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\begin{document} 
{\noindent\small {Electronic Journal of Differential Equations,
Vol. 1995(1995), No. 13, pp. 1--17.}\newline
ISSN: 1072-6691, URL: http://ejde.math.swt.edu (147.26.103.110)\newline
telnet (login: ejde), ftp, and gopher:
 ejde.math.swt.edu or ejde.math.unt.edu}
\thanks{\copyright 1995 Southwest Texas State University  and 
University of North Texas.} 
\vspace{1.5cm}
\title[\hfilneg EJDE--1995/13\hfil Dichotomy]{DICHOTOMY AND
$H^{\infty}$ FUNCTIONAL CALCULI}
\author[R. DeLaubenfels \& Y. Latushkin  \hfil EJDE--1995/13\hfilneg]
{R.~DeLaubenfels \& Y. Latushkin}
\address{R. DeLaubenfels \newline 
Scientia Research Institute\newline P.O. Box 988\newline Athens, OH 45701}
\email{72260.2403@@compuserve.com} 
\address{Y. Latushkin\newline 
Department of Mathematics\newline University of Missouri\newline
 Columbia MO 65211}
%\email{??}
\date{}
\thanks{Submitted August 1, 1995. Published September 21, 1995.}
\thanks{Second author supported in part by NSF grant DMS-9400518.}
\subjclass{47D05, 47A60}
\keywords{Abstract Cauchy problem, operator semigroups, \newline\indent
exponential dichotomy, functional calculi}

\begin{abstract}
Dichotomy for the abstract Cauchy problem
with any densely defined closed operator on a Banach space is studied.
We give conditions under which an operator with an $H^\infty$
functional calculus
has dichotomy.
For the operators with imaginary axis contained in the
resolvent set and with polynomial growth of the resolvent along the axis
we prove
the existence of  dichotomy on subspaces and superspaces.
Applications to the dichotomy of
operators on $L_p$-spaces are given.
The principle of linearized instability
for nonlinear equations is proved.
\end{abstract}
\maketitle
%
\newcommand{\re}{\operatorname{Re}}
\newcommand{\im}{{\mbox{ Im }}}
\newcommand{\gs}{g(s,u(s))}
\newcommand{\ImP}{{\mbox{ Im }P}}
\newcommand{\ImQ}{{\mbox{ Im }(I-P)}}
%
\section{INTRODUCTION}
 In the present paper we use methods from  \cite{dL2,dL3} to
study  dichotomies for solutions to
the abstract Cauchy problem
\begin{equation}\label{(ACP)}
\frac{d}{dt}u(t, x) = A u(t, x) \quad
u(0, x) = x \in X, \quad t \geq 0
\end{equation}
with a closed densely defined operator $A$ on a Banach space $X$.
By a {\it solution} of \eqref{(ACP)} we will mean a classical
solution, that is, $t \mapsto u(t, x) \in C([0, \infty),
[\cal D(A)]) \cap C^1([0, \infty), X)$.
Dichotomy means
 the existence of a bounded projection, $P$,
 such that the solutions that start in $\im(P)$
decay to zero and the solutions that start in $\im(I-P)$
are unbounded.

Dichotomy and, in particular, exponential dichotomy
is one of the main tools in the study of linear
differential equations in Banach spaces,
linearized instability for nonlinear equations, existence
of invariant and center manifolds, etc.
Due to the importance of the subject the literature on
dichotomy is vast;
besides the classical books \cite{DK,Hale,Henry,MS}, we mention here
more recent papers \cite{BGK,Chow,LatMSR,SS} and \cite{SellBibl},
where one can find further references.

Assume, for a moment, that  \eqref{(ACP)} is well-posed; that
is, $A$ generates a strongly continuous semigroup $\{e^{tA}\}_{t\ge 0}$ on $X$.
The semigroup is called hyperbolic if
$\sigma(e^{tA})\cap \Bbb T$ is empty, for $t\neq 0$,
where we write $\sigma(\cdot)$ for the spectrum and $\Bbb T$ for
the unit circle.
Suppose we know that $A$ generates a hyperbolic semigroup.
Then \eqref{(ACP)} has
dichotomy (and even uniform exponential dichotomy---
see definitions below), and $P$ is the Riesz projection
for $e^{tA}$, $t>0$ that corresponds to the part of $\sigma(e^{tA})$
in the unit disk. Also,
by the spectral inclusion theorem
$$\sigma(e^{tA})\setminus\{0\}\supseteq \exp {t\sigma(A)}, \quad t\neq 0$$
(see, e.g., \cite[p. 45]{P}),
one has
\begin{equation}\label{0sp}
\sigma(A)\cap i\Bbb R=\emptyset,
\end{equation}
 and, moreover,
\begin{equation}\label{emptysp}
\sigma(A)\cap \{z\in\Bbb C: |\re z|\leq \epsilon\} =\emptyset \mbox{ for some }
\epsilon >0.
\end{equation}

However, it is more important to know
under which additional condition on $A$ either
\eqref{0sp} or \eqref{emptysp} imply dichotomy.
If the spectral mapping theorem
\[ \sigma(e^{tA})\setminus\{0\}=\exp{t\sigma(A)},\quad t\neq 0\]
 holds for the semigroup $\{e^{tA}\}$,
then \eqref{0sp} implies the hyperbolicity of the semigroup.
This is the
case, for example, when $A$ generates an analytic semigroup; see \cite{N}.
We note that the spectral mapping theorem holds,
in fact, only provided some
condition on the growth of the resolvent $R(z,A)=(z-A)^{-1}$
is fulfilled.
If, for instance, $X$ is a Hilbert space then,
 by the Gearhart-Herbst spectral mapping theorem
(see \cite{N}),  condition \eqref{0sp}
implies the hyperbolicity of the semigroup $\{e^{tA}\}$
provided
$\|R(z,A)\|$ is
bounded along $i\Bbb R$. For any Banach space by a spectral
mapping theorem from \cite{LMS} this implication is
true also provided a certain condition
on the boundedness of the resolvent holds.

Another way to obtain $P$ under conditions
\eqref{0sp} or \eqref{emptysp}
is to integrate $R(z,A)$ along $i\Bbb R$.
If $A$ is a bounded operator with \eqref{0sp}, then
the Riesz-Dunford functional calculus for $A$
gives the dichotomy projector $P$.
If $A$ is unbounded this way does  not work
without additional conditions on the decay of $\|R(z,A)\|$ along $i\Bbb R$.
The necessary and sufficient conditions for a semigroup
with  \eqref{emptysp} to be hyperbolic are given in
\cite{KVL}. These conditions include, in particular, the integrability of
$R(z,A)$
along $i\Bbb R$ in Ces\`aro sense.

The present paper has two  goals.
First, we would like to consider dichotomy for {\it non}-well-posed
abstract Cauchy problems
\eqref{(ACP)}.
That is, we do {\it not} assume that $A$ generates
a strongly continuous semigroup.
Second, we study dichotomy under very mild conditions
on $R(z,A)$,  $z\in i\Bbb R$. We require only
a polynomial growth of the resolvent. Our main technical tool is
to use an $H^\infty$ functional calculus for $A$
to obtain the dichotomy
projection $P$.

