%% Y. Latushkin
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%% A remark on invariant manifolds:
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\newcommand{\Lip}{\operatorname{Lip}}
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\begin{document}

\title{A remark on invariant manifolds:\\
the optimal gap condition.}

\author{Y.~Latushkin}
\address{ Department of  Mathematics\\
University of Missouri\\ Columbia, MO 65211}
\thanks{Supported by the NSF grant DMS-9622105 and by the Research Board
of the University of Missouri}
\keywords {Invariant manifolds, spectral gap conditions}
\subjclass {34CXX, 34DXX, 58FXX, 35BXX}
\email{yuri@@math.missouri.edu}

\maketitle

\begin{abstract}
We give optimal gap conditions
on Lipschitz constant of the nonlinear term and growth bounds of the
linear term that imply
the existence of Lipschitz invariant manifolds
for semilinear equations on Banach spaces. This
result improves recent theorems by C.~Foias and
by N.~Casta\~{n}eda and R.~Rosa.
\end{abstract}

\section{Main Results}

 {}From its inception in now classic works,
\cite{Carr,DK,Hale,Henry,HPS}, the invariant  manifold theory
plays an important role in the modern theory of infinite dimensional
dynamical systems. We refer the reader to
 \cite{AW,BJ,CHT,ChLu,ChLuSell,DapLu,KP,Palmer,Scarp,Si,ZW}
and to the literature cited therein.
Recently, several advances in the invariant
manifold theory
have been
made to obtain optimal gap conditions, see \cite{CR,Fo,Mik}.

In this paper we consider semilinear autonomous differential
equations on Banach spaces with globally Lipschitz nonlinearities.
 We give new sharp gap conditions on the spectrum of the linear part
 and Lipschitz constants for the nonlinear part under which the
 equation has Lipschitz invariant manifolds. Descriptively,
 our result says the following.
 Suppose the real part of the spectrum of the linear term
  of the equation has several disjoint
 spectral segments. Fix a spectral segment. Add the Lipschitz constant
 of the nonlinear term that correspond to this
 segment and the Lipschitz constant of the nonlinear
 term that corresponds to all segments to the right (respectively,
 to the left) of this fixed segment.
 Lipschitz invariant manifolds exist provided
 this {\it sum} is strictly less then the length of the spectral gap
 that lies to the right (respectively, to the left) of the segment.


First, we give a precise formulation of this result for the case of two
spectral segments. Let $Z$ be a Banach
space, and $X, Y\subset Z$ are subspaces of $Z$ such that
$Z=X\oplus Y$. We equip $Z$ with the norm $|z|_Z=|x+y|_Z=\max
\{|x|_X,|y|_Y\}$. Consider a system of
differential equations on $Z$ of the form
\begin{eqnarray}\label{1.2}
\dot{x}&=&Ax+F(x+y),\\
\label{1.3} \dot{y}&=&
By+G(x+y).\end{eqnarray}
Let us assume that the operator $A$
is an infinitesimal generator of a strongly continuous
semigroup on $X$, the operator
$-B$ is a generator of a strongly
continuous semigroup on $Y$, the maps
$F:Z\to X$ and $G:Z\to Y$ are
globally Lipschitz, and
$F(0)=G(0)=0$. Further, let us assume there
exist constants
$a<b$ such that
\begin{equation}\label{1.4}
\|e^{tA}\| \leq e^{ta},\quad t\geq
0,\quad \|e^{tB}\|\leq
e^{tb},\quad t\leq
0.\end{equation}
We stress that (\ref{1.4}) implies the inequalities
$\text{Re }\sigma(A)\le a$ and
$\text{Re }\sigma(B) \ge b$ for the spectra of
$A$ and $B$.
Denote $\Lip(F)=\sup
\{|F(z_1)-F(z_2)|/
|z_1-z_2|:z_1\neq z_2\}$, and let $\Lip_1^0(X,Y):=\{\varphi:X\to Y|\,
\varphi(0)=0,\, \Lip(\varphi)\le 1\}$.

We prove the following sharp results on the existence of
``stable''- and ``unstable''-like invariant manifolds for \sys.
Assume gap condition
\begin{equation}\label{1.5}
\Lip(F)+\Lip (G) <b-a.\end{equation}
\begin{thm}\label{main} There exists a function $\varphi :X\to Y$,
$\varphi \in\Lip_1^0(x,Y)$ such that its graph
${\cal M}_\varphi =\{x+\varphi (x)|x\in
X\}$ is a forward invariant
manifold for \sys and
\begin{equation}\label{estlipphi}
\Lip(\varphi )\leq
\frac{\Lip(G)}{b-a-\Lip(F)}.
\end{equation}
If, in addition, $A$ generates a strongly continuous
{\it group} on $X$, then ${\cal M}_\varphi$
is also backward invariant.
\end{thm}
By reversing time, one obtains:
\begin{cor}\label{back}
There exists a function $\psi :Y\to X$, $\psi\in\Lip_1^0(Y,X)$
 such that its graph
${\cal M}_\psi =\{\psi (y)+y|y\in
Y\}$ is a backward invariant
manifold for \sys and
$$\Lip (\psi )\leq \frac{\Lip
(F)}{b-a-\Lip (G)}.$$
If, in addition, $B$ generates a strongly continuous
{\it group} on $Y$, then ${\cal M}_\psi$
is also forward invariant.\end{cor}
By renorming, conditions \eqref{1.4} can be relaxed as follows.
 Assume,
that there exist $M_A$, $M_B>1$ such
that
$$\|e^{tA}\|\leq
M_Ae^{ta},\quad t\leq 0,\quad
\|e^{tB}\|\leq
M_Be^{tb},\quad t<0.$$
Assume the gap condition
$$M_A \Lip (F)+M_B \Lip (G)<b-a.$$
\begin{cor}\label{renorm}
There exists $\varphi :X\to
Y$ with a forward invariant graph ${\cal
M}_\varphi$ such that
$$\Lip (\varphi )\leq M_A\frac{M_B
\Lip (G)}{b-a-M_A \Lip
(F)}.$$ \end{cor}
Similar fact holds for $\psi:Y\to X$.

The proofs of these results are given in Section~2.
The main ingredients are: a connection of Lyapunov-Perron operators
on spaces of exponentially bounded functions on axis,
and Lyapunov-Perron operators on spaces of Lipschitz
maps; a contraction mapping theorem for the
Lyapunov-Perron operator; Gronwall's inequality.

In fact, our method is just a refinement of the technique
developed in \cite{CR,Fo}, where the case of {\it groups}
generators $A$ and $B$ was considered. We note, that in
\cite{Fo} the gap condition
\[ 4 \max \{\Lip(F)\,,\Lip(G)\} < b-a \]
was used instead of \eqref{1.5}. In \cite{CR} this condition was
improved to state
\begin{equation}\label{crgap}
 2 \max \{\Lip(F)\,,\Lip(G)\} < b-a,
 \end{equation}
 which is still stronger then our condition \eqref{1.5}.
 To see, that conditions \eqref{crgap} and \eqref{1.5}
 are optimal, consider the following system of scalar
 linear equations (cf. the example in Section 6 of \cite{CR}):
 \begin{equation}\label{exsharp}
 \dot{x}=-x+\epsilon_1 y,\quad
 \dot{y}=y-\epsilon_2 x,\quad \epsilon_{1}, \epsilon_{2}\in {\Bbb R}.
 \end{equation}
 For $A=-1$, $B=1$, $F(x+y)=\epsilon_1y$,
 $G(x+y)=\epsilon_2y$ one has $\Lip(F)=|\epsilon_1|$,
 $\Lip(G)=|\epsilon_2|$, $b-a=2$. Let ${\cal E}$ be the
 set of those points $(\epsilon_1,\epsilon_2)$
 in $\epsilon_1\epsilon_2$-plane
 where \eqref{exsharp} has an invariant manifold (that is, could be
 decoupled).
 By computing eigenvalues,
 it is easy to see that ${\cal E}$  is exactly the set enclosed
 between the branches
 of the hyperbola $\epsilon_1\epsilon_2=-1$. The square $|\epsilon_1|+
 |\epsilon_2|<2$, given by our gap condition \eqref{1.5},
 is tangent to this hyperbola. This shows that \eqref{1.5}
 is optimal. The points of tangency belong to the
 line $\epsilon_1=-\epsilon_2$. This shows that \eqref{exsharp}
 is also optimal.


Further, we consider the case of $N\ge 3$ spectral segments.
Let $Z$ be a Banach space, and $X_i\subset Z$, $i=1,\ldots, N$ are
subspaces of $Z$ such that $Z=\bigoplus^{N}_{i=1}X_i$. We equip
$Z$ with the norm
$|z|_Z=|x_1+\ldots+x_N|_Z=\max\{|x_i|_{X_i}: i=1,\ldots,N\}$.
 Assume that operators
$-A_1$, $A_2,\ldots ,A_N$ are
generators of strongly continuous
semigroups on Banach spaces $X_1,X_2,\ldots
,X_N$, respectively. Assume, there exists constants $a_i$, $b_i$
 such that $a_N<\ldots<b_i\le a_i<\ldots b_1$, and
\begin{eqnarray}\|e^{tA_i}\|&\leq
&e^{ta_i},\quad t\geq 0,\quad
i=2,\ldots ,N,\label{17.1}\\
\|e^{tA_i}\|&\leq & e^{tb_i},\quad
t\leq 0,\quad i=1,2,\ldots
,N-1.\label{17.2}\end{eqnarray}
Under these conditions the spectra $\sigma(A_i)$ satisfy:
\[\text{Re }\sigma(A_N)\le a_N,\quad
b_i\le \text{Re }\sigma(A_i)\le a_i,\,i=2,\ldots, N-1,\quad
b_1\le \text{Re }\sigma(A_1).\]
For $i=1,\ldots, N-1$ let $\ell_i=b_i-a_{i+1}$ denote the
length of the $i$-th ``spectral gap''.
Let $F_i:Z\to X_i$ denote given Lipschitz maps such that $F_i(0)=0$,
$i=1,\ldots ,N$.

Consider the following system of equations:
\begin{equation}\label{sysmany}
\dot{x}_i=A_ix+F_i(x_1+\ldots
+x_N),\quad i=1,\ldots , N.
\end{equation}
Denote
 $\displaystyle\hat{X}_i=\bigoplus_{j=1,\,j\neq i}^N X_j$, $i=1,\dots,N$.
Also, let
\[
\bl_i  =  \max_{j=i+1,\ldots,N}\Lip(F_j), \quad i=1,\ldots,N-1,
\quad\text{and}\quad
 \br_i  = \max_{j=1,\ldots,i-1}\Lip(F_j), \quad i=2,\ldots,N
\]
denote the Lipschitz constant of the nonlinear term that corresponds,
respectively, to the spectral segments to the left and to the right
from the $i$-th spectral gap. The following result gives the
existence of ``center''-like manifolds, tangent to $X_i$.


