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\section{Continuous Functions on Metric Spaces}

\subsection{The Definition of Continuity}
Up to this point we have studied functions $f:D \to E$ where $D$ and $E$
are subsets of $\R$ or $\R^n$.  We now extend our discussion to functions
$f:M \to N$ where $M$ and $N$ are metric spaces.  Many concepts and
proofs that apply to Euclidean spaces can be easily transferred to general
metric spaces.

\begin{definition}
A function $f:M \to N$ is
\emph{continuous} if $f^{-1}(A)$ is open for all open sets $A\subset N$.
\end{definition}

This definition is compatible with our previous definition of continuous
functions on Euclidean spaces, see Theorem~\ref{01:CONTINUOUS}.

\begin{theorem}
The following are equivalent.
\begin{enumerate}
\item $f:M \to N$ is continuous.
\item $\forall\, x_0\in M$, and $\forall\, 1/m$, $\exists\, 1/n$ such that
\[ d_M(x,x_0) < \frac{1}{n} \Rightarrow d_N(f(x),f(x_0)) < \frac{1}{m} \]
\item $\forall\, \{x_j\}\subset M$, if $x_j\to x_0$ in $M$, then $f(x_j)\to
f(x_0)$ in $N$.
\end{enumerate}
\end{theorem}

\proof
1 $\Rightarrow$ 2:  Let $x_0\in M$ and let $1/m$ be given. By assumption,
the inverse image of an open ball is open, so 
\begin{eqnarray*}
f^{-1}(B_n(f(x_0),\frac{1}{m}))
&=& \{ x\in M\ |\ f(x) \in B_N(f(x_0),\frac{1}{m}) \}\\
&=& \{ x\in M\ |\ d_N(f(x),f(x_0)) < \frac{1}{m} \}
\end{eqnarray*}
is \emph{open} and contains~$x_0$. Thus, $\exists\, 1/n$ such that
\[ B_M(x_0,1/n) \subset f^{-1}(B_N(f(x_0),1/m)) \]
i.e., if $d_M(x,x_0) < 1/n$ then $d_N(f(x),f(x_0)) < 1/m$.

2 $\Rightarrow$ 3: Let $x_j\to x_0$ in $M$, and let $1/m$ be given.
By assumption, $\exists\, 1/n$ such that $D_n(f(x_j),f(x_0)) < 1/m$
whenever $d_M(x_j,x_0) < 1/n$. However, since $x_j\to x_0$, $\exists\, J$
such that $d_M(x_j,x_0) < 1/n$ whenever $j \ge J$. So, if $j \ge J$,
then $d_N(f(x_j),f(x_0)) < 1/m$ showing that $f(x_j)\to f(x_0)$.

3 $\Rightarrow$ 1: Let $A \subset N$ be open and suppose  $f^{-1}(A)$
is \emph{not} open. Then $\exists\, x_0 \in f^{-1}(A)$ such that no open
ball about $x_0$ is contained in $f^{-1}(A)$. Therefore, $\forall\, 1/j$,
$\exists\, x_j \in B_M(x_0,1/j)$ such that $x_j \notin f^{-1}(A)$. Clearly,
$x_j \to x_0$, so by assumption, $f(x_j) \to f(x_0)$. Since $A$ is open,
$\exists\, 1/m$ such that $B(f(x_0),1/m)\subset A$. But since $f(x_j)\to
f(x_0)$, $\exists\, J$ such that $f(x_j)\in B(f(x_0),1/m)\subset A$,
$\forall\, j \ge J$, and this implies $x_j \in f^{-1}(A)$, a contradiction.
\qed

\subsubsection{Examples}
\

1) Let $M$ be $C[a,b]$ with the sup-norm. Then $I:M\to \R$
defined by $I(f) = \int_a^b f(x)dx$ is continuous. To prove this, let
$f_n\to f$ with respect to the sup-norm. As we have seen this means that
$f_n\to f$ uniformly. Therefore, $f$ is integrable and $I(f_n)\to I(f)$
by Theorem~\ref{02:INTEGRALS}, showing that $I$ is continuous.

