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\section{Structures on Euclidean Space}

Euclidean space is the set of ordered $n$-tuples of real numbers,
\[ \R^n = \{(x_1,\dots,x_n)\ | \ x_i \in \R\} \]
In this section we shall examine various ``structures'' on $\R^n$:
\begin{itemize}
\item vector space
\item metric space
\item normed space (Banach space)
\item inner product space (Hilbert space)
\end{itemize}

\subsection{Vector Spaces}

\begin{definition}
A \emph{vector space} over a field $\F$ is a set $V$
equipped with two operations:
\begin{description}
\item[Vector Addition] $\forall\, v,w\in V$, $\exists\, v+w\in V$;
commutative, associative, identity element (0), and inverses exist
($v+(-v) = 0$).
\item[Scalar Multiplication] $\forall\, a\in\F$, $v\in V$, $\exists\,
a\cdot v\in V$; associative and distributive.
\end{description}
\end{definition}

\emph{Example:} Euclidean space becomes a vector space by defining
addition and scalar multiplication by
\begin{eqnarray*}
x + y &=& (x_1 + y_1, \dots, x_n + y_n) \\
   ax &=& (ax_1,\dots, ax_n)
\end{eqnarray*}
for $x = (x_1,\dots,x_n), y=(y_1,\dots,y_n)\in\R^n$ and $a\in\R$.

A standard \emph{basis} for $\R^n$ is $e_1=(1,0,\dots,0)$, $e_2 =
(0,1,0,\dots,0)$, \dots, $e_n = (0,\dots,0,1)$. Thus $x\in\R^n$ can
be written $x = x_1e_1 + \dots + x_ne_n$. The dimension of $\R^n$ is
$\dim\R^n = n$, but a vector space can have infinite dimension.

\emph{Example:} $V = C[a,b]$ with normal addition of functions  and
scalar multiplication is an infinite dimensional vector space over $\R$.

\subsection{Metric Spaces}

\begin{definition}
A \emph{metric space} is a set $M$ equipped with a distance function
$d:M\times M\to\R$ satisfying
\begin{enumerate}
\item $d(x,y) \ge 0$ with ``='' iff $x = y$ (\emph{positivity})
\item $d(x,y) = d(y,x)$ (\emph{symmetry})
\item $d(x,z) \le d(x,y) + d(y,z)$ (\emph{triangle inequality})
\end{enumerate}
$\forall\, x,y,z\in M$.
\end{definition}

\emph{Example:} $\R^n$ with Euclidean distance
\[ d(x,y) = \sqrt{(x_1-y_1)^2 + \cdots + (x_n-y_n)^2} \]
Properties 1 and 2 are obvious. The proof of property 3 will be a
consequence of a general theorem proved later.

\emph{Example:} $C[a,b]$ with distance
\[ d(f,g) = \int_a^b|f(x) - g(x)|\,dx \]
All three properties are easy to verify.

\subsection{Normed Spaces}

\begin{definition}
A \emph{normed space} (or \emph{Banach space}) is a vector space $V$ over
$\R$ equipped with a norm $\|\cdot\|:v\to\R$ satisfying:
\begin{enumerate}
\item $\|x\| \ge 0$ with ``='' iff $x = 0$ (\emph{positivity})
\item $\|a x\| = |a|\cdot\|x\|$ (\emph{homogeneity})
\item $\|x+y\| \le \|x\| + \|y\|$ (\emph{triangle inequality})
\end{enumerate}
$\forall\, x,y\in V$, $a\in\R$.
\end{definition}

\emph{Example:} $\R^n$ is a normed space with the Euclidean norm
\[ |x| = \sqrt{x_1^2 + \dots + x_n^2} \]
This norm is related to the Euclidean distance by $d(x,y) = |x - y|$.

A norm measures \emph{length}.

\emph{Remark:}
If $V$ is a normed space, then it has a metric structure induced
by the norm
\[ d(x,y) = \|x - y\| \]
\begin{itemize}
\item positivity: $d(x,y) = \|x-y\|\ge 0$
\item homogeneity $\Rightarrow$ symmetry:
\begin{eqnarray*}
d(x,y) &=& \|x-y\| = \|(-1)(y-x)\|\\
&=& |-1|\cdot\|y-x\|\\
&=& \|y-x\| = d(y,x)
\end{eqnarray*}
\item triangle inequality: 
\begin{eqnarray*}
d(x,z) &=& \|x-z\| = \|(x-y)+(y-z)\|\\
&\le&\|x-y\| + \|y-z\|\\
&=& d(x,y)+d(y,z)
\end{eqnarray*}
\end{itemize}

\subsubsection{Other norms on $\R^n$}
\begin{itemize}
\item $\|x\|_1 = \sum_{j=1}^n|x_i|$\quad  ($\|x-y\|_1$ is the ``taxi-cab
distance'' between $x$ and $y$)
\item $\|x\|_{p} = \Big(\sum_{j=1}^n |x_j|^p\Big)^{1/p}$, $1 \le p < \infty$
\item $\|x\|_{\sup} = \max_j \{|x_j|\}$
\end{itemize}