In the first
part of the paper, similarly to
the stability theory for semigroups, cf. \cite{N},
we define strong and uniform dichotomy for $A$ in \eqref{(ACP)}.
We show that $A$ has uniform dichotomy provided both $A|_{\im P}$ and
$-A|_{\im(I-P)}$ generate uniformly stable semigroups.
The operators that satisfy these assumptions
are called the bigenerators
and were studied in \cite{BGK}.
We show that $A$ has
strong dichotomy provided these semigroups are strongly stable and
$\sigma(A)\cap i\Bbb R$ is finite. Next, we assume that $A$ has
an $H^\infty(\Omega)$ functional calculus and prove that $A$ has strong
(resp. uniform) dichotomy provided $\overline\Omega$ is disjoint
from $i\Bbb R$ (resp. from a vertical strip, containing $i\Bbb R$). This
corresponds to conditions \eqref{0sp} and \eqref{emptysp},
respectively. We apply these results for two classes
of operators $A$ on $L^p$-spaces having $H^\infty$ calculi:
when $iA$ generates
a bounded group \cite{HP} and
when $A$ is an elliptic differential
operator \cite{[AHS]}.

In the second part of the paper we assume that \eqref{0sp} holds
and $\|R(z,A)\|$ has no more than polynomial growth along $i\Bbb R$.
Under these mild assumptions $A$, generally, does not have
the dichotomy on the entire space $X$. We are able to prove, however, the
existence
of the Banach spaces $Z$ and $W$ such that $Z \hookrightarrow X
 \hookrightarrow W$ and the restriction and extension of $A$ on
 $Z$ and $W$, respectively, have strong dichotomy.

 To comment on the last result,
 let us assume, for a moment, that $A$ generates a continuous
 semigroup, condition
 \eqref{0sp}
 holds and $\|R(z,A)\|$ is bounded for $z\in i\Bbb R$.
 If $X$ is a  Hilbert space,
 the  Gearhart-Herbst spectral mapping theorem
 implies that the semigroup $\{e^{tA}\}$ is hyperbolic.
 This means that $A$ has uniform
 dichotomy on the entire space $X$. If $X$ is a Banach space then,
 generally, $\{e^{tA}\}$
 is not hyperbolic, and, by our result, $A$ has strong dichotomy
 only on a subspace $Z \hookrightarrow X$.


 In the last section of the paper we consider a semilinear
 equation with a linear part that satisfies
 the condition of polynomial growth of the resolvent. Using the result
 on dichotomy on subspaces, we prove the ``principle of linearized instability''
 for the equation. This generalizes some results from \cite{Henry}.


 We use the following notation: $\sigma(A)$, $\rho(A)$,
 $R(z,A)$, $\cal D(A)$ - the spectrum, resolvent set, resolvent,
 domain of an operator $A$, $\cal L(X)$ - the set of bounded linear
 operators on a Banach space $X$.






\section{  DICHOTOMY AND SEMIGROUPS  }

\noindent In the theory of stable strongly continuous semigroups
(see  \cite[p. 99]{N}) the following terminology is used.
A strongly continuous semigroup $\{T(t)\}_{t \geq 0}$
is called {\it stable} if
\[\lim_{t \to \infty} T(t)x = 0
\mbox{ for all } x \in X. \]
 The semigroup is called {\it uniformly exponentially
stable} if there exists positive $\epsilon$ so that
\[
\lim_{t \to \infty} \|e^{\epsilon t}T(t)\| = 0.
\]
Similarly, we define
dichotomy for a densely defined closed operator
$A$ in \eqref{(ACP)} as follows:
\begin{defn}\label{1.2} We will say that an operator $A$
{\it has strong dichotomy} if there exists
a bounded projection, $P$,
such that $PA \subseteq AP$,
$A|_{\im P}$ generates a stable strongly continuous
semigroup, and all nontrivial solutions of
\eqref{(ACP)} such that $x \in \im(I-P)$ are unbounded.

We will say that an operator $A$
{\it has uniform exponential dichotomy} if the semigroup
generated by $A|_{\im P}$ is
uniformly exponentially stable and  there exists positive
$\epsilon$ such that
\begin{equation}\label{diver}
\underline{\lim}_{t \to \infty}
\| e^{-\epsilon t}u(t,x) \| > 0
\end{equation}
for every solution $u$ of
\eqref{(ACP)} with $x \in \im(I - P)$.
\end{defn}
The following proposition shows that \eqref{(ACP)} has
uniform exponential dichotomy provided
$A$ is, in the terminology
of \cite{BGK}, a bigenerator.
\begin{prop}\label{1.3} Suppose there exists a
bounded projection $P$ such that $PA \subseteq AP$ and both $A|_{\im P}$
and $-A|_{\im(I - P)}$ generate strongly continuous
uniformly exponentially stable
semigroups.  Then $A$ has uniform exponential
dichotomy.
\end{prop}
\begin{pf}
Suppose $u$ is a nontrivial solution of \eqref{(ACP)},
with $x \in \im(I - P)$.
We must show that $u$ satisfies \eqref{diver}.
 Let $G \equiv A|_{\im(I - P)}$.
Since $t \mapsto (I - P)u(t, x)$ is a solution of \eqref{(ACP)}, it follows
by the uniqueness of the solutions of \eqref{(ACP)}
that $u(t, x) \in \im(I - P)$, for all $t \geq 0$.  Thus we may
define
$$
w(t) \equiv e^{-tG}u(t, x) \, \, (t \geq 0).
$$
Since $\frac{d}{dt}w(t) = 0$, for all $t \geq 0$,
it follows that $w(t) = w(0) = x$, for all $t \geq 0$.
Thus
$$
\|x\| \leq \|e^{-tG}\| \|u(t, x)\|, \, \, \forall t \geq 0,
$$
so that
$$
\|u(t, x)\| \geq \|e^{-tG}\|^{-1}\|x\|, \, \, \forall t \geq 0,
$$
as desired.
\end{pf}

In order to characterize strong dichotomy in terms of
strong stability of the semigroups generated by
$A|_{\im P}$
and $-A|_{\im(I - P)}$,
we need to introduce the Hille-Yosida space (see
\cite{KLC,K,dLK}, or  \cite[Chapter V]{dL3}).
\begin{defn}\label{1.4}
Suppose $A$ is a closed operator, such that the only solution
of \eqref{(ACP)}, with $x = 0$, is trivial.
The {\it Hille-Yosida space,} $Z(A)$, for $A$,
is defined to be the set of all $x$ for which a bounded
uniformly continuous mild solution of \eqref{(ACP)} exists.
\end{defn}
\noindent We define a norm on $Z(A)$ by
\[
\|x\|_{Z(A)} \equiv \sup_{t \geq 0} \|u(t, x)\|.
\]

In the following lemma the Hille-Yosida spaces for $A$ and $-A$
were used
to find a maximal subspace on which $A$ generates a bounded group
(see \cite{K} and  \cite[Chapter V]{dL3} for the proof).
\begin{lem}\label{1.5}
Suppose that $A$ is as in Definition~\ref{1.4} and define
$Z \equiv Z(A) \cap Z(-A)$.
 Then the following holds:
\begin{itemize}
\item[(1)] $Z$ is the maximal
continuously embedded Banach subspace of $X$ such that $A|_Z$
generates a bounded strongly continuous group;
\item[(2)]$\sigma(A|_{Z}) \subseteq \sigma(A)$.
\end{itemize}
\end{lem}