2) Let $M = C^1[a,b]$ with norm $\|f\|_{*} = \|f\|_\infty + \|f'\|_\infty$
and let $N = C[a,b]$ with the sup-norm. Then $D:M\to N$ defined by $D(f) =
f'$ is continuous. To see this, note that if $f_n$ converges to $f$
with respect to the $*$-norm, then $f_n\to f$ and $f_n'\to f'$ uniformly,
see Theorems~\ref{02:UNIFORMLIMITS} and \ref{02:DERIVATIVES}. Therefore,
$D(f_n)\to D(f)$ with respect to the sup-norm, proving that $D$ is
continuous.

Recall that the uniform convergence of functions does not necessarily
imply the convergence of their derivatives. Hence, $D$ would not be
continuous with respect to the sup-norm on $C^1[a,b]$. Continuity depends
on the distance function under consideration.

3) Let $M$ be a metric space and fix $x_0\in M$. Then $f:M\to \R$
defined by $f(x) = d(x,x_0)$ is continuous: If $x_n\to x$ in $M$,
then, by definition, $\forall\, 1/m$, $\exists\, N$ such that $d(x_n,x) <
1/m$, $\forall\, n\ge N$. Since $d(x_n,x_0) \le d(x_n,x) + d(x,x_0)$
we find that
\[ |f(x_n) - f(x)| = |d((x_n,x_0) - d(x,x_0)| \le d(x_n,x) <
\frac{1}{m},\quad\forall n \ge N \]
so $f(x_n)\to f(x_0)$. Therefore, $f$ is continuous.

\emph{Remark:} The usual ways of combining continuous functions lead to
new continuous functions. For example, if $f:M\to N$ and $g:N\to P$
are continuous, then it is easy to prove that the composition, $f\circ
g:M\to P$, $f\circ g(x) = f(g(x))$, is continuous. If $f,g:M\to \R^n$ are
continuous, then $f+g:M\to \R^N$, $(f+g)(x) = f(x)+g(x)$, is continuous.
If $f:M\to\R^n$ and $g:M\to \R$ are continuous, then $f\cdot g:M\to \R^n$,
$(f\cdot g)(x) = f(x)g(x)$, and $f/g:M\to \R^n$, $(f/g)(x) = f(x)/g(x)$
(for $x$ such that $g(x)\ne 0$), are continuous.  If $f_1,\dots,f_n:M\to
\R$ are continuous, then $f:M\to \R^n$, $f(x) = (f_1(x),\dots,f_n(x))$ is
continuous and conversely.

\subsection{Continuous Functions on Compact Domains}

Let $M$ and $N$ be metric spaces with $M$ compact. In this section
we establish theorems for continuous functions $f:M \to N$ that are
analogous to familiar theorems about continuous real-valued functions
on compact intervals.

\begin{definition}
$f:M\to N$ is \emph{uniformly continuous} if $\forall\, 1/m$, $\exists\, 1/n$
such that $\forall\, x,y\in M$
\[ d_M(x,y) <\frac{1}{n} \Rightarrow d_N(f(x),f(y)) < \frac{1}{m} \]
\end{definition}

\begin{theorem}
If $M$ is compact and $f:M\to N$ is continuous, then $f$ is uniformly
continuous.
\end{theorem}

\proof Given $1/m$ and $x\in M$, $\exists\, 1/n_x$ (depending on $1/m$
and $x$) such that
\begin{equation}\label{08:1}
d_M(x,y)<\frac{2}{n_x} \Rightarrow d_N(f(x),f(y))<\frac{1}{2m}
\end{equation}
by the continuity of $f$. The open balls $B_M(x,1/n_x)$, for $x\in
M$, cover $M$, and since $M$ is compact, a finite number will cover
$M$:
\[ M = B_M(x_1,1/n_1)\cup\cdots B_M(x_t,1/n_t) \]
Let $n = \max\{n_1,\dots,n_t\}$.

