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}
\section{The Riemann mapping theorem} By using a very
beautiful physical argument, Riemann obtained the
following result.
\begin{theorem}[Riemann mapping theorem]%
\label{Riemann mapping theorem}
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}
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 normal families) 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} $a\in D(0,1)$
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}
Theorem~\ref{unique conformal}~(ii) can be modified in various
simple but useful ways.
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' is also
interesting and provide a useful revision of some results
from earlier courses on complex variable. (If these results
are strange to you, the lecturer can give a supplementary
lecture.)
\begin{example} 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{two points} we see the the `natural rigidity'
of complex analysis reassert itself.
\section{Normal families} Consider an open set $U$
in ${\mathbb C}$ and the collection ${\mathcal F}$
of analytic functions $f:U\rightarrow{\mathbb C}$.
What is the `natural topology' on ${\mathcal F}$
or, in the more old fashioned language of this course,
what is the `natural mode of convergence' for ${\mathcal F}$?
Looking at the convergence of power series and
at results like Morera's theorem suggests the following
approach.
\begin{definition} If $U$ is an open set
in ${\mathbb C}$ and $f_{n}:U\rightarrow {\mathbb C}$
we say that $f_{n}\rightarrow f_{0}$
\emph{uniformly on compacta} if, whenever $K$ is
a compact subset of $U$, $f_{n}|K\rightarrow f_{0}|K$
uniformly on $K$.
\end{definition}
We shall prove the chain of equivalences in the next lemma,
but the proof (and, sometimes, the explicit statement)
of similar chains will be left to the reader.
Here and elsewhere $\overline{E}$ is the closure
of $E$.
\begin{lemma} Let $U$ be an open set
in ${\mathbb C}$ and $f_{n}:U\rightarrow {\mathbb C}$
a sequence of functions. The following four
statements are equivalent.
(i) $f_{n}\rightarrow f_{0}$ uniformly on compacta.
(ii) Whenever
$\overline{D(w,\delta)}\subseteq U$
then $f_{n}|\overline{D(w,\delta)}
\rightarrow f_{0}|\overline{D(w,\delta)}$
uniformly on $\overline{D(w,\delta)}$.
(iii) Whenever $D(w,\delta)$ is
an open disc with $\overline{D(w,\delta)}\subseteq U$
then $f_{n}|D(w,\delta)
\rightarrow f_{0}|D(w,\delta)$
uniformly on $D(w,\delta)$.
(iv) If $w\in U$ we can find a $\delta>0$ such that
$D(w,\delta)\subseteq U$ and $f_{n}|D(w,\delta)
\rightarrow f_{0}|D(w,\delta)$
uniformly on $D(w,\delta)$.
\end{lemma}
We shall make use of the following result.
\begin{lemma} If $U$ is an open subset of ${\mathbb C}$
we can find compact sets $K_{1}\subseteq K_{2}
\subseteq K_{3}\subseteq\dots$ such that
$U=\bigcup_{j=1}^{\infty}\Int K_{j}=\bigcup_{j=1}^{\infty}K_{j}$.
\end{lemma}
\begin{exercise}\label{uniform on compacta metric}
(i) If $d_{1}$, $d_{2}$, \dots are
metrics on a space $X$ show that
\[d(x,y)=\sum_{j=1}^{\infty}2^{-j}\min(1,d_{j}(x,y))\]
defines a metric on $X$. Show that, if each $d_{j}$
is a complete metric and each $d_{j}$ defines the same
topology (i.e. has the same open sets)
then $d$ is complete.
(ii) If $U$ is an open set in ${\mathbb C}$ show that
there is a complete metric $d$ on the space $C(U)$
of continuous functions $f:U\rightarrow{\mathbb C}$
such that if $f_{n}\in C(U)$ then $d(f_{n},f_{0})\rightarrow 0$
if and only if $f_{n}\rightarrow f_{0}$ uniformly on
compacta.
\end{exercise}
We also note the following simple but important
consequence of Morera's theorem.
\begin{lemma} Let $U$ be an open set
in ${\mathbb C}$ and $f_{n}:U\rightarrow {\mathbb C}$
a sequence of functions with the property that
$f_{n}\rightarrow f_{0}$ uniformly on compacta.
If $f_{n}$ is analytic for each $n\geq 1$
then $f$ is analytic.
\end{lemma}
\begin{exercise}\label{exercise compacta} If $U$ is an open set
in ${\mathbb C}$ show that
there is a complete metric $d$ on the space $A(U)$
of analytic functions $f:U\rightarrow{\mathbb C}$
such that if $f_{n}\in A(U)$ then $d(f_{n},f_{0})\rightarrow 0$
if and only if $f_{n}\rightarrow f_{0}$ uniformly on
compacta.
\end{exercise}
We now talk about what in modern terms would be called
compactness.
\begin{definition} Let $U$ be an open set
in ${\mathbb C}$ and let $\mathcal{G}$ be a collection
of analytic functions on $U$. We say that
$\mathcal{G}$ is \emph{normal} if given any sequence
$f_{n}\in\mathcal{G}$ we can find a subsequence
$f_{n(j)}$ which converges uniformly on compacta
to a limit.
\end{definition}
Note that we do not demand that the limit lies
in $\mathcal{G}$.
Fortunately normal families have a simpler characterisation.
\begin{definition} Let $U$ be an open set
in ${\mathbb C}$ and let $\mathcal{G}$ be a collection
of functions on $U$. We say that $\mathcal{G}$
is \emph{uniformly bounded on compacta} if
given any compact subset $K$ of $U$ we can find
a constant $C_{K}$ such that $|f(z)|\leq C_{K}$
for all $f\in \mathcal{G}$ and all $z\in K$.
\end{definition}
\begin{theorem}\label{Normal} Let $U$ be an open set
in ${\mathbb C}$ and let $\mathcal{G}$ be a collection
of analytic functions on $U$. Then $\mathcal{G}$
is normal if and only if it is uniformly bounded on compacta.
\end{theorem}
We shall prove Theorem~\ref{Normal} via the following
lemma.
\begin{lemma} Let $\mathcal{G}$ be a collection
of analytic functions on the unit disc $D(0,1)$
with the property that $|f(z)|\leq 1$ for all $z\in D(0,1)$
and $f\in \mathcal{G}$. Then given any sequence
$f_{n}\in\mathcal{G}$ we can find a subsequence
$f_{n(j)}$ which converges uniformly on $D(0,1/2)$.
\end{lemma}
\begin{exercise} The following is a slightly different
treatment of Theorem~\ref{Normal}.
Recall that that we call a metric space $(X,d)$
\emph{sequentially compact} if given any sequence
$x_{n}\in X$ we can find a convergent subsequence
$x_{n(j)}$. (It can be shown that, for a metric space,
sequential compactness is equivalent to compactness
but we shall not need this.)
(i) Show that if we adopt the notation
of Exercise~\ref{exercise compacta} a subset
$\mathcal{G}$ of $A(U)$ is normal if and only
if its closure $\overline{\mathcal{G}}$
is sequentially compact with respect to the metric $d$.
(ii) {\bf (Arzeli-Ascoli theorem)}
Let $K$ be a compact subset of ${\mathbb C}$. Consider the
space $C(K)$ of continuous functions $f:K\rightarrow{\mathbb C}$
with the uniform norm. If $\mathcal{F}$ is a subset of
$C(K)$ show that $\overline{\mathcal{F}}$
is sequentially compact if and only if
\ \ (1) $\mathcal{F}$ is bounded, that is, there exists
a constant $\lambda$ such that $\|f\|\leq\lambda$
for all $f\in \mathcal{F}$.
\ \ (2) $\mathcal{F}$ is equicontinuous, that is, given any
$\epsilon>0$ there exists a $\delta(\epsilon)>0$
such that whenever $z,w\in U$, $|z-w|<\delta(\epsilon)$
and $f\in\mathcal{F}$ then $|f(z)-f(w)|<\epsilon$.
(iii) Let $U$ be a compact subset of ${\mathbb C}$. Consider the
space $C(U)$ of continuous functions $f:U\rightarrow{\mathbb C}$
under the metric $d$ defined in
Exercise~\ref{uniform on compacta metric}.
If $\mathcal{F}$ is a subset of
$C(K)$ show that $\overline{\mathcal{F}}$
is sequentially compact if and only if
\ \ (1) $\mathcal{F}$ is bounded on compacta,
that is, whenever $K$ is a compact subset of $U$,
there exists
a constant $\lambda_{K}$ such that $|f(z)|\leq\lambda_{K}$
for all $z\in K$ and $f\in \mathcal{F}$.
