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\begin{document}
\title{Topics in Analysis\\Part III, Autumn 2010}
\author{T.~W.~K\"{o}rner}
\maketitle
\begin{footnotesize}
\noindent
{\bf Small print}
This is just a first draft for the course.
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.
\end{footnotesize}
\tableofcontents
\section{Introduction} This course splits into two parts.
The first part takes a look at the Baire category
theorem, Tychonov's theorem
the Hahn Banach theorem together with some
of their consequences. There will be two
or three lectures of fairly abstract set theory
but the the rest of the course is pretty
concrete. The second half of the course will
look at the theory of distributions.
(The general approach is that of~\cite{Friedlander}
but the main application is different.)
I shall therefore assume that you know what is a normed space,
and what is a
a linear map and that you can do the following exercise.
\begin{exercise} Let $(X,\|\ \|_{X})$ and $(Y,\|\ \|_{Y})$
be normed spaces.
(i) If $T:X\rightarrow Y$ is linear, then $T$ is continuous
if and only if there exists a constant $K$ such that
\[\|Tx\|_{Y}\leq K\|x\|_{X}\]
for all $x\in X$.
(ii) If $T:X\rightarrow Y$ is linear and $x_{0}\in X$,
then $T$ is continuous at $x_{0}$
if and only if there exists a constant $K$ such that
\[\|Tx\|_{Y}\leq K\|x\|_{X}\]
for all $x\in X$.
(iii) If we write ${\mathcal L}(X,Y)$ for the space of continuous
linear maps from $X$ to $Y$ and write
\[\|T\|=\sup\{\|Tx\|_{Y}\,:\, \|x\|_{X}=1,\ x\in X\}\]
then $({\mathcal L}(X,Y),\|\ \|)$ is a normed space.
\end{exercise}
I also assume familiarity with the concept of a metric
space and a complete metric space. You should be able
to do at least parts~(i) and~(ii) of the following exercise
(part~(iii) is a little harder).
\begin{exercise} Let $(X,\|\ \|_{X})$ and $(Y,\|\ \|_{Y})$
be normed spaces.
(i) If $(Y,\|\ \|_{Y})$ is complete then
$({\mathcal L}(X,Y),\|\ \|)$ is.
(ii) Consider the set $s$ of sequences $x=(x_{1},x_{2},\dots)$
in which only finitely many of the $x_{j}$ are non-zero.
Explain briefly how $s$ may be considered as a vector space.
(iii) If $(X,\|\ \|_{X})$ is complete does it follow that
$({\mathcal L}(X,Y),\|\ \|)$ is? Give a proof or a counter-example.
\end{exercise}
The reader will notice that I have not distinguished
between vector spaces over ${\mathbb R}$ and those over ${\mathbb C}$.
I shall try to make the distinction when it matters but,
if the two cases are treated in the same way, I shall
often proceed as above.
Although I shall stick with metric spaces as much
as possible, there will be points where we shall need
the notions of a topological space, a compact
topological space and a Hausdorff topological space.
I would be happy, if requested, to give a supplementary
lecture introducing these notions. (Even where I use them,
no great depth of understanding is required.)
I shall also use, without proof, the famous
Stone-Weierstrass theorem.
\begin{theorem} (A) Let $X$ be a compact space and
$C(X)$ the space of real valued continuous functions
on $X$. Suppose $A$ is a subalgebra of $C(X)$ (that is a
subspace which is algebraically closed under multiplication)
and
(i) $1\in A$,
(ii) Given any two distinct points $x$ and $y$
in $X$ there is an $f\in A$ with $f(x)\neq f(y)$.
Then $A$ is uniformly dense in $C(X)$.
(B) Let $X$ be a compact space and
$C(X)$ the space of complex valued continuous functions
on $X$. Suppose $A$ is a subalgebra of $C(X)$
and
(i) $1\in A$,
(ii) Given any two distinct points $x$ and $y$
in $X$ there is an $f\in A$ with $f(x)\neq f(y)$.
Then $A$ is uniformly dense in $C(X)$.
\end{theorem}
The proof will not be examinable, but if you have not
met it, you may wish to request a supplementary lecture
on the topic. I may mention some measure theory
but this is for interest only and will not be
examinable\footnote{In this course, as in other Part~III
courses you should assume that everything in the
lectures and nothing outside them is examinable
unless you are explicitly told to the contrary. If you
are in any doubt, ask the lecturer.}. I intend the course
to be fully accessible without measure theory.
\section{Baire category} If $(X,d)$ is a metric space
we say that a set $E$ in $X$ has dense
complement\footnote{If the lecturer uses the words
`nowhere dense' correct him for using an old fashioned
and confusing terminology} if,
given $x\in E$ and $\delta>0$, we can find a $y\notin E$
such that $d(x,y)<\delta$.
\begin{exercise} Consider the space $M_{n}$ of $n\times n$
complex matrices with an appropriate norm. Show that
the set of matrices which do not have $n$ distinct
eigenvalues is a closed set with dense complement.
\end{exercise}
\begin{theorem}[Baire's theorem]\label{T;Baire 2}
If $(X,d)$ is a complete metric
space and $E_{1}$, $E_{2}$, \dots are
closed sets with dense complement
then $X\neq \bigcup_{j=1}^{\infty}E_{j}$.
\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}
We call the countable union of closed
sets with dense complement
a set of first category. 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}
\begin{exercise}
If $(X,d)$ is a complete metric space
and $X$ is countable show that there
is an $x\in X$ and a $\delta>0$ such that
the ball $B(x,\delta)$ with centre $x$ and
radius $\delta$ consists of one point.
\end{exercise}
The following exercise is a standard application of
Baire's theorem.
\begin{exercise} Consider the space $C([0,1])$
of continuous functions under the uniform norm
$\|\ \|$. Let
\begin{align*}
E_{m}=\{f\in C([0,1])\,:&\, \text{there exists an $x\in [0,1]$
with}\\
&\text{$|f(x+h)-f(x)|\leq m|h|$ for all $x+h\in[0,1]$}\}.
\end{align*}
(i) Show that $E_{m}$ is closed in $(C([0,1],\|\ \|_{\infty})$.
(ii) If $f\in C([0,1])$ and $\epsilon>0$ explain why we can
find an infinitely differentiable function $g$ such that
$\|f-g\|_{\infty}<\epsilon/2$. By considering the function $h$
given by
\[h(x)=g(x)+\tfrac{\epsilon}{2}\sin Nx\]
with $N$ large show that $E_{m}$ has dense complement.
(iii) Using Baire's theorem show that there exist continuous
nowhere differentiable functions.
\end{exercise}
\begin{exercise} (This is quite long and not very
central.)
(i) Consider the space ${\mathcal F}$ of non-empty
closed sets in $[0,1]$. Show that if we write
\[d_{0}(x,E)=\inf_{e\in E}|x-e|\]
when $x\in[0,1]$ and $E\in{\mathcal F}$
and write
\[d(E,F)=\sup_{f\in F}d_{0}(f,E)+\sup_{e\in E}d_{0}(e,F)\]
then $d$ is a metric on ${\mathcal F}$.
(ii) Suppose $E_{n}$ is a Cauchy sequence in
$({\mathcal F},d)$. By considering
\[E=\{x\,:\,\text{there exist $e_{n}\in E_{n}$ such that
$e_{n}\rightarrow x$}\},\]
or otherwise, show that $E_{n}$ converges.
Thus $({\mathcal F},d)$ is complete.
(iii) Show that the set
\[\mathcal{A}_{n}=\{E\in{\mathcal F}\,:\,\text{there exists
an $x\in E$ with $(x-1/n,x+1/n)\cap E=\{x\}$}\}\]
is closed with dense complement in $({\mathcal F},d)$.
Deduce that the set of elements of ${\mathcal F}$ with
isolated points is of first category. (A set $E$ has
an isolated point $e$ if we can find a $\delta>0$
such that $(e-\delta,e+\delta)\cap E=\{e\}$.)
(iv) Let $I=[r/n,(r+1)/n]$ with $0\leq r\leq n-1$ and
$r$ and $n$ integers. Show that the set
\[\mathcal{B}_{r,n}=
\{E\in{\mathcal F}\,:\,E\supseteq I\}\]
is closed with dense complement in $({\mathcal F},d)$.
Deduce that the set of elements of ${\mathcal F}$
containing an open interval is of first category.
(v) Deduce the existence of non-empty closed sets
which have no isolated points and contain no intervals.
\end{exercise}
\section{Non-existence of functions of several variables}
This course is very much a penny plain rather than tuppence
coloured\footnote{And thus suitable for those
`who want from books plain cooking made still
plainer by plain cooks'.}. One exception is the
theorem proved in this section.
\begin{theorem}\label{Hilbert; Theorem} 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{Hilbert; Theorem}
may be expressed as follows.
\begin{theorem}\label{Gilbert; Theorem}
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!
(We shall explain why this statement is slightly
less shocking than it seems at the end of this section.)
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{Hilbert; Theorem}
than one like Theorem~\ref{Gilbert; Theorem}.
Our first step is to observe that Theorem~\ref{Hilbert; Theorem}
follows from the apparently simpler result that follows.
\begin{lemma}\label{Hilbert A; Lemma} 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 continuous
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}
Next we make the following observation.
\begin{lemma}\label{Hilbert B; Lemma} 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{Hilbert A; Lemma}
from a much more specific result.
\begin{lemma}\label{Hilbert C; Lemma} Let $\lambda$ be irrational
and let the $f_{n}$ be as in Lemma~\ref{Hilbert B; Lemma}.
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 continuous
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}
Now that we have reduced the matter to
satisfying a countable set of conditions,
we can use a Baire category argument.
We need to use the correct metric space.
\begin{lemma}\label{Lemma 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{Lemma Kahane space}
\end{exercise}
\begin{lemma}\label{Hilbert D; Lemma} 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 continuous
$g:{\mathbb R}\rightarrow{\mathbb R}$ 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{Hilbert D; Lemma}.)
Lemma~\ref{Hilbert D; Lemma} is the heart of the proof
and once it is proved we can easily retrace our steps
and obtain Theorem~\ref{Hilbert; Theorem}.
By using appropriate notions of information
Vitushkin was able to show that we can not replace
continuous by continuously differentiable
in Theorem~\ref{Gilbert; Theorem}. Thus
Theorem~\ref{Hilbert; Theorem} is an `exotic'
rather than a `central' result.
\section{The principle of uniform boundedness}
We start with a result which is sometimes useful
by itself but which, for us, is merely a stepping
stone to Theorem~\ref{Theorem, Banach Steinhaus}.
\begin{lemma}[Principle of uniform boundedness]
Suppose that $(X,d)$ is a complete metric space
and we have a collection ${\mathcal F}$ of continuous
functions $f:X\rightarrow{\mathbb R}$ which
are pointwise bounded, that is, given any $x\in X$
we can find a $K(x)>0$ such that
\[|f(x)|\leq K(x)\ \text{for all $f\in{\mathcal F}$}.\]
Then we can find a ball $B(x_{0},\delta)$ and a $K$
such that
\[|f(x)|\leq K\ \text{for all $f\in{\mathcal F}$
and all $x\in B(x_{0},\delta)$ }.\]
\end{lemma}
\begin{exercise} (i) Suppose that $(X,d)$ is a complete metric space
and we have a sequence of continuous
functions $f_{n}:X\rightarrow{\mathbb R}$ and
a function $f:X\rightarrow{\mathbb R}$
such that $f_{n}$ converges pointwise
that is
\[f_{n}(x)\rightarrow f(x)\ \text{for all $x\in X$}.\]
Then we can find a ball $B(x_{0},\delta)$ and a $K$
such that
\[|f_{n}(x)|\leq K\ \text{for all $n$
and all $x\in B(x_{0},\delta)$ }.\]
(ii) (This is elementary but acts as a hint for (iii).)
