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\begin{center}

FUNCTIONAL ANALYSIS

HAHN-BANACH THEOREM

\end{center}

\begin{description}

\item[]
If $M$ is a linear subspace of a normal linear space $X$ and if
$F$ is a bounded linear functional on $M$ then $F$ can be extended
to $M+[x_0]$ without changing its norm.

\item[Proof]
We first suppose that $X$ is a vector space over the real numbers.
We may suppose without loss of generality $\|F\|=1$. We may define
$F(x_0)=\alpha$ and $F(m+\lambda x_0)=F(m)+\lambda\alpha\ m\in M$.

We must have

$$|F(m)+\lambda\alpha|\leq\|m+\lambda x_0\|$$

$$\textrm{i.e. }|F(m)+\alpha|\leq\|m+x_0\|(m\textrm{ arbitrary
$\Rightarrow\frac{m}{\lambda}$ arbitrary}$$

If $m_1\ m_2\in M$

\begin{eqnarray*}
F(m_1)+\alpha&amp;\leq&amp;\|m_1+x_0\|\\ F(m_2)+\alpha&amp;\geq&amp;-\|m_2+x_0\|\\
-F(m_2)-\|m_2-x_0\|&amp;\leq&amp;\alpha\leq-F(m_1)+\|m_1+x_0\|-\ (I)
\end{eqnarray*}

\begin{eqnarray*}
F(m_1)-F(m_2)&amp;=&amp;F(m_1-m_2)\leq\|m_1-m_2\|\\
&amp;\leq&amp;\|m_1+x_0\|+\|m_2+x_0\|
\end{eqnarray*}

therefore $\exists\alpha$ satisfying (I).

Now if $X=X({\cal{C}})$

\begin{eqnarray*}
F(x)&amp;=&amp;G(g)+iH(x)\\ iF(x)&amp;=&amp;F(ix)=iG(x)-H(x)\textrm{ therefore
}H(x)=-G(ix)\\ &amp;=&amp;G(ix)+iH(x)\textrm{ therefore}\\
F(x)&amp;=&amp;G(x)-iG(ix)
\end{eqnarray*}

and $G$ can be extended by the first part.

By Zorn's lemma there will be a maximal subspace $N$ to which $M$
can be extended and $N=X$, applying the theorem.

We can embed $X$ in $X^{**}$ as follows:

Let $x\in X$ and define $f(\lambda x)=\lambda\|x\|$. Then
$\|f\|=1$.

$f$ can be extended to the whole space without changing its norm.

\begin{eqnarray*}
|\tilde{x}(f)|&amp;=&amp;|f(x)|=\|x\|\\ \textrm{therefore
}\|\tilde{x}\|\geq\|x\|
\end{eqnarray*}

Also $|\overline{x}(f)|=|f(x)|\leq\|f\|\|x\|$ therefore
$\|\overline{x}\|\leq\|x\|$.

\item[Adjoint of an operator]
Let $T$ be a continuous linear transformation from $X\to Y$.

The adjoint $T^*$ of $T$ is a linear transformation from $Y^*$ to
$X^*$ defined as follows:

Let $f\in Y^*$.

We define $T^*(f)\in X^*$ by $(T*f)x=f(TX)$

\begin{eqnarray*}
\|T^*(f)\|&amp;=&amp;\sup_{\|x\|=1}|fT(x)|\\
&amp;\leq&amp;\|f\|\sup_{\|x\|=1}\|Tx\|\\ &amp;\leq&amp;\|F\|\|T\|
\end{eqnarray*}

Therefore $T^*$ is continuous and $\|T^*\|\leq \|T\|$.

Now $T^{**}$ maps $X^{**}$ to $Y^{**}$.

If $X$ is regarded as a subspace of $X^{**}$ then $T^{**}$ is an
extension of $T$ therefore $\|T\|\leq\|T^{**}\|\leq\|T^*\|$
therefore $\|T^*\|=\|T\|$.

\item[Weak topology]
Let $X$ be a normed vector space and let $X^*$ be the dual of $X$.
We define a topology on $X$, called the weak topology, by taking
the sets

$$V(x)_{f_1\ldots f_n\varepsilon}=\{y\in
X:|f_i(x)-f_i(y)|&lt;\varepsilon\ i=1,\ldots,n\}$$

as a basis of neighbourhoods of the point $x$, where $f_1\ldots
f_n$ are any functionals in $X^*$ and $\varepsilon$ is any
positive number.

