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\AtBeginDocument{{\noindent\small \emph{
Electronic Journal of Differential Equations},
Monograph 09, 2009, (90 pages).\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2009 by  Robert Brooks and Klaus Schmitt.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/Mon. 09\hfil The contraction mapping principle]
{The contraction mapping principle and some applications}

\author[R. M. Brooks, K.Schmitt\hfil EJDE-2009/Mon. 09\hfilneg]
{Robert M. Brooks, Klaus Schmitt}  % in alphabetical order

\address{Robert M. Brooks \newline
Department of Mathematics\\
University of Utah\\
Salt Lake City, UT 84112, USA}
\email{brooks@math.utah.edu}

\address{Klaus Schmitt \newline
Department of Mathematics\\
University of Utah\\
Salt Lake City, UT 84112, USA}
\email{schmitt@math.utah.edu}

\thanks{Submitted May 2, 2009. Published May 13, 2009.}
\subjclass[2000]{34-02, 34A34, 34B15, 34C25, 34C27, 35A10,
\hfill\break\indent 35J25, 35J35, 47H09, 47H10, 49J40, 58C15}
\keywords{Contraction mapping principle; variational inequalities;
\hfill\break\indent
  Hilbert's projective metric; Cauchy-Kowalweski theorem; 
  boundary value problems; \hfill\break\indent
differential and integral equations}

\begin{abstract}
  These notes contain various versions
  of the contraction mapping principle. Several applications to
  existence theorems in the theories of differential and integral
  equations and variational inequalities are given. Also discussed are
  Hilbert's projective metric and iterated function systems
\end{abstract}

\maketitle
\tableofcontents
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{example}[theorem]{Example}


\part{Abstract results}


\section{Introduction}\label{chapI} %\include{contintro}

\subsection{Theme and overview} %\label{secIV1}
The contraction mapping principle is one of the most useful tools in
the study
of nonlinear equations, be they algebraic equations, integral or
differential equations.
The  principle is a fixed point theorem which
guarantees that a contraction mapping of a complete metric space to
itself has a unique fixed point which may be obtained as the
limit
of an iteration scheme defined by repeated images under the mapping of
an arbitrary starting point in the  space.
As such,  it is a constructive fixed
point theorem and, hence, may be implemented for the  numerical computation
of the fixed point.

Iteration schemes have been used since the antiquity of mathematics
(viz., the ancient schemes for computing square roots of numbers) and
became particularly useful in Newton's method for solving polynomial
or systems of algebraic equations and also in the Picard iteration
process
for solving initial value  and boundary value problems for
nonlinear ordinary differential equations (see, e.g.
\cite{picard:ta93}, \cite{picard:lqp30}).


The principle was first stated and proved by Banach
 \cite{banach:oea22}
 for contraction
mappings in complete normed linear spaces (for the many
consequences of Banach's  work see \cite{pietsch:hbs07}). At about
the same time the concept of an abstract metric space was
introduced by Hausdorff, which then provided the general framework
for   the principle for contraction mappings in a complete metric
space, as was done by
 Caccioppoli
\cite{caccioppoli:tge30} (see also   \cite{weissinger:tai52}).  It
appears in the various  texts on real analysis (an early one
being, \cite{natanson:tfr60}).

In these notes we shall develop the contraction mapping principle in
several forms and present a host of useful applications which appear
in  various places in the mathematical literature. Our purpose
is
to introduce the reader to several different areas of analysis where
the principle has been found useful. We shall discuss among others:
the convergence of
Newton's method; iterated function systems
and how certain fractals are fixed points
of set-valued contractions; the Perron-Frobenius theorem
for positive matrices using Hilbert's metric, and the extension of this
theorem to infinite dimensional spaces (the theorem of Krein-Rutman);
the basic existence and uniqueness
theorem of the theory
of ordinary differential equations (the Picard-Lindel\"of theorem)
and various related results;
applications to the theory of integral equations of Abel-Liouville
type;
 the implicit function theorem;
the basic existence and  uniqueness theorem of variational inequalities
and hence a Lax-Milgram type result for not necessarily
symmetric quadratic forms;
 the basic existence theorem of Cauchy-Kowalevsky for partial
differential equations with analytic terms.

These notes have been collected over several years and have, most
recently, been used as a basis for an REU seminar which has been part
of the VIGRE program of our department. We want to thank here
 those undergraduate students who participated in the seminar
 and gave us their valuable feedback.

\section{Complete metric spaces} \label{chapII}
% Chapter II \include{contmetric}

In this section  we  review briefly some very basic concepts which are
part of most  undergraduate mathematics curricula. We shall assume these as
requisite knowledge and refer to any basic text,
e.g., \cite{bruckner:ra97}, \cite{fitzpatrick:acc96}, \cite{royden:ra88}.

\subsection{Metric spaces}
\label{ms}
 Given a set  $\mathbb{M}$, a {\it metric}\index{metric} on
 $\mathbb{M}$ is a function (also called a  {\it distance}\index{distance})
\[
{\rm d}:\mathbb{M} \times \mathbb{M} \to \mathbb{R} _+=[0,\infty),
\]
 that
satisfies
\begin{equation}
\label{distance}
\begin{gathered}
{\rm d}(x,y)={\rm d}(y,x),\quad \forall x,y\in \mathbb{M}\\
{\rm d}(x,y)=0,\quad \text{if, and only if,}\quad  x=y\\
{\rm d}(x,y)\leq {\rm d}(x,z)+{\rm d}(y,z),
\quad \forall x,y,z\in \mathbb{M},
\end{gathered}
\end{equation}
(the last requirement is called the {\it triangle
inequality}\index{triangle inequality}).
We call the pair
$(\mathbb{M},{\rm d})$
 a metric space\index{metric space} (frequently we use $\mathbb{M}$ to
 represent the pair).

A sequence $\{x_n\}_{n=1}^{\infty}$ in $ \mathbb{M}$ is said to
converge to   $x\in \mathbb{M}$ provided that
\[
\lim _{n\to \infty}{\rm d}(x_n,x)=0.
\]
This we also write as
\[
\lim _{n}x_n=x,\;\text{or}~x_n\to x~\text{as}~n\to \infty.
\]

We call a  sequence $\{x_n\}_{n=1}^{\infty}$ in $ \mathbb{M}$  a {\it Cauchy
 sequence}\index{Cauchy sequence}
 provided that for all
$\epsilon >0$, there exists $n_0=n_0(\epsilon )$, such that
\[
{\rm d}(x_n,x_m)\leq \epsilon,\quad \forall n,m\geq n_0.
\]
A metric space $\mathbb{M}$ is said to be {\it complete}\index{complete}
if, and only if,  every Cauchy sequence in $ \mathbb{M}$
converges to a point in $ \mathbb{M}$.

Metric spaces form a useful-in-analysis subfamily of the family of
topological spaces. We need to discuss some of the concepts met in
studying these spaces. We do so, however, in the context of metric
spaces rather than in the more general setting. The following concepts
are normally met in an advanced calculus or foundations of analysis
course. We shall simply list these concepts here and refer the reader
to appropriate texts (e.g. \cite{fitzpatrick:acc96} or
\cite{thomson:era01})
for the formal definitions.
We consider a fixed metric space $(\mathbb{M},{\rm d})$.
\begin{itemize}
\item
Open balls\index{open ball}
$B(x,\epsilon):=\{y\in \mathbb{M}:{\rm d}(x,y)<\epsilon \}$
 and closed balls\index{closed ball}
 $B[x,\epsilon]:=\{y\in \mathbb{M}:{\rm d}(x,y)\leq
 \epsilon \}$;
\item open\index{open set} and closed\index{closed set} subsets of
$\mathbb{M}$;
\item
bounded\index{bounded set} and totally bounded\index{totally bounded}
 sets in $\mathbb{M}$;
\item
limit\index{limit point} point (accumulation point\index{accumulation point}) of a subset of $\mathbb{M}$;
\item the closure\index{closure} of a subset of $\mathbb{M}$ (note that the closure of
an open ball is not necessarily the closed ball);
\item the diameter\index{diameter} of a set;
\item the notion of one set's being dense\index{dense} in another;
\item the distance between a point and a set (and between two sets).
 \end{itemize}

Suppose $(\mathbb{M},{\rm d})$ is a metric space  and
$\mathbb{M}_1\subset \mathbb{M}$. If  we restrict ${\rm d}$
to $\mathbb{M}_1\times \mathbb{M}_1$, then $\mathbb{M}_1$ will be
a metric space  having  the ``same'' metric as $\mathbb{M}$. We
note the important fact that if $\mathbb{M}$ is complete and
$\mathbb{M}_1$ is a closed subset of $\mathbb{M}$, then
$\mathbb{M}_1$ is also a complete metric space (any Cauchy
sequence in $\mathbb{M}_1$ will be a Cauchy sequence in $\mathbb{M}$; hence it will converge to some point in $\mathbb{M}$; since
$\mathbb{M}_1$ is closed in $\mathbb{M}$ that limit must be in
$\mathbb{M}_1$).

The notion of {\it compactness} is a crucial one. A metric space
$\mathbb{M}$ is said to be {\it compact}\index{compact} provided
that  given any family $\{G_{\alpha}:\alpha \in A\}$ of open sets
whose union is $\mathbb{M}$, there is a finite subset $A_0\subset
A$ such that the union of $\{G_{\alpha}:\alpha \in A_0\}$ is
$\mathbb{M}$. (To describe this situation one usually says that
every open cover\index{open cover} of $\mathbb{M}$ has a finite
subcover.) We may describe compactness more ``analytically'' as
follows. Given any sequence $\{x_n\}$ in $\mathbb{M}$ and a point
$y\in \mathbb{M}$; we say that $y$ is a {\it cluster
point}\index{cluster point} of the sequence
 $\{x_n\}$ provided that for any $\epsilon >0$ and any positive integer
 $k$,
there exists $n\geq k$ such that $x_n\in B(y,\epsilon)$. Thus in
any open ball, centered at $y$, infinitely many terms of the
sequence $\{x_n\}$ are to be found. We then have that $\mathbb{M}$
is compact, provided that every sequence in $\mathbb{M}$ has a
cluster point (in $\mathbb{M}$).


In the remaining  sections of this chapter we briefly list and describe some useful examples
of metric spaces.

\subsection{Normed vector spaces}
\label{nls} Let $\mathbb{M}$ be a vector space over  the real or
complex numbers (the scalars). A mapping $\|\cdot \|:\mathbb{M}\to
\mathbb{R}_+$ is called a {\it norm}\index{norm} provided that the
following conditions hold:
\begin{equation}
\label{norm}
\begin{gathered}
\|x\|=0, \quad \text{if, and only if, }x=0 \;(\in \mathbb{M})\\
\|\alpha x\|=|\alpha |\|x\|,\quad \forall \text{ scalar }\alpha,\;
\forall x\in \mathbb{M} \\
\|x+y\|\leq \|x\| +\|y\|,\quad \forall x,y\in \mathbb{M}.
\end{gathered}
\end{equation}
If $\mathbb{M}$ is a vector space and $\|\cdot \|$
 is a norm on $\mathbb{M}$, then
the pair $(\mathbb{M},\|\cdot \|)$ is called a {\it normed vector
space}\index{normed vector space}. Should no ambiguity arise we simply
abbreviate this by saying that $\mathbb{M}$ is a normed vector space.
If $\mathbb{M}$ is a vector space and $\|\cdot \|$ is a
 norm on $\mathbb{M}$, then
$\mathbb{M}$ becomes a metric space if we  define
the  metric ${\rm d}$ by
\[
{\rm d}(x,y):=\|x-y\|,\;\forall x,y\in \mathbb{M}.
\]
A normed vector space which is a complete metric space, with respect
to the  metric
${\rm d}$ defined  above,
is called a {\it Banach space}\index{Banach space}. Thus, a closed
subset of a Banach space may always be regarded as a complete metric
space;
hence,  a closed subspace of a Banach space is also a Banach space.

We pause briefly in our general discussion to put together, for future
reference, a small catalogue of Banach spaces. We shall consider only
real Banach spaces, the complex analogues being defined similarly.


In all cases the verification that these spaces are normed linear
spaces is straightforward, the verification of completeness, on the
other hand usually is more difficult.
Many of the examples that will be discussed later will have their
setting in complete metric spaces which are subsets or subspaces
of Banach spaces.

\subsection*{Examples of Banach spaces}

\begin{example}\rm
$(\mathbb{R}, |\cdot |)$ is a simple example of a  Banach space.
\end{example}
\begin{example}\rm
We fix $N\in \mathbb N$ (the natural numbers) and denote by
$\mathbb{R}^N$ the set
\[
\mathbb{R}^N:=\{x:x=(\xi _1,\dots ,\xi _N),\;\xi _i\in \mathbb{R},\;
i=1,\dots ,N\}.
\]
There are many useful norms with which we can equip $\mathbb{R}^N$.
\begin{enumerate}
\item For $1\leq p <\infty $ define $\|\cdot \|_p:\mathbb{R}^N\to \mathbb{R}_+$
by
\[
\|x\|_p:=\Big(\sum _{i=1}^N|\xi _i|^p\Big)^{1/p},\quad
x\in \mathbb{R}^N.
\]
These spaces are finite dimensional $l^p-$ spaces. Frequently used
norms are $\|\cdot \|_1$, and $\|\cdot \|_2$.
\item We define $\|\cdot \|_{\infty}:\mathbb{R}^N\to \mathbb{R}_+$
by
\[
\|x\|_{\infty}:=\max  \{|\xi _i|:1\leq i\leq N\}.
\]
This norm is called the {\it sup norm}\index{sup norm} on $\mathbb{R}^N$.
\end{enumerate}
\end{example}
The next example extends the example just considered to the
infinite dimensional setting.

\begin{example}\rm
We let
\[
\mathbb{R}^{\infty}:=\{x:x=\{\xi _i\}_{i=1}^{\infty},\;
\xi _i\in \mathbb{R},\;i=1,2,\dots \}.
\]
Then $\mathbb{R}^{\infty}$, with coordinate-wise addition and scalar
 multiplication, is a vector space, certain subspaces of which can
  be equipped with norms, with respect to which they are complete.
\begin{enumerate}
\item For  $1\leq p <\infty $ define
\[
l^p:=\{x=\{\xi _i\}\in \mathbb{R}^{\infty}:\sum _i|\xi _i|^p<\infty \}.
\]
Then $l^p$ is a subspace of $\mathbb{R}^{\infty}$ and
\[
\|x\|_p:=\Big( \sum _{i=1}^{\infty}|\xi _i|^p \Big)^{1/p}
\]
defines a norm with respect to which $l^p$ is complete.
\item
We define
\[
l^{\infty}:=\{x=\{\xi _i\}\in \mathbb{R}^{\infty}:\sup _i|\xi _i|<\infty \}.
\]
and
\[
\|x\|_{\infty}:=\sup _{i}\{|\xi _i|,\;x\in l^{\infty}\}.
\]
With respect to this ({\it sup norm}) $l^{\infty} $ is complete.
\end{enumerate}
\end{example}

\begin{example}\rm
Let $H$ be a complex (or real) vector space. An inner product\index{inner product} on $H$ is a mapping $(x,y)\mapsto \langle x, y \rangle $
($H\times H \to \mathbb C$) which satisfies:
\begin{enumerate}
\item for each $z\in H$ $\langle\cdot ,z\rangle :H\to \mathbb C$ is a linear mapping,
\item $\langle x, y \rangle =\overline{\langle y, x\rangle}$
for $x,y\in H$ ( $\langle x, y \rangle ={\langle y, x\rangle}$
if $H$ is a real vector space and the inner product is a real
valued function),
\item  $\langle x, x\rangle \geq 0$, $x\in H$, and equality holds if,
and only if, $x=0$. If one defines
\[
\|x\|:=\sqrt  {\langle x, x \rangle },
\]
then $(H,\|\cdot \|)$ will be a normed vector space.
\end{enumerate}
If it is complete, we refer to $H$ as a {\it Hilbert
space}\index{Hilbert space}. We note that
$(\mathbb{R}^N,\|\cdot \|_2)$ and $l^2$ are (real) Hilbert spaces.
\end{example}

Spaces of continuous functions are further examples of important
spaces in analysis. The following is a brief discussion of such
spaces.

\begin{example} \label{cont1} \rm
We fix $I=[a,b],\;a,b\in \mathbb{R},\;a<b$, and $k\in
\mathbb N\cup \{0\}$. Let $\mathbb{K}=\mathbb{R}, ~\text{or}~\mathbb
C$ (the reader's choice). Define
\[
C^k(I):=\{f:I\to \mathbb{K}:f, f',\dots ,f^{(k)},\;\text{exist and are
continuous on }I\}.
\]
We note that
\[
C^0(I):=C(I)=\{f:I\to \mathbb{K}:f\text{ is continuous on }I\}.
\]
For $f\in C(I)$, we define
\[
\|f\|_{\infty}:=\max _{x\in [a,b]}|f(t)|
\]
(the sup-on-$I$ norm). And for $f\in C^k(I)$
\[
\|f\|:=\sum _{i=0}^k\|f^{(i)}\|_{\infty}.
\]
With the usual pointwise definitions of $f+g$ and $\alpha f$ ($\alpha
\in \mathbb{K}$) and with the norm  defined as above, it follows that
$C^k(I)$ is a normed vector space. That the space is also complete
follows from the completeness of $C(I)$ with respect to the sup-on-$I$
norm
(see, e.g., \cite{bruckner:ra97}).

Another useful norm is
\[
\|f\|^*:=\sup _{x\in I}\sum _{i=0}^k|f^{(i)}(x)|.
\]
which is equivalent to the norm defined above; this  follows from the
inequalities
\[
\|f\|^*\leq \|f\|\leq (k+1)\|f\|^*,\quad \forall f\in C^k(I).
\]
\end{example}

Equivalent norms\index{equivalent norm} give us the same idea of
closeness and one may, in a given application, use, of equivalent
norms, that which makes calculations or verifications easier or gives
us more transparent conclusions.

\begin{example}\label{cont2} \rm
Let $\Omega$ be an open subset of $\mathbb{R}^N$, and let $\mathbb{K}$
be as above; define
\[
C(\Omega ):=C^0(\Omega ): = \{f: \Omega \to \mathbb{K}
\text{ such that $f$ is continuous on } \Omega\}.
\]
Let
\[
\| f\|_{\infty}: = \sup_{x\in\Omega}|f(x)|.
\]


Since the uniform limit of a sequence of continuous functions
 is again continuous, it follows that the space
\[
E: = \{f \in C(\Omega ): \|f\|_{\infty}  < \infty\}
\]
is a Banach space.

If $\Omega$ is as above and $\Omega'$ is an
 open set with $\bar{\Omega} \subset \Omega'$, we let
\[
C(\bar{\Omega}):=\{\text{the  restriction to }\bar{\Omega} \text{ of } f \in
 C(\Omega')\}.
\]
 If $\Omega$ is bounded and $f \in C(\bar{\Omega})$,
 then $\|f\|_{\infty}  < + \infty$. Hence $C(\bar{\Omega})$ is a Banach space.
\end{example}


\begin{example}\label{cont3} \rm
Let $\Omega$ be an open subset of $\mathbb{R}^N$.
 Let $I = (i_1,\dots,i_N)$ be a multiindex\index{multiindex}, i.e.
 $i_k \in \mathbb{N}\cup \{0\}$ (the nonnegative integers),
 $1 \leq k \leq N$. We let $|I| = \sum_{k=1}^N i_k$.
 Let $f: \Omega \to \mathbb{K}$. Then the partial derivative of
$f$ of order $I $, $ D^{I}f(x)$, is given by
\[
D^{I}f(x): = \frac{\partial^{|I |}f(x)}{\partial^{i_1}x_1 \dots
\partial^{i_N}x_N},
\]
where $x = (x_1,\dots,x_N)$. Define
\[
C^j(\Omega ): = \{f:
 \Omega \to \mathbb{K}\text{ such that } D^{I}f \in C(\Omega),\;
|I| \leq j\}.
\]
Let
\[
\|f\|_j := \sum_{k=0}^j \max_{|I|\leq k} \|D ^{I}f\|_{\infty}.
\]
Then, using further convergence results for families of
 differentiable functions it follows that the space
\[
E: = \{f \in C^j(\Omega ): \|f\|_j  < + \infty\}
\]
is a Banach space.

The space $C^j(\bar{\Omega})$ is defined in a
 manner similar to the space $C(\bar{\Omega} )$
 and if $\Omega$ is bounded $C^j(\bar{\Omega})$ is a Banach space.
\end{example}
%exm 3.

\subsection{Completions}
\label{completions}
In this section we shall briefly discuss the concept of the
completion\index{completion} of a metric space and its application to
completing normed vector spaces.

\begin{theorem} \label{completion theorem}
 If $(\mathbb{M}, {\rm d})$ is a metric space, then there
exists a complete metric space  $(\mathbb{M} ^*, {\rm d}^*)$
and a mapping $h:\mathbb{M} \to \mathbb{M}^*$ such that
\begin{enumerate}
\item $h$ is an isometry\index{isometry}
${\rm d}^*(h(x),h(y))={\rm d}(x,y),\;x,y\in \mathbb{M})$
\item $h(\mathbb{M})$ is dense in $\mathbb{M}^*$.
\end{enumerate}
\end{theorem}

 We give a short sketch of the proof.  We let $C$ be the set of all
 Cauchy sequences in $\mathbb{M}$. We observe that if $\{x_n\}$ and
 $\{y_n\}$ are elements of $\mathbb{M}$, then $\{{\rm d}(x_n,y_n)\}$ is a
 Cauchy sequence (hence, convergent) sequence in $\mathbb{R}$, as
 follows from the triangle inequality. We define
${\rm d}_C:C\times C\to \mathbb{R}$,
by
\[
{\rm d}_C(\{x_n\},\{y_n\}):=\lim _{n\to \infty}
{\rm d}(x_n,y_n).
\]
The mapping ${\rm d}_C$ is a pseudo-metric\index{pseudo-metric} on $C$
(lacking only the condition
\[
{\rm d}_C (\{x_n\},\{y_n\})=0\;\Rightarrow \;\{x_n\}=\{y_n\}
\]
from the definition of a metric). The relation $R$ defined on $C$ by
\[
\{x_n\}R\{y_n\},\text{ if, and only if, } \lim _{n\to
\infty}{\rm d}(x_n,y_n)=0,
\]
or equivalently,
\[
{\rm d}_C(\{x_n\},\{y_n\})=0,
\]
is an equivalence relation\index{equivalence relation} on $C$. The
space $C/R$, the set of all equivalence classes\index{equivalence
classes} in $C$, shall be denoted by $\mathbb{M}^*$. If we denote
by $R\{x_n\}$, the class of all $\{z_n\}$ which are $R-$
equivalent to $\{x_n\}$, we may define
\[
{\rm d}^*(R\{x_n\},R\{y_n\}):={\rm d}_C(\{x_n\},\{y_n\}).
\]
This defines a metric on $\mathbb{M}^*$.

We next connect this to $\mathbb{M}$. There is a natural mapping of
$\mathbb{M}$ to $C$ given by
\[
x\mapsto \{x\}
\]
(the sequence, all of whose entries are the same element $x$). We
clearly
have
\[
{\rm d}_C(\{x\},\{y\})={\rm d}(x,y)
\]
and thus the mapping
$x\mapsto R\{x\}$
(which we now call $h$) is an isometry of $\mathbb{M}$ to $\mathbb{M}^*$. That the image $h(\mathbb{M})$ is dense in $\mathbb{M}^*$,
follows easily from the above construction.

\begin{remark} \rm
We observe that $(\mathbb{M}^*,{\rm d}^*)$ is ``essentially
unique''. For, if $(\mathbb{M}_1,{\rm d}_1)$ and $(\mathbb{M}_2,{\rm d}_2)$ are completions of $\mathbb{M}$ with mappings $h_1$,
respectively $h_2$, then there exists a mapping $g:\mathbb{M}_1 \to
\mathbb{M}_2$ such that
\begin{enumerate}
\item
$g$ is an isometry of $\mathbb{M}_1$ onto $ \mathbb{M}_2$,
\item  $h_2=g\circ h_1$.
\end{enumerate}
\end{remark}

The above is summarized in the following theorem, whose proof is
similar to the proof of the previous result (Theorem \ref{completion
theorem})
recalling that a norm defines a metric and where we define (using the
notation of that theorem)
\[
\|\{x_n\}\|_C:=\lim_{n\to \infty}\|x_n\|,
\]
for a given Cauchy sequence,
 and
\[
\|R\{x_n\}\|^*:=\|\{x_n\}\|_C.
\]
It is also clear that the set $X^*$ becomes a vector space by defining
addition and scalar multiplication in  a natural way.


\begin{theorem} \label{thm10}
If $(X,\|\cdot\|)$ is a normed vector space, then there exists a
(essentially unique) complete normed vector space (a Banach space)
 $(X^*,\|\cdot\|^*)$ and an isomorphism\index{isomorphism} (which is an isometry\index{isometry})
$h:X\to X^*$
such that $h(X)$ is dense in $X^*$.
\end{theorem}


\subsection{Lebesgue spaces}
\label{lebesgue}
In this section we shall discuss briefly Lebesgue
spaces\index{Lebesgue spaces} generated by spaces of continuous
functions whose domain is $\mathbb{R}^N,\;N\in \mathbb N$.
If
\[
f:\mathbb{R}^N\to \mathbb{K},\quad\mathbb{K}=\mathbb{R} ~\text{or}~\mathbb C
\]
we define the support\index{support} of $f$ to be the closed set
\[
\mathop{\rm supp}(f):=\overline{\{x:f(x)\ne 0\}}.
\]
We say that $f$ has compact support\index{compact support}
whenever supp$(f)$ is a compact (i.e., closed and bounded) set and
denote by $C_0(\mathbb{R}^N)$ the set of all continuous $\mathbb{K}-$ valued functions defined on $\mathbb{R}^N$ having compact
support. (More generally, if $\Omega $ is an open set in $\mathbb R^N,$ one denotes by $C_0^j(\Omega )$ the set of all $C^j-$ functions having compact support in $\Omega .$) This is a vector subspace of $C(\mathbb{R}^N)$, the space
of   all continuous $\mathbb{K}-$ valued functions defined on
$\mathbb{R}^N$.

We first need to define the Riemann integral\index{Riemann integral} on
 $C_0(\mathbb{R}^N)$.
To do this, without getting into too many details of this procedure,
we assume that the reader is familiar with this concept for the integral
defined on closed rectangular boxes
\[
B:=\{x=(\xi _1,\dots ,\xi _N):\alpha _i\leq \xi _i \leq \beta _i,\;
1\leq i\leq N\},
\]
($B=\prod _{i=1}^N[\alpha _i,\beta _i]$), where the numbers
$\alpha _i,\;\beta _i,\;1\leq i\leq N$, are fixed real numbers (for
each box). We observe that if $f\in C_0(\mathbb{R}^N)$ and if $B_1$
and $B_2$ are such boxes, each of which contains $\mathop{\rm supp}(f)$,
then
\[
\int_{B_1}f=\int_{B_1\cap B_2}f=\int_{B_2}f,
\]
$B_1\cap B_2$ also being a box containing $\mathop{\rm supp}(f)$. This allows us
to define the Riemann integral of $f$ over $\mathbb{R}^N$ by
\[
\int f\Big( =\int_{\mathbb{R}^N}f\Big):=\int_{B}f,
\]
where $B$ is any closed box containing $\mathop{\rm supp}(f)$.

The mapping
$f\mapsto \int f$
is a linear mapping (linear functional\index{linear functional})
from $C_0(\mathbb{R}^N)$ to $\mathbb{K}$, which, in addition,
satisfies
\begin{itemize}
\item if $f$ is non-negative on $\mathbb{R}^N$, then $\int f \geq 0$,
\item If $\{f_n\}$ is a sequence of non-negative functions in $C_0(\mathbb{R}^N)$
which is monotonically decreasing (pointwise) to zero, i.e.,
\[
f_n(x)\geq f_{n+1}(x),\;n=1,\dots, ~\lim_{n\to \infty}f_n(x)=0,\;x\in \mathbb{R}^N,
\]
then $\int f_n\to 0$.
\end{itemize}

\begin{definition} \rm  \label{def11} \rm
For $f\in C_0(\mathbb{R}^N)$ we define
\[
\|f\|_1:=\int |f|.
\]
\end{definition}

It is easily verified that $\|\cdot \|_1$ is a norm - called the
$L^1$-norm\index{$L^1-$ norm}- on $C_0(\mathbb{R}^N)$.
We now sketch the process for completing the normed vector space
$\left (C_0(\mathbb{R}^N), \|\cdot \|_1\right )$ in such a way that we may
regard the vectors in the completion as {\it functions} on $\mathbb{R}^N$.

\begin{definition}   \label{def12} \rm
A subset $S\subset \mathbb{R}^N$ is called a set of measure
zero\index{measure zero} provided that for any $\epsilon >0$ there exists
a sequence of boxes $\{B_n\}_{n=1}^{\infty}$ such that
\begin{gather*}
S\subset \cup _{n=1}^{\infty} B_n, \\
\sum _{n=1}^{\infty}\mathop{\rm vol}(B_n)<\epsilon ,
\end{gather*}
where $\mathop{\rm vol}(B)=\prod _{i=1}^N(\beta _i-\alpha _i)$ for the box
$B=\prod _{i=1}^N[\alpha _i,\beta _i]$.

We say that a property holds ``almost everywhere''\index{almost everywhere}
(``a.e.'') if the set of points at which it fails to hold has measure zero.
\end{definition}

The proofs of the following theorems may be found in a very complete
discussion of $C_0(\mathbb{R}^N)$ and its $L^1-$ completion
in \cite[Chapter 7]{hoffman:aes75}.

\begin{definition} \rm  \label{def13} \rm
A sequence $\{x_n\}$ in a normed vector space $(X,\|\cdot \|)$ is said to
be a fast Cauchy sequence\index{fast Cauchy sequence} if
\[
\sum_{n=1}^{\infty}\|x_{n+1}-x_n\|
\] converges.
\end{definition}

\begin{theorem} \label{thm14}
If $\{f_n\}$ is a fast Cauchy sequence in
$(C_0(\mathbb{R}^N),\|\cdot \|_1)$, then  $\{f_n\}$ converges pointwise a.e.
in $\mathbb{R}^N$.
\end{theorem}

\begin{definition} \rm  \label{def15} \rm
A Lebesgue integrable function on $\mathbb{R}^N$ is a function $f$ such
that:
\begin{itemize}
\item $f$ is a $\mathbb{K}$ valued function defined a.e. on $\mathbb{R}^N$,
\item there is a fast Cauchy sequence in $(C_0(\mathbb{R}^N),\|\cdot \|_1)$
which converges to $f$ a.e. in $\mathbb{R}^N$.
\end{itemize}
\end{definition}

\begin{theorem} \label{tm16}
If $f$ is a Lebesgue integrable function and if $\{f_n\}$ and
$\{g_n\}$ are fast Cauchy sequences in $(C_0(\mathbb{R}^N),\|\cdot
\|_1)$ converging a.e. to $f$, then
\[
\lim_{n\to \infty}\int f_n=\lim_{n\to \infty}\int g_n.
\]
\end{theorem}

In light of this result we may define $\int f$ by
\[
\int f:=\lim_{n\to \infty}\int f_n,
\]
where $\{f_n\}$ is any fast Cauchy sequence in $(C_0(\mathbb{R}^N),\|\cdot \|_1)$, converging a.e. to $f$ on $\mathbb{R}^N$. The
resulting map
\[
f\mapsto \int f
\]
 is then defined on the space of  all Lebesgue integrable functions
$L^1(\mathbb{R}^N)$ and is a linear functional on this space which also
satisfies
\[
|\int f|\leq \int |f|, \quad \forall f\in L^1(\mathbb{R}^N).
\]

\begin{theorem} \label{thm17}
The mapping
\[
f\mapsto \|f\|_1:=\int |f|,\quad f\in L^1(\mathbb{R}^N),
\]
is a seminorm on $L^1(\mathbb{R}^N)$, i.e., it satisfies all the
conditions of a norm, except that $\|f\|_1=0$ need not imply that
$f$ is the zero of $L^1(\mathbb{R}^N)$. Further $L^1(\mathbb{R}^N)$
is complete with respect to this seminorm and $C_0(\mathbb{R}^N)$
is a dense subspace of $L^1(\mathbb{R}^N)$.
\end{theorem}

Usually we identify two elements of $L^1(\mathbb{R}^N)$ which agree a.e.;
i.e., we define an equivalence relation on $L^1(\mathbb{R}^N)$
\[
f\sim g
\]
whenever the set $A\cup B$ has measure zero, where
\begin{gather*}
A:=\{x: f(x)~\text{or}~g(x)~\text{fail to be defined}\},\\
B:=\{x: f(x),g(x)~\text{are defined, but}~f(x)\ne g(x)\}.
\end{gather*}
This equivalence relation respects the operations of addition and scalar
multiplication and two equivalent functions have the same seminorm.
The vector space of all equivalence classes then becomes a complete
normed linear space (Banach space). This space, we again call
$L^1(\mathbb{R}^N)$,

\begin{remark} \rm
1. We again refer the reader to \cite{hoffman:aes75} for a complete
discussion of this topic and others related to it, e.g., convergence
theorems for Lebesgue integrals, etc.

2. The ideas above may equally well be employed to define integrals
on open regions $\Omega \subset \mathbb{R}^N$ starting with
\[
C_0(\Omega):=\{f\in C(\Omega ):\mathop{\rm supp}(f)
\text{ is a compact subset of } \Omega \}.
\]
The resulting space being $L^1(\Omega )$.

