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\begin{document}
\centerline{Math 322 lecture for Friday, Week 6}
\bigskip

\centerline{\sc the contraction mapping principle}
\medskip

Let~$(V,\|\ \|)$ be a normed vector space over~$F=\R$ or~$\C$.  Recall this
means that for all~$v,w\in V$ and~$\alpha\in F$, 
\begin{enumerate}
  \item $\|v\|\geq0$ with equality if and only if~$v=0$;
  \item $\|\alpha v\|=|\alpha|\|v\|$;
  \item $\|v+w\|\leq\|v\|+\|w\|$.
\end{enumerate}
If every Cauchy sequence in~$V$ converges (in~$V$), then we say~$V$ is {\em
complete}, and in that case~$(V,\|\ \|)$ is called a {\em Banach space}.
We have already used the fact that~$\R^n$ and~$\C^n$ are Banach spaces, for
example, when considering the convergence of~$e^{At}$.  We will
soon need to consider a Banach space whose elements consist of potential
solutions to systems of differential equations.
\medskip

\textbf{Definition.}
Let~$(V,\|\ \|)$ be a Banach space, and let~$X\subseteq V$.  Let~$T\colon X\to
X$.
\begin{enumerate}
  \item A point~$u\in X$ is a {\em fixed point} for~$T$ if~$T(u)=u$.
  \item The function~$T$ is a {\em contraction mapping} if there is a
    constant~$c\in[0,1)\subset\R$ such that 
    \[ 
    \|T(u)-T(v)\|\leq c\|u-v\| 
    \] 
    for all~$u,v\in X$.
\end{enumerate}
\medskip

\textbf{Theorem.}  Let~$(V,\|\ \|)$ be a Banach space, and let~$X\subseteq
V$ be a closed subset of~$V$ (hence, it contains all of its limit points).
Suppose that~$T\colon X\to X$ is a contraction mapping and fix a
constant~$c\in[0,1)\subset\R$ so that
\[
  \|T(u)-T(v)\|\leq c\|u-v\|
\] 
for all~$u,v\in X$.
Then~$T$ has a unique fixed point~$\tilde{u}\in X$.  
Let~$u_0\in X$ and consider the sequence of iterates
\[
  u_0,T(u_0),T^2(u_0), T^3(u_0),\dots
\]
(For example,~$T^3(u_0)=T(T(T(u_0)))$.)  We have, for all~$m\geq0$,
\[
  \|\tilde{u}-T^m(u_0)\|\leq\frac{c^m}{1-c}\|T(u_0)-u_0\|.
\]
In particular, the sequence of iterates converges to the fixed
point,~$\tilde{u}$.
\begin{proof}
 We first show uniqueness.  Suppose that~$T(u)=u$ and~$T(v)=v$. We
 have
 \[
   \|u-v\|=\|T(u)-T(v)\|\leq c\|u-v\|,
 \]
 which implies
 \[
   (1-c)\|u-v\|\leq 0.
 \]
 Since~$1-c\geq0$, it follows that~$\|u-v\|=0$, and hence,~$u=v$.
 \medskip
 
