.
We need a definition to make things precise.

\begin{definition}\label{D;superposition} Let $n>m\geq 1$.
Define $c_{j},\,s_{j}\in C({\mathbb T}^{n})$ by
\[s_{j}({\mathbf t})=\sin t_{j},\ c_{j}({\mathbf t})=\cos t_{j}
\ [1\leq j\leq n].\]
and write
\[{\mathcal E}_{0}=\{s_{j}\,:\,1\leq j\leq n\}
\cup\{c_{j}\,:\,1\leq j\leq n\}.\]
If $E$ is a subset of $C({\mathbb T}^{m}$  
define ${\mathcal E}_{r}(E)$, inductively
by setting ${\mathcal E}_{0}(E)={\mathcal E}_{0}$
and taking ${\mathcal E}_{r}(E)$ 
to be the set of all functions 
$f\in C({\mathbb T}^{n})$
given by
\[f({\mathbf t})=g(u_{1}({\mathbf t}),u_{2}({\mathbf t}),
\ldots,u_{m}({\mathbf t})\]
with $u_{l}\in {\mathcal E}_{r-1}(E)$ $[1\leq l\leq m]$ and $g\in E$.
We say that an $f\in C_{\mathbb R}({\mathbb T}^{n})$
is written in terms of functions in $E$
if $f\in {\mathcal E}_{r}(E)$ for some $r\geq 1$.
\end{definition}

We can now state our theorem.
\begin{theorem}\label{T;Vitushkin}  
If $n>m\geq 1$, $p\geq q\geq 1$ and $n/p>m/q$
there exists an 
$f\in C^{p}({\mathbb T}^{n})$ which cannot be written in terms
of functions in $C^{q}({\mathbb T}^{m})$. 
\end{theorem}

Our proof depends on the notion of $\epsilon$-entropy
introduced by Kolmogorov.
\begin{definition}\label{D;entropy} 
We work in $C_{\mathbb R}({\mathbb T}^{n})$
equipped with the uniform norm. Let $E$ be a subset
of $C_{\mathbb R}({\mathbb T}^{n})$ and $\epsilon>0$.
If $E$ cannot be covered by a finite set of 
closed balls
\[\tilde{B}(f,\epsilon)=\{g\in C_{\mathbb R}({\mathbb T}^{n})\,
:\,\|f-g\|_{\infty}\leq \epsilon\}\]
we take $H(\epsilon,E)=\infty$. If $E$ can be covered by
a finite set of such balls,
we write $N(\epsilon,E)$ for the least number of balls required
and define $H(\epsilon,K)$,
\emph{the $\epsilon$-entropy of $K$} by
\[H(\epsilon,K)=\log N(\epsilon,K).\]
\end{definition}
Suppose that we are using the functions $f\in E$ as messages
but we cannot distinguish two messages $f_{1}$ and $f_{2}$
if their uniform distance is less than about $\epsilon$.
Then \emph{very roughly speaking} we can only distinguish
about $N(\epsilon,E)$ messages and the amount of information
we can send (defined \emph{roughly speaking} as the logarithm
of the number of possible distinct messages) is about $H(\epsilon,K)$.

We need the following simple observation. (Here
$\Cl_{\infty}E)$ denotes the closure in the uniform norm.)
\begin{exercise}\label{E;easy close} Let $E$ be a subset
of $C_{\mathbb R}({\mathbb T}^{n})$ and $\epsilon>0$.
Then 
\[H(\epsilon,E)=\log N(\epsilon,\Cl_{\infty}E).\]
\end{exercise}
Since we are interested in the behaviour of $H(\epsilon,E)$
as $\epsilon\rightarrow 0$, we shall only be interested
in those $E$ whose uniform closure is compact
in $(C_{\mathbb R}({\mathbb T}^{n},\|\ \|)$, that
is to say those $E$ which are bounded and uniformly equicontinuous.
(If you have not met uniform equicontinuity before will will
talk about it later.)

The sets $E$ we shall consider are balls in
$C^{p}({\mathbb T}^{n})$ with an the norm $\|\ \|_{(p)}$.
defined when we introduced
Theorem~\ref{T;Jackson many}.

The key inequality is given by the next theorem.
\begin{theorem}\label{T;key Vitushkin} Let $B_{p,n}$ be the 
closed unit ball in
$(C^{p}({\mathbb T}^{n}),\|\ \|_{(p)})$.
Then there exist constants $C_{p,n}$ and $C_{p,n}'$ such that
\[C_{n}\epsilon^{-n/p}\leq H(\epsilon,B_{p,n})
\leq C_{n}'\epsilon^{-n/p}\log\epsilon^{-1}\]
for $0<\epsilon<1/2$.
\end{theorem}

Notice that this theorem is a \emph{quantitative} version
of the much simpler observation that $B_{n}$ is uniformly 
equicontinuous.
Notice also that we will be considering two sorts of balls:-
\emph{uniform balls}, that is to say balls in
$(C_{\mathbb R}({\mathbb T}^{n},\|\ \|)$,
and $C^{p}$-\emph{balls}, that is to say balls in
$(C_{\mathbb R}^{(p)}({\mathbb T}^{n},\|\ \|_{(p)})$. 

The  next lemma brings us closer to Vistu{\v s}kin's theorem
\begin{lemma}\label{L;large key Vitushkin} 
Let $n>m\geq 1$. Let $B_{q,m}$ be the unit ball in
$(C^{q}({\mathbb T}^{m}),\|\ \|_{(1)})$.
If $r$ and $u$ are strictly positive integers
then, using the notation of Definitions~\ref{D;superposition}
and~\ref{D;entropy}, we know that there is a constant
$C(q,r,u,m,n)$ such that
\[H(\epsilon,{\mathcal E}_{r}(uB_{q,m})\leq C(q,r,um,n)
\epsilon^{-m}\log\epsilon^{-1}\]
for all $0<\epsilon<1/2$.
\end{lemma}
We can now prove a Baire category version of  Theorem~\ref{T;Vitushkin}.  
\begin{theorem}  
If $n>m\geq 1$, $p,\,q\geq 1$
$p/n\geq q/m$ and we work in $(C^{p}({\mathbb T}^{n}),\|\ \|_{(p)})$, 
then quasi-all functions in $(C^{p}({\mathbb T}^{n}),\|\ \|_{(p)})$
cannot be written in terms
of functions in $C^{q}({\mathbb T}^{m})$. 
\end{theorem}
\begin{exercise} Modify the discussion above to show that
there exists an 
$f\in C^{1}({\mathbb T}^{2})$ which cannot be written in terms
of functions in $C^{1}({\mathbb T})$ and the 
addition function function $(x,y)\mapsto x=y$.
\end{exercise} 
\section{Simple connectedness and the logarithm}
The rest of these lectures
form a short second course in
complex variable theory with an emphasis on technique
rather than theory. None the less I intend to be
rigorous and you should feel free to question any
`hand waving' that I indulge in.

But where should rigour start? It is neither necessary nor
desirable to start by reproving all the results of 
a first course. Instead I shall proceed on the assumption
that all the standard theorems (Cauchy's theorem, Taylor's
theorem, Laurent's theorem and so on) have been proved
rigorously for analytic functions\footnote{Analytic functions
are sometimes called `holomorphic functions'. We shall call
a function which is `analytic except for poles' a
`meromorphic function'.} on an open disc
and extend them as necessary.