Assume the following gap conditions:
\begin{eqnarray}\label{17.3}
\Lip(F_i)+\max_{j>i} \Lip (F_j) &<&
b_i-a_{i+1},\quad i=1,\ldots ,
N-1,\\ \Lip (F_i)+\max_{j<i} \Lip
(F_j) & <& b_{i-1}-a_i,\quad
i=2,\ldots ,
N.\label{17.4}\end{eqnarray}
\begin{thm}\label{manygaps}
There exist maps $\varphi _i:X_i\to
\hat{X}_i$,
$i=1,\ldots ,N$ such that $\varphi_i\in\Lip_0^1(X_i,\hat{X}_i)$,
and the graphs
${\cal M}_i=\{x_i+\varphi
_i(x_i)|x_i\in X_i\}$ of $\varphi _i$ are invariant
for the system \eqref{sysmany} (backward for
$i=1$, backward and forward for
$i=2, \ldots ,N-1$, and forward
for $i=N$).
Moreover, for the Lipschitz constants  $\Lip(\varphi_i)$
the following inequalities hold:
\begin{eqnarray}\label{bigestphilip}
\Lip (\varphi_1) & \le & \bl_1 / (\ell_1-\Lip(F_1), \quad
\Lip (\varphi_N)  \le  \br_N / (\ell_{N-1}-\Lip(F_N), \nonumber\\
\Lip(\varphi_i) &\le &
\max\left\{
\frac{\bl_i}{\ell_i-\Lip(F_i)},\,
\frac{\br_i}{\ell_{i-1}-\Lip(F_i)}
\right\},\, i=2\ldots,N-1.
\end{eqnarray}
\end{thm}
This theorem is proved in Section~3.
We remark, that the proof of \eqref{bigestphilip} in
Theorem~\ref{manygaps}
{\it cannot} be reduced just to applications of Theorems~\ref{main} and
Corollary~\ref{back}, and requires a rather long argument
(see Theorem~\ref{central} for $N=3$ in Section~3).

\noindent{\bf Acknowledgements.}
The author thanks C.~Foias for suggestions and discussions.
This paper was finished when the author
visited Georgia Institute of Technology. The author thanks
S.-N.~Chow, J.~Hale and Y.~Yi for numerious discussions.

\section{Proofs: two spectral bands}

\noindent The plan of this section is as follows. First, we explain in which
sense
we understand solutions of \sys. Next, we define a Lyapunov-Perron operator
${\cal L}$ acting on functions $\varphi:X\to Y$ and show that
$\text{graph }\varphi$ is an invariant manifold if and only if $\varphi$
is a fixed point of ${\cal L}$. Then, we introduce Lyapunov-Perron
operators $T_\xi$, $\xi\in X$ on a Banach space
$E_\sigma$ of exponentially bounded on axis functions, and prove that
each $T_\xi$, $\xi\in X$
is a strict contraction. Finally, we construct a fixed
point $\varphi$ for ${\cal L}$ out of fixed points for $T_\xi$.

 Solutions to
\eqref{1.2}-\eqref{1.3} are
understood in the following sense.
For any $-\infty<t_1<t_2<\infty$
we say that
$z(\cdot )\in C([t_1,t_2],Z)$ given by
$z(t)=x(t)+y(t)$, $t\in [t_1,t_2]$
is a solution to \sys
provided
\begin{eqnarray} \label{1.5.1}
x(t)&=&e^{(t-t_1)A}x(t_1)+\int^t_{t_1}
e^{(t-s)A}F(x(s)+y(s))ds,\\
\label{1.5.2}
y(t)&=&e^{(t-t_2)B}y(t_2)-\int^{t_2}_t
e^{(t-s)B}G(x(s)+y(s))ds\end{eqnarray}
holds for $t\in [t_1,t_2]$. We say
$z$ is a solution on $[t_1,t_2]$
for $-\infty \leq t_1<t_2\leq
\infty$ if $z$ is solution on each
finite subinterval $[t_1,t_2]$.
\begin{lem}\label{ExUniq} For
every $t_1<t_2$ and $\xi \in X$,
$\eta \in Y$ there exists a unique
$z\in C([t_1,t_2];Z)$,
$z(t)=x(t)+y(t)$, such that
\begin{eqnarray*}x(t)&=&e^{(t-t_1)A}
\xi
+\int^t_{t_1}e^{(t-s)A}F(x(s)+y(s))ds,
\\ y(t)&=&e^{(t-t_2)B}\eta
-\int^{t_2}_te^{(t-s)B}G(x(s)+y(s))ds,
\quad t\in
[t_1,t_2].\end{eqnarray*}\end{lem}
\noindent{\bf Proof.} Let $\alpha
=\sup \{\|e^{tA}\|:t\in
[0,t_2-t_1]\}$, $\beta =\sup
\{\|e^{-tB}\|: t\in
[0,t_2-t_1]\}$. Define an operator
${\cal F}$ on $C=C([t_1,t_2];Z)$ by
$$({\cal F}z)(t)=e^{(t-t_1)A}\xi
+\int^t_{t_1}e^{(t-s)A}F(z(s))ds+
e^{(t-t_2)B}\eta-\int^{t_2}_t
e^{(t-s)B}G(z(s))ds.$$
For
$z_1,z_2\in C$ and $t\in [t_1,t_2]$ one has:
$$|{\cal F}z_1(t)-{\cal
F}z_2(t)|\leq \max\{\alpha \Lip (F)
(t-t_1),\quad \beta \Lip (G)
(t_2-t)\}\|z_1-z_2\|_C.$$ By
induction,
\[|{\cal
F}^nz_1(t)-{\cal F}^nz_2(t)|\leq
\frac{1}{n!}[\max \{\alpha
\Lip(F)(t-t_1),\,\beta \Lip
(G)(t_2-t)\}]^n\|z_1-z_2\|_C\]
and
$\|{\cal F}^nz_1-{\cal
F}^nz_2\| \leq
c^n\|z_1-z_2\|/n!$ implies the result,
see
\cite{Pazy}, p. 184 for more
details.\quad $\Box$



Fix any $\varphi \in
\Lip_1^0(X,Y)$. For a $\xi \in X$
let $x(t;\xi )$, $t\geq 0$ denote
the solution to the equation
\begin{equation}
\label{2.1}x(t;\xi)=e^{tA}\xi
+\int^t_0e^{(t-s)A}F(x(s;\xi)+\varphi
(x(s;\xi )))ds.\end{equation}
Since $\xi
\mapsto F(\xi +\varphi (\xi ))$ is
Lipschitz, and $\{e^{tA}\}_{t\geq
0}$ is a strongly continuous
semigroup on
$X$, the unique solution
$x(t;\xi)$ exists for all $t\in
{\Bbb R}_+$ (\cite{Pazy}, Sec. 6.1).
\begin{prop}\label{growthx}
$|x(t,s)|\leq \exp [a+\Lip
(F)]t$ for all  $t\geq 0$.
\end{prop}
\noindent{\bf Proof.} From
\eqref{2.1}, \eqref{1.4} and
$F(0)=\varphi (0)=0$, $\Lip(\varphi
)\leq 1$ one has:
\begin{eqnarray*}|x(t,\xi )|&\leq
& e^{ta}|\xi
|+\int^t_0e^{(t-s)a}|F(x(s,\xi
)+\varphi (x(s,\xi )))-F(0)|ds\\
&\leq & e^{ta}|\xi |+\Lip
(F)\int^t_0e^{(t-s)a} \max
\{|x(s,\xi )|,|\varphi
(x(s,\xi))-\varphi (0)|\}ds\\
&\leq & e^{ta}|\xi |+\Lip
(F)\int^t_0e^{(t-s)a}\max
\{|x(s;\xi )|, \quad \Lip (\varphi
)x(s;\xi )|\}ds\\ &\leq
&e^{ta}|\xi |+\Lip
(F)\int^t_0e^{(t-s)a}|x(s;\xi
)|ds,\end{eqnarray*} and the
Gronwalls' inequality implies the
result.\quad $\Box$
\begin{prop}\label{Lipx} For any
$\xi_1,\xi_2\in X$,
$$|x(t;\xi_1)-x(t;\xi_2)|\leq
e^{[a+\Lip
(F)]t}|\xi_1-\xi_2|,\quad t\leq
0,$$\end{prop}
\noindent{\bf Proof.} Similarly to
Proposition~\ref{growthx}.\quad
$\Box$

Define the Lyapunov-Perron
operator ${\cal L}$ on Lipschitz
functions $\varphi:X\to Y$ as
follows:
\begin{equation}\label{LP} ({\cal
L}\varphi)(\xi
)=-\int^\infty_0e^{-sB}G(x(s;\xi)+
\varphi (x(s,\xi )))ds.\end{equation}
The integral exists, since, due to
\eqref{1.4} and $G(0)=\varphi
(0)=0$, $\Lip (\varphi )\leq 1$, one
has:
\begin{eqnarray*}|{\cal L}\varphi
(\xi )|&\leq
&\int^\infty_0\|e^{-sB}\|
|G(x(s,\xi )+\varphi
(x(s,\xi))-G(0)|ds\leq\\ &\leq &
\int^\infty_0e^{-sb}\Lip (G)\max
\{|x(s,\xi )|,|\varphi (x(s,\xi
))-\varphi (0)|\}ds\leq\\ &\leq &
\Lip (G)\int^\infty_0e^{-sb}\max
\{|x(s,\xi )|,\Lip (\varphi
)|x(s,\xi )|\}ds\leq\\ &\leq & \Lip
(G)\int^\infty _0e^{-sb}|x(s;\xi
)|ds.\end{eqnarray*} Using
Proposition \ref{growthx}, this is
$$ \leq \Lip (G) \int^\infty_0\exp
[(a-b+\Lip
(F))s]ds=\frac{\Lip(G)}{b-a-\Lip
(F)}\leq 1$$ by the gap condition
\eqref{1.5}.



\begin{prop}\label{Preserves}
${\cal L}$ preserves $\Lip
^0_1(X,Y)$, that is, if $\Lip
(\varphi )\leq 1$, then
$$\Lip ({\cal L}\varphi )\leq
\frac{\Lip (G)}{b-a-\Lip
(F)}<1.$$\end{prop}

