\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

\begin{center}

FUNCTIONAL ANALYSIS

THE STONE-WEIERSTRASS THEOREM

\end{center}

\begin{description}

\item[]
Let $X$ be a compact space and let $A$ be an algebra of
real-valued continuous functions on $X$ which separates the points
of $X$ (i.e. $x\neq y\Rightarrow\exists f\in A:f(x)\neq f(y)$) and
which has the property that there is no point of $X$ at which all
the functions of $A$ vanish. Then $A$ is uniformly dense on the
set of all the real-values continuous functions defined on $X$.

\item[Lemma 1]
The set of all continuous functions forms a lattice. Let $A$ be a
set of real-valued continuous functions on a compact space $X$
which is closed under the lattice operations $f\vee g$ and
$f\wedge g$. Then the uniform closure of $A$ contains every
continuous function on $X$ which can be approximated at every pair
of points by a function of $A$.

[The uniform closure of $A$ means the set of functions to which
functions of $A$ converge uniformly.]

\item[Proof]
Let $f$ be any function which can be so approximated and let
$\varepsilon>0$.

Given $x,y\in X$ let $f_{xy}\in A$ be such that
$|f_{xy}(x)-f(x)|<\varepsilon$ and $|f_{xy}(y)-f(y)|<\varepsilon$

Fixing $y$, let $U_{xy}=\{z:f_{xy}(z)<f(z)+\varepsilon\}$

Let $V_{xy}=\{z:f_{xy}(z)>f(z)-\varepsilon\}$

Now $x\in U_{xy}$ therefore $\cup_xU_{xy}\supset X$.

hence there are a finite number of these sets say $U_{x_1y}\ldots
U_{x_ny}$ whose union covers $X$.

We put $f_y=f_{x_1}y\wedge f_{x_2}y\wedge\ldots\wedge f_{x_n}y$

$f_y\in A$ as $A$ is closed under $\wedge$.

Write $Vy=Vx_1y\cap Vx_2y\cap\ldots\cap V_{x_n}y$.

$Vy$ is then a neighbourhood of $y$ and $fy<f+\varepsilon$
everywhere on $X$ $fy>f-\varepsilon$ on the neighbourhood $Vy$ of
$y$ as $y\in Vy$. $\cup_y V_y\supset X$, so there is a finite
number of these sets, say $V_{y_1}\ldots,V_{y_k}$ whose union
covers $X$.

We put $g=fy_1\vee fy_2\vee\ldots\vee fy_k$.

Then $g\in A$ and $f-\varepsilon<g<f+\varepsilon$ everywhere on
$X$.

\item[Lemma 2]
A uniformly closed algebra $A$ of bounded real-valued functions on
a set is also closed for the lattice operations.

\item[Proof]

\begin{eqnarray*}
f\vee g&=&\frac{f+g+|f-g|}{2}\\ f\vee g&=&\frac{f+g-|f-g|}{2}
\end{eqnarray*}

Hence it suffices to show that $f\in A\Rightarrow |f|\in A$.

We may suppose without loss of generality

$$\|f\|=\sup\{|f(x)|:x\in X\}\leq 1$$

The Taylor series for $(t+\varepsilon^2)^\frac{1}{2}$ about
$t=\frac{1}{2}$ converges uniformly in $0\leq t\leq 1$ therefore
putting $t=x^2$ there is a polynomial $P(x^2)$ in $x^2$ such that

$$|P(x^2)-(x^2+\varepsilon^2)^\frac{1}{2}|<\varepsilon \textrm{ on
}[-1\ 1]$$

If $Q=P-P(0)$ then since $|P(0)|\leq 2\varepsilon$ we have

$$|Q(x^2)-(x^2-\varepsilon^2)^\frac{1}{2}|<3\varepsilon\textrm{ on
}[-1\ 1]$$

Now $0<(x^2+\varepsilon^2)^\frac{1}{2}-|x|<\varepsilon$ so

$$|Q(x^2)-|x||<4\varepsilon\textrm{ on }[-1\ 1]$$

Since $Q(f^2)\in A$ and $|Q(f^2)-|f||<4\varepsilon$ everywhere on
$X$ therefore $|F|\in A$ as $A$ is uniformly closed.