Almost all the members of the audience are
already familiar with one sort of extension.
\begin{definition} An open set $U$ in ${\mathbb C}$
is called \emph{disconnected}
if we can find open sets $U_{1}$ and $U_{2}$ such that

(i) $U_{1}\cup U_{2}=U$,

(ii) $U_{1}\cap U_{2}=\emptyset$,

(iii) $U_{1},U_{2}\neq\emptyset$.

An open set which is not disconnected is called
\emph{connected}.
\end{definition}
\begin{theorem}\label{no isolated}
If $U$ is an open connected set
in ${\mathbb C}$ and $f:U\rightarrow {\mathbb C}$
is analytic and not identically zero then
all the zeros of $f$ are isolated that is,
given $w\in U$ with $f(w)=0$ we can find a $\delta>0$
such that $D(w,\delta)\subseteq U$ and $f(z)\neq 0$
whenever $z\in D(w,\delta)$ and $z\neq w$.
\end{theorem}
Here and elsewhere
\[D(w,\delta)=\{z:|w-z|<\delta\}.\]
The hypothesis of connectedness is exactly what we need
in Theorem~\ref{no isolated}.
\begin{theorem} If $U$ is an open set then
$U$ is connected if and only if the zeros of
every non-constant analytic function on $U$
are isolated.
\end{theorem}
If necessary, I shall quote results along the lines of
Theorem~\ref{no isolated} without proof but I will be
happy to give proofs in supplementary lectures if requested.
\begin{exercise}[Maximum principle]\label{Maximum principle}
(i) Suppose that $a,b\in{\mathbb C}$
with $b\neq 0$ and $N$ is an integer with $N\geq 1$.
Show that there is a $\theta\in{\mathbb R}$ such that
\[|a+b(\delta\exp i\theta)^{N}|=|a|+|b|\delta^{N}\]
for all real $\delta$ with $\delta\geq 0$.

(ii) Suppose that
$f:D(0,1)\rightarrow{\mathbb C}$ is analytic.
Show that
\[f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\]
where there is some constant $M$ such that $|a_{n}|\leq M2^{n}$
(we can make much better estimates). Deduce that
either $f$ is constant or we can find $N\geq 1$
and $a_{N}\neq 0$ such that
\[f(z)=a_{0}+(a_{N}+\eta(z))z^{N}\]
with $\eta_{z}\rightarrow 0$ as $z\rightarrow 0$.

(iii) If $U$ is a connected open subset of ${\mathbb C}$
and $f$ is a non-constant analytic function on $U$, show
that $|f|$
has no maxima.

(iv) Does the result of (iii) mean that $f$ is unbounded on $U$?
Give reasons.

(v) Show that if is an open set which is not connected then
there exists a non-constant analytic function $f$ on $U$
such that $|f|$ has a maximum.
\end{exercise}
\begin{exercise}\label{Open mapping theorem}
(i) Suppose $f:D(0,1)\rightarrow{\mathbb C}$ is
a non-constant analytic function with $f(0)=0$.
Show that we can find a $\delta$ with $0<\delta<1$
such that $f(z)\neq 0$ for all $|z|=\delta$ and
an $\epsilon>0$ such that $|f(z)|\geq\epsilon$ for all
$|z|=\delta$. Use Rouch\'{e}'s theorem to deduce
that $f(D(0,1))\supseteq D(0,\epsilon)$.

(ii) {\bf (Open mapping theorem)}
If $U$ is a connected open subset of ${\mathbb C}$
and $f$ is a non-constant analytic function on $U$ show
that $f(U)$ is open.

(iii) Deduce the result of Exercise~\ref{Maximum principle}.
(Thus the maximum principle follows from the open
mapping theorem.)
\end{exercise}
\begin{exercise} Let $D$ be the open unit disc. 
Suppose $f:D\rightarrow{\mathbb C}$
is analytic and $f(0)=f'(0)=0$ but $f$ is not
identically zero. Use Rouch{\'e}'s theorem
(and the fact that the zeros of $f'$ are isolated) to
show that there exists $\eta_{1},\eta_{2}>0$ such that
if $0<|w|<\eta_{1}$ the equation $f(z)=w$ has at least
two distinct solutions with $|z|<\eta_{2}$.

Deduce that if $\Omega_{1}$ and $\Omega_{2}$ are open
subsets of ${\mathbb C}$ and $f$ is a \emph{conformal 
mapping} of $\Omega_{1}$ into $\Omega_{2}$ then
the inverse map $f^{-1}$ is analytic and so a conformal
map of $\Omega_{2}$ into $\Omega_{1}$. 
\end{exercise}
 
It can be argued that much of complex analysis reduces
to the study of the logarithm and this course is no
exception. We need a general condition on an open set
which allows us to define a logarithm. Recall
that we write ${\mathbb T}={\mathbb R}/2\pi{\mathbb Z}$.
\begin{definition} An open set $U$ in ${\mathbb C}$
is said to be \emph{simply connected} if it is connected
and given any continuous function
$\gamma:{\mathbb T}\rightarrow U$ we can find
a continuous function
$G:[0,1]\times{\mathbb T}\rightarrow U$
such that
\begin{align*}
G(0,t)&=\gamma(t)\\
G(1,t)&=G(1,0)
\end{align*}
for all $t\in {\mathbb T}$.
\end{definition}
In the language of elementary algebraic topology
a connected open set is simply connected if every
loop can be homotoped to a point.
\begin{theorem}\label{logarithm}
If $U$ is an open simply connected
set in ${\mathbb C}$ that does not contain $0$
we can find an analytic function $\log:U\rightarrow{\mathbb C}$
such that $\exp(\log z)=z$ for all $z\in U$. The
function $\log$ is unique up to the addition of
integer multiple of $2\pi i$.
\end{theorem}


From an elementary viewpoint, the most direct
way of proving Theorem~\ref{logarithm} is to show
that any piece wise smooth loop can be homotoped
\emph{through piecewise smooth loops} to
a point and then use the integral definition of
the logarithm. However, the proof is a little
messy and we shall use a different approach
which is longer but introduces some useful ideas.
 
\begin{theorem}\label{logarithm exists}
(i) If  $0<r<|w|$
we can find an analytic function
$\log:D(w,r)\rightarrow{\mathbb C}$
such that $\exp(\log z)=z$ for all $z\in D(w,r)$. The
function $\log$ is unique up to the addition of
integer multiple of $2\pi i$.

(ii) If $\gamma:[a,b]\rightarrow{\mathbb C}\setminus\{0\}$
is continuous we can find a continuous function
$\tilde{\gamma}:[a,b]\rightarrow{\mathbb C}$
such that $\exp\circ\tilde{\gamma}=\gamma$.

(iii) Under the hypotheses of (ii), if
$\tilde{\tilde{\gamma}}:[a,b]\rightarrow{\mathbb C}$
is a continuous function such that
$\exp\circ\tilde{\tilde{\gamma}}=\gamma$
then we can find an integer $n$ such that
$\tilde{\tilde{\gamma}}=\tilde{\gamma}+2\pi i n$.

(iv) If $U$ is a simply connected open set
not containing $0$ then, if
$\gamma:[a,b]\rightarrow U$ is continuous,
$\gamma(a)=\gamma(b)$,
and $\tilde{\gamma}:[a,b]\rightarrow{\mathbb C}$
is a continuous function such that
$\exp\circ\tilde{\gamma}=\gamma$
then $\tilde{\gamma}(a)=\tilde{\gamma}(b)$.
\end{theorem}
Theorem~\ref{logarithm} is now relatively easy to prove.