\ \ (2) $\mathcal{F}$ is equicontinuous on compacta,
that is, given any $K$ is a compact subset of $U$ and any
$\epsilon>0$ there exists a $\delta(\epsilon,K)>0$
such that whenever $z,w\in K$, $|z-w|<\delta(\epsilon,K)$
and $f\in\mathcal{F}$ then $|f(z)-f(w)|<\epsilon$.
(iv) Suppose that $\mathcal{F}$ is a collection
of analytic functions
$f:D(0,1)\rightarrow{\mathbb C}$ with $|f(z)|\leq \lambda$
for all $z\in D(0,1)$. Show, by using Cauchy's
formula, or otherwise that $\mathcal{F}$
is equicontinuous on $\overline{D(0,1-\epsilon)}$
for every $\epsilon$ with $1>\epsilon>0$.
(v) Prove Theorem~\ref{Normal} using the ideas of~(iii)
and~(iv).
\end{exercise}
\section{Proof of the Riemann mapping theorem}
We now embark on a series of lemmas which together
prove the Riemann mapping theorem
stated in Theorem~\ref{Riemann mapping theorem}.
\begin{lemma}
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 a set $\Omega'$
such that ${\mathbb C}\setminus\Omega'$
contains a disc $D(w,\delta)$ with $\delta>0$.
\end{lemma}
\begin{lemma}
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 a set $\Omega''$
such that $\Omega''\subseteq D(0,1)$.
\end{lemma}
It is worth remarking that the condition
`$\Omega$ has non-empty complement' cannot be removed.
\begin{lemma} There does not exist a conformal map
$f:{\mathbb C}\rightarrow D(0,1)$.
\end{lemma}
Thus the Riemann mapping theorem follows from
the following slightly simpler version.
\begin{lemma}
If $\Omega$ is an non-empty,
open, simply connected subset
of $D(0,1)$
then there exists a conformal
map of $\Omega$ to the unit disc $D(0,1)$.
\end{lemma}
We expect the proof of the Riemann mapping theorem
to involve a maximisation argument and
Theorem~\ref{unique conformal}~(ii) suggests
one possibility.
\begin{lemma}\label{right familly}
Suppose $\Omega$ is a open, non-empty,
simply connected subset of $D(0,1)$.
If $a\in \Omega$ then the set ${\mathcal F}$
of injective
analytic functions $f:\Omega\rightarrow D(0,1)$
with $f(a)=0$, $f'(a)$ real and $f'(a)\geq 0$
then $\mathcal{F}$ is a non-empty normal set.
\end{lemma}
\begin{lemma} With the hypotheses and notation
of Lemma~\ref{right familly} there exists a
$\kappa$ such that $f'(a)\leq\kappa$ for
all $f\in\mathcal{F}$.
\end{lemma}
\begin{lemma} Suppose that $U$ is an open connected subset
of $\mathbb{C}$ and we have a sequence
of analytic functions $f_{n}$ on $U$ with $f_{N}\rightarrow f$
uniformly. If each $f_{n}$ is injective then either
$f$ is constant or $f$ is injective.
\end{lemma}
\begin{lemma} With the hypotheses and notation
of Lemma~\ref{right familly} there exists
a $g\in\mathcal{F}$ such that $g'(a)\geq f'(a)$
for all $f\in\mathcal{F}$.
\end{lemma}
It may be worth remarking that though $g$ is, in fact,
unique we have not yet proved this.
All we need now is a little ingenuity and this is
supplied by an idea of Koebe.
\begin{lemma} (i) If $U$ is an open simply connected
subset of $D(0,1)$ containing $0$ but with $U\neq D(0,1)$
then we can find a bijective analytic function
$h:U\rightarrow D(0,1)$ such that $h(0)=0$, $h'(0)$
is real and $h'(0)>1$.
(ii) With the hypotheses and notation
of Lemma~\ref{right familly}, if $g\in\mathcal{F}$
and $g(\Omega)\neq D(0,1)$ we can find an $f\in\mathcal{F}$
with $f'(0)>g'(0)$.
\end{lemma}
Theorem~\ref{Riemann mapping theorem} now follows at
once. Combining Theorem~\ref{Riemann mapping theorem}
with Theorem~\ref{unique conformal}~(ii) we obtain
the following mild sharpening
without further work.
\begin{theorem} If $U$ be a simply connected open set
with $U\neq{\mathbb C}$ and $a\in U$.
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}
By reviewing the proof of the Riemann mapping theorem
that we have given it becomes clear that we have
in fact proved Lemma~\ref{logarithm exact}.
\begin{exercise} Check that we can prove
Lemma~\ref{logarithm exact} by the method used to prove
the Riemann mapping theorem.
\end{exercise}
It is important to realise that the intuition
we gather from the use of simple conformal transforms
in physics and elsewhere may be an unreliable guide
in the more general context of the Riemann mapping theorem.
\begin{example}\label{bad boundary}
There is a bounded non-empty
simply connected open set $U$ such that if
we have conformal map $f:D(0,1)\rightarrow U$
there does not exist a continuous bijective map
$\tilde{f}:\overline{D(0,1)}\rightarrow\overline{U}$
with $\tilde{f}|D(0,1)=f$.
\end{example}
Riemann's mapping theorem is a beginning and not an
end. Riemann stated his result in a more general context
than we have done here and the continuation of
Riemann's ideas leads to Klein's uniformisation
theorem. On the other hand, if we continued the
development suggested here we would look at topics
like
Green's functions, boundary behaviour and the Picard
theorems. If time permits I shall look at some
of these ideas later but I am anxious not to hurry
through the topics from number theory which
will be discussed in the next few lectures.
\section{Infinite products} Our object in the next few
lectures will be to prove the following remarkable
theorem of Dirichlet on primes in arithmetic
progression.
\begin{theorem}[Dirichlet]\label{Dirichlet}
If $a$ and $d$ are strictly
positive coprime integers then there are infinitely
many primes of the form $a+nd$ with $n$ a positive integer.
\end{theorem}
(Obviously the result must fail if $a$ and $d$ are not
coprime.)
There exist a variety of proofs of special cases
when $d$ has particular values but, so far as I know,
Dirichlet's proof of his theorem remains, essentially,
the only approachable one. In particular there is no
known reasonable\footnote{In the sense that most
reasonable people would call reasonable. Selberg produced
a (technically) elementary proof which may be found in his
collected works.} elementary
(in the technical sense of not using analysis)
proof.
Dirichlet's method starts from an observation of Euler.
\begin{lemma}\label{Euler prime start}
If $s$ is real with $s>1$ then
\[\prod_{\substack{\text{$p$ prime}\\p\leq N}}
\left(1-\frac{1}{p^{s}}\right)^{-1}\rightarrow
\sum_{n=1}^{\infty}\frac{1}{n^{s}}.\]
\end{lemma}
Using this result, we get a new proof of the existence
of an infinity of primes.
\begin{theorem}[Euclid] There exist an infinity of primes.
\end{theorem}
This suggests that it may be worth investigating
infinite products a bit more.
\begin{definition} Let $a_{j}\in{\mathbb C}$. If
$\prod_{n=1}^{N}(1+a_{n})$ tends to a limit $L$
as $N\rightarrow\infty$ we say that the
\emph{infinite product} $\prod_{n=1}^{\infty}(1+a_{n})$
\emph{converges} to a value $L$ and write
\[\prod_{n=1}^{\infty}(1+a_{n})=L.\]
If the infinite product $\prod_{n=1}^{\infty}(1+|a_{n}|)$
converges then we say that $\prod_{n=1}^{\infty}(1+a_{n})$
is \emph{absolutely convergent}.
\end{definition}
The next result was removed from the first year of
the Tripos a couple of years before I took it.
\begin{lemma}\label{absolute convergence}
Let $a_{j}\in{\mathbb C}$.
(i) $\prod_{n=1}^{\infty}(1+a_{n})$ is absolutely
convergent if and only if $\sum_{n=1}^{\infty}a_{n}$
is.
(ii) If $\prod_{n=1}^{\infty}(1+a_{n})$ is absolutely
convergent and $1+a_{n}\neq 0$ for each $n$ then
\[\prod_{n=1}^{\infty}(1+a_{n})\neq 0.\]
\end{lemma}
\begin{exercise} Find $a_{j}\in{\mathbb C}$ such that
$\prod_{n=1}^{\infty}(1+a_{n})$ is not absolutely
convergent but is convergent to a non-zero value.
\end{exercise}
We shall only make use of absolute convergent infinite
products.
\begin{exercise} If $\prod_{n=1}^{\infty}(1+a_{n})$ is
absolutely convergent and
$\sigma:{\mathbb N}\rightarrow{\mathbb N}$
is a bijection
(that is, $\sigma$ is a permutation of ${\mathbb N}$)
show that $\prod_{n=1}^{\infty}(1+a_{\sigma(n)})$ is
absolutely convergent and
\[\prod_{n=1}^{\infty}(1+a_{\sigma(n)})
=\prod_{n=1}^{\infty}(1+a_{n})\]
\end{exercise}
Whilst this is a useful result to know, we shall make no essential
use of it. When we write
$\sum_{\text{$p$ prime}}$ or $\prod_{\text{$p$ prime}}$
we mean the primes $p$ to be taken in order of increasing
size.
