%% Y. Latushkin %% %% A remark on invariant manifolds: %% the optimal gap condition. %% AMS-LaTeX \documentstyle[12pt]{amsart} \setlength{\textwidth}{6.5in} \setlength{\textheight}{9in} \setlength{\abovedisplayskip}{14pt} \setlength{\belowdisplayskip}{14pt} \setlength{\abovedisplayshortskip} {14pt} \setlength{\belowdisplayshortskip} {14pt} \setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in} \setlength{\topmargin}{-.5in} \theoremstyle{plain} \newtheorem{thm}{Theorem} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \newtheorem{claim}[thm]{Claim} \newtheorem{cor}[thm]{Corollary} \theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \newcommand{\Lip}{\operatorname{Lip}} \renewcommand{\Pr}{\operatorname{Pr}} \newcommand{\bi}{\bibitem} \newcommand{\sys}{{(\ref{1.2})--(\ref{1.3})\,}} \newcommand{\bl}{{\bold l}} \newcommand{\br}{{\bold r}} \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 $a1$ 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 $$M_A \Lip (F)+M_B \Lip (G)i} \Lip (F_j) &<& b_i-a_{i+1},\quad i=1,\ldots , N-1,\\ \Lip (F_i)+\max_{j0 there exists a unique solution z(\cdot) of \sys on [0,t_2] in the sense of \eqref{1.5.1}-\eqref{1.5.2} such that z(0)=\xi +\varphi (\xi ) and z(t)\in {\cal M}_\varphi  for t\in [0,t_2]. \begin{lem}\label{invM} {\cal M}_\varphi is forward invariant for \sys if and only if \varphi = {\cal L}(\varphi). \end{lem} \noindent{\bf Proof.} Indeed, \begin{eqnarray}\label{4.1} ({\cal L}\varphi )(x(t;\xi ))&=&-\int^\infty_0e^{-sB}G(x(s;x (t;\xi ))+\varphi (x(s;x(t;\xi))))ds=\nonumber\\ &=&-\int^\infty_0e^{-sB}G(x(s+t;\xi)+ \varphi (x(s+t;\xi)))ds=\\ &=&-\int^\infty_te^{(t-s)B}G(x(s;\xi )+\varphi (x(s;\xi )))ds,\quad t\geq 0.\nonumber\end{eqnarray} Assume \varphi ={\cal L}\varphi. Then for each \xi \in X and every t_2>0 there exists a solution z on [0,t_2] to \eqref{1.2}-\eqref{1.3} such that z(0)=\xi +\varphi (\xi) and z(t)=x(t;\xi)+\varphi (x(t;\xi )) for each t\in [0,t_2]. Indeed, by the definition of x(t;\xi ) one has$$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,$$satisfies the same equation. Thus, \varphi={\cal L}\varphi.\quad \Box Fix \sigma \in {\Bbb R} such that a+\Lip (F)<\sigma \Lip (F)\geq 0 and b-\sigma >\Lip (G)\geq 0 by the choice of \sigma  in \eqref{5.1}. For any f_{1,2}\in E_\sigma  one has: \begin{eqnarray*}|T_\xi f_1(t)-T_\xi f_2(t)|&\leq &\max \{\int^t_0e^{(t-s)a}\Lip (F)|f_1(s)-f_2(s)|ds,\\ &\quad & \int^\infty_te^{(t-s)b}\Lip (G)|f_1(s)-f_2(s)|ds\}.\end{eqnarray*} Thus, \begin{eqnarray*}\|T_\xi f_1-T_\xi f_2\|_\sigma &\leq &\sup_{t\geq 0}\max \{ \Lip (F)\int^t_0e^{(t-s)(a-\sigma)}ds,\\ &\quad & \Lip (G)\int^\infty_te^{(t-s)(b-\sigma)} ds\}\|f_1-f_2\|_ \sigma \leq\\ &\leq &\max \left\{ \frac{\Lip (F)}{\sigma -a},\frac{\Lip (G)}{b-\sigma}\right\} \| f_1-f_2\|_\sigma .