Suppose $y\in [0,1]$. Show that we can find
a sequence of continuous functions $f_{n}:[0,1]\rightarrow{\mathbb R}$
such that $1\geq f_{n}(x)\geq 0$ for all $x$ and $n$,
$f_{n}$ converges pointwise to $0$ everywhere,
$f_{n}$ converges uniformly on $[0,1]\setminus (y-\delta,y+\delta)$
and fails to converge uniformly on
$[0,1]\cap (y-\delta,y+\delta)$ for all $\delta>0$.
(iii) State with reasons whether the following statement
is true or false. Under the conditions of (i) we can obtain the
stronger conclusion that
we can find a ball $B(x_{0},\delta)$ such that
\[f_{n}(x)\rightarrow f(x)\ \text{uniformly on}
\ B(x_{0},\delta).\]
\end{exercise}
\begin{exercise}
Suppose that $(X,d)$ is a complete metric space
and $Y$ is a subset of $X$ which is of first category
in $X$.
Suppose further that we have a collection ${\mathcal F}$ of continuous
functions $f:X\rightarrow{\mathbb R}$ which
are pointwise bounded on $X\setminus Y$,
that is, given any $x\notin Y$,
we can find a $K(x)>0$ such that
\[|f(x)|\leq K(x)\ \text{for all $f\in{\mathcal F}$}.\]
Show that
we can find a ball $B(x_{0},\delta)$ and a $K$
such that
\[|f(x)|\leq K\ \text{for all $f\in{\mathcal F}$
and all $x\in B(x_{0},\delta)$ }.\]
\end{exercise}
We now use the principle of uniform boundedness to prove
the Banach-Steinhaus theorem\footnote{You should be warned that
a lot of people, including the present writer, tend to
confuse the names of these two theorems. My research supervisor
took the simpler course of referring to all the theorems
of functional analysis as `Banach's theorem'.}.
\begin{theorem}{\bf (Banach-Steinhaus theorem)}%
\label{Theorem, Banach Steinhaus}
Let $(U,\|\ \|_{U})$ and $(V,\|\ \|_{V})$ be normed
spaces and suppose $\|\ \|_{U}$ is complete.
If we have a collection ${\mathcal F}$ of continuous
linear maps from $U$ to $V$ which are pointwise bounded
then we can find a $K$ such that $\|T\|\leq K$
for all $T\in{\mathcal F}$.
\end{theorem}
Here is a typical use of the Banach-Steinhaus theorem.
\begin{theorem} There exists a continuous
$2\pi$ periodic function $f:{\mathbb R}\rightarrow{\mathbb R}$
whose Fourier series fails to converge at a given point.
\end{theorem}
The next exercise contains results that most
of you will have already met.
\begin{exercise} (i) Show that the set $l^{\infty}$ of
bounded sequences over ${\mathbb F}$ (with
${\mathbb F}={\mathbb R}$ or ${\mathbb F}={\mathbb C}$)
\[{\mathbf a}=(a_{1},\ a_{2},\ \dots)\]
can be made into a vector space in a natural manner.
Show that $\|{\mathbf a}\|_{\infty}=\sup_{j\geq 1}|a_{j}|$
defines a complete norm on $l^{\infty}$.
(ii) Show that $s$, the set of convergent sequences
and $s_{0}$ the set of sequences convergent to $0$
are both closed subspaces of $(l^{\infty},\|\ \|_{\infty})$.
(iii) Show that the set $l^{1}$ of
sequences
\[{\mathbf a}=(a_{1},\ a_{2},\ \dots)
\ \text{such that $\sum_{j=1}^{\infty}|a_{j}|$ converges}\]
can be made into vector space in a natural manner.
Show that $\|{\mathbf a}\|_{1}=\sum_{j=1}^{\infty}|a_{j}|$
defines a complete norm on $l^{1}$.
(iv) Show that, if ${\mathbf a}\in l^{1}$, then
\[T_{\mathbf a}({\mathbf b})=\sum_{j=1}^{\infty}a_{j}b_{j}\]
defines a continuous linear map from $l^{\infty}$ to ${\mathbb F}$
and that $\|T_{\mathbf a}\|=\|{\mathbf a}\|_{1}$.
\end{exercise}
Here is another use of the Banach-Steinhaus theorem.
\begin{lemma}\label{Lemma, summation}
Let $a_{ij}\in{\mathbb R}$ $[i,j\geq 1]$.
We say that the $a_{ij}$ constitute a summation
method if whenever $c_{j}\rightarrow c$ we have
$\sum_{j=1}^{\infty}a_{ij}c_{j}$ convergent for each $i$
and
\[\sum_{j=1}^{\infty}a_{ij}c_{j}\rightarrow c\]
as $i\rightarrow\infty$.
The following conditions are necessary and sufficient
for the $a_{ij}$ to constitute a summation method:-
(i) There exists a $K$ such that
\[\sum_{j=1}^{\infty}|a_{ij}|\leq K\ \text{for all $i$}.\]
(ii) ${\displaystyle
\sum_{j=1}^{\infty}a_{ij}\rightarrow 1\ \text{as $i\rightarrow\infty$}}$.
(iii) $a_{ij}\rightarrow 0$ as $i\rightarrow\infty$ for each $j$.
\end{lemma}
\begin{exercise} Ces\`{a}ro's summation method takes
a sequence $c_{0},\ c_{1},\ c_{2},\dots $ and replaces it
with a new sequence whose $n$th term
\[b_{n}=\frac{c_{1}+c_{2}+\dots+c_{n}}{n}\]
is the average of the first $n$ terms of the old sequence.
(i) By rewriting the statement above along the lines
of Lemma~\ref{Lemma, summation} show that if the old
sequence converges to $c$ so does the new one.
(ii) Examine what happens when $c_{j}=(-1)^{j}$.
Examine what happens if $c_{j}=(-1)^{k}$ when
$2^{k}\leq j<2^{k+1}$.
(iii) Show that, in the notation of Lemma~\ref{Lemma, summation},
taking $a_{n,2n}=1$, $a_{n,m}=0$, otherwise, gives a summation
method. Show that
taking $a_{n,2n+1}=1$, $a_{n,m}=0$, otherwise,
also gives a summation
method. Show that the two methods disagree
when presented with the sequence $1$, $-1$, $1$, $-1$, \dots.
\end{exercise}
Another important consequence of the Baire
category theorem is the open mapping theorem.
(Recall that a complete normed space is called
a Banach space.)
\begin{theorem}[Open mapping theorem]
Let $E$ and $F$ be Banach spaces and
$T:E\rightarrow F$ be a continuous linear
surjection. Then $T$ is an open map
(that is to say, if $U$ is open in $E$
we have $TU$ open in $F$.)
\end{theorem}
This has an immediate corollary.
\begin{theorem}[Inverse mapping theorem]
Let $E$ and $F$ be Banach spaces and let
$T:E\rightarrow F$ be a continuous linear
bijection. Then $T^{-1}$ is continuous.
\end{theorem}
The next exercise is simple, and if you can not
do it this reveals a gap in your knowledge
(which can be remedied by asking the lecturer)
rather than in intelligence.
\begin{exercise} Let $(X,d)$ and $(Y,\rho)$
be metric spaces with associated topologies
$\tau$ and $\sigma$. Then the product topology
induced on $X\times Y$ by $\tau$ and $\sigma$
is the same as the topology given by the metric
\[\triangle((x_{1},y_{1}),(x_{2},y_{2}))
=d(x_{1},x_{2})+\rho(y_{1},y_{2}).\]
\end{exercise}
The inverse mapping theorem has the
following useful consequence.
\begin{theorem}[Closed graph theorem]
Let $E$ and $F$ be Banach spaces and let
$T:E\rightarrow F$ be linear.
Then $T$ is continuous if and only
the graph
\[\{(x,Tx)\,:\,x\in E\}\]
is closed in $E\times F$ with the product topology.
\end{theorem}
\section{Countable choice and Baire's theorem}
Most of this course is `practical' but the next
couple of sections deal with `foundational matters'.
We discuss
\noindent{\bf The axiom of choice} \emph{Let ${\mathcal A}$
be a non-empty collection of non-empty sets. Then there
exists a function
\[f:{\mathcal A}\rightarrow\bigcup_{A\in{\mathcal A}}A\]
such that $f(A)\in A$ for all $A\in {\mathcal A}$.}
Note that we do not require a specific axiom
if ${\mathcal A}$ only contains one set,
or, indeed, a finite set of sets.
However we do need an axiom to make a countable set
of choices.
\noindent{\bf The axiom of countable choice}
\emph{Let
$A_{1}$, $A_{2}$, \dots be non-empty sets. Then there
exists a function
\[f:{\mathbb Z}^{++}\rightarrow\bigcup_{j=1}^{\infty}A_{j}\]
such that $f(j)\in A_{j}$ for all $j\geq 1$.}
The standard proof of the next lemma requires the axiom
of countable choice.
\begin{lemma}\label{L;dense choice}
A closed subset of a separable metric
space is separable.
\end{lemma}
Note that it may not be necessary to use
the axiom
of countable choice
to
prove \emph{specific} cases of Lemma~\ref{L;dense choice}
when $(X,d)$ carries other structures.
\begin{lemma}\label{L;dense interval}
A non-empty closed subset of $[0,1]$
(the closed interval with the usual metric)
is separable.
\end{lemma}
In fact, logicians have shown that several elementary
theorems of analysis require a strictly stronger
axiom than the axiom of countable choice. The
axiom of countable choice enables us to use
arguments of the type `choose $a_{1}\in A_{1}$,
choose $a_{2}\in A_{2}$, choose $a_{3}\in A_{3}$
and so on'. However, sometimes we want
to use
arguments of the type `choose $a_{1}\in A_{1}$,
choose $a_{2}\in A_{2}$ depending on $a_{1}$,
choose $a_{3}\in A_{3}$ depending on $a_{1}$ and $a_{2}$
and so on'. For this we require an axiom which logicians
have shown to be strictly stronger than the axiom of
countable choice.
\noindent{\bf The axiom of countable dependent choice}
\emph{Suppose that $X$ and $R$ are sets with $R\subseteq X\times X$
and such that
\[\{y\in X\,:\,(x,y)\in R\}\neq\emptyset\]
for all $x\in X$. Then there exists a function
$g:{\mathbb Z}^{++}\rightarrow X$ such that
$\big(g(n),g(n+1)\big)\in R$ for all $n\geq 1$.}
We can easily obtain the version of the axiom
which we usually use.
\begin{lemma}
Let
$A_{1}$, $A_{2}$, \dots and $Y_{1}$, $Y_{2}$, \dots
be sets with the following properties.
(i) $A_{1}=Y_{1}$ and $A_{1}$ is non-empty.
(ii) $Y_{n}\subseteq A_{1}\times A_{2}\times \dots\times A_{n}$
for all $n\geq 1$.
(iii) If $a_{j}\in A_{j}$ for $1\leq j\leq n$ then the set
\[\{{\mathbf y}\in Y_{n+1}\,:\,y_{j}=a_{j}
\ \text{for all $1\leq j\leq n$}\}\neq\emptyset.\]
for all $n\geq 1$
Then there
exists a function
\[f:{\mathbb Z}^{++}\rightarrow\bigcup_{j=1}^{\infty}A_{j}\]
such that
\[\big(f(1),\,f(2),\,\ldots,\,f(n)\big)\in Y_{n}\]
for all $n\geq 1$.