It is clear that $Z$, from Lemma~\ref{1.5},
is the set of all bounded, uniformly continuous mild
solutions of the reversible abstract Cauchy problem
\begin{equation}
\frac{d}{dt}u(t, x) = Au(t, x),\quad
u(0, x) = x, \quad t\in\Bbb R.
\label{(1.6)}
\end{equation}
Under natural conditions on $\sigma(A)$,
this abstract Cauchy problem
 cannot have solutions bounded on the entire line:
\begin{lem}\label{1.7}
Suppose that $A$ is as in Definition~\ref{1.4},
$\sigma_p(A) \cap i\Bbb R$ is empty,
and
\[\sigma(A) \cap i\Bbb R \quad\mbox{ is countable.} \]
Then all nontrivial solutions of \eqref{(1.6)} are unbounded.
\end{lem}
\begin{pf}
Suppose $u$ is a bounded solution of \eqref{(1.6)}.
Fix $\lambda \in \rho(A)$.
Then
$$
(\lambda - A)^{-1}u(0) \in Z \equiv Z(A) \cap Z(-A),
$$
since $t \mapsto
(\lambda - A)^{-1}u(t)$ has a bounded derivative, hence
is uniformly continuous.  By Lemma~\ref{1.5}(2), $\sigma(A|_Z)
\cap i\Bbb R$ is countable.  But since $A|_Z$ generates a
bounded strongly continuous group, $\sigma(A|_Z) \subseteq i\Bbb R$.
Thus $\sigma(A|_Z)$ is a countable subset of $i\Bbb R$.
If $\sigma(A|_Z)$ is nonempty, then it follows that it must
contain an isolated point. This isolated point is an
imaginary eigenvalue for $A|_Z$  (see  \cite[Chapter 8]{Da}),
hence for $A$.  Since $\sigma_p(A) \cap i\Bbb R$ is empty, this
would be a contradiction. Thus $\sigma(A|_Z)$ is empty, which implies that
$Z$ is trivial (see  \cite[Chapter 8]{Da}).
Thus $(\lambda - A)^{-1}u(0) = 0$, so that $u$ is trivial.
\end{pf}

When $\sigma(A) \cap i\Bbb R$ is empty, Lemma 2.5 may be found
in  \cite{dLV} and \cite{Huang}.


The following proposition is the analogue of Proposition~\ref{1.3}
for the case of strong dichotomy.
\begin{prop}\label{1.8}
Suppose there exists a
bounded projection $P$ such that $PA \subseteq AP$, both $A|_{\im P}$
and $-A|_{\im(I - P)}$ generate strongly continuous
stable semigroups, and
\begin{equation}\label{finite}
\sigma(A) \cap i\Bbb R \mbox{
is countable. }
\end{equation}
Then $A$ has strong dichotomy.
\end{prop}
\begin{pf}
Note first that $\sigma_p(A) \cap i\Bbb R$ is empty.
Indeed,
if $Ax = i\lambda x$ for some real $\lambda$, then
 $APx = i\lambda Px$, so that, since $A|_{\im P}$ generates
a stable strongly continuous semigroup, $Px = 0$; similarly,
$(I - P)x = 0$.

Suppose $u$ is a bounded solution of \eqref{(ACP)}, with $x \in \im(I - P)$.
We must show that $u$ is trivial.
Clearly $u$ extends to a bounded solution of (1.6), by defining
\[
u(t) \equiv e^{tA|_{\im(I - P)}}u(0) \, \, (t \leq 0).
\]
By Lemma~\ref{1.7}, $u$ is trivial.
\end{pf}



The following example shows that
hypothesis \eqref{finite}
in Proposition~\ref{1.8} is necessary.  That is, it is not
sufficient, for $A$ to have strong dichotomy,  to have
both $A|_{\im P}$ and $-A|_{\im(I - P)}$ generate strongly continuous
stable semigroups.
\begin{exmp}
Take $X \equiv L^p(\Bbb R, g(s) \, ds)$, $1 \leq p < \infty$,
where $g$ is a nondecreasing positive function on $\Bbb R$,
and take $A$ to be $-\frac{d}{ds}$. That is,
$$
\|f\|^p \equiv \int_{\Bbb R} |f(s)|^p g(s) \, ds,
$$
and $-A$ is the generator of the
strongly continuous contracting semigroup of left-translations
\[
e^{-tA}f(s) \equiv f(s + t), \quad s \in \Bbb R, \, t \geq 0,\,
f \in X.\]
It is not hard to see that, for $f$ bounded and of compact support,
$$
\lim_{t \to \infty} \|e^{-tA}f\|^p = g(-\infty)
\int_{\Bbb R} |f(s)|^p \, ds.
$$
Thus, if we choose $g$ such that $g(-\infty) = 0$,
then $-A$ generates a stable strongly continuous semigroup.
Except for condition \eqref{finite},
we have the hypotheses in Proposition~\ref{1.8}, with $P \equiv 0$.
Strong dichotomy is thus equivalent to \eqref{(ACP)} having no
nontrivial bounded solutions.

If we assume that $g$ is exponentially bounded, then translation
becomes a strongly continuous group,
$$
e^{tA}f(s) \equiv f(s - t), \quad s, t \in \Bbb R, \, f \in X.
$$
It is again clear that
$$
\lim_{t \to \infty} \|e^{tA}f\|^p = g(\infty)
\int_{\Bbb R} |f(s)|^p \, ds
$$
for any $f \in X$.
Thus, if $g$ is bounded, we do not have strong dichotomy;
in fact, \eqref{(ACP)} has a bounded solution for all initial data
in the domain of $A$.
\end{exmp}

An analogue of this example, for incomplete second-order Cauchy
problems, is in  \cite[Example 2.15]{dL1}.  See
\cite[Section II]{dL1}
for the relationship between different versions of such
Cauchy problems and stable or bounded strongly continuous semigroups.
In the language of \cite[Definition 2.7]{dL1}, the operator
$-A|_{\im(I - P)}$, from Proposition~\ref{1.8}, generates a bounded,
nowhere-reversible strongly continuous semigroup.

\section{ DICHOTOMY AND $H^\infty$ FUNCTIONAL
CALCULI}

\noindent In this section we will study dichotomy for \eqref{(ACP)}
for  operators that have an $H^\infty$ functional calculus.
Examples of operators with this property and applications of
our dichotomy results are given in the next section.
\begin{defn}\label{2.1}
If $\Omega$ is an open subset of the complex plane,
not equal to the entire plane, we will say that an operator $A$ {\it
has an
$H^\infty(\Omega)$
functional calculus}
if $\sigma(A) \subseteq \overline{\Omega}$ and
there exists a continuous algebra homomorphism,
$f \mapsto f(A)$, from $H^{\infty}(\Omega)$ into $\cal L(X)$,
such that $f_0(A) = I$ and  $g_{\lambda}(A) = (\lambda - A)^{-1}$,
for all $\lambda \notin \overline{\Omega}$,
where $f_0(z) \equiv 1, g_{\lambda}(z) \equiv (\lambda - z)^{-1}$.
\end{defn}

The main tool in the proof of the next proposition is the ABLV-Theorem
(Arendt-Batty-Lyubich-V\~u; see
\cite{LV} and \cite {AB}), that gives the best available condition for
a strongly continuous semigroup to be stable.
\begin{prop}\label{2.2}
Suppose $\Omega$ is an open set
contained in the left half-plane, such that
$\overline{\Omega} \cap i\Bbb R$ is countable,
$\sigma_p(A) \cap i\Bbb R$ is empty and
$A$ is densely defined and has an $H^{\infty}(\Omega)$
functional calculus.  Then $A$ generates a stable
strongly continuous semigroup, if either
\begin{itemize}
\item[(1)] $X$ is reflexive, or
\item[(2)]
$\overline{\Omega} \cap i\Bbb R$ is empty.
\end{itemize}

\noindent If $\Omega \subseteq \{z \in \bold C \, | \, \re (z) <
-\epsilon \}$, for some positive $\epsilon$, then
the semigroup is uniformly exponentially stable.
\end{prop}
\begin{pf}
 Since $A$ has an $H^{\infty}
(\Omega)$ functional calculus, and $\Omega$ is contained
in the left half-plane, a short calculation
shows that $\{\|\lambda^n(\lambda - A)^{-n}\| \, | \,
\lambda > 0, n \in \bold N \}$ is bounded.  By the Hille-Yosida
theorem, since $\cal D(A)$ is dense, $A$ generates a
bounded strongly continuous semigroup.