3. One also may imitate this procedure to obtain the other
Lebesgue spaces $L^p(\mathbb{R}^N),\;1\leq p <\infty $, by replacing
the original norm in $C_0(\mathbb{R}^N)$ by
\[
\|f\|_p:=\Big(\int |f|^p\Big)^{1/p},\quad f\in C_0(\mathbb{R}^N).
\]
And, of course, in  similar vein, one can  define
$L^p(\Omega ),\;1\leq p <\infty $.

4.
For given $f\in L^1(\mathbb{R} )$ define the functional $T_f$ on
$C_0^{\infty }(\mathbb{R} )$ as follows
\[
T_f(\phi ):=\int f\phi .
\]
The functional $T_f$ is called the {\it distribution}\index{distribution}
defined by $f$. More generally, the set of all linear functionals on
$C_0^{\infty }(\mathbb{R} )$
is called the set of distributions on $\mathbb{R} $ and if $T$ is such,
its {\it distributional derivative}\index{distributional derivative}
$\partial T$ is defined by
\[
\partial T(\phi ):=-T(\phi '), ~\forall \phi \in C_0^{\infty }(\mathbb{R} ),
\]
hence for $f\in L^1(\mathbb{R} )$ and $T_f$, the distribution
determined by $f$,
\[
 \partial T_f(\phi )=\int f \phi ', \quad
 \forall \phi \in C_0^{\infty }(\mathbb{R} ).
\]
We henceforth, for given $f\in L^1(\mathbb{R})$, we denote by $f$
the distribution $T_f$ determined by $f$, as well.

5. The Cartesian product
\[
E:=\prod _{i=1}^{2}L^1(\mathbb{R} )
\]
may be viewed as a normed linear space with norm defined as
\[
\|(u_1 , u_2)\|:=\sum _{i=1}^{2}\|u_i\|_1\quad
\forall u_i\in L^1(\Omega ),\;i=1,2,
\]
and the space $C^1_0(\mathbb{R} )$ may be viewed as a subspace of $E$ by
identifying $f\in  C^1_0(\mathbb{R})$ with $ ( f, f') $. The completion of
the latter  space in $E$ is called the {\it Sobolev space}\index{Sobolev space}
$W^{1,1}(\mathbb{R} )$. Where we think of $W^{1,1}(\mathbb{R})$ as a space of
 tuples of
$L^1$ functions. On the other hand, if $F=(f,g)$ is such an element, then
there exists a sequence $\{f_n\}\subset C^1_0(\mathbb{R} )$ such that
\[
f_n\to f, \quad f'_n\to g,
\]
with respect to the $L^1$ norm. It follows that
\begin{gather*}
\int f_n\phi \to \int f\phi, \quad
 \forall \phi \in C_0^{\infty }(\mathbb{R} ),\\
\int f'_n\phi \to \int g\phi, \quad
 \forall \phi \in C_0^{\infty }(\mathbb{R} ).
\end{gather*}
On the other hand, using integration by parts,
\[
\int f'_n\phi =-\int f_n\phi '\to -\int f\phi '
\]
and therefore
\[
-\int f\phi '=\int g\phi ,\quad
\forall \phi \in C_0^{\infty }(\mathbb{R} ).
\]
I.e., in the sense of distributions $\partial f=g $. This may be
summarized as follows: The space $W^{1,1}(\mathbb{R})$ is the set of
all $L^1$ functions whose distributional derivatives are $L^1$ functions,
as well.

If, instead of the $L^1$ norm, we use the $L^2$ norm in the above
process, one obtains the space $W^{1,2}(\mathbb{R})$ which is
usually denoted by $H^1(\mathbb{R})$. Using $L^p$ as an underlying
space, one may define the Sobolev spaces $W^{1,p}(\mathbb{R})$, as
well. In the case of functions of $N$ variables and open regions
$\Omega \subset \mathbb{R}^N $, analogous procedures are used to
define the Sobolev spaces $W^{1,p}(\Omega )$. Of particular
interest to us later in these notes will be  the space
$H^1_0(\Omega )$ which is the closure in $H^1(\Omega )$ of the
space $C^{\infty}_0(\Omega )$. We refer the interested reader to
the book by Adams \cite{adams:ss75} for detailed developments and
properties of Sobolev spaces.
\end{remark}


\subsection{The Hausdorff metric}
\label{haus} Let $\mathbb{M}$ be a  metric space with metric
${\rm d}$. For $x\in \mathbb{M}$ and $\delta >0$, we set,
as before,
\begin{gather*}
B(x,\delta ):=\{y\in \mathbb{M}:{\rm d}(x,y)<\delta \},\\
{B}[x,\delta ]:=\{y\in \mathbb{M}:{\rm d}(x,y)\leq\delta \},
\end{gather*}
 the open and closed balls with center
at $x$ and radius $\delta $. As pointed out before, the closed ball is closed,
but need not
be the closure of the open ball.

Let $A$ be a nonempty
closed subset of $\mathbb{M}$. For $\delta >0$ we
define
\begin{align*}
A_{\delta}:&=\cup \{{B}[y,\delta ]:y\in A\}\\
&=\{x\in \mathbb{M}:{\rm d}(x,y)\leq \delta,\text{ for some }
y\in A\}.
\end{align*}
We observe that
\[
A_{\delta}\subset \{x\in \mathbb{M}:{\rm d}(x,A)\leq \delta\},
\]
where
\[
{\rm d}(x,A):=\inf \{{\rm d}(x,a):a\in A\}.
\]
If $A$ is  compact these sets are equal; if $A$ is not compact,  the
containment  may be proper.

\begin{definition} \rm
 We let
\[
\mathcal{H}:=\mathcal{H} ({\mathbb {M}}) =\{A\subset \mathbb{M}:A\ne
 \emptyset,\; A\text{ is closed and bounded }\}.
\]
For each pair of sets
 $A,B$ in  $ \mathcal{H}$
we define
\begin{gather}
\label{hmetric1}
{\rm D}_1(A,B):=\sup\{{\rm d}(a,B):a\in A\}, \\
\label{hmetric2}
{\rm D}_2(A,B):=\inf\{\epsilon >0:A\subset B_{\epsilon},\}.
\end{gather}
\end{definition}
It is a straightforward exercise to prove the following proposition.

\begin{proposition}
For $A,B\in \mathcal{H}$,
 ${\rm {D}_1}(A,B)={\rm {D}_2}(A,B)$.
\end{proposition}
We henceforth denote the common value
\begin{equation}
\label{hmetric21}
{\rm D}(A,B):={\rm D}_1(A,B)={\rm D}_2(A,B).
\end{equation}

\begin{proposition}
For $A,B\in \mathcal{H}$  let
$ {\rm h}:\mathcal{H}\times \mathcal{H}\to [0,\infty )$
be defined  by
\begin{equation}
\label{hmetric3}
{\rm h}(A,B):={\rm D}(A,B)\vee {\rm D}(B,A)
:=\max\{{\rm D}(A,B),{\rm D}(B,A)\}.
\end{equation}
Then ${\rm h}$ is a metric on $\mathcal{H}$ (the Hausdorff
metric\index{Hausdorff metric}).
\end{proposition}

We briefly sketch the proof.

That ${\rm h}$  is symmetric with
respect to its arguments and that
\[
{\rm h}(A,B)=0
\]
if, and only if,
$A=B$, follow easily.

To verify that the triangle inequality holds, we let $A,B,C\in
\mathcal{H}$
and let
\[
x\in A,\;y\in B,\;z\in C .
\]
Then
\[
{\rm d}(x,z)\leq {\rm d}(x,y)+{\rm d}(y,z),
\]
and hence,
\[
{\rm d}(x,C)\leq {\rm d}(x,y)+{\rm d}(y,z),\;\forall y\in B,\;\forall z\in C .
\]
Therefore,
\[
{\rm d}(x,C)\leq {\rm d}(x,B)+{\rm D}(B,C),
\]
which implies that
\[
{\rm D}(A,C)\leq {\rm D}(A,B)+{\rm D}(B,C)\leq {\rm h}(A,B)+{\rm h}(B,C),
\]
and similarly
\[
{\rm D}(C,A)\leq {\rm h}(A,B)+{\rm h}(B,C).
\]

The following corollary, which is an immediate consequence of the
 definitions,
 will be of use later.

\begin{proposition} \label{dh-relation}
Let $A,B\in \mathcal{H}$, $a\in A$, and $\eta >0$ be given. Then
there exists $b\in B$ such that
\[
{\rm d}(a,b) \leq {\rm h}(A,B) +\eta
\]
\end{proposition}

The following examples will serve to illustrate the computation of the
 Hausdorff distance between two closed sets.

\begin{example}\rm  Let
\[
A:=[0,1]\times \{0\},\quad B:=\{0\}\times [1,2]\subset {\mathbb{R}^2},
\]
then
\[
{\rm D}(A,B)=\sqrt {2},\quad
{\rm D}(B,A)= {2},
\]
so
${\rm h}(A,B)=2$.
\end{example}

\begin{example}\rm
Let
\[
A:={B}[a,r],\quad B:={B}[b,s],\quad a,b\in \mathbb{M},\quad 0<r\leq s,
\]
then
\[
{\rm h}(A,B)=d+s-r,
\]
where
$d={\rm d}(a,b)$.
\end{example}

There is a natural mapping associating points of $\mathbb{M}$ with
elements
of $\mathcal{H}$ given by
\[
x\mapsto \{x\}.
\]
This mapping, as one easily verifies, is an isometry, i.e.,
\[
{\rm d}(x,y)={\rm h}(\{x\},\{y\}),\;\forall x,y \in \mathbb{M}.
\]

We next establish that
  $(\mathcal{H},{\rm h})$ is a complete metric space, whenever
  $(\mathbb{M},{\rm d})$ is a complete metric space
(see also \cite{dugundji:t73},
which contains many very good exercises concerning the
Hausdorff metric and the hierarchy of metric spaces constructed in the
above manner).

Let $\{A_n\}\subset \mathcal{H}$ be a sequence of sets such that
\[
{\rm h}(A_n,A_{n+1})<2^{-n},\quad n=1,2, \dots .
\]
We call a sequence $\{x_n\}\subset {\mathbb{M}},\;x_n\in
A_n,\;n=1,2,\dots $ a {\it fast convergent}\index{fast convergent}
 sequence, provided that
\[
{\rm d}(x_n,x_{n+1})<2^{-n},\;n=1,2, \dots .
\]
 We have the following lemma whose proof follows immediately from
the definition of the Hausdorff metric.

\begin{lemma}
Let  $(\mathbb{M},{\rm d})$ be a complete metric space and
$\{A_n\}\subset \mathcal{H}$ be a sequence of sets such that $
{\rm h}(A_n,A_{n+1})<2^{-n},\;n=1,2, \dots . $ If $j$ is a given
positive integer and $y\in A_j $, then there exists a fast
convergent sequence $\{x_n\}\subset {\mathbb{M}},\;x_n\in
A_n,\;n=1,2,\dots $ with $x_j=y $.
\end{lemma}

To see the above one proceeds as follows: Let $x_i=y $ and induct
backwards to $x_1$ and then induct forward through $x_{i+1}, \dots $.

\begin{theorem} \label{thm26}
If  $(\mathbb{M},{\rm d})$ is a complete metric space, then
$(\mathcal{H},{\rm h})$ is a complete metric space.
\end{theorem}

\begin{proof}
Let  $\{A_n\}\subset \mathcal{H}$ be a Cauchy sequence. Then, by passing
to a subsequence, we may assume that
\[
{\rm h}(A_n,A_{n+1})<2^{-n},\quad n=1,2, \dots .
\]
Let
\[
A:=\{x\in \mathbb{M}:x=\lim _{i\to \infty }x_i\},
\] where $\{x_i\}\subset \mathbb{M},\;x_i\in A_i,\;i=1,2,\dots $,
is a fast convergent sequence. We claim that the closure of $A$,
$\overline A$, is an element of $\mathcal{H}$ and
\[
A_i\to \overline A .
\]
with respect to the Hausdorff metric.
 To establish that $\overline A$ is an element of $\mathcal{H}$,
it suffices to show that $A$ is a bounded set. Thus let $x,y\in A$ and
 let
$\{x_i\}\subset \mathbb{M},\;x_i\in A_i,\;i=1,2,\dots $, and
$\{y_i\}\subset \mathbb{M},\;y_i\in A_i,\;i=1,2,\dots $ be fast
 convergent sequences with
\[
\lim _{i\to \infty }x_i=x,\quad \lim _{i\to \infty }y_i=y.
\]
Then
\[
{\rm d}(x,x_1)=\lim _{n\to \infty}{\rm d}(x_n,x_1)\leq \lim _{n\to
\infty}
\sum_{k=1}^{k=n-1}{\rm d}(x_k,x_{k+1})<1,
\]
and similarly
${\rm d}(y,y_1)<1$.
Hence
\[
{\rm d}(x,y)\leq {\rm d}(x,x_1)+{\rm d}(y,y_1)+{\rm d}(x_1,y_1) ,
\]
or
\[
{\rm d}(x,y)\leq 2+\sup \{{\rm d}(v,w):v,w\in A_1\} <\infty .
\]

We next note that
${\rm h}(A_n,\overline{A})\to 0$,
if, and only if,
\[
{\rm D}(A_n,\overline{A})\to 0 \quad\text{and}\quad
{\rm D}(\overline{A},A_n)\to 0,
\]
which, in turn, is equivalent to
\[
\sup _{y\in A_n}{\rm d}(y,{A})\to 0 \quad\text{and}\quad
\sup _{z\in A}{\rm d}({A},A_n)\to 0.
\]
For given $y\in A_n$, there exists a fast convergent sequence
$\{x_i\},\;x_n=y,\;x_i\in A_i$, with, say, $x_i\to x \in A $. Hence,
\[
{\rm d}(y,A)\leq {\rm d}(y,x)={\rm d}(x_n,x)\leq 2^{-n+1},
\]
so $\sup _{y\in A_n}{\rm d}(y,{A})\to 0 $, as $n\to \infty
$.

 Let $z\in
A $. Then there exists a fast convergent sequence $\{x_i\},\;x_i\in
A_i$, with, say $x_i\to z $. Thus, for each $n=1,2, \dots $,
\[
{\rm d}(z,A_n)\leq {\rm d}(z,x_n)\leq 2^{-n+1},
\]
and consequently
$\sup _{z\in A}{\rm d}(z,A_n)\to 0$ as $n\to \infty$.
\end{proof}

As a consequence of this result we also obtain the following theorem.

\begin{theorem} \label{thm27}
If  $(\mathbb{M},{\rm d})$ is a  metric space which is compact, then
$(\mathcal{H},{\rm h})$ is a compact metric space.
\end{theorem}

\begin{proof}

Since $\mathbb{M}$ is compact it is
complete
(see, e.g., \cite{royden:ra88}). It follows from the
previous theorem that
$\mathcal{H}$ is complete. Thus we need to show that $\mathcal{H}$ is
totally bounded\index{totally bounded}.

Fix $\epsilon >0, $ and choose $0<\delta <\epsilon $. Since
$\mathbb{M}$
is compact, there exists a finite subset $S\subset \mathbb{M}$
such that
\[
\mathbb{M}=\cup \{B(x,\delta ):x\in S\}.
\]
If we denote by $\mathcal{S}:=2^S\backslash {\emptyset}$, the set
of nonempty subsets of $S$, then $\mathcal{S}$ is a finite set and
one can easily show that
\[
\mathcal{H}=\cup \{B(A,\epsilon ):A\in \mathcal{S}\},
\]
where $B(A,\epsilon )$ is the ball, centered at $A\in \mathcal{H}$
with Hausdorff metric radius $\epsilon $. Hence $\mathcal{H}$ is totally
bounded and, since complete, also compact.
\end{proof}

\section{Contraction mappings} \label{chapIII} % Chapter III \include{contmap}

Let $(\mathbb{M},{\rm d})$ be a complete metric space and let
\[
T:\mathbb{M}\to \mathbb{M}
\]
be a mapping. We call $T$ a {\it Lipschitz
mapping}\index{Lipschitz mapping} with {\it Lipschitz constant}
$k\geq 0$, provided that
\begin{equation}
\label{lip}
{\rm d}(T(x),T(y))\leq k{\rm d}(x,y),\;\forall x,y\in \mathbb{M}.
\end{equation}
We note that Lipschitz mappings are necessarily continuous
mappings and that the product of two Lipschitz  mappings (defined
by composition of mappings) is again a Lipschitz mapping. Thus for
a Lipschitz mapping $T$, and  for all positive integers $n$, the
mapping $T^n=T\circ \dots \circ T$, the mapping $T$ composed with
itself $n $ times, is a  Lipschitz mapping, as well. We call a
Lipschitz mapping $T$ a {\it nonexpansive}\index{nonexpansive}
mapping provided the constant $k$ may be chosen so that $k\leq 1$,
 and a {\it contraction
mapping}\index{contraction mapping} provided the Lipschitz constant
$k$
may be chosen so that
 $0\leq k<1$.
In this case the Lipschitz constant $k$ is also called the  {\it contraction
constant}\index{contraction constant} of $T$.

\subsection{The contraction mapping principle}
\label{tcmp}
In this section we shall discuss the {\it contraction mapping
principle}\index{contraction mapping principle} or what is often
also called the {\it Banach fixed point theorem}\index{Banach fixed
point theorem}.  We shall also give
some extensions and examples.

We have the following theorem.

\begin{theorem} \label{contraction}
Let $(\mathbb{M},{\rm d})$ be a complete metric space and
let $T:\mathbb{M}\to \mathbb{M}$ be a contraction mapping with
contraction constant $k$. Then $T$ has a unique fixed point $x\in
\mathbb{M}$. Furthermore, if $y\in \mathbb{M}$ is arbitrarily
chosen, then the iterates $\{x_n\}_{n=0}^{\infty}$, given by
\begin{gather*}
x_0 = y\\
x_n = T(x_{n-1}),\;n\geq 1,
\end{gather*}
have the property that
$\lim _{n\to \infty}x_n=x$.
\end{theorem}

\begin{proof}
Let $y\in \mathbb{M}$ be an arbitrary point of $\mathbb{M}$ and
consider the sequence  $\{x_n\}_{n=0}^{\infty}$, given by
\begin{gather*}
x_0 = y\\
x_n = T(x_{n-1}),\;n\geq 1.
\end{gather*}
We shall prove that $\{x_n\}_{n=0}^{\infty}$ is a Cauchy sequence in
$\mathbb{M}$.
For $m<n$ we use the triangle inequality and note that
\[
{\rm d}(x_m,x_n)\leq {\rm d}(x_m,x_{m+1})
+{\rm d}(x_{m+1},x_{m+2})+\dots
+{\rm d}(x_{n-1},x_{n}).
\]
Since $T$ is a  contraction mapping, we have that
\[
{\rm d}(x_p,x_{p+1})={\rm d}(T(x_{p-1}),T(x_p))
\leq k{\rm d}(x_{p-1},x_p),
\]
for any integer $p\geq 1$. Using this inequality repeatedly, we obtain
\[
{\rm d}(x_p,x_{p+1})\leq k^{p}{\rm d} (x_0,x_1);
\]
 hence,
\[
{\rm d}(x_m,x_n)\leq \left (k^m+k^{m+1}+\dots +k^{n-1}\right ){\rm d}
(x_0,x_1),
\]
i.e.,
\[
{\rm d}(x_m,x_n)\leq \frac{k^m}{1-k}{\rm d}
(x_0,x_1),
\]
whenever $m\leq n $.
 From this we deduce that  $\{x_n\}_{n=0}^{\infty}$ is a Cauchy sequence in
$\mathbb{M}$. Since $\mathbb{M} $ is complete, this sequence has a limit, say $x\in
\mathbb{M}$.
On the other hand, since $T$ is continuous, it follows that
\[
x=\lim_ {n\to \infty }x_n=\lim_{n\to \infty }T(x_{n-1})=T(\lim_ {n\to
\infty }x_{n-1})=T(x),
\]
and, thus, $x$ is a fixed point of $T$.

If $x$ and $z$ are both fixed points of $T$, we get
\[
{\rm d}(x,z)={\rm d} (T(x),T(z))\leq k{\rm d} (x,z).
\]
Since $k<1$, we must have that $x=z$.
\smallskip

The following is an alternate proof.
It follows (by induction) that for any $x\in \mathbb{M}$ and any natural number
  $m$
\[
{\rm d}(T^{m+1}(x),T^{m}(x))\leq k^m{\rm d}(T(x),x).
\]
 Let
\[ \alpha :=\inf _{x\in \mathbb{M}}{\rm d}(T(x),x).
\]
Then, if $\alpha >0$, there exists ${x\in \mathbb{M}}$ such that
\[
{\rm d}(T(x),x)<\frac{3}{2}\alpha
\]
and hence for any $m$
\[
{\rm d}(T^{m+1}(x),T^{m}(x))\leq k^m\frac{3}{2}\alpha .
\]
On the other hand,
\[
\alpha \leq {\rm d}(T(T^m(x)),T^{m}(x))={\rm d}(T^{m+1}(x),T^{m}(x))
\]
and thus, for any $m\geq 1$,
\[
\alpha \leq k^m\frac{3}{2}\alpha
\]
which is impossible, since $k<1$. Thus $\alpha =0$.

 We choose a sequence $\{x_n\}$ (a minimizing sequence) such that
\[
\lim _{n\to \infty }{\rm d}(T(x_n),x_n))=\alpha =0.
\]
For any $m,n$ the triangle inequality implies that
\[
{\rm d}(x_n,x_m)\leq {\rm d}(T(x_n),x_n))+{\rm d}(T(x_m),x_m))
+{\rm d}(T(x_n),T(x_m)).
\]
And hence
\[
(1-k){\rm d}(x_n,x_m)\leq {\rm d}(T(x_n),x_n))+{\rm d}(T(x_m),x_m)).
\]
Which implies that  $\{x_n\}$ is a Cauchy sequence and hence has a
limit $x$ in $\mathbb{M}$. One now concludes that
\[
{\rm d}(T(x),x)=0
\]
and thus $x$ is a fixed point of $T$.
\end{proof}

It may be the case that $T:\mathbb{M} \to \mathbb{M}$ is not a
contraction on the whole space $\mathbb{M}$, but rather a
contraction on some neighborhood of a given point. In this case we
have the following result:

\begin{theorem}\label{bcontraction}
Let $(\mathbb{M},{\rm d})$ be a complete metric space and let
\[
B=\{x\in \mathbb{M}:{\rm d} (z,x)<\epsilon\},
\]
where $z\in \mathbb{M}$ and $\epsilon >0$ is a positive number
and let $T:B\to \mathbb{M}$ be a
 mapping
such that
\[
{\rm d} (T(y),T(x))\leq k{\rm d} (x,y),\;\forall ~  x,y\in B,
\]
with contraction constant $k<1$. Furthermore assume that
\[
{\rm d}(z,T(z))<\epsilon (1-k).
\]
Then $T$ has a
unique fixed point $x\in B$.
\end{theorem}

\begin{proof}
 While the hypotheses do not assume that $T$
is defined on the closure $\overline {B}$ of $B$, the uniform
continuity of $T$ allows us to extend $T$ to a mapping defined on
$\overline {B}$ which is a contraction mapping having the same
Lipschitz constant as the original mapping.  We also note that for
$x\in \overline { B}$,
\[
{\rm d} (z,T(x))\leq {\rm d} (z,T(z))
 +{\rm d} (T(z),T(x))
<\epsilon(1-k)+k\epsilon=\epsilon,
\]
and hence
$T: \overline{ B}\to  B$.
Hence, by Theorem \ref{contraction},
since $ \overline{ B}$ is a complete metric space, $T$ has a unique
fixed point in $\overline { B}$ which, by the above calculations, must,
in fact, be in $B$.
\end{proof}


\subsection{Some extensions}\label{se}

\begin{example}\rm  Let us consider the space
\[
\mathbb{M}=\{x\in \mathbb{R}:x\geq 1\}
\]
 with metric
\[
{\rm d} (x,y)=|x-y|,\quad \forall x,y\in \mathbb{M},
\]
and let
$T:\mathbb{M}\to \mathbb{M}$
be given by
\[
T(x):=x+\frac{1}{x}.
\]
Then, an easy computation shows that
\[
{\rm d} (T(x),T(y))=\frac{xy-1}{xy}|x-y|<|x-y|={\rm d} (x,y).
\]
On the other hand, there does not exist $0\leq k<1$
such that
\[
{\rm d} (T(x),T(y))\leq k{\rm d} (x,y),\quad
\forall x,y\in \mathbb{M},
\]
and one may verify that $T$ has no fixed points in $\mathbb{M}$.
\end{example}

This shows that if we replace the assumption of the theorem that
$T$ be a contraction mapping by the less restrictive hypothesis that
\[
{\rm d} (T(x),T(y))<{\rm d} (x,y),\;\forall x,y\in \mathbb{M},
\]
then $T$ need not have a fixed point.
On the other hand, we have the following result of Edelstein
\cite{edelstein:ebc61}
(see also \cite{edelstein:fpp62}):

\begin{theorem}\label{wcontraction}
Let $(\mathbb{M},{\rm d})$ be a  metric space and let $T:\mathbb{M}\to \mathbb{M}$ be a
 mapping such that
\[
{\rm d} (T(x),T(y))<{\rm d} (x,y),\quad
\forall x,y\in \mathbb{M},\;x\ne y.
\]
 Furthermore assume that there exists $z\in \mathbb{M}$ such that
 the iterates $\{x_n\}_{n=0}^{\infty}$, given by
\begin{gather*}
x_0=z\\
x_n=T(x_{n-1}),\;n\geq 1,
\end{gather*}
have the property that there exists a subsequence
$\{x_{n_j}\}_{j=0}^{\infty}$ of $\{x_n\}_{n=0}^{\infty}$, with
\[
\lim_{j\to \infty}x_{n_j}=y\in \mathbb{M}.
\]
Then $y$ is a fixed point of $T$ and this fixed point is unique.
\end{theorem}

\begin{proof}
 We note  from the definition of the iteration
process that we may write
\[
x_n=T^n(x_0),
\]
where, as before,  $T^n$ is the mapping $T$ composed with itself $n$ times.
We abbreviate by
\[
y_j=T^{n_{j}}(x_0)=
T^{n_{j}}(z),
\]
where the sequence $\{n_{j}\}$ is given by the theorem.
 Let us assume $T$ has no fixed points.
Then the function
$f:\mathbb{M} \to \mathbb{R}$
defined by
\[
x\mapsto \frac{{\rm d}(T^2(x),T(x))}{{\rm d}(T(x),x)}
\]
is a continuous function.
 Since the sequence
$\{y_j\}_{j=1}^{\infty}$ converges to $y$, the set $K$ given by
\[
K=\{y_j\}_{j=1}^{\infty}\cup \{y\}
\]
 is compact and, hence, its image under $f$ is compact.

On the other hand, since,
\[
f(x){\rm d}
(T(x),x)={\rm d} (T^{2}(x),T(x))
<{\rm d}
(T(x),x),\;\forall ~x\in \mathbb{M},
\]
it follows that $f(x)<1,\;\forall~ x\in \mathbb{M}$ and, since $K$
is compact, there exists a
positive  constant
 $ k <1$ such that
\[
f(x)\leq k,\quad \forall x\in K.
\]
We now observe that for any positive integer $m$ we have that
\[
{\rm d}(T^{m+1}(z),T^m(z))
=\Big(\prod _{i=0}^{m-1}f(T^i(z))\Big){\rm d} (T(z),z).
\]
Hence, for $m=n_j$, we have
\[
{\rm d}(T(T^{n_j}(z)),T^{n_j}(z))= \Big(\prod
_{i=0}^{n_j-1}f(T^{i}(z))\Big)
{\rm d} (T(z),z),
\]
and, since $ f(T^{i}(z))\leq k <1$, we obtain that
\[
{\rm d}(T(y_j),y_j)\leq k^{j-1}{\rm d} (T(z),z).
\]
On the other hand, as $j\to \infty $,
$y_j\to y$
 and by continuity
\[
T(y_j)\to T(y),
\]
and also
\[
 k^{j-1}\to 0,
\]
we obtain a contradiction to the assumption that
$T(y)\ne y$.
\end{proof}

The above result has the following important consequence.

\begin{theorem} \label{ccontraction}
Let $(\mathbb{M},{\rm d})$ be a  metric space and let $T:\mathbb{M}\to \mathbb{M}$ be a
 mapping such that
\[
{\rm d} (T(x),T(y))<{\rm d} (x,y),\quad
\forall x,y\in \mathbb{M},\;x\ne y.
\]
Further assume that
\[
T:\mathbb{M}\to K,
\]
where $K$ is a compact subset of $\mathbb{M}$.
Then $T$ has a unique fixed point in $\mathbb{M}$.
\end{theorem}

\begin{proof}
 Since $K$ is compact, it follows that for every
$z\in \mathbb{M}$ the sequence $\{T^n(z)\}$ has a convergent subsequence.
Hence Theorem \ref{wcontraction} may be applied.

A direct way of seeing the above is the following. By hypothesis we
have that
$T:K\to K$,
and the function
\[
x\mapsto {\rm d}(T(x),x)
\]
is a continuous function on $K$ and must assume its minimum, say,
at a point $y\in K$. If $T(y) \ne y$, then
\[
{\rm d}(T^2(y),T(y))<{\rm d}(T(y),y),
\]
 contradicting that ${\rm d}(T(y),y) $ is the minimum value.
Thus $T(y) = y$.
\end{proof}

In some applications it is the case that the mapping $T$ is a
Lipschitz mapping which is not necessarily a contraction,
 whereas some power of $T$
is a contraction mapping (see e.g. the result of \cite{weissinger:tai52}).
In this case we have the following theorem.

\begin{theorem} \label{pcontraction}
Let $(\mathbb{M},{\rm d})$ be a complete metric space and let
$T:\mathbb{M}\to \mathbb{M}$ be a
 mapping such that
\[
{\rm d} (T^m(x),T^m(y))\leq k{\rm d} (x,y),\quad
\forall x,y\in \mathbb{M},
\]
for some $m\geq 1$, where $0\leq k <1$ is a constant. Then $T$ has
a unique fixed point in $\mathbb{M}$.
\end{theorem}

\begin{proof}
It follows from Theorem \ref{contraction} that $T^m$ has a unique
fixed point $z\in \mathbb{M}$. Thus
\[
z=T^m(z)
\]
implies that
\[
T(z)=TT^m(z)=T^m(T(z)).
\]
Thus $T(z)$ is a fixed point of $T^m$ and hence by uniqueness of such
fixed points
$z=T(z)$.
\end{proof}

\begin{example}\rm
 Let the metric space $\mathbb{M}$ be given by
\[
\mathbb{M} =C[a,b],
\]
 the set of continuous real valued functions defined on the compact
interval
$[a,b]$. This set is a Banach space with respect to the norm
\[
\|u\| =\max _{t\in [a,b]}|u(t)|,\quad u\in \mathbb{M}.
\]
We define
$T:\mathbb{M}\to \mathbb{M}$
by
\[
T(u)(t)=\int _a^tu(s)ds.
\]
Then
\[
\|T(u)-T(v)\|\leq (b-a)\|u-v\|.
\]
(Note that $b-a$ is the best possible Lipschitz constant for $T.$)
On the other hand, we compute
\[
T^2(u)(t)=\int _a^t\int _a^su(\tau)d \tau ds=\int _a^t(t-s)u(s)ds
\]
and, inductively,
\[
T^m(u)(t)=\frac{1}{(m-1)!}\int _a^t(t-s)^{m-1}u(s)ds.
\]
It hence follows that
\[
\|T^m(u)-T^m(v)\|\leq \frac{(b-a)^m}{m!}\|u-v\|.
\]
It is  therefore the case that $T^m$ is a contraction mapping
 for all values of $m$ for which
\[
\frac{(b-a)^m}{m!}<1.
\]
It, of course, follows that $T$ has the unique fixed point
$u=0$.
\end{example}

\subsection{Continuous dependence upon parameters}
\label{cdup}
It is often the case in applications that a contraction mapping
depends
upon other variables (parameters) also. If this dependence is continuous, then
the fixed point will depend continuously upon the parameters, as well.
This is the content of the next result.

\begin{theorem} \label{parameter}
 Let $(\Lambda , {\rm \rho})$ be a metric space and
 $(\mathbb{M},{\rm d})$  a complete metric space and let
 \[
T:\Lambda \times \mathbb{M}\to \mathbb{M}
\] be a family of
contraction mappings with uniform contraction constant $k$, i.e.,
\[
{\rm d}\left (T(\lambda ,x),T(\lambda ,y)\right )\leq k{\rm d}(x,y),\;
\forall \lambda \in \Lambda,\;\forall x,y \in \mathbb{M} .
\]
Further more assume that for each $x\in \mathbb{M}$ the mapping
\[
\lambda \mapsto T(\lambda ,x)
\]
is a continuous mapping from $\Lambda$  to $\mathbb{M}$. Then for
each $\lambda \in \Lambda $, $T(\lambda ,\cdot )$ has a unique
fixed point $x(\lambda )\in \mathbb{M}$, and the mapping
\[
\lambda \mapsto x(\lambda ),
\]
is a continuous mapping from $\Lambda $ to $\mathbb{M}$.
\end{theorem}

\begin{proof}
The contraction mapping principle may be applied for each $\lambda
\in \Lambda $, therefore the mapping
$\lambda \mapsto x(\lambda )$,
is well-defined.
For $\lambda _1, \lambda _2 \in \Lambda $ we have
\begin{align*}
{\rm d}\left (x(\lambda _1), x(\lambda _2)\right )
&= {\rm d}\left (T(\lambda _1,x(\lambda _1)), T(\lambda _2,x(\lambda _2))
\right )\\
&\leq {\rm d}\left (T(\lambda _1,x(\lambda _1)),
T(\lambda _2,x(\lambda _1)) \right )
 +{\rm d}\left (T(\lambda _2 ,x(\lambda _1)),
  T(\lambda _2,x(\lambda _2)) \right )\\
&\leq {\rm d}\left (T(\lambda _1,x(\lambda _1)),
 T(\lambda _2,x(\lambda _1)) \right )
+k{\rm d}\left (x(\lambda _1)), x(\lambda _2)
\right ).
\end{align*}
Therefore
\[
(1-k){\rm d}\left (x(\lambda _1), x(\lambda _2)\right )
\leq {\rm d}\left (T(\lambda _1,x(\lambda _1)), T(\lambda _2,x(\lambda
_1))\right ).
\]
The result thus follows from the continuity of $T$ with respect to
$\lambda $
for each fixed $x$.
\end{proof}

\subsection{Monotone Lipschitz mappings}\label{mlm}

In this section we shall assume that $\mathbb{M}$ is a
Banach space with norm $\|\cdot \|, $ which also a Hilbert
space\index{Hilbert space}, i.e, that $\mathbb{M}$ is  an inner
product space (over the field of complex numbers) (see
\cite{rudin:rca66}, \cite{schechter:pfa71}) with inner product
$(\cdot , \cdot )$, related to the norm by
\[
\|u\|^2=(u,u),\quad \forall u\in \mathbb{M}.
\]
We call a mapping
$T:\mathbb{M}\to \mathbb{M}$,
a {\it monotone mapping}\index{monotone mapping} provided that
\[
{\rm Re}\left ((T(u)-T(v),u-v)\right )\geq 0,\quad
\forall u,v\in \mathbb{M} ,
\]
where
${\rm Re( c)}$ denotes the real part of a complex number $c$.

