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\begin{center}

FUNCTIONAL ANALYSIS

PRODUCT SPACES

\end{center}

\begin{description}

\item[]
Suppose we have $\{X_\alpha\}_{\alpha\in A}$.

We define a new space called the product space.

$$x\in\prod_{\alpha\in
A}X_\alpha\textrm{if}x=\{x_\alpha\}_{\alpha\in A}\ x_\alpha\in
X_\alpha$$

Alternatively we could define $\prod_{\alpha\in A}X_\alpha$ as the
set of all functions whose domain is $A$ and such that
$f(\alpha)\in X_\alpha$.

We can define a topology as follows.

Let $G=\{x\in X:x_{\alpha_i}\in U_{\alpha_i}\
i=1,2,\ldots,n\textrm{ for some }n\}$

$U_{\alpha_i}$ open in $X_{\alpha_i}$

We take all such sets $G$ as a basis for the topology in $X$.

A directed set $\{X_\beta\}$ is a set in which

\begin{description}

\item[(i)]
there is a partial ordering of the indices $\beta$,

\item[(ii)]
given $\beta_1\beta_2\exists\beta_3$ such that $\beta_3&gt;\beta_1,\
\beta_3&gt;\beta_2$.

\end{description}

A directed set $\{x_\beta\}$ is said to converge to a limit point
$x$ if, given any neighbourhood $U$ of $x\exists$ an index $\beta$
such that $x_\gamma\in U$ whenever $\gamma&gt;\beta$.

Convergence in a product space is co-ordinate-wise convergence
i.e.

$$\{x^\beta\}\to x\in X\Leftrightarrow x^\beta_\alpha\to
x_\alpha\in X_\alpha\textrm{ for every }\alpha$$

\item[Vector Spaces]
$E$ is called a vector space over the field $F$ if $E$ is a set
with two operations of addition and scalar multiplication, the
first mapping $E\times E$ to $E$ and the second mapping $F\times
E$ to $E$ in such a way that the following conditions hold.

\begin{description}

\item[(i)]
$E$ is an Abelian group under addition,

\item[(ii)]
$\alpha(x+y)=\alpha x+\alpha y$,

\item[(iii)]
$(\alpha+\beta)x=\alpha x+\beta x$,

\item[(vi)]
$\alpha(\beta x)=(\alpha\beta)x$,

\item[(v)]
$1.x=x$.

\end{description}

\item[Normed Vector Spaces]
Let $E$ be a vector space over the real or complex numbers. $E$ is
a normed space if every $x\in V$ is associated with a non-negative
real number $\|x\|$ which has the properties:

\begin{description}

\item[(i)]
$\|X\|=0\Leftrightarrow x=0$,

\item[(ii)]
$\|\lambda x\|=|\lambda|\|X\|$,

\item[(iii)]
$\|x+y\|=\|x\|+\|Y\|$.

\end{description}

We can always define a metric on a normed vector space by
$p(x,y)=\|x-y\|$ but the converse is not necessarily true.

\item[Example]
Consider $\ell^\infty$, the space of all bounded sequences
$\{x_i\}$. Define $p(x,y)=\sum\frac{|y_i-x_i|}{2^4}$. Then this is
a metric.

\item[]
in a topological space a set $B$ is said to be bounded if, given
any neighbourhood $U$ of the origin, for some $n,\ nU\supset B$.

In a normed vector space there are always bounded neighbourhood of
the origin e'g' $B_1=\{x:\|x\|\leq1\}$

In $\ell^\infty$ a basic neighbourhood $U$ is defined as follows.
Let $\varepsilon&gt;0$. Let $n_1n_2\ldots n_k$ be a finite sequence
of integers let $U=\{\{x_n\}:|x_{n_i}|&lt;\varepsilon\
i=1,2,\ldots,k\}$.

Now choose $m\not\in\{n_1n_2\ldots n_k\}$ and let
$V=\{\{x_n\}:|x_m|&lt;1\}$.

Then $NV\supset U$ for any $N$ i.e. no neighbourhood of the origin
is bounded.

\item[A Banach Space]
is a normed vector space which is complete for the metric defined
by the norm.

The classical examples of Banach spaces are

$\ell^P$- the space of all sequences $\{x_n\}$ such that
$\sum_{n=1}^\infty|x_n|^p&lt;\infty$.

$L^P(0,1)$- the space of all measurable functions $f(x)$ such that
$\int_0^1|f(x)|^p\,dx&lt;\infty$.

These are examples of more general spaces $L^P(X,\mu)$.

The norms usually defined on them are

$\ds\ell^P:\|x\|=\left(\sum|x^n|^p\right)^\frac{1}{p}$

$\ds L^P:\|f\|=\left(\int_0^1|f(x)|^P\,dx\right)^\frac{1}{p}$
where $f=g$ means $f\equiv g$ p.p.

\item[Lemma]
If $\alpha$ and $\beta$ are non-negative real numbers and if
$0&lt;\lambda&lt;1$ then
$\alpha^\lambda\beta^{1-\lambda}\leq\lambda\alpha+(1-\lambda)\beta$
with equality $\Leftrightarrow\alpha=\beta$.