For any $x,y\in M$, $x$ must lie in one of the open balls,
$x\in B_M(x_i,1/n_i)$ for some~$i$. If $d_M(x,y)<1/n$ then
\[ d_M(x_i,y) \le d_M(x_i,x) + d_M(x,y) < \frac{1}{n_i} + \frac{1}{n} \le
\frac{2}{n_i} \]
so
\[ d_N(f(x),f(y)) \le d_N(f(x),f(x_i)) + d_N(f(x_i),f(y)) < \frac{1}{2m} +
\frac{1}{2m} = \frac{1}{m} \]
by~(\ref{08:1}). \qed

\begin{theorem}
If $M$ is compact and $f:M\to \R$ is continuous, then
$f$ has a maximum and minimum in $M$, that is,
there exists points $u,v\in M$ such that
\[ f(u) = \sup_{x\in M} f(x) \mbox{\quad and\quad } f(v) = \inf_{x\in M} f(x) \]
\end{theorem}

\proof There is a sequence $\{x_n\}\subset M$ such that $\lim f(x_n)=\sup
f(x)$ (or to $\infty$ if $\sup f(x) = \infty$). Since $M$ is compact,
there is a convergent subsequence, $x_{n'}\to b$. Then $\sup f(x) =
\lim f(x_{n'}) = f(b)$ by the continuity of $f$, showing
that the supremum is finite and achieved at a point $b\in M$. The proof
for the infimum is similar. \qed

\begin{theorem}
Suppose $f:M\to N$ is continuous. If $A\subset M$ is compact, then
$f(A)\subset N$ is compact.
\end{theorem}

\proof The argument is word-for-word the same as given in
Theorem~\ref{01:COMPACT}: Let $f(A) \subset \bigcup U_\alpha$ be an
open cover of $f(A)$. Then $\bigcup f^{-1}(U_\alpha)$ is an open cover
of $A$.  Since $A$ is compact, there exists a finite subcover, $A \subset
f^{-1}(U_{\alpha_1})\cup\dots\cup f^{-1}(U_{\alpha_n})$. But then $f(A)
\subset U_{\alpha_1}\cup\dots\cup U_{\alpha_n}$ is a finite subcover
of~$f(A)$ proving that $f(A)$ is compact. \qed

\subsection{Connectedness}

Recall the Intermediate Value Theorem:
\begin{quote}
\emph{If $f:[a,b]\to\R$ is continuous, then $f([a,b]) = [c,d]$
where $c = \min f(x)$ and $d =\max f(x)$.}
\end{quote}
The point of this theorem is that if there are no gaps in the domain
of a continuous function, then there are no gaps in the values of the
function.  To generalize this theorem to metric spaces we need a way to
specify that there are ``no gaps.'' There are two ways to define such
a concept: being ``connected,'' and  being ``arc-wise connected.'' The
first is the more general and is defined in terms of open sets. The
second concept is more restrictive, but is often more practical to apply.

\begin{definition}
Let $M$ be a metric space.
\begin{enumerate}
\item $M$ is \emph{connected} if there does not exist a pair of
disjoint, non-empty, open sets $A$ and $B$ with $M = A \cup B$.
\item $M$ is \emph{arc-wise connected} is there is a curve connecting
any two points in $M$. More precisely, given any two points $x,y\in M$,
there exists a continuous function $f:[a,b]\to M$ such that $f(a) = x$
and $f(b) = y$.
\end{enumerate}
\end{definition}

\begin{proposition}\label{08:INTERVAL}
Any interval of real numbers is connected.
\end{proposition}

\proof Suppose $I$ is an interval and $I = A\cup B$ with $A$ and $B$
disjoint, non-empty, open subsets of $I$.  Fix $a\in A$ and $b\in B$. We
may assume $a < b$. To produce a contradiction, we look for a dividing
point between $A$ and $B$.  Let
\[ d = \sup \{x\in A\ |\ x \le b\} \]
Let $\{x_n\}\subset A$ such that $x_n\to d$. Since $A$ is the complement
of the open set $B$ in $I$, $A$ is closed in $I$. Therefore, $d\in A$
unless $d\notin I$. The latter is only possible if $d$ is an endpoint
of the interval~$I$, which is not the case here since $a \le d \le
b$. Therefore, $d\in A$. But $A$ is also open, so there exists an open
interval around $d$ contained in $A$, which contradicts $d$ being the
supremum of points in $A$ that are $\le b$. \qed

\begin{theorem}
Let $M$ be a metric space.
\begin{enumerate}
\item $M$ is arc-wise connected $\Rightarrow$ $M$ is connected.
\item $M$ is connected $\iff$ the only subsets of $M$ that are both
open and closed are $\emptyset$ and $M$.
\end{enumerate}
\end{theorem}

\proof
1) Suppose $M$ is arc-wise connected but not connected. Then $M = A\cup
B$ where $A$ and $B$ are disjoint, non-empty and open. Let $x\in A$ and
$y\in B$ and let $f:[a.b]\to M$ be a continuous function such that $f(a)
= x$ and $f(b) = y$.  Then
\[ [a,b] = f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B) \]
decomposes $[a,b]$ into disjoint, non-empty, open sets, contradicting
Proposition~\ref{08:INTERVAL}. Therefore, $M$ must be connected.