The following theorem, due to Zarantonello (see \cite{saaty:mne81}),
gives the existence of unique fixed points of mappings which are  perturbations of the identity mapping by monotone
Lipschitz mappings, without the assumption that they be contraction
mappings.

\begin{theorem} \label{monotone}
 Let $\mathbb{M}$ be a Hilbert space and let
\[
T:\mathbb{M}\to \mathbb{M},
\]
be a monotone mapping such that for some constant $\beta >0$
\[
\|T(u)-T(v)\|\leq \beta\|u-v\|,\quad \forall u,v\in \mathbb{M} .
\]
Then for any $w\in \mathbb{M} $, the equation
\begin{equation}
\label{monotone1}
u+T(u)=w
\end{equation}
has a unique solution $u=u(w)$, and the mapping
$w\mapsto u(w)$
is continuous.
\end{theorem}

\begin{proof}
If $\beta <1$, then the mapping
\[
u\mapsto w-T(u),
\]
is a contraction mapping and the result  follows from the
contraction mapping principle. Next, consider the case that $\beta
\geq 1$. We note that for $\lambda \ne 0$, $u$ is a solution of
\begin{equation}
\label{monotone2}
u=(1-\lambda )u -\lambda T(u)+\lambda w,
\end{equation}
if, and only if, $u$ solves (\ref{monotone1}). Let us denote by
\[
T_{\lambda}(u)=(1-\lambda )u -\lambda T(u)+\lambda w .
\]
It follows that
\[
T_{\lambda}(u)-T_{\lambda} (v)=(1-\lambda )(u-v) -\lambda (T(u)-T(v)).
\]
Using properties of the inner product, we obtain
\begin{align*}
\| T_{\lambda}(u)-T_{\lambda} (v)\|^2
&\leq \lambda ^2\beta ^2\|u-v\|^2
 -2{\rm Re}\left (\lambda (1-\lambda )(T(u)-T(v),u-v)\right )\\
&\quad  +(1-\lambda )^2\|u-v\|^2.
\end{align*}
Therefore, if $0<\lambda <1$, the monotonicity of $T$ implies that
\[
\| T_{\lambda}(u)-T_{\lambda} (v)\|^2\leq (\lambda ^2\beta ^2+(1-\lambda )^2)\|u-v\|^2.
\]
Choosing
\[
\lambda =\frac{1}{\beta ^2+1},
\]
We obtain that
 $T_\lambda $ satisfies a Lipschitz condition with Lipschitz constant
$k$ given by
\[
k^2=\frac{\beta ^2}{\beta ^2+1},
\]
hence is a contraction mapping. The result thus follows by an
application of the contraction mapping principle. On the other
hand, if $u$ and $v$, respectively, solve (\ref{monotone1}) with
right hand sides $w_1$ and $w_2$, then we may conclude that
\[
\|u-v\|^2+2{\rm Re}\left ( (T(u)-T(v),u-v)\right
)+\|T(u)-T(v)\|^2=\|w_1-w_2\|^2.
\]
The monotonicity of $T$, therefore implies that
\[
\|u-v\|^2+\|T(u)-T(v)\|^2\leq \|w_1-w_2\|^2,
\]
from which the continuity of the mapping
$w\mapsto u(w)$ follows.
\end{proof}

\subsection{Multivalued mappings}
\label{mm}

 Let $\mathbb{M}$ be a  metric space
with metric ${\rm d}$ and let
\begin{equation}
\label{multi1}
T :\mathbb{M}\to \mathcal{H} ({\mathbb {M}}),
\end{equation}
which is a contraction with respect to the Hausdorff metric ${\rm
h}$, i.e.,
\begin{equation}
\label{multi2}
{\rm h}(T (x),T (y))\leq k{\rm d}(x,y), \quad
\forall x,y\in \mathbb{M},
\end{equation}
where $0\leq k<1 $ is a constant. Such a mapping is called a {\it
contraction
correspondence}\index{contraction correspondence}.

For such mappings we have the following extension of the contraction
mapping
principle. It is due to Nadler \cite{nadler:mcm69}. We note  that the
theorem
is an existence theorem, but that uniqueness of fixed points cannot be
guaranteed (easy examples are provided by  constant mappings).

\begin{theorem} \label{multi3}
Let
$T :\mathbb{M}\to \mathcal {H}({\mathbb {M}})$
with
\[
{\rm h}(T (x),T (y))\leq k{\rm d}(x,y), \quad
\forall x,y\in \mathbb{M},
\]
be a contraction correspondence. Then there exists $x\in \mathbb{M}$
such that $x\in T (x)$.
\end{theorem}

\begin{proof}
The proof uses the Picard iteration scheme. Choose any point $x_0\in
\mathbb{M}, $ and $x_1\in T(x_0)$. Then choose $x_2\in T(x_1)$ such that
\[
{\rm d}(x_2,x_1)\leq {\rm h}(T(x_1),T(x_0))+k,
\]
where $k$ is the contraction constant of $T$
(that this may  be done follows from Proposition \ref{dh-relation} of
Chapter \ref{chapII}).  We then construct
inductively
the sequence $\{x_n\}_{n=0}^{\infty}$ in $\mathbb{M}$ to satisfy
\[
x_{n+1}\in T(x_n),\;{\rm d}(x_{n+1},x_n)\leq {\rm
h}(T(x_n),T(x_{n-1}))+k^n.
\]
We then obtain, for $n\geq 1$,
\begin{align*}
{\rm d}(x_{n+1},x_n)
&\leq {\rm h}(T(x_n),T(x_{n-1}))+k^n\\
&\leq k{\rm d}(x_n,x_{n-1}) +k^n\\
&\leq k\left ({\rm h}(T(x_{n-1},T(x_{n-2})) +k^{n-1}\right )+k^n\\
&\leq  k^2{\rm d}(x_{n-1},x_{n-2}) +2k^n\\
&\dots  \\
&\leq  k^n{\rm d}(x_1,x_0) +nk^n.
\end {align*}
Using the triangle inequality for the metric d, we obtain, using the
above computation
\begin{align*}
{\rm d}(x_{n+m},x_n)
&\leq \sum _{i=n}^{n+m-1}{\rm d}(x_{i+1},x_i)\\
&\leq \sum _{i=n}^{n+m-1}\left (k^i{\rm d}(x_{1},x_0)+ik^i\right
)\\
&\leq  \Big(\sum _{i=n}^{\infty }k^i\Big){\rm d}(x_{1},x_0)+
\Big(\sum _{i=n}^{\infty }ik^i\Big).
\end{align*}
Since both $\sum _{i=0}^{\infty }k^i$ and $ \sum _{i=0}^{\infty
}ik^i$ converge, it follows that $\{x_n\}_{n=0}^{\infty}$ is a
Cauchy sequence in $\mathbb{M}$, hence has a limit $x\in \mathbb{M}$. We next recall the definition of the Hausdorff metric (see
 Chapter \ref{chapII}, Section
\ref{haus})
and compute
\[
{\rm d}(x_{n+1}, T(x))\leq {\rm h}(T(x_n), T(x))\leq k {\rm d}(x_n,x).
\]
Since $\{T(x)\}$ is a closed  set and $\lim _{n\to \infty }x_n=x$,
it follows that
\[
{\rm d}(x,T(x))=0,
\]
i.e.,
$x\in T(x)$.
\end{proof}

\subsection{Converse to the theorem}

In this last section of the chapter we  discuss a result of Bessaga \cite{bessaga:cbf59} which provides a converse to the contraction mapping principle\index{converse}. We follow the treatment given in \cite{jachymski:spc00}, see also
\cite{deimling:nfa85} (this last reference is also a very good reference to fixed point theory, in general, and to the topics of these notes, in particular).
 We shall establish the following theorem.

\begin{theorem} \label{converse}
 Let $\mathbb{M} \ne \emptyset $ be a set, $k\in (0,1)$ and let
\[
F:\mathbb{M} \to \mathbb{M} .
\]
Then:
\begin{enumerate}
\item If $F^n$ has at most one fixed point for every $n=1,2,\dots $,
there exists a metric ${\rm d}$ such that
\[
{\rm d}(F(x),F(y))\leq k{\rm d}(x,y), \quad
\forall x,y\in \mathbb{M} .
\]
\item If, in addition, some $F^n$ has a fixed point, then there is a metric
${\rm d}$ such that
\[
{\rm d}(F(x),F(y))\leq k{\rm d}(x,y), ~\forall x,y\in \mathbb{M}
\]
and $(\mathbb{M}, {\rm d})$ is a complete metric space.
\end{enumerate}
\end{theorem}

The proof of Theorem \ref{converse} will make use of the following lemma.

\begin{lemma}
Let $F$ be a selfmap of $\mathbb{M} $ and $k\in (0,1)$. Then the
following statements are equivalent:
\begin{enumerate}
\item There exists a metric ${\rm d}$ which makes $\mathbb{M}$
a complete metric space such that
\[
{\rm d}(F(x),F(y))\leq k{\rm d}(x,y), \quad
\forall x,y\in \mathbb{M} .
\]
\item There exists a function
$\phi :\mathbb{M} \to [0,\infty )$
such that $\phi ^{-1}(\{0\}) $ is a singleton and
\begin{equation}
\label{functional}
\phi (F(x))\leq k\phi (x),\quad \forall x\in \mathbb{M} .
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
($1.\Rightarrow 2.$) The contraction mapping principle implies that $F$ has a unique fixed point $z\in \mathbb{M} $.
Put
\[
\phi (x):={\rm d}(x,z),\;\forall x\in \mathbb{M} .
\]

($2. \Rightarrow 1.$) Define
\begin{gather*}
{\rm d}(x,y):=\phi (x) +\phi (y),\quad x\ne y \\
{\rm d}(x,x):= 0.
\end{gather*}
Then, one easily notes that ${\rm d}$ is a metric on $\mathbb{M} $ and that $F$ is a contraction with contraction constant
$k$. Let $\{x_n\}\subset \mathbb{M} $ be a Cauchy sequence. If this sequence has only finitely many distinct terms, it clearly converges. Hence, we may assume it to contain infinitely many distinct terms. Then there exists a subsequence $\{x_{n_k}\}$ of distinct elements, and hence, since

\[
{\rm d}(x_{n_k},x_{n_m})=\phi (x_{n_k}) +\phi (x_{n_m}),
\]
it follows that
\[
\phi (x_{n_k})\to 0.
\]
Since there exists $z\in \mathbb{M} $ such that $\phi (z)=0$, it
follows that
\[
{\rm d}(x_{n_k},z)\to 0,
\]
and therefore
$x_n\to z$.
\end{proof}

To give a proof of Theorem \ref{converse} it will therefore suffice
to produce such a function $\phi$. To do this, we  will rely on the
use of the Hausdorff maximal principle\index{Hausdorff maximal
principal} (see \cite{royden:ra88}).

Let $z\in \mathbb{M}$ be a fixed point of $F^n $, as guaranteed by
part 2. of the theorem. Uniqueness then implies that
\[ z=F(z),
\] as well.

For a given function $\phi $ defined on a subset of $\mathbb{M}$ we denote by $D_{\phi}$ its domain of definition
and we let
\[
\Phi:=\{\phi :D_{\phi }\to [0,\infty ):z\in D_{\phi }\subset
\mathbb{M},\;\phi^{-1}(\{0\})=z, ~F(D_{\phi })\subset D_{\phi }\}.
\]
We note that, for the  given $z$, if we put
\[
D_{\phi ^*}:=\{z\},\quad \phi ^*(z):=0,
\]
then $\phi ^* \in \Phi$. Hence the collection is not empty. One next defines a partial order on the set $\Phi $ as follows:
\[
\phi _1 :\preceq \phi _2 \iff D_{\phi _1}\subset D_{\phi _2} ~{\text{and}}~ \phi _2|_{D_{\phi _1}}=\phi _1.
\]
If $\Phi _0$  is a chain in $(\Phi, \preceq )$, then the set
\[
D:=\cup _{\phi \in \Phi _0}D_{\phi }
\]
is a set which is invariant under $F$, it contains $z$ and if we
define
\[
\psi :D\to [0,\infty )
\]
by
\[
\psi (x):=\phi (x),\;x\in D_{\phi },
\]
then $\psi $ is an upper bound for $\Phi _0$ with domain
$D_{\psi }:=D$.
Hence, by the Hausdorff maximal principle, there exists a maximal element
\[
\phi _0:D_0:=D_{\phi _0}\to [0,\infty)
\]
in $(\Phi, \preceq )$. We next show that $D_0=\mathbb{M} $, hence
completing the proof.

This we proof indirectly. Thus, let $x_0\in \mathbb{M}\setminus D_0$
and consider the set
\[
O:=\{F^n(x_0):n=0,1,2, \dots \}.
\]
If it is the case that
\[
O\cap D_0 =\emptyset ,
\] then the elements $F^n(x_0):n=0,1,2, \dots $ must be distinct; for, otherwise $z\in O $.
We define
\[
D_{\phi }:=O\cup D_0, ~\phi |_{D_0}:=\phi _0,\;\phi (F^n(x_0)):=k^n, ~n=0,1,2,\dots .
\]
Then
\[
\phi \in \Phi ,\quad \phi \ne \phi _0,\quad \phi _0\preceq \phi ,
\]
contradicting the maximality of $\phi _0 $. Hence
\[O\cap D_0 \ne \emptyset .
\]
Let us set
\[
m:=\min\{n:F^n(x_0)\in D_0\},
\]
then $F^{m-1}(x_0)\notin D_0$.
Define
\[
D_{\phi}:=\{F^{m-1}(x_0)\}\cup D_0.
\]
Then
\[
F(D_{\phi})=\{F^{m}(x_0)\}\cup F(D_0)\subset D_0\subset D_{\phi }.
\]
So $D_{\phi } $ is $F$ invariant and contains $z$.
Define $\phi :D_{\phi }\to [0,\infty )$ as follows:
\begin{itemize}
\item
$\phi |_{D_0}:\phi _0$.

\item If $F^m(x_0)=z$, put $\phi (F^{m-1}(x_0)):=1$.
\item If $F^m(x_0)\ne z$, put $\phi (F^{m-1}(x_0)):=\frac{\phi _0(F^m(x_0))}{k}$.
\end{itemize}
With this definition we obtain again a contradiction to the maximality of $\phi _0 $ and hence must conclude that
$D_0=\mathbb{M}$.

\part{Applications}

\section{Iterated function systems} \label{chapIV} % Chapter IV. \label{ifs}

In this chapter we shall discuss an application of the contraction
mapping principle to the study of  iterated function
systems\index{iterated function system}.
The presentation follows the work of Hutchinson
\cite{hutchinson:fss81}
who established that a finite number of contraction mappings on a
complete metric space $\mathbb{M}$ define in a natural way a contraction mapping
on a subspace of $\mathcal{H}(\mathbb{M})$ with respect to the Hausdorff
metric\index{Hausdorff metric} (see Chapter \ref{chapII}, Section
\ref{haus}).

\subsection{Set-valued contractions}
\label{svc}
Let $\mathbb{M}$ be a complete metric space with metric $\rm d$
and let
\[
\mathcal{C}(\mathbb{M}) \subset \mathcal{H}(\mathbb{M})
\]
 be the metric space of nonempty
compact  subsets of $\mathbb{M}$ endowed
 with the Hausdorff metric $\rm h$. Then if $\{A_n\}$ is a Cauchy
 sequence
in $\mathcal{C}(\mathbb{M})$, its limit $A$ belongs to $\mathcal{H}
(\mathbb{M})$ and hence is a closed and thus complete set. On the
other hand, since $A_n\to A$, for given $\epsilon >0$, there
exists $N$, such that for $ n\geq N$, $A\subset (A_{n})_\epsilon$,
and hence $A$ is totally bounded and therefore compact. Thus
$\mathcal{C}(\mathbb{M})$ is a
 closed subspace of $\mathcal{H}(\mathbb{M})$, hence a complete metric
 space in its own right.

We have the following theorem.

\begin{theorem}\label{frac1}
Let
$f_i:\mathbb{M}\to \mathbb{M}$, $i=1,2,\dots k $,
be $k$ mappings which are Lipschitz continuous with Lipschitz
constants
$L_1,L_2,\dots,L_k$,
i.e.,
\begin{equation}
\label{frac2}
{\rm d}\left (f_i(x),f_i(y)\right )\leq L_i{\rm
d}(x,y),\quad i=1,2,\dots, k,\;x,y\in \mathbb{M} .
\end{equation}
Define
\begin{equation}
\label{frac3}
F:\mathcal{C}(\mathbb{M})\to \mathcal{C}(\mathbb{M})
\end{equation}
by
\begin{equation}
\label{frac4}
F(A):=\cup _{i=1}^{k}f_i(A),\quad A\in \mathcal{C}(\mathbb{M}).
\end{equation}
Then $F$ satisfies a Lipschitz condition, with respect to the
Hausdorff metric,
with Lipschitz constant
\[
L:=\max_{i=1,2,\dots , k}L_i,
\]
i.e.,
\begin{equation}
\label{frac4*}
{\rm h}\left (F(A),F(B)\right )\leq L {\rm h}(A,B),\quad
\forall A,B\in \mathcal{C}(\mathbb{M}).
\end{equation}
In particular, if $f_i,\;i=1,2,\dots ,k$, are contraction mappings
on $\mathbb{M}$, then $F$, given by (\ref{frac4}), is a contraction
mapping on $\mathcal{C}(\mathbb{M})$ with respect to the Hausdorff
metric, and $F$ has a unique fixed point $A\in \mathcal{C}(\mathbb{M})$.
\end{theorem}

\begin{proof}
We present two arguments based on the two equivalent definitions of the
Hausdorff metric. In both cases we establish the result for the case
of two mappings. The general case will follow using an induction
argument.

We first observe that for any $A\in \mathcal{C}(\mathbb{M})$,
because of the compactness of $A$ and the Lipschitz continuity of
$f_i,\;i=1,2$ it is the case that $f_i(A)\in \mathcal{C}(\mathbb{M}),\; i=1,2$ and hence $F(A)\in \mathcal{C}(\mathbb{M})$.

Let us recall the definition of the Hausdorff metric which may
equivalently be stated as
\[
{\rm h}(A,B)=\sup \{{\rm d}(a,B),{\rm d}(b,A),\;a\in A,\;b\in B\}.
\]
Suppose then that $A_1,A_2,B_1,B_2$ are compact subsets of
$\mathbb{M}$. Letting $A=A_1\cup A_2,\;B=B_1\cup B_2$, we claim that
\begin{equation}
\label{frac6}
{\rm h}(A,B)\leq \max _{i=1,2}{\rm h}(A_i,B_i)=:m.
\end{equation}
To see this, we let $a\in A,b\in B$ and show that
\[
{\rm d}(a,B),{\rm d}(b,A)\leq m.
\]
Now
\begin{align*}
{\rm d}(a,B)
&= {\rm d}(a,B_1\cup B_2)=\min \{{\rm d}(a,B_i), i=1,2\}\\
&\leq  {\rm d}(a,B_i)\leq D(A_i,B_i)\\
&\leq {\rm h}(A_i,B_i)\leq m.
\end{align*}
and similarly
${\rm d}(b,A)\leq m$,
which establishes the claim.

Let $A,B \in \mathcal{C}(\mathbb{M})$ and let
${\rm h}(A, B)=\epsilon$.
Then
\[
A\subset B_{\epsilon}, \quad
B\subset A_{\epsilon},
\]
where $A_{\epsilon},B_{\epsilon}$ are defined at the beginning of
Section
\ref{haus} of Chapter \ref{chapII}.
We therefore have
\begin{equation}
\label{col}
\begin{gathered}
f_1(B)\subset f_1(A_{\epsilon}),\quad f_2(B)\subset f_2(A_{\epsilon})\\
f_1(A)\subset f_1(B_{\epsilon}),\quad f_2(A)\subset f_2(B_{\epsilon}).
\end{gathered}
\end{equation}
It also follows that
\begin{equation}
\label{col1}
\begin{gathered}
f_i(A_{\epsilon})\subset \left (f_i(A)\right) _{L_i\epsilon},\;i=1,2\\
f_i(B_{\epsilon})\subset \left (f_i(B)\right) _{L_i\epsilon},\;i=1,2.
\end{gathered}
\end{equation}
Further, we obtain
\begin{equation}
\label{col2}
\begin{gathered}
f_i(A)\subset \left (f_1(B)\cup f_2(B)\right )_{L\epsilon},\;i=1,2 \\
f_i(B)\subset \left (f_1(A)\cup f_2(A)\right )_{L\epsilon},\;i=1,2.
\end{gathered}
\end{equation}
Using again the definition of Hausdorff distance, it follows from
(\ref{col2}) that
\begin{equation}
\label {col3}
{\rm h}\left (F(A), F(B)\right )\leq L\epsilon=L{\rm h}(A, B) .
\end{equation}

An alternate argument is contained in the following.
It follows from the discussion in Chapter \ref{chapII} (see formula
(\ref{hmetric3}) there) that
${\rm h}\left (F(A),F(B)\right )$ is given by
\begin{equation}
\label{frac5}
{\rm h}\left (F(A),F(B)\right )=
{\rm D}\left (F(A),F(B))\vee {\rm D}(F(B),F(A)\right ).
\end{equation}

 Formula (\ref{frac6}), on the other hand implies, since,
\begin{equation}
\label{frac7}
{\rm h}\left (F(A),F(B)\right )={\rm h}\left (f_1(A)\cup f_2(A),f_1(B)\cup
f_2(B)\right ),
\end{equation}
that
\begin{equation}
\label{frac8}
{\rm h}\left (F(A),F(B)\right )\leq
{\rm h}\left (f_1(A),f_1(B)\right ) \vee {\rm h}\left (f_2(A),f_2(B)\right ) .
\end{equation}
Then, using the definition of the Hausdorff metric, we find that
\begin{equation}
\label{frac9}
{\rm h}\left (f_i(A),f_i(B)\right )\leq L_i {\rm h}(A,B), \quad i=1,2 .
\end{equation}
Combining (\ref{frac9}) with (\ref{frac8}), we obtain
(\ref{frac4*}) for $k=2$.
\end{proof}

\begin{remark} \rm
Given the contraction mappings $f_1,f_2,\dots , f_k$,
the mapping $F$, defined by
\[
F(A):=\cup _{i=1}^{k}f_i(A),\quad A\in \mathcal{C}(\mathbb{M})
\]
has become known as the {\it Hutchinson operator}\index{Hutchinson
operator}
and the iteration scheme
\begin{equation}
\label{frac10}
A_{i+1}=F(A_i),\quad i=0,1,\dots
\end{equation}
an {\it iterated function system}\index{iterated function system}.
The iteration scheme (\ref{frac10}), of course, has, by the
contraction mapping principle, a unique limit $A$, which is
independent of the choice of the initial set $A_0$ and satisfies
\begin{equation}
\label{frac11}
A=F(A)=f_1(A)\cup f_2(A)\cup \dots \cup f_k(A).
\end{equation}
If it is the case that that $\mathbb{M} $ is a compact subset of
$\mathbb{R}^N$ and $f_1,f_2,\dots ,f_k$ are similarity
transformations\index{similarity transformations},
formula (\ref{frac11}) says that the fixed set $A$ is the union of $k$
similar
copies of itself.
\end{remark}

\subsection{Examples}

In this section we shall gather some examples which will illustrate
the utility of the above theorem in the study and use of
fractals\index{fractals}.

\subsection*{The Cantor set}
Let
$ \mathbb{M}=[0,1]\subset \mathbb{R}$,
with the metric given by the absolute value.
 We define
\[
F:\mathcal{H}(\mathbb{M})\to \mathcal{H}(\mathbb{M})=\mathcal{C}(\mathbb{M})
\]
by
\[
F(A)=f_1(A)\cup f_2(A),
\]
where
\[
f_1(x)=\frac{1}{3}x,\quad
f_2(x)=\frac{1}{3}x+\frac{2}{3},\quad 0\leq x\leq 1.
\]
Then $f_1$ and $f_2$ are contraction mappings with the same  contraction
constant
$\frac{1}{3}$. Hence, $F$ has the same contraction constant, also.

The unique fixed point of $F$ must satisfy
\[
A=F(A)=f_1(A)\cup f_2(A).
\]
Considering the nature of the two transformations
(similarity transformations)\index{similarity transformation},
one deduces from
the last equation, that the fixed point set $A$ must be the Cantor
subset\index{Cantor set} of $[0,1]$.

It is apparent how other types of Cantor subsets of an interval may
be constructed using other types of linear contraction mappings.

\subsection*{The Sierpinski triangle}
Let
$ \mathbb{M}=[0,1]\times [0,1] \subset \mathbb{R}$,
with metric given by the Euclidean distance.
 We define
\[
F:\mathcal{H}(\mathbb{M})\to \mathcal{H}(\mathbb{M})=\mathcal{C}(\mathbb{M})
\]
by
\[
F(A)=f_1(A)\cup f_2(A)\cup f_3(A),
\]
where
\begin{gather*}
f_1(x,y)=\frac{1}{2}(x,y), \\
f_2(x,y)=\frac{1}{2}(x,y)+\big(\frac{1}{2},0\big), \\
f_3(x,y)=\frac{1}{2}(x,y)+\big(\frac{1}{4},\frac{1}{2}\big).
\end{gather*}
Here, again, all three contraction constants are equal to
$\frac{1}{2}$, hence, the Hutchinson operator\index{Hutchinson
operator} is a contraction mapping with the same contraction
constant. All three mappings are similarity transformations and,
since the fixed point $A$ of $F$ satisfies
\[
A=F(A)=f_1(A)\cup f_2(A)\cup f_3(A),
\]
$A$ equals a union of three similar copies of itself, i.e., it is a
self-similar\index{self-similar} set, which in this case is, what has become known
the {\it Sierpinski triangle}\index{Sierpinski triangle}.

For many detailed examples of the above character, we refer to
\cite{barnsley:fe88},
\cite{peitgen:cfn92}.


\section{Newton's method} \label{chapV} % {newt}   Chapter V.  %\include{contnewt}

One of the important numerical methods for computing solutions of
nonlinear equations is Newton's  method\index{Newton's method}, often
also referred to as the Newton-Raphson method\index{Newton-Raphson method}. It is an iteration
scheme,
whose convergence may easily be demonstrated by means of the
contraction mapping principle. Many other numerical methods contain
Newton's method as one of their subroutines (see, e.g.,
\cite{allgower:ncm90}).

Let $G$ be a domain in $\mathbb{R}^N$ and let
\[
F:G\to \mathbb{R}^N
\]
be $C^2$ mapping (i.e., all first and second partial
derivatives of all components of $F$ are continuous on $G$).

Let us assume that the equation
\begin{equation}
\label{newt1}
F(x)=0,
\end{equation}
has a solution $x^*\in G$ such that the Jacobian
matrix\index{Jacobian matrix} $F'(x^*)$ has full rank\index{full
rank} (i.e., the matrix $F'(x^*)$ is a nonsingular matrix). It
then follows by a simple continuity argument that  $F'(x)$ has full
rank in a closed neighborhood of $x^*$, say
\[
B_r:=\{x:\|x^*-x\|\leq r,\;r >0 \},
\]
where $\|\cdot \|$, is a given norm in $\mathbb{R}^N$, and that
$x^*$ is the unique solution of \eqref{newt1} there. The mapping
\begin{equation}
\label{newt2}
x\mapsto x-\left (F'(x)\right )^{-1}F(x)=:N(x)
\end{equation}
is therefore defined  in that neighborhood and we note that $x^*$ is a solution of
\eqref{newt1} in that neighborhood if, and only if, $x^*$ is a fixed
point of $N$ in $B_r$.

The Newton iteration scheme \index{Newton iteration scheme} is then defined by:
\begin{equation}
\label{newt3}
x_{n+1}=N(x_n),\quad x_1\in B_r, \quad n=1,2,\dots .
\end{equation}

The following theorem holds.

\begin{theorem} \label{thm1N}
Assume the above conditions hold. Then, for all $r>0$,
sufficiently small, the Newton iteration scheme, given by
(\ref{newt3}), converges to the solution $x^*$ of \eqref{newt1}.
\end{theorem}

\begin{proof}
We use Taylor's theorem to write
\[
N(x)=N(x^*)+N'(x^*)(x-x^*) +O(\|x-x^*\|^2).
\]
On the other hand, because $F$ is a $C^2$ mapping and $F'(x^*)$ is
nonsingular,
we obtain that
\begin{align*}
N(x^*+y)&= N(x^*)+y-\left (F'(x^*+y)\right )^{-1}F(x^*+y)\\
&= N(x^*)+y-\left (F'(x^*)\right )^{-1}F(x^*+y) \\
 &\quad +\left (\left ( F'(x^*)\right)^{-1}-\left (F'(x^*+y)\right )^{-1}\right )F(x^*+y)\\
&= N(x^*)+O(\|y\|^2)
 +\left (F'(x^*+y)\right )^{-1}\Big(F'(x^*+y)\\
&\quad -F'(x^*)\Big)\left (F'(x^*)\right )^{-1}
 \left(F'(x^*)y+O(\|y\|^2)\right )\\
&= N(x^*)+O(\|y\|^2).
\end{align*}
 From which it follows that
\[
N'(x^*)=0,
\]
the zero matrix, and thus, there exists $r>0$, such that the
matrix norm of $N'(x)$ satisfies
\[
\|N'(x)\|\leq \frac{1}{2},\;\text{for}~\|x-x^*\|\leq r.
\]
Hence, for  $x, y\in B_r$
\[
\|N(x)-N(y)\|\leq \int _0^1\|N'((1-t)y+tx)\|dt\|y-x\|\leq
\frac{1}{2}\|x-y\|
\]
and for $y=x^*$
\[
\|x^*-N(x)\|\leq \int _0^1\|N'((1-t)x^*+tx)\|dt\|x^*-x\|\leq r.
\]
Hence,
$N:B_r\to B_r$ and $N$ is a contraction mapping for such $r$.
The assertion of the theorem
then follows from the contraction mapping principle.
\end{proof}


\section{Hilbert's metric}  \label{chapVI} % hmapo Chapter VI.  %\include{conthilbert}

This chapter is concerned with a fundamental result of matrix theory,
the theorem of Perron-Frobenius\index{Perron-Frobenius theorem} about
the existence of positive eigenvectors\index{positive eigenvectors} of positive matrices\index{positive matrix}. Upon the
introduction of Hilbert's metric, the result may be deduced via the
contraction mapping principle. Since the approach also works in
infinite dimensions, a version of the celebrated theorem of
Krein-Rutman\index{Krein-Rutman
theorem} may be
established, as well. The approach to establishing these important results
using Hilbert's projective metric\index{Hilbert's projective metric}
goes back to Birkhoff \cite{birkhoff:ejt57}. Here we also rely on the work in
\cite{bushell:hmp73} and \cite{kohlberg:cma82}.

\subsection{Cones} \label{cones}
Let $E$ be a real Banach space with norm $\|\cdot \|$.
A closed subset $K$ in $E$ is called a {\it   cone}\index{cone}
provided that:
\begin{enumerate}
\item
for all $\lambda ,\mu \geq 0$,  and all $u,v \in K$
$\lambda u +\mu v \in K$,

\item if $u\in K$, $-u\in K$, then $u=0\in K$.
\end{enumerate}

If it is the case that the interior of $K$, $\mathop{\rm int}K$, is not
empty, the cone is called {\it solid.}\index{solid cone} In this
chapter we shall always assume that the cone $K$ is solid,
 even though several of the results presented are valid in the absence
 of this assumption.

 A cone $K$ induces a partial order\index{partial order} $\leq $ by:
\[
u\leq v \quad \text{if, and only if,}\quad  v-u\in K,
\]
and if the cone $K$ is solid, another partial order $<$
by:
\[
u<v \quad \text{if, and only if}\quad  v-u\in \mathop{\rm int}K.
\]
Since we assume  that  a  cone is closed, it
follows that it is also {\it
Archimedean}\index{Archimedean cone}, i.e.,
\[
\text{if } nu\leq v, \quad n=1,2, \dots ,\text{ then } u\leq 0.
\]
We shall denote by $K^+$ the set of all nonzero elements of $K$.