Using Lemma~\ref{absolute convergence} we obtain
the following strengthening of Euclid's theorem.
\begin{theorem}[Euler]
${\displaystyle \sum_{\text{$p$ prime}}\frac{1}{p}=\infty.}$
\end{theorem}
Since we wish to consider infinite products of functions
it is obvious that we shall need an analogue
of the Weierstrass M-test for products, obvious what that
analogue should be and obvious how to prove it.
\begin{lemma} Suppose $U$ is an open subset of ${\mathbb C}$
and that we have a sequence of functions
$g_{n}:U\rightarrow{\mathbb C}$ and a sequence of positive
real numbers $M_{n}$ such that $M_{n}\geq |g_{n}(z)|$
for all $z\in U$. If $\sum_{n=1}^{\infty}M_{n}$
converges then $\prod_{n=1}^{N}(1+g_{n}(z))$ converges
uniformly on $U$.
\end{lemma}
Later we shall need to consider $\sum n^{-s}$ with $s$ complex.
To avoid ambiguity, we shall take $n^{-s}=\exp(-s\log n)$
where $\log n$ is the real logarithm of $n$.
\begin{lemma} If $\Re s>1$ we have
\[\prod_{\text{$p$ prime}}(1-p^{-s})^{-1}=\sum_{n=1}^{\infty}n^{-s}\]
both sides being absolutely convergent for each $s$
and uniformly convergent for $\Re s>1+\epsilon$ for
each fixed $\epsilon>0$.
\end{lemma}
We now detour briefly from the main argument to show
how infinite products can be used to answer a
very natural question. `Can we always find an analytic
function
with specified zeros?' (We count multiple zeros
multiply in the usual way.) Naturally we need to
take account of the following fact.
\begin{lemma} If $z_{1}$, $z_{2}$, \dots are distinct zeros
of an analytic function which is not identically
zero then $z_{n}\rightarrow\infty$ as $n\rightarrow\infty$.
\end{lemma}
A little thought suggests the path we ought to take
though we may not see how to reach it. A way to
reach the path is provided by the Weierstrass
primary function $E(z,m)$.
\begin{definition} If $m$ is a strictly positive integer
\[E(z,m)=(1-z)e^{z+z^{2}/2+z^{3}/3+\dots+z^{m}/m}.\]
\end{definition}
\begin{lemma} The function $E(\ ,m):\mathbb{C}\rightarrow\mathbb{C}$
is analytic with a unique zero at 1. If $|z|\leq 1$
then
\[|1-E(z,m)|\leq |z|^{m+1}.\]
\end{lemma}
(It is nice to have such a neat result but for our purposes
$|1-E(z,m)|\leq A|z|^{m+1}$ for $|z|\leq R$ with any $A$
and $R$ would be just as good.)
\begin{theorem}[Weierstrass] If $k$ is a positive integer
and $z_{1},z_{2},\dots$ is a sequence of non-zero complex
numbers with $z_{n}\rightarrow\infty$ then
\[F(z)=z^{k}\prod_{j=1}^{\infty}E(z/z_{j},j)\]
is a well defined
analytic function with a zero of order $k$ at $0$,
and zeros at the $z_{j}$ (multiple zeros counted multiply)
and no others.
\end{theorem}
\begin{lemma} If $f_{1}$ and $f_{2}$ are analytic functions
on ${\mathbb C}$ with the same zeros (multiple zeros counted
multiply) then there exists an analytic function $g$
such that
\[f_{1}(z)=e^{g(z)}f_{2}(z).\]
\end{lemma}
\begin{lemma} If $z_{1},z_{2},\dots$ and $w_{1},w_{2},\dots$
are sequences of complex numbers with
$z_{j},w_{j}\rightarrow\infty$ as $j\rightarrow\infty$
and $z_{j}\neq w_{k}$ for all $j,k$ then there exists
a meromorphic function with zeros at the $z_{j}$
and poles at the $w_{k}$ (observing the usual multiplicity
conventions).
\end{lemma}
\begin{exercise} (It may be helpful to attack parts
of this question non-rigourously first and then tighten up
the argument second.)
(i) If $C_{N}$ is the contour consisting of the square
with vertices $\pm (N+1/2)\pm (N+1/2)i$ described
anti-clockwise show that there is a constant $K$
such that
\[|\cot \pi z|\leq K\]
for all $z\in C_{N}$ and all integers $N\geq 1$.
(ii) By integrating an appropriate function round the
contour $C_{N}$, or otherwise,
show that, if $w\notin{\mathbb Z}$,
\[\sum_{n=-N}^{n=N}\frac{1}{w-n}\rightarrow \pi\cot\pi w.\]
(iii) Is it true that, if $w\notin{\mathbb Z}$,
\[\sum_{n=-M}^{n=N}\frac{1}{w-n}\rightarrow \pi\cot\pi w,\]
as $M,N\rightarrow\infty$? Give reasons.
(iv) Show that
\[P(z)=z\prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right)\]
is a well defined analytic function and that there exists
an analytic function $g$ such that
\[\sin\pi z=e^{g(z)}P(z).\]
(v) Find a simple expression for
$P'(z)/P(z)$. [Hint: If $p(z)=\prod_{j=1}^{N}(z-\alpha_{j})$,
what is $p'(z)/p(z)$?] Find a related expression for
$\frac{d\ }{dz}\sin \pi z/\sin \pi z$.
(vi) Show that
\[\sin\pi z=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right).\]
(vii) Find a similar expression for $\cos\pi z$. (These
results are due to Euler.)
\end{exercise}
\begin{exercise} (This makes use of some of the
techniques of the previous exercise.) (i) Show that the infinite product
\[g(z)=\prod_{n=1}^{\infty}\left(1-\frac{z}{n}\right)\]
exists and is analytic on the whole complex plane.
(ii) Show that
\[g'(z)=g(z)
\sum_{n=1}^{\infty}\left(\frac{1}{z-n}+\frac{1}{n}\right).\]
Explain why $\sum_{n=1}^{\infty}(\frac{1}{z-n}+\frac{1}{n})$
is indeed a well defined everywhere analytic function.
(iii) By using (ii), or otherwise, show that
\begin{equation*}
g(z+1)=-Azg(z) \tag*{(*)}
\end{equation*}
for some constant $A$.
(iv) By considering a particular value of $z$, or otherwise,
show that $A$ is real and positive and
\[\sum_{n=1}^{N}\frac{1}{n}-\log N\rightarrow \log A\]
as $N\rightarrow\infty$.
Deduce the existence of Euler's constant
$\gamma=i(\lim_{N\rightarrow\infty}\sum_{n=1}^{N}n^{-1}-\log N)$
and rewrite $(*)$ as
\begin{equation*}
g(z+1)=-e^{\gamma}zg(z)
\end{equation*}
(v) Find a simple expression for $zg(z)g(-z)$. Use $(*)$
to show that $\sin \pi z$ is periodic.
\end{exercise}
\section{Fourier analysis on finite Abelian groups} One
of Dirichlet's main ideas is a clever extension of
Fourier analysis from its classical frame. Recall
that classical Fourier analysis deals with formulae
like
\[f(t)=\sum_{n=-\infty}^{\infty}\hat{f}(n)e_{n}(t)\]
where $e_{n}(t)=\exp(int)$. The clue to further extension
lies in the following observation.
\begin{lemma} Consider the Abelian group
${\mathbb T}={\mathbb R}/(2\pi{\mathbb Z})$ and the
subgroup $S=\{z:|z|=1\}$ of $({\mathbb C}\setminus\{0\},\times)$.
The continuous homomorphisms
$\theta:{\mathbb T}\rightarrow S$ are precisely
the functions $e_{n}:{\mathbb T}\rightarrow S$ given
by $e_{n}(t)=\exp(int)$ with $n\in{\mathbb Z}$.
\end{lemma}
\begin{exercise} (i) Find (with proof) all the
continuous homomorphisms
$\theta:({\mathbb R},+)\rightarrow (S,\times)$.
What is the connection with Fourier transforms?
(ii) (Only for those who know Zorn's lemma%
\footnote{And, particularly, those who only know Zorn's lemma.}.)
Assuming Zorn's lemma show that any linearly independent
set in a vector space can be extended
to a basis. If we consider ${\mathbb R}$ as a vector
space over ${\mathbb Q}$ show that there exists
a linear map $T:{\mathbb R}\rightarrow{\mathbb R}$ such
that $T(1)=1$, $T(\surd 2)=0$. Deduce the existence
of a function $T:{\mathbb R}\rightarrow{\mathbb R}$
such that $T(x+y)=T(x)+T(y)$ for all $x,y\in {\mathbb R}$
which is not continuous (with respect to the usual metric).