 Since $\sigma(A) \cap i\Bbb R$ is contained in
$\overline{\Omega} \cap i\Bbb R,$ the ABVL-Theorem
(\cite{LV} and \cite {AB}) guarantees that either (1) or (2) above
implies
that the semigroup generated by $A$ is stable.

If there exists positive $\epsilon$ such that
$\Omega \subseteq
\{z \in \bold C \, | \, \re (z) < - \epsilon \}$,
then, exactly as argued at the beginning of the proof,
$(A + \epsilon)$ generates a bounded strongly continuous semigroup,
so that the semigroup generated by $A$ is uniformly exponentially
stable.
\end{pf}

To obtain dichotomy, we need to apply this result for
both ``stable'' and ``unstable'' parts of $A$ as follows.
\begin{cor}\label{2.3}
Suppose $\sigma_p(A) \cap i\Bbb R$ is empty and
$A$ is densely defined and has an
 $H^{\infty}(\Omega)$
functional calculus, where $\Omega$ is an open subset of
the complex plane such that $\overline{\Omega} \cap i\Bbb R$
is countable.
Then there exists a bounded projection $P$ such that
$PA \subseteq AP$ and $A|_{\im P}$ and $-A|_{\im(I-P)}$ generate
stable strongly continuous
semigroups, if either
\begin{itemize}
\item[(1)] $X$ is reflexive, or
\item[(2)]
$\overline{\Omega} \cap i\Bbb R$ is empty.
\end{itemize}


\noindent If there exists positive $\epsilon$ such that
$\Omega \cap \{z \in \bold C \, | \, |\re (z)| < \epsilon \}$
is empty, then ``stable'' may be replaced by ``uniformly
exponentially stable.''
\end{cor}
\begin{pf}
Let
$$
\Omega_1 \equiv \Omega \cap \{z \in \bold C \, |
\, Re(z) < 0 \}, \, \,
\Omega_2 \equiv \Omega \cap \{z \in \bold C \, |
\, Re(z) > 0 \}.
$$
Let $P \equiv 1_{\Omega_1}(A)$ for the characteristic function
$1_{\Omega_1}(\cdot)$ of $\Omega_1$.
Then $I - P = 1_{\Omega_2}(A)$, thus we may apply Proposition~\ref{2.2}
 to both $A|_{\im P}$ and
$-A|_{\im(I-P)}$.

If there exists positive $\epsilon$ such that
$\Omega \cap \{z \in \bold C \, | \, |\re (z)| < \epsilon \}$
is empty, then replace $Re(z) < 0$ with $Re(z) < -\epsilon$
and $Re(z) > 0$ with $Re(z) > \epsilon$, and again use Proposition
3.2.
\end{pf}

We are ready to prove the main result of this section.
For $0 < \theta \leq \pi$ let
$S_{\theta} \equiv \{re^{i\phi} \, | \, r > 0, | \phi | < \theta \}$
denote a sector of angle $\theta$.

\begin{thm}\label{2.5}
Suppose $\Omega$ is an open subset
of the complex plane such that $\overline{\Omega} \cap
i\Bbb R$ is countable, $\sigma_p(A) \cap i\Bbb R$ is empty and
$A$ is densely defined and has an $H^{\infty}(\Omega)$
functional calculus.
Then $A$ has strong dichotomy, if either
\begin{itemize}
\item[(1)] $X$ is reflexive, or
\item[(2)]
$\overline{\Omega} \cap i\Bbb R$ is empty.
\end{itemize}

\noindent If, in addition to (2), either
\begin{itemize}
\item[(3)]  there exists $\epsilon > 0$ such that
$\Omega$ is disjoint from $\{z \in \Bbb C \, | \,
|\re (z)| < \epsilon \}$,
or
\item[(4)] $0 \in \rho(A)$ and $\Omega$ is contained in a cone
$(S_{\theta} \cup -S_{\theta})$, for some $\theta < \frac{\pi}{2}$,
\end{itemize}
then $A$ has uniform exponential dichotomy.
\end{thm}
\begin{pf}
The assertion about strong dichotomy follows from
Corollary~\ref{2.3} and Proposition~\ref{1.8}, since
$\sigma(A)$ is contained in $\overline{\Omega}$.

Under hypothesis (3), uniform exponential dichotomy follows from
Corollary~\ref{2.3} and Proposition~\ref{1.3}.

Under hypothesis (4), it is straightforward to show, analogously
to the proof of Proposition~\ref{2.2}, that, for $P$ as in
Corollary~\ref{2.3}, both $A|_{\im P}$ and $-A|_{\im(I - P)}$
generate bounded holomorphic strongly continuous semigroups.
Since $0 \in \rho(A)$, so that $0 \in \rho(A|_{\im P})$ and
$\rho(-A|_{\im(I - P)})$, these semigroups are both uniformly
exponentially stable (see \cite[Theorem 4.4.3]{P}).
 Thus we may again
apply Proposition~\ref{1.3}.
\end{pf}
\begin{rem} Let us stress, that under hypothesis (3)
both $A|_{\im P}$ and $-A|_{\im(I-P)}$ generate uniformly stable
strongly continuous semigroups. We will use this fact in the last section.
\end{rem}

\section{EXPONENTIAL DICHOTOMY ON $L^P$ SPACES}

\noindent In this section we will apply Theorem~\ref{2.5} for two
classes of operators on $L^p$-spaces having an $H^\infty$
functional calculus.

\noindent {\bf 1. Bounded groups.}
We cite the following result from \cite{HP}.
Let $X = L^p(\Omega, \mu)$,
for $1 < p < \infty$, $(\Omega, \mu)$ be a measure space.
\begin{lem}\label{3.1} If $iA$ generates a bounded
strongly continuous group, $A$ is injective
and $0 < \theta < \frac{\pi}{2}$, then $A$ has an
$H^{\infty}(S_{\theta} \cup -S_{\theta})$ functional
calculus.
\end{lem}
Theorem~\ref{2.5} now implies the following.
\begin{cor}\label{3.2}
If $iA$ generates a bounded
strongly continuous group and $A$ is injective, then
$A$ has strong dichotomy.

If, in addition, $0 \in \rho(A)$, then $A$ has uniform
exponential dichotomy.
\end{cor}

\noindent {\bf 2. Differential operators.}
Our next goal is to combine Theorem~\ref{2.5} and  results
from \cite{[AHS]} to study the dichotomy of elliptic differential operators
acting on vector valued $L^p$-functions over $\Bbb R^n$
with sufficiently large zero order term and certain regularity conditions
on the coefficients.