For solid cones we have the following lemma (see
\cite{schaefer:tvs67}
for most of the details).

\begin{lemma} \label{solid}
Let $K$ be a solid cone and let $v\in \mathop{\rm int}K$,
$u\in K$.
Then:
\begin{enumerate}
\item
$\{w:w=(1-t)v+tu,\;0\leq t<1\}\subset \mathop{\rm int}K$.

\item
If $u\in \partial K$, then for $t>1$,
\[
w=(1-t)v+tu \not\in K.
\]
\item
\[
K=\overline{\mathop{\rm int}K}.
\]
\item
\[
K+v\subset \mathop{\rm int}K.
\]
\end{enumerate}
\end{lemma}

\begin{proof}
We shall establish the last part of the lemma and leave the remaining
parts to the reader for verification.

If $v\in \mathop{\rm int}K$ and $z \in K$, then $v+z=u \in K$. If it were
the case that $u\in \partial K$, then by the first two assertions
of the lemma $t=1$ is the maximal number such that
\[
w=(1-t)v+tu \in K.
\]
On the other hand
\[
w=(1-t)v+tu =(1-t)v+t(v+z)=v+tz \in K,\;\forall t\geq 0,
\]
yielding a contradiction.
\end{proof}

\subsection{Hilbert's metric} \label{hm}

We
define the mappings
\[
m(\cdot , \cdot),\;M(\cdot , \cdot),\;[\cdot , \cdot ]:E\times K^+\to [-\infty ,\infty ]
\]
as follows:
\begin{gather}
\label{hil1}
m(u,v):=\sup\{\lambda :\lambda v\leq u\}
\\
\label{hil2}
M(u,v):=\inf\{\lambda : u\leq \lambda v\},
\end{gather}
with the interpretation that $m(u,v)=-\infty $, if the set
$\{\lambda  : \lambda v\leq u\}$ is empty, and
 $M(u,v)=\infty $,
if the set $\{\lambda  : u\leq \lambda v\}$ is empty,
and
\begin{equation} \label{hil2*}
[u,v]:=M(u,v)-m(u,v).
\end{equation}
The last quantity is called the {\it
$v-$oscillation}\index{oscillation} of $u$.
We remark that the Archim- edean property immediately implies that
\[
m(u,v)< \infty ,\;M(u,v)>-\infty ,
\]
which makes (\ref{hil2*}) well-defined in the extended real numbers.

In what is to follow many of the statements are to be interpreted in
the extended real numbers. If this should be the case, we shall not
remark
so explicitly; it will be clear from the context.
The following lemma is easy to prove and we leave the details to the
reader.
We remark that (\ref{hi14*}) follows from
 Lemma \ref{solid} (above).

\begin{lemma} \label{hil2**}
For  $u,v,w \in K^+$, the following hold:
\begin{gather}
\label{hil3}
m(u,v)v\leq u\leq M(u,v)v,\quad \text{provided }~M(u,v)<\infty ,
\\
\label{hil4}
0\leq m(u,v)\leq M(u,v)\leq \infty ,
\\
\label{hi14*}
m(u,v)>0,\quad \text{if }u\in \mathop{\rm int}K,\text{ and }
u-m(u,v)v\in \partial K,
\\
\label{hi14***}
m(u,v)=0,\quad \text{if }v\in \mathop{\rm int}K,\text{ and } u\in \partial K,
\\
\label{hi14**}
M(u,v)<\infty ,\quad \text{if } v\in \mathop{\rm int}K,
\\
\label{hil5}
M(u,w)\leq M(u,v)M(v,w),
\\
\label{hil6}
m(u,w)\geq m(u,v)m(v,w),
\\
\label{hil7}
m(u,v) M(v,u)=1,
\\
\label{hil8}
M(\lambda u+\mu v,v)=\lambda  M(u,v) +\mu,\quad
\forall \lambda ,\mu \geq 0 ,
\\
\label{hil9}
m(\lambda u+\mu v,v)=\lambda  m(u,v) +\mu,\quad \forall \lambda ,\mu \geq
0 ,
\\
\label{hil9*}
[\lambda u+\mu v,v]= \lambda [u,v],\quad \forall \lambda ,\mu \geq 0 .
\\
\label{hil9*0}
[ u,v]=0,\text{ implies } u=\lambda v,\quad \text{for some } \lambda \geq 0.
\end{gather}
\end{lemma}

Using the properties in the previous lemma, one may establish the
following result.

\begin{lemma} \label{hil9**}
For  $u,v \in K^+$, the following hold:
\begin{equation} \label{hil9*1}
\begin{aligned}
M( u, u+v)&=  \frac{1}{m(u+v,u)}\\
&=  \frac{1}{1+m(v,u)}\\
&=  \frac{M(u,v)}{1+M(u,v)}\leq 1,
\end{aligned}
\end{equation}
and
\begin{equation} \label{hil9*2}
\begin{aligned}
m( u, u+v)&=  \frac{1}{M(u+v,u)}\\
&=   \frac{1}{1+M(v,u)}\\
&=  \frac{m(u,v)}{1+m(u,v)}\leq 1.
\end{aligned}
\end{equation}
Hence
\begin{equation}
\label{hil9*3}
M(u,u+v)+m(v,u+v)=1.
\end{equation}
\end{lemma}

We now define {\it Hilbert's projective metric}\index{Hilbert's
projective metric}
\[
{\rm d}:\mathop{\rm int}K\times \mathop{\rm int}K\to [0,\infty )
\]
as follows:
\begin{equation}
\label{hil10}
{\rm d}(u,v):=\log \frac{M(u,v)}{m(u,v)}.
\end{equation}
We have the following theorem.

\begin{theorem} \label{hmetric}
The function ${\rm d}$ defined by (\ref{hil10}) has the
following properties: If $u,v,w\in \mathop{\rm int}K$, then
\begin{gather}
\label{hil12*}
{\rm d}(u,v)={\rm d}(v,u), \;
{\rm d}(\lambda u,\mu v)={\rm d}(u,v),\quad
\forall \lambda >0,\mu >0,
\\
\label{hil12}
{\rm d}(u,v)=0,\quad \text{if, and only if, }\quad
u=\lambda v,\quad \text{for some }\lambda >0,
\\
\label{hil13}
{\rm d}(u,v)\leq {\rm d}(u,w)+{\rm d}(w,v).
\end{gather}
Let
\begin{equation} \label{hil14}
\mathbb{M}:=\{u\in \mathop{\rm int}K:\|u\|=1\},
\end{equation}
then
 $( \mathbb{M},{\rm d})$
is a metric space.
\end{theorem}

\begin{proof}
 The symmetry property (\ref{hil12*})
and the triangle inequality (\ref{hil13}) follow immediately  from
Lemma \ref{hil9**}. That (\ref{hil12}) holds follows from the fact
that ${\rm d}(u,v)=0$, if, and only if, $m(u,v)=M(u,v)$,
which is the case, if, and only if, $u=M(u,v)v$. The properties
together imply that $\rm d$ is a metric on $\mathbb{M}$.
\end{proof}

The following examples will serve to illustrate these concepts.
In all examples we shall assume that $u,v\in\mathop{\rm int}K$.

\begin{example}\label{hilbexm1} \rm
Let
\[
E:=\mathbb{R}^N,\quad
K:=\{(u_1,u_2, \dots , u_N): u_i\geq 0,\;i=1,2, \dots , N\}.
\]
Then
\begin{gather*}
\mathop{\rm int} K=\{(u_1,u_2, \dots , u_N): u_i> 0,\;i=1,2, \dots , N\},
\\
m(u,v)=\min_{i}\frac{u_i}{v_i},\quad
M(u,v)=\max_{i}\frac{u_i}{v_i},
\\
{\rm d}(u,v)=\log \max _{i,j}\frac{u_iv_j}{u_jv_i}.
\end{gather*}
\end{example}

\begin{example}\rm
\label{hilbexm2}
 Let
\[
E:=\mathbb{R}^N,\quad
K:=\{(u_1,u_2, \dots , u_N): 0\leq u_1\leq u_2 \leq \dots \leq u_N\}.
\]
Then
\begin{gather*}
\mathop{\rm int}K=\{(u_1,u_2,
\dots , u_N): 0< u_1< u_2 <\dots < u_N\},
\\
m(u,v)=\min_{i< j}\frac{u_j-u_i}{v_j-v_i},\quad
M(u,v)=\max_{i<j}\frac{u_j-u_i}{v_j-v_i},
\\
{\rm d}(u,v)=\log \max _{i<j,k<l}\frac{(u_j-u_i)(v_l-v_k)}
{(u_l-u_k)(v_j-v_i)}.
\end{gather*}
\end{example}

\begin{example}\rm  Let
\[
E:=C[0,1],\quad K:=\{ u\in E:u(x)\geq 0,\;0\leq x\leq 1 \}.
\]
Then
\begin{gather*}
\mathop{\rm int}K=\{ u\in E:u(x)> 0,\;0\leq x\leq 1 \},\\
\\
m(u,v)=\min _{x\in [0,1]}\frac{u(x)}{v(x)},\quad
M(u,v)=\max _{x\in [0,1]}\frac{u(x)}{v(x)},
\\
{\rm d}(u,v)=\log \max _{(x,y)\in
[0,1]^2}\frac{u(x)v(y)}{u(y)v(x)}.
\end{gather*}
\end{example}

It is an instructive exercise to compute the various quantities
$m(u,v),M(u,v)$, etc., also in the cases that $u,v$ are not
necessarily interior elements to the cone $K$.

\subsection{Positive mappings} \label{pm}
A mapping $T:E \to E$ is called a {\it positive}
mapping\index{positive mapping} (with respect to the cone $K$)
provided that
\[
T({K}^+)\subset  {K}^+.
\]

A positive mapping $T$ is called {\it homogeneous of degree
p}\index{homogeneous mapping}, $p\geq 0$, whenever
\[
T(\lambda  u)=\lambda ^pT(u),\quad \forall \lambda >0,\;u\in {K}.
\]

A positive mapping is called {\it monotone}\index{monotone}
provided that
\[
u,v \in K,\;u\leq v ,\quad \text{imply}\quad  T(u)\leq T(v).
\]


In the following we are interested to see under what conditions
positive mappings are contractions with respect to Hilbert's
projective metric. In order to achieve this, we shall derive some
properties of positive mappings with respect to the functions
introduced above.

We have the following lemma.

\begin{lemma}\label{lemhil15}
Let $T$ be a positive monotone mapping which is
homogeneous of degree $p$. Then for any $u,v \in K^+$
\begin{equation}
\label{hil15}
m(u,v)^p\leq  m(T(u),T(v))\leq
 M(T(u),T(v))\leq {M(u,v)}^p,
\end{equation}
and if
\begin{equation}
\label{hil16}
k(T):=\inf \{\lambda :{\rm d}(T(u),T(v))\leq \lambda {\rm
d}(u,v),\;{\rm d}(u,v)<\infty\},
\end{equation}
where ${\rm d}$ is Hilbert's projective metric,
then
\[
k(T)\leq p.
\]
In particular:
\begin{enumerate}
\item If $p<1$, then $T$ is a contraction with respect
to the projective metric.
\item If $T$ is linear, then
\[
{\rm d}(T(u),T(v))\leq  {\rm d}(u,v),\quad
\forall u,v \in K^+.
\]
\end{enumerate}
\end{lemma}

\begin{proof}
Since
\[
m(u,v)v\leq u \leq M(u,v)v,
\]
Inequality (\ref{hil15}) follows from the monotonicity and homogeneity
of $T$.
Using the definition of Hilbert's projective metric and (\ref{hil15})
we obtain that
\[
 {\rm d}(T(u),T(v))\leq \log \Big(\frac
{M(u,v)}{m(u,v)}\Big)^p=p{\rm d}(u,v),
\]
from which the result follows.
\end{proof}

\begin{remark} \rm
The constant $k(T)$, above, is called the {\it contraction
ratio}\index{contraction ratio} of the mapping $T$.
\end{remark}

We next concentrate on computing the contraction ratio for positive
linear mappings $T$. We define the following constants.
\begin{equation}
\label{hil17}
\begin{gathered}
\Delta (T):=\sup \{{\rm d}(T(u),T(v)):u,v \in K^+\},\\
\Gamma (T):= \frac{e^{\frac{1}{2}{\Delta (T)}}-1}
{e^{\frac{1}{2}{\Delta (T)}}+1}
\end{gathered}
\end{equation}
($\Delta (T)$ is called the {\it projective diameter}\index{projective
 diameter}  of $T$) and
\begin{equation}
\label{hil18}
N (T):=\inf \{\lambda :[T(u),T(v)]\leq \lambda [u,v],\;u,v \in K^+\}.
\end{equation}

We next establish  an extension of a
result
originally proved by Hopf (\cite{hopf:ipl63}, \cite{hopf:rpi63})
in his studies of integral equations and extended by Bauer
\cite{bauer:eph65} to the general setting.

\begin{theorem} \label{hil19}
Let $T:E\to E$ be a linear mapping which is positive with respect to
the cone $K$. Let $u,v \in K^+$  be such that $[u,v]<\infty $. Then
\begin{equation}
\label{hil20}
[T(u),T(v)]\leq \Gamma (T)[u,v] ,
\end{equation}
where $\Gamma (T)$ is given by (\ref{hil14}), i.e.
$N(T)\leq \Gamma (T)$.
Furthermore
\begin{equation}
\label{hil20*}
k(T)\leq N(T).
\end{equation}
\end{theorem}

\begin{proof}
If  $[u,v]=0$ (which is the case, if, and only if,
$u$ and $v$ are co-linear), then $[T(u),T(v)]=0$ and the result holds
trivially.

In the contrary case, $0<[u,v]<\infty $, and, since $T$ is a
positive operator, we have that for any $u,v \in K^+$, the images
of the elements
\begin{gather*}
p=u-(m(u,v))v,\\
q=(M(u,v)) v-u,
\end{gather*}
$T(p)$ and $T(q)$
belong to $K^+$.
Then
\[
p+q =[u,v]v,
\]
and (see Lemma \ref{hil2**})
\begin{equation}
\label{hil21}
\begin{aligned}
m(T(u),T(v))&= [u,v]m(T(p),T(p)+T(q))+m(u,v)\\
&=  \nu M(u,v)+(1-\nu )m(u,v),
\end{aligned}
\end{equation}
where
\[
\nu =m(T(p),T(p)+T(q)).
\]
Since $T$ is a positive mapping, it follows from Lemma \ref{hil9**}
that
\[
 \nu = \frac{1}{1+M(T(q),T(p))}.
\]
We similarly obtain
\begin{equation} \label{hil22}
M(T(u),T(v))= \mu M(u,v)+(1-\mu )m(u,v),
\end{equation}
where
\[
\mu =M(T(p),T(p)+T(q)) = \frac{1}{1+m(T(q),T(p))}.
\]
Hence
\begin{equation}
\label{hil23}
[T(u),T(v)]=(\mu -\nu)[u,v],
\end{equation}
where
\begin{equation}
\label{hil24}
\mu - \nu = \frac{M(T(p),T(q))M(T(q),T(p))-1}
{(1+M(T(p),T(q)))(1+M(T(q),T(p)))}=:\phi (T(p),T(q)).
\end{equation}
We next observe that
\begin{equation}
\label{hil25}
\begin{aligned}
\phi (T(p),T(q)&\leq  \frac{M(T(p),T(q))M(T(q),T(p))-1}
{\left (\sqrt{M(T(p),T(q))M(T(q),T(p))}+1 \right )^2}\\
&=  \frac{e^{{\rm d}(T(p),T(q))}-1}{ \left (\sqrt{e^{{\rm
d}(T(p),T(q))}}+1 \right )^2}\\
&= \frac{ \sqrt{e^{{\rm
d}(T(p),T(q))}}-1 }{ \sqrt{e^{{\rm
d}(T(p),T(q))}}+1 }\\
&\leq  \Gamma (T).
\end{aligned}
\end{equation}
proving (\ref{hil20}).

To verify (\ref{hil20*}), we use the above together with the
identities of Lemmas \ref{hil2**} and \ref{hil9**}.
Since
\[
[T(u),T(v)]\leq N(T)[u,v],
\]
we have
\[
 \frac{1}{m(T(v),T(u))}-\frac{1}{M(T(v),T(u))}\leq
N(T)\Big(\frac{1}{m(v,u)}-\frac{1}{M(v,u)}\Big).
\]
We now replace $v$ by $cv+u$, $c >0$ and use Lemma \ref{hil2**}
to find
\begin{align*}
&\frac{c[T(v),T(u)]}{\left (cM(T(v),T(u))+1\right )\left
(cm(T(v),T(u))+1\right )}\\
&\leq
N(T)\frac{c[v,u]}{\left (cM(v,u)+1\right )\left
(cm(v,u)+1\right )}.
\end{align*}
We integrate this inequality with respect to $c$
and obtain
\[
 \log \frac{cM(T(v),T(u))+1}{cm(T(v),T(u))+1}
\leq N(T)\log \frac{cM(v,u)+1}{cm(v,u)+1}.
\]
We let $c\to \infty $
and obtain
\[
{\rm d}(T(v),T(u))\leq N(T){\rm d}(v,u),
\]
or, equivalently
\[
{\rm d}(T(u),T(v))\leq N(T){\rm d}(u,v),
\]
i.e., (\ref{hil20*}) holds, which completes the proof of the theorem.
\end{proof}

We summarize the above results in the following theorem.

\begin{theorem} \label{hil}
Let $T$ be a positive monotone mapping which is homogeneous of
degree $p$. Then for any $u,v \in K^+$, with $\mathop{\rm
d}(u,v)<\infty $,
\begin{equation}
\label{hil26}
{\rm d}(T(u),T(v))\leq p {\rm
d}(u,v),
\end{equation}
where ${\rm d}$ is Hilbert's projective metric.

In particular, if $p<1$, then $T$ is a contraction with respect to
the projective metric. If $T$ is linear, then
\begin{equation}
\label{hil27}
{\rm d}(T(u),T(v))\leq \Gamma (T) {\rm d}(u,v),\;\forall u,v \in K^+.
\end{equation}
 Thus, in particular,
if $\Delta (T)<\infty $, where $\Delta (T) $ and $\Gamma (T)$ are
defined in (\ref{hil17}), then $T$ is a contraction mapping with
respect to Hilbert's projective metric.
\end{theorem}


\subsection{Completeness criteria} \label{cc}

It follows from Theorem \ref{hil} that if $T$ is a mapping,
satisfying the hypotheses there, it will be a contraction mapping
with respect to the projective metric and, hence, if the mapping leaves
the unit spheres
of the cone $K$ invariant and $ \mathbb{M}$ is complete
with respect to the topology defined by the metric, then the
contraction mapping principle may be applied.
We shall now describe situations where completeness prevails.

We shall discuss one such situation, namely, the case that
the Banach space $E$ is a Banach space whose norm is
monotone\index{monotone norm}
 with
respect to the cone $K$, i.e.,
\[
u,v \in K,\; u\leq v, ~\text{then}~ \|u\|\leq \|v\|
\]
(e.g. if $E$ is a Banach lattice\index{Banach
lattice},
see \cite{schaefer:tvs67},
 with respect to the partial order induced by the cone $K $).
Each of the cones in the examples discussed earlier generates such a Banach
space,
as do the cones of nonnegative functions in all $L^p-$ spaces.

We have the following result.

\begin{theorem}\label{normal}
Let $E$ be a real Banach space whose norm is monotone
with respect to a solid cone
  $K$.
Then
\[
\mathbb{M}:=\{ u\in \mathop{\rm int} K:\|u\|=1\}
\] is complete with respect to Hilbert's projective  metric $\rm d$.
\end{theorem}

\begin{proof}
Assume that  $\{u_n\} $ is  a Cauchy sequence in $\mathbb{M} $ with
respect to the metric $\rm d $, then for $\epsilon >0$, given,
there exists an integer $N$, such that
\[
n,m \geq N,\;\text{implies that } ~
1 \leq \frac{M(u_n,u_m)}{m(u_n,u_m)}\leq 1+\epsilon .
\]
Furthermore, we have (see the definitions of $m$ and $M$),
\begin{equation}
\label{hil28}
m(u_n,u_m)u_m\leq u_n\leq M(u_n,u_m)u_m\leq (1+\epsilon  )m(u_n,u_m)u_m,
\end{equation}
and therefore (using the monotonicity of the norm),
\[
\frac {1}{1+\epsilon}\leq m(u_n,u_m)\leq 1, ~n,m\geq N.
\]
We next use (\ref{hil28}) to conclude that
\[
0\leq u_n-m(u_n,u_m)u_m\leq {m(u_n,u_m)}\left (e^{{\rm
d}(u_n,u_m)}-1\right )u_m,
\]
and, therefore
\[
\|u_n-m(u_n,u_m)u_m\|\leq (e^{{\rm d}(u_n,u_m)}-1).
\]
Thus

\begin{equation}
\label{hil29}
\begin{aligned}
\|u_n-u_m\|&\leq  \|u_n-m(u_n,u_m)u_m\|+\|m(u_n,u_m)u_m-u_m\|\\
&\leq  (e^{{\rm d}(u_n,u_m)}-1)+ (1-m(u_n,u_m))\\
 &\leq \epsilon  +\frac{\epsilon}{1+\epsilon},
\end{aligned}
\end{equation}
proving that $\{u_n\}$ is  a Cauchy sequence in $E$. Since $E$
is complete and the unit sphere of $E$ and $K$ are closed, this sequence will
have a limit $u\in K$ of norm 1. We next show that $u\in \mathop{\rm int}K$.
This follows from the  fact
that the boundary of $K$ may be characterized by
\[
\partial K=\{ v\in K:m(v,w)=0 ,\;\forall ~w\in \mathop{\rm int}K\},
\]
(see Lemma \ref{hil2**}) and that for each $u\in \overline
{\mathbb{M}}$, the mapping
\begin{gather*}
v \mapsto   m(u,v)\\
\overline {\mathbb{M}}\to  [0,1]
\end{gather*}
is an upper semicontinuous function with respect to the norm.
 To see this, it suffices to show the sequential upper
semicontinuity of $m$.
Thus, let $\{v_n\}\subset \overline {\mathbb{M}} $
be a sequence with
\[
v_n\to v,
\]
and let $~u\in\overline {\mathbb{M}}$.
Let
\[
\alpha =\limsup _{n\to \infty} m(u,v_n),
\]
then $0\leq \alpha \leq 1$, and given $\epsilon \in (0,1)$
\[
(1-\epsilon )\alpha\leq m(u,v_n)\leq 1.
\]
Hence,
\[
(1-\epsilon )\alpha v_n\leq u,
\]
and consequently,
\[
(1-\epsilon )\alpha v\leq u,
\]
i.e.,
\[
(1-\epsilon )\alpha \leq m(u,v),
\]
showing that $\alpha \leq m(u,v)$,
proving the upper semicontinuity of $m$.

Returning to the sequence $\{u_n\}$, above with $u_n\to u$, we see
that for fixed $m$,
\[
m(u,u_m)\geq \limsup _{n\to \infty}m(u_n,u_m)\geq \frac{1}{1+\epsilon},
\]
and, therefore, $u\in \mathop{\rm int}K$. One may similarly verify that
the mapping $M$, and hence, $d$ are lower semicontinuous
functions, which will further imply that
\[
\lim _{n\to \infty }{\rm d}(u,u_n)=0.
\]
\end{proof}


\subsection{Homogeneous operators}

 In this section we shall establish an eigenvalue  theorem for
monotone positive operators which are homogeneous of degree less than
one.
We have.

\begin{theorem} \label{homogeneous}
Let $E$ be a real Banach space whose norm is monotone with respect to
the cone $K.
$ Let
\[
T:K^+\to K^+
\]
be a monotone operator which is homogeneous of degree $p<1 $ and
leaves the interior of the cone, $ \mathop{\rm int}K$, invariant. Then
for any positive number $\mu $, there exists $u\in \mathop{\rm int}K$
such that
\begin{equation}
\label{hil29*}
T(u)=\mu u.
\end{equation}
\end{theorem}


\begin{proof}
Let
\[
f(u):= \frac{T({u})}{\|T(u)\|},
\]
then
$f:\mathbb{M} \to \mathbb{M}$.
We have, by the properties of Hilbert's projective metric that
\[
{\rm d}(f(u),f(v))\leq p {\rm d}(u,v),\quad
\forall u,v \in \mathbb{M}.
\]
Hence $f$ is a contraction mapping. Since $\mathbb{M}$ is complete
with respect to this metric, it follows from the contraction
mapping theorem that $f$ has a unique fixed point $u$ in $\mathbb{M}$, i.e.
\[
u= \frac{T({u})}{\|T(u)\|}
\]
or
\[
T(u) ={\|T(u)\|} u.
\]
We let $u=\lambda y$ and obtain
\[
T(y)=\lambda ^{1-p}r y, \quad r=\|T(u)\|,
\]
and for given $\mu $ choose $\lambda $ such that
$\mu = \lambda ^{1-p}r$.
\end{proof}

To provide an example illustrating the above result, we consider the
following.
Let
\[
E:=C[0,1],
\] with the usual maximum norm, and let
\[
G:[0,1]^2\to [0,\infty)
\] be a nontrivial continuous function. Let
$T:E\to E$
be given by
\begin{equation}
\label{hom}
T(u)(t):=\int _0^1G(t,s)|u(s)|^{p-1}u(s)ds,
\end{equation}
where $0<p<1$ is a constant.
In $E$ we may consider the solid cone
\[
K:=\{u\in E:u(t)\geq 0, ~0\leq t\leq 1 \}.
\]
Then $T$ is a monotone operator which is homogeneous of degree
$p$ and the norm of $E$
is monotone with respect to $K$. We hence have the following result.

\begin{example}\rm
Let the above assumptions hold. Then for any
$\mu \in (0,\infty)$ there exists a continuous function
\[
u:[0,1]\to [0,\infty)
\]
with
$u:(0,1)\to (0,\infty)$
solving the integral equation
\begin{equation}
\label{hom1}
\mu u(t)=\int _0^1G(t,s)|u(s)|^{p-1}u(s)ds.
\end{equation}
\end{example}

\subsection{On positive eigenvectors and eigenvalues}

In this section we shall assume that
$T:E\to E$
is a linear operator which is positive with respect to the cone
$K$ and satisfies
\begin{equation}
\label{pos}
T(K^+)\subset K^+, \quad
T(\mathop{\rm int}K)\subset \mathop{\rm int}K.
\end{equation}
We shall establish the classical
results of Perron-Frobenius\index{Perron-Frobenius} and Krein-Rutman\index{Krein-Rutman}
(see, for example \cite{krasnoselskii:pso64})
about principal eigenvalues of a special class of such operators.

We call a  positive linear operator $S$ uniformly
positive\index{uniformly
positive}, provided there exists $u_0\in \mathop{\rm int}K$ and a constant $\beta
>1$
such that
\begin{equation}
\label{hil30}
\lambda (u)u_0\leq S(u)\leq \beta \lambda (u)u_0,\quad
u\in \mathop{\rm int}K,
\end{equation}
where $\lambda (u)$ is a positive constant depending upon $u$.

We have the following theorem.

\begin{theorem} \label{perron}
Let the norm of  $E$ be monotone with respect to the cone $K$ and
let $T$ be a linear positive operator satisfying (\ref{pos}) such
that for some integer $n$, the operator $T^n$
 is uniformly positive. Then there exists a unique pair $(\mu ,u)\in
(0,\infty )\times \mathbb{M}$
 such that
\begin{equation}
\label{hil31}
T(u)=\mu u.
\end{equation}
\end{theorem}

\begin{proof}
We define the mapping
$g:\mathbb{M}\to \mathbb{M}$ by
\[
g(u):= \frac{T(u)}{\|T(u)\|},
\]
and let
\[
f:=\underbrace{g\circ \dots \circ g}_{n},
\]
i.e., $g$ composed with itself $n$ times.
Then
\[
f(u)= \frac{S(u)}{\|S(u)\|},
\]
where $S=T^n$.
It follows from the properties of Hilbert's projective metric,
that
\[
{\rm d}(f(u),f(v))={\rm d}(S(u),S(v)),\quad u,v\in \mathbb{M},
\]
and that $\mathbb{M}$ is complete, since the norm of $E$ is monotone
with respect to the cone $K$.
Thus, $f$ will have a unique fixed point, once we show that
$f$ is a contraction mapping, which will follow from Theorem \ref{hil} once we establish that
 $\Delta (S)<\infty $.

To compute $\Delta (S)$, we recall the definition of projective
diameter\index{projective diameter} (see (\ref{hil17})) and find
that for any $u,v\in \mathbb{M}$,
\[
{\rm d}(S(u),S(v))\leq {\rm d}(S(u),u_0)+{\rm d}(S(v),u_0)
\]
and, therefore, by the uniform positivity of $T^n$,
\[
 {\rm d}(S(u),u_0),\;{\rm d}(S(v),u_0)\leq \log\beta ,
\]
implying that
\[
{\rm d}(S(u),S(v))\leq 2\log \beta .
\]
Thus $S$, and hence, $f$, are contraction mappings with respect to
the projective metric and therefore, there exists a unique $u\in
\mathbb{M}$ such that
\[
f(u)=u,
\]
i.e. $S(u)=u$,
or
\[
T^n(u)=\|T^n(u)\| u,
\]
and the direction $u$ is unique. Furthermore, since $f$ has a
unique fixed point in $\mathbb{M} $, $g$ will have a unique fixed
point also, as follows from Theorem \ref{pcontraction} of Chapter
\ref{chapIII}. This also implies the uniqueness of the  eigenvalue
with corresponding unique eigenvector $u\in \mathop{\rm int}K,\;\|u\|=1$.
\end{proof}

In the following we provide two examples to illustrate the above
theorem.
The first example illustrates part of the Perron-Frobenius theorem and
the second is an extension of this result to operators on spaces of
continuous functions. We remark here that the second result concerns
an integral equation which is not given by a compact linear operator
(see also \cite{birkhoff:ejt57}).


\begin{example} \label{hilbexm5} \rm
Let
\[
E=\mathbb{R}^N,\quad
K=\{(u_1,u_2, \dots , u_N): u_i\geq 0,\;i=1,2, \dots , N\}.
\]
Let
$T:K\to K$
be a linear transformation whose $N\times N$ matrix representation
is irreducible. Then there exists a unique pair $(\lambda , u)\in
(0,\infty )\times \mathop{\rm int}K$, $\|u\|=1 $, such that
\[
Tu =\lambda u.
\]
\end{example}

\begin{proof}
An $N\times N$ matrix is irreducible (see \cite{leon:laa98}), provided
 there does not
 exist a
permutation matrix $P$ such that
\[
PTP^T=\begin{pmatrix}
B&O\\
C&D
\end{pmatrix},
\]
where $B$ and $D$ are square submatrices. This is equivalent to
saying, that for some positive integer $n$, the matrix $T^n=\left
(t_{i,j}\right )$ has only positive entries $t_{i,j},i,j=1,\dots
, N$. Since,
\[
\mathop{\rm int}K=\{(u_1,u_2,
\dots , u_N): u_i> 0,\;i=1,2, \dots , N\},
\]
if we let
\[
m=\min _{i,j}t_{i,j},\quad
M=\max _{i,j}t_{i,j},\quad u_0=(1,1,\dots ,1),
\]
then for any $u\in K^+$,
\[
m\|u\|_1u_0\leq T^nu\leq M\|u\|_1u_0,
\]
where
\[
\|u\|_1=\sum _{i=1}^N|u_i|,
\]
is the $l_1$ norm of the vector $u$.
This shows that $T^n$ is a uniformly positive operator. Hence Theorem
\ref{perron} may be applied.
\end{proof}

For many applications of positive matrices (particularly to  economics)
we refer to \cite{leon:laa98}, \cite{strang:laa88}.
The following example is discussed in \cite{birkhoff:ejt57}.

Let again $E:=C[0,1]$,
with the usual maximum norm and $K$ the cone of nonnegative
functions.
Let
\[
p:[0,1]^2\to (0,\infty)
\] be a continuous function.
Let
\[
0<I:=\inf _{[0,1]^2}p(x,y)\leq \sup _{[0,1]^2}p(x,y)=:\mu I.
\]
Suppose
\[
g:[0,1]\to [0,1]
\] is a continuous function
and define
$T:E\to E$
by
\begin{equation}
\label{perron1}
T(u)(x):=\int _0^1p(x,y)u(y)dy +au(g(x)),
\end{equation}
where $a$ is a positive constant.

We have the following example.

\begin{example}\label{perron2} \rm
Let $T$ be defined by (\ref{perron1}), where $p,g,a$ satisfy the above
conditions. Then there exists a unique positive number $\lambda $ and
a continuous function
\[
u:[0,1]\to (0,1],\quad \max _{x\in [0,1]}u(x)=1,
\] such that
\begin{equation}
\label{perron3}
\lambda u(x)=\int _0^1p(x,y)u(y)dy +au(g(x)),\quad 0\leq x\leq 1.
\end{equation}
\end{example}

To see how the result of Example \ref{perron2} follows from Theorem
\ref{perron} we proceed as follows.

We replace the cone $K$ by the following subcone, which we denote
by
$K_1$
\[
K_1:=\{u\in K: \max u\leq \nu \min u\},
\]
where
\[
\max u=\max_{x\in [0,1]}u(x),\quad
\min u=\min_{x\in [0,1]}u(x),
\]
and
$\nu > \mu$.