\quad\Box\end{eqnarray*} Therefore, for each \xi \in X there exists a unique f_\xi \in E_\sigma such that f_\xi =T_\xi f_\xi. We also remark that the operator$$(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$$is a Lipschitz map on E_\sigma with $$\Lip (T)\leq \max \left\{ \frac{\Lip (F)}{\sigma -a},\frac{\Lip (G)}{b-\sigma }\right\}<1.\label{7.1}$$ \begin{rem}\label{VanAppr} Define S:X\to E_\sigma  such that S(\xi )(t):=e^{tA}\xi. Then T_\xi f=S(\xi )+Tf for each f\in E_\sigma. By a general result on Lipschitz maps (see, e.g. \cite{Shub}, Appendix I, Ch.5, p. 49) I-T:E_\sigma \to E_\sigma  is a Lipschitz homeomorphism with \Lip ((I-T)^{-1})=[1-\Lip (T)]^{-1}. Equation f_\xi =T_\xi f_\xi  has the form f_\xi =S(\xi)+Tf_\xi. Thus, f_\xi =(I-T)^{-1}S(\xi). \end{rem} Define \varphi :X\to Y as follows: $$\varphi (\xi ):=\Pr_Yf_\xi (0)=-\int^\infty_0e^{-sb}G(f_\xi (s))ds\label{7.2}$$ \begin{prop}\label{Bigsmall} \varphi={\cal L}\varphi for the Lyapunov-Perron operator \eqref{LP}.\end{prop} \noindent{\bf Proof.} For f_\xi =T_\xi f_\xi let us denote: \begin{eqnarray*}x(t)&:=&\Pr_Xf_\xi (t)=e^{tA}\xi +\int^t_0e^{(t-s)A}F(f_\xi (s))ds,\\ y(t)&:=&\Pr_Yf_\xi(t)=-\int^\infty_t e^{(t-s)B}G(f_\xi (s))ds.\end{eqnarray*} Since f_\xi =T_\xi f_\xi for each \xi \in X, for any t\geq 0 and s\geq 0 one has: $$\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.$$ Also, one has: \begin{eqnarray*}f_\xi (s+t)&=&e^{(s+t)A}\xi +\int^{s+t}_0e^{(s+t-\tau)A} F(f_\xi (\tau ))d\tau -\int^\infty_{s+t}e^{(s+t-\tau)B} G(f_\xi (\tau ))d\tau\\ &=&e^{sA}\left[e^{tA}\xi +\int^t_0e^{(t-\tau)A}F(f_\xi (\tau ))d\tau\right]+\int^{t+s}_te^{(s+ t-\tau)A}F(f_\xi (\tau))d\tau\\ &\quad & -\int^\infty _{s+t}e^{(s+t-\tau)B}G(f_\xi (\tau ))d\tau =\\ &=&e^{sA}x(t)+\int^s_0e^{(s-\tau)A} F(f_\xi (\tau+t))d\tau -\int^\infty_se^{(s-\tau)B}G(f_\xi (\tau+t))d\tau\end{eqnarray*} Since \eqref{3*} has the unique solution f_{x(t)}(\cdot), we conclude f_\xi (s+t)=f_{x(t)}(s), s,t\geq 0. Using our definition \eqref{7.2} we obtain: \begin{eqnarray*}\varphi (x(t))&=&-\int^\infty_0e^{-sB}G( f_{x(t)}(s))ds=-\int^\infty_0e^{-sB} G(f_\xi (s+t))ds\\ &=&-\int^\infty_te^{(t-s)B}G(f_\xi (s))ds=y(t).\end{eqnarray*} Therefore, x(t)=x(t;\xi ) satisfies $$x(t)=e^{tA}\xi +\int^t_0e^{(t-s)a}f(x(s)+\varphi (x(s)))ds,\label{**}$$ and \begin{equation*}\varphi (\xi )=-\int^\infty_0e^{-sB}G(x(s;\xi )+\varphi (x(s; \xi )))ds=({\cal L}\varphi )(\xi)\quad\Box\end{equation*} \begin{rem} If \varphi ={\cal L}\varphi then g_\xi (t)=x(t;\xi )+\varphi (x(t;\xi )), t\geq 0 satisfies T_\xi g_\xi =g_\xi. Indeed, as in \eqref{4.1},$$\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.