\item[Proof]
Let $\phi(t)=(1-\lambda)+\lambda t-t^\lambda$ then
$\phi'(t)=\lambda(1-t^{\lambda-1})$ therefore $\phi'(t)&lt;0$ if
$t&lt;1$, $=0$ if $t=1$ and $&gt;0$ if $t&gt;1$. Therefore
$\phi(1)\leq\phi(t)$ for all $t&gt;0$ with equality $\Leftrightarrow
t=1$

$(1-\lambda)+\lambda t=t^\lambda\geq0$ with equality $
\Leftrightarrow t=1$.

$(1-\lambda)+\lambda\frac{\alpha}{\beta}-
\frac{\alpha^\lambda}{\beta^\lambda}\geq0$ with equality
$\Leftrightarrow\alpha=\beta$

i.e. $\alpha^\lambda\beta^{1-\lambda}\leq
\lambda\alpha+(1-\lambda)\beta$.

\item[Holder's Inequality]
let $p&gt;2$ and let $\frac{1}{p}+\frac{1}{q}=1$

\begin{description}

\item[(i)]
for any $\{x_n\}\in\ell^p$ and $\{y_n\}\in\ell^q$

$$\sum|x_ny_n|\leq\left(\sum|x_n|^p\right)^
\frac{1}{p}\left(\sum|y_n|^q\right)^\frac{1}{q}$$

\item[(ii)]
for any $f\in L^P\ g\in L^q$

$$\int|fg|\,dx\leq\left(\int|f|^p\,dx\right)^\frac{1}{p}
\left(\int|g|^q\,dx\right)^\frac{1}{q}$$

\end{description}

\item[Proof]

\begin{description}

\item[(i)]
First suppose that $\sum|x_n|^p=1\ \ \sum|y_n|^q=1$. By the Lemma,
taking $\lambda=\frac{1}{p},\ 1-\lambda=\frac{1}{q}$,

$$|X_ny|n|\leq\lambda|x_n|^P+(1-\lambda)|y_n|^q$$

therefore $\sum|x_ny_n|\leq1$.

Now consider the sequences

$$\xi_n=\frac{x_n}{\left(\sum|x_n|^p\right)^\frac{1}{p}}\ \
\eta_n=\frac{y_n}{\left(\sum|y_n|^q\right)^\frac{1}{q}}$$

Then $\sum|\xi_n|^p=1\ \ \sum|\eta_n|^q=1$ therefore
$\sum|\xi_n\eta_n|\leq1$. Hence the result.

\item[(ii)]
Proved in a similar way.

\end{description}

\item[Minkowski's Inequality]

\begin{description}

\item[(i)]
Let $p&gt;1$. If $\{x_n\}\in\ell^p\{y_n\}\in\ell^p$ then
$\{x_n+y_n\}\in\ell^P$ and

$$\left(\sum|x_n+y_n|^P\right)^\frac{1}{p}\leq
\left(\sum|x_n|^p\right)^\frac{1}{p}+
\left(\sum|y_n|^p\right)^\frac{1}{p}$$

\item[(ii)]
If $f,g\in L^P$ then $f+g\in L^p$ and

$$\left(\int|f+g|^p\,dx\right)^\frac{1}{p}\leq
\left(\int|f|^p\,dx\right)^\frac{1}{p}+
\left(\int|g|^p\,dx\right)^\frac{1}{p}$$

\end{description}

\item[Proof]

\begin{description}

\item[(i)]
For any $N$

\begin{eqnarray*}
\sum_1^N|x_n+y_n|^p&amp;\leq&amp;\sum_1^N|x_n+y_n|^{p-1}
|x_n|+\sum_1^N|x_n+y_n|^{p-1}|y_n|\\
&amp;\leq&amp;\left(\sum_1^N|x_n|^p\right)^\frac{1}{p}
\left(\sum_1^N|x_n+y)n|^{q(p-1)}\right)^\frac{1}{q}\\
&amp;&amp;+\left(\sum_1^N|x_n|^p\right)^\frac{1}{p}
\left(\sum_1^N|x_n+y_n|^{q(p-1)}\right)^\frac{1}{q}\\
&amp;=&amp;\left(\sum|x_n+y_n|^p\right)^\frac{1}{q}
\left[\left(\sum_1^N|x_n|^p\right)^\frac{1}{p}+
\left(\sum_1^N|y_n|^p\right)^\frac{1}{p}\right]\\
\left(\sum_1^N|x_n+y_n|^P\right)^\frac{1}{p}&amp;\leq&amp;
\left(\sum_1^N|x_n|^p\right)^\frac{1}{p}+
\left(\sum_1^N|y_n|^p\right)^\frac{1}{p}\\
&amp;\leq&amp;\left(\sum_1^\infty|x_n|^p\right)^\frac{1}{p}+
\left(\sum_1^\infty|y_n|^p\right)^\frac{1}{p}
\end{eqnarray*}

Hence the result.

\item[(ii)]
Similar proof.