As all the $f$ are continuous this set will be open in the
original topology and so this topology is weaker than the original
one.

\item[Example]
Let $\xi=(x_n)\in\ell^2$

Let $f=(y_n)$

$$f(\xi)=\sum x_ny_n=(\xi,\eta).$$

$\xi_\alpha\to0$ in the weak topology $\Leftrightarrow
(|xi_\alpha,\eta)\to0$.

let $\varepsilon_1=(1,0,0,\ldots)\ \varepsilon_2=(0,1,0,\ldots)$
etc.

$\|\varepsilon_m-\varepsilon_n\|=\sqrt{2}\ m\neq n$. But for the
weak topology this sequence converges to zero as $\varepsilon_n\
\eta)=y_n\to0$ as $\sum|y_n|^2&lt;\infty$.

\item[Example]
Consider the space of all real valued functions defined on $[0\
1]$.

Consider the topology given by $f_\alpha\to f\Leftrightarrow
f_\alpha(x)\to f(x)$ for each $x$ in $[0\ 1]$.

Basic neighbourhoods:

Given $x_1\ldots x_n$ and $\varepsilon&gt;0$

$$N=\{g:|f(x_i)-g(x_i)|&lt;\varepsilon\ i=1\ldots n$$

Let $C$ be the subspace of continuous functions. Let
$d(x)=\left\{\begin{array}{ll}1&amp;x\textrm{ irrational}\\
0&amp;x\textrm{ rational}\end{array}\right.$

$d\in\overline{C}$ for this topology, for given $x_1\ldots x_n$ we
can find $f\in C$ such that

$$f(x_i)=d(x_i)\ i=1,\ldots,n$$

and so $f\in N(d)$.

But no sequence of continuous functions converges to $d$ in this
topology for, given $\{f_n\}\in C$ and $f_n(x)\to d(x)$ at every
$x$.

Let $H_n=\cap_{r\geq n}\{x:f_r(x)\geq\frac{1}{2}\}$. $H_n$ is
closed. $H_n$ contains no rational and so is nowhere dense,
therefore $\cup_{n=1}^\infty H_n$ is of the first category. But
$\cup_{n=1}^\infty H_n$= irrationals - of second category.

\item[Theorem]
Suppose $X$ is a Banach space, than the unit sphere of $X^*$ is
compact in the weak $*$ topology.

\item[Proof]
For each $x\in X$ define

$$C_x=\{z:|z|\leq\|x\|\}$$

$c_x$ is a compact set therefore $C=\prod_x C_x$ is compact.

$\prod_{x\in X}C_x=$ set of all functions mapping $X$ to the
complex plane with the property that $|F(x)|\leq\|x\|$. Hence the
unit sphere of $X^{*}$ can be regarded as a subspace and so will
be compact as it is closed.

\item[Theorem]
If $X$ is a Banach space, $X$ is reflexive $\Leftrightarrow$ its
unit sphere is weakly compact.

\item[Proof]
If $X$ is reflexive $X^{**}=X$ the weak topology of the unit
sphere of $X$ is the same as the weak $*$ topology which is
compact by th previous theorem.

\item[Closed Graph Theorem]

$T$ linear

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Let $G(T)=\{(X,T(X)\}\subset X\times Y$. If $T$ is continuous
$G(T)$ is closed. The theorem states that the converse is true.

\item[Lemma]
Let $T$ be a bounded linear transformation of a Banach space $X$
into a Banach sphere $Y$. If the image under $T$ of the unit
sphere $S_1=S(0,1)$ is dense in some sphere $U_r=S(0,r)$ about the
origin of $Y$ than it includes the whole of $U_r$.

\item[Proof]
Let $\delta&gt;0$. We wish to define a sequence $\{y_n\}$ such that

\begin{equation}
y_{n+1}-y_n\in \delta^nS,\ \|y_{n+1}-\overline{y}\|&lt;\delta^{n+1}r
\end{equation}

We define $y_0=0$. Suppose that $y_0\ldots y_n$ have been defined.

Since $\overline{y}\in
S(\overline{y},\delta^{n+1}r)\cap(y_n+\delta^nU_r)$ this is a
non-empty  open subset of $y_n+\delta^n U_r$ therefore there is an
element $y_{n+1}$ of $y_n+\delta^nT(S_1)$ which belongs to this
set, and $y_{n+1}$ satisfies the conditions (1).