2) ($\Rightarrow$) Suppose $M$ is connected and $A\subset M$
is both open and closed. Then $B = A^c$ is open and closed by
Theorem~\ref{01:COMPLEMENTS}. Both $A$ and $B$ cannot be non-empty,
since otherwise $M = A\cup B$ with $A$ and $B$ disjoint, non-empty, open
subsets. Therefore, either $A = \emptyset$ and $B = M$, or $B=\emptyset$
and $A = M$.

($\Leftarrow$) Conversely, suppose the only subsets of $M$ that are
both open and closed are $\emptyset$ and~$M$.  If $M = A\cup B$ with $A$
and $B$ disjoint and open in $M$, then $A = B^c$ and $B = A^c$ are
also be closed by Theorem~\ref{01:COMPLEMENTS}. By assumption, either
$A = \emptyset$ and $B = M$, or $B = \emptyset$ and $A = M$, proving
that $M$ cannot be decomposed into a disjoint union of \emph{non-empty}
open sets. Therefore, $M$ is connected.  \qed

\emph{Example:} It is possible for a set to be connected but not arc-wise
connected.  The graph of the function
\[ f(x) = \left\{\begin{array}{ll}
\sin(1/x) & x>0\\
0         & x = 0
\end{array}\right. \]
is a connected subset of $\R^2$: Any decomposition of the graph, $G =
\{(x,f(x)\ | \ x \ge 0\}$, into disjoint, non-empty, open subsets would
give a similar decomposition of the part of the graph, $G_0$, for $x>0$
which is impossible since $G_0$ is clearly arc-wise connected. However,
there is no curve from $(0,0)$ to any other point on the graph. Suppose
$f:[0,1]\to G$ is continuous with $f(0) = (0,0)$ and $f(1)=(a,\sin(1/a))$
for some $a>0$.  Then $f(t) = (x(t), \sin(x(t))) \to (0,0)$ as
$t\to 0$. This implies $x(t)\to 0$ and $\sin(1/x(t))\to 0$, which
is impossible since $\lim_{u\to 0} \sin(1/u)$ does not exist.
\[ \resizebox{!}{1in}{\includegraphics{Pictures/08.fig1.eps}} \]

\begin{theorem}
Let $M$ and $N$ be metric spaces and suppose $f:M\to N$ is continuous. If
$M$ is connected then $f(M)$ is connected. If $M$ is arc-wise connected,
then $f(M)$ is arc-wise connected.
\end{theorem}

\proof Suppose $M$ is connected. If $f(M) = A\cup B$ with $A$ and $B$
disjoint, non-empty, open subsets, then
\[ M = f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B) \]
decomposes $M$ into disjoint, non-empty, open subsets, a contradiction.
Therefore, no such decomposition of $f(M)$ exists and $f(M)$ is connected.

Next assume $M$ is arc-wise connected. Let $x,y\in f(M)$. Then there are
$u,v\in M$ such that $x = f(u)$ and $y = f(v)$. Let $g:[a,b]\to M$ be
a continuous function such that $g(a) = u$ and $g(b) = v$. Then $f\circ
g:[a,b]\to f(M)$ is continuous with $f\circ g(a) = x$ and $f\circ g(b)
= y$, proving that $f(M)$ is arc-wise connected. \qed

\goodbreak
\subsection{The Contractive Mapping Principle}

In this section we consider a continuous function of a complete metric
space $M$ to itself, $f:M\to M$, that ``shrinks'' distance.  We shall
prove that any such function must have a unique \emph{fixed point}, that
is, a point $x_0\in M$ such that $f(x_0) = x_0$. This theorem has
many applications in analysis. In fact, we shall use it to prove two
fundamentally important theorems:
\begin{itemize}
\item The existence of solutions to ordinary differential equations.
\item The Implicit Function Theorem.
\end{itemize}

\begin{definition}
Let $M$ and $N$ be metric spaces.