It would be nice to show that simple connectedness is
the correct condition here.
The following result, although not the best possible,
is hard enough and shows that this is effectively the case.
\begin{lemma}\label{logarithm exact}
Suppose that $U$ is a  non-empty open connected
set in ${\mathbb C}$ with non-empty complement.
The following two conditions are equivalent.

(i) The set $U$ is simply connected.

(ii) If $f:U\rightarrow{\mathbb C}$ is a
non-constant analytic
function with no zeros
then we can find an analytic function
$\log:f(U)\rightarrow{\mathbb C}$
with $\exp(\log f(z))=f(z)$ for all $z\in U$
\end{lemma}
(In looking at condition~(ii), recall that the open mapping
theorem given in Exercise~\ref{Open mapping theorem}
tells us that $f(U)$ is open.)
The reader is invited to try and prove this result
directly but
we shall obtain it only after a long chain of arguments
leading to the Riemann mapping theorem.

The following result is trivial but worth noting.
\begin{lemma} If $U$ and $V$ are open subsets of ${\mathbb C}$
such that there exists a homeomorphism $f:U\rightarrow V$
then if $U$ is simply connected so is $V$.
\end{lemma}
\begin{exercise} In the next two exercises
we develop an alternative approach to
Theorem~\ref{logarithm} along the lines suggested above.

(i) Suppose that $U$ is an open set in ${\mathbb C}$
and that $G:[0,1]\times{\mathbb T}\rightarrow U$
is a continuous function. Show, by using compactness
arguments or otherwise, that there exists an $\epsilon>0$
such that $N(G(s,t),\epsilon)\subseteq U$ for all
$(s,t)\in [0,1]\times{\mathbb T}$, and that we can find
an integer $N\geq 1$ such that if
\[(s_{1},t_{1}),(s_{2},t_{2})\in [0,1]\times{\mathbb T}
\ \text{and}\ |s_{1}-s_{2}|<4N^{-1},
\ |t_{1}-t_{2}|<8\pi N^{-1}\]
then $|G(s_{1},t_{1})-G(s_{2},t_{2})|<\epsilon/4$.

(ii) Continuing with the notation and hypotheses of
of (i) show that if
$\gamma_{1},\gamma_{2}:{\mathbb T}\rightarrow{\mathbb C}$
are the piecewise linear functions\footnote{Strictly speaking
the simplest piecewise linear functions.} with
\begin{align*}
\gamma_{0}(2\pi r/N)&=G(0,2\pi r/N)\\
\gamma_{1}(2\pi r/N)&=G(1,2\pi r/N)
\end{align*}
for all integers $r$ with $0\leq r\leq N$
then there exists a constant $\lambda$
and a continuous function
$H:[0,1]\times{\mathbb T}\rightarrow U$
with
\begin{align*}
H(0,t)=\gamma_{0}(0,t)\\
H(1,t)=\gamma_{1}(1,t)
\end{align*}
for all $t\in [0,1]$, such that, for each fixed $t$,
$H(s,t)$ is a piecewise
linear function of $s$ and the curve
$H(\ ,t):{\mathbb T}\rightarrow U$ is of length less than
$\lambda$.

(iii) Continuing with the notation and hypotheses of
of (i) show that if $G(s,1)$ and $G(s,0)$ are piecewise
smooth functions of $s$
then there exists a constant $\lambda$ and
a continuous function
$F:[0,1]\times{\mathbb T}\rightarrow U$
with
\begin{align*}
F(0,t)=\gamma_{0}(0,t)\\
F(1,t)=\gamma_{1}(1,t)
\end{align*}
for all $t\in [0,1]$, such that, for each fixed $t$,
$F(s,t)$ is a piecewise
smooth function of $s$ and the curve
$F(\ ,t):{\mathbb T}\rightarrow U$ is of length less than
$\lambda$.

(iv) Show that in any simply connected open set
any piece wise smooth loop can be homotoped
through piecewise smooth loops of bounded length to
a point.
\end{exercise}
\begin{exercise} (i) Suppose that $U$ is an open set
in ${\mathbb C}$ and
$F:[0,1]\times{\mathbb T}\rightarrow U$ is a continuous
function such that, for each fixed $t$,
$F(s,t)$ is a piecewise
smooth function of $s$ and the curve
$F(\ ,t):{\mathbb T}\rightarrow U$ is of length less than
$\lambda$. We write $\Gamma_{s}$ for the contour
defined by $F(\ ,t)$. Show by a compactness argument, or
otherwise, that if $f:U\rightarrow{\mathbb C}$ is continuous
then
$\int_{\Gamma_{s}}f(z)\,dz$ is a continuous function of $s$.

(ii) If $0<\delta<|w|$ show that if $\Gamma$ is a contour
lying entirely within $N(w,\delta)$ joining
$z_{1}=r_{1}e^{i\theta_{1}}$ to $z_{2}=r_{2}e^{i\theta_{2}}$
[$r_{1},r_{2}>0$, $\theta_{1},\theta_{2}\in{\mathbb R}$]
show that
\[\int_{\Gamma}\frac{1}{z}\,dz=(\log r_{2}-\log r_{1})
+i(\theta_{1}-\theta_{2})+2n\pi i\]
for some integer $n$.

(iii) By using compactness arguments to split
$\Gamma$ into suitable bits, or otherwise,
show that if $U$ is any open set not containing $0$
and $\Gamma$ is any closed contour (i.e. loop)
lying entirely within $U$ then
\[\int_{\Gamma}\frac{1}{z}\,dz=2N\pi i\]
for some integer $N$.

(iv) Use results from this exercise and its
predecessor to show that if $U$ is any simply
connected open set not containing $0$
and $\Gamma$ is any closed contour
lying entirely within $U$ then
\[\int_{\Gamma}\frac{1}{z}\,dz=0.\]
Hence, prove Theorem~\ref{logarithm}.
\end{exercise}
\begin{exercise} Let us say that two open subsets  
of  ${\mathbb C}$, $\Omega_{1}$ and $\Omega_{2}$
are \emph{conformally equivalent} if there is 
a conformal 
mapping of $\Omega_{1}$ into $\Omega_{2}$.
Show that conformal equivalence is indeed an
equivalence relation.
\end{exercise} 



\section{The Riemann mapping theorem} By using a very
beautiful physical argument,  Riemann obtained the
following result.
\begin{theorem}[Riemann mapping theorem]%
\label{T;Riemann map}
If $\Omega$ is an non-empty,
open, simply connected subset
of $\mathbb C$ with non-empty complement
then there exists a conformal
map of $\Omega$ to the unit disc $D(0,1)$.
\end{theorem}
Notice that we can reduce this result to a version
which is easier to think about.
\begin{theorem}\label{T;Riemann blob} If $\Omega$ is an
open simply connected subset of $D$ and $0\in \Omega$,
then there exists a conformal map $f:\Omega\rightarrow D$.
\end{theorem}

Unfortunately his argument depended on the assumption
of the existence
of a function which minimises a certain energy.
Since Riemann was an intellectual giant
and his result is correct
it is often suggested that all that was needed was
a little rigour to be produced by pygmies.
However, Riemann's argument actually fails in
the related three dimensional case  so (in
the lecturer's opinion) although Riemann's
argument certainly showed that a  very wide class
of sets could be conformally transformed into
the unit disc the extreme generality of the final
result could not reasonably have been expected
from his argument alone.