To formulate the results from \cite{[AHS]} we will need some notations.
Let $$\cal A=\sum_{|\alpha|\leq m}a_{\alpha}D^{\alpha}$$ be a
linear differential operator of order $m$ on $X=L^p(\Bbb R^n,\Bbb R^k)$,
$1<p<\infty$,
with $\cal L(\Bbb R^k)$-valued coefficients:
\[a_\alpha:\Bbb R^n\to\cal L(\Bbb R^k), \quad
\alpha\in\Bbb N^n,\quad |\alpha|\leq m.\]
Fix any $M>0$ and $\theta_0\in[0,\pi/2]$. We will say that
$\cal A$ is {\it uniformly $(M,\theta_0)$-elliptic} if
$\max_{|\alpha|=m}\|a_\alpha\|_\infty\le M$
and for its principal symbol
\[\cal A_\pi(x,\xi)\equiv\sum_{|\alpha|=m}a_\alpha(x)\xi^\alpha,\quad
(x,\xi)\in\Bbb R^n\times\Bbb R^n\]
the following conditions hold:
\[\sigma(\cal A_\pi(x,\xi))\subset S_{\theta_0}\setminus\{0\},\quad
\| [\cal A_\pi(x,\xi)]^{-1} \|\le M,
\quad x\in\Bbb R^n,\, \|\xi\|=1.\]

To formulate the regularity conditions on the coefficients,
for fixed $p\in (1,\infty)$ and $m\in\Bbb N$ choose any $q_\alpha$
such that
\[q_\alpha=p \mbox{ if } |\alpha|<m-n/p \mbox{ and }
q_\alpha> n/(m-|\alpha|) \mbox{ if } m-n/p\le |\alpha|\le m.\]
Let $\omega:\Bbb R\to\Bbb R$ be a modulus of continuity
that satisfies the condition
\[\int\limits_0^1\frac{\omega^{1/3}(t)}{t}\,dt<\infty.\]
Let $BUC(\Bbb R^n,\cal L(\Bbb R^k);\omega)$ denote the set of
bounded uniformly continuous functions with the finite norm
\[\|a\|_{C(\omega)}\equiv\|a\|_\infty+\sup_{x\neq y}
\frac{|a(x)-a(y)|}{\omega(|x-y|)}.\]
Also, let
\begin{align*}
L^q_{\mbox{ unif }} & (\Bbb R^n,\cal L(\Bbb R^k))   \equiv \\
&  \left\{
a\in L^1_{\mbox{ loc }}
(\Bbb R^n,\cal L(\Bbb R^k)):
 \|a\|_{q,{\mbox{ unif }}}  \equiv
\sup_{x\in\Bbb Z^n}\|a(\cdot - x)\|_{L^q((-1,1)^n,
\cal L(\Bbb R^k))}<\infty\right\}.
\end{align*}

We impose the following regularity conditions on the coefficients:
\begin{align}\label{regcond1}
& a_\alpha\in BUC(\Bbb R^n,\cal L(\Bbb R^k);\omega)
\mbox{ if } |\alpha|=m, \nonumber \\
& a_\alpha\in L^q_{\mbox{unif}}(\Bbb R^n,\cal L(\Bbb R^k))
\mbox{ if } |\alpha|\le m-1,
\end{align}
and
\begin{equation}\label{regcond2}
\max_{|\alpha|\leq m-1}\|a_\alpha\|_{q_\alpha,{\mbox{ unif }}}+
\max_{|\alpha|=m}\|a_\alpha\|_{C(\omega)}\le M.
\end{equation}
The following result was proved in \cite{[AHS]}.
\begin{lem}\label{ahs}
There exists a constant $\mu>0$ such that for each
$(M,\theta_0)$-elliptic operator $\cal A$ on $\Bbb R^n$, satisfying
\eqref{regcond1}--\eqref{regcond2}, the operator
$\mu+\cal A$ has $H^\infty(S_\theta\setminus\{0\})$ functional calculus
for $0\leq\theta_0<\theta<\pi/2$.
\end{lem}
Theorem~\ref{2.5} now gives the following fact.
\begin{cor}\label{ahscor}
Assume $\cal A$ and $\mu$ are as in Lemma~\ref{ahs}. If
$\mu+\cal A$ is injective, then $\mu+\cal A$
has strong dichotomy. If $\mu+\cal A$ is invertible, then
$\mu+\cal A$ has uniform exponential dichotomy.
\end{cor}

\section{EXPONENTIAL DICHOTOMY ON SUBSPACES AND SUPERSPACES}

\noindent In this section we will assume that $\sigma(A)\cap i\Bbb R=\emptyset$
and the resolvent of $A$
grows no faster than a polynomial along $i\Bbb R$. Under these conditions
$A$, generally, does not have dichotomy on $X$. However, we will
identify Banach spaces $Z$ and $W$ such that $Z \hookrightarrow X
 \hookrightarrow W$ and the restriction and extension of $A$ on
 $Z$ and $W$, respectively, have dichotomy.
 Our main tool is the existence of an $H^\infty$ functional calculus
 on $Z$ and $W$.


\begin{lem}\label{4.1}
 Suppose
$\Omega$ is an open subset of the complex plane whose complement
contains a half-line and whose boundary is a positively
oriented countable system of piecewise smooth, mutually
nonintersecting (possibly unbounded) arcs, $\sigma(A)
\subseteq \Omega$, $A$ is densely defined
 and $\|(w - A)^{-1}\|$ is $O((1 + |w|)^N)$,
for $w \notin \Omega$.

Then there exist Banach spaces $Z$, $W$,
and an operator $B$, on $W$, such that
$$
[ \cal D(A^{N+2}) ] \hookrightarrow Z =
[ \cal D(B^{N+2}) ] \hookrightarrow X \hookrightarrow W,
$$
$A|_Z$ and $B$ are densely defined and have $H^{\infty}(\Omega)$
functional calculi and $A = B|_X$.
\end{lem}
\begin{pf}
The existence of $Z$ is proven in  \cite[Theorem 7.1]{dL2},
except that the density of $\cal D(A|_Z)$ is not addressed.
This density follows by observing that, since $\cal D(A)$ is dense,
it follows that $\cal D(A^{N+3})$ is dense in $[ \cal D(A^{N+2}) ]$,
hence is dense in $Z$; it is clear that $\cal D(A^{N+3})$ is
contained in $\cal D(A|_Z)$.

Define $W$ to be the completion of $Z$ with respect to
the norm
$$
\|x\|_W \equiv \|A^{-(N+2)}x\|_Z.
$$
We construct a functional calculus as follows.
For any $f \in H^{\infty}(\Omega)$, $x \in W$, define
$$
(\Lambda f)x \equiv \lim_{n \to \infty} f(A|_Z)x_n,
$$
where the limit is taken in $W$, and $\{x_n\}$ is any
sequence in $Z$ converging to $x$ in $W$.  Note that
the existence and uniqueness of $\lim_{n \to \infty}
f(A|_Z)x_n$ follows from the boundedness of $f(A|_Z)$
and the fact that $A^{-(N+2)}$ commutes with $f(A|_Z)$.

It is clear that $f \mapsto \Lambda f$ is a continuous
algebra homomorphism from $H^{\infty}(\Omega)$ into $\cal L(W)$.
Let us show that this homomorphism is as in Definition~\ref{2.1}.