\begin{enumerate}
\item A \emph{mapping} of $M$ to $N$ is a continuous function
$f:M\to N$.
\item A mapping $f:M\to M$ is \emph{contractive}
if $\exists\, r$, $0 \le r < 1$, such that
\[ d(f(x),f(y)) \le r\, d(x,y),\quad \forall\, x,y\in M \]
\end{enumerate}
\end{definition}

We introduce the notation $f^n$ for the iterated mapping $f\circ
f\circ\cdots\circ f$ ($n$ times).

\begin{theorem}[Contractive Mapping Principle]\label{08:CMP}
Let $M$ be a complete metric space and assume $f:M\to M$ is a contractive
mapping. Then there exists a unique fixed point $x_0\in M$, $f(x_0) = x_0$.
Moreover, $\forall\, x\in M$, $\exists\, c>0$ such that
\[ d(f^n(x), x_0) \le c r^n \]
In particular, $\lim_{n\to\infty}f^n(x) = x_0$.
\end{theorem}

\proof
Let $x\in M$. We first show the sequence $\{f^n(x)\}$ is Cauchy.
The definition of a contractive mapping implies 
\[ d(f^{n+1}(x),f^n(x)) \le r\, d(f^n(x),f^{n-1}(x)) \le \cdots
\le r^n d(f(x),x) \]
For any $j >n$, the above inequality, along with the triangle
inequality, implies
\begin{eqnarray*}
d(f^j(x),f^n(x)) &\le& d(f^j(x),f^{j-1}(x)) + \cdots + d(f^{n+1}(x),f^n(x))\\
&\le& (r^{j-1} + \cdots + r^n)d(f(x),x)\\
&\le& \frac{r^n}{1-r}d(f(x),x)
\end{eqnarray*}
Since $r<1$, $r^n d(f(x),x)/(1-r)\to 0$ as $n\to \infty$, proving that the
sequence $\{f^n(x)\}$ is Cauchy.

Because $M$ is complete, the sequence $\{f^n(x)\}$ must converge to some
point $x_0\in M$. This point is a fixed point of $f$:
\[ f(x_0) = f(\lim_{n\to\infty} f^n(x)) =
\lim_{n\to\infty}f^{n+1}(x) = x_0 \]
The previous inequality gives us the rate of convergence:
\[ d(x_0,f^n(x)) = \lim_{j\to\infty}d(f^j(x),f^n(x)) \le c r^n \]
where $c = d(f(x),x)/(1-r)$.

The fixed point is unique since if $f(x_1) = x_1$ then
\[d(x_0,x_1) = d(f(x_0),f(x_1)) \le r d(x_0,x_1)\]
But $r<1$, so $d(x_0,x_1) = 0$ and hence $x_1 = x_0$.
\qed

It can be difficult to apply the Contractive Mapping Principle. In order
for a mapping $f:M\to M$ to have the strong property of shrinking
distances, one must often first restrict the domain to a smaller subset
$M_0\subset M$. But then one must verify that $f(M_0) \subset M_0$
in order to apply the theorem. For example, if $f(x) = x^2 + 1$, then
$|f(x) - f(y)| = |x^2 - y^2| = |x+y|\cdot|x-y|$. If we restrict the domain
to $[-1/3,1/3]$, then $|f(x)-f(y)| \le (2/3)|x-y|$ which looks like the
condition for a contractive mapping. However, $f([-1/3,1/3]) = [1,10/9]$,
so $f$ does not map $[-1/3,1/3]$ to itself.

\emph{Example:} Suppose $f:[a,b]\to[a,b]$ satisfies
$|f'(x)| < 1$ for all $x\in[a,b]$. Then $f$ is contractive because
$\forall\, x,y\in[a,b]$, the Mean Value Theorem implies there is a $z$
between $x$ and $y$ such that
\[ |f(x) - f(y)| = |f'(z)|\cdot|x - y| \le r\cdot|x - y|,\quad r =
\max_{z\in[a,b]} |f'(z)| < 1\]
The Contractive Mapping Principle tells us that $f$ has a fixed point
in $[a,b]$ that can be found by repeatedly applying $f$ to any starting
number in $[a,b]$. To get a numerical approximation to the fixed point,
one could program $f$ into a calculator, evaluate $f$ on $a$, then
repeatedly evaluate $f$ on the previous output until the numbers on the
display do not change.