\noindent{\bf Proof.} By
Proposition~\ref{Lipx}, and using
$\Lip (\varphi ) \leq 1$,
\begin{eqnarray*}|({\cal L}\varphi
)(\xi_1)-({\cal L}\varphi )(\xi
_2)|&\leq &
\int^\infty_0e^{-sb}\Lip (G)
\max
\{|x(s;\xi_1)-x(s;\xi_2)|,\\
|\varphi (x(s;\xi_1))-\varphi
(x(s;\xi_2))|\}ds&\leq & \Lip
(G)\int^\infty_0e^{-sb}|x(s;\xi_1)-
x(s;\xi_2)|ds\\ &\leq &\Lip (G)
\int^\infty_0\exp [(a-b+\Lip (F))s]ds
|\xi_1-\xi_2|\\
&=&\frac{\Lip(G)}{b-a-\Lip(G)}|\xi
_1-\xi_2|.\quad
\Box\end{eqnarray*}

A set ${\cal M}_\varphi =\{\xi
+\varphi (\xi )\{\xi \in X\}$, the graph of a function
$\varphi :X\to Y$, is called
forward invariant for
\sys if for each $\xi \in
X$ and $t_2>0$ there exists a
unique solution $z(\cdot)$ of
\sys on
$[0,t_2]$ in the sense of
\eqref{1.5.1}-\eqref{1.5.2} such
that $z(0)=\xi +\varphi (\xi )$
and $z(t)\in {\cal M}_\varphi $
for $t\in [0,t_2]$.
\begin{lem}\label{invM} ${\cal
M}_\varphi$ is forward invariant
for \sys if and only if
$\varphi = {\cal L}(\varphi)$.
\end{lem}
\noindent{\bf Proof.} Indeed,
\begin{eqnarray}\label{4.1} ({\cal
L}\varphi )(x(t;\xi
))&=&-\int^\infty_0e^{-sB}G(x(s;x
(t;\xi ))+\varphi
(x(s;x(t;\xi))))ds=\nonumber\\
&=&-\int^\infty_0e^{-sB}G(x(s+t;\xi)+
\varphi (x(s+t;\xi)))ds=\\
&=&-\int^\infty_te^{(t-s)B}G(x(s;\xi
)+\varphi (x(s;\xi )))ds,\quad
t\geq 0.\nonumber\end{eqnarray}
Assume
$\varphi ={\cal L}\varphi$. Then
for each $\xi \in X$ and every
$t_2>0$ there exists a solution
$z$ on
$[0,t_2]$ to
\eqref{1.2}-\eqref{1.3} such that
$z(0)=\xi +\varphi (\xi)$ and
$z(t)=x(t;\xi)+\varphi (x(t;\xi ))$
for each
$t\in [0,t_2]$. Indeed, by the
definition of $x(t;\xi )$ one has
$$x(t;\xi )=e^{tA}\xi
+\int^t_0e^{(t-s)A}F(x(s;\xi
)+\varphi (x(s;\xi )))ds,
\quad t\geq 0.$$ Let
$g(s)=G(x(s,\xi )+\varphi (x(s,\xi
)))$, $s\geq 0$. Denote $y(t):=\varphi (x(t;\xi ))$.
Then, by the
assumption,
$$y(t)=({\cal
L}\varphi)(x(t,\xi))=-\int^\infty_t
e^{(t-s)B}g(s)ds,\quad t\geq 0,$$ and $y$
satisfies for $t\in [0,t_2]$ the
following equation:
\begin{eqnarray}y(t)&=&-\int^\infty_t
e^{(t-s)B}g(s)ds=-e^{(t-t_2)B}
\int^\infty_{t_2}
e^{(t_2-s)B}g(s)ds-\int^{t_2}_t
e^{(t-s)B}g(s)ds\nonumber\\
&=&e^{(t-t_2)B}y(t_2)-\int^{t_2}_t
e^{(t-s)B}g(s)ds.\label{*0}
\end{eqnarray} Thus,
$z(t)=x(t;\xi)+\varphi (x(t;\xi))$ is
the unique solution to
\sys in the sense
\eqref{1.5.1}-\eqref{1.5.2} on
$[0,t_2]$ with the required
properties.

Conversely, assume that the graph
${\cal M}_\varphi$ of a $\varphi
\in \Lip_1^0(X,Y)$ is forward
invariant for \sys. This
means that for each $\xi \in X$
and $t_2>0$ there exists a unique
solution $z$ to
\eqref{1.2}-\eqref{1.3} on
$[0,t_2]$ such that $z(0)=\xi
+\varphi (\xi )$ and $z(t)\in
{\cal M}_\varphi$ for each $t\in
[0,t_2]$. For $z(t)=x(t)+y(t)$ equation
\eqref{1.5.1} yields
$$x(t)=e^{tA}\xi
+\int^t_0e^{(t-s)A}F(x(s)+\varphi
(x(s)))ds,\quad t>0,$$ that is,
$x(t)=x(t;\xi )$, $t\geq 0$. Also,
$y(t)=\varphi (x(t;\xi))$ is the
unique solution to
$$y(t)=e^{(t-t_2)}y(t_2)-\int^{t_2}_t
e^{(t-s)B}G(x(s;\xi)+\varphi (x(s,\xi
)))ds,\quad t\in [0,t_2].$$ But, as we
have seen in
\eqref{*0}, the function
$$\tilde{y}(t):=({\cal
L}\varphi)(x(t;\xi
))=-\int^\infty_te^{(t-s)B}g(s)ds,
\quad t\geq 0,$$ satisfies the same
equation. Thus, $\varphi={\cal
L}\varphi$.\quad $\Box$


Fix $\sigma \in {\Bbb R}$ such that
\begin{equation} a+\Lip (F)<\sigma
<b-\Lip
(G).\label{5.1}\end{equation}
Define the Banach space $E_\sigma$ as follows:
\[E_\sigma =\{f\in C({\Bbb
R};Z)|\quad\|f\|_\sigma :=\sup_{t\geq
0}e^{-t\sigma} |f(t)|<\infty \}.\]
For each $\xi \in X$ define on
$E_\sigma $ the operator $T_\xi$ by the rule
$$(T_\xi f)(t)=e^{tA}\xi
+\int^t_0e^{(t-s)A}F(f(s))ds-
\int^\infty_te^{(t-s)B}G(f(s))ds,
\quad t\geq 0.$$
\begin{lem}\label{smallT} For each
$\xi \in X$ the operator $T_\xi$
maps $E_\sigma$ in
$E_\sigma$, is a strict
contraction, and
$$\| T_\xi f_1-T_\xi f_2\|_\sigma
\leq \max \left\{\frac{\Lip
(F)}{\sigma
-a},\frac{\Lip(G)}{b-\sigma }
\right\}\|f_1-f_2\|_\sigma.$$\end{lem}
\noindent{\bf Proof.}\quad  Using
$F(0)=G(0)=0$, one has:
$$|(T_\xi f)(t)| \leq \max \{
e^{ta}|\xi |+\int^t_0 e^{(t-s)a}
\Lip (F)|f (s)|ds,\quad
\int^\infty_t e^{(t-s)b} \Lip
(G)|f(s)|ds\}.$$ Thus,
\begin{eqnarray*}\| T_\xi
f\|_\sigma &\leq &\sup_{t\ge 0} \max \{ e^{t(a-\sigma)}|\xi
| + \Lip (F)\int^t_0
e^{(t-s)(a-\sigma)}\|f\|_\sigma
ds,\\ &\quad \qquad & \Lip
(G)\int^\infty_t
e^{(t-s)(b-\sigma)}\|f\|_\sigma
ds\}
\leq\\ &
\leq &\max \{ |\xi |+\frac{\Lip
(G)}{\sigma -a}\|f\|_\sigma,
\frac{\Lip
(G)}{b-\sigma}\|f\|_\sigma\}
<\infty ,
\end{eqnarray*} since $\sigma
-a>\Lip (F)\geq 0$ and $b-\sigma
>\Lip (G)\geq 0$ by the choice of
$\sigma $ in
\eqref{5.1}.

For any $f_{1,2}\in E_\sigma $ one
has:
\begin{eqnarray*}|T_\xi
f_1(t)-T_\xi f_2(t)|&\leq &\max
\{\int^t_0e^{(t-s)a}\Lip
(F)|f_1(s)-f_2(s)|ds,\\  &\quad &
\int^\infty_te^{(t-s)b}\Lip
(G)|f_1(s)-f_2(s)|ds\}.\end{eqnarray*}
Thus,
\begin{eqnarray*}\|T_\xi f_1-T_\xi
f_2\|_\sigma &\leq &\sup_{t\geq
0}\max \{ \Lip
(F)\int^t_0e^{(t-s)(a-\sigma)}ds,\\
&\quad & \Lip
(G)\int^\infty_te^{(t-s)(b-\sigma)}
ds\}\|f_1-f_2\|_
\sigma
\leq\\ &\leq &\max \left\{
\frac{\Lip (F)}{\sigma
-a},\frac{\Lip
(G)}{b-\sigma}\right\} \|
f_1-f_2\|_\sigma
.\quad\Box\end{eqnarray*}
Therefore, for each
$\xi \in X$ there exists a unique
$f_\xi
\in E_\sigma$ such that $f_\xi
=T_\xi f_\xi$.

We also remark that the operator
$$(Tf)(t)=\int^t_0e^{(t-s)A}F(f(s))
ds-\int^\infty_te^{(t-s)B}G(f(s))
ds,\quad t\geq 0$$ is a Lipschitz map
on
$E_\sigma$ with
\begin{equation}\Lip (T)\leq \max
\left\{ \frac{\Lip (F)}{\sigma
-a},\frac{\Lip (G)}{b-\sigma
}\right\}<1.\label{7.1}\end{equation}
\begin{rem}\label{VanAppr} Define
$S:X\to E_\sigma $ such that
$S(\xi )(t):=e^{tA}\xi$. Then
$T_\xi f=S(\xi )+Tf$ for each
$f\in E_\sigma$. By  a general
result on Lipschitz maps (see,
e.g. \cite{Shub}, Appendix I,
Ch.5, p. 49)
$I-T:E_\sigma \to E_\sigma $ is a
Lipschitz homeomorphism with $\Lip
((I-T)^{-1})=[1-\Lip (T)]^{-1}$.
Equation
$f_\xi =T_\xi f_\xi $ has the form
$f_\xi =S(\xi)+Tf_\xi$. Thus,
$f_\xi =(I-T)^{-1}S(\xi)$.
\end{rem}
Define $\varphi :X\to Y$ as
follows:
\begin{equation}\varphi (\xi
):=\Pr_Yf_\xi
(0)=-\int^\infty_0e^{-sb}G(f_\xi
(s))ds\label{7.2}
\end{equation}
\begin{prop}\label{Bigsmall}
$\varphi={\cal L}\varphi$ for the
Lyapunov-Perron operator
\eqref{LP}.\end{prop}