Easy computations show that the norm is monotone with respect to
the new cone $K_1 $, and that for any $x\in [0,1]$
\[
(I+1)\min u\leq (Tu)(x)\leq \nu (\mu I+a)\min u .
\]
This inequality shows that the operator $T$ is a uniformly
positive operator, as required by the theorem. Furthermore,
letting $v=Tu$, we obtain that
\begin{gather*}
\max v\leq \mu I\int _0^1udx+a\nu \min u,
\\
\min v\geq I \int _0^1udx+a \min u,
\end{gather*}
and therefore
\[
 \frac{\max v}{\min v}\leq  \frac {\mu I\int
_0^1udx+a\nu \min u}{I \int _0^1udx+a \min u}\leq \nu,
\]
showing that
$T:K_1\to K_1$.
We may, hence, apply Theorem \ref{perron}. We remark  that the
operator $T$, above, is not a compact operator and hence
techniques based on Leray-Schauder degree and fixed point theory
may not be applied here.


Let us consider another situation, to which the
results derived above apply.

\begin{theorem} \label{krein}
Let the norm of $E$ be monotone with respect to the solid cone $K$
  and let
\[
T:K^+\to \mathop{\rm int}K,
\]
be a positive, linear, and compact operator.
 Then $T$ has a unique eigenvector in $\mathbb{M}$.
\end{theorem}

\begin{proof}
Let  $v\in \mathbb{M}$ be a fixed element.
Then $T(v) \in \mathop{\rm int}K$. Hence, there exist positive
numbers $\epsilon >0$ and
$\alpha $, depending on $v$, such that
\[
\alpha v\leq T(v),\;\text{and}~ \overline{B}(v,\epsilon)\subset
\mathop{\rm int}K .
\]
For each integer $n=1,2, \dots $, we define the mapping
$S_n:\mathbb{M}\to \mathbb{M}$, by
\[
S_n(u):=\frac{T(u+\frac{1}{n}v)}{\|T(u+\frac{1}{n}v)\|}.
\]
It follows from earlier considerations, that $S_n$ will be a
contraction mapping with respect to Hilbert's metric, once we show
that
\[
{\rm d}(S_n(u),v)\leq c,
\]
where $c$ is a constant, independent of $u\in \mathbb{M}$.
To see this, we observe that
\[
S_n(u)\geq \frac{T(\frac{1}{n}v)}{\|T(u+\frac{1}{n}v)\|}
\geq \frac{\alpha }{n(1+\frac{1}{n})\|T\|}v,
\]
hence,
\[
m(S_n(u),v)\geq \frac{\alpha }{n(1+\frac{1}{n})\|T\|}.
\]
Furthermore, since
$v-\epsilon S_n(u)\in \mathop{\rm int}K$,
it follows that
\[
M(S_n(u),v)\leq \frac{1}{\epsilon} .
\]
Thus,
\[
{\rm d}(S_n(u),v)
\leq \log \frac{n(1+\frac{1}{n})\|T\|}{\epsilon \alpha}.
\]
Hence, for each $n=1,2, \dots $, there exists a unique $u_n\in
\mathbb{M} $ such that
\[
S_n(u_n)=u_n,
\]
i.e.,
\begin{equation} \label{eigen}
T(u_n+\frac{1}{n}v)=\lambda _nu_n,
\end{equation}
where
\[
\lambda _n=\|T(u_n+\frac{1}{n}v)\|.
\]
This implies that
$\lambda _n\leq 2\|T\|$.
Since,
\[
m(u_n,v)v\leq u_n,
\]
and $m(u_n,v)>0$  is the maximal number $\lambda $
such that
$\lambda v\leq u_n$,
we obtain that
\begin{align*}
 u_n&=  \frac{1}{\lambda _n}T(u_n+\frac{1}{n}v)\\
 &\geq  \frac{1}{\lambda _n}T(
m(u_n,v)v+\frac{1 }{n}v)\\
 &\geq  \frac{\alpha}{\lambda _n}(
m(u_n,v)+\frac{1 }{n})v .
\end{align*}
Therefore, by the maximality of $m(u_n,v)$, we obtain
\[
\frac{\alpha}{\lambda _n}\Big(m(u_n,v)+\frac{1}{n}\Big)\leq
m(u_n,v),
\]
i.e.,
\[
\alpha \Big(1+\frac{1}{nm(u_n,v)}\Big)\leq \lambda _n.
\]
The sequence $\{\lambda _n\}$ is therefore uniformly bounded away from
zero and, as has been shown above,  also bounded above.
 It therefore  has a convergent
subsequence, $\{\lambda _{n_i}\}$, converging, say, to $\lambda
>0$.
 We now use equation
(\ref{eigen}) and the compactness of $T$ to obtain that the
sequence $\{u_{n_i}\}$ has a convergent subsequence, converging,
to, say, $u,\;\|u\|=1$, and, since $T$ is continuous,
\[
T(u)=\lambda u .
\]
Since it must be the case that $u\in K^+$, we see that, in fact,
$u\in \mathop{\rm int}K$, and hence, $u\in \mathbb{M}$.

If $u_1,u_2 \in \mathbb{M}$ are such that
\[
T(u_1)=\lambda _1u_1, ~T(u_2)=\lambda _2u_2 ,
\]
then
\[
u_1\geq m(u_1,u_2)u_2,\;m(u_1,u_2)>0,
\]
and
\[
\lambda _1u_1=T(u_1)\geq m(u_1,u_2)T(u_2)=m(u_1,u_2)\lambda _2 u_2,
\]
hence,
\[
u_1\geq \Big(m(u_1,u_2)\frac{\lambda _2}{\lambda _1}\Big)u_2.
\]
On the other hand
\[
m(u_1,u_2)=\sup \{\alpha :\alpha u_2\leq u_1\},
\]
which implies that $ \lambda _1 \geq \lambda _2$. Reversing the
roles of $u_1$ and $u_2$, we obtain $\lambda _1 \leq \lambda _2$,
and thus,
\[
\lambda _1 = \lambda _2=\lambda .
\]
Since
\[
u_1\geq m(u_1,u_2)u_2,\;m(u_1,u_2)>0,
\]
we have
\[
T(u_1- m(u_1,u_2)u_2)\in \mathop{\rm int}K,
\]
unless
\[
u_1- m(u_1,u_2)u_2=0.
\]
On the other hand
\[
T(u_1- m(u_1,u_2)u_2)=\lambda (u_1- m(u_1,u_2)u_2),
\]
and thus,
if
\[
u_1- m(u_1,u_2)u_2\in K^+,
\]
then
\[
u_1- m(u_1,u_2)u_2\in \mathop{\rm int}K,
\]
 and we obtain a contradiction to the maximality of
$m(u_1,u_2)$. Hence,
\[
u_1= m(u_1,u_2)u_2
\]
and since, $\|u_1\|=\|u_2\|=1$, we  have that $m(u_1,u_2)=1 $ and
we have proved that $u_1=u_2$.
\end{proof}


\section{Integral equations} \label{chapVII} % plt  Chapter VII  include{contexst}



In this chapter, we shall present the basic existence and
 uniqueness theorem for solutions of initial value
 problems\index{initial value problem} for systems of ordinary differential equations. We shall
 also discuss the existence of mild solutions\index{mild solutions} of integral equations
which under additional assumptions provide the existence of solutions
 of initial value problems for parabolic partial differential
 equations.
We conclude the chapter by presenting some results about functional
 differential equations and integral equations.

\subsection{Initial value problems}\label{secV1}
 To this end let $D$ be an open connected subset of
 $\mathbb{R}\times E$, where $E$ is a Banach space, and let
\[
f:D\to E
\]
be a continuous and bounded mapping, i.e., it maps bounded sets in $D$
to bounded sets in $E$.

We consider the differential equation
\begin{equation}\label{chIVeq1}
u'=f(t,u),\quad '=\frac{d}{dt}.
\end{equation}
and seek sufficient conditions for the existence of
 solutions of \eqref{chIVeq1}, where
 $u\in C^1(I,E)$, with $I$ an interval, $I\subset \mathbb{R}$,
is called a solution, if $(t,u(t))\in D$, $t\in I$ and
\[
u'(t)=f(t,u(t)),\quad t\in I.
\]
 By an {\it initial value problem\index{initial value
problem}} we mean the following:

 Given a point
$(t_0,u_0)\in D$ we seek a solution
 $u$ of \eqref{chIVeq1} defined on some open interval $I$ such that
\begin{equation}\label{chIVeq2}
u(t_0)=u_0,\quad t_0\in I.
\end{equation}
We have the following proposition whose proof is straightforward:

\begin{proposition}\label{pro41}
A function $u\in C^1(I,E)$, with $I\subset \mathbb{R}$,
 and $I$ an interval containing $t_0$ is a solution of
 the initial value problem \eqref{chIVeq1},
 satisfying the initial condition \eqref{chIVeq2}
 if, and only if, $(t,u(t))\in D,\;t\in I$, and
\begin{equation}\label{chIVeq3}
u(t)=u(t_0)+\int _{t_0}^tf(s,u(s))ds.
\end{equation}
\end{proposition}

The integral in (\ref{chIVeq3}) is a Riemann integral of a continuous function.

We shall now, using Proposition \ref{pro41}, establish one of
  the classical and basic
existence and uniqueness theorems.

\subsection{The Picard-Lindel\"{o}f theorem}\label{secIV2}

We say that $f$ satisfies a {\it local Lipschitz}
condition\index{local Lipschitz condition}
 on the domain $D$, provided for every closed and bounded
 set $K\subset D$, there exists a constant $L=L(K)$,
 such that for all $(t,u_1),\;(t,u_2)\in K$
\[
\|f(t,u_1)-f(t,u_2)\|\leq L\|u_1-u_2\|,
\]
where $\|\cdot \|$ is the norm in the space $E$.
For such functions, one has the following existence and uniqueness
theorem.
This result is usually called the Picard-Lindel\"of
theorem.\index{Picard-Lindel\"of theorem}

\begin{theorem}\label{thm41}
Assume that $f:D\to E$ is a continuous and bounded mapping
which satisfies a
 local Lipschitz condition on  the domain $D$.
 Then for every $(t_0,u_0)$ in $D$, equation \eqref{chIVeq1}
 has a unique solution on some interval $I$ satisfying the initial condition
\eqref{chIVeq2}.
\end{theorem}

We remark that the theorem as stated
 is a {\it local} existence and uniqueness theorem,
in the sense that the interval $I$, where the solution exists will
depend upon the initial
 condition.

\begin{proof}
Let $(t_0,u_0)\in D$, then, since $D$ is open,
 there exist positive constants $a$ and $b$ such that
\[
Q=\{(t,u):|t-t_0|\leq a,\;\|u-u_0\|\leq b\}\subset D.
\]
Let $L$ be the Lipschitz constant for $f$ associated
 with the set $Q$. Further let
\begin{gather*}
m:= \sup_{(t,u)\in Q}\|f(t,u)\|,\\
\alpha := \min\{a,\frac{b}{m}\}.
\end{gather*}
Let $\tilde L$ be any constant, $\tilde{L}>L$, and define
$I=[t_0-\alpha , t_0+\alpha]$, and
\[
\mathbb{M}:=\{u\in C\left (I, E\right ):u(I)\subset B_r(u_0) \}.
\]
In $C\left (I, E\right )$
we define a new norm as follows:
\[
\|u\|_{\mathbb{M}}:=\max_{|t-t_0|\leq \alpha}e^{-\tilde{L}|t-t_0|}\|u(t)\|.
\]
And we let ${\rm d} (u,v)=\|u-v\|_{\mathbb{M}}$, then
$(\mathbb{M},{\rm d})$
 is a complete metric
 space.
 Next define the operator $T$ on $\mathbb{M}$ by:
\begin{equation}\label{chIVeq4}
(Tu)(t):=u_0+\int _{t_0}^tf(s,u(s))ds,\;|t-t_0|\leq \alpha.
\end{equation}
Then
\[
\|(Tu)(t)-u_0\|\leq \big|\int _{t_0}^t\|f(s,u(s))\|ds\big|,
\]
and, since $u\in \mathbb{M}$,
\[
\|(Tu)(t)-u_0\|\leq \alpha m\leq b.
\]
Hence
$T:\mathbb{M}\to \mathbb{M}$.
Computing further, we obtain, for $u,v\in \mathbb{M}$ that
\begin{align*}
\|(Tu)(t)-(Tv)(t)\|
&\leq \Big|\int_{t_0}^t\|f(s,u(s))-f(s,v(s))\|ds\Big|\\
&\leq L\Big|\int _{t_0}^t\|u(s)-v(s)\|ds\Big|,
\end{align*}
and hence
\begin{align*}
e^{-\tilde{L}|t-t_0|}\|(Tu)(t)-(Tv)(t)\|
&\leq e^{-\tilde{L}|t-t_0|} L\Big|\int _{t_0}^t\|u(s)-v(s)\|ds\Big|\\
 &\leq  \frac{L}{\tilde{L}}\|u-v\|_{\mathbb{M}}.
\end{align*}
Therefore,
\[
{\rm d} (Tu,Tv)\leq
\frac{L}{\tilde{L}}{\rm d} (u,v),
\]
proving that $T$ is a contraction mapping. The result therefore
follows from the contraction mapping principle.
\end{proof}

We remark that, since $T$ is a contraction mapping\index{contraction mapping},
  the contraction mapping theorem gives a constructive
 means for the  solution of the initial value problem
in Theorem \ref{thm41} and the solution may in fact be
obtained via an iteration procedure. This procedure is known
as {\it Picard iteration\index{Picard iteration}.}

\subsection{Abel-Liouville integral equations}
\label{abel}
In establishing the Picard-Lindel\"of theorem we studied an associated
integral equation
\begin{equation}\label{abel1}
u(t)=u(t_0)+\int _{t_0}^tf(s,u(s))ds.
\end{equation}
By translating the time variable appropriately and changing variables,
there is no loss in generality in assuming that $t_0=0$ in
(\ref{abel1}).
 In this
section we shall consider a generalization of this integral equation,
namely an equation of Abel-Liouville type\index{Abel-Liouville
equation}
\begin{equation}\label{abel2}
u(t)=v(t)+\frac{1}{\Gamma (\mu)}\int _{0}^t(t-s)^{\mu
-1}f(t,s,u(s))ds,\quad 0\leq t \leq a,\;a>0,
\end{equation}
where $\mu \in (0,1]$, and $\Gamma $ is the Gamma function.
Choosing $\mu =1 $ and $v(t)=\text{constant}$, one clearly obtains
(\ref{abel1}) as a special case.

While the result to follow will be valid for Banach space valued
functions,
we shall restrict ourselves to the case of real-valued functions,
remarking that the treatment will be similar in the more general setting.
The discussion below follows the paper of Rainermann and Stallbohm
\cite{reinermann:aef71}, see also \cite{jeggle:nfa79}.

We shall introduce the following notations and
make the  assumptions:
\begin{enumerate}
\item
$v:[0,a]\to \mathbb{R}$ is a continuous function.

\item
\[
S:=\{z:|v(s)-z|\leq b,\; 0\leq s \leq a\},
\]
where $b>0$ is a fixed positive constant.
\item
\[
\Delta :=\{(t,s):0\leq s\leq t\leq a \}.
\]
\item
$f:\Delta \times S \to \mathbb{R}$
is a continuous function and
\[
M:=\max \{|f(t,s,z)|: (t,s,z)\in \Delta \times S \}.
\]
\item for all
$(t,s,z_1),(t,s,z_2)\in \Delta \times S$,
\begin{equation}
\label{abel3}
s^{\mu}|f(t,s, z_1)-f(t,s, z_2)|\leq \Gamma (\mu+1 )|z_1-z_2|.
\end{equation}
\item
\[
\alpha :=\min \Big(a, \Big(\Gamma (\mu +1)\frac{b}{M}\Big)^{1/\mu}\Big).
\]
\item
\[
\mathbb{M} _1:=\{ w\in C[0,\alpha ]:w(0)=v(0),\;\max_{0\leq t\leq
 \alpha} |w(t)-v(t)|\leq b\} .
\]
\item
We define the operator
$T:\mathbb{M}_1 \to C[0,\alpha ]$
by
\begin{equation}
\label{abel4}
T(w)(t):=v(t)+\frac{1}{\Gamma (\mu)}\int _{0}^t(t-s)^{\mu
-1}f(t,s,u(s))ds,\quad 0\leq t \leq \alpha .
\end{equation}
\end{enumerate}

\begin{lemma}\label{abel5}
With the above notation and assumptions, we have
\[
T:\mathbb{M}_1 \to \mathbb{M}_1.
\]
\end{lemma}

\begin{proof}
It is clear that
$T:\mathbb{M}_1 \to C[0,\alpha ]$.
Further
\[
T(w)(0)=v(0),
\]
and for $0\leq t\leq \alpha$,
\begin{align*}
|T(w)(t)-v(t)|
&\leq M \frac{1}{\Gamma (\mu)}\int _{0}^t(t-s)^{\mu
-1}ds\\
&\leq  M\frac{t ^{\mu}}{\Gamma (\mu +1)}\\
&\leq  M\frac{\alpha ^{\mu}}{\Gamma (\mu +1)}\\
&\leq  b .
\end{align*}
We next let
 \[
\mathbb{M}:= T(\mathbb{M} _1 )
\]
and
find a metric ${\rm d}$ on $\mathbb{M}$ so that
$(\mathbb{M}, {\rm d})$ becomes a complete metric space.
The above lemma, of course, implies that
$T:\mathbb{M} \to \mathbb{M}$.

We next define
${\rm d}:\mathbb{M}\times \mathbb{M}\to [0,\infty )$
by
\[
{\rm d}( w_1,w_2):=\sup _{t\in [0,\alpha ]}|w_1(t)-w_2(t)|.
\]
Then, it is easily seen that ${\rm d}$ is a metric on
$\mathbb{M}$ and, using the continuity assumptions imposed on $f$,
that for all $u_1,u_2\in \mathbb{M}$,
\begin{align*}
|T(u_1)(t)-T(u_2)(t)|
&\leq \frac{1}{\Gamma (\mu)}\int _{0}^t(t-s)^{\mu
-1}|f(t,s,u_1(s))-f(t,s,u_2(s))|ds\\
&\leq  t^{\mu}\max_{0\leq s\leq t}|f(t,s,u_1(s))-f(t,s,u_2(s))|,
\end{align*}
implying that (recall that $u_1(0)=u_2(0)$)
\[
t^{-\mu}|T(u_1)(t)-T(u_2)(t)|\to 0.
\]
Also
\begin{align*}
|T(u_1)(t)-T(u_2)(t)|
&\leq \frac{1}{\Gamma (\mu)}\int _{0}^t(t-s)^{\mu
-1}|f(t,s,u_1(s))-f(t,s,u_2(s))|ds\\
&\leq \int _{0}^t(t-s)^{\mu
-1}s^{-\mu }|u_1(s)-u_2(s)|ds .
\end{align*}
On the other hand, if $u_1\ne u_2$, then $\mathop{\rm
d}(u_1,u_2)>0$. Thus, if  ${\rm d}(T(u_1),T(u_2))\ne 0$, we
may choose $t_1 \in (0,\alpha ]$ so that
\begin{align*}
{\rm d}(T(u_1),T(u_2))
&= t_1^{-\mu}|T(u_1)(t_1)-T(u_2)(t_1)|\\
&\leq \frac{\Gamma (\mu +1)}{\Gamma (\mu)}t_1^{-\mu}\int _{0}^t(t-s)^{\mu
-1}s^{-\mu }|u_1(s)-u_2(s))|ds\\
&< \mu t_1^{-\mu}\int _{0}^t(t-s)^{\mu -1}ds {\rm d}(u_1,u_2),
\end{align*}
where the latter strict inequality follows from the continuity of the
functions involved and the  fact that there
exists
$s\in (0,t_1)$ such that
\[
s^{-\mu }|u_1(s)-u_2(s))|<
{\rm d}(u_1,u_2).
\]

Using calculations and considerations like the above, it is
straightforward to show that the family $\mathbb{M}$ is an
equicontinuous family of functions which is uniformly bounded.
Hence, for a given $u\in \mathbb{M}$ the sequence of iterates
$\{T^n(u)\}$ will have a convergent subsequence, say
$\{T^{n_j}(u)\}$, converging to some function $v\in \mathbb{M}_1$.
The sequence $\{T^{n_j+1}(u)\}$, will therefore converge to
$T(v)\in \mathbb{M} $.

It therefore follows from Edelstein's contraction mapping
principle,
Theorem III.\ref{wcontraction}, that equation (\ref{abel2})
has a unique solution in the metric space $\mathbb{M}$.
\end{proof}


\subsection{Mild solutions}
\label{mild}

 Let $E$ be a Banach space and let
\[
S:[0,\infty )\to \mathcal{L}(E,E)
\]
($\mathcal{L}(E,E)$ are the bounded linear maps from $E$ to $E$)
be a family of bounded linear operators which form a {\it strongly
continuous}\index{strongly continuous}
 {\it semigroup}\index{semigroup} of operators, i.e.,
\begin{gather*}
S(t+s)=S(t)S(s), ~\forall t,s \geq 0,
\\
S(0)={\rm id},\;\text{the identity mapping}
\\
\lim _{t\to t_0}S(t)x=S(t_0)x,\quad \forall t_0\geq 0,x\in E.
\end{gather*}
For such semigroups it is the case that there exist constants
(see \cite{yosida:fa95})
$\beta \in \mathbb{R} $ and $M>0$ such that
\[
\|S(t)\|\leq M e^{\beta t},\quad t\geq 0.
\]
(In this section, we shall use $\|\cdot \|$ for both the norm in
$E$, and the norm in $\mathcal{L}(E,E).$)

Let
$f:[0,\infty )\to E$
be a continuously differentiable  function and let
\begin{equation}
\label{semi1}
u(t):=S(t)x+\int _0^tS(t-s)f(s)ds.
\end{equation}
The integral in (\ref{semi1}) is a Riemann integral of a
continuous function. Then for $x\in D(A)$, where
\[
D(A):=\{x:Ax:=\lim _ {t\to 0+}\frac{1}{t}\left (S(t)x-x \right )
~\text{exists}\},
\]
(the operator $A$ is called the {\it infinitesimal
generator}\index{infinitesimal generator} of the semigroup
$\{S(t);t\geq 0\}$) $u$, given by (\ref{semi1}) is the solution of
the initial value problem (\ref{semi2}) (see
\cite{showalter:hsm77}, \cite{yosida:fa95})
\begin{equation}
\label{semi2}
\frac{du}{dt}=Au(t)+f(t),\quad u(0)=x.
\end{equation}

On the other hand, if it is only assumed that $f$ is continuous,
then a solution of (\ref{semi1}) need not necessarily be a solution of
(\ref{semi2}). One calls the function $u$ defined by (\ref{semi1}) a
{\it mild
solution}\index{mild solution} of (\ref{semi2}). Thus a mild solution
 is defined for all $x\in E$, even if $D(A)$ is a proper subset of
$E$.

For example, one may easily verify that, if $A\in
\mathcal{L}(E,E)$, then
\[
S(t):=e^{At}=\sum_{n=0}^{\infty}\frac{A^nt^n}{n!}
\]
is a strongly continuous semigroups of operators and if $u$ is defined
by
(\ref{semi1}), then $u$ solves (\ref{semi2}) for any $x\in E$ and any
continuous $f:[0,\infty)\to E$.

We have the following theorem for the existence of mild solutions.

\begin{theorem} \label{mildexst}
 Let $S:[0,\infty )\to \mathcal{L}(E,E)$ be a strongly
continuous semigroup of operators and let
\[
f:[0,\infty )\times E\to E
\]
be a continuous mapping which satisfies the Lipschitz condition
\begin{equation}
\label{semi3}
\|f(t,u)-f(t,v)\|\leq L\|u-v\|,\forall t\in [0,\infty),\quad \forall u,v\in
E,
\end{equation}
where $L$ is a positive constant.
Then the equation
\begin{equation}
\label{semi4}
u(t)=S(t)x+\int _0^tS(t-s)f(s,u(s))ds
\end{equation}
has a unique continuous solution $u:[0,\infty )\to E$, i.e., the
initial value problem
\begin{equation}
\label{semi5}
\frac{du}{dt}=Au(t)+f(t,u),\quad u(0)=x,
\end{equation}
where $A$ is the generator of $\{S(t);t\geq 0\}$, has a unique
mild solution.
\end{theorem}

\begin{proof}
Given any $\tau >0$, we shall prove the existence of a unique
continuous solution of (\ref{semi4}) defined on the interval
$[0,\tau ]$. Since $\tau $ is chosen arbitrarily the result will
follow, observing that if $u$ solves (\ref{semi4}) on an interval
$[0,\tau _1 ]$ it will solve (\ref{semi4}) on an interval $[0,\tau
_2 ]$, for any $\tau _2 \leq \tau _1$.

Let us consider the space
$\mathcal{E}:=C([0,\tau ], E)$
with norm
\[
\|u\|_{\tau }:=\max _{[0,\tau ]}e^{-(\tilde {L}+\beta )t}\|u(t)\|,
\]
where $\tilde {L}>0$ is to be chosen,
and define the mapping
$T:\mathcal{E}\to \mathcal{E}$
by
\begin{equation}
\label{semi6}
T(u)(t):=S(t)x+\int _0^tS(t-s)f(s,u(s))ds.
\end{equation}
For $u,v \in \mathcal{E}$ we compute
\[
e^{-(\tilde {L}+\beta )t}\|T(u)(t)-T(v)(t)\|\leq ML e^{-\tilde {L}t}
\int _0^te^{\tilde {L}s}e^{-(\tilde {L}+\beta )s}\|u(s)-v(s)\|ds.
\]
It follows that
\begin{equation}
\label{semi7}
\|T(u)-T(v)\|_{\tau}\leq \frac{ML}{\tilde {L}}\|u-v\|_{\tau}.
\end{equation}
Therefore  $T$ is a contraction mapping provided
\begin{equation}
\label{semi8}
\frac{ML}{\tilde {L}}<1.
\end{equation}
We  choose $\tilde {L}$ this way and obtain the existence of a
unique
fixed point $u$ of $T$ in $\mathcal{E}$.
\end{proof}

\subsection{Periodic solutions of linear systems}
\label{periodic}
\subsection*{Mild periodic solutions}
\label{mps}
Let
$f:[0,\infty )\to E$
be a continuous  function which is periodic of period $\tau >0$,
i.e.
\[
f(t+\tau) =f(t),\quad t\geq 0.
\]
In this section we show that the integral equation
\begin{equation}
\label{mild1}
u(t)=S(t)x+\int _0^tS(t-s)f(s)ds,
\end{equation}
where $\{S(t):t\geq 0\}$ is a strongly
continuous\index{strongly continuous}
 semigroup\index{semigroup} of operators,
with
 infinitesimal
generator\index{infinitesimal generator} $A$, has a unique
periodic solution, provided the semigroup is a so-called {\it
asymptotically stable}\index{asymptotically stable}
 semigroup. I.e., we shall show that the problem
\begin{equation}
\label{mild2}
\frac{du}{dt}=Au(t)+f(t),
\end{equation}
has a unique mild solution\index{mild solution} which is also periodic of period $\tau $.


To this end, we shall employ the contraction mapping principle
as given by Theorem \ref{pcontraction} of Chapter \ref{chapII}.

We call the semigroup $\{S(t):t\geq 0\}$ asymptotically stable\index{asymptotically stable},
provided there exist constants $M>0, ~\beta >0$ such that
\[
\|S(t)\|\leq M e^{-\beta t},\quad t\geq 0.
\]

Well known examples of such asymptotically stable semigroups are
given in the finite dimensional case by $e^{At}$, where $A$ is an
$N\times N$ matrix all of whose eigenvalues have negative real
parts (see e.g. \cite{hartman:ode82}), or in the infinite
dimensional case by certain parabolic partial differential
equations (see e.g. \cite{evans:pde98}, \cite{showalter:hsm77}).

Let us define the operator $T:E\to E$
by
\begin{equation}
\label{mild3}
T(x):=S(\tau )x+\int _0^{\tau}S(\tau-s)f(s)ds.
\end{equation}
It then follows from Theorem \ref{mildexst} and the periodicity of
the function $f$, that
\begin{equation}
\label{mild4}
T^n(x)=S(n\tau )x+\int _0^{n\tau}S(n\tau-s)f(s)ds,
\end{equation}
for any positive integer $n$.
We have
\begin{equation}
\label{mild5}
\begin{aligned}
\|T^n(x)-T^n(y)\|&= \|S(n\tau )x-S(n\tau)(y)\|\\
&\leq \|S(n\tau )\|\|x-y\|\\
&\leq Me^{-\beta n\tau}\|x-y\|,\quad \forall x,y\in E.
\end{aligned}
\end{equation}
Since
\[
Me^{-\beta n\tau}<1,
\]
for $n$, large enough, we have that $T^n$ is a contraction mapping
for such $n$. It follows therefore that $T^n$, hence $T$, has a
unique fixed point $x\in E $. The solution
\begin{equation}
\label{mild6}
u(t)=S(t )x+\int _0^{t}S(t-s)f(s)ds,
\end{equation}
is a periodic function of period $\tau $.

Summarizing the above, we have proved the following theorem.

\begin{theorem} \label{persol}
Let $\{S(t):t\geq 0\}$ be an asymptotically stable,
strongly continuous semigroup with infinitesimal generator $A$.
 Then for any continuous $f:[0,\infty)\to E$ which is periodic of
period $\tau >0$, there exists a unique mild periodic solution
$u$, of period $\tau $ of equation
 (\ref{mild2}).
\end{theorem}

\subsection{The finite dimensional case}

We consider next the system of equations

\begin{equation}
\label{per1}
\frac{du}{dt}=Au(t)+f(t),
\end{equation}
where  $A$ is an $N\times N$ matrix and
$f:\mathbb{R} \to \mathbb{R}^N$,
is a function of period $T$.

We shall assume here that $A$ is a matrix all of whose eigenvalues
have
nonzero real part. If this is the case then there exists a nonsingular
matrix
$P$ such that (see, e.g. \cite{strang:laa88})
\[
P^{-1}AP=\begin{pmatrix}
A_1& O\\
O &A_2
\end{pmatrix},
\]
where  $A_1$ is an $N_1\times N_1$ and $A_2$ is an $N_2\times N_2
$ matrix with $N_1+N_2=N$, and all eigenvalues of $A_1$ have
negative real parts and those of $A_2$ have positive real parts,
and the matrices $O$ are zero matrices of appropriate dimensions.
Hence if we make the transformation $u=Pv$, the the system
(\ref{per1} becomes
\begin{equation}
\label{per2}
\frac{dv}{dt}=P^{-1}APv(t)+P^{-1}f(t).
\end{equation}
This is a decoupled system which we may rewrite as
\begin{equation}
\label{per3}
\begin{gathered}
\frac{dv_1}{dt} = A_1v_1(t)+f_1(t)\\
\frac{dv_2}{dt} = A_2v_2(t)+f_2(t),
\end{gathered}
\end{equation}
where
\[
v=\begin{pmatrix}
v_1\\
v_2
\end{pmatrix},\quad
P^{-1}f=\begin{pmatrix}
f_1\\
f_2
\end{pmatrix}.
\]
On the other hand $v$ is a solution of (\ref{per3})
if and only if
\[
w=\begin{pmatrix}
w_1(t)\\
w_2(t)
\end{pmatrix}
=\begin{pmatrix}
v_1(t)\\
v_2(-t)
\end{pmatrix}
\]
is a solution of
\begin{equation}\label{per4}
\begin{gathered}
\frac{dw_1}{dt}= A_1v_1(t)+f_1(t)\\
\frac{dw_2}{dt}= -A_2w_2(t)-f_2(-t).
\end{gathered}
\end{equation}
The fundamental solution $S(t)$ of the system (\ref{per4}) is given by
\[
S(t)=e^{Bt},
\]
where $B$ is the matrix
\[
B=\begin{pmatrix}
A_1& O\\
O &-A_2
\end{pmatrix},
\]
all of whose eigenvalues have negative real part. Hence there
exist constants $ M>0$, $\beta >0$, such that
\[
\|S(t)\|\leq Me^{-\beta t}.
\]
We may therefore apply Theorem \ref{persol} to conclude that
equation (\ref{per4}) and hence (\ref{per1}) have unique periodic solutions.