Show that, if we accept Zorn's lemma, there exist
discontinuous homomorphisms
$\theta:({\mathbb R},+)\rightarrow (S,\times)$.
\end{exercise}
This suggests the following definition.
\begin{definition} If $G$ is a finite Abelian group
we say that a homomorphism $\chi:G\rightarrow S$
is a \emph{character}. We write $\hat{G}$ for the
collection of such characters.
\end{definition}
In this section we shall accumulate a
substantial amount of information about $\hat{G}$
by a succession of small steps.
\begin{lemma} Let $G$ be a finite Abelian group.
(i) If $x\in G$ has order $m$ and $\chi\in\hat{G}$
then $\chi(x)$ is an $m$th root of unity.
(ii) $\hat{G}$ is a finite Abelian group under
pointwise multiplication.
\end{lemma}
To go further we consider for each finite Abelian group
$G$ the collection $C(G)$ of functions $f:G\rightarrow{\mathbb C}$.
If $G$ has order $|G|$ then $C(G)$ is a vector space of
dimension $N$ which can be made into a complex inner
product space by means of the inner product
\[\langle f,g\rangle=\frac{1}{|G|}\sum_{x\in G}f(x)g(x)^{*}.\]
\begin{exercise} Verify the statements just made.
\end{exercise}
\begin{lemma} Let $G$ be a finite Abelian group.
The elements of $\hat{G}$ form an orthonormal
system in $C(G)$.
\end{lemma}
Does $\hat{G}$ form an orthonormal basis of $C(G)$? The
next lemma tells us how we may hope to resolve this
question.
\begin{lemma} Let $G$ be a finite Abelian group.
The elements of $\hat{G}$ form an orthonormal basis
if an only if given an element $x\in G$ which is
not the identity we can find a character $\chi$
with $\chi(x)\neq 1$.
\end{lemma}
The way forward is now clear.
\begin{lemma}
Suppose that $H$ is a subgroup of a finite Abelian
group $G$ and that $\chi\in\hat{H}$. If $K$
is a subgroup of $G$ generated by $H$ and an element
$a\in G$ then we can find a $\tilde{\chi}\in\hat{K}$
such that $\tilde{\chi}|H=\chi$.
\end{lemma}
\begin{lemma} Let $G$ be a finite Abelian group
and $x$ an element of $G$ of order $m$. Then
we can find a $\chi\in\hat{G}$ with
$\chi(x)=\exp 2\pi i/m$.
\end{lemma}
\begin{theorem} If $G$ is a finite Abelian group
then $\hat{G}$ has the same number of elements
as $G$ and they form an orthonormal basis
for $C(G)$.
\end{theorem}
\begin{lemma} If $G$ is a finite Abelian group
and $f\in C(G)$ then
\[f=\sum_{\chi\in\hat{G}}\hat{f}(\chi)\chi\]
where $\hat{f}(\chi)=\langle f,\chi \rangle$.
\end{lemma}
\begin{exercise} Suppose that $G$ is a finite Abelian
group. Show that if we define
$\theta_{x}:\hat{G}\rightarrow {\mathbb C}$ by
$\theta_{x}(\chi)=\chi(x)$ for $\chi\in \hat{G}$,
$x\in G$ then the map $\Theta:G\rightarrow\hathatG$
given by $\Theta(x)=\theta_{x}$ is an isomorphism.
If we now identify $x$ with $\theta_{x}$ (and, so,
$G$ with $\hathatG$) show that
\[\hathatf(x)=|G|f(x^{-1})\]
for all $f\in C(G)$ and $x\in G$.
\end{exercise}
We have now done all that that is required to
understand Dirichlet's motivation. However, it
seems worthwhile to make a slight detour to
put `computational' bones on this section by
exhibiting the structure of $G$ and $\hat{G}$.
\begin{lemma} Let $(G,\times)$ be an Abelian group.
(i) Suppose that $x,y\in G$ have order $r$ and $s$
with $r$ and $s$ coprime. Then $xy$ has order $rs$.
(ii) If $G$ contains elements of order $n$ and $m$
then $G$ contains an element of order the least
common multiple of $n$ and $m$.
\end{lemma}
\begin{lemma}\label{maximum order}
Let $(G,\times)$ be a finite Abelian group.
Then there exists an integer $N$ and an element $k$
such that $k$ has order $N$ and, whenever $x\in G$
we have $x^{N}=e$.
\end{lemma}
\begin{lemma} With the hypotheses and notation
of Lemma~\ref{maximum order} we can write $G=K\times H$
where $K$ is the cyclic group generated by $x$
and $H$ is another subgroup of $K$.
\end{lemma}
As usual we write $C_{n}$ for the cyclic group
of order $n$.
\begin{theorem} If $G$ is a finite Abelian group we can find
$n(1)$, $n(2)$, \dots $n(m)$ with $n(j+1)|n(j)$
such that $G$ is isomorphic to
\[C_{n(1)}\times C_{n(2)}\times \dots C_{n(m)}.\]
\end{theorem}
\begin{lemma} If we have two sequences
$n(1)$, $n(2)$, \dots $n(m)$ with $n(j+1)|n(j)$
and
$n'(1)$, $n'(2)$, \dots $n'(m')$ with $n'(j+1)|n'(j)$
then
\[C_{n(1)}\times C_{n(2)}\times \dots C_{n(m)}
\ \text{is isomorphic to}
\ C_{n'(1)}\times C_{n'(2)}\times \dots C_{n'(m')}\]
if and only if $m=m'$ and $n(j)=n'(j)$ for each $1\leq j\leq m$.
\end{lemma}
It is easy to identify $\hat{G}$.
\begin{lemma} Suppose that
\[G=C_{n(1)}\times C_{n(2)}\times \dots C_{n(m)}\]
with $C_{n(j)}$ a cyclic group of order $n(j)$ generated
by $x_{j}$. Then the elements of $\hat{G}$ have the
form
$\chi_{\omega_{n(1)}^{r(1)},\omega_{n(2)}^{r(2)}},\dots
\omega_{n(m)}^{r(m)}$ with $\omega_{n(j)}=\exp(2\pi i/n(j)$
and
\[\chi_{\omega_{n(1)}^{r(1)},\omega_{n(2)}^{r(2)}},\dots
\omega_{n(m)}^{r(m)}(x_{1}^{s(1)}x_{2}^{s(2)}\dots x_{m}^{s(m)})
=\omega_{n(1)}^{r(1)s(1)}\omega_{n(2)}^{r(2)s(2)}\dots
\omega_{n(m)}^{r(m)s(m)}.\]
\end{lemma}
My readers will see that $\hat{G}$ is isomorphic to $G$
but the more sophisticated algebraists will also
see that this is \emph{not a natural isomorphism}
(whereas $G$ and $\hathatG$ are \emph{naturally isomorphic}).
Fortunately such matters are of no importance
for the present course.
\section{The Euler-Dirichlet formula} Dirichlet was interested
in a particular group. If $d$ is a positive integer consider
${\mathbb Z}/(n)$ the set of equivalence classes
\[[m]=\{r:r\equiv m \mod{d}\}\]
under the usual multiplication modulo $n$.
We set
\[G_{d}=\{[m]:\text{$m$ and $d$ coprime}\}\]
and write $\phi(d)$ for the order of $G_{d}$ ($\phi$ is
called Euler's totient function).
\begin{lemma} The set $G_{d}$ forms a finite
Abelian group under
standard multiplication.
\end{lemma}
The results of the previous section show that, if $[a]\in G_{n}$
and we define $\delta_{a}:G_{d}\rightarrow{\mathbb C}$ by
\begin{align*}
\delta_{a}([a])&=1\\
\delta_{a}([m])&=0\qquad \text{if $[m]\neq [a]$},
\end{align*}
then
\[\delta_{a}=\phi(d)^{-1}\sum_{\chi\in G_{d}}\chi([a])^{*}\chi\]
We now take up the proof of Dirichlet's theorem in earnest.
We shall operate under the standing assumption that $a$ and $d$
are positive coprime integers and our object is to show
that the sequence
\[a,\ a+d,\ a+2d,\ \dots, a+nd,\ \dots\]
contains infinitely many primes. Following
Euler's proof that there exist infinitely many primes
we shall seek to prove this by showing that
\[\sum_{\substack{\text{$p$ prime}\\p=a+nd\ \text{for some $n$}}}
\frac{1}{p}=\infty.\]
Henceforward, at least in the number theory part of the
course $p$ will be a prime, $\sum_{p}$ will mean the
sum over all primes and so on.