\end{lemma}
If we now look at our proof of the
Baire category theorem (Theorem~\ref{T;Baire 2})
she will see that we made use of the axiom of
countable dependent choice.
A clever argument of Blair show that we cannot avoid this
and that indeed \emph{the
Baire category theorem is equivalent
to the axiom!}
\begin{theorem}\label{T;Blair}
The following two statements
are equivalent.
(A) If $(E,d)$ is a complete metric space, then
$E$ cannot be written as the union of a countable collection
of closed sets with empty interior.
(B) Suppose that $X$ and $R$ are sets with $R\subseteq X\times X$
and such that
\[\{y\in X\,:\,(x,y)\in R\}\neq\emptyset\]
for all $x\in X$. Then there exists a function
$G:{\mathbb Z}^{++}\rightarrow X$ such that
$\big(G(n),G(n+1)\big)\in R$ for all $n\geq 1$.
\end{theorem}
We have seen
that use of the axiom of countable dependent choice
is deeply embedded in ordinary analysis.
Logicians have shown that
if the other rules of reasoning and axioms
of set theory are free from contradiction
then adding this axiom will not produce
contradictions. The mode of reasoning expressed
by the axiom seems so natural to almost
all mathematicians that they are
happy to accept it\footnote{It is possible
to produce a version of analysis which does not
use the axiom or any related `non-constructive
rule of reasoning'. Bishop's book \cite{Bishop} shows
that the result is of comparable elegance to
ordinary analysis but it is a very different
theory.}.
\section{Zorn's lemma and Tychonov's theorem}
In the previous section we considered `making
a countable number of choices'. In this section
we consider the full axiom of choice.
\noindent{\bf The axiom of choice} \emph{Let ${\mathcal A}$
be a non-empty collection of non-empty sets. Then there
exists a function
\[f:{\mathcal A}\rightarrow\bigcup_{A\in{\mathcal A}}A\]
such that $f(A)\in A$ for all $A\in {\mathcal A}$.}
Most mathematicians are happy to add the axiom of choice
to the standard axioms and this is what we shall do.
Note that if we prove something using
the standard axioms and the axiom of choice
then we will be unable to find a counter-example
using only the standard axioms. Note also that,
when dealing with specific systems we may be able
to prove the result for that system without using
the axiom of choice.
The axiom of choice is not very easy to use in the form
that we have stated it and it is usually more convenient
to use an equivalent formulation called Zorn's lemma.
\begin{definition} Suppose $X$ is a non-empty set.
We say that
$\succeq$ is a partial order on $X$, that is to say,
that $\succeq$ is a relation on $X$ with
(i) $x\succeq y$, $y\succeq z$ implies $x\succeq z$,
(ii) $x\succeq y$ and $y\succeq x$ implies $x=y$,
(iii) $x\succeq x$
\noindent for all $x$, $y$, $z$.
We say that a subset $C$ of $X$ is a chain
if, for every $x,\ y\in C$ at least one of
the statements $x\succeq y$, $y\succeq x$ is true.
If $Y$ is a non-empty subset of $X$ we say that $z\in X$
is an upper bound for $Y$ if $z\succeq y$ for all $y\in Y$.
We say that $m$ is a maximal element for $(X,\succeq)$
if $x\succeq m$ implies $x=m$.
\end{definition}
You must be able to do the following exercise.
\begin{exercise} (i) Give an example of a partially
ordered set which is not a chain.
(ii) Give an example of a partially ordered set
and a chain $C$ such that (a) the chain has an upper bound
lying in $C$, (b) the chain has an upper bound but no
upper bound within $C$, (c) the chain has no upper bound.
(iii) If a chain $C$ has an upper bound lying in $C$,
show that it is unique. Give an example to show that,
even in this case $C$ may have infinitely many
upper bounds (not lying in $C$).
(iv) Give examples of partially ordered sets
which have (a) no maximal elements, (b) exactly one
maximal element, (b) infinitely many maximal elements.
(v) how should a minimal element be defined?
Give examples of partially ordered sets
which have (a) no maximal or minimal elements, (b) exactly one
maximal element and no minimal element,
(c) infinitely many maximal elements and infinitely many minimal
elements.
\end{exercise}
\begin{axiom}[Zorn's lemma] Let $(X,\succeq)$ be
a partially ordered set. If every chain in $X$ has
an upper bound then $X$ contains a maximal element.
\end{axiom}
Zorn's lemma is associated with a proof routine
which we illustrate in Lemmas~\ref{Zorn to choice}
and~\ref{Hamel}
\begin{lemma}\label{Zorn to choice} Zorn's lemma implies the axiom of choice.
\end{lemma}
The converse result is less important to us
but we prove it for completeness.
\begin{lemma}\label{choice to Zorn} The axiom of choice
implies Zorn's lemma.
\end{lemma}
\begin{proof} (Since the proof we use is non-standard,
I give it in detail.)
Let $X$ be a non-empty set with a partial order $\succeq$
having no maximal elements. We show that the assumption
that every chain has a upper bound leads to a contradiction.
We write $x\succ y$ if $x\succeq y$ and $x\neq y$.
If $C$ is a chain we write
\[C_{x}=\{c\in C\,:\,x\succ c\}.\]
Observe that, if $C$ is a chain in $X$, we can
find an $x\in X$ such that $x\succ c$ for all $c\in C$.
(By assumption, $C$ has an upper bound, $x'$, say.
Since $X$ has no maximal elements, we can find an
$x\in X$ such that $x\succ x'$.) We shall take
$\emptyset$ to be a well ordered chain.
We shall look at well ordered chains, that is to say,
chains for which every non-empty subset has a minimum.
(Formally, if $S\subseteq C$ is non-empty we can
find an $s_{0}\in C$ such that $s\succeq s_{0}$
for all $s\in S$. We write $\min C=s_{0}$.)
By the previous paragraph
\[A_{C}=\{x\,:\,x\succ c\ \text{for all $c\in C$}\}\neq\emptyset.\]
Thus, if we write ${\mathcal W}$ for the set of all well
ordered chains, the axiom of choice tells us that
there is a function $\kappa:{\mathcal W}\rightarrow X$
such that $\kappa(C)\succ c$ for all $c\in C$.
We now consider `special chains' defined to be
well ordered chains $C$ such that
\[\kappa(C_{x})=x\ \text{for all $x\in C$}.\]
(Note that `well ordering' is an important general
idea, but `special chains' are an ad hoc notion
for this particular proof. Note also that if $C$
is a special chain and $x\in C$ then $C_{x}$ is a special
chain.)
The key point is that, if $K$ and $L$ are special
chains, then either $K=L$ or $K=L_{x}$ for some $x\in L$
or $L=K_{x}$ for some $x\in K$.
\newline\emph{Subproof} If $K=L$, we are done. If not, at
least one of $K\setminus L$ and $L\setminus K$ is non-empty.
Suppose, without loss of generality, that $K\setminus L\neq \emptyset$.
Since $K$ is well ordered, $x=\min K\setminus L$ exists.
We observe that $K_{x}\subseteq L$. If $K_{x}=L$, we are done.
We show that the remaining possibility $K_{x}\neq L$ leads
to contradiction. In this case, $L\setminus K_{x}\neq \emptyset$
so $y=\min L\setminus K_{x}$ exists.
If $L_{y}=K_{x}$
then
\[y=\kappa(L_{y})=\kappa(K_{x})=x\]
so $x=y\in L\cap K$ contradicting the statement that $x\in K\setminus L$.
If $L_{y}\neq K_{x}$ let $z$
be the least member of $K_{x}\setminus L_{y}$.
Observe that, since $K_{x}\subseteq L$
and so
\[w\in L_{y},\ z'\in K_{x},\ w\succ z'\Rightarrow
z'\in L,\ y\succ w\succ z'\Rightarrow z'\in L_{y}\]
whence
\[z'\notin L _{y},\ z'\in K_{x},\ w\succ z'\Rightarrow
w\notin L_{y}.\]
Thus $K_{z}=L_{y}$ and
\[y=\kappa(L_{y})=\kappa(K_{z})=z\]
so $y=z\in K\cap L$ contradicting the definition of $y$.
\newline\emph{End subproof}
We now take $S$ to be the union of all special chains.
Using the key observation, it is routine to see that:
(i) $S$ is a chain. (If $a,\,b\in S$, then $a\in L$ and $b\in K$
for some special chains. By our key observation, either $L\supseteq K$
of $K\supseteq L$. Without loss of generality, $K\supseteq L$ so $a,\,b \in K$
and $a\succeq b$ or $b\succeq a$.)
(ii) If $a\in S$, then $S_{a}$ is a special chain. (We must have
$a\in K$ for some special chain $K$. Since $K\subseteq S$, we have
$K_{a}\subseteq S_{a}$. On the other hand, if $b\in S_{a}$
then $b\in L$ for some special chain $L$ and each of the three
possible relationships given in our key observation imply $b\in K_{a}$.
Thus $S_{a}\subseteq K_{a}$, so $S_{a}=K_{a}$ and $S_{a}$ is a special chain.)
(iii) $S$ is well ordered. (If $E$ is a non empty subset of $S$,
pick an $x\in E$. If $S_{x}\cap E=\emptyset$, then $x$ is a minimum for $E$.
If not, then $S_{x}\cap E$ is a non-empty subset of the special, so
well ordered, chain $S_{x}$, so $\min S_{x}\cap E$
exists and is a minimum for $E$.)
(iv) $S$ is a special chain. (If $x\in S$, we can find a special chain $K$
such that $x\in K$. Let $y=\kappa(K)$. Then $L=K\cup\{y\}$ is a special
chain. As in (ii), $S_{y}=L_{y}$, so $S_{x}=L_{x}$ and
$\kappa(S_{x})=\kappa(L_{x})=x$.)
We can now swiftly obtain a contradiction. Since $S$ is well ordered
$\kappa(S)$ exists and does not lie in $S$. But $S$ is special,
so $S\cup \kappa(S)$ is, so $S\cup\kappa(S)\subseteq S$,
so $\kappa(S)$ lies in $S$. The required result follows
by reductio ad absurdum\footnote{To the best of my knowledge,
this particular proof is due to
Jonathon Letwin (\emph{American Mathematical Monthly},
Volume 98, {\bf 1991}, pp. 353--4). If you know about
transfinite induction, there are more direct proofs.}.
\end{proof}
\begin{lemma}[Hamel basis theorem]\label{Hamel}
(i) Every vector space
has a basis.
(ii) If $U$ is an infinite dimensional normed space
over ${\mathbb F}$ then we can find a
linear map $T:U\rightarrow{\mathbb F}$.
(iii) If $U$ is an infinite dimensional normed space
over ${\mathbb F}$ then we can find a
discontinuous linear map $T:U\rightarrow{\mathbb F}$.
\end{lemma}
[Note that we do not claim that $T$ in (ii) is continuous.]
\begin{exercise} (i) Show that, if
$f:{\mathbb R}\rightarrow{\mathbb R}$ is continuous
and satisfies the equation
\[f(x+y)=f(x)+f(y)\]
for all $x,\ y\in{\mathbb R}$, then there exists a $c$
such that $f(x)=cx$ for all $x\in{\mathbb R}$.
(ii) Show that there exists a discontinuous function
$f:{\mathbb R}\rightarrow{\mathbb R}$
satisfying the equation
\[f(x+y)=f(x)+f(y)\]
for all $x,\ y\in{\mathbb R}$.
\noindent[Hint. Consider ${\mathbb R}$ as a vector space
over ${\mathbb Q}$.]