It is non-trivial to prove the triangle inequality for $\|x\|_p$ when $p \ne
1,2$. The proof for $p=2$ follows from the Cauchy-Schwartz Inequality
below. The sup-norm can be considered the case ``$p=\infty$'':
\[ \Big(\sum_{j=1}^n |x_j|^p\Big)^{1/p} =
|x_k| \Big(\frac{|x_1|^p}{|x_k|^p}+
\cdots+1+\cdots+\frac{|x_n|^p}{|x_k|^p}\Big)^{1/p} \]
If $|x_k| = \max_j |x_j| = \|x\|_{\sup}$, then $|x_j|^p/|x_k|^p \to 0$
or $1$, so
\[ |x_k| \Big(\frac{|x_1|^p}{|x_k|^p}+
\cdots+1+\cdots+\frac{|x_n|^p}{|x_k|^p}\Big)^{1/p} \to |x_k| (1+\cdots+1)^0 =
\|x\|_{\sup}\]
In fact, the sup-norm is sometimes denoted $\|x\|_\infty$.

\[ \resizebox{1in}{!}{\includegraphics{Pictures/06.fig1.eps}} 
\qquad\raisebox{0.5in}{\begin{minipage}[b]{2in}
\[\begin{array}{lcl}
\|x\|_\infty &=& 1\quad\mbox{square}\\
\|x\|_2 &=& 1\quad\mbox{circle}\\
\|x\|_1 &=& 1\quad\mbox{diamond}\\
\end{array}\]
\end{minipage}}
\]

\emph{Example:} $C[a,b]$ is a vector space over $\R$ ($\dim = \infty$).
It is also a normed space with either
\begin{itemize}
\item $\|f\|_{\sup} = \sup\{|f(x)|\ |\ x\in[a,b] \}$
\item $\|f\|_p = \big(\int_a^b |f(x)|^p\,dx\Big)^{1/p}$, $1 \le p <\infty$
\end{itemize}
The same remarks apply to $\|f\|_p$ as apply to $\|x\|_p$ above.

\subsection{Inner Product Spaces}

\begin{definition}
An \emph{inner product space} (or \emph{Hilbert space}) is a vector space
$V$ over $\R$ with a function $\<\cdot,\cdot\>:V\times V\to \R$ satisfying
\begin{enumerate}
\item $\<x,x\> \ge 0$ with ``='' iff $x=0$ (\emph{positive definite})
\item $\<x,y\> = \<y,x\>$ (\emph{symmetry})
\item $\<ax+by,z\> = a\<x,y\> + b\<y,z\>$ (\emph{bilinear})
\end{enumerate}
$\forall\, a,b\in\R$, $x,y,z\in V$.
\end{definition}

\emph{Example:} An inner product on $\R^n$ can be defined by
\[ \<x,y\> = x_1y_1 + \cdots x_ny_n \]
This inner product is often written $x\cdot y$ and called the \emph{dot
product}. It is related to the angle $\theta$ between $x$ and $y$ by the
formula
\[ x\cdot y = |x|\,|y|\,\cos(\theta) \]
We can demonstrate this for $n=2$ using the addition formulas for sine and
cosine. If $x = (r \cos(\alpha), r\sin(\alpha))$ and $y = (R \cos(\beta),
R\sin(\beta))$, where $r = |x|$ and $R = |y|$, then
\begin{eqnarray*}
x\cdot y &=& r R (\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta))\\
&=& r R \cos(\beta - \alpha)\\
&=& |x|\,|y|\,\cos(\theta)
\end{eqnarray*}
\[ \resizebox{!}{2in}{\includegraphics{Pictures/06.fig2.eps}} \]

In $\R^n$ it is clear that the dot product is related to the Euclidean
norm: $|x| = \sqrt{x\cdot x}$. This is a special case of a general
fact: an inner product $\<\cdot,\cdot\>$ always defines a norm by $\|x\|
= \sqrt{\<x,x\>}$. To prove this we need the following important theorem.

\begin{theorem}[Cauchy-Schwartz Inequality]\label{06:CSI}
If $V$ is a vector space over $\R$ with inner product $\<\cdot,\cdot\>$, then
\[ |\<x,y\>| \le \sqrt{\<x,x\>}\sqrt{\<y,y\>} \]
with ``='' iff $x$ and $y$ are collinear (i.e., $x = cy$ or $y = cx$
for some $c\in\R$).
\end{theorem}

\proof
$\<x+y,x+y\>\ge0$ and $\<x-y,x-y\>\ge0$ so using bilinearity,
\[ \begin{array}{rcrl}
\<x,x\> + 2\<x,y\> + \<y,y\> \ge 0 &\Rightarrow&
-\<x,y\> &\le\frac{1}{2}(\<x,x\>+\<y,y\>)\\
\<x,x\> - 2\<x,y\> + \<y,y\> \ge 0 &\Rightarrow&
\<x,y\> &\le\frac{1}{2}(\<x,x\>+\<y,y\>)\\
\end{array} \]
Therefore,
\begin{equation}\label{06:A}
|\<x,y\>| \le \frac{1}{2}(\<x,x\>+\<y,y\>)
\end{equation}