In order to simplify our notation it will also be convenient
to modify the definition of a character. From now on, we say
that $\chi$ is a character if $\chi$ is a map from
${\mathbb N}$ to ${\mathbb C}$ such that there exists a character
(in the old sense) $\tilde{\chi}\in \hat{G}_{d}$
such that
\begin{alignat*}{2}
\chi(m)&=\tilde{\chi}([m])&&\qquad\text{if $m$ and $d$ are coprime}\\
\chi(m)&=0&&\text{otherwise}.
\end{alignat*}
We write $\sum_{\chi}$ to mean the sum over all characters
and take $\chi_{0}$ to be the character with
$\chi_{0}([m])=1$ whenever $m$ and $d$ are coprime.
\begin{lemma} (i) If $\chi$ is a character then
$\chi(m_{1}m_{2})=\chi(m_{1})\chi(m_{2})$
for all $m_{1},m_{2}\geq 0$.
(ii) If $\chi\neq\chi_{0}$ then
$\sum_{m=k+1}^{k+d}\chi(m)=0$.
(iii) If $\delta_{a}(m)=\phi(d)^{-1}\sum_{\chi}\chi(a)^{*}\chi(m)$
then $\delta_{a}(m)=1$ when $m=a+nd$ and $\delta_{a}(m)=0$
otherwise.
(iv) $\displaystyle{\sum_{p=a+nd}p^{-s}
=\phi(d)^{-1}\sum_{\chi}\chi(a)^{*}\sum_{p}\chi(p)p^{-s}}$.
\end{lemma}
\begin{lemma} The sum $\sum_{p=a+nd}p^{-1}$ diverges if
$\sum_{p}\chi(p)p^{-s}$ remains bounded
as $s$ tends to $1$ through real values of $s>1$
for all $\chi\neq\chi_{0}$.
\end{lemma}
We now prove a new version of Euler's formula.
\begin{theorem}[Euler-Dirichlet formula]
With the notation of this section,
\[\prod_{n=1}^{\infty}(1-\chi(p)p^{-s})^{-1}=
\sum_{n=1}^{\infty}\chi(n)n^{-s},\]
both sides being absolutely convergent for $\Re s>1$.
\end{theorem}
To link $\prod_{n=1}^{\infty}(1-\chi(p)p^{-s})^{-1}$
with $\sum_{p}\chi(p)p^{-s}$ we use logarithms.
(If you go back to our discussion of infinite products
you will see that this is not unexpected.) However,
we must, as usual, use care when choosing our logarithm
function. For the rest of the argument $\log$
will be the function on
\[{\mathbb C}\setminus\{x:\text{$x$ real and $x\leq 0$}\}\]
defined by $\log (re^{i\theta})=\log r+i\theta$
[$r>0$, $-\pi<\theta<\pi$].
\begin{lemma} (i) If $|z|\leq 1/2$ then $|\log(1-z)+z|\leq |z|^{2}$.
(ii) If $\epsilon>0$ then $\sum_{p}\log(1-\chi(p)p^{-s})$
and $\sum_{p}\chi(p)p^{-s}$ converge uniformly in
$\Re s\geq 1+\epsilon$, whilst
\[\left|\sum_{p}\log(1-\chi(p)p^{-s})+\sum_{p}\chi(p)p^{-s}\right|
\leq \sum_{n=1}^{\infty}n^{-2}.\]
\end{lemma}
We have thus shown that if $\sum_{p}\log(1-\chi(p)p^{-s})$
remains bounded as $s\rightarrow 1+$ then $\sum_{p}\chi(p)p^{-s}$
does. Unfortunately we can not equate $\sum_{p}\log(1-\chi(p)p^{-s})$
with $\log(\prod_{n=1}^{\infty}(1-\chi(p)p^{-s})^{-1})$.
However we can refresh our spirits by proving Dirichlet's
theorem in some special cases.
\begin{example} There are an infinity of primes
of the form $3n+1$ and $3n+2$.
\end{example}
\begin{exercise} Use the same techniques to show that
there are an infinity of primes
of the form $4n+1$ and $4n+3$.
\end{exercise}
\section{Analytic continuation of the Dirichlet functions}
Dirichlet completed his argument without having to consider
$\sum_{n=1}^{\infty}\chi(n)n^{-s}$ for anything other
than real $s$ with $s>1$. However, as we have already seen,
$\sum_{n=1}^{\infty}\chi(n)n^{-s}=L(s,\chi)$ is defined and well
behaved in $\Re s>1$. Riemann showed that it is advantageous
to extend the definition of analytic
functions like $L(s,\chi)$
to larger domains.
There are many ways of obtaining such
\emph{analytic continuations}. Here is one.
\begin{lemma} If $f:{\mathbb R}\rightarrow{\mathbb R}$
is bounded on ${\mathbb R}$ and
locally integrable\footnote{Riemann or Lebesgue at
the reader's choice} then
\[F(s)=\int_{1}^{\infty}f(x)x^{-s}\,dx\]
is a well defined analytic function on the set of $s$ with
$\Re s>1$.
\end{lemma}
\begin{lemma}\label{Extend Dirichlet 1}
(i) If $\chi\neq \chi_{0}$ and
$S(x)=\sum_{1\leq m\leq x}\chi(m)$ then
$S:{\mathbb R}\rightarrow{\mathbb R}$ is bounded
and locally integrable. We have
\[\sum_{n=1}^{N}\chi(n)n^{-s}
\rightarrow s\int_{1}^{\infty}S(x)x^{-s-1}\, dx\]
as $N\rightarrow\infty$ for all $s$ with $\Re s>1$.
(ii) If $S_{0}(x)=0$ for $x\leq 0$ and
$S_{0}(x)=\sum_{1\leq m\leq x}\chi_{0}(m)$ then,
writing $T_{0}(x)=S_{0}(x)-d^{-1}\phi(d)x$,
$T_{0}:{\mathbb R}\rightarrow{\mathbb R}$ is bounded
and locally integrable. We have
\[\sum_{n=1}^{N}\chi(n)n^{-s}\rightarrow
s\int_{1}^{\infty}T_{0}(x)x^{-s-1}\, dx+\frac{\phi(d)s}{d(s-1)}\]
as $N\rightarrow\infty$ for all $s$ with $\Re s>1$.
\end{lemma}
\begin{lemma}\label{Extend Dirichlet 2}
(i) If $\chi\neq\chi_{0}$ then
$\sum_{n=1}^{\infty} \chi(n)n^{-s}$ converges to an analytic
function $L(s,\chi)$, say, on $\{s\in{\mathbb C}:\Re s>0\}$.
(ii) There exists an meromorphic function $L(s,\chi_{0})$
analytic on $\{s\in{\mathbb C}:\Re s>0\}$ except for
a simple pole, residue $\phi(d)/d$ at $1$ such that
$\sum_{n=1}^{\infty} \chi_{0}(n)n^{-s}$ converges to
$L(s,\chi)$ for $\Re s>1$.
\end{lemma}
\begin{exercise}
(i) Explain carefully why
$L(\ ,\chi_{0})$ is defined uniquely by
the conditions given.
(ii) Show that
$\sum_{n=1}^{\infty} \chi_{0}(n)n^{-s}$ diverges
for $s$ real and $1\geq s >0$.
\end{exercise}
We now take up from where we left off at the end
of the previous section.
\begin{lemma} (i) If $\Re s>1$ then
$\exp(-\sum_{p}\log(1-\chi(p)p^{-s})=L(s,\chi)$.
(ii) If $\Re s>1$ then $L(s,\chi)\neq 0$.
(iii) There exists a function $\Log L(s,\chi)$ analytic
on $\{s: \Re s>1\}$ such that $\exp(\Log L(s,\chi))=L(s,\chi)$
for all $s$ with $\Re s>1$.
(iv) If $\chi\neq \chi_{0}$ and
$L(1,\chi)\neq 0$ then $\Log L(s,\chi))$ tends to
a finite limit as $s\rightarrow 1$ through real values with $s>1$.
(v) There is a fixed integer $M_{\chi}$ such that
\[\Log L(s,\chi)+\sum_{p}\log(1-\chi(p)p^{-s})=2\pi M_{\chi}\]
for all $\Re s>1$.
(vi) If $\chi\neq\chi_{0}$ and $L(1,\chi)\neq 0$ then
$\sum_{p}\chi(p)p^{-s}$ remains bounded as
$s\rightarrow 1$ through real values with $s>1$.
\end{lemma}
We mark our progress with a theorem.
\begin{theorem} If $L(1,\chi)\neq 0$ for all $\chi\neq\chi_{0}$
then there are an infinity of primes of the form
$a+nd$.