We remark here that the existence of a unique periodic solution of
(\ref{per1}) also easily follows from the fact that $A$ is assumed not
to have any eigenvalues with zero real part, and hence that
\[
\mathop{\rm id}-e^{AT},
\] where $\rm id$ is the $N\times N$ identity matrix,
is nonsingular for any $T\ne 0$. Furthermore the reduction made above
shows that, without loss in generality we may assume that all the
eigenvalues of $A$ have negative real parts, say
\begin{equation}
\label{per41}
{\rm Re}(\lambda _i)\leq - \beta <0,
\end{equation}
where $\beta $ is a a positive constant and
$\lambda _1, \dots , \lambda _N$
are the eigenvalues of $A$. Using a further change of basis
(using the Jordan canonical form of $A$) we may  assume
that $A$ has the form (the, perhaps, unconventional labeling
has been chosen for convenience's sake)
\[
A=\begin{pmatrix}
\lambda _N& a_{2,N},&\dots ,&a_{N,N}\\
0&\lambda _{N-1},&\dots &a_{N-1,N}\\
\vdots &\ddots &\vdots &\vdots \\
0 &\dots &\dots &\lambda _1
\end{pmatrix}.
\]

We thus consider the system
\begin{equation}
\label{per5}
\frac{du}{dt}=Au(t)+f(t),
\end{equation}
where
\[
u=\begin{pmatrix}
u_N\\
u_{N-1}\\
\vdots\\
u_1
\end{pmatrix},
\quad
f=\begin{pmatrix}
f_N\\
f_{N-1}\\
\vdots\\
f_1
\end{pmatrix}.
\]
If then $u$ is a $T-$ periodic solution of (\ref{per5})
it must be the case that
\[
u_1(t)=e^{\lambda _1t}\Big(c + \int _0^te^{-\lambda _1s}f_1(s)ds
\Big).
\]
Since $u_1$ is periodic, it must be bounded on the real line
and hence,
\[
c + \int _0^te^{-\lambda _1s}f_1(s)ds
\to 0,\quad \text{as } t\to -\infty ;
\]
i.e.,
\[
c =- \int _0^{-\infty}e^{-\lambda _1s}f_1(s)ds
\]
or
\[
u_1(t)=\int _{-\infty}^te^{\lambda _1(t-s)}f_1(s)ds.
\]
Therefore,
\[
|u_1(t)|\leq \frac{1}{\beta}\|f_1\|,
\]
(see (\ref{per41}) for the choice of $\beta $)
where
\[
\|f_1\|=\sup _{t\in \mathbb{R}}|f_1(t)|.
\]
We next consider the component $u_2$ of $u$.
Since $u_1$ has been found (and estimated), we can employ a similar
argument and the estimate on $u_1$ to find that
\[
|u_2(t)|\leq \frac{1}{\beta}\|a_{2,N}u_1 +f_2\|,
\]
and thus
\[
|u_2(t)|\leq \frac{c_2}{\beta}\max\{\|f_1\|,\|f_2\|\},
\]
where the constant $c_2$ only depends upon the matrix $A$.
Using an induction argument one obtains for $i=1,\dots , N$
\[
|u_i(t)|\leq \frac{c_i}{\beta}\max\{\|f_1\|,\dots ,\|f_i\|\}
\]
with constants $c_i$ depending on $A$, only. We therefore have,
letting
\[
\|u\|=\max_{i=1,\dots, N}\|u_i\|,\quad
\|f\|=\max_{i=1,\dots, N}\|f_i\|,
\]
that
\[
\|u\|\leq \frac{c}{\beta} \|f\|,
\]
where $c$ is a constant, depending upon $A$ only.
We summarize the above in the following theorem.

\begin{theorem} \label{per7}
Let $A$ be a matrix none of whose eigenvalues has zero real part.
Then for any given forcing term $f$, a periodic function of period
$T$, there exists a unique periodic solution $u$ of (\ref{per5}).
Further, there exists a constant $c$ which depends upon $A$ only,
such that
\[
\|u\|\leq \frac{c}{\beta} \|f\|,
\]
with the norms defined above.
\end{theorem}

We note that the conclusion of Theorem \ref{per7} is equally valid
if we replace the requirement that $f$ be periodic with the
 requirement that $f$ be bounded.

\subsection{Almost periodic differential equations}
\label{apde}

In this section we shall return to the equation
\begin{equation}
\label{ap1}
\frac{du}{dt}=Au(t)+f(t),
\end{equation}
and the more general nonlinear equation
\begin{equation}
\label{ap2}
\frac{du}{dt}=Au(t)+f(t,u),
\end{equation}
where it is assumed that $A$ is an $N\times N$ matrix and
either
\[
f:\mathbb{R}  \to \mathbb{R}^N\quad\text{or}\quad
f:\mathbb{R}\times \mathbb{R}^N \to \mathbb{R}^N
\]
is a function which is {\it almost periodic}\index{almost periodic}
in the $t$ variable (see definitions below). It will be assumed that
all eigenvalues of $A$ have nonzero real part which will imply that
for almost periodic $f$ equation (\ref{ap1}) has a unique almost
periodic solution. This fact will be employed for the study
of equation (\ref{ap2}) under the assumption that $f$ satisfies
a Lipschitz condition with respect to the dependent variable,
in which case (\ref{ap2}) will be shown to have a unique almost
periodic solution, provided the Lipschitz constant of $f$ is small
enough.
Our presentation relies mainly on the work of Coppel
\cite{coppel:app67},
see also \cite{corduneanu:apf89},
\cite{amerio:apf71} and \cite{fink:apd74},
and Theorem \ref{per7}, above.

\subsection{Bounded solutions}\label{bs}

We consider again the system (\ref{per5}) under the assumption that
none of the eigenvalues of $A$ have zero real part
and $f$ a continuous (not necessarily periodic) function.
 It then follows
from the superposition principle that, if (\ref{ap1}) has a bounded
solution,
no other bounded solutions may exist (the unperturbed system
\[
u'=Au,
\]
has the zero solution as the only bounded solution). Furthermore the
discussion above, proving Theorem \ref{per7}, may be used to establish
the following result.

\begin{theorem}\label{ap4}
Let $A$ be a matrix none of whose eigenvalues has zero real part.
Then for any given forcing term $f$, with $f$ bounded on $\mathbb{R}$,
there exists a unique bounded solution $u$ of (\ref{ap1}).
Further, there exists a constant $c$ which depends upon $A$ only,
such that
\[
\|u\|\leq \frac{c}{\beta} \|f\|,
\]
with the norms as defined before.
\end{theorem}

\subsection{Almost periodic functions}\label{apf}

Let us denote by $V$ the set of continuous functions
\[
V:=\{f:\mathbb{R}\to \mathbb{R}^N: \exists T>0:f(t+T)=f(t),\;t\in
\mathbb{R}\};
\]
i.e., the set of all continuous functions which are periodic of some
period,
the period not being fixed. We then define
\[
\tilde {V}:=\mathop{\rm span}(V),
\]
i.e., the smallest vector space containing $V$. We note that for
$f\in \tilde V $,
\[
\|f\|:=\sup _{t\in \mathbb{R}}\max _{i=1,\dots , N}|f_i(t)|<\infty ,
\]
and that $\|\cdot \|$ defines a norm in $\tilde V $. We then denote by
$E$ the completion of $\tilde V$ with respect to this norm
(the norm of uniform convergence on the real line).
This space is the space of {\it almost periodic}\index{almost
periodic}
functions which, by definition, includes all periodic and
quasiperiodic\index{quasiperiodic}
functions (i.e., finite linear combinations of periodic functions
having possibly
different
periods).  It follows that an almost periodic function is the
uniform limit (uniform on the real line) of a sequence of quasiperiodic functions.

For detailed studies of almost periodic functions we refer the reader
to \cite{amerio:apf71},
\cite{bohr:ff32}, \cite{favard:lfp33},
\cite{corduneanu:apf89},
 and \cite{fink:apd74}.


\subsection{Almost periodic systems}
In this section we shall first establish an extension to almost
periodic systems of Theorem \ref{persol} and then use it to establish
the existence of a unique almost periodic solution of system
(\ref{ap2}),
in case the nonlinear forcing term $f$ satisfies a Lipschitz condition
with respect to the dependent variable $u$. We have the following theorem.

\begin{theorem} \label{ap7}
Let $A$ be a matrix none of whose eigenvalues has zero real part.
Then for any given almost periodic forcing term $f$,
 there exists a unique almost periodic solution $u$ of
(\ref{ap1}).
Further, there exists a constant $c$ which depends upon $A$ only,
such that
\[
\|u\|\leq \frac{c}{\beta} \|f\|,
\]
with the norms defined above.
\end{theorem}

\begin{proof}
Let us consider system (\ref{ap1}) in case the forcing term $f$ is
quasiperiodic.
In this case $f$ may be written as a finite linear combination of
periodic functions, say
\[
f(t)=\sum _{i=1}^k f_i(t),
\]
where $f_i$ has period $T_i$, $i=1,\dots, k$. It follows from Theorem
\ref{persol} that each of the systems
\[
u'=Au +f_i(t),
\]
has a unique periodic solution $u_i(t)$, of period $T_i$,
$i=1,\dots ,k$. Hence, the superposition principle and Theorem
\ref{per7} imply that (\ref{ap2}) has the unique quasiperiodic
solution
\[
u(t)=\sum _{i=1}^k u_i(t),
\]
and
\[
\|u\|\leq \frac{c}{\beta} \|f\|.
\]
On the other hand, if $f$ is an almost periodic function, there exists
a sequence of quasiperiodic functions $\{f_n(t)\}$ such that
\[
f(t)=\lim _{n\to \infty }f_n(t),
\] where the limit is uniform on the real line. We let $\{u_n(t)\}$
be the sequence of quasiperiodic solutions of (\ref{ap1}) with $f$
replaced by $f_n$, $n=1,2,\dots $. Then
\[
\|u_n-u_m\|\leq \frac{c}{\beta }\|f_n-f_m\|.
\]
Hence, $\{u_n(t)\}$ is a Cauchy sequence of quasiperiodic functions
which is uniform on the real line and therefore must converge to an
almost periodic function $u$. Using equivalent integral equations, as
in Section 1, one shows that $u$ solves (\ref{ap1}) and, since $u $ is
bounded
it must satisfy
\[
\|u\|\leq \frac{c}{\beta} \|f\|.
\]
\end{proof}

We next consider the problem (\ref{ap2}). We have the following
theorem:


\begin{theorem} \label{ap8}
Let $A$ be a matrix none of whose eigenvalues has zero real part.
Let
\[
f:\mathbb{R}\times \mathbb{R}^N \to \mathbb{R}^N
\]
be such that
\begin{equation}
\label{lipap}
|f(t,u)-f(t,v)|\leq L|u-v|,\quad
\forall t\in \mathbb{R},\;\forall u,v\in
\mathbb{R}^N
\end{equation}
and let $f(\cdot, u)$ be almost periodic for each $u\in \mathbb{R}^N$.
Then
 there exists a constant $L_0$ such that for each
\[
L<L_0,
\]
equation (\ref{ap2} has a unique almost periodic solution $u$.
\end{theorem}

\begin{proof}
It follows from the Weierstrass approximation theorem and the
definition of almost periodicity (see e.g. \cite{corduneanu:apf89},
\cite{coppel:app67}) that for each almost periodic function $v$
the composition $f(t,v(t))$ is an almost periodic function.
We may thus define an operator
$T:E\to E$, $v\mapsto T(v)$
by setting
\[
u:=T(v),
\]
where $u$ is the unique almost periodic solution of
\[
u'=Au+f(t,v(t)),
\]
whose existence follows from  Theorem \ref{ap7}. This theorem also
implies that
\[
|T(v)(t)-T(w)(t)|\leq \frac{c}{\beta }\sup _{t\in \mathbb{R}}|f(t,v(t))-f(t,w(t))|,
\]
i.e., using the Lipschitz condition satisfied by $f$,
\[
\|T(v)-T(w)\|\leq L\frac{c}{\beta }\|v-w\|.
\]
The operator $T$ will therefore be a contraction mapping, provided
that
\[
L\frac{c}{\beta }<1.
\]
Thus, if we choose
\[
L_0=\frac {\beta }{c}
\]
the contraction mapping principle will apply and the existence
of a unique almost periodic solution follows.
\end{proof}

\section{The implicit function theorem}\label{chapVIII}
 % ift Chapter VIII \include{continverse}

In this chapter we shall prove the {\it implicit function theorem}
\index{implicit function theorem}
for mappings defined between Banach spaces which are Fr\'echet
differentiable\index{Fr\'echet differentiable}.
\subsection{Fr\'echet differentiable mappings}
Let us  assume we have Banach spaces $E,X$ and
let
\[
f: U \to X,
\]
(where $U$ is open in $E$) be a continuous mapping. Let $u_0\in
U$, then $f$ is said to be {\it Fr\'echet differentiable at} $u_0$
provided there exists
\[
L\in \mathcal{L} (E,X)
\]
 (the continuous (or bounded) linear
mappings from $E$ to $X$) such that
\[
f(u_0+v)=f(u_0)+L(v)+o(\|v\|).
\]

If $f$ is Fr\'echet differentiable for every $u\in U$, the mapping
$U\to \mathcal{L} (E,X)$
given by
\[
u\mapsto D_uf(u),
\]
where $ D_uf(u)$ is the Fr\'echet derivative of $f$ at $u$, is
then defined. We remark, that the Fr\'echet derivative is uniquely
determined (if it exists) and the above definition provides a
Taylor expansion expression\index{Taylor expansion}. (See
\cite{hoffman:aes75}, \cite{rudin:rca66}.)

\subsection{The implicit function theorem}
\label{ssecI54}
Let us  assume we have Banach spaces $E,X,\Lambda$ and
let
\[
f: U \times V \to X,
\]
(where $U$ is open in $E$, $V$ is open in
$\Lambda$) be a continuous mapping satisfying the following
condition:
\begin{quote}
For each $\lambda \in V$ the map
$f(\cdot,\lambda): U \to X$ is
Fr\'{e}chet-differentiable on $U$ with Fr\'{e}chet
derivative $
D_uf(u,\lambda )
$
 and the mapping
$(u,\lambda ) \mapsto D_u f(u,\lambda )$ is a
continuous
mapping from $ U \times V $ to $\mathcal{L} (E,X)$ (the continuous
(or bounded) linear
mappings
from $E$ to $X$).
\end{quote}

\begin{theorem}\label{thmI3}
Let $f$ satisfy the above condition and let there exist
$(u_0,\lambda _0) \in U \times V$ such that $D_uf(u_0,\lambda _0)$
is a linear homeomorphism of $E$ onto $X$ (i.e. $D_uf(u_0,\lambda
_0) \in \mathcal{L}(E,X)$ and $[D_uf(u_0,\lambda _0)]^{-1} \in
\mathcal{L}(X,E))$.  Then there exist $\delta > 0$, $r > 0$, and
unique continuous mapping $u:  B_{\delta}(\lambda _0) = \{\lambda
:  \|\lambda - \lambda _0\|_{\Lambda} \leq \delta\} \to E$
such that
\begin{equation}\label{chIeq29}
f(u(\lambda ),\lambda ) = f(u_0,\lambda _0),\quad u(\lambda _0) = u_0
\end{equation}
and
 \[
\|u(\lambda ) - u_0\| \leq r,\quad \forall \lambda \in B_{\delta
 }(\lambda _0).
\]
\end{theorem}

\begin{proof}
  Let us consider the equation
\[
f(u,\lambda) = f(u_0,\lambda_0)
\]
which is equivalent to
\begin{equation}\label{chIeq30}
T\left (f(u,\lambda ) -
f(u_0,\lambda _0)\right ) = 0,
\end{equation}
where $T=\left [D_uf(u_0,\lambda _0)\right ]^{-1}$, or
\begin{equation}\label{chIeq31}
u = u - T\left (f(u,\lambda ) -
f(u_0,\lambda _0)\right ){=:} G(u,\lambda ).
\end{equation}
The mapping $G$ has the following properties:
\begin{itemize}
\item[(i)] $G(u_0,\lambda _0) = u_0$,
\item[(ii)]  $G $ and $D_uG$ are continuous in
$(u,\lambda)$,
\item[(iii)]  $D_u G(u_0,\lambda _0) = 0$.
\end{itemize}
Hence, since
\[
G(u_1,\lambda ) - G(u_2,\lambda )=\int
_0^1D_uG\left (u_2+t(u_1-u_2),\lambda \right )(u_1-u_2)dt,
\]
we obtain
\begin{equation}\label{chIeq32}
\begin{aligned}
\|G(u_1,\lambda ) - G(u_2,\lambda )\|
& \leq
\Big(\sup_{0 \leq t \leq 1}
\|D_uG(u_1+t(u_2-u_1),\lambda )\|_\mathcal{L}\Big)
\|u_1-u_2\|\\
& \leq \frac {1}{2} \|u_1-u_2\|,
\end{aligned}
\end{equation}
 provided $\|u_1 - u_0\| \leq r$, $\|u_2 - u_0\|
\leq r$, $\|\lambda -\lambda _0\|_{\Lambda}\leq \delta$, where
$r>0$ and $\delta >0$ are small enough. Now
\begin{align*}
\|G(u,\lambda) - u_0\|
&=  \|G(u,\lambda)-G(u_0,\lambda_0)\|\\
&\leq  \|G(u,\lambda) - G(u_0,\lambda)\|
+ \|G(u_0,\lambda) - G(u_0,\lambda_0)\|\\
& \leq \frac{1}{2}
\|u - u_0\| + \| G(u_0,\lambda) - G(u_0,\lambda_0)\| \\
&\leq  \frac{1}{2}r +\frac{1}{2} r,
\end{align*}
  provided $\|\lambda - \lambda_0\|_{\Lambda} \leq \delta $,
where $\delta >0 $ has been further restricted  so that
\[
\|G(u_0,\lambda) -
G(u_0,\lambda_0)\| \leq \frac {1}{2} r.
\]
 We now think of $u$ as a continuous function
\[
u:B_{\delta}(\lambda_0)\to E
\]
 and define
\begin{align*}
\mathbb{M} &:= \big\{u: B_{\delta}(\lambda_0) \to E,
\text{ such that $u$  is continuous,}\\
&\quad u(\lambda_0) = u_0,\;
u(B_{\delta }(\lambda _0 ))\subset B_r(u_0)\big\},
\end{align*}
and equip $\mathbb{M}$ with the  norm
\[
\|u\|_{\mathbb{M}} := \sup_{\lambda \in B_{\delta}(\lambda_0)}
 \|u({\lambda})\| .
\]
  Then $\mathbb{M}$ is a closed subset of the Banach space
of bounded continuous functions defined on $B_{\delta}(\lambda_0)$ with
values in $E$. Since $E$ is a
Banach space, it is  a complete metric space.
Thus, (\ref{chIeq31}) defines an equation
\begin{equation}\label{chIeq33}
u(\cdot ) = G(u(\cdot ),\cdot )
\end{equation}
 in $\mathbb{M}$.

Define $g$ by (here we think of $u$ as an element of $\mathbb{M}$)
\[
g(u)(\lambda) := G(u(\lambda),\lambda),
\]
 then $g: \mathbb{M} \to \mathbb{M}$ and it follows from
(\ref{chIeq32}) that
\[
\|g(u) - g(v)\|_{\mathbb{M}} \leq \frac {1}{2} \|u - v\|_{\mathbb{M}},
\]
  hence $g$ has a unique fixed point by the
contraction mapping principle (Theorem \ref{contraction}
of Chapter \ref{chapII}).
\end{proof}


\begin{remark} \label{remk17} \rm
If in the implicit function theorem $f$
is $k$ times continuously differentiable with respect to
$\lambda $, then the
mapping $\lambda \mapsto u(\lambda)$ inherits this
property.
\end{remark}

\begin{proof}
 We sketch a proof for the case that $f$ is continuously
differentiable with respect to $\lambda $. It follows from the above
computation that
\[
\|D_uG(u(\lambda ),\lambda )\|_\mathcal{L}\leq \frac {1}{2},
\]
for $\|\lambda -\lambda _0\|_{\Lambda}\leq \delta $.
 It follows that
\[
\left [\text{id}-D_uG( u(\lambda ),\lambda)\right ]^{-1}D_{\lambda
 }G(u(\lambda ),\lambda )\in \mathcal{L} (\Lambda ,E).
\]
Furthermore, a quick calculation shows that
\[
u(\lambda +h)=u(\lambda )+\left [\text{id}-D_uG( u(\lambda ),\lambda)\right ]^{-1}D_{\lambda
 }G(u(\lambda ),\lambda )(h)+o(\|h\|_{\Lambda})
\]
and thus
\[
D_{\lambda }u(\lambda )=\left [\text{id}-D_uG( u(\lambda ),\lambda)\right ]^{-1}D_{\lambda
 }G(u(\lambda ),\lambda )
\]
is continuous with respect to the parameter $\lambda $.
\end{proof}

\subsection{Two examples}

\begin{example} \label{rmk18} \rm
Let us consider the nonlinear boundary value
problem
\begin{equation}\label{chIeq34}
u'' + \lambda e^u = 0, \quad 0 < x < \pi,\quad
u(0) = 0 = u(\pi).
\end{equation}
This is a  one
space-dimensional mathematical  model from the theory of
combustion
 (cf. \cite{bebernes:mpc89}), where $u$ represents a dimensionless
temperature.
\end{example}

 The problem may be explicitly solved using quadrature
methods
and was first posed by Liouville \cite{liouville:ldp53}.
We shall show, by an application of Theorem \ref{thmI3}, that for
$\lambda \in \mathbb{ R}$, in a neighborhood of $0$,
(\ref{chIeq34}) has a unique solution of small norm in
$C^2([0,\pi])$.
To this end we define
\begin{gather*}
E := C^2_0([0,\pi]):= C^2([0,\pi])\cap\{u:u(0)=0=u(\pi)\}
\\
X := C([0,\pi]), \quad
\Lambda := \mathbb{ R}.
\end{gather*}
These spaces are Banach spaces when  equipped with their usual
norms, i.e.,
\begin{gather*}
\|u\|_X:=\sup_{t\in [0,\pi]}|u(t)|, \\
\|u\|_E:=\|u\|_X+\|u'\|_X+\|u''\|_X.
\end{gather*}
and $|\cdot |$ represents absolute value.

We let
$f: E \times \Lambda \to X$
 be given by
\[
f(u,\lambda) := u'' + \lambda e^u.
\]
 Then $f$ is continuous and $f(0,0) = 0$.
(When $\lambda = 0$ (no heat generation) the unique
solution is $u \equiv 0$.)  Furthermore, for $u_0 \in E$,
$D_uf(u_0,\lambda)$ is given by (the reader should carry out the
verification)
\[
D_uf(u_0,\lambda)v = v'' + \lambda e^{u_0}v,
\]
  and, hence,  the mapping
\[
(u,\lambda) \mapsto D_uf(u,\lambda)
\]
 is continuous.  Let us consider the linear
mapping
\[
T: = D_uf(0,0): E \to X.
\]
We must show that this mapping is a linear
homeomorphism.  To see this we note that for every $h \in
X$, the unique solution of
\[
v'' = h(x),\quad 0 < x < \pi,\quad v(0) = 0 = v(\pi),
\]
is given by
\begin{equation}\label{chIeq35}
v(x) = \int_0^{\pi}G(x,s)h(s)ds,
\end{equation}
where
\begin{equation}
\label{green}
G(x,s) = \begin{cases}
- \frac {1}{\pi} (\pi - x)s, & 0 \leq s \leq x\\
- \frac {1}{\pi} x(\pi - s), &x \leq s \leq \pi .
\end{cases}
\end{equation}
 From the representation (\ref{chIeq35}) we may
conclude that there exists a constant $c$ such that
\[
\|v\|_E = \|T^{-1}h\|_E \leq c\|h\|_X;
\]
i.e. $T^{-1}$ is one to one and continuous.
Hence, all conditions of the implicit function theorem are
satisfied and we may conclude that for each
 sufficiently small $\lambda$, (\ref{chIeq34}) has a
unique small solution
$u \in C^2([0,\pi].)$,
Furthermore, the map $\lambda \mapsto u(\lambda)$ is
continuous (in fact, smooth) from a neighborhood of $0 \in \mathbb{R}$ to
$C^2([0,\pi])$. We observe  that this
`solution branch' $(\lambda,u(\lambda))$ is bounded in the $\lambda -$
direction.
  To see this, we note  that if $\lambda > 0$ is such
that (\ref{chIeq34}) has a solution, then the corresponding
solution $u$ must be positive, $u(x) > 0$, $0 < x <
\pi$.  Hence
\begin{equation}
\label{chIeq36}
0 = u'' + \lambda e^u > u'' + \lambda u.
\end{equation}
Let $v(x) = \sin x$. Then $v$ satisfies
\begin{equation}\label{chIeq37}
v'' + v = 0,\quad 0 < x < \pi,\quad v(0) = 0 = v(\pi).
\end{equation}
 From (\ref{chIeq36}) and (\ref{chIeq37}) we obtain
\[
0 > \int_0^{\pi} (u''v-v''u)dx + (\lambda - 1)
\int_0^{\pi} uvdx,
\]
and, hence, integrating by parts,
\[
0 > (\lambda - 1) \int_0^{\pi} uvdx,
\]
implying that $\lambda < 1$.


As a second example, we consider the following.

%Example 20
\begin{example}\rm  Given any $l_0\ne n^2$, $n=1,2,\dots $,
the forced nonlinear oscillator (periodic boundary value problem)
\begin{equation}\label{chIeq38}
u'' + l u + u^2 = g,\quad
u(0) = u(2\pi),\quad u'(0) = u'(2\pi)
\end{equation}
where $g$ is a continuous $2\pi-periodic$ function and $l \in
 \mathbb{R}$, is a parameter, has a unique $2\pi $ periodic solution for
 all $g$ of sufficiently small norm and $|l-l_0|$ sufficiently small.
\end{example}

 Let
\begin{gather*}
E := C^2([0,2\pi]) \cap
 \{u: u(0) = u(2\pi), ~u'(0) = u'(2\pi)\},
\\
X :=  C([0,2\pi]),
\end{gather*}
 where both spaces are equipped with the
 norms as in the previous example. As a parameter space we choose
\[
\Lambda :=\mathbb{R}\times X.
\]
The norm in $\Lambda $
 is given by $\|\lambda =(l ,g)\|_{\Lambda}=|l |+\|g\|_X$.
 We shall show that, for certain values
of $l , $ (\ref{chIeq38}) has a unique solution for all forcing
terms $g$ of small norm.

To this end
let
$f: E\times \Lambda  \to X$
be given by
\[
f(u,\lambda)=f(u,l,g) := u'' + l u + u^2-g.
\]
Then $f(0,0)=0$, and $D_uf(u,\lambda )$ is defined by
\[
(D_uf(u,\lambda ))(v) = v'' + l v + 2uv,
\]
and hence the mapping
\[
u \mapsto D_uf(u,\lambda )
\]
is a continuous mapping of $E$ to $\mathcal{L}(E;X)$, i.e. $f$ is
a $C^1$ mapping. It follows from elementary differential equations
theory (see e.g., \cite{birkhoff:ode69}) that the problem
\[
v'' + l_0 v = h,
\]
has a unique $2\pi$--periodic solution for every $2\pi$--periodic
$h$ as long as $l_0 \neq n^2$, $ n = 1,2,\dots$, and that $\|v\|
\leq C\|h\|_X$ for some constant $C$ (only depending upon $l_0$).
Hence,  $D_uf(0,l_0,0)$ is a linear homeomorphism of $E$ onto $X$
whenever  $l_0 \neq n^2$, $ n = 1,2,\dots $, and we  conclude that
for every $g \in X$ of small norm and $|l-l_0|$ sufficiently
small, (\ref{chIeq38}) has a unique solution $u \in E$ of small
norm.

We note that the above example is prototypical
 for forced nonlinear oscillators. Virtually the same arguments can be applied (the reader might carry out the necessary calculations) to conclude
that the forced pendulum equation
\[
u'' + l \sin u = g
\]
has a unique  $2\pi $- periodic response of small norm for
 every $2\pi$ - periodic forcing term $g$ of small norm,
 as long as $l \neq n^2$, $n = 1,2,\dots $.

\section{Variational inequalities} \label{chapIX}
% vi  Chapter IX \include{contineq}

In this chapter we shall discuss existence results for solutions of
variational inequalities which are defined by bilinear forms on a
Banach space. The main result proved is a Lax-Milgram type result. The
approach follows the basic paper of Lions-Stampacchia
\cite{lions:vi67} and also \cite{kinderlehrer:ivi80}.

\subsection{On symmetric bilinear forms} \label{obf}
Let $E$ be a real reflexive Banach space\index{reflexive Banach space} with norm $\|\cdot \|$ and let
\[
 a: E\times E \to \mathbb{R}
\]
be  a continuous,
  coercive\index{coercive}, symmetric\index{symmetric},
  bilinear form\index{bilinear form}, i.e.,
\[
|a(u,v)|\leq c_1\|u\|\|v\| ,\; a(u,u)\geq
 c_2\|u\|^2,\;a(u,v)=a(v,u),\;\forall u,v\in E,
\]
and $a$ is linear in each variable separately,
where $c_1$ and $c_2$ are positive constants.
As is common, we denote by $E^*$ the dual space\index{dual space} of $E$ and
for
$b\in E^*$ we denote by
$\langle b,u\rangle$
the value of the continuous linear functional $b$ at the point
$u$, the pairing\index{pairing} between $E^*$ and $E$. The norm in
$E^*$ we shall denote by $\|\cdot \|_*$.

Along with the norm topology\index{norm topology} on $E$, we shall
also have occasion to make use of the weak
 topology\index{weak topology} (see below and, e.g., \cite{royden:ra88},
\cite{taylor:ifa72}, \cite{yosida:fa95}).

 For given $b\in E^*$ and   a weakly
closed set $K$ we
 consider the functional
\begin{equation}
\label{functional-b}
f(u)=\frac {1}{2}a(u,u)-\langle b,u\rangle.
\end{equation}
An easy computation shows that
\[
f(u)\geq \frac{c_2}{2}\|u\|^2-\|b\|_{E^*}\|u\|
\]
 and, hence, that
\[
f(u)\to \infty,\;\text{as}~\|u\|\to \infty
\]
($f$ is {\it coercive})
and that $f$ is bounded below on $E$. Hence, it is the case that
\[
\alpha :=\inf_{v\in K}f(v)>-\infty.
\]
Let us choose a sequence $\{u_n\}_{n=1}^{\infty}$ in $ K$ (a
{\it minimizing sequence})\index{minimizing sequence} such that
\[
f(u_n)\to \alpha.
\]
It follows (because $f$ is coercive) that the sequence  $\{u_n\}_{n=1}^{\infty}$ is a bounded
sequence
and hence has (since $E$ is reflexive, see
 \cite{royden:ra88},
\cite{taylor:ifa72}, \cite{yosida:fa95}) a weakly convergent
subsequence,
converging weakly to, say, $u$. We denote this subsequence, after
appropriate relabeling, again by  $\{u_n\}_{n=1}^{\infty}$
and hence have
\[
u_n\rightharpoonup u
\]
($\rightharpoonup $ denotes weak convergence\index{weak convergence}),
i.e.
for any element $h\in E^*$
\[
\langle h, u_n\rangle \to \langle h , u\rangle .
\]
Since $K$ is weakly closed\index{weakly closed}, we have that $u\in K$.
Since $a$ is bilinear and nonnegative ($a$ is coercive),
we obtain
\[
a(u_n,u_n)\geq a(u_n,u)+a(u,u_n)-a(u,u),
\]
and that
\[
\liminf_{n \to \infty}a(u_n,u_n)\geq a(u,u)
\]
(the form is {\it weakly sequentially lower semicontinuous}\index{weakly
lower semicontinuous}).
We may, therefore, conclude that
\begin{equation}
\label{min}
f(u)=\min _{v\in K}f(v).
\end{equation}

Let us now assume that the set $K$ is also convex (hence, it is also
closed,
since, in reflexive Banach spaces, convex sets are closed if,
 and only if, they are weakly closed,
cf. \cite{taylor:ifa72}, \cite{yosida:fa95}). Then for any $v\in K$
and $0\leq t\leq 1$ we have that
\begin{equation}
\label{convmin}
f(u)\leq f(u+t(v-u)).
\end{equation}
Computing $f(u+t(v-u))-f(u)$, and using the properties of the form
$a$, we obtain
\[
0\leq ta(u,v-u)-t\langle b,v-u\rangle +t^2a(v-u,v-u), ~0\leq t\leq 1,\;v\in K.
\]
Hence, upon dividing the latter inequality by $t>0$, and letting
$t\to 0$, we see that there exists $ u\in K$ such that
\begin{gather}
\label{k1}
f(u)=\min_{v\in K}f(v),\\
\label{k3}
a(u,v-u)\geq \langle b,v-u\rangle,
\quad \forall v \in K.
\end{gather}
Hence, if $b_1, b_2\in E^*$ are given and $u_1,u_2$ are solutions in
$K$
of the corresponding problems (\ref{k1}), (\ref{k3}), then, denoting by
\[
Tb_i=u_i,\quad i=1,2,
\]
we easily conclude from (\ref{k3}) and the coerciveness of $a$ that
\begin{equation}\label{k4}
\|Tb_1-Tb_2\|\leq \frac {1}{c_2}\|b_1-b_2\|_{*}.
\end{equation}
Thus, we see that the problems (\ref{k1}), (\ref{k3}) have a unique solution.

We have the following theorem.

\begin{theorem}\label{symmetric}
Let $a:E\times E\to \mathbb{R}$ be a continuous,
 bilinear, symmetric,
and
coercive form and let $K$ be a closed convex subset of $E$. Then for any $b\in E^*$ the variational inequality\index{variational inequality}
\begin{equation}\label{k3*}
a(u,v-u)\geq \langle b,v-u\rangle,
\quad \forall v \in K.
\end{equation}
has a unique solution $u\in K$. Hence, equation (\ref{k3*}) defines a
solution mapping
\[
T:E^*\to K,\quad b\mapsto Tb=u,
\]
which is Lipschitz continuous with Lipschitz constant
$\frac{1}{c_2}$, where $c_2$ is the coercivity constant of $a$.
\end{theorem}

\begin{remark} \label{uniqueness} \rm
It follows from the above considerations that if $a$ satisfies the
conditions of Theorem \ref{symmetric} except that it is not
necessarily symmetric, and if inequality (\ref{k3*}) has a
solution for every $b\in E^*$, then the solution mapping $T$,
above is well-defined and satisfies the Lipschitz condition
(\ref{k4}).
\end{remark}


\subsection{Bilinear forms continued}
Let $a\colon E\times E\to \mathbb{R}$
be a continuous, coercive, bilinear form,
$b\in E^\ast$, and $K$ a closed convex set.

\subsection{The problem}
We pose the following
problem: Find (prove the existence of) $u\in K$ such that
\begin{equation}\label{k15}
a(u,v-u)\geq \langle{b},{v-u}\rangle,\quad \forall v\in K.
\end{equation}
In case $a$ is symmetric, this problem has been solved above
and Theorem \ref{symmetric} provides its solution.
Thus, it remains to be shown that the theorem remains true in case
 $a$ is not necessarily symmetric.

The development in this section follows closely the development
in \cite{lions:vi67} and \cite{kinderlehrer:ivi80}.

\subsection*{Uniqueness of the solution}

 Using properties of bilinear forms, one concludes (see Remark
\ref{uniqueness}) that for all $b\in E^\ast$, problem (\ref{k15})
has at most one solution and if $b_1,b_2\in E^\ast$ and solutions
$u_1,u_2$ exist, then
\[
\|u_1-u_2\|\leq \frac 1{c_2}\|b_1-b_2\|_{\ast},
\]
where $c_2$ is a coercivity constant of $a$.