Using the above results it is easy to verify that $\ell^p$ and
$L^p$ are normed vector spaces.

\begin{eqnarray*}
\ell^1&amp;:&amp;\{\{x_n\}:\sum|x_n|&lt;\infty\}\\ &amp;&amp;\|\{x_n\}\|=\sum|x_n|\\
L^1(0\ 1)&amp;:&amp;\{f(x):\int_0^1|f(x)|\,dx&lt;\infty\}\\
&amp;&amp;\|f\|=\int_0^1|f(x)|\,dx\\
\ell^\infty&amp;:&amp;\{\{x_n\}:\{x_n\}\textrm{bounded}\}\\
&amp;&amp;\|\{x_n\}\|=\sup\{|x_n|\}\\ L^\infty(0\ 1)&amp;:&amp;\{f(x)\ |f(x)|&lt;M\
p.p\}\\ &amp;&amp;\|f\|=\sup \{M:|f(x)|&lt;M\ p.p.\}
\end{eqnarray*}

\end{description}

\item[Lemma]
The normed vector space $E$ is complete
$\Leftrightarrow\sum_1^\infty\xi_n$ exists in $E$ whenever
$\{\xi_n\}$ is a sequence of vectors in $E$ such that
$\sum\|\xi_n\|&lt;\infty$.

\item[Proof]

\begin{description}

\item[(i)]
Let $\eta_n=\sum_{r=1}^n\xi_r$

$$\|\eta_n-\eta_m\|=\left|\left| \sum_{r=m+1}^n\xi_r\right|\right|
\leq\sum_{r=m+1}^n\|\xi_r\|&lt;\varepsilon$$

If $m$ is sufficiently large. Therefore $\{\eta_n\}$ is a cauchy
sequence therefore $\sum_1^\infty\xi_n$ exists.

\item[(ii)]
Let $\{\zeta_n\}$ be a cauchy sequence in $E$. We can find
$\{n_r\}$ such that
$\|\zeta_{n_{r+1}}-\zeta_{n_r}\|\leq\frac{1}{2}r$

Write $\xi_1=\zeta_{n_1}\ \xi_r=\zeta_{n_{r+1}}-\zeta_{n_r}$

$\sum\|\xi_r\|&lt;\infty$ therefore $\sum_{r=1}^n\xi_r\to\zeta$ as
$n\to\infty$ i.e. $\zeta_{n_r}\to\zeta$ as $r\to\infty$ therefore
$\zeta_n\to\zeta$ as $n\to\infty$.

\end{description}

\item[Theorem (Riesz-Fischer)]
The $L^p$ spaces are complete.

\item[Proof]
Let $\{f_n\}$ be a sequence in $L^p$ such that
$\sum_1^\infty\|F_n\|=m&lt;\infty$.

Put $g_n(x)=\sum_1^n|f_r(x)|$.

At every point $x$ the increasing sequence $g_n(x)$ has a limit
(finite or infinite). Denote this limit by $g(x)$ then $g(x)$ is
measurable and

$$\int|g(x)|^p\,dx=\lim_{n\to\infty}\int|g_n(x)|^p\,dx\leq m^p$$

therefore $g(x)\int L^P$ and $g(x)&lt;\infty$ p.p. therefore
$\sum_1^\infty f_n(x)$ is absolutely convergent p.p. to some
function $f(x)$.

$$\left|f(x)-\sum_1^n f_r(x)\right|^p\leq\left(\sum_{n+1}^\infty
\left|f_r(x)\right)\right)^p\leq\left(g(x)\right)^p$$

Since $(g(x))\in L^p$

$$\lim\int\left(\sum_{n+1}^\infty|f_r(x)|\right)
^p\,dx=\int\lim\sum_{n+1}^\infty|f_r(x)|^p\,dx=0$$

by Lebesgues's Dominated Convergence theorem therefore

$\int\left|f(x)-\sum_1^nf_r(x)\right|^p\to0$ as $n\to\infty$
therefore

$\left|\left|f(x)-\sum_1^nf_r(x)\right|\right|\to0$ as
$n\to\infty$.

Hence by the Lemma $L^P$ is complete.

\item[Other examples of Banach Spaces]

\begin{description}

\item[(i)]
The set $C(X)$ of all real-valued, or complex valued, functions
defined and continuous on a complex space $X$ where
$\|F\|=\sup\{|f(x)|:x\in X\}$.

Convergence in this norm is uniform.

\item[(ii)]
Set of all functions $f(z)$ analytic on the unit disc with

$$\|F\|=\sup\{|f(z)|:|z|\leq1\}$$

\item[(iii)]
Set of all functions $f(z)$ analytic on the unit disc with

$$\|F\|=\int\!\!\!\int_{|z|\leq1}|f|\,dxdy$$

\item[(iv)]
The set of all functions $f(z)$ harmonic on the unit circle with

$$\|f\|=\sup\{|f(z)|:|z|\leq1\}$$

[$f(z)$ is harmonic if $\frac{1}{\ell}\int_Cf(z)\,dz=f(z_0)\ \
C=\{z:|z-z_0|=\frac{\ell}{2\pi}\}$].

\end{description}

\end{description}

\end{document}
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