\end{theorem}
Since it is easy to find the characters $\chi$ in any given case
and since it is then easy to compute $\sum_{n=1}^{N}\chi(n)n^{-1}$
and to estimate the error $\sum_{n=N+1}^{\infty}\chi(n)n^{-1}$
to sufficient accuracy to prove that
$L(1,\chi)=\sum_{n=1}^{\infty}\chi(n)n^{-1}\neq 0$,
it now becomes possible to prove Dirichlet's theorem
for any particular coprime $a$ and $d$.
\begin{exercise} Choose $a$ and $d$ and carry out the
program just suggested.
\end{exercise}
However, we still need to show that the theorem holds
in all cases.
\section{$L(1,\chi)$ is not zero} Our first steps are easy.
\begin{lemma} (i) If $s$ is real and $s>1$ then
\[\prod_{\chi}L(s,\chi)=
\exp(-\sum_{p}\sum_{\chi}\log(1-\chi(p)p^{-s}).\]
(ii) If $s$ is real and $s>1$ then $\prod_{\chi}L(s,\chi)$
is real and $\prod_{\chi}L(s,\chi)\geq 1$.
(iii) $\prod_{\chi}L(s,\chi)\nrightarrow 0$ as
$s\rightarrow 1$.
\end{lemma}
\begin{lemma} (i) There can be at most one character
$\chi$ with $L(1,\chi)=0$.
(ii) If a character $\chi$ takes non-real values
then $L(1,\chi)\neq 0$.
\end{lemma}
We have thus reduced the proof of Dirichlet's theorem
to showing that if $\chi$ is a character
with $\chi\neq \chi_{0}$ which only
takes the values $1$, $-1$ and $0$ then $L(1,\chi)\neq 0$.
There are several approaches to this problem but
none are short and transparent. We use a proof
of de la Vall{\'{e}}e Poussin which is quite short
but not, I think, transparent.
\begin{lemma}\label{Smoke 1} Suppose that the
character $\chi\neq \chi_{0}$ and only
takes the values $1$, $-1$ and $0$. Set
\[\psi(s)=\frac{L(s,\chi)L(s,\chi_{0})}{L(2s,\chi_{0})}.\]
(i) The function $\psi$ is well defined and meromorphic
for $\Re s>\frac{1}{2}$. It analytic except, possibly
for a simple pole at $1$.
(ii) If $L(1,\chi)=0$ then $1$ is a removable singularity
and $\psi$ is analytic everywhere on $\{s:\Re s>\frac{1}{2}\}$.
(iii) We have $\psi(s)\rightarrow 0$ as $s\rightarrow \frac{1}{2}$
through real values of $s$ with $s\geq \frac{1}{2}$.
\end{lemma}
\begin{lemma}~\label{Smoke 2}
We adopt the hypotheses and notation of
Lemma~\ref{Smoke 1}. If $\Re s>1$ then the following is true.
(i) ${\displaystyle
\psi(s)=\prod_{\chi(p)=1}\frac{1+p^{-s}}{1-p^{-s}}.}$
(ii) We can find subsets $Q_{1}$ and $Q_{2}$ of $\mathbb{Z}$
such that
\begin{align*}
\prod_{\chi(p)=1}(1+p^{-s})&=\sum_{n\in Q_{1}}n^{-s}\\
\prod_{\chi(p)=1}(1-p^{-s})^{-1}&=\sum_{n\in Q_{2}}n^{-s}.
\end{align*}
(iii) There is a sequence of real positive numbers $a_{n}$
with $a_{1}=1$ such that
\[\psi(s)=\sum_{n=1}^{\infty}a_{n}n^{-s}.\]
\end{lemma}
\begin{lemma} We adopt the hypotheses and notation of
Lemmas~\ref{Smoke 1} and~\ref{Smoke 2}.
(i) If $\Re s>1$ then
\[\psi^{(m)}(s)=\sum_{n=1}^{\infty}a_{n}(-\log n)^{m}n^{-s}.\]
(ii) If $\Re s>1$ then $(-1)^{m}\psi^{(m)}(s)>0$.
(iii) If $\psi$ has no pole at $1$ then if $\Re s_{0}>1$
and $|s-s_{0}|<\Re s_{0}-1/2$ we have
\[\psi(s)=\sum_{m=0}^{\infty}
\frac{\psi^{(m)}(s_{0})}{m!}(s-s_{0})^{m}.\]
(iv) If $\psi$ has no pole at $1$ then
$\psi(s)\nrightarrow 0$ as $s\rightarrow \frac{1}{2}$
through real values of $s$ with $s\geq \frac{1}{2}$.
\end{lemma}
We have proved the result we set out to obtain.
\begin{lemma} If a character $\chi\neq\chi_{0}$
only takes real values
then $L(1,\chi)\neq 0$.
\end{lemma}
\begin{theorem} If $\chi\neq\chi_{0}$ then $L(1,\chi)\neq 0$.
\end{theorem}
We have thus proved Theorem~\ref{Dirichlet},
If $a$ and $d$ are strictly
positive coprime integers then there are infinitely
many primes of the form $a+nd$ with $n$ a positive integer.
\section{Natural boundaries} This section is included
partly for light relief between two long and tough
topics and partly to remind the reader that analytic
continuation is not quite as simple as it looks.
\begin{lemma} If
\[f(z)=\sum_{n=0}^{\infty}z^{n!}\]
for $|z|<1$ then $f:D(0,1)\rightarrow{\mathbb C}$
is analytic but if $U$ is any connected open set with
$U\cap D(0,1)\neq\emptyset$ and $U\setminus D(0,1)\neq\emptyset$
then there does not exist an analytic function
$g:U\rightarrow{\mathbb C}$ with $g(z)=f(z)$ for all
$z\in U\cap D(0,1)$.
\end{lemma}
More briefly we say that the unit circle is a natural boundary
for $f$.
Here is a much more subtle result whose central idea
goes back to Borel.
\begin{theorem}
Suppose $\sum_{n=0}^{\infty}a_{n}z^{n}$ has radius of
convergence~$1$ and suppose $Z_{1}$, $Z_{2}$, \dots
is a sequence of independent random variables with
each $Z_{j}$ uniformly distributed over
$\{z\in{\mathbb C}:|z|=1\}$. Then, with probability~$1$,
the unit circle is a natural boundary
for
\[\sum_{n=1}^{\infty}Z_{n}a_{n}z^{n}.\]
\end{theorem}
Our proof will require the following result due to
Kolmogorov.
\begin{theorem}[Kolmogorov zero-one law]
Let $P$ be a reasonable property
which any sequence of complex numbers $w_{n}$
either does or does not have and such that any
two sequences $w_{n}$ and $w'_{n}$ with $w_{n}=w'_{n}$
for $n$ sufficiently large either both have or
both do not have. Then if $W_{n}$ is a sequence
of independent complex valued random variables,
it follows that $W_{n}$ has property $P$ with
probability~0 or has property $P$ with
probability~1.
\end{theorem}
I will try to explain why this result is plausible
but for the purposes of the exam this result may be
assumed without discussion.
We remark the following simple consequence.
\begin{lemma}\label{softly} There exists a power series
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ with the unit
circle as natural boundary having the property
that $f$ and all its derivatives can be extended
to continuous functions on $\overline{D(0,1)}$.
\end{lemma}
\begin{exercise} Obtain Lemma~\ref{softly} by
non-probabilistic means. One possibility is to consider
$f(z)=\sum_{n=1}^{\infty}\epsilon_{n}g((1-n^{-1})z^{n!})$ where
$g(z)=(1-z)^{-1}$ and $\epsilon_{n}$ is a very rapidly decreasing
sequence of positive numbers.
\end{exercise}
Finally we remark that whilst it is easy to define
a natural boundary for a power series the notion
does not easily extend.
\begin{lemma}
Let $\Omega={\mathbb C}\setminus\{x\in{\mathbb R}:x\geq 0\}$
We can find an analytic function $f:\Omega\rightarrow{\mathbb C}$
with the following properties.
(i) There exists an analytic function
$F:\{z:\Re z>0\}\rightarrow{\mathbb C}$ such that $F(z)=f(z)$
whenever $\Re z>0$, $\Im z<0$.
(2) If $D$ is an open disc which contains $z_{1}$, $z_{2}$
with $\Re z_{1}>0$, $\Im z_{1}>0$ and
$\Re z_{2}>0$, $\Im z_{2}<0$ there exists no analytic
function $g$ on $D$ with $g(z)=f(z)$ for all $z\in D$
with $\Re z>0$, $\Im z>0$.
\end{lemma}
\section{Chebychev and the distribution of primes}
On the strength of numerical evidence, Gauss was
lead to conjecture that the number $\pi(n)$
of primes less than $n$
was approximately $n/\log n$. The theorem which
confirms this conjecture is known as the
prime number theorem.