\item[Proof of Theorem]
Let $\overline{A}$=uniform closure of $A$ then it is clear that
$\overline{A}$ is an algebra therefore by Lemma 2 it is closed
under the lattice operations.

Using the given properties of $A$ we can find a function $g\in A$
so that

$$g(x)\neq0\ \ g(y)\neq0\ \ g(x)\neq g(y).$$

then $g(x)g^2(y)\neq g(y)g^2(x)$. Thus we can always find $\alpha\
\beta$ satisfying

\begin{eqnarray*}
\alpha f(x)+\beta f^2(x)&=&a\\ \alpha g(y)+\beta g^2(y)&=&b
\end{eqnarray*}

for any given $a$ and $b$. Hence any $f$ can be approximated at a
pair of points by a function of $\overline{A}$ (as $\alpha g+\beta
g^2\in \overline{A}$). Hence the result follows from lemma 1.

\item[Linear Transformations]
Let $E$ and $F$ be two vector spaces. A transformation $T$ from
$E$ to $F$ is linear if $A(\
alpha x+\beta y)=\alpha T(x)+\beta
T(y)$ for any $x,y\in E$ and any $\alpha,\ \beta$.

The set of all linear transformations from $E$ to $F$ is itself a
vector space over the same field of scalars, for if $T_1,T_2$ are
two such transformations we can define $T_1$ and $T_2$ and $\alpha
T_1$ by

\begin{eqnarray*}
(T_1+T_2)x&=&T_1x+T_2x\\ (\alpha T_1)x&=&\alpha(T_1x)
\end{eqnarray*}

The linear transformations from a vector space onto itself form an
algebra, for if $T_1,\ T_2$ are 2 such transformations we can
define $T_1T_2x=T_1(T_2x)$.

\item[Continuous linear transformations between Banach Spaces]
Let $E$ and $F$ be Banach Spaces and let $T$ be a linear
transformation from $E$ to $F$. The following statements are
equivalent:

\begin{description}

\item[(i)]
$T$ is continuous.

\item[(ii)]
$T$ is continuous at one point.

\item[(iii)]
$T$ is bounded on the unit sphere.

\item[(iv)]
There is a number $N$ such that $\|Tx\|\leq N\|x\|$ for any $x\in
E$.

\end{description}

The set of all continuous linear transformations form a Banach
Sphere.

Define $\|T\|=\sup_{\|X\|=1}\|Tx\|=\sup_x\frac{\|Tx\|}{\|x\|}$.

This is clearly a norm.

Suppose $\{T_n\}$ is a Cauchy sequence. For any $x\in E\
\|T_n(x)-T_m(x)\|\leq\|T_n-T_m\|\|x\|$ therefore $\{T_n(x)\}$ for
every $x$ is a Cauchy sequence therefore $T_n(x)\to T(x)$.

Now suppose without loss of generality $T_n(x)\to0$ for every $x$.
R.T.P. $\|T_n\|\to0$.

Given $\varepsilon>0$ choose $N$ such that
$\|T_n-T_m\|<\frac{\varepsilon}{2}$ whenever $m,n>N$.

Let $\|x\|\leq1$. Let $m>N$. Since $T_n(x)\to0\exists n>N$ such
that $\|T_n(x)\|<\frac{\varepsilon}{2}$

\begin{eqnarray*}
\|T_m(x)-T_n(x)\|\leq\|T_m-T_n\|\|x\|&\leq&\frac{\varepsilon}{2}\\
\textrm{ therefore
}\|T_m(x)\|\leq\|T_n(x)-T_m(x)\|+\|T_n(x)\|&<&\varepsilon
\end{eqnarray*}

if we have bounded linear mappings from $X\to$ Complex numbers,
the Banach space of these mappings is called the dual space $X^*$
of $X$. Its elements are called functionals. We may also regard
the elements of $X$ as functions defined on $X^*$.