\noindent{\bf Proof.} For $f_\xi
=T_\xi f_\xi$ let us denote:
\begin{eqnarray*}x(t)&:=&\Pr_Xf_\xi
(t)=e^{tA}\xi
+\int^t_0e^{(t-s)A}F(f_\xi
(s))ds,\\
y(t)&:=&\Pr_Yf_\xi(t)=-\int^\infty_t
e^{(t-s)B}G(f_\xi
(s))ds.\end{eqnarray*}
Since $f_\xi =T_\xi f_\xi$ for each $\xi
\in X$, for any
$t\geq 0$ and $s\geq 0$ one has:
\begin{equation}\label{3*}
f_{x(t)}(s)=e^{sA}x(t)+\int^x_0
e^{(s-\tau )A}F(f_{x(t)}(\tau ))d\tau
-\int ^\infty_s e^{(s-\tau
)B}G(f_{x(t)}(\tau ))d\tau.
\end{equation}
Also, one has:
\begin{eqnarray*}f_\xi
(s+t)&=&e^{(s+t)A}\xi
+\int^{s+t}_0e^{(s+t-\tau)A} F(f_\xi
(\tau ))d\tau
-\int^\infty_{s+t}e^{(s+t-\tau)B}
G(f_\xi (\tau ))d\tau\\
&=&e^{sA}\left[e^{tA}\xi
+\int^t_0e^{(t-\tau)A}F(f_\xi
(\tau
))d\tau\right]+\int^{t+s}_te^{(s+
t-\tau)A}F(f_\xi (\tau))d\tau\\ &\quad
& -\int^\infty
_{s+t}e^{(s+t-\tau)B}G(f_\xi (\tau
))d\tau =\\
&=&e^{sA}x(t)+\int^s_0e^{(s-\tau)A}
F(f_\xi (\tau+t))d\tau -\int^\infty_se^{(s-\tau)B}G(f_\xi
(\tau+t))d\tau\end{eqnarray*}
Since \eqref{3*} has the unique
solution $f_{x(t)}(\cdot)$, we
conclude
$f_\xi (s+t)=f_{x(t)}(s)$,
$s,t\geq 0$. Using our definition
\eqref{7.2} we obtain:
\begin{eqnarray*}\varphi
(x(t))&=&-\int^\infty_0e^{-sB}G(
f_{x(t)}(s))ds=-\int^\infty_0e^{-sB}
G(f_\xi (s+t))ds\\
&=&-\int^\infty_te^{(t-s)B}G(f_\xi
(s))ds=y(t).\end{eqnarray*} Therefore,
$x(t)=x(t;\xi )$ satisfies
\begin{equation}x(t)=e^{tA}\xi
+\int^t_0e^{(t-s)a}f(x(s)+\varphi
(x(s)))ds,\label{**}\end{equation}
and
\begin{equation*}\varphi (\xi
)=-\int^\infty_0e^{-sB}G(x(s;\xi
)+\varphi (x(s; \xi )))ds=({\cal
L}\varphi
)(\xi)\quad\Box\end{equation*}

\begin{rem} If $\varphi ={\cal
L}\varphi$ then $g_\xi (t)=x(t;\xi
)+\varphi (x(t;\xi ))$, $t\geq 0$
satisfies $T_\xi g_\xi =g_\xi$.
Indeed, as in \eqref{4.1},
$$\varphi (x(t;\xi
))=-\int^\infty_te^{(t-s)B}G(x(s;\xi
)+\varphi (x(s;\xi )))ds.$$ By
adding this to
\eqref{**}, one gets
$T_\xi g_\xi =g_\xi$.\end{rem}

\begin{rem}\label{justLip} Proposition
\ref{Bigsmall} and Lemma~\ref{invM} already imply the
existence of a Lipschitz $\varphi
:X\to Y$ with the invariant graph
${\cal M}_\varphi$. To see that
$\varphi$, as defined in
\eqref{7.2}, is a Lipschitz map, we recall
that also $\varphi (\xi
)=(I-T)^{-1}\circ S(\xi )(0)$.
Since
$(I-T)^{-1}:E_\sigma \to E_\sigma
$ is  Lipschitz by Remark
\ref{VanAppr}, we have from
\eqref{7.1}:
 \begin{eqnarray*}|\varphi (\xi
_1)-\varphi (\xi _2)|&=&
|(I-T)^{-1}\circ S(\xi
_1)(0)-(I-T)^{-1}\circ S(\xi
_2)(0)|\leq\\
&\quad & \leq \|
(I-T)^{-1}\circ S(\xi
_1)-(I-T)^{-1}\circ S(\xi
_2)\|_\sigma \\
& \quad & \leq \Lip
((I-T)^{-1})\| S(\xi _1)-S(\xi
_2)\|_\sigma\\
&=& [1-\max \{ \Lip
(F )/(\sigma -a),\Lip (G)/(b-\sigma
)\}]^{-1}\|S(\xi _1)-S(\xi
_2)\|_\sigma .\end{eqnarray*}
Since $$\|S(\xi _1)-S(\xi
_2)\|_\sigma =\sup_{t\geq
0}|e^{-\sigma t}e^{tA}(\xi
_1-\xi_2)|\leq |\xi_1-\xi_2|,$$
we conclude that $\varphi $ is Lipschitz.
\end{rem}

However, to prove the sharp estimate \eqref{estlipphi}
for $\Lip (\varphi )$,  we will need a lemma. For fixed
$\xi_i\in X$, $\xi_1\neq \xi_2$ let
$z_i(t)=f_{\xi_i}(t)$, and set
$z_i(t)=x_i(t)+y_i(t)$, $i=1,2$,
$t\geq 0$. Recall, that we define
$|z|_z=\max
\{|x|_x,|y|_z\}$,  $z=x+y$.

\begin{lem}\label{xz}
$|z_1(t)-z_2(t)|=|x_1(t)-x_2(t)|$ for all $t\geq 0$.\end{lem}
Let us first finish the proof of the
main Theorem \ref{main}, using the
lemma. Since $z_i=T_{\xi_i}z_i$,
the functions $z_i(\cdot )$ are
solutions to \sys in the
sense \eqref{1.5.1}-\eqref{1.5.2}.
In particular,
$x_i(t)=\Pr_Xz_i(t)$ satisfy
$$x_i(t)=e^{tA}\xi_i+\int^t_0
e^{(t-s)A}F(z_i(s))ds,\quad i=1,2.$$
Using Lemma \ref{xz}, we obtain:
\begin{eqnarray*}|x_1(t)-x_2(t)|&\leq
& e^{ta}|\xi _1-\xi
_2|+\int^t_0e^{(t-s)a}\Lip
(F)|z_1(s)-z_2(s)|ds=\\
&=&e^{ta}|\xi_1-\xi_2|+\Lip
(F)\int^t_0e^{(t-s)a}|x_1(s)-x_2(s)
|ds,\end{eqnarray*} and
\begin{equation}\label{(1*)}|x_1(t)-
x_2(t)|\leq
e^{t(a+\Lip(F))}|\xi_1-\xi_2|,\quad
t\geq 0\end{equation} by Gronwall's
inequality.

Since $\varphi
(\xi_i)=y_i(0)=\Pr_Yf_{\xi_i}(0)$
by the definition \eqref{7.2}, we
have from Lemma \ref{xz} and
\eqref{(1*)}:
\begin{eqnarray*}|\varphi
(\xi_1)-\varphi
(\xi_2)|&=&|\Pr_Yf_{\xi_1}(0)-\Pr_Y
f_{\xi_2}(0)|\leq\\ &\leq &
\int^\infty_0e^{-sb}\Lip (G)
|z_1(s)-z_2(s)|ds=\\ &=&\Lip (G)
\int^\infty_0e^{-sb}|x_1(s)-x_2(s)
|ds\leq \Lip (G)\int^\infty
_0e^{(a-b+\Lip (F))s}|\xi_1-\xi_2|ds\\
&=&\frac{\Lip (G)}{b-a-\Lip (F)}
|\xi_1-\xi_2|.\end{eqnarray*} This
proves Theorem \ref{main}.\quad
$\Box$

\noindent{\bf Proof of Lemma
\ref{xz}.} We want to show that
\begin{equation}\label{11.1}|x_1(t)-
x_2(t)|\geq |y_1(t)-y_2(t)| \mbox{ for
all } t\geq 0.\end{equation}  Let us
suppose that for some $t_1>0$ one
has
\begin{equation}\label{11.2}
|y_1(t_1)-y_2(t_1)|>|x_1(t_1)-x_2
(t_1)|.\end{equation}
\begin{claim} Inequality
\eqref{11.2} implies
\begin{equation}\label{11.3}|y_1(t)-
y_2(t)|>|x_1(t)-x_2(t)|\mbox{ for all
} t\geq t_1.\end{equation}\end{claim}
To prove the claim, we suppose
there is a point $t_2>t$, such that
\begin{eqnarray}|x_1(t_2)-x_2(t_2)|
&=&|y_1(t_2)-y_2(t_2)|\quad \mbox{ but}\nonumber \\
|y_1(t)-y_2(t)|&>&|x_1(t)-x_2(t)\quad
\mbox{ for all }\quad t\in
[t_1,t_2).\label{11.4}\end{eqnarray}
Recall, that
$z_i(t)=x_i(t)+y_i(t)$, $i=1,2$
are solutions to
\eqref{1.5.1}-\eqref{1.5.2}, that
is, for $t\in (t_1,t_2)$ and
$i=1,2$ one has:
\begin{eqnarray*}x_i(t)&=&
e^{(t-t_1)A}x_i(t_1)+\int^t_{t_1}
e^{(t-s)A}F(z_i(s))ds,\\
y_i(t)&=&e^{(t-t_2)B}y_i(t_2)-
\int^{t_2}_te^{(t-s)B}G(z_i(s))
ds.\end{eqnarray*}
Let us denote, for
brevity:
\begin{eqnarray*}
u(t)&=&x_1(t)-x_2(t),\quad
v(t)=y_1(t)-y_2(t),\\ f(t)
&=&F(z_1(t))-F(z_2(t)),\quad
g(t)=G(z_1(t))-G(z_2(t)),\quad
t\in [t_1,t_2].\end{eqnarray*} Then
\begin{eqnarray}u(t)&=&e^{(t-t_1)A}
u(t_1)+\int^t_{t_1}e^{(t-s)A}f(s)ds,
\label{12.1}\\
v(t)&=&e^{(t-t_2)B}v(t_2)-
\int^{t_2}_te^{(t-s)}g(s)ds.
\label{12.2}\end{eqnarray} Since
$|z_1(t)-z_2(t)|=|u(t)|$ for $t\in
[t_1,t_2]$ by \eqref{11.4}, we
conclude:
$$|f(t)|\leq \Lip (F) |v(t)|,\quad
|g(t)|\leq \Lip (G)|v(t)|,\quad
t\in [t_1,t_2].$$ Now \eqref{12.2}
implies