The first real
progress in this direction was due to
Chebychev\footnote{His prefered transliteration
seems to have been Tchebycheff, but he has
been over-ruled.} We give his results, not out
of historical piety, but because we shall make
use of them in our proof of the prime number
theorem. (Note the obvious conventions that
$n$ is an integer with $n\geq 1$,
$\prod_{n(\log 2)n/(\log 2n)$.
(iv) There exists a constant $A>0$ such that
$\pi(n)\geq An(\log n)^{-1}$.
(v) There exists a constant $B'$ such that
$\sum_{p\leq n} \log p\leq B'n$.
(vi)There exists a constant $B$ such that
$\pi(n)\leq Bn(\log n)^{-1}$.
\end{lemma}
We restate the main conclusions of Lemma~\ref{Chebychev lemma}.
\begin{theorem}[Chebychev]\label{Chebychev theorem}
There exist constants $A$ and $B$
with $0a\}$.
(ii) We define the Heaviside function $H$ by
writing $H(t)=0$ for $t<0$ and $H(t)=1$ for $t\geq 0$.
If $a\in {\mathbb R}$ and $b\geq 0$ consider
$H_{a,b}(t)=H(t-b)e^{at}$. Show that $H_{a,b}\in {\mathcal E}_{a}$
and that ${\mathcal L}H_{a,b}(z)$ can be extended
to a meromorphic function on ${\mathbb C}$ with a simple
pole at $a$.
\end{exercise}
Engineers are convinced that the converse to
Exercise~\ref{Laplace}~(i) holds in the sense that if
$F\in {\mathcal E}_{a}$ has a Laplace transform $f$
which can be extended to a function $\tilde{f}$
analytic on $\{z\in{\mathbb C}:\Re z>b\}$
[$a$, $b$ real, $a\geq b$] then $F\in {\mathcal E}_{b}$.
Unfortunately, this is not true but it represents
a good heuristic principle to bear in mind in what follows.
Number theorists use the Mellin transform
\[{\mathcal M}F(z)=\int_{0}^{\infty}F(t)t^{z-1}\,dt\]
in preference to the Laplace transform but the
two transforms are simply related.
\begin{exercise} Give the relation explicitly.
\end{exercise}
Riemann considered the two functions
\[\Phi(s)=\sum_{p}p^{-s}\log p\]
and the \emph{Riemann zeta function}
\[\zeta(s)=\sum_{n=1}^{\infty}n^{-s}.\]
Both of these functions are defined for $\Re s>1$
but Riemann saw that they could be extended
to analytic functions over a larger domain.
The next lemma is essentially a repeat of
Lemmas~\ref{Extend Dirichlet 1}~(ii)
and~\ref{Extend Dirichlet 2}~(ii).
\begin{lemma}
(i) Let $S_{0}(x)=0$ for $x\leq 0$ and
$S_{0}(x)=\sum_{1\leq m\leq x}1$. If
$S_{0}:{\mathbb R}\rightarrow{\mathbb R}$
and $T_{0}(x)=S_{0}(x)-x$ then $T_{0}$ is bounded
and locally integrable. We have
\[\sum_{n=1}^{N}n^{-s}\rightarrow
s\int_{1}^{\infty}T_{0}(x)x^{-s-1}\, dx+\frac{s}{s-1}\]
as $N\rightarrow\infty$ for all $s$ with $\Re s>1$.
(ii) There exists an meromorphic function $\zeta$
analytic on $\{s\in{\mathbb C}:\Re s>0\}$ except for
a simple pole, residue $1$ at $1$ such that
$\sum_{n=1}^{\infty}n^{-s}$ converges to
$\zeta(s)$ for $\Re s>1$.
\end{lemma}
The use of $s$ rather than $z$ goes back to Riemann.
Riemann showed that $\zeta$ can be extended to a meromorphic
function over ${\mathbb C}$ but we shall not need this.
How does this help us study $\Phi$?
\begin{lemma}\label{extend half}
(i) We have $\prod{p1+\delta$ whenever $\delta>0$.
(ii) We have
\[\frac{\zeta'(s)}{\zeta(s)}=-\sum_{p}\frac{\log p}{p^{s}-1}\]
for all $\Re s>1$.
(iii) We have
\[\Phi(s)=-\frac{\zeta'(s)}{\zeta(s)}-
\sum_{p}\frac{\log p}{(p^{s}-1)p^{s}}\]
for all $\Re s>1$.
(iv) The function $\Phi$ can be analytically extended
to a meromorphic function on
$\{s:\Re s>\frac{1}{2}\}$. It has a simple pole at 1
with residue $1$ and simple poles at the zeros of
$\zeta$ but nowhere else.
\end{lemma}
The next exercise is long and will not be used later but is,
I think, instructive.
\begin{exercise} (i) Show by grouping in pairs
that $\sum_{n=1}^{\infty}(-1)^{n-1}n^{-s}$
converges to an analytic function $g(s)$ in the
region $\{s:\Re s>0\}$.
(ii) Find $A$ and $B$ such that $g(s)=A\zeta(s)+B2^{-s}\zeta(s)$
for all $\Re s>1$. Why does this give another proof
that $\zeta$ can be extended to an analytic function
on $\{s:\Re s>0\}$.
(iii) Show that $g(1/2)\neq 0$ and deduce that $\zeta(1/2)\neq 0$.
(iv) By imitating the arguments of Lemma~\ref{extend half}
show that we we can find an analytic function $G$ defined
on $\{s:\Re s>1/3\}$ such that
\[\Phi(s)=-\frac{\zeta'(s)}{\zeta(s)}-\Phi(2s)-G(s).\]
Deduce that $\Phi$ can be extended to a meromorphic
function on $\{s:\Re s>1/3\}$.
(v) Show, using (iii), that $\Phi$ has a pole at $1/2$.
(vi) Show that the assumption that
$|\sum_{p0$ and $A>0$ and all large enough $N$
leads to the conclusion that
$\Phi$ can be analytically extended from $\{s:\Re s>1\}$
to an everywhere analytic function
on $\{s:\Re s>1/2-\epsilon\}$.
(vii) Deduce that if $\epsilon>0$ and $A>0$
\[|\sum_{p1/2\}$ and that his conjecture
is the most famous open problem in mathematics.
The best we can do is to follow Hadamard and
de la Vall\'{e}e Poussin and show that $\zeta$
has no zero on $\{s:\Re s=1\}$. Our proof makes
use of the slightly unconventional convention
that if $h$ and $g$ are analytic in a neighbourhood of $w$,
$g(w)\neq 0$ and
and $h(z)=(z-w)^{k}g(z)$ then we say that $h$ has a zero
of order $k$ at $w$. (The mild unconventionality arises
when $k=0$.)
\begin{lemma} Suppose that $\zeta$ has a zero of order
$\mu$ at $1+i\alpha$
and a zero of order $\nu$ at $1+2i\alpha$
with $\alpha$ real and $\alpha>0$. Then
(i) $\zeta$ has a zero of order $\mu$ at $1-i\alpha$
and a zero of order $\nu$ at $1-2i\alpha$.
(ii) As $\epsilon\rightarrow 0$ through
real positive values of $\epsilon$
\begin{align*}
\epsilon\Phi(1+\epsilon \pm i\alpha)&\rightarrow -\mu\\
\epsilon\Phi(1+\epsilon \pm 2i\alpha)&\rightarrow -\nu\\
\epsilon\Phi(1+\epsilon)&\rightarrow 1.
\end{align*}
(iii) If $s=1+\epsilon$ with $\epsilon$ real and positive
then
\begin{align*}
0&\leq\sum_{p}p^{-s}\log p (e^{(i\alpha\log p)/2}+
e^{-(i\alpha\log p)/2})^{4}\\
&=\Phi(s+2i\alpha)+\Phi(s-2i\alpha)
+4(\Phi(s+i\alpha)+\Phi(s-i\alpha))
+6\Phi(s).
\end{align*}
(iv) We have $0\leq-2\nu-8\mu+6$.
\end{lemma}
\begin{theorem} If $\Re s=1$ then $\zeta(s)\neq 0$.
\end{theorem}
We note the following trivial consequence.
\begin{lemma}
If we write
\[T(s)=\frac{\zeta'(s)}{\zeta(s)}-(s-1)^{-1},\]
then given any $R>0$ we can find a $\delta(R)$
such that $T$ has no poles in
$\{z:\Re z\geq 1-\delta(R),\ Im z\leq R\}$
\end{lemma}
We shall show that the results we have obtained on
the behaviour of $\zeta$ suffice to show that
\[\int_{1}^{X}\frac{\theta(x)-x}{x^{2}}\,dx\]
tends to a finite limit as $X\rightarrow\infty$.
The next lemma shows that this is sufficient to
give the prime number theorem.
\begin{lemma} Suppose that
$\beta:[1,\infty)\rightarrow{\mathbb R}$
is an increasing (so integrable) function.