Now write
\[ \begin{array}{lll}
x = a\cdot u &\mbox{where\quad} a = \sqrt{\<x,x\>}, &u = a^{-1}\cdot x\\
y = b\cdot v &\mbox{where\quad} b = \sqrt{\<y,y\>}, &v = b^{-1}\cdot y
\end{array} \]
(We may assume $x\ne 0$ and $y\ne 0$ for otherwise the theorem is trivial.)

Note that $\<u,u\> = a^{-2}\<x,x\> = 1$ and $\<v,v\> = b^{-2}\<y,y\>
= 1$. Using bilinearity and applying (\ref{06:A}) to $u$ and $v$, we get
\begin{eqnarray*}
|\<x,y\>| &=& |\<au,bv\>| = ab|\<u,v\>|\\
&\le& ab\frac{1}{2}(\<u,u\>+\<v,v\>) =
ab\frac{1}{2}(1+1) = ab\\
&=&\sqrt{\<x,x\>}\sqrt{\<y,y\>}
\end{eqnarray*}

To prove the last assertion of the theorem, assume ``='' holds. Then
$|\<u,v\>| = 1$, where $u$ and $v$ are as above.  If $\<u,v\>=1$, then
$\<u-v,u-v\> = \<u,u\>-2\<u,v\> +\<v,v\> = 0$, while if $\<u,v\>=-1$,
then $\<u+v,u+v\> = \<u,u\>+2\<u,v\> +\<v,v\> = 0$. In either case we
get $u = \pm v$ and hence $x = \pm ab^{-1} y$.  \qed

\begin{corollary}
\[\Big|\sum_{i=1}^n x_iy_i\Big| \le
\Big(\sum_{i=1}^n x_i^2\Big)^{1/2} \Big(\sum_{i=1}^n y_i^2\Big)^{1/2} \]
\end{corollary}

\begin{corollary}
\[ \Big|\int_a^b f(x)g(x)\,dx \Big| \le
\Big(\int_a^b|f(x)|^2\,dx\Big)^{1/2}
\Big(\int_a^b|g(x)|^2\,dx\Big)^{1/2} \]
\end{corollary}

\proof $\<f,g\> = \int_a^b f(x)g(x)\,dx$ is an inner product on $C[a,b]$.
\qed

\begin{theorem}
If $\<\cdot,\cdot\>$ is an inner product on $V$, then for $x\in V$
\[ \|x\| =\sqrt{\<x,x\>} \]
is a norm.
\end{theorem}

\proof
Positivity: $\|x\| = \sqrt{\<x,x\>} \ge 0$

Homogeneity: $\|ax\| = \sqrt{\<ax,ax\>} = \sqrt{a^2\<x,x\>} = |a|\cdot\|x\|$

Triangle inequality:
\begin{eqnarray*}
\|x+y\|^2 &=& \<x+y,x+y\> = \<x,x\> + 2\<x,y\> + \<y,y\>\\
&=& \|x\|^2 + 2\<x,y\> + \|y\|^2\\
&\le& \|x\|^2 + 2\|x\|\,\|y\| + \|y\|^2 \quad\mbox{[Cauchy-Schwartz]}\\
&=& (\|x\|+\|y\|)^2
\end{eqnarray*}
Therefore, $\|x+y\| \le \|x\|+\|y\|$. \qed

The triangle inequality for the usual Euclidean distance also follows
this theorem.

\subsection{Complex Case}

Inner products and norms can be defined on vector spaces over $\C$, but
definitions must be modified.

\begin{definition}
A \emph{hermitian} inner product, $\<\cdot,\cdot\>:V\times V\to \C$,
satisfies
\begin{enumerate}
\item $\<x,x\>\in\R$ and $\<x,x\>\ge0$ with ``='' iff $x=0$.
\item $\<x,y\> = \overline{\<y,x\>}$
\item $\<ax+by,z\> = a\<x,z\>+b\<y,z\>$ and 
$\<z,ax+by,z\> = \overline{a}\<z,x\>+\overline{b}\<z,y\>$
\end{enumerate}
$\forall\, x,y,z\in V$, $a,b\in\C$.
\end{definition}

A \emph{norm} can be defined by $\|x\| = \sqrt{\<x,x\>}$ which satisfies
$\|ax\| = |a|\cdot\|x\|$ where $|a| = \sqrt{a\cdot\overline{a}}$. Moreover
the Cauchy-Schwartz inequality holds, $|\<x,y\>|\le\|x\|\cdot\|y\|$, which
implies the triangle inequality for the norm. Proofs are omitted.
 \end{document}
