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\title{Math 101\\Solution and Commentary\\for the First Group Project}
\author{Stefan Karpinski}

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\begin{document}

\maketitle

Rather than addressing each point of the project in order, I will first give
a statement and proof of the Contraction Mapping Theorem using the principle
of nested closed intervals.  Ensuing commentary should address all points of
the the assignment.

\begin{Theorem*}

Every contraction of a complete metric space has a unique fixed point.

\end{Theorem*}

This is a relatively easy generalization of the theorem you were asked to
prove, which was simply the special case where our metric space is taken to be
a closed subset of $\R$ with the standard Euclidean metric inherited from $\R$.
However neither of these two assumptions is essential to the proof, so it is
simpler and cleaner to work in a more general context.  The principle of
nested closed intervals presents the only dilemma:  what are nested closed
intervals in a general metric space?  Or more to the point, what does it mean
for a general metric space to be complete?  To answer this we simply
generalize the concept of a closed interval to the concept of a closed ball
\begin{align*}
B(x;r) = \{ y \in X : d(x,y)  \leq  r \}.
\end{align*}
Given a sequence of points $\{x_{n}\}$ and a sequence of radii (positive real
numbers) $\{r_{n}\}$ indexed over $\N$, we say that they are nested if for $m > n$ we
have $B(x_{m};r_{m}) \subseteq B(x_{n};r_{n})$. From homework we know that this is
equivalent to requiring that $\{x_{n}\}$ and $\{r_{n}\}$ satisfy
\begin{align*}
d(x_{n},x_{m})  \leq  r_{n} - r_{m}  \quad \text{whenever} \quad  m > n,
\end{align*}
which in particular tells us that $r_{m}  \leq  r_{n}$ for $m > n$.  With this
generalization of nested closed intervals, we may define completeness as
follows.

\begin{Definition*}

A metric space $(X,d)$ is said to be complete if for every pair of
sequences $\{x_{n}\}, \{r_{n}\}$ satisfying nestedness we have
\begin{align*}
\Inter_{n\in\N} B(x_{n};r_{n})  \neq  \emptyset.
\end{align*}

\end{Definition*}

With this established, let's proceed with the proof.

\begin{proof}
Suppose $f \colon X \rightarrow X$ is our contraction mapping.  We want to use the
completeness of our metric space in some way to produce a fixed point of
the contraction.  To this end, pick any $x_{0} \in X$ and define recursively $x_{n}
= f(x_{n-1})$. This gives us our sequence of points $\{x_{n}\}$ but before we can
use the priniple of nested closed intervals (or closed balls as we have
here) we need to define our sequence of radii $\{r_{n}\}$ so that we satisfy nestedness.
This is the hardest part of the proof---the rest is just applying definitions
cleverly.  Notice that not every sequence $\{x_{n}\}$ has a sequence of radii $\{r_{n}\}$
such that nestedness is satisfied.  To produce such a sequence of radii we need
to use the way that $\{x_{n}\}$ is defined and the hypothesis that $f$ is a
contraction.  Satisfying nestedness amounts to establishing an upper bound on
$d(x_{n},x_{m})$ for $m > n$ which depends only on $n$ and which decreases (sufficiently
fast) as $n$ increases.  Let us do precisely that.  Assume $m > n$.
\begin{align*}
d(x_{n},x_{m}) &  \leq  d(x_{n},x_{n+1}) +  \dots  + d(x_{m-1},x_{m}) \\
 &  \leq  C^{n} d(x_{0},x_{1}) +  \dots  + C^{m} d(x_{0},x_{1}) \\
 & =  C^{n} (1 + C +  \dots  + C^{m-n}) d(x_{0},x_{1}) \\
 &  \leq  C^{n} (1 + C + C^{2} +  \dots ) d(x_{0},x_{1}) \\
 & =  C^{n} (1-C)^{-1} d(x_{0},x_{1}).
\end{align*}
Call this last value $k_{n}$.  This bound certainly doesn't leap to mind
immediately---it requires some sneaky tricks and several strong facts and
results.  Notable among them are (in order of usage) the triangle inequality,
contractioness of $f$, non-negativity of $C$ (tells us the infinte sequence is
bigger than its finte initial segment), and finally the fact that $|C| < 1$ lets
us evaluate the infinte sequence $1 + C + C^{2} +  \dots  = (1-C)^{-1}$. Finally we
define
\begin{align*}
r_{n} & = k_{n} + k_{n+1} +  \dots  \\
 & = (C^{n} + C^{n+1} +  \dots ) (1-C)^{-1} d(x_{0},x_{1}) \\
 & = C^{n} (1-C)^{-2} d(x_{0},x_{1})
\end{align*}
so that we now have for $m > n$,
\begin{align*}
r_{n} - r_{m} &  = k_{n} + k_{n+1} +  \dots  + k_{m-1} \\
 &  \geq  k_{n} \\
 &  \geq  d(x_{n},x_{m})
\end{align*}
which means precisely that we have $\{r_{n}\}$ satisfying nestedness.  The upshot of
all of this trickery and sneaky approximation is that we now have a pair of
sequences $\{x_{n}\}$ and $\{r_{n}\}$ satisfying nestedness, so we may use the completeness
of our metric space to deduce that there is some point $x \in \inter_{n}
B(x_{n};r_{n})$. (Moreover, since $\lim_{n \rightarrow \infty} r_{n} = 0$, this point $x$ is unique but
we don't need this fact---and the uniqueness of $x$ in the above intersection is
not to be confused with the uniqueness of the fixed point.)  For arbitrary $n$
we have
\begin{align*}
d(x,f(x)) &  \leq  d(x,f(x_{n})) + d(f(x),f(x_{n})) \\
 &  \leq  d(x,x_{n+1}) + C d(x,x_{n}) \\
 &  \leq  r_{n+1} + C r_{n},
\end{align*}
but we can make this last term arbitrarily small by picking $n$ large enough, so
we must have $d(x,f(x)) = 0$ which means $x = f(x)$.  Score!  Moreover, this fixed
point is unique:  suppose $x = f(x)$ and $x' = f(x')$;  then $d(x,x') =
d(f(x),f(x'))  \leq  C d(x,x')$, where $0  \leq  C < 1$, which yeilds a contradiction
unless $d(x,x') = 0$, in which case $x = x'$.
\end{proof}