For $\lambda \notin \overline{\Omega}$, we claim that
$\Lambda g_{\lambda}$ is injective.  To see this,
suppose $x \in W$ and $\Lambda g_{\lambda}x = 0$.
Choose $\{x_n\} \subset Z$ such that $x_n \to x$
in $W$.  Then $y_n \equiv g_{\lambda}(A|_Z)x_n \to 0$ in $W$.
Since $g_{\lambda}(A|_Z) = (\lambda - A|_Z)^{-1}$,
this means that
\[ (\lambda - A|_Z)A^{-(N+2)}y_n =
A^{-(N+2)}x_n \to A^{-(N+2)}x\quad \mbox{ and } A^{-(N+2)}y_n \to 0,\]
both in $Z$.  Since $A|_Z$ is closed, this implies that
$A^{-(N+2)}x$, hence $x$, must equal $0$, proving the
claim.

Since $f \mapsto \Lambda f$ is an algebra homomorphism,
$\{ \Lambda g_{\lambda} \, | \, \lambda \notin
\overline{\Omega} \}$ is a pseudoresolvent family.
Thus $\{ \Lambda g_{\lambda} \, | \, \lambda \notin
\overline{\Omega} \}$ is a pseudoresolvent family of
injective operators.  This means that there exists
an operator $B$, on $W$, such that $(\lambda - B)^{-1} =
\Lambda g_{\lambda}$, for all $\lambda$ not in $\overline{\Omega}$.
It is clear that $f \mapsto \Lambda f$ is now an
$H^{\infty}(\Omega)$ functional calculus for $B$.

There exists a constant $M$ such that
$$
\|A^{-(N+2)}y\|_Z \leq M\|A^{-(N+2)}y\|_{[\cal D(A^{(N+2)})]}
\equiv M \|y\|,
$$
for all $y \in X$.
This implies that
$$
\|x\|_W \equiv \|A^{-(N+2)}x\|_Z \leq M \|x\|,
$$
for all $x \in Z$; that is,
$
X \hookrightarrow W.
$

To show that $A = B|_X$, it is sufficient to show
that $A^{-1} = B^{-1}|_X$.
Suppose $x \in X$.  Then $x \in W$, so there exists
$\{x_n\}\subseteq Z$ such that $x_n \to x$ and
$A^{-1}x_n \to B^{-1}x$, both in $W$.  This means that
$A^{-(N+2)}x_n \to A^{-(N+2)}x$ and
\[A^{-(N+2)}A^{-1}x_n \to A^{-(N+2)}B^{-1}x\quad\mbox{ in } Z,\]
 hence in $X$.
Thus $A^{-(N+2)}A^{-1}x = A^{-(N+2)}B^{-1}x$, so that
$A^{-1}x = B^{-1}x$, as desired.

Since $\cal D(A|_Z)$ is dense in $Z$, it is dense in $W$;
since $\cal D(A|_Z)$ is contained in $\cal D(B)$, it follows
that $B$ is densely defined.

All that remains is to show that $[ \cal D(B^{N+2})] = Z$.
For $x \in \cal D((A|_Z)^{N+2})$,
$$
\|x\|_Z = \|x\|_{[\cal D(B^{N+2)})]},
$$
thus, since $\cal D((A|_Z)^{N+2})$ is dense in $Z$,
it follows that
$$
[ \cal D(B^{N+2}) ] = Z.
$$
\end{pf}



The following lemma shows that the polynomial
growth of the resolvent
along $i\Bbb R$ automatically implies the same growth
outside some $\Omega$ as  in Lemma~\ref{4.1}
\begin{lem}\label{4.2}
 Suppose $i\Bbb R \subseteq
\rho(A)$
and $\|(iy - A)^{-1}\|$ is $O(1 + |y|^N)$,
for $y$ real.

Then there exists $\Omega$, as in Lemma 5.1,
such that $\overline{\Omega} \cap i\Bbb R$ is empty,
$\sigma(A) \subseteq \Omega$ and
$\|(z - A)^{-1}\|$ is $O(1 + |z|^N)$, for
$z$ outside $\Omega$.
\end{lem}
\begin{pf}
This follows from a power series expansion
of the resolvent.  There exists a constant $M$ so that
$$
\|(iy - A)^{-1}\| \leq M(1 + |y|^N), \forall y \in \Bbb R.
$$
For $\frac{1}{\epsilon} > 2M(1 + |y|^N)$ one has
$(\epsilon + iy) \in \rho(A)$ with
$$
(\epsilon + iy - A)^{-1} = \sum_{k=0}^{\infty}
(-\epsilon)^k(iy - A)^{-(k+1)},
$$
so that
\begin{align*}
\|(\epsilon+iy-A)^{-1}\|  & \leq
\sum_{k=0}^\infty(\epsilon)^k(M(1+|y|^N))^{k+1} \\
& =  \dfrac{M(1+|y|^N)}{1-\epsilon M(1+|y|^N)}
\leq 2M(1+|y|^N),
\end{align*}
as required.
\end{pf}
\begin{rem}\label{bddstr} The proof of Lemma~\ref{4.2}
also shows that the resolvent of $A$ is bounded
in a vertical strip around $i\Bbb R$ provided
 $i\Bbb R\subset \rho(A)$ and $\|(iy-A)^{-1}\|$, $y\in\Bbb R$,
 is bounded.
 \end{rem}

 The following theorem is an immediate consequence of Theorem~\ref{2.5}
 and Lemmas~\ref{4.1} and \ref{4.2}.
 \begin{thm}\label{4.3}
Suppose $A$ is densely defined, $i\Bbb R\subseteq\rho(A)$ and
 $\|(iy - A)^{-1}\|$ is $O(1 + |y|^N)$,
for $y$ real.
Then
\begin{itemize}
\item[(1)] there exists a Banach space $Z$ such that
$$
[ \cal D(A^{N+2}) ] \hookrightarrow Z \hookrightarrow X
$$
and $A|_Z$ has strong dichotomy, and
\item[(2)]
there exists a Banach space $W$ and an operator $B$, on $W$,
such that
$$
[ \cal D(B^{N+2}) ] \hookrightarrow X \hookrightarrow W,
$$
$A = B|_X$, and $B$ has strong dichotomy.
\end{itemize}
\end{thm}
A similar result for uniform exponential dichotomy
also follows from Theorem~\ref{2.5} and Lemma~\ref{4.2}.
\begin{thm}\label{4.4} Suppose $A$ is densely defined,
there exists positive $\epsilon$
such that $\{z\in\Bbb C: |\re z|<\epsilon\}\subseteq \rho(A)$,
and $\|(z-A)^{-1}\|$ is $O(1+|z|^N)$, for $|\re z|<\epsilon$.
Then
\begin{itemize}
\item[(1)] there exists a Banach space $Z$ such that
$$
[ \cal D(A^{N+2}) ] \hookrightarrow Z \hookrightarrow X
$$
and $A|_Z$ has uniform exponential dichotomy, and
\item[(2)]
there exists a Banach space $W$ and an operator $B$, on $W$,
such that
$$
[ \cal D(B^{N+2}) ] \hookrightarrow X \hookrightarrow W,
$$
$A = B|_X$, and $B$ has uniform exponential dichotomy.
\end{itemize}
\end{thm}