In order to rescue the Riemann mapping theorem
mathematicians embarked on two separate programmes.
The first was to study conformal mapping in more detail
and the second to find abstract principles to guarantee
the existence of minima in a wide range of general
circumstances (in modern terms, to find appropriate
compact spaces). The contents of this section
come from the first of these programmes,
the contents of the next (on equicontinuity) come
from the second. (As a point of history, the first
complete proof of the Riemann mapping theorem
was given by Poincar\'{e}.)

\begin{theorem}[Schwarz's inequality]%
\label{Schwarz's inequality}
If  $f:D(0,1)\rightarrow D(0,1)$ is analytic and
$f(0)=0$ then

(i) $|f(z)|\leq |z|$ for all $|z|<1$ and $|f'(0)|\leq 1$.

(ii) If $|f(w)|=|w|$ for some $|w|<1$ with $w\neq 0$,
or if $|f'(0)|=1$, then we can find a $\theta\in{\mathbb R}$
such that $f(z)=e^{i\theta}z$ for all $|z|<1$.
\end{theorem}

Schwarz's inequality enables us to classify the conformal
maps of the unit disc into itself. If $a\in D(0,1)$
and $\theta\in{\mathbb R}$ let us write
\begin{align*}
T_{a}(z)&=\frac{z-a}{1-a^{*}z}\\
R_{\theta}(z)&=e^{i\theta}z
\end{align*}
\begin{lemma}\label{L;disc 1A} If $a\in D(0,1)$
and $\theta\in{\mathbb R}$ then $T_{a}$ and $R_{\theta}$
map $D(0,1)$ conformally into itself. Further $T_{a}^{-1}=T_{a}$.
\end{lemma}
\begin{theorem}\label{unique conformal}
(i) If $S$ maps $D(0,1)$ conformally into itself
then we can find $a\in D(0,1)$ and $\theta\in \mathbb{R}$
such that $S=R_{\theta}T_{a}$. If $S=R_{\theta'}T_{a'}$
with $a'\in D(0,1)$ and $\theta'\in \mathbb{R}$ then
$a=a'$ and $\theta-\theta'\in 2\pi{\mathbb Z}$.

(ii) Let $U$ be a simply connected open set and $a\in U$.
If there exists a
conformal map $g:U\rightarrow D(0,1)$
then there exists precisely one conformal map
$f:U\rightarrow D(0,1)$ with $f(a)=0$ and $f'(a)$ real
and positive.
\end{theorem}
We shall prove the following version of Theorem~\ref{T;Riemann blob}.
\begin{theorem}\label{T;Riemann unique blob} If $\Omega$ is an
open simply connected subset of $D$ and $0\in \Omega$,
then there exists a conformal map $f:\Omega\rightarrow D$
with $f(0)=0$ and $f'(0)$ real and positive.
\end{theorem}
\begin{exercise}\label{E;Pick}{\bf [Pick's inequality]} 
Let $a,\,b\in{\mathbb C}$,
$R,\,S>0$. Set 
\[D_{1}=\{z\in{\mathbb C}\,:\,|z-a|<R\},
\ \text{and}
\ D_{2}=\{z\in{\mathbb C}\,:\,|z-b|<S\}.\]
If $f:D_{1}\rightarrow{\mathbb C}$ is analytic and
$f(D_{1})\subseteq D_{2}$ show that
\[\left|S\frac{f(z)-f(w)}{S^{2}-f(z)^{*}f(w)}\right|
\leq\left|R\frac{z-w}{R^{2}-z^{*}w}\right|\]
for $z\neq w$, $z,\,w\in D_{1}$ and
\[|f'(w)|\leq \frac{R}{S}\frac{S^{2}-|f(w)|^{2}}{R^{2}-|w|^{2}}\]
for $w\in D_{1}$
Show that if we have equality in the first inequality
for any $z$ and $w$ or in the second for any $w$, then
$f$ is a M{\"o}bius map.
\end{exercise}

We conclude this section with some results which are not
needed for the proof of the Riemann mapping theorem but
which show that the `surrounding scenery'.

There is one remark that needs to be made.
to add one remark. In elementary complex variable theory
we use the heuristic `see how the boundaries transform'.
This worked in Lemma~\ref{L;disc 1A} because M{\"o}bius maps
are defined on the whole plane ${\mathbb C}$ (apart, perhaps,
from one point). If we look at the more general conformal maps
considered in Riemann's theorem, we run into two linked
problems. The first is that the boundary of an open simply
connected set may be rather complicated and the second is that
the conformal maps may not extend to `nice' maps on the closure
of the sets considered. I do not wish to spend time showing
precisely how nasty things can become, but the following
simple
example should convince you that it would be ill advised
to try to include boundaries in our discussion.
\begin{exercise}\label{E;bad boundary} Find an explicit conformal
map $T$ taking
\[\Omega=D\setminus\{x\in{\mathbb R}\,0\leq x\leq 1\}\]
to $D$. Show that there is no continuous function
$S:\Cl{\Omega}\rightarrow\Cl D$ with $Sz=Tz$ for all $z\in\Omega$.

Explain why if $\tilde{T}$ is any conformal map of $\Omega$
to $D$  there is no continuous function
$\tilde{S}:\Cl{\Omega}\rightarrow\Cl D$ with 
$\tilde{S}z=\tilde{T}z$ for all $z\in\Omega$.
\end{exercise}



\begin{example}\label{E;two points} 
If $a,b\in D(0,1)$ then there exists
a conformal map
\[f:D(0,1)\setminus\{a\}\rightarrow D(0,1)\setminus\{b\}.\]
\end{example}
\begin{example}\label{two points}
If $a_{1},a_{2},b_{1},b_{2}\in D(0,1)$
then there exists
a conformal map
\[f:D(0,1)\setminus\{a_{1},a_{2}\}\rightarrow 
D(0,1)\setminus\{b_{1},b_{2}\}\]
if and only if
\[\left|\frac{a_{2}-a_{1}}{a_{1}^{*}a_{2}-1}\right|
=\left|\frac{b_{2}-b_{1}}{b_{1}^{*}b_{2}-1}\right|.\]
\end{example}
In Example~\ref{E;two points} we see the the `natural rigidity'
of complex analysis reassert itself.

We now give a more complicated example of this rigidity.
The ideas will be reused later in the proof of Picard's Little Theorem.
I assume the reader has already met.
\begin{definition} (i) Let ${\mathbf p}$ and ${\mathbf q}$
are orthonormal vectors in ${\mathbb R}^{2}$.
If ${\mathbf a}$ is a vector in ${\mathbb R}^{2}$
and $x,y\in{\mathbb R}$ the reflection of
${\mathbf a}+x{\mathbf p}+y{\mathbf q}$ in the
line through ${\mathbf a}$ parallel to ${\mathbf p}$
is ${\mathbf a}+x{\mathbf p}-y{\mathbf q}$.

(ii) If ${\mathbf a}$ and ${\mathbb b}$
are vectors in ${\mathbb R}^{2}$ and $R,r>0$
then the reflection of ${\mathbf a}+r{\mathbf b}$
in the circle centre ${\mathbf a}$ and radius $R$
is ${\mathbf a}+r^{-1}R^{2}{\mathbf b}$.
\end{definition}

\begin{lemma}[Schwarz reflection principle]%
\label{L;Schwarz reflection}
Let $\Sigma_{1}$ and $\Sigma_{2}$ be two circles
(or straight lines).
Suppose $G$ is an open set which is taken to
itself by reflection in $\Sigma_{1}$. Write $G_{+}$
for that part of $G$ on one side\footnote{There are no
topological difficulties here. The two sides of $|z-a|=r$
are $\{z:|z-a|<r\}$ and $\{z:|z-a|>r\}$.} of
$\Sigma_{1}$ and $G_{0}=G\cap\Sigma_{1}$.
If $f:G_{+}\cup G_{0}$ is a continuous function,
analytic on $G_{+}$ with $f(G_{0})\subseteq \Sigma_{2}$
then we can find an analytic function
$\tilde{f}:G\rightarrow{\mathbb C}$ with
$\tilde{f}(z)=f(z)$ for all $z\in G_{+}\cup G_{0}$.
If $f(G_{+})$ lies on one side of $\Sigma_{2}$
then we can ensure that $\tilde{f}(G_{-})$ lies on the other.
\end{lemma}
We first prove the result when $\Sigma_{1}$ and $\Sigma_{2}$
are the real axis and then use M\"{o}bius transforms
to get the full result.