 Now take~$u_0\in X$, and define~$u_{k+1}:=T(u_k)$ for~$k\geq0$.
 Thus,~$u_k=T^k(u_0)$ for all~$k\geq0$.  For all pairs of natural numbers~$m\leq n$, 
 \begin{align*}
   \|u_n-u_m\|
   &= \|T(u_{n-1})-T(u_{m-1})\|\\[5pt]
   &\leq c\|u_{n-1}-u_{m-1}\|\\[5pt]
   &= c\|T(u_{n-2})-T(u_{m-2})\|\\[5pt]
   &\leq c^2\|u_{n-2}-u_{m-2}\|\\
   &\ \ \vdots\\
   &\leq c^{m}\|u_{n-m}-u_0\|\\[5pt]
 &= c^m\|(u_{1}-u_{0})+(u_{2}-u_{1})+(u_3-u_2)\dots+(u_{n-m}-u_{n-m-1})\|\\[5pt]
 &\leq
 c^m\left(\|u_{1}-u_{0}\|+\|u_{2}-u_{1}\|+\|u_3-u_2\|+\dots+\|u_{n-m}-u_{n-m-1}\|\right)\\[5pt]
 &\leq
 c^m\left(\|u_{1}-u_{0}\|+c\|u_{1}-u_{0}\|+c^2\|u_1-u_0\|+\dots+c^{n-m-1}\|u_{1}-u_{0}\|\right)\\[5pt]
 &= c^m\|u_1-u_0\|\left(1+c+c^2+\dots+c^{n-m-1} \right)\\[5pt]
 &= c^m\frac{1-c^{n-m}}{1-c}\|u_1-u_0\|\\[5pt]
 &\leq \frac{c^m}{1-c}\|u_1-u_0\|.
 \end{align*}
 Given any~$\ve>0$, we then see that by choosing~$N$ sufficiently large, it
 follows that if~$m,n\geq N$, then~$\|u_n-u_m\|<\ve$.  So the
 sequence of iterates,~$(u_k)_{k\geq0}$ is Cauchy.  Since~$V$ is a Banach space,
 the sequence converges to some~$\tilde{u}$, and since~$X$ is
 closed,~$\tilde{u}\in X$.  Since~$T$ is a contraction mapping, it's continuous
 (exercise), and therefore commutes with limits:
 \[
   \lim_{k\to\infty}T(u_k)=T(\lim_{k\to\infty}u_k)=T(\tilde{u}).
 \]
 On the other hand, by definition of the~$u_k$, we have
 \[
   \lim_{k\to\infty}T(u_k)=\lim_{k\to\infty}u_{k+1}=\lim_{k\to\infty}u_k=\tilde{u}.
 \]
 This shows~$T(\tilde{u})=\tilde{u}$, i.e.,~$\tilde{u}$ is the unique fixed point
 of~$T$.
 
 Finally, in our calculation above, we saw that for all~$m\leq n$,
 \[
 \|u_n-u_m\|= \|T^{n}(u_0)-T^m(u_0)\|\leq \frac{c^m}{1-c}\|u_{1}-u_0\|\leq \frac{c^m}{1-c}\|T(u_0)-u_0\|.
 \]
 Since the norm function and~$T$ are continuous, they both commute with limits.
 Therefore, taking the limit as~$n\to\infty$ on both sides of the above
 inequality yields
 \[
   \|\tilde{u}-T^m(u_0)\|\leq\frac{c^m}{1-c}\|T(u_0)-u_0\|.
 \]
\end{proof}


\textbf{Method of successive approximations.} We are interested in applying the
contraction mapping principle to the operator
\[
  T(u)=x_0+\int_{s=0}^tf(u(s))\,ds,
\]
discussed in the previous lecture. So we need to find the appropriate Banach
space and find conditions under which~$T$ is a contraction mapping.
\medskip

\textbf{Definition.} If~$I\subset\R$ is a closed bounded interval, let~$C(I)$
denote the~$\R$-vector space of continuous functions on~$I\to\R^n$ (where~$n$ is
fixed).  For each~$u\in C(I)$, define
\[
  \|u\|:=\sup_{t\in I}|u(t)|=\max_{t\in I}|u(t)|.
\]
(The last equality is due to the fact that the continuous image of a compact set
is compact---a generalization of the extreme value theorem of one-variable
calculus.)  Geometrically, $\|u\|$ is the maximum distance from the origin
reached by $u(t)$.
\medskip

\textbf{Proposition.}~$(C(I),\|\ \|)$ is a Banach space.
\begin{proof}
 Math 321.
\end{proof}

Thus, the method of successive approximations is an operator
\[
  T\colon C(I)\to C(I)
\]
on the Banach space of continuous functions on~$I$.
Under what conditions is it a contraction mapping?  We have
\begin{align*}
  |(Tu)(t)-(Tv)(t)|
  &=\left| \left( x_0+\int_{s=0}^tf(u(s))\,ds \right)
  -\left( x_0+\int_{s=0}^tf(v(s))\,ds \right)\right|\\[10pt]
  &=\left|\int_{s=0}^tf(u(s))-f(v(s))\,ds \right|\\[10pt]
  &\leq\int_{s=0}^t\left|f(u(s))-f(v(s))\right|\,ds \\[10pt]
  &\leq t\max_{s\in [0,t]}\{\left|f(u(s))-f(v(s))\right|\}.
\end{align*}
From this, we can see two things that will help to control the size
of~$|T(u)-T(v)|$: first, restrict to a small enough region around~$x_0$ so
that~$f$ does not vary much on that region, and second, make the interval in
which~$t$ varies small.  We address the first problem below by
considering the derivative of~$f$.
\end{document}