$$|v(t)|\leq
e^{(t-t_2)b}|v(t_2)|+\int^{t_2}_t
e^{(t-s)b}\Lip (G)|v(s)|ds,$$ and, by
the Gronwall's inequality,
\begin{equation} |v(t)|\leq
|v(t_2)|e^{(\Lip
(G)-b)(t_2-t)},\quad t\in
[t_1,t_2]\label{12.4}\end{equation}
We use the last inequality to have
the following estimate in
\eqref{12.1} for $t=t_2$:

\begin{eqnarray}|u(t_2)|&\leq &
e^{(t_2-t_1)a}|u(t_1)|+\Lip
(F)|v(t_2)|\int^{t_2}_{t_1}
e^{(t_2-s)a}e^{(\Lip
(G)-b)(t_2-s)}ds\nonumber\\
&=&e^{(t_2-t_1)a}|u(t_1)|+ \frac{\Lip
(F)}{b-a-\Lip (G)}|v(t_2)|[1-
e^{(b-a-\Lip
(G))(t_1-t_2)}].
\label{12.3}\end{eqnarray}  However,
by the first assumption in
\eqref{11.4} one has
$|v(t_2)|=|u(t_2)|$. Denote, for
brevity, $\delta =b-a-\Lip (G)$,
$\ell =\Lip (F)/\delta$. Solve
\eqref{12.3} for $|v(t_2)|$ to get:

$$|v(t_2)|\leq
e^{(t_2-t_1)a}|u(t_1)|\{1-\ell
(1-e^{\delta (t_1-t_2)})\}^{-1}.$$
Use again \eqref{12.4}, this time
for $t=t_1$ to get:

$$|v(t_1)|\leq
|u(t_1)|\left\{\frac{e^{-\delta
(t_2-t_1)}}{1-\ell (1-e^{-\delta
(t_2-t_1)})}\right\}<|u(t_1)|.$$
Indeed, the expression in \{\quad \}
is strictly less then 1 due to the
gap condition \eqref{1.5}. But the
inequality
$|v(t_1)|<|u(t_1)|$ contradicts
\eqref{11.2}, and the claim is
proved.

To finish the proof of the lemma,
we stress that \eqref{11.3}
implies $|z_1(t)-z_2(t)|=|v(t)|$
for all $t\in [t_1,\infty )$.
Therefore, using \eqref{12.4} with
$t=t_1$, we conclude that for an
arbitrarily large
$t_2\geq t_1$ one has:
\begin{equation}\label{13.1}
|v(t_1)|\leq
|v(t_2)|e^{(\Lip(G)-b)
(t_2-t_1)}.\end{equation} Recall, that
$z_1-z_2=f_{\xi_1}-f_{\xi_2}\in
E_\sigma$ with $\sigma<b-\Lip (G)$
by \eqref{5.1}. Thus, for $t\geq
t_1$, by the definition of
$\|\cdot \|_\sigma$,
$$|v(t)|=|z_1(t)-z_2(t)|\leq
\|f_{\xi_1}-f_{\xi_2}\|_\sigma
e^{\sigma t}=ce^{\sigma t}.$$ In
particular, using this inequality
for $t=t_2$ in \eqref{13.1}, one
has:
$$|f_{\xi_1}(t_1)-f_{\xi_2}(t_1)|=
|v(t_1)|\leq ce^{(b-\Lip (G))t_1}\cdot
e^{(\sigma +\Lip (G)-b)t_2}.$$

Letting $t_2\to +\infty$ we find
$f_{\xi _1}(t_1)=f_{\xi_2}(t_1)$,
since $\sigma +\Lip (G)-b<0$ by
\eqref{5.1}. Equality
$f_{\xi_1}(t_1)=f_{\xi_2}(t_1)$
implies, in particular, that
$y_1(t_1)=y_2(t_1)$ which
contradicts the strict inequality
in
\eqref{11.2}.\quad $\Box$
\begin{rem}\label{groupA} Assume,
in addition to
\eqref{1.4}-\eqref{1.5}, that $A$
generates a strongly continuous
{\it group} on $X$. We claim, that
$\varphi :X\to Y$ obtained in
Theorem \ref{main} is such that
${\cal M}_\varphi$ is also
backward invariant. To see this,
take $\varphi$ from Theorem
\ref{main} and notice that
solution $x(t;\xi)$ to the equation
\begin{equation}x(t)=e^{tA}\xi+
\int^t_0e^{(t-s)A}F(x(s)+\varphi
(x(s)))ds,\quad t\in {\Bbb R}
\label{15a.1}\end{equation} is
defined for $t\in {\Bbb R}$ provided
$A$ is a group generator. By Proposition
\ref{Bigsmall}, $\varphi ={\cal
L}\varphi$ for the Lyapunov-Perron
operator
$$({\cal L}\varphi )(\xi
)=-\int^\infty_0e^{-sB}G(x(s;\xi
)+\varphi (x(s;\xi ))ds,\quad \xi
\in X.$$ Since
$\xi \mapsto x(t;
\xi )$, $t\in {\Bbb R}$ is a flow,
equation \eqref{4.1} holds now for
all $t\in {\Bbb R}$, that is,
$$({\cal L}\varphi )(x(t;\xi
))=-\int^\infty_te^{(t-s)B}G(x(s;\xi
)+\varphi (x(s;\xi )))ds,\quad
t\in {\Bbb R}.$$ Also,
$g(s)=G(x(s,\xi )+\varphi (x(s;\xi
))$ is defined for all $s\in {\Bbb
R}$. Let $y(t)=\varphi (x(t;\xi
))$,
$t\in {\Bbb R}$. Then $z(t)=x(t;
\xi )+\varphi (x(t;\xi ))\in {\cal
M}_\varphi $ for all $t\in {\Bbb
R}$. But $\varphi ={\cal L}\varphi
$ implies that for any $t<0$
\begin{eqnarray*}y
(t)&=&-\int^\infty_te^{(t-s)B}g(s)
ds=-e^{tB}\int^\infty_0e^{-sB}g(s)
ds-\int^0_te^{(t-s)B}g(s)ds\\
&=&e^{tB}y(0)-\int^0_te^{(t-s)B}
g(s)ds.\end{eqnarray*} Fix any
$t_1<0$. Using
\eqref{15a.1} for $t=t_1$ on
checks that for
$t\in [t_1,0]$
$$e^{(t-t_1)A}x(t_1)+\int^t_{t_1}
e^{(t-s)A}F(z(s))ds=e^{tA}\xi
+\int^t_0e^{(t-s)A}F(z(s))ds=x(t).$$
Therefore, $z(t)=x(t)+y(t)$ is the
unique solution on $[t_1,0]$ to
\sys in the sense of
\eqref{1.5.1}-\eqref{1.5.2}, and
${\cal M}_\varphi$ is backward invariant.

For a $\psi \in \Lip ^0_1(Y,X)$ let
${\cal M}_\psi =\{\psi (y)+y|y\in
Y\}$. The set ${\cal M}_\psi$ for a
$\psi :Y\to X$ is called backward
invariant for \sys if for
each $\eta \in Y$ and $t_1<0$
there exists a solution $z(\cdot
)$ of
\sys on $[t_1,0]$ in the
sense of
\eqref{1.5.1}-\eqref{1.5.2} such
that $z(0)=\psi (\eta )+\eta$ and
$z(t)\in {\cal M}_\psi $ for
$t\leq 0$.
\end{rem}



\noindent{\bf Proof of Corollary~\ref{back}.} For any
$t_1<t_2$ a function
$z:[t_1,t_2]\to Z:t\mapsto
x(t)+y(t)$ is a solution to
\eqref{1.5.1}-\eqref{1.5.2} if and
only if, for $\tau_1=-t_2$, $\tau
_2=-t_1$, the function
$$\tilde{z}:[\tau_1,\tau_2]\to
Z:\tau \mapsto \tilde{x}(\tau
)+\tilde{y}(\tau ):=x(-\tau
)+y(-\tau )$$ is the solution to
the system
\begin{eqnarray} \tilde{x}(\tau
)&=& e^{(\tau -\tau
_2)(-A)}\tilde{x}(\tau
_2)-\int^{\tau _2}_\tau e^{(\tau
-s)(-A)}(-F)(\tilde{x}(s)+
{\tilde{y}}(s))ds,\label{*2}\\
\tilde{y}(\tau )&=&e^{(\tau
-\tau_1)(-B)}\tilde{x}(\tau
_1)+\int^\tau _{\tau _1}e^{(\tau
-s)(-B)}(-
G)(\tilde{x}(s)+\tilde{y}(s)ds.
\nonumber\end{eqnarray} Here
$(-F)(z)=-F(z)$,
$(-G)(z)=-G(z)$. A set
${\cal M}_\psi$ is backward
invariant for
\sys if
and only if ${\cal M}_\psi$ is
forward-invariant for \eqref{*2}.
Note, that \eqref{*2} is a system
of type
\eqref{1.5.1}-\eqref{1.5.2}, but
with $X$ in
\eqref{1.5.1}-\eqref{1.5.2}
replaced by $Y$ in \eqref{*2}, $A$
replaced by $-B$,
$B$ replaced by $-A$, $F$ replaced
by $-G$, $G$ replaced by $-F$.
Thus, Theorem \ref{main} gives the
result.\quad $\Box$

\noindent{\bf Proof of Corollary~\ref{renorm}.} Let
$$|x|_{M_A}=\sup_{t\geq
0}e^{-ta}|e^{tA}x|_X,\quad
|y|_{M_B}=\sup_{t\leq
0}e^{-tb}|e^{tB}y|_Y$$ be the
equivalent norms on $X$ and $Y$:
$$|x|_X\leq |x|_{M_A}\leq
M_A|x|_X,\quad |y|_Y\leq
|y|_{M_B}\leq M_B|y|_Y.$$ Then,
for the Lipschitz constants $\Lip'$
in the new norms,
$$\Lip '(F)+\Lip'(G)\leq M_A \Lip
(F)+M_B \Lip (G) <b-a$$ implies the
existence of $\varphi :X\to Y$
such that
$$\frac{1}{M_A}\Lip (\varphi )\leq
\Lip '(\varphi )\leq \frac{\Lip
'(G)}{b-a-\Lip '(F)}\leq \frac{M_B
\Lip (G)}{b-a-M_A \Lip (F)}<1.\quad
\Box$$

\section{Proofs: several spectral bands}

 \noindent In this section we prove Theorem~\ref{manygaps}. Note, that the
 existence of $\varphi_i$ and estimates for
 $\Lip(\varphi_i)$ for $i=1$ and $i=N$ is an immediate
 consequence of Theorem~\ref{main} and Corollary~\ref{back}. These
 assertions also give the existence of $\varphi_i$ for $i=2,\ldots,N-1$.
 However, to prove the estimates \eqref{bigestphilip} we need to consider
 separately the case of $N=3$ spectral bands.
 This is done in Theorem~\ref{central}.