The next theorem is less powerful than it appears
(The theorem we would wish for is true, but will not be proved
here.)
\begin{theorem} There exists
a homeomorphism 
\[f:\{z\,:\,a\leq |z|\leq b\}\rightarrow\{z\,:\,A\leq |z|\leq B\}\]
whose restriction
\[\tilde{f}:\{z\,:\,a<|z|< b\}\rightarrow\{z\,:\,A<|z|<B\}\]
is conformal if and only if $a/b=A/B$.
\end{theorem}
\begin{exercise} If $a/b=A/B$ find all the maps of the type
described in the previous theorem.
\end{exercise}
\section{Equicontinuity} (The reader is welcome to take a different
approach but if she derives the results from more general theorems
she must be able to prove those theorems.)
We start with a general remark.
\begin{lemma} If $\Omega$ is an open subset of ${\mathbb R}^{n}$
we can find a sequence $K_{1}$, $K_{2}$, \ldots
of compact subsets of $\Omega$ such that
\[\bigcup{n=1}^{\infty}K_{n}=\Omega.\]
\end{lemma}
\begin{definition}\label{D;uniform eqicontinuity} 
Let $K$ be a compact subset of ${\mathbb R}^{n}$
with the standard metric $d$. 
A subset ${\mathcal F}\subseteq C(K)$ is said to be
\emph{uniformly equicontinuous}\footnote{Traditionally 
the word `equicontinuous'
was used instead of the phrase `uniformly equicontinuous'
and the majority of mathematicians follow this older usage.
See  Exercise~\ref{E;equicontinuity}.} 
if, given $\epsilon>0$, we can
find a $\delta(\epsilon)>0$ such that $|f(x)-f(y)|<\epsilon$ 
whenever $d(x,y)<\delta(\epsilon)$ and $f\in{\mathcal F}$.
\end{definition}
\begin{exercise}\label{E;equicontinuity} 
We use the notation and hypotheses of 
A subset ${\mathcal F}\subseteq C(K)$ is said to be
\emph{equicontinuous at the point $x$} if, given $\epsilon>0$, we can
find a $\delta(\epsilon)>0$ such that $|f(x)-f(y)|<\epsilon$ 
whenever $d(x,y)<\delta(\epsilon)$ and $f\in{\mathcal F}$.
Show that, if ${\mathcal F}$ is eqicontinuous at every point of $K$,
then  ${\mathcal F}$ is uniformly eqicontinuous.
\end{exercise}
\begin{definition} Let $X$ be a compact subset of ${\mathbb R}^{n}$.
A subset ${\mathcal F}\subseteq C(X)$ is said to be
\emph{uniformly bounded} if we can
find a $C$ such that $\|f\|_{\infty}\leq C$ 
whenever $f\in{\mathcal F}$.
\end{definition}
\begin{theorem}{\bf [The Arzel{\'a}--Ascoli theorem]}\label{T;Ascoli}
Let $K$ be a compact subset of ${\mathbb R}^{n}$.
Then ${\mathcal F}\subseteq C(K)$ is compact if and
only if ${\mathcal F}$ is closed, uniformly
bounded and uniformly equicontinuous.
\end{theorem}
The natural mode of convergence for analytic functions
on an open set is `converging uniformly on compacta'.
\begin{definition}\label{D;uniform on compacta}
Let $\Omega$ be an open set in ${\mathbb C}$. 
Consider a sequence of $f_{n}:\Omega\rightarrow{\mathbb C}$
and an $f:\Omega\rightarrow{\mathbb C}$. We say that
$f_{n}\rightarrow f$ uniformly on compacta if
whenever $K$ is a compact subset of $\Omega$
$f_{n}\rightarrow f$ uniformly on $K$.
\end{definition}
\begin{exercise} We use the notation and hypotheses of
Definition~\ref{D;uniform on compacta}.

(i) If  $f_{n}\rightarrow f$ uniformly on compacta
and each $f_{n}$ is continuous on $\Omega$,
show that $f$ is continuous on $\Omega$.

(ii) If  $f_{n}\rightarrow f$ uniformly on compacta
and each $f_{n}$ is analytic on $\Omega$,
show that $f$ is analytic on $\Omega$.
\end{exercise}

\begin{theorem}{\bf [Montel's theorem]} 
Let $\Omega$ be an open set in ${\mathbb C}$
and ${\mathcal F}$ a set of analytic 
function $f:{\Omega}\rightarrow{\mathbb C}$.
Then every sequence of functions in ${\mathcal F}$ contains a subsequence 
which is uniformly convergent on compacta 
if and only if ${\mathcal  F}$ is 
uniformly bounded on compacta.
\end{theorem}

The next exercise is not needed, but may help put things in perspective.
\begin{exercise}\label{E;complex Frechet} 
Let $\Omega$ be an open set in ${\mathbb C}$
and suppose that  $K_{1}$, $K_{2}$, \dots are compact sets
such that $K_{m}\subseteq\Omega$ and   
$\bigcap_{m=1}^{\infty}\Int K_{j}=\Omega$. 

(i) Show that
the equation
\[d_{K}(f,g)=\sum_{m=1}^{\infty}2^{-m}\max\{1,\sup_{z\in K_{m}}|f(z)-g(z)|\}\]
defines a metric on $A(\Omega)$, the space of analytic functions
$f:{\Omega}\rightarrow{\mathbb C}$. Show that $d_{K}$ is complete.

(ii) If $f,\,f_{n}\in A(\Omega)$ show that $d_{K}(f,f_{n})\rightarrow 0$
if and only if $f_{n}\rightarrow f$ uniformly on compacta.
If $g_{n}\in A(\Omega)$ show that $g_{n}$ is Cauchy for $d$
if and only if $g_{n}-g_{m}\rightarrow 0$ uniformly on compacta.

Observe that Montel's theorem may be restated as saying 
that a closed subset ${\mathcal F}$ of $\big(A(\Omega),d_{K}\big)$
is compact 
if and only if it is uniformly bounded on compacta.