\end{exercise}
The rest of this section is devoted to a proof of Tychonov's
theorem. We recall a definition.
\begin{definition}\label{D;weak}
Let $(X_{\alpha},\tau_{\alpha})$
be a topological space for each $\alpha\in A$.
The product (or Tychonov or weak) topology $\tau$ on
$\prod_{\alpha\in A}X_{\alpha}$ is the collection of sets
$U$ such that if ${\mathbf u}\in U$ we can find
$\alpha_{1},\,\alpha_{2},\,\dots,\,\alpha_{n}\in A$
and $O_{\alpha_{j}}\in\tau_{\alpha_{j}}$ such that
$u_{\alpha_{j}}\in O_{\alpha_{j}}$ for $1\leq j\leq n$ and
${\mathbf x}\in U$ whenever $x_{\alpha_{j}}\in O_{\alpha_{j}}$
for $1\leq j\leq n$.
\end{definition}
\begin{exercise} We retain the notation of
Definition~\ref{D;weak} and write
$\pi_{\alpha}{\mathbf x}=x_{\alpha}$.
(i) Show that $\tau$ is indeed a topology
and that, with this topology, the maps
$\pi_{\alpha}:X\rightarrow X_{\alpha}$
are continuous.
(ii) Show that $\tau$ is the weakest topology
for which the $\pi_{\alpha}$ are continuous.
[Thus, if $\sigma$ is a topology for which
the $\pi_{\alpha}$ are continuous, we have
$\sigma\supseteq\tau$.]
(iii) Show that if all the $\tau_{\alpha}$ are Hausdorff
so is $\tau$.
(iv) Suppose that $A=[0,1]$ and $X_{t}={\mathbb R}$
and $\tau_{t}$ is the usual Euclidean topology
on $X_{t}$. Explain how an $f\in\prod_{t\in[0,1]}X_{t}$
can be identified in a natural manner with a function
$f:[0,1]\rightarrow{\mathbb R}$. With this identification
show that the sequence of functions $f_{n}\rightarrow f$
pointwise if and only if, given any $U\in\tau$ with $f\in U$
we can find an $N$ such that $f_{n}\in U$ for all $n\geq N$.
\end{exercise}
\begin{theorem}[Tychonov] The product of compact
spaces is itself compact under the weak topology.
\end{theorem}
We follow the presentation in~\cite{Bollobas}.
(The method of proof is due to Bourbaki.)
The following result should be familiar to almost
all of my readers.
\begin{lemma}[Finite intersection property]
(i) A topological
space is compact if and only if whenever a non-empty collection
of closed sets ${\mathcal F}$ has the property
that $\bigcap_{j=1}^{n}F_{j}\neq\emptyset$,
for any $F_{1}$, $F_{2}$, \dots, $F_{n}\in {\mathcal F}$
it follows that $\bigcap_{F\in{\mathcal F}}F\neq\emptyset$.
(ii) A topological
space is compact if and only if whenever a non-empty collection
of sets ${\mathcal A}$ has the property
that $\bigcap_{j=1}^{n}A_{j}\neq\emptyset$
for any $A_{1}$, $A_{2}$, \dots, $A_{n}\in {\mathcal A}$
it follows that $\bigcap_{A\in{\mathcal A}}\bar{A}\neq\emptyset$.
\end{lemma}
\begin{definition} A system ${\mathcal F}$
of subsets of a given set $S$ is said to be of finite character
if
(i) whenever every finite subset of a set $A\subseteq S$
belongs to ${\mathcal F}$ it follows that $A\in{\mathcal F}$
and
(ii) whenever $A\in{\mathcal F}$ every finite subset
of $A$ belongs to ${\mathcal F}$.
\end{definition}
\begin{lemma} [Tukey's lemma] If a system ${\mathcal F}$
of subsets of a given set $S$ has finite character
and $F\in{\mathcal F}$ then ${\mathcal F}$ has a maximal
(with respect to inclusion) element containing $F$.
\end{lemma}
We now prove Tychonov's theorem.
\begin{exercise} If $(X,\tau)$ is a Hausdorff
space and ${\mathcal G}$ is a maximal collection
of sets with the finite intersection property
explain why $\bigcap_{G\in{\mathcal G}}\bar{G}$
consists of one point.
If you are interested, examine how the second
(but not the first) appeal to the axiom of choice
may be avoided in our proof of Tychonov's theorem
if all our spaces are Hausdorff.
\end{exercise}
The reason why Tychonov's theorem demands the axiom
of choice is made clear by the final result
of this section.
\begin{lemma} Tychonov's theorem implies the axiom
of choice.
\end{lemma}
\section{The Hahn-Banach theorem} A good example
of the use of Zorn's lemma occurs when we ask
if given a Banach space $(U,\|\ \|)$ (over ${\mathbb C}$,
say) there exist
any non-trivial continuous linear maps $T:U\rightarrow{\mathbb C}$.
For any space that we can think of, the answer is
obviously yes, but to show that the result is
always yes we need Zorn's lemma\footnote{In fact the statement
is marginally weaker than Zorn's lemma but you need to
be logician either to know or care about this.}.
Our proof uses the theorem of Hahn-Banach.
One form of this theorem is the following.
\begin{theorem}{\bf (Hahn--Banach)}\label{T;Hahn--Banach} Let $U$
be a real vector space. Suppose
$p:U\rightarrow{\mathbb R}$ is such that
\[p(u+v)\leq p(u)+p(v)\ \text{and}\ p(au)=ap(u)\]
for all $u,\ v\in U$ and all real and positive $a$.
If $E$ is a subspace of $U$ and there exists
a linear map $T:E\rightarrow{\mathbb R}$
with $Tx\leq p(x)$ for all $x\in E$
then there exists
a linear map $\tilde{T}:U\rightarrow{\mathbb R}$
with $Tx\leq p(x)$ for all $x\in U$
and $\tilde{T}(x)=Tx$ for all $x\in E$.
\end{theorem}
\noindent[Note that we do not assume that
the vector space $U$ is normed but we do assume
that the vector space is real.]
\begin{exercise} We say that a function
$f:{\mathbb R}\rightarrow{\mathbb R}$ is convex if
\[f\big(\lambda x+(1-\lambda)y\big)\leq \lambda f(x)+(1-\lambda)f(y)\]
for all $x,\,y\in{\mathbb R}$. Using the ideas of the
proof of the Hahn--Banach theorem but not the result itself
show that a convex function $f$ is continuous
and that given any $a\in{\mathbb R}$ we can find
a $c$ such that
\[f(x)\geq f(a)+c(x-a)\]
for all $x\in{\mathbb R}$. Give an example to show that
$f$ need not be differentiable.
\end{exercise}
\begin{exercise} Let $X$ be a real vector space and
$p,\,q:X\rightarrow {\mathbb R}$ be functions such that
$p(\lambda x)=\lambda p(x)$, $q(\lambda x)=\lambda q(x)$
for all $\lambda\in{\mathbb R}$ with $\lambda\geq 0$
and all $x\in X$, whilst
\[p(x+y)\leq p(x)+p(y),\ q(x)+q(y)\leq q(x+y)\]
for all $x,\,y\in X$.
(i) Suppose that $Y$ is a subspace of $X$ and
$S:Y\rightarrow{\mathbb R}$ a linear function such that
\[S(y)\leq p(x+y)-q(x)\]
for all $x\in X$, $y\in Y$. Show that
\[S(y')-p(x'+y'-z)+q(x')\leq -S(y)+p(x+y+z)-q(x)\]
for all $x,\,x',\,z\in X$ and $y,\,y' \in Y$.
(ii) Suppose that $Y_{0}$ is a subspace of $X$ and
$T_{0}:Y\rightarrow{\mathbb R}$ a linear function such that
\[T_{0}(y)\leq p(x+y)-q(x)\]
for all $x\in X$, $y\in Y_{0}$. Show that there exists a linear
function $T_{0}:X\rightarrow{\mathbb R}$ such that
\[T(y)\leq p(x+y)-q(x)\]
for all $x,\,y\in X$ and $Tu=Tu_{0}$ for all $u\in Y_{0}$.
Show that
\[q(x)\leq T(x)\leq p(x)\]
for all $x\in X$.
(iii) Suppose $p(x)\geq q(x)$ for all $x\in X$.
Show that there exists
a linear function (possibly the zero function)
$U:X\rightarrow{\mathbb R}$
such that
\[q(x)\leq U(x)\leq p(x)\]
for all $x\in X$.
(iv) Let $X={\mathbb R}^{2}$, ${\mathbf n}$ be a unit vector
and $p({\mathbf x})=|{\mathbf n}.{\mathbf x}|$
(the absolute value of the usual inner product)
and $q({\mathbf x})=-|{\mathbf n}.{\mathbf x}|$.
Show that $p$ and $q$ obey the conditions of the introductory
paragraph and part~(iii). What can you say about $U$?
\end{exercise}
We have the following important corollary
to Theorem~\ref{T;Hahn--Banach}.
\begin{theorem} Let $(U,\|\ \|)$
be a real normed vector space.
If $E$ is a subspace of $U$ and there exists
a continuous linear map $T:E\rightarrow{\mathbb R}$,
then there exists
a continuous linear map $\tilde{T}:U\rightarrow{\mathbb R}$
with $\|\tilde{T}\|=\|T\|$.
\end{theorem}
The next result is famous as `the result that
Banach did not prove'.
\begin{theorem} Let $(U,\|\ \|)$
be a complex normed vector space.
If $E$ is a subspace of $U$ and there exists
a continuous linear map $T:E\rightarrow{\mathbb C}$
then there exists
a continuous linear map $\tilde{T}:U\rightarrow{\mathbb C}$
with $\|\tilde{T}\|=\|T\|$.
\end{theorem}
We can now answer the question posed in the first
sentence of this section.
\begin{lemma} If $(U,\|\ \|)$ is
normed space over the field ${\mathbb F}$
of real or complex numbers
and $a\in U$ with $a\neq 0$, then
we can find a continuous linear map
$T:U\rightarrow{\mathbb F}$ with $Ta\neq 0$.
\end{lemma}
The importance of this lemma becomes more obvious if
we state it in reverse. If $Ta=0$ for all continuous linear
maps then $a=0$.
Some analysts are consciously or unconsciously
unwilling to use the Hahn--Banach theorem because
of its association with the axiom of choice.
This is a MISTAKE. If we have a `concretely presented'
normed vector space then we can prove the
appropriate Hahn--Banach using, at worst, countable
dependent choice\footnote{This sentence is a slogan
but a good one.}.
\begin{exercise} If Let $(U,\|\ \|)$
be a real separable normed vector space.
If $E$ is a subspace of $U$ and there exists
a continuous linear map $T:E\rightarrow{\mathbb R}$,
show using only countable dependent choice
that there exists
a continuous linear map $\tilde{T}:U\rightarrow{\mathbb R}$
with $\|\tilde{T}\|=\|T\|$.
\end{exercise}
Here are a couple of results proved by Banach using the full
power of his theorem.
\begin{theorem}[Generalised limits]\label{Theorem, Banach limits}
Consider the vector space $l^{\infty}$ of bounded real
sequences. There exists a linear map $L:l^{\infty}\rightarrow {\mathbb R}$
such that
(i) If $x_{n}\geq 0$ for all $n$ then $L{\mathbf x}\geq 0$.
(ii) $L((x_{1},x_{2},x_{3},\dots))=L((x_{0},x_{1},x_{2},\dots))$.
(iii) $L((1,1,1,\dots))=1$.