\noindent{\bf Proof of Theorem~\ref{manygaps}.} For $i=1$ let
$Y=X_1$,
$X=\bigoplus^N_{i=z}X_i$,
$B=A_1$, $A=\mbox{diag}
(A_2,\ldots , A_N)$,
$F(z)=\sum^N_{i=2}F_i(z)$,
$G(z)=F_1(z)$, $b=b_1$, $a=a_2$.
Note that $\Lip (F)=\max_{j>1} \Lip
(F_j)$, $\Lip (G)=\Lip (F_1)$ and
the gap condition \eqref{1.5} of
Corollary~\ref{back} gives the
existence of the desired
$\varphi _1=\psi :Y\to X$.
Similarly, for $i=N$ Theorem
\ref{main} gives $\varphi
_N=\varphi :X\to Y$ for
$X=X_N$, $Y=X_1+\ldots +X_{N-1}$.

Fix $i=2,\ldots , N-1$. We will show here only
the existence of
a map $\varphi_i:X_i\to\hat{X}_i$ with the invariant graph.
The estimates on the $\Lip(\varphi_i)$ as in \eqref{bigestphilip}
will require more work; they will follow from Theorem \ref{central}
below.

Define subspaces $Y_-=X_1+\ldots +X_{i-1}$,
$Y_+=X_{i+1}+\ldots +X_N$,
$Y=Y_++Y_-$ and operators $B_-=\mbox{diag}
(A_j)_{j=1}^{i-1}$,
$B_+=\mbox{diag} (A_j)^N_{j=i+1}$.
First, we apply Theorem \ref{main}
for $X=Y_++X_i$, $Y=Y_-$ and
$A=B_++A_i$,
$B=B_-$ to get
$\varphi :Y_++X_i\to Y_-$ with
(forward) invariant graph ${\cal
M}_\varphi =\{ y_++x_i+\varphi
(y_++x_i)|y_++x_i\in Y_++X_i\}$.
Next, we apply Corollary~\ref{back}
for $X=Y_+$, $Y=X_i+Y_-$ and $A=B_+$,
$B=A_i+B_-$ to get $\psi
:X_i+Y_-\to Y_+$ with (backward)
invariant graph ${\cal M}_\psi
=\{\psi
(x_i+y_-)+x_i+y_-|x_i+y_-\in
X_i+Y_-\}$. Note, that $M={\cal
M}_\varphi
\cap {\cal M}_\psi$ is invariant
in both directions, and
$$M=\{y_++x_i+y_-|x_i\in
X_i,y_+=\psi (x_i+\varphi
(y_++x_i)),y_-=\varphi (\psi
(x_i+y_-)+x_i\}.$$ Fix
$x_i\in X_i$. Define a map
$\rho_{x_i}:Y_++Y_-\to Y_++Y_-$ as
follows:
$$\rho_{x_i}:y_++y_-\mapsto \psi
(x_i+\varphi (y_++x_i))+\varphi
(\psi (x_i+y_-)+x_i).$$ Since $\Lip
(\psi )<1$ and $\Lip (\varphi) <1$, one
concludes $\sup_{x_i}\Lip (\rho
_{x_i})<1$. Thus, there exists a
unique $y(x_i)$ such that $\rho
_{x_i}(y(x_i))=y(x_i)$. Define:
$$\varphi _i(x_i):=y(x_i)=(I-\rho
_{x_i})^{-1}(0).$$ Clearly,
$\varphi _i$ is Lipschitz,
$M=\mbox{graph }\varphi _i$.\quad $\Box$

The main step in the proof of estimates
\eqref{bigestphilip} is to consider a system of type  \sys,
but with three equations. Consider \sys, and  assume that the space
$Y=Y_++Y_-$, the operator
$B=\mbox{diag}(B_+,B_-)$, and the nonlinearity
$G(z)=G_+(z)+G_-(z)$, where $G_\pm
:Z\to Y_\pm$, and $B_\pm$ are
operators on $Y_\pm$,
respectively. In fact, system \sys
takes now the following form:
\begin{eqnarray}\dot{y}_+&=&
B_+y_++G_+(y_++x+y_-),\nonumber\\
\dot{x} &=& Ax
+F(y_++x+y_-),\label{3S}\\
\dot{y}_-&=&B_-y_-
+G_-(y_++x+y_-).\nonumber
\end{eqnarray}

We assume that $A$
generates a strongly continuous
group on $X$, and $B_+$ and $-B_-$
generates a strongly continuous
semigroup on $Y_+$ and $Y_-$,
respectively. Also, for constants $\alpha< b\le a <\beta$
we assume that the following
estimates hold:
\begin{eqnarray}\|e^{tA}\|&\leq
&e^{ta},\quad \|e^{tB_+}\|\leq
e^{t\alpha},\quad t\geq
0\nonumber\\
\|e^{tA}\| &\leq & e^{tb},\quad
\|e^{tB_-}\|\leq e^{t\beta },\quad
t\leq 0.\label{19.1}\end{eqnarray}
Solutions to \eqref{3S} are
understood in a mild sense, as in
\eqref{1.5.1}-\eqref{1.5.2},  if
applied for $A$ replaced by
$B_++A$ and $B$ replaced by $B_-$.

Assume gap conditions:
\begin{equation}\Lip (G_+)+\Lip
(F)<b-\alpha ,\quad \Lip (F)+\Lip
(G_-)<\beta
-a\label{19.2}\end{equation}
The following theorem gives the existence of a ``central-like'' manifold,
tangent to $X$.
\begin{thm}\label{central} There
exists $\Phi :X\to Y$, $\Phi\in\Lip_1^0(X,Y)$ such
that $M_\Phi =\{x+\Phi (x)|x\in
X\}$ is invariant (backward and
forward) for \sys, and
$$\Lip (\Phi )\leq
\max\left\{\frac{\Lip
(G_+)}{b-\alpha -\Lip (F)},\quad
\frac{\Lip (G_-)}{\beta -a-\Lip
(F)}\right\}.$$\end{thm}

\noindent{\bf Proof.}\quad We
stress, that for a fixed $\Phi
:X\to Y$, $\Phi \in \Lip
^1_0(X, Y)$, the equation
\begin{equation*} x(t)=e^{tA}\xi
+\int^t_0e^{(t-s)A}f(x(s)+\Phi
(x(s)))ds,\quad t\in {\Bbb
R}\end{equation*} has a unique
solution $x(t;\xi)$ for all $t\in
{\Bbb R}$, since $A$ is a group generator.
Similarly to Propositions
\ref{growthx} and
\ref{Lipx} one has:
\begin{prop}\label{centralx} For
$\Phi \in \Lip^1_0(X,Y)$,
\begin{eqnarray*}|x(t;\xi )|&\leq
& e^{(a+\Lip (F))t},\quad t\geq
0,\\ |x(t;\xi )|&\leq & e^{(b-\Lip
(F))t},\quad t\leq 0,\\ |x(t;\xi
_1)-x(t;\xi _2)| &\leq & e^{(a+\Lip
(F))t}|\xi _1-\xi_2|,\quad t\geq
0,\\ |x(t;\xi _1)-x(t;\xi
_2)|&\leq & e^{(b-\Lip
(F))t}|\xi_1-\xi_2|,\quad t\leq
0.\end{eqnarray*}\end{prop}

\noindent{\bf Proof.}\quad
Gronwall's inequality.\quad $\Box$

The Lyapunov-Perron operator
${\cal L}$ on functions $\Phi :X\to Y$ is
defined as
\begin{eqnarray*}({\cal L}\Phi
)(\xi
)&=&\int^0_{-\infty}e^{-sB_+}G_+
(x(s;\xi)+\Phi (x(s,\xi )))ds-\\
&\quad &
-\int^\infty_0e^{-sB_-}G_-(x(s;\xi
)+\Phi (x(s;\xi )))ds.\end{eqnarray*}
As in Lemma
\ref{invM},
$\Phi ={\cal L}(\Phi )$ if and
only if $M_\Phi $ is invariant.
\begin{prop}\label{centrpves}
${\cal L}$ preserves $\Lip
^0_1(X,Y)$, that is, if $\Lip(\Phi)\le 1$ then
$$\Lip ({\cal L}\Phi )\leq
\max\left\{\frac{\Lip
(G_+)}{b-\alpha -\Lip
(F)},\frac{\Lip (G_-)}{\beta -a-\Lip
(F)}\right\}\le 1.$$
\end{prop}

\noindent{\bf Proof.}\quad Since
$\Lip (\Phi )\leq 1$ and $|z|=\max
\{|x|,|y_+|,|y_-|\}$, one has
$|{\cal
L}\Phi (\xi _1)-{\cal L}\Phi (\xi
_2)|\leq \max (I_+,I_-)$, where,
using Proposition \ref{centralx},
\begin{eqnarray*}I_+&=&\int^0
_{-\infty}e^{-\alpha s}\Lip (G_+)\cdot
e^{(b-\Lip (F))s}|\xi _1-\xi
_2|ds=\frac{\Lip (G_+)}{b-\alpha -\Lip
(F)},\\ I_-&=&\int^\infty_0e^{-s\beta
}\Lip (G_-)\cdot e^{(a+\Lip (F))s}|\xi
_1-\xi _2|ds=\frac{\Lip
(G_-)}{\beta -a-\Lip (F)}.\quad
\Box\end{eqnarray*}

 Select
$\sigma_+>\sigma_-$ as follows:
\begin{eqnarray}a+\Lip (F)
&<&\sigma_+<\beta -\Lip
(G_-),\nonumber\\
\alpha +\Lip (G_+)&<&\sigma
_-<b-\Lip
(F).\label{21.1}\end{eqnarray}
Define Banach spaces
$$E_{\sigma_\pm}=\{f\in C({\Bbb
R};Z)|\quad \| f\|_{\sigma_\pm}:=\max
\{\sup_{t\leq
0}e^{-t\sigma_-}|f(t)|,\quad
\sup_{t\geq
0}e^{-t\sigma_+}|f(t)|\}<\infty
\},$$ and operator
\begin{eqnarray*}(Tf)(t)&=&\int^t_0
e^{(t-s)a}F(f(s))ds+\int^t_{-\infty}
e^{(t-s)B_+}G_+(f(s))ds-\\ &\quad &
-\int^\infty_te^{(t-s)B_-}G_-(f(s))
ds,\quad t\in {\Bbb R}.\end{eqnarray*}
Also, for any $\xi \in X$, let
$(T_\xi f)(t)=e^{tA}\xi +(Tf)(t)$,
$t\in {\Bbb R}$.