(iii) If $L_{1}$, $L_{2}$, \dots are compact sets
such that $L_{m}\subseteq\Omega$ and   
$\bigcap_{m=1}^{\infty}\Int L_{j}=\Omega$ and we set
\[d_{L}(f,g)=\sum_{m=1}^{\infty}2^{-m}\max\{1,\sup_{z\in L_{m}}|f(z)-g(z)|\}\]
show that the identity map 
$\iota:\big(A(\Omega),d_{L})\rightarrow\big(A(\Omega),d_{K})$
is a homeomorphism which preserves Cauchy sequences.
Is it true that we can always find a $C>1$ such that
\[Cd_{L}(f,g)\geq d_{K}(f,g)\geq C^{-1}d_{L}(f,g)?\] 
\end{exercise}

We shall be dealing with the
limit of of injective analytic functions and will
make use of the following result. 
\begin{theorem} {\bf[Hurwitz's theorem]}\label{T;Hurwitz} 
Suppose that $\Omega$ is a pathwise connected 
open set. If $f_{n}:\Omega\rightarrow {\mathbb C}$ is
an injective analytic function and $f_{n}\rightarrow f$
uniformly on compacta, then either $f$ is a constant
function or $f$ is injective.
\end{theorem}
We are now in a position to embark on a proof of
the Riemann mapping theorem.
\begin{lemma}\label{L;Riemann one} 
If $\Omega$ is an
open simply connected subset of $D$ and $0\in \Omega$,
then there exists an injective analytic function $f:\Omega\rightarrow D$
with $f(0)=0$, $f'(0)$ real and positive such that,
if $g:\Omega\rightarrow D$ is an injective 
analytic function 
with $g(0)=0$ and $g'(0)$ real and positive,
then $f'(0)\geq g'(0)$.
\end{lemma}
\begin{lemma}\label{L;Riemann two} Suppose that $\Omega$ is an
open simply connected subset of $D$ 
and $f:\Omega\rightarrow D$ is an injective analytic function
with $f(0)=0$ and $f'(0)$ real and positive. If $f$ is
not surjective, we can find 
an injective analytic function $g:\Omega\rightarrow D$
with $g(0)=0$, $g'(0)$ real and $g'(0)>f'(0)$
\end{lemma}
Lemmas~\ref{L;Riemann one} and~\ref{L;Riemann two}
yield Theorem~\ref{T;Riemann unique blob} and by the
earlier this gives Theorem~\ref{T;Riemann map}
\begin{exercise} Reprove Lemma~\ref{L;Riemann two} using
$n$th roots rather than square roots.
\end{exercise}
\section{Boundary behaviour of conformal maps}%
\label{Good boundary}
We now return to the boundary behaviour
of the Riemann mapping. (Strictly speaking we should
say, a Riemann mapping but we have seen that it
is `essentially unique'. We saw in Exercise~\ref{E;bad boundary}
that there is no general theorem but the following result
is very satisfactory.
\begin{theorem}\label{Jordan boundary}
If $\Omega$ is a simply connected
open set in ${\mathbb C}$ with boundary a Jordan curve
then any bijective analytic map $f:D(0,1)\rightarrow \Omega$
can be extended to a bijective continuous map
from $\overline{D(0,1)}\rightarrow\overline{\Omega}$.
\end{theorem}
Recall\footnote{In the normal weasel-worded mathematical
sense.} that a \emph{Jordan curve} is
a continuous injective
map $\gamma:{\mathbb T}\rightarrow{\mathbb C}$.
We say that $\gamma$ is the boundary of $\Omega$
if the image of $\gamma$ is
$\overline{\Omega}\setminus\Omega$.

I shall use the proof in Zygmund's magnificent
treatise~\cite{Zygmund}
(see Theorem~10.9 of Chapter~VII)
which has the advantage of minimising the topology
but the minor disadvantage of using measure theory
(students who do not know measure theory may take
the results on trust) and the slightly greater disadvantage
of using an idea from Fourier analysis
(the \emph{conjugate} trigonometric sum $\tilde{S}_{N}(f,t)$)
which
can not be properly placed in context here.

We shall use the following simple consequence
of Fej{\'e}r's theorem (Theorem~\ref{Fejer convergence}).
\begin{lemma} If $f:{\mathbb T}\rightarrow{\mathbb R}$
is integrable but $S_{N}(f,x)\rightarrow\infty$
as $N\rightarrow\infty$ then
$f$ can not be continuous at $x$.
\end{lemma}
\begin{exercise} If $f:{\mathbb T}\rightarrow{\mathbb R}$
is integrable and there exist $\delta>0$ and $M>0$
such that $|f(t)|\leq M$ for all $|t|<\delta$ show that
it is not possible to have $S_{N}(f,0)\rightarrow\infty$
as $N\rightarrow\infty$.
\end{exercise}

\begin{lemma} (i) If $f:{\mathbb T}\rightarrow{\mathbb C}$
is integrable and $f$ is continuous at $x$ then
\[\frac{\tilde{S}_{N}(f,x)}{\log N}\rightarrow 0\]
as $N\rightarrow 0$.

(ii) If $h(t)=\sgn(t)-t/\pi$ then there is a non-zero
constant $L$ such that
\[\frac{\tilde{S}_{N}(h,0)}{\log N}\rightarrow L\]
as $N\rightarrow \infty$.

(iii) If $f:{\mathbb T}\rightarrow{\mathbb C}$
is integrable and $f(x+\eta)\rightarrow f(x+)$,
$f(x-\eta)\rightarrow f(x-)$ as $\eta\rightarrow 0$
through positive values then
\[\frac{\tilde{S}_{N}(f,x)}{\log N}\rightarrow 
\frac{L(f(x+)-f(x-))}{2}\]
as $N\rightarrow 0$.
\end{lemma}
We now come to the object of our Fourier analysis.
\begin{lemma}\label{jump} If $f:{\mathbb T}\rightarrow{\mathbb C}$
is integrable with $\hat{f}(n)=0$ for $n<0$. If
$f(x+\eta)\rightarrow f(x+)$,
$f(x-\eta)\rightarrow f(x-)$ as $\eta\rightarrow 0$
through positive values then $f(x+)=f(x-)$.
\end{lemma}
In other words, power series cannot have `discontinuities
of the first kind'.
\begin{exercise} Give an example of a discontinuous
function with no discontinuities
of the first kind.
\end{exercise}

Once Lemma~\ref{jump} has been got out of the way
we can return to the proof of Theorem~\ref{Jordan boundary}
on the boundary behaviour of the Riemann mapping.
The proof turns out to be long but reasonably clear.
We start with a very general result.
\begin{lemma}
If $\Omega$ is a simply connected
open set in ${\mathbb C}$
and $f:D(0,1)\rightarrow \Omega$ is a bijective
bicontinuous map then given any compact subset $K$ of $\Omega$
we can find an $1>r_{K}>0$ such that, whenever
$1>|z|>r_{K}$, $f(z)\notin K$.
\end{lemma}
Any bounded open set $\Omega$ has an area $|\Omega|$
and a simple application of the Cauchy-Riemann equations
yields the following result.
\begin{lemma}
Suppose that $\Omega$ is a simply connected bounded
open set in ${\mathbb C}$
and $f:D(0,1)\rightarrow \Omega$ is a bijective
analytic map. Then
\[|\Omega|=\int_{0\leq r<1}\int_{0}^{2\pi}
|f'(r e^{i\theta})|^{2}r\,d\theta\,dr.\]
\end{lemma}
\begin{lemma}\label{Jordan almost}
Suppose that $\Omega$ is a simply connected bounded
open set in ${\mathbb C}$
and $f:D(0,1)\rightarrow \Omega$ is a bijective
analytic map. The set $X$
of $\theta [0,2\pi)$ such that $f(r e^{i\theta})$ tends
to a limit as $r\rightarrow 1$ from below
has complement of Lebesgue
measure $0$.
\end{lemma}

From now on until the end of the section we operate
under the standing hypothesis that
$\Omega$ is a simply connected
open set in ${\mathbb C}$ with boundary a Jordan curve.
This means that $\Omega$ is bounded (we shall accept
this as a topological fact). We take $X$ as
in Lemma~\ref{Jordan almost} and write
$f(e^{i\theta})=\lim_{r\rightarrow 1-}f(r e^{i\theta})$
whenever $\theta\in X$. We shall assume
(as we may without loss of generality) that $0\in X$.
\begin{lemma}~\label{increasing Jordan}
Under our standing hypotheses we
can find a continuous bijective
map $g:{\mathbb T}\rightarrow{\mathbb C}$
such that $g(0)=f(1)$ and
such that, if $x_{1}$, $x_{2}\in X$ with
$0\leq x_{1}\leq x_{2} <2\pi$ and $t_{1}$, $t_{2}$
satisfy $g(t_{1})=x_{1}$, $g(t_{2})=x_{2}$
and $0\leq t_{1}, t_{2}<2\pi$ then $t_{1}\leq t_{2}$.
\end{lemma}
(The reader will, I hope, either excuse
or correct the slight abuse of notation.)