\subsection*{Existence of the solution}
We write
$a=a_e+a_o$,
where
\begin{gather*}
a_e(u,v):=\frac 1{2}(a(u,v)+a(v,u)), \\
a_o(u,v):=\frac 1{2}(a(u,v)-a(v,u)),
\end{gather*}
then $a_e$ is a continuous, symmetric, coercive, bilinear form and $a_o$ is
continuous and bilinear.

Consider the family of problems
\begin{equation}\label{k16}
a_e(u,v-u)+ta_o(u,v-u)\geq \langle{b},{v-u}\rangle, \quad \forall v \in K, \quad 0\leq t\leq 1,
\end{equation}
and let us denote by
\[
a_t:=a_e+ta_0.
\]
We have the following lemma.

\begin{lemma}
Let $t\in [0,\infty)$ be such that the problem
\begin{equation}
\label{k17}
a_t(u,v-u)\geq \langle{b},{v-u}\rangle , \quad \forall v \in K,
\end{equation}
has a unique solution for all $b\in E^*$. Then there exists a
constant $c>0$, depending only on the continuity and coercivity
constants of $a$, such that problem
\begin{equation}
\label{k17*}
a_{t+\tau}(u,v-u)\geq \langle{b},{v-u}\rangle , \quad \forall v \in K,
\end{equation}
has a unique solution for all $b\in E^*$ and $0\leq \tau \leq c$.
\end{lemma}

\begin{proof}
For $w\in K$ and $t\geq 0$, consider
\begin{equation}\label{k18}
a_t(u,v-u)\geq \langle{b},{v-u}\rangle -\tau a_o(w,v-u), \quad
 \forall v \in K.
\end{equation}
Note that for fixed $w\in K$,
\[
b_w:=b-\tau a_o(w,\cdot)\in E^\ast,
\]
hence, there exists a unique $u=Tw$ solving (\ref{k18}) and
\[
\|Tw_1-Tw_2\|\leq \frac 1{c_2}\|b_{w_1}-b_{w_2}\|_{\ast}.
\]
On the other hand
\[
\|b_{w_1}-b_{w_2}\|_{\ast}
=\sup_{\|u\|=1}\tau |a_o(w_1,u)- a_o(w_2,u)|\leq \tau c_1\|w_1-w_2\|,
\]
and hence
\[
\|Tw_1-Tw_2\|\leq \frac{\tau c_1}{c_2}\|w_1-w_2\|,
\]
and $T\colon K\to K$ is a contraction mapping provided
$\frac{\tau c_1}{c_2}<1$. Therefore, there is a unique
solution of (\ref{k17*})
 as long as $\tau <\frac{c_2}{c_1}$, and we  may  choose
$c=\frac{c_2}{2c_1}$, for example.
\end{proof}

We may apply the above lemma with $t=0$, since $a_0=a_e$, and
$a_e$ is symmetric, and   obtain that
\begin{equation}\label{k19}
a_{t}(u,v-u)\geq \langle{d},{v-u}\rangle, \quad \forall v \in K
\end{equation}
has a unique solution for all $d\in E^\ast$, for $0\leq t\leq c$.
Hence by the lemma, we obtain that (\ref{k19}) has a unique
solution for $0\leq t\leq 2c$, and continuing in this manner we
obtain a unique solution of (\ref{k19}) for all $t\in [0,\infty)$,
and in particular for $t=1$, and we have shown that problem
(\ref{k15}) is uniquely solvable.


We therefore have the following theorem.

\begin{theorem}\label{general}
Let $a:E\times E\to \mathbb{R}$ be a continuous, bilinear,
and coercive form and let $K$ be a closed convex subset of $E$.
Then for any $b\in E^*$ the variational inequality\index{variational inequality}
\begin{equation}\label{k3**}
a(u,v-u)\geq \langle b,v-u\rangle,
\quad \forall v \in K.
\end{equation}
has a unique solution $u\in K$. Hence equation (\ref{k3*}) defines a
solution mapping
\[
T:E^*\to K,\quad b\mapsto Tb=u,
\]
which is Lipschitz continuous with Lipschitz constant
$\frac{1}{c_2}$, where $c_2$ is a coercivity constant of $a$.
\end{theorem}

Using this result one may immediately obtain an existence result for
solutions
of nonlinearly
perturbed variational inequalities of the form
\begin{equation}\label{k5}
a(u,v-u)\geq \langle F(u),v-u \rangle,\quad \forall v\in K,
\end{equation}
where
$F:E\to E^\ast$,
is a Lipschitz continuous mapping, say,
\[
\|F(u_1)-F(u_2)\|_*\leq k\|u_1-u_2\|.
\]
We have the following result.

\begin{theorem}\label{nonlinear}
Let $a,K,F$ be as above. Then the variational inequality \eqref{k5}
has a unique solution, provided that
\[
k<c_2,
\]
where $k$ is the Lipschitz constant for $F$ and $c_2$ is the
 coercivity constant for $a$.
\end{theorem}

\begin{proof}
It follows from Theorem \ref{general} that the variational
 inequality
\eqref{k5}
 is equivalent to the fixed point problem
\begin{equation}\label{k6}
u=TF(u).
\end{equation}
Since
\[
TF:K\to K,
\]
and $K$ is a closed convex subset of $E$, hence a complete metric
space, the result then follows from the contraction mapping
principle and Theorem \ref{general}, once we observe that for any
$u_1,u_2\in K$
\[
\|TF(u_1)-TF(u_2)\|\leq \frac{1}{c_2}\|F(u_1)-F(u_2)\|_{*}\leq
\frac{k}{c_2}\|u_1-u_2\|.
\]
\end{proof}

\subsection{Some examples}
\subsection*{An obstacle problem}
Let $\Omega$ be a bounded domain in $\mathbb{R}^N$ and let $E=L^2(\Omega)$. Let
$\psi\in E$ be given and let
\[
K:=\{u\in E: u(x)\geq \psi(x) ,\; \text{a.e.  in }~ \Omega\}.
\]
Then $K$ is a closed, convex subset of $E$.
We let
$a:E\times E\to \mathbb{R}$
 be defined by
\[
a(u,v):=\int _{\Omega}uvdx=\langle u, v \rangle .
\]
Then
\[
a(u,u)=\|u\|^2,
\]
where $\|\cdot \|$ is the norm in the space $E$. Thus we see that $a$
is a continuous, symmetric, coercive, and bilinear form.

Let $b\in E$ and define $f: E\to \mathbb{R}$ as
\[
f(u)=\frac
1{2}\|u\|^2-\langle b,u\rangle .
\]
Then there exists a unique $u\in K$ such that
\[
f(u)=\min_{v\in K}f(v)
\]
 and,
furthermore, $u$ solves the variational inequality
\[
a(u,v-u)-\langle{b},{v-u}\rangle \geq 0, \quad \forall v\in K;
\]
i.e.,
\begin{equation}\label{k14}
\int_{\Omega}(u-b)(v-u)dx\geq 0,  \quad \forall v\in K,
\end{equation}
and the latter must have a unique solution. The natural candidate
for this solution is $u=\max(\psi,b)$, as one easily verifies by
substituting into (\ref{k14}).

\subsection*{Another example}
Let
$E:=L^2(0,1)$, $ K:=\{u: \int_0^1 udx=1 \}$. Then $K$ is
closed and convex (hence weakly closed). Let
\[
a(u,v):=\int_0^1 uv\,dx= \langle u, v \rangle,
\]
then, as above,  $a$ is a continuous, coercive, symmetric, and
bilinear form. Hence
there exists a unique $u\in K$ such that
\[
a(u,v-u)\geq 0, \quad \forall v\in K,
\]
i.e.
\[
\langle u,v-u\rangle \geq 0, \quad \forall v\in K,
\]
or
\[
\langle u,v\rangle \geq \langle u,u\rangle , \quad \forall v\in K,
\]
i.e.
\[
\int_0^1 uv dx \geq \int_0^1 u^2dx , \quad \forall v\in K.
\]
On the other hand
\[
\Big|\int_0^1 udx \Big|\leq \Big( \int_0^1 u^2 dx
\Big)^{1/2},
\]
and hence
\[
\int_0^1 uv dx \geq 1,\quad \forall v \in K.
\]
Clearly $u=1$ solves the inequality.


\subsection{A second order boundary value problem}

In this section we shall provide an example of a boundary value
problem
for a second order ordinary differential equation on the interval
$(0,\infty )$ which may be solved by the methods developed here. It
furnishes an example where the associated quadratic form is not
symmetric.

Let us denote by $E=H^1_0(0,\infty )$ (the closure of the space
$C^{\infty}_0(0,\infty )$ in the Sobolev space\index{Sobolev
space} of
functions
$u:(0,\infty )\to \mathbb{R}$ which together with their first
distributional derivatives\index{distributional derivative} are square integrable on $(0,\infty )$; see
Chapter \ref{chapII}, section 2.3).
The norm in $E$ is given by
\[
\|u\|^2 :=\int _0^{\infty}u^2dx +\int _0^{\infty}(u')^2dx.
\]
Let the quadratic form $a:E\times E\to \mathbb{R}$, be given by
\begin{equation}
\label{form}
a(u,v):=\int _0^{\infty}u'v'dx +\int _0^{\infty}uv'dx+\int
_0^{\infty}uvdx.
\end{equation}
One quickly may check that $a$ is continuous, bilinear, and coercive
(with coercivity constant $\frac{1}{2}$) but, it is clearly not
symmetric.
It follows that for any $b\in L^2(0,\infty )$  the variational
inequality
\begin{equation}
\label{kode}
a(u,v-u)-\langle{b},{v-u}\rangle \geq 0, \quad \forall v\in E,
\end{equation}
has a unique solution, and, hence, since in this problem the
convex set $K$ is the whole space $E$, the equation
\begin{equation}
\label{kode*}
a(u,v)-\langle{b},{v}\rangle = 0, \quad \forall v\in E,
\end{equation}
has a unique solution. I.e., there exists a unique $u\in E$ such that
\begin{equation}
\label{kode**}
\int _0^{\infty}u'v'dx +\int _0^{\infty}uv'dx+\int
_0^{\infty}uvdx=\int
_0^{\infty}bvdx, \quad \forall v\in E.
\end{equation}
Since $C_0^{\infty }(0,\infty )$ (the infinitely smooth functions
with compact support, or test functions) is dense in $E$, we may
interpret (\ref{kode**}) in the sense of distributions and obtain
\begin{equation}
\label{kode***}
-\partial ^2u(v) -\partial u(v)+\int
_0^{\infty}uvdx=\int
_0^{\infty}bvdx, \quad \forall v\in C_0^{\infty }(0,\infty ),
\end{equation}
and hence
\begin{equation}
\label{kode&}
-\partial ^2u-\partial u+u=b,
\end{equation}
has a unique solution $u\in E$ for any $b\in L^2(0,\infty )$ (here
$\partial ^2u, \partial u $ are the second, respectively, first
distributional derivatives of the function $u$
(see again Chapter \ref{chapII},
section 2.3)).

\subsection{An obstacle problem}\index{obstacle problem}

Let us consider here once more the quadratic form $a$ of the previous
section
and pose the obstacle problem
\begin{equation}
\label{kode4}
a(u,v-u)-\langle{b},{v-u}\rangle \geq 0, \quad \forall v\in K,
\end{equation}
where $K$ is the closed convex set
\[
K:=\{u\in H^1_0(0,\infty): u(x)\geq 0,\; 0\leq x\leq 1\}.
\]
Again this problem will have a
 a unique solution for any $b\in L^2(0,\infty)$. Rewriting
 (\ref{kode4})
as
\begin{equation}
\label{kode5}
\int _0^{\infty}u'v'dx +\int _0^{\infty}uv'dx+\int
_0^{\infty}uvdx\geq \int
_0^{\infty}bvdx, \quad \forall v\in K.
\end{equation}
we may conclude that
\begin{equation}
\label{kode6}
\int _0^{\infty}u'v'dx +\int _0^{\infty}uv'dx+\int
_0^{\infty}uvdx= \int
_0^{\infty}bvdx, \quad \forall v\in C_0^{\infty }(0,\infty ),
\end{equation}
with
\[
v(x)=0,\quad 0\leq x\leq 1
\]
(note that, for such $v$, both $v$ and $-v$ belong to $K$). This
implies
\begin{equation}
\label{kode6*}
\int _1^{\infty}u'v'dx +\int _1^{\infty}uv'dx+\int
_1^{\infty}uvdx= \int
_1^{\infty}bvdx, \quad \forall v\in C_0^{\infty }(1,\infty ),
\end{equation}
and, as above, we conclude that $u$ is a solution of
\[
-\partial ^2u-\partial u+u=b,\quad 1\leq x\leq \infty .
\]
The latter equation will also be satisfied at those points $x\in
(0,1)$, for which $u(x)>0$. Combining these statements, we
conclude that $u\in K$ solves
\begin{gather*}
-\partial ^2 u-\partial u+u = b,\quad 1\leq x\leq \infty\\
u(-\partial ^2 u-\partial u+u-b) = 0 ,\quad 0\leq x\leq 1.
\end{gather*}

\subsection{Elliptic boundary value problems}\index{elliptic boundary
value problem}
Let $\Omega$ be a bounded open set (with smooth boundary) in $\mathbb{R}^N$,
let $\{a_{ij}\}_{i,j = 1}^N
\subset L^{\infty}(\Omega)$ satisfy
\[
\sum_{i,j} a_{ij}(x)\xi_i\xi_j\geq c_0|\xi|^2, ~\forall  \xi \in
\mathbb{R}^N,
\quad \forall x\in \Omega, \;
c_0>0 \text{ a constant},
\]
where $|\cdot |$ is a norm in $\mathbb{R}^N$.
Let $E:=H_0^1(\Omega)$ with
 \[\|u\|^2=\|u\|^2_{H_0^1(\Omega)}=\int
_\Omega|\nabla u|^2dx,
\]
(this is a norm, equivalent to the $H^1$ norm, as follows from an
inequality due to Poincar\'{e}\index{Poincar\'{e}}, see \cite{adams:ss75}),
and let $a(u,v)$ be given by
\[
a(u,v)=\sum_{i,j} \int
_\Omega  a_{ij}(x)\partial_i u\partial_j vdx =\int
_\Omega A\nabla u\cdot \nabla vdx,
\]
where $A$ is the $N\times N$ matrix whose $ij$ entry is $a_{ij}$ and
$\partial _iu,\;i=1,\dots, N$ are the partial distributional
derivatives of
$u$ and $\nabla $ is the distributional gradient\index{distributional gradient}.
Then
\begin{gather*}
|a(u,v)|\leq c_1\| v\|\|u\|,\quad
c_1=\max_{ij}\|a_{ij}\|_{L^{\infty}(\Omega)},
\\
|a(u,u)|\geq c_0 \| u\|^2.
\end{gather*}
For $b\in L^2(\Omega)\subset H_0^1(\Omega)^\ast$ we obtain the existence
of a unique $u\in H_0^1(\Omega)$ such that
\[
a(u,v-u)\geq \int_\Omega b(v-u),\quad \forall v \in H_0^1(\Omega),
\]
hence,
\[
a(u,v)= \int_\Omega bvdx ,\quad \forall v \in H_0^1(\Omega).
\]
In particular, this will hold for  all $v\in
C_0^{\infty}(\Omega)$, and therefore the partial differential
equation
\[
-\sum_{i,j} \partial_j(a_{ij}\partial_i u)=b
\]
has  a unique solution $u\in  H_0^1(\Omega)$, in the sense of
distributions (see also \cite{brezis:af83}, \cite{evans:pde98},
\cite{gilbarg:epd83}, \cite{lions:hbv72}).
 As a special case we obtain that  the partial differential equation
\begin{equation}
\label{poisson}
-\Delta u=b,
\end{equation}
has, in the sense of distributions, a unique solution $u\in
H_0^1(\Omega )$
for every $b\in L^2(\Omega )$ and
\begin{equation}
\label{poisson1}
\int _{\Omega}|\nabla u |^2dx\leq \int _{\Omega}b^2dx.
\end{equation}


If $\Omega $ is a not necessarily bounded open set, one may, using arguments like the above establish the
unique solvability in $H^1_0(\Omega )$
of the elliptic equation
\begin{equation}
\label{poisson2}
-\Delta u+u=b,
\end{equation}
for every
$b\in L^2(\Omega )$ and
\begin{equation}
\label{poisson3}
\int _{\Omega}|\nabla u |^2dx+ \int _{\Omega} u^2dx\leq \int _{\Omega}b^2dx.
\end{equation}

For additional and more detailed examples see the following:
\cite{brezis:pu72,chipot:vif84,ciarlet:evk80,
duvaut:imp72, friedman:vpf83,kinderlehrer:ivi80,
le:gbv97,lions:hbv72}.


\section{Semilinear elliptic equations} \label{chapX}
% see Chapter X \include{contell.tex}

In this chapter we shall discuss how the contraction mapping
principle may be used to deduce the existence of solutions of
Dirichlet problems\index{Dirichlet problem} for
 semilinear\index{semilinear} elliptic partial differential
equations.
Such results have their origin in work of Picard \cite{picard:lqp30}, Lettenmeyer
\cite{lettenmeyer:upa44}.
Our derivation is based on work contained in \cite{hai:eur94},
and \cite{mawhin:tpb80}, using the results established in Chapter \ref{chapIX}.

\subsection{The boundary value problem}

Let $\Omega$ be a bounded open set (with smooth boundary) in
$\mathbb{R}^N$,
and let
\[
f:\Omega \times \mathbb{R}\times \mathbb{R}^N\to \mathbb{R}
\]
be a continuous function which satisfies the Lipschitz condition
\begin{equation}
\label{see1}
|f(x,u_1,v_1)-f(x,u_2,v_2)|\leq L_1|u_1-u_2|+L_2|v_1-v_2|,
\end{equation}
for all $(x,u_1,v_1),\; (x,u_2,v_2)\in \Omega \times \mathbb{R}\times \mathbb{R}^N$, here $|\cdot |$ denotes both absolute value
and the Euclidean norm in $\mathbb{R}^N$. We shall also assume that
$f(\cdot ,0,0)\in L^2(\Omega )$.

We consider the following boundary value problem (in the sense of distributions)
\begin{equation}
\label{see2}
-\Delta u=f(x,u,\nabla u),\;u\in H^1_0(\Omega)=:H.
\end{equation}
We note that if $u\in H^1(\Omega )$, then
\[
|f(x,u,\nabla u)|\leq |f(x,0,0)|+L_1|u|+L_2|\nabla u|.
\]
Hence $f$ may be thought of as a mapping
\[
f:H^1(\Omega )\to L^2(\Omega ),
\]
which, because of the Lipschitz condition (\ref{see1})
is, in fact, a continuous mapping.

Let us denote by $T$, the solution operator
\begin{equation} \label{see3}
\begin{gathered}
T:L^2(\Omega )\to H^1_0(\Omega )\\
T(w)= u,
\end{gathered}
\end{equation}
where $u\in H^1_0(\Omega )$ solves
$-\Delta u=w$
(see Chapter \ref{chapIX}).
Inequality (\ref{poisson3})  implies
\begin{equation}
\label{see4}
\|u\|^2_H=\int _{\Omega}|\nabla u |^2dx
\leq \int _{\Omega}w^2dx=\|w\|^2_{L^2}.
\end{equation}
We, therefore, find that problem (\ref{see2})
is equivalent to the fixed point problem in $L^2(\Omega )$
\begin{equation}
\label{see5}
w=f(\cdot, T(w),\nabla T(w)).
\end{equation}
Define the operator
\[
S:L^2(\Omega )\to L^2(\Omega )
\]
by
\begin{equation}
\label{see6}
S(w)=f(\cdot, T(w),\nabla T(w)).
\end{equation}
Using the Lipschitz condition imposed on $f$, we find
\[
|S(w_1)(x)-S(w_2(x)|
\leq L_1|T(w_1)(x)-T(w_2(x)|+L_2|\nabla T(w_1)(x)-\nabla T(w_2(x)|
\]
and thus
\begin{equation}
\label{see7}
\|S(w_1)-S(w_2\|_{L^2}
\leq L_1\|T(w_1)-T(w_2\|_{L^2}+L_2\||\nabla T(w_1)-\nabla T(w_2)|\|_{L^2}
\end{equation}
We now recall the Poincar\'e inequality\index{Poincar\'e's inequality}
 for $H^1_0(\Omega )$
(see \cite{evans:pde98})
\[
\|u\|_{L^2}\leq \frac{1}{\lambda _1}\|u\|_H,\quad
\forall u\in H^1_0(\Omega ),
\]
where $\lambda _1^2$ is the smallest eigenvalue\index{eigenvalue}
of $-\Delta $
on $H^1_0(\Omega )$ (see \cite{adams:ss75}, \cite{gilbarg:epd83}).
Inequalities (\ref{see7}) and (\ref{see4}) imply
\begin{equation}
\label{see8}
\begin{aligned}
\|T(w_1)-T(w_2)\|_{L^2}
&= \|T(w_1-w_2)\|_{L^2}\\
&\leq \frac{1}{\lambda _1}\|T(w_1-w_2)\|_H\\
&\leq  \frac{1}{\lambda _1}\|w_1-w_2\|_{L^2}.
\end{aligned}
\end{equation}
We next use Green's identity\index{Green's identity}
(see \cite{gilbarg:epd83}) to compute
\begin{equation} \label{see9}
\begin{aligned}
\||\nabla T(w_1)-\nabla T(w_2)|\|^2_{L^2}
&= -\langle w_1-w_2,T(w_1)-T(w_2)\rangle \\
&\leq \|w_1-w_2\|_{L^2}\|T(w_1-w_2)\|_{L^2}\\
&\leq  \frac{1}{\lambda _1}\|w_1-w_2\|^2_{L^2},
\end{aligned}
\end{equation}
where $\langle \cdot, \cdot \rangle $ is the $L^2$ inner product.
Combining (\ref{see8}) and (\ref{see9}) we obtain
\begin{equation}
\label{see10}
\|S(w_1)-S(w_2\|_{L^2}\leq \Big(\frac{L_1}{\lambda _1}
+\frac{L_2}{\sqrt{\lambda _1}}\Big)\|w_1-w_2\|_{L^2}.
\end{equation}
The operator $S:L^2(\Omega )\to L^2(\Omega )$, therefore, has a
unique fixed point provided that
\begin{equation}
\label{see11}
\frac{L_1}{\lambda _1}+\frac{L_2}{\sqrt{\lambda _1}}<1.
\end{equation}
As observed earlier, this fixed point is a solution of (\ref{see5})
and setting $u=T(w)$ we obtain the solution of (\ref{see2}).

We summarize the above in the following theorem.

\begin{theorem}\label{ell1}
Let $\Omega$ be a bounded open set (with smooth boundary) in
$\mathbb{R}^N$,
and let
\[
f:\Omega \times \mathbb{R}\times \mathbb{R}^N \to \mathbb{R}
\]
be a continuous function which satisfies the Lipschitz condition
\[
%\begin{equation}
%\label{see1}
|f(x,u_1,v_1)-f(x,u_2,v_2)|\leq L_1|u_1-u_2|+L_2|v_1-v_2|,
%\end{equation}
\]
for all $(x,u_1,v_1),\; (x,u_2,v_2)\in \Omega \times \mathbb{R}\times \mathbb{R}^N$, further
 assume that $f(\cdot ,0,0)\in L^2(\Omega )$.
Then the  boundary value problem
\[
-\Delta u=f(x,u,\nabla u),\;u\in H^1_0(\Omega).
\]
has a unique solution provided that
\[
\frac{L_1}{\lambda _1}+\frac{L_2}{\sqrt{\lambda _1}}<1,
\]
where $\lambda _1^2$ is the principal eigenvalue of $-\Delta $
with respect to $H^1_0(\Omega )$.
\end{theorem}

In case $f$ is independent of $\nabla u$ such a result, under the
assumption that $L_1$ be sufficiently small,
was already
established for the two-dimensional case by Picard
\cite{picard:lqp30}.

\subsection{A particular case}

In this section we shall consider  problem (\ref{see2})
in the case of one space dimension, $N=1$. We shall study this problem
for the case
\[
\Omega =(0,\pi).
\]
The case of an arbitrary finite interval $(a,b)$ may easily be deduced
for this one.

In this case
$\lambda _1 =1$
and condition (\ref{see11}) becomes
\begin{equation}
\label{see12}
{L_1}+{L_2}<1.
\end{equation}
Given the boundary value problem
\begin{equation}
\label{see13}
 u''=f(x,u, u'),\quad u\in H^1_0(0,\pi),
\end{equation}
with
$f$ satisfying the above assumptions, it is natural, also, to seek a
solution
$u\in C^1[0,\pi]$ and to formulate an integral equation  equivalent to
(\ref{see13}) in this space, rather than $L^2(0,1)$.
This may be accomplished by using the Green's function   $G$ given by
formula
(\ref{green}) of Chapter \ref{chapVIII} (see e.g \cite{hartman:ode82}). I.e., we have that problem
(\ref{see13}) is equivalent to the integral equation problem
\begin{equation}
\label{see14}
u(x) = \int_0^{\pi}G(x,s)f(s,u(s),u'(s))ds,\;u\in E:= C^1[0,\pi].
\end{equation}
As usual, we use
\[
\|u\|_E:=\|u\|+\|u'\|,
\]
where
\[
\|u\|:=\max_{[0,\pi]}|u(x)|.
\]
We  define
$T:E\to E$
by
\[
T(u)u(x) := \int_0^{\pi}G(x,s)f(s,u(s),u'(s))ds,\quad
u\in E= C^1[0,\pi].
\]
An easy computation shows that
\[
\|T(u_1)-T(u_2\|_E\leq \frac{\pi ^2L_1}{8}\|u_1-u_2\|+\frac{\pi L_2}{2}
\|u_1'-u_2'\|,
\]
and therefore
\begin{equation}
\|T(u_1)-T(u_2\|_E\leq \Big(\frac{\pi ^2L_1}{8}
+\frac{\pi L_2}{2}\Big)\|u_1-u_2\|_E,
\end{equation}
i.e., $T$ is a contraction mapping, whenever
\begin{equation}
\label{see14*}
\frac{\pi ^2L_1}{8}+\frac{\pi L_2}{2}<1.
\end{equation}
Clearly the requirement (\ref{see14*}) is different from the
requirement
(\ref{see12}). Thus, for a given problem, several different metric
space settings may be possible, with the different settings yielding
different requirements.

(The condition (\ref{see14*}) was already derived by Picard
\cite{picard:lqp30}
and later by Lettenmeyer \cite{lettenmeyer:upa44}; many different
types of requirements using different approaches  may be found, e.g.,
in \cite{coles:csa67} and \cite{hai:eur94}.)

We remark that in the above considerations we could equally
have assumed that $f$ is a mapping
\[
f:(0,1)\times E \times E \to E,
\]
where $E$ is a Banach space.

\subsection{Monotone solutions}

In this section we shall discuss some recent work in \cite{dube:nnt04}
concerning the nonlinear second order equation
\begin{equation}
\label{mon1}
u''+F(t,u)=0, ~t\in [0,\infty ),
\end{equation}
where
$F:[0,\infty )\times [0,\infty )\to [0, \infty )$
is continuous and  satisfies the Lipschitz condition
\begin{equation}
\label{mon2}
|F(t,u)-F(t,v)|\leq k(t)|u-v|,
\end{equation}
with
\[
k:[0,\infty )\to [0,\infty )
\]
a continuous function, satisfying
\begin{equation}
\label{mon3}
\int _0^{\infty }tk(t)dt <1.
\end{equation}

We have the following theorem on monotone\index{monotone} solutions
of \eqref{mon1}.

\begin{theorem}
Let the above conditions hold and assume that there exists $M>0$
such that for any $u\in X$, where
\[
X:=\{u\in C[0,\infty ):0\leq u(t) \leq M,\;t\in [0,\infty )\},
\]
we have
\begin{equation}
\label{mon4}
\int _0^{\infty }tF(t,u(t))dt \leq M .
\end{equation}
Then \eqref{mon1} has a monotone solution
$u:[0, \infty )\to [0,M]$
such that
\[
\lim _{t \to \infty }u(t)= M.
\]
\end{theorem}

\begin{proof}
Let
 \[
E:=\{u\in C[0,\infty ):\|u\|:=\sup _{t\in [0,\infty )}|u(t)|<\infty \},
\]
then $E$ is a Banach space and $X$ is a closed subset of $E$ and hence
a complete metric space with respect to the metric defined by the norm
in $E$.

Next consider the mapping $T$ on $X$ defined by
\begin{equation}
\label{mon5}
(Tu)(t):=M-\int _t^{\infty }(\tau -t)F(\tau ,u(\tau ))d\tau .
\end{equation}
Then, since for $u\in X$
\begin{equation}
\label{mon6}
0\leq \int _t^{\infty }(\tau -t)F(\tau ,u(\tau ))d\tau \leq \int
_0^{\infty }\tau F(\tau ,u(\tau ))d\tau \leq M ,
\end{equation}
it follows that
$T:X\to X $.

On the other hand, for $u,v\in X$ we have
\begin{align*}
|(Tu)(t)-T(v)(t)|&
\leq \int _t^{\infty }(\tau -t)|F(\tau ,u(\tau
 ))-F(\tau ,v(\tau ))|d\tau \\
&\leq  \int _t^{\infty }(\tau -t)k(\tau )|u(\tau
 )-v(\tau )|d\tau \\
&\leq \int _0^{\infty }\tau k(\tau )d\tau \|u-v\|.
\end{align*}
Hence $T$ is a contraction on $X$ and therefore has a unique fixed
point. If $u\in X$ is the fixed point of $T$, then it easily
follows that $u$ is monotone and $ \lim _{t \to \infty }u(t)= M$.
\end{proof}

For applications of this result to nonoscillation theory of second
order differential equations, see \cite{dube:nnt04}


\section{A mapping theorem in Hilbert space} \label{chapXI}
% mth %Chapter XI \include{contcons}

In this chapter we shall discuss a result of Mawhin \cite{mawhin:cmp76}
on nonlinear mappings in Hilbert spaces which has, among others,
several interesting applications to existence questions about periodic
solutions of nonlinear conservative systems (see \cite{mawhin:cmp76}
and \cite{ward:eps79} for many references to this interesting problem
area; see also \cite{amann:uss82} and \cite{schmitt:fan82} for further directions).

\subsection{A mapping theorem}
\label{mapping}
Let $H$ be a real Hilbert space with scalar product
$\langle \cdot ,\cdot \rangle$ and norm $\| \cdot \| $. We assume that
\[
L:{\rm dom}L\subset H\to H,
\] is a linear, self-adjoint\index{self-adjoint} operator
 (${\rm dom}L$ is the domain of $L$) and
\[
N:H\to H
\] is a nonlinear mapping which is Fr\'echet differentiable\index{Fr\'echet differentiable} there,
with symmetric\index{symmetric} Fr\'echet derivative $N'$ (see \cite{rudin:rca66}, \cite{schechter:pfa71}).

For a given linear operator
\[
A:{\rm dom}A\subset H\to H,
\]
we denote by $\rho (A), ~\sigma (A),\; r(A)$, respectively the {\it
resolvent set}\index{resolvent set}, {\it
spectrum}\index{spectrum}, and the {\it spectral
radius}\index{spectral radius} of the operator $A$ (see
\cite{kato:ptl66}). Also one writes, for a given linear operator
$A$,
\[
A\geq 0,\;\text{\rm if, and only if},\;\langle Au,u\rangle \geq 0,\;\forall u\in {\rm dom}A.
\]
and
\[
A\geq B,\;\text{\rm if, and only if}, ~A-B\geq 0,
\]
(here, of course, $A-B$ is an operator defined on the intersection of the domains of $A$ and $B$).