(i) If $\lambda>1$, $y>1$ and $y^{-1}\beta(y)>\lambda$
then
\[\int_{y}^{\lambda y}\frac{\beta(x)-x}{x^{2}}\,dx
\geq A(\lambda)\]
where $A(\lambda)$ is a strictly positive number
depending only on $\lambda$.
(ii) If $\int_{1}^{X}\frac{\beta(x)-x}{x^{2}}\,dx$
tends to limit as $X\rightarrow\infty$ then
$x^{-1}\beta(x)\rightarrow 1$ as $x\rightarrow\infty$.
\end{lemma}
We need a couple of further preliminaries.
First we note a simple consequence of the Chebychev
estimates (Theorem~\ref{Chebychev theorem}).
\begin{lemma} There exists a constant $K$
such that
\[\frac{|\theta(x)-x|}{x}\leq K\]
for all $x\geq 1$.
\end{lemma}
Our second step is to translate our results into
the language of Laplace transforms. (It is
just a matter of taste whether to work with Laplace
transforms or Mellin transforms.)
\begin{lemma} Let $f(t)=\theta(e^{t})e^{-t}-1$ for
$t\geq 0$ and $f(t)=0$ otherwise. Then
\[\mathcal{L}f(z)=\int_{-\infty}^{\infty}f(t)e^{-tz}\,dt\]
is well defined and
\[\mathcal{L}f(z)=\frac{\Phi(z-1)}{z}-\frac{1}{z}\]
for all $\Re z>0$.
The statement $\int_{1}{\infty}(\theta(x)-x)/x^{2}\,dx$
convergent is equivalent to the statement that
$\int_{-\infty}^{\infty}f(t)\,dt$ converges.
\end{lemma}
We have reduced the proof of the prime number theorem
to the proof of the following lemma.
\begin{lemma} Suppose $\Omega$ is an open set
with $\Omega\supseteq \{z:\Re z\geq 0\}$,
$F:\Omega\rightarrow{\mathbb C}$ is an analytic
function and $f:[0,\infty]\rightarrow{\mathbb R}$
is bounded locally integrable function such that
\[F(z)=\mathcal{L}f(z)=\int_{0}^{\infty}f(t)e^{-tz}\,dt\]
for $\Re z>0$. Then $\int_{0}^{\infty}f(t)\,dt$ converges.
\end{lemma}
This lemma and its use to prove the prime number theorem
are due to D.~Newman. (A version will be found in~\cite{Newman}.)
\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 Example~\ref{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.
\begin{definition} If $f:{\mathbb T}\rightarrow{\mathbb C}$
is an integrable\footnote{Use whichever integral you
are happiest with.} function we define
\[\hat{f}(n)=\frac{1}{2\pi}\int_{\mathbb T}f(t)\exp(-int)\,dt.\]
We set
\begin{align*}
S_{N}(f,t)&=\sum_{n=-N}^{N}\hat{f}(n)\exp(int)\\
\tilde{S}_{N}(f,t)&=-i\sum_{n=-N}^{N}\sgn(n)\hat{f}(n)\exp(int)\\
\sigma_{N}(f,t)&=(N+1)^{-1}\sum_{n=0}^{N}S_{n}(f,t)
\end{align*}
\end{definition}
Recall that
\[f*g(x)=\frac{1}{2\pi}\int_{\mathbb T}f(t)g(x-t)\,dt.\]
\begin{lemma} We have
\[S_{N}(f)=D_{N}*f,\ \tilde{S}_{N}(f)=\tilde{D}_{N}*f,
\sigma_{N}(f)=K_{N}*f,\]
with
\begin{align*}
D_{N}(t)&=\sum_{n=-N}^{N}\exp int=\frac{\sin(N+\tfrac{1}{2})t}
{\tfrac{t}{2}}\\
\tilde{D}_{N}(t)&=2\sum_{n=1}^{N}\sin nt
=\frac{\cos\tfrac{1}{2}t-\cos(N+\tfrac{1}{2})t}{\sin\tfrac{1}{2}t}\\
\ K_{N}(t)&=\frac{1}{N+1}\left(\tfrac{\sin\frac{N+1}{2}t}
{\sin\tfrac{1}{2}t}\right)^{2}.
\end{align*}
\end{lemma}
(Formally speaking, we have not defined $D_{N}(t)$,
$K_{N}(t)$ and $\tilde{D}_{N}(t)$ when $t=0$.
By inspection
$D_{N}(0)=2N+1$, $K_{N}(0)=N+1$, $\tilde{D}_{N}(0)=0$.)
By looking at the properties of the \emph{kernels}
$K_{N}(t)$ and $\tilde{D}_{N}(t)$ we obtain results
about the associated sums.
\begin{lemma} We have
\begin{alignat*}{2}
K_{N}(t)&>0&&\qquad\text{for all $t$}\\
K_{N}(t)&\rightarrow 0&&\qquad
\text{uniformly for $|t|\geq \delta$
whenever $\delta>0$}\\
\frac{1}{2\pi}\int_{\mathbb T}K_{N}(t)\,dt&=1.
\end{alignat*}
\end{lemma}
\begin{theorem}[F\'{e}jer] If $f:{\mathbb T}\rightarrow{\mathbb C}$
is integrable and $f$ is continuous at $x$ then
\[\sigma_{N}(f,x)\rightarrow f(x)\ \text{as $N\rightarrow\infty$}.\]
\end{theorem}
We shall only use the following simple consequence.
\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.
is very satisfactory.
\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 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.)
In the case that we require,
Lemma~\ref{pull up} gives the following result.
\begin{lemma} If $\lambda$ is as in Theorem~\ref{Picard cover}
and $f:{\mathbb C}\rightarrow{\mathbb C}\setminus\{0,1\}$
is analytic we can find $F:{\mathbb C}\rightarrow D(0,1)$
such that $\lambda(F(z))=f(z)$.
\end{lemma}
Picard's little theorem follows on considering Louiville's
theorem that a bounded analytic function on ${\mathbb C}$
is constant.
The proof of Picard's theorem thus reduces to the
construction of the function
$\lambda$ of Theorem~\ref{Picard cover}.
We make use ideas concerning reflection which
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]
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 $\{z:|z-a|r\}$.} 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.
We now use the work of section~\ref{Good boundary}
on boundary behaviour.
\begin{lemma} Let $\mathcal{H}$ be the upper half plane $\{z:\Im z>0$
and $V$ the region bounded by the lines $C_{1}=\{iy:y\geq 0\}$,
$C_{3}=\{1+iy:y\geq 0\}$,
and the arc $C_{2}=\{z:|z-\frac{1}{2}|=\frac{1}{2},\ \Im z\geq 0\}$
and containing the point $\frac{1}{2}+i$. There
is a continuous bijective map
$f:\overline{V}\rightarrow \overline{\mathcal{H}}$
which is analytic on $V$, takes $0$ to $0$, $1$ to $1$,
$C_{1}$ to $\{x:x\leq 0\}$, $C_{2}$ to $\{x:0\leq x\leq 1\}$,
and $C_{3}$ to $\{x:x\geq 1\}$.
\end{lemma}
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{References and further reading} 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 is an excellent treatment of Dirichlet's
theorem and much more in Davenport's
\emph{Multiplicative Number Theory}~\cite{Davenport}
[The changes between the first and second editions
are substantial but do not affect that part which
deals with material in this course.]
If you wish to know more about the Riemann zeta-function
you can start with~\cite{Patterson}.
In preparing this course I have also used~\cite{Korner1}
and~\cite{Korner2} since I find the author sympathetic.
\begin{thebibliography}{99}
\bibitem{Newman} J.~Bak and D.~J.~Newman
\emph{Complex Analysis}
Springer, New York, 1982.
\bibitem{Davenport} H.~Davenport
\emph{Multiplicative Number Theory}
(2nd Edition), Springer, New York, 1980.
\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{Korner1} T.~W.~K\"{o}rner
\emph{Fourier Analysis}
CUP, 1988.
\bibitem{Korner2} T.~W.~K\"{o}rner
\emph{Exercises for Fourier Analysis}
CUP, 1993.
\bibitem{Nevanlinna} R.~Nevanlinna and V.~Paatero
\emph{Introduction to Complex Analysis}
(Translated from the German), Addison-Wesley
(Reading, Mass), 1969.
\bibitem{Patterson} S.~J.~Patterson
\emph{An Introduction to the Theory of the Riemann zeta-function}
CUP, 1988.
\bibitem{Veech} W.~A.~Veech
\emph{A Second Course in Complex Analysis}
Benjamin, New York, 1967.
\bibitem{Zygmund} A.~Zygmund
\emph{Trigonometric Series} (2 Volumes)
CUP, 1959.
\end{thebibliography}
\end{document}