We now need a simple lemma.
\begin{lemma} Suppose $G:D(0,1)\rightarrow{\mathbb C}$
is a bounded analytic function such that
$G(r e^{i\theta})\rightarrow 0$ as $r\rightarrow 1-$
for all $|\theta|<\delta$
and some $\delta>0$. Then $G=0$.
\end{lemma}
Using this we can strengthen Lemma~\ref{increasing Jordan}
\begin{lemma}~\label{strictly increasing Jordan}
Under our standing hypotheses we
can find a continuous bijective
map $\gamma:{\mathbb T}\rightarrow{\mathbb C}$
such that $\gamma(0)=f(1)$ and
such that, if $x_{1}$, $x_{2}\in X$ with
$0\leq x_{1}< x_{2} <2\pi$ and $t_{1}$, $t_{2}$
satisfy $g(t_{1})=x_{1}$, $g(t_{2})=x_{2}$
and $0\leq t_{1}, t_{2}<2\pi$ then $t_{1}< t_{2}$.
\end{lemma}
From now on we add to our standing hypotheses
the condition that $\gamma$ satisfies the conclusions
of Lemma~\ref{strictly increasing Jordan}.

We now `fill in the gaps'.
\begin{lemma} We can find a strictly increasing function                     $w:[0,2\pi]\rightarrow [0,2\pi]$ with $w(0)=0$
and $w(2\pi)=2\pi$,
such that $\gamma(w(\theta))=f(e^{i\theta})$ for all
$\theta\in X$.
\end{lemma}
We now set $f(e^{i\theta})=\gamma(w(\theta))$
and $F(\theta)=f(e^{i\theta})$ for all $\theta$.
A simple use of dominated convergence gives us the next lemma.
\begin{lemma} If $f(z)=\sum_{n=1}^{\infty}c_{n}z^{n}$
for $|z|<1$ then,
we have $\hat{F}(n)=c_{n}$ for $n\geq 0$ and
$\hat{\gamma}(n)=0$ for $n<0$.
\end{lemma}
However increasing functions
can only have discontinuities of the first kind.
Thus $w$ and so $F$ can only have discontinuities
of the first kind. But, using our investment in Fourier
analysis (Lemma~\ref{jump}) we see that $F$ can have no
discontinuities
of the first kind..
\begin{lemma} The function $F:{\mathbb T}\rightarrow{\mathbb C}$
is continuous.
\end{lemma}
Using the density of $X$ in ${\mathbb T}$ we have the required
result.
\begin{lemma} The function
$f:\overline{D(0,1)}\rightarrow\overline{\Omega}$
is continuous and bijective.
\end{lemma}
This completes the proof of Theorem\ref{Jordan boundary}.

Using a little analytic topology we may restate
Theorem\ref{Jordan boundary} as follows.
\begin{theorem}\label{Jordan boundary both}
If $\Omega$ is a simply connected
open set in ${\mathbb C}$ with boundary a Jordan curve
then any bijective analytic map $f:D(0,1)\rightarrow \Omega$
can be extended to a bijective continuous map
from $\overline{D(0,1)}\rightarrow\overline{\Omega}$.
The map $f^{-1}\overline{\Omega}\rightarrow\overline{D(0,1)}$
is continuous on $\overline{\Omega}$.
\end{theorem}.
\section{Picard's little theorem} The object of this section
is to prove the following remarkable result.
\begin{theorem}[Picard's little theorem] If
$f:{\mathbb C}\rightarrow{\mathbb C}$ is analytic and
non-constant, then
${\mathbb C}\setminus f({\mathbb C})$ contains
at most one point.
\end{theorem}
The example of $\exp$ shows that
${\mathbb C}\setminus f({\mathbb C})$
may contain one point.

The key to Picard's theorem is the following result.
\begin{theorem}\label{Picard cover}
There exists an analytic map
$\lambda:D(0,1)\rightarrow {\mathbb C}\setminus\{0,1\}$
with the property that given $z_{0}\in {\mathbb C}\setminus\{0,1\}$,
$w_{0}\in D(0,1)$ and $\delta>0$ such that
$\lambda(w_{0})=z_{0}$ and
$D(z_{0},\delta)\subseteq {\mathbb C}\setminus\{0,1\}$
we can find
an analytic function
$g:D(z_{0},\delta)\rightarrow D(0,1)$
such that $\lambda (g(z))=z$ for all $z\in D(z_{0},\delta)$.
\end{theorem}

We combine this with a result whose proof differs hardly at all
from that of Theorem~\ref{logarithm}.
\begin{lemma}\label{pull up}
Suppose that $U$ and $V$ are
open sets and that $\tau:U\rightarrow V$ is a
analytic map with the following property.
Given $u_{0}\in U$
and $v_{0}\in V$ such that
$\tau(u_{0})=v_{0}$ then, given any $\delta>0$ with
$D(v_{0},\delta)\subseteq V$, we can find
an analytic function
$g:D(v_{0},\delta)\rightarrow U$
such that $\lambda (g(z))=z$ for all $z\in D(v_{0},\delta)$.
Then if $W$ is an open simply connected set
and $f:W\rightarrow U$
is analytic we can find an analytic
function $F:W\rightarrow U$
such that $\tau(F(z))=f(z)$ for all $z in W$.
\end{lemma}
(The key words here are `lifting' and `monodromy'. It is at points
like this that the resolutely `practical' nature of the presentation
shows its weaknesses. A little more theory about analytic
continuation for its own sake would turn a `technique'
into a theorem.)

We now combine the Schwarz reflection principle
(given in Lemma~\ref{L;Schwarz reflection})
with the work of section~\ref{Good boundary}
on boundary behaviour.
By repeated use of the Schwarz reflection principle
we continue $f$ analytically to the whole of $\mathcal{H}$.
\begin{lemma}\label{Picard cover upper}
Let $\mathcal{H}$ be the upper half plane.
There exists an analytic map
$\tau:\mathcal{H}\rightarrow {\mathbb C}\setminus\{0,1\}$
with the property that given $z_{0}\in {\mathbb C}\setminus\{0,1\}$
and $w_{0}\in \mathcal{H}$ such that
$\tau(w_{0})=z_{0}$ we can find $\delta>0$ with
$D(z_{0},\delta)\subseteq {\mathbb C}\setminus\{0,1\}$ and
an analytic function
$g:D(z_{0},\delta)\rightarrow \mathcal{H}$
such that $\tau (g(z))=z$ for all $z\in D(z_{0},\delta)$.
\end{lemma}
Since $\mathcal{H}$ can be mapped conformally to $D(0,1)$
Theorem~\ref{Picard cover} follows at once and
we have proved Picard's little theorem.
\section{Picard's great theorem} The object of this
section is to prove the following remarkable
generalisation of the Casorati--Weierstrass theorem.
\begin{theorem}{\bf Picard's great theorem}
Let $\Omega$ be an open subset of ${\mathbb C}$
and let $w_{0}\in\Omega$. If $f:{\Omega}\setminus\{w_{0}\}$
is analytic with $w$ as an essential singularity,
then we can find an $\omega_{0}\in{\mathbb C}$ such that,
given any $\delta>0$ and any $\omega\neq \omega_{0}$
we can find a $w\in\Omega\setminus\{w_{0}\}$ with $|w-w_{0}|<\delta$
and $f(w)=\omega$.
\end{theorem} 
\begin{exercise} (i) If $f:{\mathbb C}\rightarrow{\mathbb C}$ is analytic
and $|z^{-n-1}f(z)|\rightarrow 0$ as $|z|\rightarrow 0$ show,
by looking at the coefficients of the Taylor expansion,or
otherwise, that $f$ is a polynomial of degree at most $n$.