We establish the following surjectivity theorem.

\begin{theorem} \label{surjective}
Suppose $L$ and $N$ are as above and there exist real numbers
$\lambda$, $\mu$, $p$, $q$ such that
\[
 \lambda <q\leq p<\mu , \quad [\lambda , \mu ]\subset \rho (L),
\]
and that
\[
qI\leq N'(u)\leq pI, \quad \forall u\in H,
\]
where $I:H\to H$ is the identity mapping.
Then
\[
L-N:{\rm dom}L\to H
\]
is a bijection\index{bijection}.
\end{theorem}

\begin{proof}
For any $\nu \in (\lambda ,\mu )$, and $v\in H $ the equation
\[
Lu-N(u)=v,
\]
is equivalent to the equation
\[
(L-\nu I)u-(N-\nu I )(u)=v,
\]
or
\begin{equation}
\label{A,B}
Au-B(u)=v,
\end{equation}
where
\[
A:=L-\nu I, \quad
B:=N-\nu I.
\]
It follows from the assumptions that $B$ has the symmetric Fr\'echet
 derivative given by
\[
B'(u)=N'(u)-\nu I.
\]
Since $\nu \in \rho (L)$, it follows that
\[
A^{-1}=(L-\nu I)^{-1}
\]
exists and is a bounded operator and further that
\[
[\lambda -\nu, \mu -\nu ]\subset \rho (A).
\]
Hence, since $A$ is self-adjoint,
\[
\sigma (A)\subset (-\infty ,\lambda -\nu)\cup (\mu -\nu , \infty ).
\]
One may further deduce that
\[
\sigma (A^{-1})\subset \left ((\lambda -\nu )^{-1},(\mu -\nu )^{-1}\right ).
\]
Hence
\begin{equation}
\label{alph}
\|A^{-1}\|=r(A^{-1})\leq \max  \{(\nu -\lambda  )^{-1},(\mu -\nu )^{-1}\}=:\alpha .
\end{equation}
All of the above follow from properties of linear operators
(see \cite{kato:ptl66}).
Next, we note that
\begin{align*}
\|B(u)-B(v)\|
&\leq  \sup _{\tau \in [0,1]}\|B'(u+\tau (v-u)\|\|u-v\|\\
&\leq \sup _{w\in H}\|N'(w)-\nu I\|\|u-v\|.
\end{align*}
On the other hand
\[
(q-\nu )I\leq N'(u)-\nu I =B'(u) \leq (p-\nu ) I,
\]
and  hence for each $u\in H$,
\[
\langle (q-\nu )Iv,v\rangle \leq \langle B'(u)v,v\rangle
\leq \langle (p-\nu ) Iv,v\rangle ,\forall v\in H,
\]
or
\[
(q-\nu )\|v\|^2\leq \langle B'(u)v,v\rangle \leq (p-\nu )\|v\|^2.
\]
The latter implies (see again \cite{kato:ptl66})
\begin{equation}
\label{bet}
\|B'(u)\|\leq \max (|q-\nu |,|p-\nu |)=:\beta .
\end{equation}

Now, equation (\ref{A,B}) is equivalent to the equation
\begin{equation}
\label{B,A}
u=A^{-1}(B(u)+v).
\end{equation}
We next note that
\begin{equation}
\label{A,B,v}
\begin{aligned}
\|A^{-1}(B(u)+v)-A^{-1}(B(w)+v)\|
&\leq \|A^{-1}\|\|B(u)-B(w)\|\\
&\leq  \alpha \beta \|u-w\|
\end{aligned}
\end{equation}
(see (\ref{alph}), (\ref{bet})),
and hence, for every $v$,
 the mapping
\[
u\mapsto A^{-1}(B(u)+v)
\]
 will be a contraction mapping provided
$\alpha \beta <1$,
will be satisfied, whenever
\[
\frac{p+\lambda }{2}<\nu <\frac{q+\mu }{2}.
\]
The latter will hold, for example if we choose
\[
\nu =\frac{p+q}{2}\quad\text{or}\quad
\nu =\frac{\lambda +\mu}{2}.
\]
\end{proof}

We remark that it follows from the results in \cite{kato:ptl66} that
\[
\|A^{-1}\|=\frac{1}{{\rm dist}(\nu ,\sigma (L))}
\] and hence
(\ref{A,B})
may be rewritten as
\begin{equation}
\label{A,B,C}
\|A^{-1}(B(u)+v)-A^{-1}(B(w)+v)\|\leq \|A^{-1}\|\|N(u)-N(v)-\nu (u-v)\|.
\end{equation}
Using this, one obtains the following more general version
of Theorem \ref{surjective}.


\begin{theorem} \label{surjective1}
 Suppose $L$ and $N$ are as above and there exists a real
 number $\nu \in \rho (L)$, such that
$0<{\rm dist}(\nu ,\sigma (L))$
and
\[
\|N(u)-N(v)-\nu (u-v)\|\leq k \|u-v\|,\quad \forall u,v\in H ,
\]
where
\[
k<{\rm dist}(\nu ,\sigma (L)).
\]
Then
$L-N:{\rm dom}L\to H$
is a bijection.
\end{theorem}

\subsection{Periodic solutions of conservative systems}
\label{cons1}
In this section we shall discuss the existence of $2\pi$-periodic
solutions of the system of second order differential equations
\begin{equation}
\label{cons}
u''+n(u)=h(t),
\end{equation}
where $n:\mathbb{R}^N \to \mathbb{R}^N$
 is a $C^1$  function
such that its Fr\'echet derivative $n'(u)$ satisfies
\[
rI\leq n'(u)\leq sI ,
\]
where, for some $m$,
\[
m^2<r\leq s<(m+1)^2,\quad m\in \{0,1,2,\dots \},
\]
  $r$ and $s$ are given and
 \[
h:(-\infty , \infty )\to \mathbb{R}^N
\]
 is a  $2\pi -$ periodic function with $h\in L^2(0,2\pi )$.

The following result (for more general versions;
see, e.g. \cite{mawhin:cmp76}
or \cite{ward:eps79}) is valid:

\begin{theorem}
Let the above assumptions hold. Then there exists a unique
$2\pi -$ periodic solution of \eqref{cons}.
\end{theorem}

\begin{proof}
We let
$ H=L^2(0,2\pi ) $ with inner product
\[
\langle u,v\rangle :=\frac{1}{2\pi }\int _0^{2\pi }uvdx.
\]
Let
\[
{\rm dom}L=\{ u\in H: u,\;u'~\text{absolutely continuous, periodic,}~u''\in H\}
\]
\[
L:{\rm dom}L\to H,\;u\mapsto  u''.
\]
Then $L$ is self-adjoint and
it follows, using elementary ordinary differential equations results, that
\[
\sigma (L)=\{-n^2,\;n=0,1,\dots \}.
\]
It is also an easy exercise to verify that, if we define by
\[
Nu:=-n(u(\cdot )),
\]
then
$N:H\to H$,
 and
\[ (N'(u))v(t)=-n'(u(t))v(t).
\]
Using the other assumptions, one sees that all hypotheses of Theorem 1
hold, proving the theorem.
\end{proof}


\section{The theorem of Cauchy-Kowalevsky} \label{chapXII}
% see  Chapter XII \include{contcauchy}

In this chapter we provide one of the fundamental theorems
of the theory of partial differential equations, the existence theorem
of Cauchy-Kowalevsky\index{Cauchy-Kowalevsky theorem}. It marks
 the starting point for the general theory of partial
 differential equations, both from the point of view
 of analysis and as well as from the historical point of view.
 It basically asserts that the noncharacteristic Cauchy problem with
holomorphic coefficients and holomorphic initial data is
 well-posed. The proof is
based on an elegant paper of Walter \cite{walter:epc85} and follows
the treatment provided in \cite{schmitt:nad98}.

\subsection{The  setting}

Let $\mathbb{C}$ denote the set of complex numbers and $\mathbb{C}^n$
denote the $n$-dimensional space of $n$-tuples of complex numbers
$z=(z_1, \dots z_n)$. For $z\in \mathbb{C}^n$, we define the norm of
$z$ by $|z|=\max_{1\le j\le n} |z_j|$.

\begin{definition} \rm
Let $G$ be a domain in $ \mathbb{C}^n$. A function
 $f:G\to \mathbb{C}$ will be said to be
 \emph{holomorphic}\index{holomorphic} in $G$, if $f$
 and $\partial f/\partial z_j=f_{z_j}$ , $j=1, \dots n$,
 are continuous in $G$.
\end{definition}


Let $\Omega$ be an open set in $\mathbb{C}^n$
 such that the boundary $\Gamma =\partial\Omega$
 is nonempty.  Let
\[
d(z):=\text{\rm dist}(z,\Gamma )
\]
 denote the distance from $z$ to the boundary of $\Omega $,
\[
\text{\rm dist}(z,\Gamma ):=\inf_{\zeta\in\Gamma} |z-\zeta |.
\]
Let $\eta $ be a positive real number and define the set $\Omega ^\eta$ by
\[
\Omega ^\eta :=\{ (t,z): z\in\Omega , |t|<\eta d(z)\},
\]
 and either $t\in \mathbb{R}$ (real case) or $t\in \mathbb{C}$
 (complex case). In geometrical terms in the real case, i.e.,
$\Omega ^\eta$ is the double cone, with base $\Omega$ whose sides
 have slope $\eta $.
We  denote by $\Omega _t\subset \mathbb{C}^n$ the set
\[
\Omega _t=\{z\in \Omega : (t,z)\in \Omega ^{\eta}\},
\]
and by
$\Gamma_t$ the boundary of $\Omega_t$.
 If $z\in \Omega_t$, then $d(z)> |t|/\eta$.
 Geometrically, in the real case, the set $\Omega_t$
is the projection onto $\mathbb{C}^n$ of
 the base of that part of $\Omega $ which
 lies above $t$ ($t>0$) or below $t$ ($t<0$).
  For $z\in \Omega_t$, we define $d(t,z)$ by
\begin{equation}\label{distform}
 d(t,z):= d(z)-\frac{|t|}{\eta}.
\end{equation}
The function $d(t,z)$ is positive and represents the distance
 from $z\in \Omega_t$ to $\Gamma_t$. The
 following property of $d(t,z)$ will be needed
 later in the proof of the theorem.

\begin{lemma}\label{distprop}
If $z'\in \mathbb{C}^n$ satisfies $|z-z'|=r<d(t,z)$
for some $z\in\Omega_t$, then $z'\in\Omega_t$ and
\begin{equation}\label{distineq}
d(t,z')\ge d(t,z)-r.
\end{equation}
\end{lemma}

\subsection{The linear case}

We will first give a proof of the Cauchy-Kowalevsky Theorem
 in the case where the equation is linear. I.e.,
 we consider the  initial value problem
\begin{equation}\label{ckeqnlin}
\begin{gathered}
u_t =  A(t,z)u + \sum_{j=1}^n B_j(t,z)u_{z_j}+ C(t,z),\quad z\in \Omega,\\
u(0,z)= f(z)\quad z\in\Omega.
\end{gathered}
\end{equation}
Here, $z\in \Omega $, $t\in \mathbb{R}$ (real case), or $t\in
\mathbb{C}$ (complex case) and  $u$, $u_t$, $u_{z_j}$, and
$C(t,z)$ are complex
 valued column vectors in
 $\mathbb C^m$; and $A(t,z)$ and $B_j(t,z)$ are complex
 valued $m\times m$ matrices. The set $G:=\Omega^\eta$ is the set
defined
above. If we integrate both sides of equation
 (\ref{ckeqnlin}) with respect to $t$ and use
the initial condition to evaluate the integral
 of the left hand side, we obtain the
 equivalent integral formulation of the problem
\begin{equation} \label{ckintlin}
\begin{aligned}
u(t,z) & =  f(z)+\int_0^t C(\tau ,z)\, d\tau  \\
& \quad +\int_0^t \Big[ A(\tau ,z) u(\tau ,z) +
 \sum_{j=1}^n B_j(\tau ,z)u_{z_j}
(\tau ,z)\Big]\, d\tau.
\end{aligned}
\end{equation}

\begin{remark} \rm
In the complex case ($t\in \mathbb{C}$), the integration in equation
 \eqref{ckintlin} is taken along the straight line
 which connects $0$ to $t$ in $\mathbb{C}$.
\end{remark}

\begin{definition} \rm
By a solution to equation \eqref{ckintlin}, we mean a
 function $u(t,z)$ which is continuous in $G$,
 holomorphic in $z$ for fixed $t$ (real case) or holomorphic in $(t,z)$
(complex case), and which satisfies \eqref{ckintlin}.
\end{definition}

Under suitable assumptions on the coefficients
 $A(t,z)$, $B_j(t,z)$, $C(t,z)$, and $f(z)$, we shall see that
 if $u(t,z)$ is a solution to one of the two
 problems \eqref{ckintlin} or (\ref{ckeqnlin}),
then $u$ is of class $C^1$ and is a solution
 to the other of the two problems, i.e.,
the two formulations of the problem are equivalent.
 Throughout  we shall use
 the operator norm for matrices, $|A|=\max_{1\le k\le m}\sum_{j=1}^n
 |a_{jk}|$,
induced by the vector norm $|\cdot |$.

Before stating the main result of the section, we will first prove a
lemma which gives a crucial bound on the  derivatives
of holomorphic functions. The result is usually referred to as
Nagumo's lemma\index{Nagumo's lemma}.

\begin{lemma}\label{nagumo}
Let $\Omega $ be a domain in $\mathbb{C}^n$ and let $f:\Omega
\to \mathbb{C}^m$ be holomorphic and let $p\ge 0$. If, for
$z\in \Omega $,
\[
|f(z)|\le \frac{c}{d^p(z)},
\]
then
\[
|f_{z_j}(z)|\le C_p\frac{c}{d^{p+1}(z)},
\]
where
\[
 C_p=(1+p)\Big( 1+\frac{1}{p}\Big)^p< e(p+1),\quad C_0=1.
\]
\end{lemma}

\begin{proof}
Consider first the case of a single function of a single complex
 variable. Let $\psi :\mathbb{C}\to \mathbb{C}$
 be holomorphic in the disk $\{ \zeta': |\zeta -\zeta' |\le r\}$.
Then by the Cauchy integral formula\index{Cauchy integral formula},
see \cite{rudin:rca66}
\[
 \psi '(\zeta )=\frac{1}{2\pi i}\oint_{|\zeta - \zeta'
 |=r}
 \frac{\psi (\zeta' )}{(\zeta' -\zeta )^2}\, d\zeta' .
\]
Thus, we obtain
\begin{align*}
|\psi '(\zeta)| & = { \frac{1}{2\pi}
\Big|\oint_{|\zeta -\zeta' |=r} \frac{\psi (\zeta' )}{(\zeta' -\zeta
)^2}\,  d\zeta' \Big|} \\
 & \le  { \frac{1}{2\pi} \oint_{|\zeta -\zeta' |=r}
\frac{|\psi (\zeta' )|}{|\zeta' -\zeta |^2} \, |d\zeta' |} \\
 & \le  { \frac{1}{2\pi r^2}
\max_{|\zeta -\zeta' |=r} |\psi (\zeta' )|\cdot (2\pi r)} \\
 & =  { \frac{1}{r} \max_{|\zeta -\zeta' |=r}
 |\psi (\zeta' )|} .
\end{align*}
Now let $f:\Omega\to \mathbb{C}^m$ be
 holomorphic. We apply the result that we have
just obtained to $f(z)$, with $\zeta =z_j$. If $0<r<d(z)$, then
\begin{align*}
|f_{z_j}(z)|
& \le  { \frac{1}{r}
 \max_{|z -z' |=r} |f (z' )|} \\
 & \le  { \frac{c}{r} \max_{|z-z' |=r} \frac{1}{d^p(z')}} \\
 & \le  { \frac{c}{r(d(z)- r)^p}}.
\end{align*}
Where $d(z')\ge d(z)-r$ because of Lemma~\ref{distprop}.
 The choice $r=d(z)/(p+1)$ gives the (optimal) value stated in the lemma.
\end{proof}

We next establish the linear Cauchy-Kowalevsky theorem.

\begin{theorem}\label{cklin}
Let the functions $A(t,z)$, $B_j(t,z)$ and $C(t,z)$
 be continuous in $(t,z)$ and holomorphic in $z$ for fixed
 $t$ and let $f(z)$ be holomorphic in $z$. Suppose that
 there exist constants $\alpha$, $\beta_j$, $\gamma$, $\delta$
and $p$ such that for every $(t,z)$ in $G$,
\begin{equation}\label{coefbnd52}
\begin{gathered}
|A(t,z)|\le\frac{\alpha}{d(t,z)},\quad |B_j(t,z)| \le \beta_j,\\
 |C(t,z)|\le \frac{\gamma}{d^{p+1}(t,z)} ,\quad |f(z)|\le
\frac{\delta}{d^p(z)}.
\end{gathered}
\end{equation}
Suppose further that $\eta>0$ is such that
\begin{equation}\label{eta}
\frac{\alpha}{p} + \Big( 1+ \frac{1}{p}\Big)^{p+1}
 \sum_{j=1}^n \beta_j <\frac{1}{\eta}.
\end{equation}
 Then  \eqref{ckintlin} has a unique solution
 $u(t,z)$ in $G$ and for some constant $c$ this solution satisfies
\[
|u(t,z)|\le \frac{c}{d^p(t,z)}.
\]
\end{theorem}

Note that condition (\ref{eta}) is a smallness condition on $\eta $
and thus makes the theorem a local one, i.e., guarantees the existence
of solutions for small time. The condition, as will be seen, is
imposed in order for the contraction mapping principle to be applicable.

\begin{proof}
Let $E$ be the normed linear space defined by
\[
E:=\{ u\in C^0(G,\mathbb{C}^m):
\text{ $u$ is holomorphic in $z$ and } \|u\|<\infty \},
\]
where the norm on $E$ is defined by
\[
\|u\|:= \sup_G \{d^p(t,z)|u(t,z)|\}.
\]
Note that convergence in this norm implies
 uniform convergence on compact subsets of $G$;
 hence limit functions are holomorphic in $z$ and $E$ is complete.

We write equation \eqref{ckintlin} in the form
\[
u=g+Tu,
\]
where $g$ is given by the equation
\[
g(t,z):=f(z)+\int_0^t C(\tau ,z)\, d\tau,
\]
and $T$ is the linear operator given by
\[
(Tu)(t,z):=\int_0^t\Big[ A(\tau ,z)u(\tau ,z)
+\sum_{j=1}^nB_j(\tau ,z)u_{z_j}(\tau ,z)\Big]\, d\tau.
\]
We first show that $g+Tu\in E$ if $u\in E$.
 Consider each of the terms in turn. For the
 function $f(z)$ we have, using (\ref{coefbnd52}),
\[
d^p(t,z)|f(z)|  \le   d^p(t,z) \frac{\delta}{d^p(z)}   \le  \delta.
\]
The last inequality follows from
 the fact that $d(z)\ge d(z)- |t|/\eta=d(t,z)$.
 Taking the supremum of the left-hand side of this inequality, we get
$\| f\|\le \delta$.

Before estimating the next term in $g$, we
 note that a direct integration gives
\begin{equation}\label{est152}
\Big|\int_0^t \frac{d\tau}{d^{p+1} (\tau z)}\Big|
\le \int_0^{|t|}
 \frac{ds}{\big( d(z)- \frac{s}{\eta}\big)^{p+1}}
\le \frac{\eta} {pd^p(t,z)}.
\end{equation}
Therefore, for the second term in $g$, we have
\begin{align*}
\Big|\int_0^t C(\tau ,z)\, d\tau \Big|
& \le   \Big| \int_0^t |C(\tau ,z)|\, d\tau \Big| \\
& \le  \gamma\Big|\int_0^t \frac{d\tau}{d^{p+1} (\tau , z)}\Big| \\
& <   \frac{\gamma\eta}{pd^p(t,z)}.
\end{align*}
Again, we have used the bound from (\ref{coefbnd52}).
 Multiplying both sides of this inequality by $d^p(t,z)$
 and taking the supremum over $G$, we obtain
\[
\big\| \int_0^t C(\tau ,z)\, d\tau \big\| \le \frac{\gamma\eta}{p}.
\]
It follows from this last inequality and the bound on $f$, that $g\in E$.

According to the definition of the norm on $E$, we have the inequality
\[
|u(t,z)|\le \frac{\| u\|}{d^p(t,z)}.
\]
Starting from this inequality, we may now apply Nagumo's
 lemma, Lemma~\ref{nagumo}, to the region $\Omega_t$,
using the distance function $d(t,z)$. As a result, we get the estimate
\[
|u_{z_j}(t,z)|\le C_p\frac{\| u\|}{d^{p+1}(t,z)}.
\]
Combining (\ref{coefbnd52}) with the bounds
just obtained on $u$ and $u_{z_j}$, we get the estimates
\begin{gather*}
|A(t,z)u(t,z)|\le  \frac{\alpha \| u\|}{d^{p+1}(t,z)},
\\
|B_j(t,z)u_{z_j}(t,z)|\le
\frac{\| u\|}{d^{p+1}(t,z)}\beta_jC_p.
\end{gather*}
Hence, with $\beta :=\sum_{j=1}^n \beta_j$ and
 using the estimate (\ref{est152}), we have
\begin{align*}
|(Tu)(t,z)| & \le  \| u\|  (\alpha +\beta C_p)
 \Big| \int_0^t \frac{d\tau} {d^{p+1}(\tau ,z)}\Big| \\
 & \le   \frac{1}{p}
(\alpha +\beta C_p)\eta \frac{\| u\|}{d^p(t,z)}.
\end{align*}
If we multiply both sides of this inequality by $d^p(t,z)$
and take the supremum of the left hand side, we get the final estimate
\begin{equation}\label{est252}
\| Tu\| \le q\| u\|,
\end{equation}
where
\[
q=\Big(\frac{\alpha}{p} +\beta \big( 1+\frac{1}{p}\big)^{p+1}\Big)\eta .
\]
This shows that $Tu\in E$. Furthermore, the constant $q$
 satisfies $q<1$ by the hypotheses of the theorem, hence,
the contraction mapping principle may be applied to obtain a unique solution
of the equation $u=g+Tu$.
\end{proof}


\begin{example}\rm
Consider the equation
\[
u_t=bu_z,
\]
(here $n=1$, $b=$constant) subject to the initial condition
 \[
u(0,z)=\phi (z).
\]
 In this example $\alpha =0$ and $\beta =|b|$.
 The solution of this initial value problem is
 given by $u(t,z)=\phi (bt+z)$.  Suppose that $\Omega$
 is the disk $|z|<1$ and that $\phi$ is holomorphic in $\Omega$.
Then the solution exists for $|bt+z|<1$, and if $\phi$
 cannot be continued analytically beyond the unit circle,
then this is best possible. If we vary $b$, but
keep $|b|=\beta$ fixed, then the largest region common to
all those regions of existence is the circular
cone $\beta |t|<1-|z|$, i.e., $\eta =1/\beta$ is best possible.
\end{example}

An added advantage to the approach that has been
adopted for the Cauchy-Kowalevsky Theorem,
is that the fixed point of a contractive
mapping $S$ is not only unique, it also
depends continuously on the operator $S$.
Thus we can easily obtain  results on
continuous dependence of the solution
on the coefficients and the initial data.

\subsection{The quasilinear case}\index{quasilinear}

We are now prepared to present the result for the general case.
 It may be  shown that the Cauchy problem\index{Cauchy problem}
 for a general nonlinear system with holomorphic coefficients
 can be reduced, by a change of variables,
 to an equivalent initial value problem for a first order system
which is quasilinear.
Therefore, it is sufficient to state and prove the result for the initial
value problem for a first order quasilinear system.
 Furthermore, if $u$ satisfies an initial condition of the
 form $u(0,z)=f(z)$, then
the substitution of the function
$u(t,z)=f(z)+v(t,z)$ gives a new
differential equation for $v$ which
is again a first order quasi-linear equation
of the same type and satisfies the initial
condition $v(t,z)=0$. Taking these observations into account,
we shall consider the following initial value problem
\begin{equation}\label{ckeqnq}
\begin{gathered}
u_t = \sum_{j=1}^n B_j(t,z,u)u_{z_j}+C(t,z,u),\quad (t,z)\in G,\\
u(0,z)= 0,\quad z\in \Omega.
\end{gathered}
\end{equation}
As was the case for linear equations,
the $B_j$ are $m\times m$ matrices and
$C$ is a column vector of length $m$. The set $G=\Omega^\eta$ is
again defined by the inequality $|t|<\eta d(z)$,
where $\eta$ is defined later in the theorem
and $d(z)$ is an appropriately determined ``distance'' function.
The set $B_R:=B_R(0)$ is the open ball $\{ z: |z|<R \}$
 in $\mathbb{C}^n$.  We shall assume that $d(z)$ is bounded
on $\Omega$ and satisfies the inequalities
\[
0< d(z)\le \text{\rm dist}(z,\Gamma ),\quad
 |d(z)-d(z')|\le |z-z'|,\quad  z,z'\in \Omega.
\]
If $G$, $\Omega_t$, and $d(t,z)$ are defined as before, using the
new function $d(z)$, then Lemma~\ref{distprop} remains valid.
Moreover, $0<d(t,z)<\text{\rm dist}(z,\Gamma_t)$, for
$z\in\Omega_t$. As a consequence, Lemma~\ref{nagumo} remains true.
Although the region $G$ of existence obtained is less than optimal
under this assumption, we derive the benefit that the proof of the
theorem is greatly simplified. We now state and prove the
Cauchy-Kowalevsky theorem for the  quasilinear case.

\begin{theorem}\label{ckql}
Let the functions $B_j(t,z,u)$ and $C(t,z,u)$ be continuous
in $\Omega ^\eta\times B_R$ and holomorphic with respect
 to $z$ and $u$ (real case),
or holomorphic in $\Omega ^\eta\times B_R$ (complex case).
 Suppose further, that the following estimates hold on
 $\Omega ^\eta\times B_R$:
\begin{equation}\label{coefbnd53}
\begin{gathered}
|B_j(t,z,u)|\le \beta_j,\\
 \sqrt{d(t,z)} |B_j(t,z,u) -B_j(t,z v)|\le \beta_j'|u-v|,\quad
j=1,\dots , n,\\
 |C(t,z,u)|\le \frac{\gamma}{\sqrt{d(t,z)}},\quad
d(t,z)|C(t,z,u)-C(t,z,v)|\le \gamma' |u- v|.
\end{gathered}
\end{equation}
If $\eta >0$ is such that
\[
2\eta\sqrt{d_0}(\beta +\gamma )<R,\quad
\eta(3\sqrt{3}(\beta +\gamma )+2\beta )\le 1,\quad
\eta e\beta' < 1,
\]
where $d_0=\sup d(z)<\infty$, $\beta =\sum_{j=1}^n \beta_j$, and
 $\beta' =\sum_{j=1}^n \beta_j'$.
 Then the initial value problem \eqref{ckeqnq}
 has a unique solution which exists in $\Omega^\eta$.
\end{theorem}

\begin{proof}
  The proof proceeds along lines similar to the linear case.
 We consider an equivalent integral formulation of \eqref{ckeqnq} given by
\begin{equation}\label{ckintq}
u(t,z)=\int_0^t \Big[\sum_{j=1}^n B_j(\tau ,z,u(\tau ,z))u_{z_j}
(\tau, z)+C(\tau ,z,u(\tau ,z))\Big]\, d\tau
\end{equation}
and treat it as a fixed point equation of the form $u=Su$ to which
the contraction mapping principle, can be applied.
As the underlying Banach space $E$, we use the
same space as in the proof of Theorem~\ref{cklin}.
However, in contrast to the linear case, the operator $S$ is not
globally defined on $E$ and it will be necessary to
define a proper closed subset $F$ of $E$ which is mapped  into itself by $S$.

Let $F$ be the closed subset of $E$ defined by
\[
F:=\big\{ u\in E: |u(t,z)|\le \rho ,\; |u_{z_j}(t,z)|\le 1/\sqrt{d(t,z)}\big\},
\]
where $\rho :=2\eta\sqrt{d_0}(\beta +\gamma )<R$. Let $u\in F$ and set
$v=Su$. Then, using the bounds in (\ref{coefbnd53}), we obtain
\begin{align*}
|v_t(t,z)| & =  \big|\sum_{j=1}^n B_j(t,z,u(t,z))u_{z_j}(t,z)
 +C(t,z,u(t,z))\big| \\
 & \le  \sum_{j=1}^n |B_j(t,z,u(t,z))||u_{z_j}(t,z)| +
|C(t,z,u(t,z))| \\
 & \le   \frac{\beta +\gamma}{\sqrt{d(t,z)}}.
\end{align*}
 From this inequality and the fundamental theorem of
calculus\index{fundamental theorem of calculus}, it
follows that
\[
|v(t,z)|=\Big|\int_0^t v_t(\tau ,z)\, d\tau \Big|
\le (\beta +\gamma )\int_0^{|t|}
\frac{ds}{\sqrt{d(s,z)}}.
\]
A direct integration then yields
\begin{align*}
\int_0^{|t|} \frac{ds}{\sqrt{d(s,z)}}
& = \int_0^{|t|} \frac{ds}{\sqrt{d(z)-{\frac{s}{\eta}}}} \\
& =   -2\eta\sqrt{d(z)-
{\frac{s}{\eta}}}\Big|_0^{|t|} \\
 & \le  2\eta\sqrt{d(z)} \\
 & \le  2\eta\sqrt{d_0}.
\end{align*}
Therefore,
\[
|v(t,z)|\le 2\eta (\beta +\gamma )\sqrt{d_0} =\rho ,
\]
which is the first inequality in the definition of the set $F$.
 In order to verify the second inequality in the
definition of $F$, we estimate the derivatives of $v$.
Using the inequalities in (\ref{coefbnd53}) and
applying Nagumo's Lemma, Lemma~\ref{nagumo}, we obtain the bounds
\begin{align*}
 \big|\frac{\partial}{\partial z_k}C(t,z,u(t,z))\big|
& \le   \frac{\gamma
C_{1/2}}{d^{3/2}(t,z)} \\
 \big|\frac{\partial}{\partial z_k} B_j(t,z,u(t,z))\big|
& \le \frac{\beta_j}{d^{3/2}(t,z)},
\end{align*}
and from the second inequality in the definition of $F$ and Nagumo's Lemma,
\[
|u_{z_jz_k}| \le  \frac{C_{1/2}}{d(t,z)}.
\]
Here $C_{1/2}=3\sqrt{3}/2$. The product formula for derivatives,
leads to the the inequality
\begin{align*}
|v_{t,z_k}| & =  \Big|\sum_{j=1}^n [(B_j)_{z_k}u_{z_j} +
B_ju_{z_jz_k}] +C_{z_k}\Big| \\
 & \le  \sum_{j=1}^n |(B_j)_{z_k}||u_{z_j}| + |B_j||u_{z_jz_k}|
+|C_{z_k}| \\
 & \le  \frac{(\beta +\gamma )C_{1/2}+\beta}{d^{3/2}(t,z)}.
\end{align*}
A direct integration with respect to $s$ establishes the inequality
\[
\int_0^{|t|} \frac{ds}{d^{3/2}(s,z)}\le \frac{2\eta}{\sqrt{d(t,z)}}.
\]
Therefore, using the fundamental theorem of calculus, we have the estimate
\begin{align*}
|v_{z_j}|&\le \big|\int_0^t v_{t,z_j}(\tau ,z)\, d\tau\big|\\
& \le ((\beta +\gamma )C_{1/2}+\beta)\int_0^{|t|}\frac{1}{(d^{3/2}(\tau ,z))}\,
d\tau \\
&\le \frac{1}{\sqrt{d(t,z)}}.
\end{align*}
This last inequality verifies that $v$ satisfies the second
inequality in the definition of $F$, as well. It follows from the
above that $Su\in F$ whenever $u\in F$, i.e., $S$ maps $F$ into
itself.

To complete the proof, we need to show that $S$ is a contraction
mapping. Let $u$ and $v$ be in $F$ and consider the $t$
derivative of the difference
\begin{align*}
(Su-Sv)_t  &=  \sum_{j=1}^n [B_j(t,z,u)-B_j(t,z,v)]u_{z_j}\\
&\quad +\sum_{j=1}^n B_j(t,z,v)[u_{z_j}-v_{z_j}]
+ C(t,z,u)-C(t,z,v).
\end{align*}
The definition of the Banach space $E$ gives us the inequality
\[
|u-v|\le \frac{\| u-v\|}{d^p(t,z)}.
\]
Applying Nagumo's lemma to this inequality, gives the estimate
\[
|u_{z_j} -v_{z_j}|\le \frac{C_p\| u- v\|}{d^{p+1}(t,z)}.
\]
Combining these results with the
hypotheses (\ref{coefbnd53}), we obtain the following
inequality for the difference in the derivatives
\begin{align*}
|(Su-Sv)_t| & \le  \Big( \sum_{j=1}^n \beta_j'\Big)
\frac{|u-v|}{d(t,z)} +\sum_{j=1}^n
\beta_j|u_{z_j} - v_{z_j}| +\gamma'\frac{|u-v|}{d(t,z)} \\
& \le  \big(\beta' +\beta C_p+\gamma'\big)\frac{\| u- v\|}{d^{p+1}(t,z)}.
\end{align*}
Thus, making use of (\ref{est152}), and the
fundamental theorem of calculus, we get
\[
|Su-Sv|\le \big|\int_0^t (Su-Sv)_t\, dt\big|
\le \frac{\eta}{p}(\beta' +C_p\beta +\gamma' )\frac{\| u- v\|}{d^p(t,z)}.
\]
Hence, when we multiply by $d^p(t,z)$ and then take
the supremum over $\Omega ^\eta$, we get
\[
\| Su-Sv\|\le \frac{\eta}{p}(\beta' + C_p\beta +\gamma' )\| u-v\|.
\]
It follows from our hypotheses on $\eta$ that $S$ is a
contraction mapping for $p$ sufficiently large
and  $S$ has a unique fixed point $u\in F$ which
 is the solution to (\ref{ckintq}).
\end{proof}

An example to which the above, for example may be applied, is the
 the initial value problem for a Burger's type
 equation\index{Burger's equation}
\begin{gather*}
u_t= -(1+z^2+u)u_z-2z(1+z^2+u),\quad z\in\Omega,\; 0<t\\
u(0,z)= 0,\quad z\in \Omega ,
\end{gather*}
where $\Omega =\{ z: |z|<1\}$ is the unit disk, centered at the
 origin.  In the terms of the notation of Theorem~\ref{ckql},
we have $B(t,z,u)=-(1+z^2+u)$ and $C(t,z,u)=-2z(1+z^2+u)$,
which are entire functions. Note also that $B(t,z,u)$ and $C(t,z,u)$
are real valued for real values of $t$, $z$, and $u$.
 The distance function $d(z)$ is given by $d(z)=1-|z|$, $d_0=1$, and
 $d(t,z)=1-|z|-|t|\eta$. Let $R>0$ be given.
Then the conditions (\ref{coefbnd53}) are satisfied with the constants
\begin{gather*}
\beta = 2+R,\quad \beta' = 1, \\
\gamma = 2(2+R),\quad \gamma' = 2.
\end{gather*}
If $\eta$ is chosen such that
\[
\eta\le \min\big\{\frac{R}{6(2+r)},\, \frac{1}{(9\sqrt{3}+2)(2+R)},\,
\frac{1}{e}\big\} ,
\]
then the hypotheses of Theorem \ref{ckql} are satisfied and the
existence of a unique holomorphic solution defined on $\Omega ^\eta$
follows.


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\printindex
\end{document}