$y_n\to\overline{y}$ as $n\to\infty$ provided $\delta&lt;1$.

$\exists x_n$ such that $x_n\in \delta^nS_1$ and
$T(x_n)=y_{n+1}-y_n$.

We can define $\overline{x}=\sum_1^\infty x_n$ provided
$\delta&lt;1$. Since $T$ is bounded

\begin{eqnarray*}
T(\overline{X})&amp;=&amp;\lim_{N\to\infty}\sum_1^NT(x_n)\\
&amp;=&amp;\lim_{N\to\infty} y_{N+1}=\overline{y}\\
\|\overline{x}\|&amp;\leq&amp;|sum\delta^n=\frac{1}{1-\delta}\textrm{ for
all}\delta\\ \textrm{therefore }\|\overline{x}\|&amp;\leq&amp;1\\
U_{r(1-\delta)}&amp;\subset&amp; T(S_1)\textrm{ for every }\delta\\
\textrm{and } U_r&amp;=&amp;\cup_\delta U_{r(1-\delta}\subset T(S_1)
\end{eqnarray*}

\item[Proof of Theorem]
Let $N(x)=\|x\|+\|T(x)\|$

If $\{x_n\}$ is a Cauchy sequence for the norm $N$ then it is a
Cauchy sequence for $\|X\|$ and also $\{T(x_n)\}$ is a Cauchy
sequence for $\|T_x\|$.

Therefore $x_n\to x$ and $Tx_n\to y$ as $n\to\infty$ as $G(T)$ is
closed $(x,y)\in G(T)$ therefore $y=T(x)$.

$$N(x-x_n)=\|x-x_n\|+\|Tx-Tx_n\|\to0\textrm{ as }n\to\infty$$

Therefore $X$ is a Banach space for the new norm $N$. Now the
identity mapping from $X$ with norm $N$ to $(X,\|\ \|)$ is bounded
since $\|X\|\leq N(x)$.

If $S_1$ denotes the unit sphere defined by $N$, it follows from
Baire's category theorem that $S_1$ is dense in some sphere $U_r$
about the origin defined by $\|\ \|$.

Thus by the lemma applied to the identity mapping $U_r\subset S_1$

i.e. if $\|X\|&lt;r\Rightarrow N(x)&lt;1$

i.e. $N(X)\leq\frac{1}{r}\|x\|$

so $\|T(x)\|\leq N(x)\leq\frac{1}{r}\|x\|$ and so $T$ is
continuous.

\item[Hilbert Space]
A pre-Hilbert Space is a real or complex vector spade in which an
inner product $(x,y)$ is defined having the following properties.

\begin{description}

\item[(i)]
$(x,x)&gt;0$ unless $x=0$

\item[(ii)]
$(x,y)=\overline{(y,x)}$

\item[(iii)]
$(x+y,z)=(x,z)+(y,z)$

\item[(iv)]
$\lambda x,y)=\lambda(x,y)$

\end{description}

A pre-Hilbert space can be normed by defining
$\|x\|=(x,x)^\frac{1}{2}$.

A Hilbert space is a pre-Hilbert space which is complete for this
norm.

A Banach space is a Hilbert space

$$\Leftrightarrow \|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$$

\item[Schwarz inequality]
$|(x,y)|\leq\|x\|\ \|Y\|$

\item[Proof]

\begin{eqnarray*}
(\lambda x-y,\lambda
x-y)&amp;=&amp;|\lambda|^2\|x\|^2-2R\lambda(x,y)+\|y\|^2\\
2R\lambda(x,y)&amp;\leq&amp;|\lambda|^2\|x\|^2+\|y\|^2
\end{eqnarray*}

Choose $\lambda$ so that $|\lambda|=\frac{\|Y\|}{\|X\|}$ and
arg$\lambda=-$arg$(x,y)$.

$$2\frac{\|Y\|}{\|X\|}|(x,y)|\leq2\|y\|^2$$

Hence the result.

\item[Minkowski Inequality]
$\|x+y\|\leq\|x\|+\|y\|$

\item[Proof]
\begin{eqnarray*}
|\|x+y\|^2|&amp;=&amp;|(x+y,x+y)|\\ &amp;=&amp;|\|X\|^2+2R(x,y)+\|y\|^2|\\
&amp;\leq&amp;\|x\|^2+\|Y\|^2+2|(x,y)|\leq(\|x\|+\|Y\|)^2
\end{eqnarray*}

using Schwarz.