(ii) Continuing with the notation of (i), define
$g:{\mathbb C}\setminus\{0\}\rightarrow{\mathbb C}$
by $g(z)=f(1/z)$. Show that either $f$ is a polynomial
or $0$ is an essential singularity.

(iii) Deduce Picard's little theorem from
Picard's great theorem.
\end{exercise}
We introduce a couple of definitions. We could
avoid using them, but the reader may find
them helpful in later work.
\begin{definition}
Let $\Omega$ be an open set in ${\mathbb C}$
and $f_{n}:{\Omega}\rightarrow{\mathbb C}$
a sequence of analytic functions.
We say that $f_{n}$ \emph{diverges to infinity
uniformly on compacta} if, given $K$ a compact
subset of $\Omega$ and $C>0$, we can find
an $N$ such that $|f_{n}(z)|\geq C$
for all $z\in K$ and $n\geq N$.
\end{definition}
\begin{definition} 
Let $\Omega$ be an open set in ${\mathbb C}$
and ${\mathcal F}$ a set of analytic 
function $f:{\Omega}\rightarrow{\mathbb C}$.
We say that ${\mathcal F}$ is \emph{normal}
if every sequence of functions in ${\mathcal F}$ contains a subsequence 
which is either uniformly convergent on compacta
or diverges to infinity uniformly on compacta. 
\end{definition}
In view of our earlier discussion of uniformly convergence 
on compacta the next result is closer to an exercise than a lemma.
\begin{exercise} Let $\Omega$ be a connected open 
set in ${\mathbb C}$
and ${\mathcal F}$ a set of analytic 
function $f:{\Omega}\rightarrow{\mathbb C}$.
Then ${\mathcal F}$ is normal if and
only if, given $w\in\Omega$, we can find a $\delta>0$
such that $D(w,\delta)\subseteq\Omega$ and
\[{\mathcal F}_{D(w,\delta)}
=\{f|_{D(w,\delta)}\,:\,f\in{\mathcal F}\}\]
is normal. (More briefly a `locally normal' family
is normal.)

Give a simple counterexample to show that we need
$\Omega$ connected.
\end{exercise}
If $\Omega$ is an open subset of ${\mathbb C}$
and $a\neq b$,
let us write ${\mathcal F}_{a,b}(\Omega)$ for the set
of analytic functions on $\Omega$ which do not take the values $a$ or $b$.
As usual we write $D$ for the open unit disc.
\begin{theorem}{\bf [Schottky]}
Given any $\epsilon>0$ and any $1>\rho>0$ we can find a $\delta>0$
such that the following is true for an $f\in{\mathcal F}_{0,1}(D)$.

(i) $|f(0)|\leq\delta\Rightarrow |f(z)|\leq\epsilon
\ \text{for all $|z|\leq\rho$}.$

(ii) $|f(0)-1|\leq\delta\Rightarrow |f(z)-1|\leq\epsilon
\ \text{for all $|z|\leq\rho$}.$

(iii) $|f(0)|\geq\delta^{-1}\Rightarrow |f(z)|\geq\epsilon^{-1}
\ \text{for all $|z|\leq\rho$}.$
\end{theorem}
\begin{lemma} Let $U$ be an open disc 
If $f_{n}\in{\mathcal F}_{0,1}(U)$
and there exists a $w\in D$ such that $f_{n}(w)\rightarrow 0$
then $f_{n}$ converges uniformly on compacta to $0$.
Similar results hold with $0$ replaced by $1$ and $\infty$.
\end{lemma}
\begin{theorem}{\bf [A theorem of Montel]}
If $\Omega$ is a connected open set and $a\neq b$
${\mathcal F}_{a,b}(\Omega)$ is a normal family.
\end{theorem}   
We can now prove Picard's great theorem.
\section{References and further reading} 
There is not a great deal on Kolmogorov's and
Vitu{\v s}kin's theorem. I will hand out some
notes intended for different purposes which
(if only because they are available)
are probably as satisfactory as the references
that I used to prepare them (\cite{Kahane}, \cite{Lorenz1},
\cite{Vitushkin}). Jackson's theorems are dealt with
in~\cite{Lorenz1} and elsewhere.

There exist many
good books on advanced classical complex variable
theory which cover what is in this course and much more.
I particularly like~\cite{Veech} and~\cite{Epstein}.
For those who wish to study from the masters there
are Hille's two volumes~\cite{Hille} and
the elegant text of Nevanlinna~\cite{Nevanlinna}.

There are also many excellent books on Fourier analysis.
I used~\cite{Zygmund} but those who want a first
introduction are recommended to look at~\cite{Katznelson}.
\begin{thebibliography}{9}
\bibitem{Epstein} L.-S.~Hahn and B.~Epstein
\emph{Classical Complex Analysis}
Jones and Bartlett, Sudbury, Mass, 1996.
\bibitem{Hille} E. Hille
\emph{Analytic Function Theory} (2 Volumes)
Ginn and Co (Boston), 1959.
\bibitem{Kahane}
J.-P. Kahane,
\emph{Sur le treizi{\`e}me probl{\`e}me de {H}ilbert,
le th{\'e}or{\`e}me de superposition de {K}olmogorov
et les sommes alg{\'e}briques d'arcs croissants}
in the conference proceedings
\emph{Harmonic analysis, Iraklion 1978}
Springer, 1980.
\bibitem{Katznelson} Y. Katznelson
\emph{An Introduction to Harmonic Analysis}
3rd Edition, CUP, 2004. (The earlier editions
with different publishers are equally satisfactory.)
\bibitem{Lorenz1} G. G. Lorenz,
\emph{Approximation of functions}
Chelsea Publishing Co, 1986
(A lightly revised second edition. The first edition
appeared in 1966.)

\bibitem{Nevanlinna} R.~Nevanlinna and V.~Paatero
\emph{Introduction to Complex Analysis}
(Translated from the German), Addison-Wesley
(Reading, Mass), 1969.
\bibitem{Veech} W.~A.~Veech
\emph{A Second Course in Complex Analysis}
Benjamin, New York, 1967.
\bibitem{Vitushkin} A. G. Vistu{\v s}kin
\emph{On the representation of functions by
superpositions and related topics}
in L'Enseignement Math{\'e}matique,
1977, Vol 23,
pages 255--320.


\bibitem{Zygmund} A.~Zygmund
\emph{Trigonometric Series} (2 Volumes)
CUP, 1959.
\end{thebibliography}

\end{document}
