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FUNCTIONAL ANALYSIS

METRIC SPACES

\end{center}

\begin{description}

\item[]
A metric space is a space in which is defined a distance function
$p$ with the following properties:

\begin{enumerate}

\item
$p(x,y)\geq0$ with equality $\Leftrightarrow x=y$.

\item
$p(x,y)=p(y,x)$.

\item
$p(x,y)\leq p(x,z)+p(y,z)$.

\end{enumerate}

A sequence $\{x_n\}$ is a cauchy sequence if, given
$\varepsilon&gt;0\exists N$ such that $p(x_n,x_m)&lt;\varepsilon$
whenever $n,m&gt;N$.

\item[Definition]
A complete metric space is a metric space in which every Cauchy
sequence converges.

\item[Contraction Mapping Theorem]
Let $X$ be a complete metric space and $T$ a continuous mapping of
$X$ into $X$ such that $p(Tx,Ty)\leq\theta(p(x,y))$ where
$0&lt;\theta&lt;1$. Then $T$ has a unique fixed point.

\item[Proof]
Let $x$ be an arbitrary point in $X$. Let $x_0=x,\ x_1=T(x),\
x_2=T(T(x))\ldots$.

The sequence $\{x_n\}$ is a cauchy sequence for
$p(x){n+1},x_n)\leq\theta^np(x_0,x_1)$ and so

\begin{eqnarray*}
p(x_{n+p},x_n)&amp;\leq&amp;p(x_{n+1},x_n)+\ldots+p(x_{n+p},x_{n+p-1})\\
&amp;\leq&amp;(\theta^n+\theta^{n+1}+\ldots+\theta^{n+p-1})p(x_0,x_1)
\end{eqnarray*}

and therefore the coefficient of $p(x_0,x_1)$ can be made
arbitrarily small provided only that $n$ is large.

Therefore there is a point $\xi\in X$ such that $x_n\to\xi$ as
$n\to\infty\ \ x_0x_1x_2\ldots\to\xi$.

Therefore $Tx_0Tx_1\ldots\to T\xi$ by continuity of $T$ i.e.
$x_1x_2\ldots\to T\xi$ therefore $T\xi=\xi$.

Suppose $\eta=T\eta,\ \xi=T\xi$.

$$p(\xi,\eta)=p(T\xi T\eta)\leq\theta p(\xi,\eta)\
\theta&lt;1\textrm{therefore $p(\xi,\eta)=0$ therefore }\xi=\eta.$$

\item[Exercise]
$$4\frac{dx}{dt}+\sin x+\int_\frac{t}{2}^t1+x^2(s)\sin s\,ds=0$$

prove that there is a unique function $x(t)$ defined and
continuous for $0\leq t\leq1$ which is a solution of this
equation.

\item[Definition]
A set $H$ is nowhere dense if $(\overline{H})^0=\emptyset$.

\item[Baire's Theorem]
Let $X$ be a complete metric space and $\{H_n\}$ a sequence of
subsets of $X$ such that $\cup H_n=X$. Then it is impossible for
every one of the $H_n$ to be nowhere dense.

\item[Proof]
Suppose the $H_n$ are all nowhere dense. Then the complement of
$\overline{H}_1$ contains some non-empty sphere $S_1$ where
$r(S_1)&lt;1$. Since $S_1$ is not contained in $\overline{H}_2$ we
can find a non-empty closed sphere $S_2$ contained in the
complement of $\overline{H}_2$ and in $S_1$.

We may suppose $r(S_2)&lt;\frac{1}{2}$.

Proceeding in this way we can find a sequence of non-empty closed
spheres $S_1\supset S_2\supset\ldots$ with $\overline{H}_n\cap
S_n=\emptyset$ and $r(S_n)&lt;\frac{1}{n}$.