Since $\|(iy-A)^{-1}\|=O(1/|y|)$ provided $iA$ generates
a bounded strongly continuous group, the following
result holds.
\begin{cor}\label{4.5}
If $iA$ generates a bounded
strongly continuous group and $0 \in \rho(A)$, then
\begin{itemize}
\item[(1)] there exists a Banach space $Z$ such that
$$
[ \cal D(A) ] \hookrightarrow Z \hookrightarrow X
$$
and $A|_Z$ has uniform exponential dichotomy, and
\item[(2)]
there exists a Banach space $W$ and an operator $B$, on $W$,
such that
$$
[ \cal D(B) ] \hookrightarrow X \hookrightarrow W,
$$
$A = B|_X$, and $B$ has uniform exponential dichotomy.
\end{itemize}
\end{cor}
 \begin{exmp}
To illustrate the effect of ``dichotomy on subspaces''
in Theorems~\ref{4.3}--\ref{4.4},
 let us consider the operator
 \[A\equiv i\dfrac{d}{dx}\quad \mbox{ on }
 \quad X\equiv\{f\in L^p[0,1]: \int_0^1f(x)\,dx=0\}, \quad 1 \leq p < \infty.\]
 For $1 < p < \infty$ the operator $A$ has the
 uniform dichotomy with the bounded projector
 \[ P: f\sim \sum_{k\neq 0}a_ke^{ikx} \mapsto \sum_{k>0}a_ke^{ikx}. \]
 For
 $p=1$ this projector is unbounded, and $A$ does not have dichotomy on the
entire space $X$.
 Note that $iA$ generates a bounded strongly continuous group and
 $\|R(iy,A)\|=O(1/|y|)$.
  Theorem~\ref{4.3} gives a dense
 subspace $Z$ in $X$ such that $A|_Z$ has strong dichotomy.
\end{exmp}



\section {NONLINEAR ABSTRACT CAUCHY PROBLEM}

\noindent In this section we assume that $A$ generates a strongly
continuous semigroup on $X$.
Let $g$ be a nonlinear function,
\[ g: \Bbb R\times U\to \cal D(A^{N+2})\quad\mbox{ for an
open set }\quad U\subset \cal D(A^{N+2}), \quad 0\in U,\]
 such that $g(t,0)=0$.
Assume that $g$ is H\"{o}lder:
\begin{eqnarray}\label{H}
 &&\|g(t,x_1)-g(t,x_2)\|_{\cal D(A^{N+2})}\leq k(r) \|x_1-x_2\|_X,\\
&&\mbox{ for } x_i\in U, \, \|x_i\|_X\leq r, \, i=1,2,
 \mbox{ and } k(r)\to 0 \mbox{ as } r\to 0.\nonumber
\end{eqnarray}

For $t_0\in\Bbb R$  consider the following semilinear abstract
Cauchy problem:
\begin{eqnarray}\label{Cp}
\dfrac{d}{dt}u(t,x)=Au(t,x)+g(t,u(t,x)),\quad
u(t_0,x)=x\in X.
\end{eqnarray}
We say, that $u(\cdot,x)$ is a mild solution of \eqref{Cp} on $(t_0,\tau)$,
if it satisfies the integral equation
 \begin{equation}\label{inteq}
u(t,x)=e^{A(t-t_0)}x+\int\limits_{t_0}^t e^{A(t-s)}g(s,u(s,x))\, ds, \quad
t\in (t_0,\tau).
\end{equation}

For $A$ as in Theorem~\ref{4.4}
 we will prove the following ``principle of linearized instability''
(see \cite[Th.~5.1.3]{Henry}, \cite{Kato,Sh} and references therein
for similar results on sectorial operators $A$).
Recall (see Remark~\ref{bddstr}) that, for instance, the condition
$\|(iy-A)^{-1}\|=O(1)$, $y\in\Bbb R$
implies the hypothesis of Theorem~\ref{4.4} with $N=0$.

\begin{thm} Suppose there exists positive $\epsilon$
such that $\{z\in\Bbb C: |\re z|<\epsilon\}\subseteq \rho(A)$,
and $\|(z-A)^{-1}\|$ is $O(1+|z|^N)$, for $|\re z|<\epsilon$.
Assume $\sigma(A)\cap \{z: \re z>0\}\neq \emptyset$.
Then the zero solution of \eqref{Cp} is unstable in $X$.
That is, for some positive $\epsilon$ and a sequence $x_n\in X$
such that
$\|x_n\|_X\to 0$  there exist solutions $u_*(\cdot)=u_*(\cdot,x_n)$ of
\eqref{inteq} with $x=x_n$ so that $\|u_*(t_n,x_n)\|_X\ge\epsilon$ for some
$t_n\ge t_0$.
 \end{thm}
\begin{pf}
 By Theorem~\ref{4.4} there exists a
Banach space $Z$ with
$\cal D(A^{N+2}) \hookrightarrow Z \hookrightarrow X$
such that, for  constants $M,M_1>0$,
\begin{equation}\label{imb}
\|x\|_X\le M \|x\|_Z,\, x\in Z,\quad\mbox{ and }\quad
\|x\|_Z\le M_1\|x\|_{\cal D(A^{N+2})},\,x\in\cal D(A^{N+2}),
\end{equation}
and $A|_Z$ generates a hyperbolic strongly continuous semigroup on $Z$. This
means that for a projection $P$ bounded on $Z$  and
some $\beta>0$ and $C>1$ one has
\begin{equation}\label{est}
\|e^{tA_-}x\|_Z\le Ce^{-t\beta}\|x\|_Z,\quad
\|e^{-tA_+}x\|_Z\le Ce^{-t\beta}\|x\|_Z, \quad t>0.
\end{equation}
Here $A_-$ and $A_+$ denote the restrictions of $A$ on $\Im(P)$ and
$\Im(Q)$, respectively, where $Q \equiv I-P$.

For $C$ and $\beta$ from \eqref{est} choose $\tilde{x}_0\in \ImQ$,
$\tilde{x}_0\in Z$, so that
\begin{equation}\label{1}
0<\|\tilde{x}_0\|_Z \leq \dfrac{1}{2CM},
\end{equation}
and $r$ small enough, so that
\begin{equation}\label{main0}
\dfrac{2}{\beta}CMM_1k(r)(\|P\|_{\cal L(Z)}+\|I-P\|_{\cal L(Z)})
\leq\dfrac{\|\tilde{x}_0\|_X}{8}.
\end{equation}
By \eqref{1} and \eqref{imb} with $C>1$
one has
\begin{equation}\label{main1}
\dfrac{2}{\beta}CMM_1k(r)(\|P\|_{\cal L(Z)}+\|I-P\|_{\cal L(Z)})
\leq\dfrac{1}{16C}\le\dfrac{1}{2}.
\end{equation}
Denote $x_0=r\tilde{x}_0$. Then \eqref{1} gives:
\begin{equation}\label{2}
\|x_0\|_Z=\|r\tilde{x}_0\|_Z\le\dfrac{r}{2CM}.
\end{equation}