\end{theorem}
The theorem is illustrated by the following lemma.
\begin{lemma} Let $L$ be as in Theorem~\ref{Theorem, Banach limits}.
Then
\[\limsup_{n\rightarrow\infty}x_{n}\geq L({\mathbf x})
\geq \liminf_{n\rightarrow\infty}x_{n}.\]
In particular, if $x_{n}\rightarrow x$ then $L({\mathbf x})=x$.
\end{lemma}
\begin{exercise} (i) Show that, even though the sequence
$x_{n}=(-1)^{n}$ has no limit, $L({\mathbf x})$ is uniquely
defined.
(ii) Find, with reasons, a sequence ${\mathbf x}\in l^{\infty}$
for which $L({\mathbf x})$ is not uniquely
defined.
\end{exercise}
Banach used the same idea to prove the following
odd result.
\begin{lemma}\label{Banach integral}
Let ${\mathbb T}={\mathbb R}/{\mathbb Z}$
be the unit circle and let $B({\mathbb T})$ be the
vector space
of real valued bounded functions.
Then we can find a linear map
$I:B({\mathbb T})\rightarrow{\mathbb R}$ obeying the following conditions.
(i) $I(1)=1$.
(ii) $If\geq 0$ if $f$ is positive.
(iii) If $f\in B({\mathbb T})$, $a\in{\mathbb T}$ and
we write $f_{a}(x)=f(x-a)$ then $If_{a}=If$.
\end{lemma}
\begin{exercise} Show that if $I$ is as in
Lemma~\ref{Banach integral} and $f$ is Riemann
integrable then
\[If=\int_{\mathbb T}f(t)\,dt.\]
\end{exercise}
However, Lemma~\ref{Banach integral} is put in context
by the following.
\begin{lemma}\label{Two generators}
Let $G$ be the group freely generated by
two generators and $B(G)$ be the
vector space
of real valued bounded functions on $G$.
If $f\in B(G)$ let us write $f_{c}(x)=f(xc^{-1})$
for all $x,\ c\in G$.
There exists a function $f\in B(G)$
and $c_{1},\ c_{2},\ c_{3}$ such that
$f(x)\geq 0$ for all $x\in G$ and
\[f(x)+f_{c_{1}}(x)-f_{c_{2}}(x)-f_{c_{3}}(x)\leq -1\]
for all $x\in G$.
\end{lemma}
\begin{exercise} If $G$ is as in Lemma~\ref{Two generators}
then there is no linear map $I:B(G)\rightarrow{\mathbb R}$
obeying the following conditions.
(i) $I(1)=1$.
(ii) $If\geq 0$ if $f$ is positive.
(iii) $If_{c}=If$ for all $c\in G$.
\end{exercise}
It can be shown that there is a finitely additive,
congruence respecting integral for ${\mathbb R}$ and
${\mathbb R}^{2}$ but not ${\mathbb R}^{n}$ for $n\geq 3$.
\section{Three more uses of Hahn-Banach} The following exercise
provides background for our first discussion but is not examinable.
For the moment $C([a,b])$ will be the set of real valued
continuous functions.
\begin{exercise} We say that a function $G:[a,b]\rightarrow{\mathbb R}$
is of bounded variation if there exists a $K$ such that
whenever we have a dissection
\[{\mathcal D}=\{x_{0},\ x_{1},\ x_{2},\ \dots,\ x_{n}\}\]
$a=x_{0}0$ we can find a $\delta>0$
such that, given any
\[{\mathcal D}=\{x_{0},\ x_{1},\ x_{2},\ \dots,\ x_{n}\}\]
with $|x_{j-1}-x_{j}|<\delta$ $[1\leq j\leq n]$ we have
\[|S({\mathcal D},f,G)-I(f,G)|<\epsilon.\]
We write
\[I(f,G)=\int_{a}^{b}f(t)\,dG(t).\]
(i) Let $[a,b]=[0,1]$. Find elementary expressions
for $\int_{a}^{b}f(t)\,dG(t)$
in the three cases when $G(t)=t$, when $G(t)=-t$
and when $G(t)=0$ for $t<1/2$, $G(t)=1$ for $t\geq 1/2$.
(ii) Show that the map $T:(C([a,b]),\|\ \|_{\infty})\rightarrow{\mathbb R}$
given by
\[Tf=\int_{a}^{b}f(t)\,dG(t)\]
is linear and continuous with $\|T\|\leq\|G\|_{BV}$.
\end{exercise}
(In order to obtain the more satisfactory result $\|T\|=\|G\|_{BV}$
we must put an extra condition on $G$ such as left continuity.)
\begin{exercise} (i) Suppose $G:{\mathbb R}\rightarrow{\mathbb R}$
is of bounded variation in every interval $[a,b]$.
Show that if we write
\[G_{+}(t)=G(a)+\sup\left\{\sum_{j=1}^{m}G(b_{j})-G(a_{j})
\,:\,a\leq a_{1}\leq b_{1}\leq a_{2}\leq b_{2}\leq\dots\leq
a_{m}\leq b_{m}\leq t,\ m\geq 1\right\}\]
and
\[G_{-}(t)=\sup\{\sum_{j=1}^{m}G(a_{j})-G(b_{j})
\,:\,a\leq a_{1}\leq b_{1}\leq a_{2}\leq b_{2}\leq\dots\leq
a_{m}\leq b_{m}\leq t,\ m\geq 1\}\]
then $G_{+},\,G_{-}:[a,b]\rightarrow{\mathbb R}$
are increasing functions with
\[G(t)=G_{+}(t)-G_{-}(t)\]
and $\|G\|_{BV}=(G_{+}(b)-G_{+}(a))+(G_{-}(b)-G_{-}(a))$.
(ii) By first considering increasing functions, or otherwise,
show that if $G$ is a function of bounded variation in every interval
then the left and right limits
\[G(t+)=\lim_{h\rightarrow 0,h>0}G(t+h),
G(t-)=\lim_{h\rightarrow 0,h>0}G(t-h)\]
exist everywhere.
\end{exercise}
\begin{theorem} If $T:C([a,b])\rightarrow{\mathbb R}$
is a continuous linear function then we can find
a left continuous function $G:{\mathbb R}\rightarrow{\mathbb R}$
of bounded variation in every interval such that
\[Tf=\int_{a}^{b}f(t)\,dG(t)\]
for all $f\in C([a,b])$.
\end{theorem}
If you know a little measure theory you can restate
the theorem in more modern language.
\begin{theorem} {\bf (The Riesz representation theorem.)}
The dual of $C([a,b])$ is the space of Borel
measures on $[a,b]$.
\end{theorem}
The method used can easily be extended to all compact
spaces.
Our second result is more abstract. We require
Aloaoglu's theorem.
\begin{theorem} The unit ball of the dual of a normed
space $X$ is compact in the weak star topology.
\end{theorem}
Our proof of the Riesz representation theorem
used the Hahn-Banach theorem as a convenience.
Our proof of the next result uses it as basic
ingredient.
\begin{theorem} Every Banach space is isometrically
isomorphic to some subspace of $C(K)$ for some
compact space $K$.
\end{theorem}
(In my opinion this result looks more interesting
than it is.)
Our third result requires us to recast
the Hahn Banach theorem in a geometric form.
\begin{lemma}\label{L;geometrical Banach}
If $V$ is a real normed space
and $E$ is a convex subset of $V$ containing
$B({\boldsymbol 0},\epsilon)$ for some $\epsilon>0$,
then, given any ${\mathbf x}\notin E$ we can find
a continuous linear map $T:V\rightarrow{\mathbb R}$
such that $T{\mathbf x}=1\geq T{\mathbf e}$ for
all ${\mathbf e}\in E$.
\end{lemma}
\begin{theorem}
If $V$ is a real normed space
and $F$ is a closed convex subset of $V$,
then, given any ${\mathbf x}\notin F$ we can find
a continuous linear map $T:V\rightarrow{\mathbb R}$
and a real $\alpha$
such that $T{\mathbf x}>\alpha>T{\mathbf k}$ for
all ${\mathbf k}\in F$.
\end{theorem}
\begin{definition} Let $V$ be a real or complex
vector space. If $K$ is a non-empty subset of $V$
we say that $E\subseteq K$ is an \emph{extreme set}
of $K$ if, whenever $u,\ v\in K$, $1>\lambda>0$
and $\lambda u+(1-\lambda)v\in E$, it follows that
$u,\ v\in E$. If $\{e\}$ is an extreme set we
call $e$ an extreme point.
\end{definition}
\begin{exercise} Define an extreme point directly.
\end{exercise}
\begin{exercise} We work in ${\mathbb R}^{2}$.
Find the extreme points, if any, of the following sets
and prove your statements.
(i) $E_{1}=\{{\mathbf x}\,:\,\|{\mathbf x}\|<1\}$.
(ii) $E_{2}=\{{\mathbf x}\,:\,\|{\mathbf x}\|\leq 1\}$.
(iii) $E_{3}=\{(x,0)\,:\,x\in{\mathbb R}\}$.
(iv) $E_{4}=\{(x,y)\,:\,|x|,|y|\leq 1\}$.
\end{exercise}
\begin{theorem}{\bf (Krein--Milman).} A non-empty
compact convex subset $K$ of a normed vector space
has at least one extreme point.
\end{theorem}
\begin{theorem} A non-empty
compact convex subset $K$ of a normed vector space
is the closed convex hull of its extreme points
(that is, is the smallest closed convex set
containing its extreme points).
\end{theorem}
Our hypotheses in our version of the Krein--Milman theorem
are so strong as to make the conclusion practically useless.
However the hypotheses can be much weakened as is indicated
by the following version.
\begin{theorem}{\bf (Krein--Milman).} Let $E$
be the dual space of a normed vector space.
A non-empty
convex subset $K$ which is compact in the weak star topology
has at least one extreme point.
\end{theorem}
\begin{theorem} Let $E$
be the dual space of a normed vector space.
A non-empty
convex subset $K$ which is compact in the weak star topology
is the weak star closed convex hull of its extreme points.
\end{theorem}
The results follow at once from a new version of
Lemma~\ref{L;geometrical Banach}
\begin{lemma}
If $V$ is a real normed space
and $E$ is a convex subset of the dual space $V'$ containing
an open (in the wek topology)
neighbourhood of ${\boldsymbol 0}$,
then, given any ${\mathbf x}\notin E$ we can find
a continuous (in the weak topology) linear map $T:V'\rightarrow{\mathbb R}$
such that $T{\mathbf x}=1\geq T{\mathbf e}$ for
all ${\mathbf e}\in E$.
\end{lemma}
\begin{lemma} The extreme points of the closed
unit ball of the dual of $C([0,1])$ are the delta masses
$\delta_{a}$ and $-\delta_{a}$ with $a\in[0,1]$.
\end{lemma}
\section{The Rivlin-Shapiro formula} In this section
we give an elegant use of extreme points due to
Rivlin and Shapiro.
\begin{lemma}{\bf Carath\'{e}odory}
We work in ${\mathbb R}^{n}$. Suppose
that ${\mathbf x}\in{\mathbb R}^{n}$ and we are given
a finite set of points ${\mathbf e}_{1}$,
${\mathbf e}_{2}$, \dots, ${\mathbf e}_{N}$ and positive
real numbers $\lambda_{1}$, $\lambda_{2}$, \dots, $\lambda_{N}$
such that
\[\sum_{j=1}^{N}\lambda_{j}=1,
\ \sum_{j=1}^{N}\lambda_{j}{\mathbf e}_{j}={\mathbf x}.\]
Then {\bf after renumbering} the ${\mathbf e}_{j}$
we can find positive
real numbers $\lambda_{1}'$, $\lambda_{2}'$, \dots, $\lambda_{m}'$
with $m\leq n+1$ such that
\[\sum_{j=1}^{m}\lambda_{j}'=1,
\ \sum_{j=1}^{m}\lambda_{j}'{\mathbf e}_{j}={\mathbf x}.\]
\end{lemma}
\begin{exercise} Show by means of an example
that we can not necessarily
take $m=n$ in Carath\'{e}odory's lemma.