If $x\in X$ and $x^*\in X^*$ we use $\left<x,x^*\right>$ for
$x^*(x)$.

If $E$ is a vector space, $E$ is a linear subspace of deficiency 1
if $\exists x\in E$ such that $H+[x]=E$.

Suppose $f$ is any functional $f:X\to{\cal{C}}$.

$H=\{x\in E:f(x)-0\}$ is a hyperplane if $f\not\equiv0$.

Chooses $x_0$ such that $f(x_0)\neq0$.

Let $y\in E$. Then $y-\frac{f(y)}{f(x_0)}x_0\in H,\ \ y=h+\lambda
x_0$.

Conversely given any hyperplane $H\exists x$ such that $H+[x]=E$
i.e. $y=h+\lambda x\ \lambda$ is unique.

Define $f(y)=\lambda\alpha\ \ \alpha\neq0$ fixed.

Two functionals have the same null space $\Leftrightarrow$ one is
a multiple of the other. The continuous functionals are those for
which the null space is a closed hyperplane.

If $X$ is a finite dimensional space $X^*$ is the same as $X$.

\item[Example]
$$(\ell^P)^*=\ell^p$$

Let $\{b_n\}\in\ell^q$ then we can define

$$f(\{a_n\})=\sum a_nb_n\leq\left(\sum|a_n|^P\right)^\frac{1}{p}
\left(|sum|b_n|^q\right)^\frac{1}{q}<\infty.$$

$|f(\{a_n\})\leq \|a_n\|_p\|b_n\|_q$ therefore $\|f\|\leq
\|b_n\|_q$.

To show $\|f\|=\|\{b_n\}\|\_q$:

Choose any $N$ and define

$$A_n=|B_n|^{q-1}\frac{\overline{b}_n}{|b_n|}\ n\leq N\ 0,n>N$$

Then

\begin{eqnarray*}
|f(\{a_n\})|&=&\sum_1^N|b_n|^q\leq\|f\|\|a_n\|_p\\
&=&\|f\|\left(\sum_1^N|b_n|^{qp-p}\right)^\frac{1}{p}=
\|f\|\left(\sum_1^N|b_n|^q\right)^\frac{1}{p}\\ \textrm{therefore
}\left(\sum_1^N|b_n|^q\right)^\frac{1}{q}&\leq&\|f\|\\
\textrm{therefore
}\left(\sum_1^\infty|b_n|^q\right)^\frac{1}{q}\leq\|f\|
\end{eqnarray*}

Now let $f\in(\ell^p)^*$.

Define $b_n=f(\{0,0,\ldots,0,,1,0,0\ldots\})$ where the 1 is in
the $n$th place.

By linearity

$$|f(\{a_1a_2\ldots
a_n0,0,\ldots\})|=\left|\sum_1^Na_nb_n\right|\leq
\|f\|\left(\sum_1^N|a_n|^P\right)^\frac{1}{p}$$

Now choose $a_n=b_n^{q-1}\frac{\overline{b_n}}{|B_n|}$. Then

$$\left|\sum_1^Na_nb_n\right|=\sum|b_n|^q\leq\|f\|\left(\sum_1^N|b_n|^{qp-p}\right)^\frac{1}{p}$$

Therefore

$$\left(\sum_1^N|b_n|^q\right)^\frac{1}{q}\leq\|f\|$$

therefore letting $N\to\infty$ it follows that $\{b_n\in \ell^q$
and is that sequence from which $f$ arises.

$(\ell^P)^{**}=(\ell^q)^*=\ell^p$ - reflexive.

$(\ell^1)^*=\ell^\infty$.

\end{description}

\end{document}