\item[Theorem]
A closed convex subset $C$ of a Hilbert space contains a unique
element of smallest norm.

\item[Proof]
Let $d=\inf\{\|x\|:x\in C\}$.

Then $\exists\{x_n\}\subset C$ such that $\|x_n\|\to d$. Since $C$
is convex $\frac{x_n+x_m}{2}\in C$ therefore $\|x_n+x_m\|\geq2d$.

\begin{eqnarray*}
\|x_n-x_m\|^2&amp;=&amp;2\|x_n\|^2+2\|x_m\|^2-\|x_n+x_m\|^2\\
&amp;\leq&amp;2\{\|x_n\|^2-d^2\}+2\{\|x_m\|-d^2\}&lt;\varepsilon
\end{eqnarray*}

if $n$ and $m$ are sufficiently large.

Hence the sequence is a Cauchy sequence and has a limit point $x$
which belongs to $C$ as $C$ is closed, and $\|x\|=d$.

If $y\in C$ and $\|y\|=d$ then $\|x+y\|\geq 2d=\|x\|+\|y\|$ and so
$y=\lambda x$ where $\lambda&gt;0\Rightarrow
\|y\|=\lambda\|x\|\Rightarrow \lambda=1$ therefore $x=y$.

\item[Theorem]
Let $M$ be a closed subspace  of a Hilbert space ${\cal{H}}$. Then
any $x=x_1+x_2$ where $x_1\in M$ and $x_2$ perpendicular $M$ (i.e.
$(x_2,y)=0$ for all $y\in M$).

\item[Proof]
Suppose $x\in M$. Let $x_2$ be the element in the closed convex
set $x+M$ which is closest to 0.

Put $x_1=x-x_2\in M$.

If $y\in M$ then for any scalor $\lambda$

$$\|x_2+\lambda y\|^2\geq\|x_2\|^2$$

Since $2R\overline{\lambda}(x_2,y)+|\lambda|^2\|y\|^2\geq0$

Put $\lambda=-\frac{(x_2\ y)}{\|y\|^2}$.

Then $-\frac{|(x_2\ y)|^2}{\|y\|^2}\geq0$ therefore $(x_2\ y)=0$.

Suppose $x=x_1'+x_2'=x_1+x_2$ therefore $x_1-x_1'=x_2'-x_2=0$.

Hence uniqueness.

If $M$ is closed ${\cal{H}}=M+M^{\bot}$

If $M$ is closed and $x\in M^{\bot\bot}$

\begin{eqnarray*}
x&amp;=&amp;x_1+x_2\ \ x_1\in M\ x_2\in M^{\bot}\\ (x\ x_2)&amp;=&amp;(x_1\
x_2)+(x_2\ x_2)
\end{eqnarray*}

Therefore $(x_2\ x_2)=0$ therefore $x_2=0$ therefore $x\in M$.

\item[Theorem]
Suppose ${\cal{H}}$ is any Hilbert Space and let $f\in X^*$. Then
there is an element $y\in H$ such that $f(x)=(x,y)$ for every
$x\in{\cal{H}}$.

\item[Proof]
Let $M$= null space of $f$.

$\exists y_0\bot M$ such that if $x\in{\cal{H}}$

\begin{eqnarray*}
x&amp;=&amp;m+\lambda y_0\ m\in M\\ f(x)&amp;=&amp;\lambda f(y_0)\\ (x,\
y_0)&amp;=&amp;\lambda\|y_0\|^2\\ f(x)&amp;=&amp;\frac{(x,\
y_0)}{\|y_0\|^2}f(y_0)=\left(x,\frac{\overline{f(y_0)}}{\|y_0\|^2}y_0\right)
\end{eqnarray*}

Write $y=\frac{\overline{f(y_0)}}{\|y_0\|^2}y_0$.

If $M$ is any closed subspace and $x\in{\cal{H}}$

\begin{eqnarray*}
x&amp;=&amp;x_1+x_2\ x_1\in M\ x_2\in M^{\bot}\\ x_1&amp;=&amp;\textrm{Proj}_Mx
\end{eqnarray*}

If $T(x)=x_1\ T $ is a linear operator from ${\cal{H}}$ to itself,
and $\|T\|=1$.

\begin{eqnarray*}
TT'&amp;=&amp;T\\ (Tx,y)=(x_1\ y)=(x_1\ y_1)\\ (x,Ty)&amp;=&amp;(x\ y_1)=(x_1\
y_1) \end{eqnarray*}

Therefore $T$ is self-adjoint.