With this general form of the CMT we can immediately address parts $4$ and $7$ of
the assignment:  CMT does not apply to $\Q$ because $\Q$ is not complete and
connectedness is completely irrelevant to the CMT.  It happens that $\Q$ is not
connected, but that's just a coincidence---CMT applies to all sorts of
disconnected sets such as $[0,1]\union[2,3]$, or even something more complicated
like $\Union_{n\in\Z} [2n,2n+1]$ and anything else you can cook up, as long as its
a complete metric space.  As to a function on $A = \{ x \in \Q : x  \geq  1 \}$ which
is a contraction but has no fixed point in $A$, there are of course infinitely
many such functions but one example would be the function
\begin{align*}
f(x) = 1 + \frac{1}{1+x}.
\end{align*}
A much more challenging question is to produce a contraction on all of $\Q$
which has no fixed point.  (I'm not even certain that one exists but I'm
pretty sure that one must.  Think about it and let me know if you come up
with anything.)

We can also address the last question of part $3$.  In the more restricted
version of the theorem that you were asked to prove, we need $A$ to be closed
because if $(X,d)$ is a complete metric space and $A$ is a subspace of $X$, then $A$
is complete if and only if $A$ is closed.  The simplest example of a contraction
on $(0,1)$ without a fixed point is just $f(x) = \frac{1}{2} x$.

Now for part $5$.  While we know that given any particular distinct $x,y \in A$ if
$d(f(x),f(y)) < d(x,y)$, we can find some $C$ such that $0  \leq  C < 1$ and
$d(f(x),f(y))  \leq  C d(x,y)$, we cannot guarantee that this will be the same $C$ for
all $x,y$ and moreover, that we can put an upper bound on such $C$ which is
strictly less than $1$.  However, if our set $A$ is compact then we can actually
find some maximal $C$ strictly less than $1$.  This is since continuous functions
attain their maxima on compact sets.  (What function am I actually applying
this fact to?  Hint it's not just $f$ and it involves the metric (twice) and
division.) What that means is that when, for example, $A = [0,1] \union [2,3]$,
all ``pseudo-contractions'' (functions $f \colon A \rightarrow A$ that satisfy the
``pseudo-contraction'' condition $d(f(x),f(y)) < d(x,y)$---this is not a real
term, I just made it up) actually do have to have a fixed point since we can
find some contraction constant that works for all pairs $x,y$ because $A$ is
compact. What I'm saying is that in compact spaces, pseudo-contractions are
real contractions.  But if we choose $A$ to be something non-compact such as all
of $\R$, then we can find functions which satisfy $d(f(x),f(y)) < d(x,y)$ but have
no fixed points.  So for non-compact spaces, there are pseudo-contractions
which are not real contractions.  To find such a function on $\R$ all we need to
do is find a function whose graph never touches the line $x = y$ (so it has no
fixed points) and such that the absolute value of the slope between any two
points on the graph of the function is always less than $1$ (this is equivalent
to it satisfying the pseudo-contraction condidtion---why?). One such function
is the half-hyperbola $f(x) = -(1+x^{2})^{1/2}$. You should verify that this
function satisfies the pseudo-contraction condition $d(f(x),f(y)) < d(x,y)$.

And finally, part $6$.  If $f$ is a contraction allowing contraction constant $C$
then it is immediate that $g = f^{2}$ is also a contraction allowing the still
smaller contraction constant $C^{2}$. On the other hand, $g = f^{2}$ may be a
contraction without $f$ being a contraction. For example, define $f \colon \R \rightarrow \R$ by
\begin{align*}
f(x) = \begin{cases} -x, & x < 0, \\ 0, & x  \geq  0. \end{cases}
\end{align*}
Then $g = f^{2} = 0$, which is clearly a contraction while $f$ is blatantly not.  Of
course, since $f^{2} = 0$, we have $f^{n} = 0$ for all higher $n$ as well, so in
particular, $f^{n}$ is always a contraction.  If $f$ is a contraction with fixed
point $x$ then $f^{n}$ will always be a contraction and will moreover have the same
fixed point.  (More generally, if $f$ is any function then the fixed points of
$f^{m}$ will always include the fixed points of $f^{n}$ when $m > n$.)

That seems to cover all of the material of the second group project.




\end{document}