\begin{prop} \label{centralsmallT}
$T$ and $T_\xi $ map
$E_{\sigma_\pm}$ on
$E_{\sigma_\pm}$ and are strict
contractions:
\begin{eqnarray*}\|Tf_1-Tf_2
\|_{\sigma_\pm}&\leq & \max
\left\{\frac{\Lip
(F)}{\sigma_+-a},\frac{\Lip
(F)}{b-\sigma_-},\frac{\Lip
(G_+)}{\sigma _- -\alpha},\frac{\Lip
(G_-)}{\beta -\sigma_+}\right\}
\|f_1-f_2\|_{\sigma
_\pm}.\end{eqnarray*}\end{prop}

\noindent{\bf Proof.}\quad For
$t\geq 0$
$$e^{-\sigma_+t}|Tf_1(t)-Tf_2(t)|\leq
\max \{I_1,I_2,I_3\}\|
f_1-f_2\|_{\sigma_\pm},$$ where
\begin{eqnarray*}
I_1&=&\int^t_0e^{-t\sigma_+}\|
e^{(t-s)A}\|\Lip (F)e^{s\sigma_+}ds,\\
I_2&=&\int^0_{-\infty}
e^{-t\sigma_+}\|e^{(t-s)B_+}\|\Lip
(G_+)e^{s\sigma_-}ds+\int
^t_0e^{-t\sigma_+}\|e^{(t-s)B_+}\|\Lip
(G_+)e^{s\sigma _+}ds,\\
I_3&=&\int^\infty_te^{-t\sigma_+}\|
e^{(t-s)B_-}\|\Lip (G_-)e^{s\sigma
_+}ds.\end{eqnarray*} Using
estimates \eqref{19.1} and the
choice of $\sigma_\pm$ in
\eqref{21.1}, one has:
\begin{eqnarray*}I_1&\leq &
\frac{\Lip
(F)}{\sigma_+-a}(1-
e^{-t(\sigma_+-a)})\leq
\frac{ \Lip (F)}{\sigma_+-a};\\
I_2&\leq &
\frac{e^{-t(\sigma _+-\alpha)}\Lip
(G_+)}{\sigma_--\alpha}+\frac{(1-
e^{-t(\sigma _+-\alpha )})\Lip
(G_+)}{\sigma_+-\alpha}\leq
\frac{\Lip
(G_+)}{\sigma_--\alpha};\\
I_3&\leq & \frac{\Lip (G_-)}{\beta
-\sigma_+}.\end{eqnarray*}
Similarly, for $t=-\tau <0$,
$$e^{-\sigma
_-t}|Tf_1(t)-Tf_2(t)|\leq \max \{
J_1,J_2,J_3\}\|f_1-f_2
\|_{\sigma_\pm},$$ where
\begin{eqnarray*}J_1&=&\int^\tau
_0 e^{\tau
\sigma_-}\|e^{(-\tau+s)A}\|\Lip
(F)e^{-\sigma_-s}ds,\quad \tau
>0\\
J_2&=&\int^t_{-\infty}e^{-t\sigma_-}
\|e^{(t-s)B_+}\| \Lip
(G_+)e^{\sigma_-s}ds,\\ J_3&=&\int^0_t
e^{-t\sigma _-}\|e^{(t-s)B_-}\|\Lip
(G_-)e^{\sigma_-s}ds %+\\ &\quad &
+\int
^\infty_0e^{-t\sigma_-}\|e^{(t-s)B_-}
\|\Lip (G_-)e^{\sigma
_+s}ds,\end{eqnarray*} and
\begin{eqnarray*}J_1 &\leq &
\frac{\Lip (F)}{a-\sigma_-}\leq
\frac{\Lip (F)}{b-\sigma_-},\\ J_2
&\leq &\frac{\Lip (G_+)}{\sigma_-
-\alpha},\\ J_3 &\leq &\frac{\Lip
(G_-)(1-e^{t(\beta
-\sigma_-)})}{\beta
-\sigma_-}+\frac{e^{t(\beta
-\sigma_-)}\Lip (G_-)}{\beta
-\sigma_+}\leq \frac{\Lip
(G_-)}{\beta -\sigma_+}.\quad
\Box\end{eqnarray*} Thus, for each
$\xi \in X$ there exists a unique
$f_\xi \in E_{\sigma _\pm}$ such
that $T_\xi f_\xi =f_\xi$. Define:
\begin{equation}\label{24.1}
\Phi(\xi ):=\Pr_Yf_\xi
(0)=\int^0_{-\infty}e^{-sB_+}G_+(f_\xi
(s))ds-\int^\infty_0e^{-sB_-}G_-(f_\xi
(s))ds.\end{equation} As in Remark~\ref{justLip},
one already has  that
$\Phi $ is Lipschitz. We proceed
to estimate $\Lip (\Phi )$. Fix $\xi
_1\neq
\xi_2$ in $X$, let
$$z_i(t)=f_{\xi_i}(t),\quad
z_i(t)_=y^+_i(t)+x_i(t)+y^-_i(t),\quad
t\in {\Bbb R}, \quad i=1,2.$$
Note, that
$z_i$ are solutions to
\sys, and $z_1-z_2\in
E_{\sigma_\pm}$.

\begin{lem}\label{centralxz}
$|z_1(t)-z_2(t)|=|x_1(t)-x_2(t)|$ for all $t\in {\Bbb R}$.
\end{lem}

Let us first finish the proof of
Theorem \ref{central} using the
lemma.
Since
$$x_i(t)=e^{tA}\xi_i+\int^t_0
e^{(t-s)A}F(z_i(s))ds,\quad t\in
{\Bbb R},\quad i=1,2,$$ Lemma
\ref{centralxz} implies, for
$t\leq 0$:
$$|x_1(t)-x_2(t)|\leq
e^{ta}|\xi_1-\xi_2|+\Lip
(F)\int^t_0e^{(t-s)a}|x_1(s)-x_2(s)
|ds.$$ The Gronwall's inequality
implies, for
$t\geq 0$:
\begin{equation}\label{24.2}
|x_1(t)-x_2(t)|\leq e^{t(a+\Lip
(F))}|\xi _1-\xi _2|,\quad t\geq
0.\end{equation} Similarly, for
$t\leq 0$:
$$|x_1(t)-x_2(t)|\leq
e^{tb}|\xi_1-\xi_2|+\Lip
(F)\int^0_te^{(t-s)b}|x_1(s)-x_2(s)
|ds,$$ and
\begin{equation}|x_1(t)-x_2(t)|\leq
e^{t(b-\Lip
(F))}|\xi_1-\xi_2|,\quad t\leq
0.\label{24.3}\end{equation} We
use \eqref{24.1} and Lemma
\ref{centralxz} to estimate:
\begin{eqnarray*}|\Phi (\xi
_1)-\Phi (\xi_2)|&\leq & \max
\{\int^0_{-\infty}e^{-s\alpha} \Lip
(G_+)|z_1(s)-z_2(s)|ds,\\ &\quad &
\int^\infty_0e^{-s\beta}\Lip
(G_-)|z_1(s)-z_2(s)|ds\}=\\&=&\max
\{\Lip
(G_+)\int^0_{-\infty}e^{-s\alpha}|
x_1(s)-x_2(s)|ds,\\ &\quad & \Lip
(G_-)\int^\infty_0e^{-s\beta}|x_1(s)-
x_2(s)|ds\}.\end{eqnarray*} Now apply
\eqref{24.2}-\eqref{24.3} to get:
\begin{eqnarray*}\Lip (\Phi )&\leq
& \max \{ \Lip
(G_+)\int^0_{-\infty}e^{-s\alpha
}e^{s(b-\Lip (F))}ds,\\ &\quad \qquad &
\Lip (G_-)\int^\infty_0e^{-s\beta
}e^{s(a+\Lip (F))}ds\}=\\
&=&\max\left\{\frac{\Lip
(G_+)}{b-\alpha -\Lip
(F)},\frac{\Lip (G_-)}{\beta -a-\Lip
(F)}\right\}.\quad
\Box\end{eqnarray*}

\noindent{\bf Proof of Lemma
\ref{centralxz}.}
Some preparations are in order.
Denote $z(t)=z_1(t)-z_2(t)$,
$x(t)=x_1(t)-x_2(t)$, $y^\pm
(t)=y^\pm _1(t)-y^\pm _2(t)$. We
want to prove that
\begin{equation}\label{25.1}
|x(t)|\geq \max \{|y^+(t)|,
|y^-(t)|\},\quad t\in {\Bbb
R}.\end{equation}

\begin{prop}\label{contrprop} For
each $t_0\in {\Bbb R}$ {\bf neither}  one of the
following inequalities is possible:
\begin{enumerate}\item[(a)]
$\max\{|x(t_0)|,|y^-(t_0)|\}<|y^+
(t_0)|$;
\item[(b)] $\max
\{|x(t_0)|,|y^+(t_0)|\}<|y^-(t_0)|$;
\item[(c)]
$|x(t_0)|<|y^+(t_0)|=|y^-(t_0)|$.
\end{enumerate}\end{prop}

To derive \eqref{25.1} from the
proposition, let us suppose that,
say, $|x(t_0)|<|y^+(t_0)|$ for
some $t_0\in {\Bbb R}$. Clearly,
$|y^-(t_0)|\neq |y^+(t_0)|$,
otherwise we get (c). By
$|y^-(t_0)|>|y^+(t_0)|$ yields
(b), while $|y^-
(t_0)|<|y^+(t_0)|$ yields (a), a
contradiction that proves the
lemma.\quad $\Box$