\end{exercise}
\begin{lemma}\label{Shapiro 1}
Consider ${\mathcal P}_{n}$, the subspace
of $C([-1,1])$ consisting of real polynomials of degree
$n$ or less. If $S:{\mathcal P}_{n}\rightarrow{\mathbb R}$
is linear then we can find an $N\leq n+2$
and
distinct points $x_{1}$, $x_{2}$, \dots, $x_{N}\in[-1,1]$
and non-zero
real numbers $\lambda_{1}$, $\lambda_{2}$, \dots, $\lambda_{N}$
such that
\[\sum_{j=1}^{N}|\lambda_{j}|=1,
\ \|S\|\sum_{j=1}^{N}\lambda_{j}P(x_{j})=SP\]
for all $P\in{\mathcal P}_{n}$.
\end{lemma}
\begin{lemma}\label{Shapiro 2}
We continue with the hypotheses
and notation of Lemma~\ref{Shapiro 1}
There exists a $P_{*}\in{\mathcal P}_{n}$ such that
\[P_{*}(x_{j})=\|P_{*}\|_{\infty}\sgn\lambda_{j}\]
for all $j$ with $0\leq j\leq N$.
Further, if $P\in{\mathcal P}_{n}$ satisfies
\[P(x_{j})=\|P\|_{\infty}\sgn\lambda_{j}\]
then $\|P\|_{\infty}\|S\|=SP$.
\end{lemma}
The following
results are of considerable interest
in view of Lemma~\ref{Shapiro 2}.
\begin{lemma} We have $\cos n\theta=T_{n}(\cos\theta)$
where $T_{n}$ is a real polynomial of degree $n$.
Further
(i) $|T_{n}(x)|\leq 1$ for all $x\in [-1,1]$.
(ii) There exist $n+1$
distinct points $x_{1}$, $x_{2}$, \dots, $x_{n+1}\in[-1,1]$
such that $|T_{n}(x_{j})|=1$ for all $1\leq j\leq n+1$.
\end{lemma}
\begin{lemma}\label{L;Tchebychev extreme}
Suppose that $P$ is a real polynomial of degree
$n$ or less such that
(i) $|P(x)|\leq 1$ for all $x\in [-1,1]$ and
(ii) there exist $n+1$
distinct points $x_{1}$, $x_{2}$, \dots, $x_{n+1}\in[-1,1]$
such that $|P(x_{j})|=1$ for all $1\leq j\leq n+1$.
\noindent Then $P=\pm T_{n}$.
\end{lemma}
Note that Lemma~\ref{L;Tchebychev extreme} tells
us that there is no real polynomial of degree
$n$ or less which takes its extreme absolute value
on $[-1,1]$ at $n+2$ points. (Thus we can replace the condition
$N\leq n+2$ in Lemma~\ref{Shapiro 1} by $N\leq n+1$.)
\begin{theorem}If
$P$ is a real polynomial of degree at most $n$
and $t\notin[-1,1]$, then
\[|P(t)|\leq \sup _{|x|\leq 1}|P(x)||T_{n}(t)|.\]
\end{theorem}
\begin{exercise} If
$P$ is a real polynomial of degree at most $n$
and $t\notin[-1,1]$, then
then
\[|P^{(r)}(t)|\leq |T^{(r)}(t)|\sup_{|x|\leq 1}|P(x)|.\]
\end{exercise}
\begin{exercise} (This exercise is part of the course.)
(i) Show that if $n\geq 1$ the coefficient of $t^{n}$ in
$T_{n}(t)$ is $2^{n-1}$.
(ii) Show that if $n\geq 1$ and $P$ is a real polynomial
of degree $n$ or less with $|P(t)|\leq 1$ then
the coefficient of $t^{n}$ in
$P(t)$ has absolute value at most $2^{n-1}$.
(iii) Find, with proof, a polynomial $P$ of degree at
most $n-1$ which minimises
\[\sup_{t\in[-1,1]}|t^{n}-P(t)|.\]
Show that $P$ is unique. (Tchebychev introduced his polynomials
$T_{n}$ in this context.)
\end{exercise}
\section{Uniqueness of Fourier series}
The subject of distributions has its roots
in the study of partial differential equations
and the study of trigonometric series. Most of
its applications lie in the
study of partial differential equations
but I shall consider one from harmonic analysis.
Recall that ${\mathbb T}={\mathbb R}/2\pi{\mathbb Z}$,
that the Fourier coefficient $\hat{f}$ of
of a continuous function $f:{\mathbb T}\rightarrow{\mathbb C}$
is given by
\[\hat{f}(n)=
\frac{1}{2\pi}\int_{{\mathbb T}}f(t)\chi_{-n}(t)\,dt\]
where $\chi_{n}(t)=\exp(2\pi int)$.
We are
used to the idea of studying the Fourier sum
\[\sum_{n=-\infty}^{\infty}\hat{f}(n)\chi_{n}\]
where $f$ is some continuous function.
\begin{lemma} (i) If $f:{\mathbb T}\rightarrow{\mathbb C}$
is continuous and $\hat{f}(n)=0$ for all $n\in{\mathbb Z}$
then $f=0$.
(ii) If $f:{\mathbb T}\rightarrow{\mathbb C}$
is continuous and
$\sum_{n=-\infty}^{\infty}|\hat{f}(n)|<\infty$
then
\[\sum_{n=-N}^{N}a_{n}\chi_{n}(t)\rightarrow f(t)\]
as $N\rightarrow\infty$ for all $t\in{\mathbb T}$
\end{lemma}
What happens if we study general trigonometric sums
\[\sum_{n=-\infty}^{\infty}a_{n}\chi_{n}\]
with $a_{n}\in{\mathbb C}$? One of the first
questions about such sums is the problem of uniqueness.
If
\[\sum_{n=-N}^{N}a_{n}\chi_{n}(t)\rightarrow 0\]
as $N\rightarrow\infty$ for all $t\in{\mathbb T}$
does it follow that $a_{n}=0$ for all $n$?
\begin{exercise} (Easy.) Show that,
if $\sum_{n=-N}^{N}a_{n}\chi_{n}(t)\rightarrow 0$
as $N\rightarrow\infty$ for all $t\in{\mathbb T}$ then
$a_{n}\rightarrow 0$ as $|n|\rightarrow\infty$.
\end{exercise}
Riemann had the happy idea of considering the effect
of formally integrating twice to obtain
\[F(t)=A+Bt+\frac{a_{0}t^{2}}{2}
-\sum_{n=-\infty}^{\infty}\frac{a_{n}}{n^{2}}\chi_{n}(t).\]
\begin{exercise} (Easy.)
Suppose that $a_{n}\rightarrow 0$ as $|n|\rightarrow$.
Explain why $F$ is a well defined continuous function.
\end{exercise}
When $\sum_{n=-N}^{N}a_{n}\chi_{n}(t)$ converges to a certain
value we can recover that value by looking at the
`generalised second derivative'
\[\lim_{h\rightarrow 0}\frac{F(+h)-2F(t)+F(t-h)}{4h^{2}}.\]
\begin{exercise} If $f:{\mathbb R}\rightarrow{\mathbb R}$
is twice differentiable at $0$ with $f(0)=f'(0)=f''(0)=0$
use the mean value theorem to show that
\[\frac{f(h)-2f(0)+f(-h)}{4h^{2}}\rightarrow 0\]
as $h\rightarrow 0$.
Deduce that if $g:{\mathbb R}\rightarrow{\mathbb R}$
is twice differentiable at $0$, then
\[\frac{g(h)-2g(0)+g(-h)}{4h^{2}}\rightarrow g''(0)\]
as $h\rightarrow 0$.
\end{exercise}
\begin{exercise} Suppose that $a_{n}\in{\mathbb C}$,
$a_{n}\rightarrow 0$ as $|n|\rightarrow\infty$. If
\[F(t)=A+Bt+\frac{a_{0}t^{2}}{2}
-\sum_{n=-\infty}^{\infty}\frac{a_{n}}{n^{2}}\chi_{n}(t),\]
show that
\[\frac{F(x+h)-2F(x)+F(x-h)}{4h^{2}}
=a_{0}+\sum_{n\neq 0}a_{n}\chi_{n}(x)
\left(\frac{\sin 2\pi nh}{nh}\right)^{2}.\]
\end{exercise}
\begin{lemma}\label{L;Riemann summation} If
$\sum_{n=0}^{\infty}b_{n}$ converges then
\[b_{0}+\sum_{n=1}^{\infty}b_{n}\left(\frac{\sin nh}{nh}\right)^{2}
\rightarrow \sum_{n=0}^{\infty}b_{n}\]
as $h\rightarrow 0$.
\end{lemma}
\begin{exercise}\label{E;reverse Riemann} Deduce from
Lemma~\ref{L;Riemann summation} that,
if
\[\sum_{n=-N}^{N}a_{n}\chi_{n}(t)\rightarrow 0\]
as $N\rightarrow\infty$ for all $t\in{\mathbb T}$
and we set
\[F(t)=\frac{a_{0}t^{2}}{2}
-\sum_{n=-\infty}^{\infty}\frac{a_{n}}{n^{2}}\chi_{n}(t)\]
then
\[\frac{F(t+h)-2F(t)+F(t-h)}{4h^{2}}\rightarrow 0\]
as $h\rightarrow 0$ for all $t\in{\mathbb T}$
\end{exercise}
Part of the proof of Lemma~\ref{L;Riemann summation}
rests on ideas which are now familiar.
\begin{exercise}\label{E;Toeplitz}
(i) Suppose that $\gamma_{n}(h)\in{\mathbb C}$
satisfies the following two conditions.
(A) $\gamma_{n}(h)\rightarrow 0$ as $h\rightarrow 0$.
(B) There exists a $C$ such that
\[\sum_{n=0}^{\infty}|\gamma_{n}(h)|\leq C\]
for all $h$.
Then, if $t_{n}\rightarrow 0$ as $n\rightarrow\infty$ it follows
that
\[\sum_{n=0}^{\infty}\gamma_{n}(h)t_{n}\rightarrow 0\]
as $h\rightarrow 0$.
(ii) Suppose, in addition, that
(C) ${\displaystyle \sum_{n=1}^{\infty}\gamma_{n}(h)\rightarrow 1}$
as $h\rightarrow 0$.
Then, if $s_{n}\rightarrow t$ as $n\rightarrow\infty$, it follows
that
\[\sum_{n=0}^{\infty}\gamma_{n}(h)s_{n}\rightarrow t\]
as $h\rightarrow 0$.
\end{exercise}
We combine the result of Lemma~\ref{E;reverse Riemann}
with a very neat result of Schwarz.
\begin{lemma}\label{L;Schwarz concave}
Let $f:[a,b]\rightarrow{\mathbb R}$ be continuous.
(i) Suppose that $f(a)=f(b)$ and
\[\limsup_{h\rightarrow 0}\frac{f(x+h)-2f(x)+f(x-h)}{4h^{2}}>0\]
for all $x\in (a,b)$. Then $f(x)\leq 0$ for all $x\in[a,b]$.