For example, consider $f(x) = e^{-x}$ on the interval $[.3,.8]$. Since $f$
is decreasing, $f([.3,.8]) = [f(.8),f(.3)] = [e^{-.8},e^{-.3}] \subset
[.3,.8]$. Moreover, $|f'(x)| \le e^{-.3}  < 1$ for $x \ge .3$, so $f$ is
contractive.  The Contractive Mapping Principle implies that there exists a
unique number $x_0\in [.3,.8]$  such that $f(x_0) = x_0$, i.e., such that
$e^{-x_0} = x_0$.  If we start with 0.5 and repeatedly press the ``button''
for the function $e^{-x}$ on the calculator, we obtain the sequence:
 0.500000, 0.606531, 0.545239, 0.579703, 0.560065,
 0.571172, 0.564863, 0.568438, 0.566409, 0.567560,
 0.566907, 0.567277, 0.567067, 0.567186, 0.567119,
 0.567157, 0.567135, 0.567148, 0.567141, 0.567145,
 0.567142, 0.567144, 0.567143, \dots.
The digits stabilize at 0.567143, and we find that $e^{-0.567143} =
0.567143$.

Actually, we do not need a contractive mapping $f:[a,b]\to[a,b]$ in order
to find a fixed point.
\begin{quote}
\emph{Any mapping $f:[a,b]\to[a,b]$ has a fixed point.}
\end{quote}
To prove this, consider $g(x) = f(x)-x$. Since $g(a) = f(a) - a \ge 0$ and
$g(b) = f(b) - b \le 0$, the Intermediate Value Theorem implies that there
exists $c\in[a,b]$ such that $g(c) = 0$ and hence $f(c) = c$. This is a
special case of a more general fact known as the \emph{Brouwer Fixed Point
Theorem}:
\begin{quote}
\emph{Any mapping of a closed ball in $\R^n$ into itself has
a fixed point.}
\end{quote}
It is necessary to have a \emph{closed} ball for this theorem, even in the
case of intervals. The function $f(x) = x/2$ maps $(0,1)$ to itself, but
does not have a fixed point in $(0,1)$. The disadvantage of Brouwer's
theorem over the Contractive Mapping Principle is that it is not
constructive. We know the fixed point must exist, but the theorem does
not tell us how to find it.

%\subsection{The Stone-Weierstrass Theorem}
%
%The Weierstrass Approximation Theorem (Theorem~\ref{04:WAT}) says that
%any continuous function on a closed interval can be uniformly approximated
%by polynomials on that interval. This theorem has a generalization to
%continuous functions on a compact metric space. In order to formulate
%this generalization, we need a replacement for polynomials which do no
%exist on a general compact metric space. We can however, determine the
%properties of polynomials that are needed in the proof and define them
%in an abstract way. Let us denote the continuous real-valued functions on
%a metric space $M$ by $C[M] = \{ f:M\to \R\ |\ f\ \mbox{continuous}\}$.
%
%\begin{definition}
%\begin{enumerate}
%\item A collection $\mathcal{A}\subset C[M]$ forms an
%\emph{algebra} if
%\[ \forall\, f, g\in \mathcal{A}, \forall\, a,b\in \R \Rightarrow
%af + bg\in \mathcal{A}, f\cdot g\in \mathcal{A} \]
%\item $\mathcal{A}\subset C[M]$ \emph{strongly separates points} if
%  \begin{itemize}
%  \item $\forall\, x\in M$, $\exists\, f\in \mathcal{A}$ such that $f(x) \ne 0$
%  \item $\forall\, x,y\in M$, $\exists, f\in\mathcal{A}$ such that $f(x)
%  \ne f(y)$
%  \end{itemize}
%\end{enumerate}
%\end{definition}
%
%\begin{theorem}[Stone-Weierstrass]
%Let $M$ be a compact metric space and assume $\mathcal{A}\subset C[M]$ is an
%algebra that strongly separates points. For any $f\in C[M]$ there is a
%sequence $\{f_n\}\subset\mathcal{A}$ such that $f_n\to f$ uniformly.
%\end{theorem}

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