\noindent{\bf Proof of Proposition~\ref{contrprop}.}
Let us suppose that (a) holds. We
claim, that (a) implies
\begin{equation}\label{26.1} \max
\{ |x(t)|,
|y^-(t)|\}<|y^+(t)|\mbox{ for all
}t\leq t_0.\end{equation} to prove
the claim
\eqref{26.1}, let us denote:
$u(t)=x(t)+y^-(t)$, $v(t)=y^+(t)$.
Thus, (a) means that
$|u(t_0)|<|v(t_0)|$. Suppose,
there exists a point $t_1<t_0$
such that $|u(t)|<|v(t)|$ for all
$t\in (t_1,t_0]$, but
$|u(t_1)|=|v(t_1)|$. We note that
$|z_1(t)-z_2(t)|=|z(t)|=|v(t)|$
for $t\in [t_1,t_0]$. Remind,
that $z_i(\cdot )=y^+_i(\cdot
)+x_i(\cdot )+y^-_i(\cdot )$ are
solutions to \eqref{3S}. Denote
\[f(t)=F(z_1(t))-
F(z_2(t))+G_-(z_1(t))-G_-(z_2(t)),\,
g(t)=G_+(z_1(t))-G_+(z_2(t)).
\] By subtracting
corresponding equations in \eqref{3S},
we see that for $t\in [t_1,t_0]$
\begin{eqnarray}
u(t)&=&e^{(t-t_0)(A+B_-)}u(t_0)-
\int^{t_0}_te^{(t-s)(A+B_-)}f(s)ds
\label{26.2}\\
v(t)&=&e^{(t-t_1)B_+}v(t_1)+
\int^t_{t_1}e^{(t-s)B_+}g(s)ds.
\label{26.3}\end{eqnarray} Also,
$|z(t)|=|v(t)|$ implies that
$|f(t)|\leq \max (\Lip (F)$, $\Lip
(G_-))|v(t)|$ and
\begin{equation}|g(t)|\leq \Lip
(G_+)|v(t)|,\quad t\in [t_1
,t_0].\label{27.1}\end{equation}
Thus, the Gronwall's inequality
applied in \eqref{26.3} gives:
\begin{equation}\label{27.2}
|v(t)|\leq e^{(t-t_1)(\alpha +\Lip
(G_+))}|v(t_1)|.\end{equation} In
particular, for $t=t_0$ one gets:
\begin{equation}\label{27.3}
|v(t_0)|\leq
e^{(t_0-t_1)(\alpha+\Lip
(G_+))}|v(t_1)|.\end{equation} Let
us apply the estimate
\eqref{27.2} for $|v(t)|$ in
$u$-equation \eqref{26.2}. For its
$x$-component we get:
\begin{eqnarray*}|x(t_1)|&\leq &
e^{(t_1-t_0)b}|x(t_0)|+
\int^{t_0}_{t_1}e^{(t_1-s)b}\Lip
(F)|v(s)|ds\leq\\ &\leq &
e^{(t_1-t_0)b}|x(t_0)|+\Lip
(F)\left(\int^{t_0}_{t_1}
e^{(t_1-s)b}e^{(s-t_1)(\alpha +\Lip
(G_+))}ds\right)|v(t_1)|.
\end{eqnarray*} Thus,
\begin{equation}\label{27.4}|x(t_1)|
\leq e^{(t_1-t_0)b}|x(t_0)|+\frac{\Lip
(F)}{b-\alpha -\Lip
(G_+)}\left(1-e^{(t_1-t_0)(b-\alpha
-\Lip
(G_+))}\right)|v(t_1)|.\end{equation}
Similarly, for the
$y^-$-component in the
$u$-equation \eqref{26.2},
\begin{equation}\label{27.5}|y^-
(t_1)|\leq e^{(t_1-t_0)\beta
}|y^-(t_0)+\frac{\Lip (G_-)}{\beta
-\alpha -\Lip
(G_+)}(1-e^{(t_1-t_0)(\beta
-\alpha -\Lip
(G_+))})|v(t_1)|\end{equation} By
the gap condition \eqref{19.2} one
has:
\begin{eqnarray*}\Lip (G_-)+\Lip
(G_+)&\leq & \Lip (G_-)+\Lip (F)+\Lip
(F)+\Lip(f_+)<\\ &<& \beta
-a+b-\alpha \leq
\beta -\alpha.\end{eqnarray*}
Therefore, $\max \{\gamma ,\gamma
_-\}<1$, where $\gamma$ and
$\gamma_-$ are defined as
\begin{eqnarray*}\gamma :&=&
\frac{\Lip (F)}{b-\alpha -\Lip
(G_+)}(1-e^{(t_1-t_0)(b-\alpha
-\Lip (G_+))}),\\
\gamma _-:&=& \frac{\Lip
(G_-)}{\beta -\alpha -\Lip
(G_+)}(1-e^{(t_1-t_0)(\beta
-\alpha -\Lip
(G_+))}).\end{eqnarray*} Recall,
that by our assumption,
$$ |v(t_1)|=|u(t_1)|=\max
\{|x(t_1)|,\quad |y^-(t_1)|\}.$$
If $|v(t_1)|=|x(t_1)|$, then
\eqref{27.4} gives
$|v(t_1)|\leq
e^{(t_1-t_0)b}|x(t_0)|/(1-\gamma)$,
and estimate \eqref{27.3} implies
$$|v(t_1)|\leq
\frac{e^{(t_1-t_0)(b-\alpha -\Lip
(G_+))}}{1-\gamma}|v(t_1)|.$$ But
this is a contradiction, since
$e^{(t_1-t_0)(b-\alpha -\Lip
(G_+))}<1-\gamma$. If
$|v(t_1)|=|y^-(t_1)|$, then
\eqref{27.5} and estimate
\eqref{27.3} imply:
$$|v(t_1)|\leq
\frac{e^{(t_1-t_0)(\beta -\alpha
-\Lip (G_+)}}{1-\gamma
_-}|v(t_1)|,$$  a contradiction
again. Therefore, claim
\eqref{26.1} is proved.

Now \eqref{26.1} shows that
$|u(t)|<|v(t)|$ for $t\in
[t_1,t_0]$ for any $t_1$. Thus,
estimate \eqref{27.3} holds.
Remind, that
$z(\cdot )=z_1(\cdot )-z_2(\cdot
)\in E_{\sigma _\pm}({\Bbb R})$.
Since $|z(t_1)|=|v(t_1)|$ for any
$t_1\leq t_0$, we conclude that
$|v(t_1)|\leq \|z\|_{\sigma
_\pm}e^{\sigma _-t_1}$ for
$t_1<\min \{0,t_0\}$. Together
with \eqref{27.3} we have:
\begin{equation}\label{29.0}
e^{-t_0(\alpha +\Lip
(G_+))}|v(t_0)|\leq
\|z\|_{\sigma_\pm}e^{t_1(\sigma _-
-\alpha -\Lip (G_+))}\to
0\end{equation} as $t_1\to
-\infty$ since $\sigma _- -\alpha
-\Lip (G_+)>0$ by the choice of
$\sigma_-$ in \eqref{21.1}. So, we
conclude that
$0=|v(t_0)|=|y^+(t_0)|$, which  contradicts to the strict
inequality  in (a).

Let us suppose that (b) holds. We
claim that (b) implies
\begin{equation}\max\{|x(t)|,|y^+(t)|
\}<|y^-(t)|\mbox{ for all }t\geq
t_0.\label{29.1}\end{equation}
Indeed, denote
$v(t)=x(t)+y^+(t)$,
$v(t)=|y^-(t)|$,
$f(t)=F(z_1(t))-F(z_2(t))+G_+(z_1
(t))-G_+(z_2(t))$,
$g(t)=G_-(z_1(t))-G_-(z_2(t))$.
Suppose, there exists $t_2>t_0$
such that $|u(t)|<|v(t)|$ for all
$t\in [t_0,t_2)$, but
$|u(t_2)|=|v(t_2)|$. Now, for
$t\in [t_0,t_2]$,
\begin{eqnarray}
u(t)&=&e^{(t-t_0)(A+B_+)}u(t_0)+
\int^t_{t_0}e^{(t-s)(A+B_+)}f(s)ds
\label{29.2}\\
v(t)&=&e^{(t-t_2)B_-}v(t_2)-
\int^{t_2}_te^{(t-s)B_-}g(s)ds.
\label{29.3}\end{eqnarray} Gronwall's
inequality and the estimate
$|g(s)|\leq \Lip (G_-)|v(s)|$
applied in \eqref{29.3} give
\begin{equation} |v(t)|\leq
e^{(t-t_2)(\beta -\Lip
(G_-))}|v(t_2)|.\label{30.1}
\end{equation} In particular, for
$t=t_0$ one gets:
\begin{equation}|v(t_0)|\leq
e^{(t_0-t_2)(\beta -\Lip
(G_-))}|v(t_2)|.\label{30.2}
\end{equation} As before, apply
\eqref{30.1} in
$u$-equation \eqref{29.2} at
$t=t_2$. For the $x$-component one
has:
\begin{equation}|x(t_2)|\leq
e^{(t_2-t_0)a}|x(t_0)|+\frac{\Lip
(F)}{\beta -a-\Lip
(G_-)}\left(1-e^{(t_0-t_2)(\beta
-a-\Lip
(G_-))}\right)|v(t_2)|.
\label{30.3}\end{equation} By our
assumption
$|v(t_2)|=|u(t_2)|=\max
\{|x(t_2)|,\quad |y^+(t_2)|\}$.
If $|v(t_2)|=|x(t_2)|$, similarly
to the case (a),
\eqref{30.3} leads to a
contradiction. Analogous argument
works for the
$y^+$-component in the
$u$-equation \eqref{29.2}, and the
claim \eqref{29.1} is proved.

Now, since $z(\cdot )=z_1(\cdot
)-z_2(\cdot )\in E_{\sigma _\pm }$
and $|z(t_2)|=|v(t_2)|$ for any
$t_2>\max
\{0,t_0\}$, we have
$|v(t_2)|\leq \|z \|_{\sigma_\pm
}e^{\sigma _+t_2}$. Thus, by
\eqref{30.2},
$$e^{(\Lip (G_-)-\beta
)t_0}|v(t_0)|\leq
e^{t_2(\sigma_+-\beta +\Lip
(G_-))}\to 0\mbox{ as } t_2\to
\infty$$ by the choice of
$\sigma_+$ in \eqref{21.1}. But
$0=|v(t_0)|=|y^-(t_0)|$
contradicts the strict inequality
in (b).

Finally, let us suppose that (c)
holds. First, we note that there
exists $t_1<t_0$ such that
$|x(t)|<|y^-(t)|=|y^+(t)|$ for all
$t\in (t_1,t_0]$. Indeed, by the
continuity of the solutions, there
is $t_1$ such that
$|x(t)|<|y^-(t)|$ and
$|x(t)|<|y^+(t)|$ for $t\in
(t_1,t_0]$. However, then
$|y^-(t)|=|y^+(t)|$, otherwise we
are in the situation (a) or (b)
above.

Let us suppose that there exists
$t_1<t_0$ such that
$|x(t_1)|=|y^-(t_1)|=|y^+(t_1)|$
but
$|x(t)|<|y^-(t)|=|y^+(t)|$ for all
$t\in (t_1,t_2]$. Since
$|z(t)|=|y^+(t)|$ for $t\in
[t_1,t_0]$, one can apply
arguments used to consider case (a).
That is, for
$v (t)=y^+(t)$, $u(t)=x(t)+y^-(t)$
one has estimates \eqref{27.2} and
\eqref{27.3} and also inequality
\eqref{27.5}. Since
$|v(t_1)|=|y^-(t_1)|$, inequality
\eqref{27.5} leads to a
contradiction.

Thus, we know that
$|x(t)|<|y^-(t)|=|y^+(t)|$ for all
$t\leq t_0$. Now we can apply the
argument as in
\eqref{29.0} to see that
$0=|u(t_0)|=|y^-(t_0)|>|x(t_0)|$,
which is a contradiction.\quad
$\Box$


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