(ii) Suppose that $f(a)=f(b)$ and
\[\limsup_{h\rightarrow 0}\frac{f(x+h)-2f(x)+f(x-h)}{4h^{2}}\geq 0\]
for all $x\in (a,b)$. Then $f(x)\leq 0$ for all $x\in[a,b]$
(iii) Suppose that $f(a)=f(b)$ and
\[\frac{f(x+h)-2f(x)+f(x-h)}{4h^{2}}\rightarrow 0\]
as $h\rightarrow 0$
for all $x\in (a,b)$. Then $f(x)=0$ for all $x\in[a,b]$.
(iv) Suppose that
\[\frac{f(x+h)-2f(x)+f(x-h)}{4h^{2}}\rightarrow 0\]
as $h\rightarrow 0$
for all $x\in (a,b)$. Then there exist constants $A$ and $B$
such that $f(x)=Ax+B$ for all $x\in[a,b]$.
\end{lemma}
Putting our results together we obtain the following
uniqueness theorem.
\begin{theorem} If
\[\sum_{n=-N}^{N}a_{n}\chi_{n}(t)\rightarrow 0\]
as $N\rightarrow\infty$ for all $t\in{\mathbb T}$
then $a_{n}=0$ for all $n$.
\end{theorem}
If we try to push matters further we arrive at a natural definition.
\begin{definition} A subset $E$ of ${\mathbb T}$ is called
a \emph{set of uniqueness} if
\[\sum_{n=-N}^{N}a_{n}\chi_{n}(t)\rightarrow 0\]
as $N\rightarrow\infty$ for all $t\notin E$
implies that $a_{n}=0$ for all $n$.
\end{definition}
We shall use the theory of distributions to
show that every countable closed set is a set of uniqueness.
\section{A first look at distributions} We will work
in ${\mathbb R}^{m}$ but we will also keep in mind
the simpler case of ${\mathbb T}^{m}$.
\begin{definition} If $f:{\mathbb R}^{m}\rightarrow{\mathbb R}$
is a continuous function we write
\[\supp f=\Cl\{x\,\,:\,f(x)\neq 0\}.\]
\end{definition}
\begin{definition} We write ${\mathcal D}({\mathbb R}^{m})$
(or $C_{c}({\mathbb R}^{n})$)
for the set of smooth (ie infinitely differentiable)
functions $f:{\mathbb R}^{n}\rightarrow{\mathbb R}$
of compact support. If $f_{n},\,f\in{\mathcal D}$
we say that
\[f_{n}\underset{\mathcal D}{\rightarrow}f\]
if and only if there exists a compact set $K$
such that $\supp f_{n}\subseteq K$ for each $n$,
and
\[sup_{x\in K}|f_{n}^{({\mathbf p})}(x)-f^{({\mathbf p})}(x)|\rightarrow 0\]
for every ${\mathbf p}\in{\mathbb Z}_{n}^{+}$.
\end{definition}
Here
\[f_{n}^{({\mathbf p})}(x)=\partial^{{\mathbf p}}f(x)
=\frac{\partial^{p_{1}+p_{2}+\ldots+p_{m}}}
{\partial^{p_{1}}x_{1}\partial^{p_{2}}x_{2}\ldots\partial^{p_{m}}x_{n}}\]
and ${\mathbb Z}_{+}$ is the set of non-negative integers.
Before going any further we need to establish that
${\mathcal D}$ is reasonably rich.
To do this we go back to a function introduced by
Cauchy as a counterexample.
\begin{exercise} (i) Let $F:{\mathbb R}\rightarrow{\mathbb R}$
be given by
\[F(x)=\begin{cases}\exp(-1/x^{2})&\text{if $x>0$,}\\
0&\text{if $x\leq 0$}.\end{cases}\]
Show inductively that $F$ is $n$ times differentiable
with
\[F^{(n)}(x)=\begin{cases}P_{n}(x)\exp(-1/x^{2})&\text{if $x>0$,}\\
0&\text{if $x\leq 0$.}\end{cases}\]
for some polynomial $P_{n}$.
(ii) By considering functions of the form $F(x-a)F(a-x)$
show that, given any $\delta>0$ we can find an
infinitely differentiable function $G:{\mathbb R}\rightarrow{\mathbb R}$
with
$\supp g\subseteq[-\epsilon,\epsilon]$, $G(x)\geq 0$
for all $x$ and $G(0)>0$.
(iii) By considering functions of the form
\[C\int_{-\infty}^{x}G(t+a)-G(t-a)\,dt\]
show that, given any $\eta>0$, we can find an
infinitely differentiable function $K:{\mathbb R}\rightarrow{\mathbb R}$
with
\[K(x)=\begin{cases}1&\text{if $|x|\leq 1$,}\\
0&\text{if $1+\eta\leq |x|$.}\end{cases}\]
and $0\leq K(x)\leq 1$ for all $x$.
(iv) Show that given any $\eta>0$, we can find an
infinitely differentiable function
$E:{\mathbb R}^{n}\rightarrow{\mathbb R}$
with
\[K(x)=\begin{cases}1&\text{if $\|x\|\leq 1$,}\\
0&\text{if $1+\eta\leq \|x\|$.}\end{cases}\]
and $0\leq K(x)\leq 1$ for all $x$.
\end{exercise}
\begin{exercise} Identify with reasons the set
of compactly supported analytic functions
$f:{\mathbb C}\rightarrow{\mathbb C}$.
\end{exercise}
If ${\mathbf p}\in{\mathbb Z}_{+}^{m}$ we write
$|{\mathbf p}|=p_{1}+p_{2}+\ldots+p_{m}$.
\begin{exercise} Let $K$ be a compact set. We write
${\mathcal D}_{k}$ for the set of $f\in{\mathcal D}$
such that $\supp f\subseteq K$.
(i) If $f_{n},\,f\in{\mathcal D}_{K}$
show that
\[f_{n}\underset{\mathcal D}{\rightarrow}f\]
if and only if
\[\sup_{\|x\|\leq q}|\partial^{\mathbf p}f_{n}(x)
-\partial^{\mathbf p}f(x)|\rightarrow 0\]
for all integers $p\in{\mathbb Z}_{+}^{m}$, $q\geq 0$.
(ii) Show that
\[d(f,g)=\sum_{q=0}^{\infty}
\sum_{{\mathbf p} \in{\mathbb Z}_{+}^{m}}
|{\mathbf p}|^{-m}2^{-|\mathbf p|-q}
\sup_{\|x\|\leq q}
|\partial^{\mathbf p}f_{n}(x)-\partial^{\mathbf p}f(x)|\]
gives a well defined metric on ${\mathcal D}_{K}$.
(iii) If $f_{n},\,f\in{\mathcal D}_{K}$ show that
\[f_{n}\underset{\mathcal D}{\rightarrow}f
\Leftrightarrow d(f_{n},f)\rightarrow 0\]
\end{exercise}
\begin{exercise} We show that convergence in distribution
cannot be derived from convergence in norm even
for ${\mathcal D}({\mathbb T}$ (with the obvious
modified definitions).
Suppose that $\|\ \|_{D}$ is a norm on ${\mathcal D}$
such that
\[\|f_{n}-f\|_{D}\rightarrow 0\Rightarrow f_{n}
\underset{\mathcal D}{\rightarrow}f.\]
(i) By reductio ad absurdum, or otherwise, show that, for each
integer $j\geq 0$, there exists an $\epsilon_{j}>0$ such that
we have
$\|f\|_{D}\geq \epsilon_{j}$ whenever $\sup_{x\in[0,1]}|f^{(j)}(x)|\geq 1$.
(ii) Deduce that
$\|g\|_{D}\geq \epsilon_{j}\sup_{x\in[0,1]}|g^{(j)}(x)|$
for all $g\in C^{\infty}([0,1])$ and all $j\geq 0$.
(iii) Show that, by choosing $\delta_{k}$ and $N_{k}$
appropriately, and setting $f_{k}(x)=\delta_{k}\sin\pi N_{k}x$,
or otherwise, that we can find $f_{k}\in C^{\infty}([0,1])$
such that
\begin{align*}
\sup_{x\in[0,1]}|f_{k}^{(j)}(x)|&\leq 2^{-k}\qquad
\text{when $0\leq j\leq k-1$},\\
\sup_{x\in[0,1]}|f_{k}^{(k)}(x)|&\geq 2^{j}\epsilon_{j}.
\end{align*}
(iv) Show that $\|f_{n}\|_{D}\rightarrow \infty$ as $n\rightarrow\infty$,
but $f_{n}^{(j)}(x)\rightarrow 0$ uniformly
on $[0,1]$ for each $j$. Thus
\[f_{n}
\underset{\mathcal D}{\rightarrow}f\not\Rightarrow
\|f_{n}-f\|_{D}\rightarrow 0.\]
\end{exercise}
The fact that ${\mathcal D}$ is not a normed space
led to investigations of the much more general
`topological vector spaces' but ${\mathcal D}({\mathbb T}$
only just fails to be normed since its topology
is given by a countable collection of norms.
We can now introduce the notion of a distribution.
\begin{definition} We say that
a linear map $T:{\mathcal D}\rightarrow{\mathbb C}$
is a distribution if
\[\phi_{n}\underset{\mathcal D}{\rightarrow}\phi
\Rightarrow T\phi_{n}\rightarrow T\phi.\]
\end{definition}
The next lemma gives some equivalent formulations
of the definition.
\begin{lemma} If $T:{\mathcal D}\rightarrow{\mathbb C}$
then the following statements are equivalent.
(i) $\phi_{n}\underset{\mathcal D}{\rightarrow}\phi
\Rightarrow T\phi_{n}\rightarrow T\phi$.
(ii) $\phi_{n}\underset{\mathcal D}{\rightarrow}0
\Rightarrow T\phi_{n}\rightarrow 0$.
(iii) If $K$ is a compact set
then there exists a constant $C(K)$
and an integer $N(K)\geq 0$ such that
whenever $\phi\in{\mathcal D}$ and $\supp\phi\subseteq K$
we have
\[|T\phi|\leq C(K,{\mathbf p})
\sum_{|{\mathbf p}|\leq N(K)}\|\partial^{\mathbf p}\phi\|_{\infty}.\]
\end{lemma}
We often write $T\phi=\langle T,\phi\rangle$ and say that
$\phi\in{\mathcal D}$ is a \emph{test function}.
\begin{lemma} If $a\in{\mathbb R}^{m}$, then
\[\langle \delta_{a},\phi\rangle=\phi(a)\]
defines a distribution.
\end{lemma}
We call $\delta_{a}$ the \emph{Dirac delta function} at $a$
\begin{lemma}\label{L;smooth distribution}
If $f\in C({\mathbb R}^{m})$ then
\[\langle T_{f},\phi\rangle=\int_{{\mathbb R}^{m}}f(t)\phi(t)\,dt\]
defines a distribution.
\end{lemma}
\begin{exercise} (If you know some measure theory.) We
write $f\in L^{1}_{loc}({\mathbb R}^{m})$ if
$f$ is measurable and $f{\mathbb I}_{K}\in L^{1}$
(ie $\int_{K}|f(t)|\,dt<\infty$) for all compact sets,
show that
\[\langle T_{f},\phi\rangle=\int_{{\mathbb R}^{m}}f(t)\phi(t)\,dt\]
defines a distribution.
\end{exercise}
We often write $T_{f}=f$ (whence the name expression `generalised
function for a distribution) and for this reason we must careful
to make our definitions consistent when do so.
\begin{lemma} (i) If $T, S\in{\mathcal D}$
and $\lambda,\,\mu\in{\mathbb C}$, then
\[\langle\lambda T+\mu S,\phi\rangle=
\lambda\langle T,\phi\rangle+
\mu\langle S,\phi\rangle\]
for $\phi\in{\mathcal D}$ defines a distribution
$\lambda T+\mu T$.