For each $n$ suppose $x_n\in S_n$. Then $\{x_n\}$ is a Cauchy
sequence therefore $x_n\to x$ as $n\to\infty$ therefore
$x\in\cap_{n=1}^\infty S_n$ therefore $x\in\cup_{n=1}^\infty H_n$
which is a contradiction.

Baire's theorem is also true for locally compact spaces.

\item[Definition]
A set of the first category is a set which is the join of an
enumerable number of nowhere dense sets.

A set of second category is one which is not of the first
category.

\item[Example]
The rational numbers cannot be expressed as an intersection of
open sets.

\item[Proof]
The real line is a complete metric space and so by Baire's Theorem
is of the second category. The rationals, being a countable set,
are a set of the first category. Suppose that $\cap_{i=1}^\infty
G_i={\cal{Q}}$ (the rationals).

Then $F_1'$ is closed and $\cup_{i=}^\infty G_i'={\cal{r-Q}}$.

$)\overline{G}_2')^0=(G_i')^0=\emptyset$, for otherwise $G_2'$
would contain an interval, and so would contain points of
$\cal{Q}$. But this tells us that $\cal{R-Q}$ is of the first
category which is a contradiction.

\item[Example]
The set of continuous functions which are differentiable at even
one point is a set of the fist category.

\item[Proof]
Let $f$ be defined and continuous on $[0\ 1]$. Define
$p(f,g)=\sup_{x\in[0\ 1]}\{|f(x)-g(x)|\}$. Then this is a complete
metric space, denoted by $L^p[0\ 1]$.

Consider the set of functions which are differentiable at 0 say,
i.e. $\frac{f(x)-f(0)}{x}\to$ limit as $n\to\infty$.

Define $H_{m,n}=\left\{f:\left|\frac{f(x)-f(0)}{x}\right|\leq
n\textrm{ whenever }x&lt;\frac{1}{m}\right\}$.

Each $H_{m,n}$ is closed and is nowhere dense, and
$\cup_{m,n}H_{m,n}$ contains the set of all functions
differentiable at the origin. Hence the set of functions
differentiable at the origin is in the first category.

The argument can be extended as follows. Let $U_1 u_2\ldots$ be an
enumerable basis for the open sets e.g. the rational spheres with
rational centres, and define

$$H_{m,n}=\left\{f:\left|\frac{f(x)-f(y)}{x-y}\right|\leq
n\textrm{ whenever }x,y\in U_m\right\}$$

\item[Zorn's Lemma (A form of the axioms of choice)]
Any partially ordered set in which every simply ordered subset has
a maximum element, has a maximum element.

A Hamel basis for the real numbers is a set $B$ such that

\begin{description}

\item[(a)]
Any subset of $B$ is rationally independent.

\item[(b)]
Any real number is a t.c. of a finite number of the $B$.

\end{description}

Consider all subsets $Y$ of $R_1$ with the property that
$y_1\ldots y_n\in Y\Rightarrow r_1y_1+\ldots +r_ny_n\neq0$ unless
$r_1=\ldots=r_n=0$, $r$ rational.

We order the sets $Y$ by inclusion.

Let $\{Y_\alpha\}$ be a simply ordered class of these sets.

If $Y=\cup_\alpha$ then $Y$ has the required independence
property, for if $y_{\alpha_i}\in Y\ (i=1,\ldots,n)$ then
$y_{\alpha_i}\in Y_i\ i=1,\ldots,n$ and
$\exists\max_{i=1,\ldots,n}Y_i$ for $\left\{Y_\alpha\right\}$ is
simply ordered so the $y_{\alpha_i}$ are rationally independent.

Hence $y$ is a maximum element, therefore by Zorn's Lemma
$\exists$ a set $B$ which is a maximal element.

Suppose $r$ cannot be expressed as a r.c. of elements in $B$. Then
we add $x$ to $B$ to get a greater set with the same independence
property therefore as $B$ is maximal it follows that every real
$x$ can be expressed as a r.c. of elements in $B$.

The same method of proof also shows the existence of a basis for
any vector space whatsoever.

\end{description}

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