\item[Theorem]
If $M_1,\ldots,M_n$ are $n$ mutually perpendicular closed
subspaces of a Hilbert space ${\cal{H}}$ and if $x\in {\cal{H}}$
and $x_i,\ldots,x_n$ are the projections of $x$ on $M_1,\ldots
M_n$ respectively, then

$$\sum\|x_i\|^2leq\|x\|^2$$

\item[Proof]
Put $M=M_1++M_n\ x=x_1+\ldots+x_n+y,\ y\in M^{\bot}$, then
$\|x\|^2=\sum\|x_1\|^2+\|y\|^2$.

\item[Theorem]
Let $\{M_\alpha\}$ be a family, possibly uncountable, of pairwise
orthogonal closed subspaces of ${\cal{H}}$, and let $M$ be the
closure of their direct sum.

If $x_\alpha=\textrm{proj}_{M_\alpha}x\ x\in{\cal{H}}$ then
$x_\alpha=0$ except for a countable set of indices $\alpha_n$.

$\sum x_{\alpha_n}$ is convergent and its sum is the projection of
$x$ on $M$.

\item[Proof]
$$\sum_{i=1}^r\|x_{\beta_i}\|\leq\|x\|^2$$

Hence for any $n$ the number of indices satisfying
$\|x_\alpha\|\geq\frac{1}{n}$ is finite therefore the number of
indices satisfying $\|x_\alpha\|&gt;0$ is countable.

$\displaystyle\sum_1^N\|x_{\alpha_n}\|^2\leq\|x\|^2$ for each $N$
therefore $\displaystyle\sum_1^\infty\|x_{\alpha_n}\|^2&lt;+\infty$.

If $\ds y_n=\sum_1^Nx_{\alpha_n}$

$$\|y_n-y_m\|^2\leq\sum_{m+1}^n\|x_{\alpha_i}\|^2&lt;\varepsilon$$

if $m$ is sufficiently large. Therefore $\{y_n\}$ is a Cauchy
sequence which tends to a limit $\ds y=\sum_1^\infty x_{\alpha_n}$
in $M$, as $M$ is closed.

It remains to prove that $x-y\bot M$.

It is sufficient to prove that

$$w_{\beta_1}+w_{\beta_2}+\ldots+w_{\beta_r}\bot x-y$$

where $w_{\beta_i}\in M_{\beta_1}$ as the class of all such
vectors is everywhere dense in $M$.

If $\beta_1$ as an $\alpha_n$

\begin{eqnarray*}
(x-y,w_{\beta_1})&amp;=&amp;(x\ w_{\beta_1})-(x_{\beta_1}\ w_{\beta_1})\\
&amp;=&amp;(x_{\beta_1}\ w_{\beta_1})-(x_{\beta_1}\ w_{\beta_1})=0
\end{eqnarray*}

If $\beta_1$ is not an $\alpha_n$ then $w_{\beta_1}\bot x$ and
$\bot y$ and so to $x-y$.

\item[Orthonormal vectors]
A set $N$ of vectors in a Hilbert space ${\cal{H}}$ is said to be
orthonormal if $\|x\|=1$ for every $x$ in $N$, and $(x,y)=0$ for
all $y$ in $N\neq x$.

An orthonormal set $N$ of vectors is conplete if $N^{\bot}=\{0\}$.

Let $M_x$ be the 1-dimensional subspace generated by $x$ in $N$.

If $y\in{\cal{H}}$

$$\textrm{proj}_{M_x}y=\frac{(y\ x)}{\|x\|}.x=(y,x).x$$

as $\|x\|=1\ (y,x)=0$ except for a sequence $\{x_n\}|subset N$ and
for this sequence

\begin{eqnarray*}
y&amp;=&amp;\sum(y\ x_n)x_n\\ \|y\|^2&amp;=&amp;\sum|(y\ x_n)|^2
\end{eqnarray*}

This condition of completeness is equivalent to:

\begin{description}

\item[(i)]
for any $y$ in $\ds {\cal{H}}\ y=\sum_{x\in N}(y\ x)x$

\item[(ii)]
for any $y$ in $\ds {\cal{H}}\ \|y\|^2=z\sum_{x\in N}|(y\ x)|^2$.

\end{description}


\end{description}

\end{document}
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