Fix $\tau\ge t_0$. Denote $\cal C=C((-\infty,\tau],Z)$ the space of
$Z$-valued continuous functions with $\sup$-norm. Consider a subset
$\cal B=\cal B_{\tau,x_0}$ of $\cal C$, defined as follows:
\begin{equation}\label{defB}
\cal B=\{ u\in\cal C:
\|u(t)\|_Z\le \dfrac{r}{M} \cdot e^{\frac{\beta}{2}(t-\tau)},\quad
t\le\tau,\quad
(I-P)u(\tau)=x_0\}.
\end{equation}
Define  a nonlinear operator $T=T_{\tau,x_0}$ in $\cal C$ as follows:
\begin{eqnarray*}
(Tu)(t)\equiv e^{-A_+(\tau-t)} x_0 & - & \int\limits_t^\tau
e^{-A_+(s-t)}(I-P) \gs \, ds\\
& + & \int\limits_{-\infty}^t e^{A_-(t-s)}P \gs \, ds,\quad t\le\tau.
\end{eqnarray*}
 \begin{claim} \label{cl1}
$T$ preserves $\cal B$.
\end{claim}
\begin{pf}
By \eqref{est} and \eqref{2} one has:
\begin{equation}\label{11}
\|
e^{-A_+(\tau-t)}
x_0\|_Z
\leq Ce^{-\beta(\tau-t)}\|x_0\|_Z\leq
\dfrac12\dfrac{r}{M}e^{\frac{\beta}{2}(t-\tau)}.
\end{equation}
Fix $u\in\cal B$. Then \eqref{est} gives:
\begin{equation}\label{21}
\left\|\int\limits_{-\infty}^t
e^{A_-(t-s)} P g(s,u(s)) ds\right\|_Z
\le C\|P\|_{\cal L(Z)}
\int\limits_{-\infty}^t e^{-\beta(t-s)}\|\gs\|_Z ds.
\end{equation}
Since $u\in\cal B$,  one has from \eqref{imb}:
\[\|u(s)\|_X\le M\|u(s)\|_Z\le
re^{\frac{\beta}{2}(s-\tau)}\le r,\quad s\le \tau.\]
Then \eqref{H} can be applied in \eqref{21}, and we use \eqref{imb}
to continue the estimate in \eqref{21}:
\begin{equation}\label{22}
\le C\|P\|_{\cal L(Z)} M_1k(r) r\int\limits_{-\infty}^t
e^{\frac{\beta}{2}(s-\tau)}e^{-\beta(t-s)}\, ds=
\dfrac{2}{3\beta}CMM_1k(r)\|P\|\cdot\dfrac{r}{M}
e^{\frac{\beta}{2}(t-\tau)}.
\end{equation}
Similarly,
\begin{equation}\label{23}
\left\| \int\limits_t^\tau
e^{-A_+(s-t)}(I-P) \gs \, ds\right\|\leq
\dfrac{2}{\beta}CMM_1k(r)\|I-P\|\cdot\dfrac{r}{M}
e^{ \frac{\beta}{2} (t-\tau)}.
\end{equation}
Adding \eqref{11},\eqref{22} and \eqref{23}, and taking into account
the inequality \eqref{main1}, we get the desired estimate, as in
\eqref{defB}.
 \end{pf}
 \begin{claim}\label{cl2} $T$ is a strict contraction on $\cal B$.
\end{claim}
\begin{pf} Indeed, similarly to Claim~\ref{cl1},
for $u_1,u_2\in\cal B$ one has :
\begin{eqnarray*} &&\|Tu_1-Tu_2\|_{\cal C}\le\\
\max_t
\{
&&\left\|
\int\limits_{t}^\tau e^{-A_+(s-t)}(I-P) [g(s,u_1(s))-g(s,u_2(s))]\, ds
\right\|_Z +\\
&&\left\|
\int\limits_{-\infty}^t e^{A_-(t-s)}P [g(s,u_1(s))-g(s,u_2(s))]\, ds
\right\|_Z
\}\\
&&\leq
\dfrac{1}{\beta} C M M_1 k(r) (\|P\|+\|I-P\|)\|u_1-u_2\|_{\cal C}\le
\dfrac14 \|u_1-u_2\|_{\cal C}
\end{eqnarray*}
by \eqref{est} and \eqref{main1}.
\end{pf}

Therefore, the equation $u=Tu$ has a unique solution $u_*(\cdot)\equiv
u_{\tau,x_0}(\cdot)$ in $\cal B$.
\begin{claim}\label{cl3}
$u_*$ is a  solution of \eqref{inteq} with $x\equiv u_{\tau,x_0}(t_0)$.
\end{claim}
\begin{pf}Indeed, we project $u_*=Tu_*$ on $\ImP$  to
obtain:
\begin{eqnarray}\label{Pcomp}
&&Pu_*(t)=
\int\limits_{-\infty}^t e^{A_-(t-s)}P g(s,u_*(s))\, ds=\\
&&e^{A_-(t-t_0)}\left[
\int\limits_{-\infty}^{t_0} e^{A_-(t_0-s)}P g(s,u_*(s))\, ds\right]
+\int\limits_{t_0}^t e^{A_-(t-s)}P g(s,u_*(s))\, ds.\nonumber
\end{eqnarray}
Similarly,
\begin{eqnarray}\label{Qcomp}
(I-P)u_*(t) & = & e^{-A_+(\tau-t)}x_0-
\int\limits_{t}^\tau e^{-A_+(s-t)}(I-P) g(s,u_*(s))\, ds\\
& = & e^{A_+(t-t_0)}\left[
e^{A_+(t_0-\tau)}x_0-
\int\limits_{t_0}^\tau e^{A_+(t-s)}(I-P) g(s,u_*(s))\, ds
\right]\nonumber\\
& + & \int\limits_{t_0}^t e^{A_+(t-s)}(I-P) g(s,u_*(s))\, ds.\nonumber
\end{eqnarray}
Since
\begin{eqnarray*}
 x=u_{\tau,x_0}(t_0) & = & e^{A_+(t_0-\tau)}x_0-
\int\limits_{t_0}^\tau e^{A_+(t-s)}(I-P) g(s,u_*(s))\, ds\\
& + & \int\limits_{-\infty}^{t_0} e^{A_-(t_0-s)}P g(s,u_*(s))\, ds,
\end{eqnarray*}
we see that $u_*$ satisfies \eqref{inteq} just by adding
\eqref{Pcomp} and \eqref{Qcomp}.
\end{pf}

To finish the proof of the theorem, let $\epsilon=\frac78\|x_0\|_X$.
For $n\in\Bbb N$ and $\tau=t_0+n$ construct
$u_*(\cdot)\equiv u_{t_0+n,x_0}(\cdot)$ as above and denote
$x_n=u_{t_0+n,x_0}(t_0)$. Since $u_*\in\cal B_{t_0+n,x_0}$, one has
\[\|x_n\|_X\le M\|x_n\|_Z=M\|u_{t_0+n,x_0}(t_0)\|_Z
\le re^{-\frac{\beta}{2}n}\to 0 \mbox{ as } n\to\infty.\]
By Claim~\ref{cl3},  $u_*=u_*(\cdot,\,x_n)$ is
a mild solution for \eqref{Cp} with
$x=x_n$.

It remains to show that, for $t_n\equiv\tau$,
\begin{equation}\label{final}
\|u_*(t_n,x_n)\|_X=
\|u_{\tau,x_0}(\tau)\|_X\ge \epsilon=\dfrac78\|x_0\|_X.
\end{equation}
Indeed, as in Claim~\ref{cl1}, one has:
\begin{eqnarray*}\label{last}
 \|u_*(\tau)-x_0\|_Z& = &\left\|
\int\limits_{-\infty}^{\tau} e^{A_-(t_0-s)}P g(s,u_*(s))\, ds
\right\|_Z \mbox{ (using \eqref{main0}) }  \\
& \le & \dfrac{\|\tilde{x}_0\|_X}{8}
\cdot\dfrac{r}{M}=\dfrac{\|x_0\|_X}{8M}.
\nonumber
\end{eqnarray*}
Now the estimate
\[\|x_0\|_X -\|u_*(\tau)\|_X\le
\|u_*(\tau)-x_0\|_X\le M
\|u_*(\tau)-x_0\|_Z\le\dfrac{\|x_0\|_X}{8}\] gives \eqref{final}.
 \end{pf}


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\end{document}