(ii) ${\mathcal D}'$ is a vector space with these operations.
(iii) If $f$ and $g$ are continuous functions
$T_{\lambda f+\mu g}=\lambda T_{f}+\mu T_{g}$.
\end{lemma}
The next result is more interesting.
\begin{lemma} (i) If $T\in{\mathcal D}$, and
${\mathbf p}$ has all entries zero except one which has value $1$
\[\langle {\partial}_{\mathbf p} T,\phi\rangle=
-\langle T,{\partial}_{\mathbf p} f\rangle\]
for $\phi\in{\mathcal D}$ defines a distribution
$T'$.
(iii) If $f$ is a continuous function with continuous
partial derivative,then
$T_{\lambda f+\mu g}=\lambda T_{f}+\mu T_{g}$.
\end{lemma}
We call ${\partial}_{\mathbf p} T$ the partial derivative
of $T$.
\begin{exercise} (i) Distributions are infinitely
differentiable with
\[\langle{\partial}_{\mathbf p} T,\phi\rangle
=(-1)^{|{\mathbf p}|}
\langle T,{\partial}_{\mathbf p}\phi\rangle.\]
(ii) If we work on ${\mathbb R}$,
$\langle \delta'_{0},\phi\rangle=-\phi'(0)$.
(If what you have been told in the past
contradicts this, forget it. This is the correct
sign.)
\end{exercise}
We can also multiply distributions by test functions.
\begin{lemma}\label{L;multiply smooth}
(i) If $T\in{\mathcal D}$, and
$\psi\in{\mathcal D}$ then
${\mathbf p}$ has all entries zero except one which has value $1$
\[\langle \psi T,\phi\rangle=
\langle T,\phi\psi\rangle\]
for $\phi\in{\mathcal D}$ defines a distribution
$\psi T$.
(ii) If $T\in{\mathcal D}$, and
$\psi\in{\mathcal D}$ then we have the following form of
Leibniz's rule
\[{\partial}_{\mathbf p}\phi T
=\sum_{{\mathbf r}+{\mathbf s}={\mathbf p}}\frac{p!}{r!s!}
\partial^{\mathbf r}\phi\partial^{\mathbf s}T.\]
\end{lemma}
Here and elsewhere we write
$(r_{1},r_{2},\ldots,r_{m})!=r_{1}!r_{2}!\ldots r_{m}!$.
\begin{exercise} Show that, in Lemma~\ref{L;multiply smooth},
we can replace $\psi\in {\mathcal D}$
by $\psi\in C^{\infty}$ that is to say, $\psi$
a smooth function).
\end{exercise}
\section{The support of distribution} The statement that $T$
is a distribution seems very abstract but in this section
we shall see that $T$ actually `lives on a well defined
region' in ${\mathbb R}^{m}$.
We use the useful idea of a \emph{partition of unity}.
\begin{lemma} (i) Let $K$ be a compact subset and $\Omega$
an open subset of ${\mathbb R}^{m}$ such that $K\subseteq\Omega$.
Then we can find a
$\phi\in{\mathcal D}$ such that $0\leq\phi(x)\leq 1$
for all $x$, $\phi(x)=1$ for $x\in K$ and
$\phi(x)=0$ for $x\notin\Omega$.
(ii) Let $K$ be a compact subset and $\Omega_{j}$
open subsets of ${\mathbf R}^{m}$ $[1\leq j\leq J]$
such that $K\subseteq\bigcup_{j=1}^{J}\Omega_{j}$.
Then we can find
$\phi_{j}\in{\mathcal D}$ such that $0\leq\phi_{j}(x)\leq 1$
for all $x$, $\phi(x)=1$
$\phi_{j}(x)=0$ for $x\notin\Omega_{j}$ $[1\leq j\leq J]$
and
\[\sum_{j=1}^{J}\phi_{j}(x)=1\]
for all $x\in K$.
\end{lemma}
\begin{definition} If $T\in{\mathcal D}$ then $\supp T$ is the complement
of the set of points $x$ with the following property. There
exists an $\epsilon>0$ such that if $\phi\in{\mathcal D}$
and $\supp\phi\subseteq B(x,\epsilon)$ (the open ball centre
$x$ radius $\epsilon$) then $\langle T,\phi\rangle=0$.
\end{definition}
\begin{exercise}
(i) If $T$ is a distribution $\supp T$ is closed.
(ii) If $f$ is continuous then
\[\supp T_{f}=\Cl\{x\,:\,f(x)=0\}.\]
In other words the support of $f$ as a distribution
coincides with the support of $f$ as a function.
\end{exercise}
\begin{exercise} (i) If $T$ is a distribution, show that
$\supp\partial^{\mathbf p}T\subseteq\supp T$. Give an example
for ${\mathbf R}$ with $p=1$ to show that the inclusion may be proper.
(ii) If $T\in{\mathcal D}'$ and $\phi\in{\mathcal D}$, show that
\[\supp \phi T\subseteq\supp T\cap\supp\phi.\]
\end{exercise}
\begin{lemma} If $T\in{\mathcal D}'$, $\phi\in{\mathcal D}$
$\supp T\cap\supp\phi=\emptyset$, then
\[\langle T,\phi\rangle=0.\]
\end{lemma}
The following example is extremely important.
\begin{example} We work on ${\mathbb R}$.
We have $\supp\delta_{0}'=\{0\}$ but we can find a
$\phi\in{\mathcal D}$ such that $\phi(0)=0$
and $\langle \delta_{0}',\phi\rangle\neq 0$.
\end{example}
Thus
\[\phi(x)=0\ \text{for all $x\in\supp T$}\not\Rightarrow
\langle T,\phi\rangle=0.\]
\begin{exercise} If $T\in{\mathcal D}'$, $\phi\in{\mathcal D}$
and $\supp T\cap\supp\phi=\emptyset$, show that
$\phi T=0$.
\end{exercise}
We now find all distributions with support consisting of a single
point. to do this we require a \emph{local}
form of Taylor's theorem
\begin{lemma} If $f:{\mathbb R}^{m}\rightarrow{\mathbb R}$
is infinitely differentiable then
\[f(h)=\sum_{p_{1}+p_{2}+\ldots+p_{m}\leq N}
\frac{(p_{1}+p_{2}+\ldots+p_{m})!}{p_{1}!p_{2}!\ldots p_{m}!}
\partial^{{\mathbf p}}f(0)
h_{1}^{p_{1}}h_{2}^{p_{2}}\ldots h_{m}^{p_{m}}
+A(h)\|h\|^{N+1}\]
where $A(h)$ remains bounded as $\|h\|\rightarrow 0$.
\end{lemma}
\begin{theorem} If $T\in{\mathcal D}$ and
$\supp T\subseteq\{a\}$ then we can find
an $N$ and $c_{\mathbf p}$ such that
\[T=\sum_{|{\mathbf p}|\leq N}c_{\mathbf p}\partial^{\mathbf p}\delta_{a}.\]
\end{theorem}
\section{Distributions on ${\mathbb T}$} In
this section we work work on ${\mathbb T}$.
To maintain consistency we modify one of earlier definitions
(see Lemma~\ref{L;smooth distribution} and the associated discussion).
by introducing a scaling factor.
\begin{definition}
If $f\in C({\mathbb T})$ then
\[\langle T_{f},\phi\rangle=
\frac{1}{2\pi}\int_{{\mathbb T}}f(t)\phi(t)\,dt.\]
\end{definition}
If we work on ${\mathbb T}$
the space of distributions takes a particularly simple form.
\begin{definition} If $T\in{\mathcal D}({\mathbb T})$
we write
\[\hat{T}(r)=\langle T,\chi_{-r}\rangle.\]
\end{definition}
\begin{exercise} If $f\in C{(\mathbb T})$ check that
the new and old definitions of $\hat{f}$ coincide.
\end{exercise}
\begin{theorem} (i) If $T\in{\mathcal D}'({\mathbb T})$
and then there exists a $N$ such that
$|r|^{-N}\hat{T}(r)\rightarrow 0$ as $|r|\rightarrow\infty$.
(ii) If $T\in{\mathcal D}'({\mathbb T})$ and
$\phi\in {\mathcal D}({\mathbb T})$, then
\[\langle T,\phi\rangle
=\sum_{r=-\infty}^{\infty}\hat{T}(r)\hat{\phi}(-r)\]
the convergence being uniform.
(iii) If $T,\,S\in{\mathcal D}'({\mathbb T})$,
\[T=S\Leftrightarrow\hat{T}(r)=\hat{S}(r)
\ \text{for all $r\in{\mathbb Z}$}.\]
(iv) If $a_{r}\in{\mathbb C}$ and there exists an $N$
such that
$|r|^{-N}a_{r}\rightarrow 0$ as $|r|\rightarrow\infty$,
then
\[\langle T,\phi\rangle
=\sum_{r=-\infty}^{\infty}a_{r}\hat{\phi}(-r)\]
defines a distribution $T$ with $\hat{T}(r)=a_{r}$.
\end{theorem}
We could say that we have identified the distributions
with sequences of \emph{polynomial growth}.
We link the theory of distributions with the theory
of sets of uniqueness via a version of the Riemann localisation
theorem.
\begin{theorem} Suppose that $T$ is a distribution
on ${\mathbb D}$ with $\hat{T}(n)\rightarrow 0$
and $\phi\in{\mathcal D}$. If
\[\sum_{j=-n}^{n}\hat{T}(j)\chi_{j}(t)\rightarrow 0\]
as $n\rightarrow\infty$
on an open interval $I$ with $I\supseteq \supp\phi$
then writing $S=\phi T$ we have
\[\sum_{j=-n}^{n}\hat{S}(j)\chi_{j}(t)\rightarrow 0\]
everywhere.
\end{theorem}
\begin{lemma} Suppose that $a_{n}\rightarrow 0$ as
$|n|\rightarrow\infty$. If $T$ is the distribution
with $\hat{T}(n)=a_{n}$ then
\[\supp T
=\Cl\{t\,:\,\ \sum_{j=-n}^{n}\hat{T}(j)\chi_{j}(t)
\ \text{fails to converge to $0$}\}.
\]
\end{lemma}
We now make the following observation.
\begin{lemma} Any closed countable subset of ${\mathbb T}$
must contain an isolated point.
\end{lemma}
Combining our results gives the following classical theorem
\begin{theorem} Every closed countable subset of
${\mathbb T}$ is a set of uniqueness.
\end{theorem}
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\bibitem{Bollobas} B. Bollob\'as \emph{Linear Analysis:
an Introductory Course} (CUP 1991)
\bibitem{Friedlander} F. G. Friedlander
and M. S. Joshi \emph{Introduction to Theory of Distributions}
(CUP 1998. This is the second edition with extra material.
The first edition by Friedlander alone is ample for this course.)
\bibitem{Gofman}
C. Gofman and G. Pedrick \emph{A First
Course in Functional Analysis} (Prentice Hall 1965,
available as a Chelsea reprint from the AMS)
\bibitem{Pryce} J. D. Pryce \emph{Basic Methods of Linear Functional
Analysis} (Hutchinson 1973)
\bibitem{Rudin 1}
W. Rudin \emph{Real and Complex Analysis} (McGraw Hill, 2nd Edition, 1974)
\bibitem{Rudin 2}
W. Rudin \emph{Functional Analysis} (McGraw Hill 1973)
\end{thebibliography}
\end{document}