$$Using Lemma \ref{xz}, we obtain: \begin{eqnarray*}|x_1(t)-x_2(t)|&\leq & e^{ta}|\xi _1-\xi _2|+\int^t_0e^{(t-s)a}\Lip (F)|z_1(s)-z_2(s)|ds=\\ &=&e^{ta}|\xi_1-\xi_2|+\Lip (F)\int^t_0e^{(t-s)a}|x_1(s)-x_2(s) |ds,\end{eqnarray*} and $$\label{(1*)}|x_1(t)- x_2(t)|\leq e^{t(a+\Lip(F))}|\xi_1-\xi_2|,\quad t\geq 0$$ by Gronwall's inequality. Since \varphi (\xi_i)=y_i(0)=\Pr_Yf_{\xi_i}(0) by the definition \eqref{7.2}, we have from Lemma \ref{xz} and \eqref{(1*)}: \begin{eqnarray*}|\varphi (\xi_1)-\varphi (\xi_2)|&=&|\Pr_Yf_{\xi_1}(0)-\Pr_Y f_{\xi_2}(0)|\leq\\ &\leq & \int^\infty_0e^{-sb}\Lip (G) |z_1(s)-z_2(s)|ds=\\ &=&\Lip (G) \int^\infty_0e^{-sb}|x_1(s)-x_2(s) |ds\leq \Lip (G)\int^\infty _0e^{(a-b+\Lip (F))s}|\xi_1-\xi_2|ds\\ &=&\frac{\Lip (G)}{b-a-\Lip (F)} |\xi_1-\xi_2|.\end{eqnarray*} This proves Theorem \ref{main}.\quad \Box \noindent{\bf Proof of Lemma \ref{xz}.} We want to show that $$\label{11.1}|x_1(t)- x_2(t)|\geq |y_1(t)-y_2(t)| \mbox{ for all } t\geq 0.$$ Let us suppose that for some t_1>0 one has $$\label{11.2} |y_1(t_1)-y_2(t_1)|>|x_1(t_1)-x_2 (t_1)|.$$ \begin{claim} Inequality \eqref{11.2} implies $$\label{11.3}|y_1(t)- y_2(t)|>|x_1(t)-x_2(t)|\mbox{ for all } t\geq t_1.$$\end{claim} To prove the claim, we suppose there is a point t_2>t, such that \begin{eqnarray}|x_1(t_2)-x_2(t_2)| &=&|y_1(t_2)-y_2(t_2)|\quad \mbox{ but}\nonumber \\ |y_1(t)-y_2(t)|&>&|x_1(t)-x_2(t)\quad \mbox{ for all }\quad t\in [t_1,t_2).\label{11.4}\end{eqnarray} Recall, that z_i(t)=x_i(t)+y_i(t), i=1,2 are solutions to \eqref{1.5.1}-\eqref{1.5.2}, that is, for t\in (t_1,t_2) and i=1,2 one has: \begin{eqnarray*}x_i(t)&=& e^{(t-t_1)A}x_i(t_1)+\int^t_{t_1} e^{(t-s)A}F(z_i(s))ds,\\ y_i(t)&=&e^{(t-t_2)B}y_i(t_2)- \int^{t_2}_te^{(t-s)B}G(z_i(s)) ds.\end{eqnarray*} Let us denote, for brevity: \begin{eqnarray*} u(t)&=&x_1(t)-x_2(t),\quad v(t)=y_1(t)-y_2(t),\\ f(t) &=&F(z_1(t))-F(z_2(t)),\quad g(t)=G(z_1(t))-G(z_2(t)),\quad t\in [t_1,t_2].\end{eqnarray*} Then \begin{eqnarray}u(t)&=&e^{(t-t_1)A} u(t_1)+\int^t_{t_1}e^{(t-s)A}f(s)ds, \label{12.1}\\ v(t)&=&e^{(t-t_2)B}v(t_2)- \int^{t_2}_te^{(t-s)}g(s)ds. \label{12.2}\end{eqnarray} Since |z_1(t)-z_2(t)|=|u(t)| for t\in [t_1,t_2] by \eqref{11.4}, we conclude:$$|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,$$and, by the Gronwall's inequality, $$|v(t)|\leq |v(t_2)|e^{(\Lip (G)-b)(t_2-t)},\quad t\in [t_1,t_2]\label{12.4}$$ We use the last inequality to have the following estimate in \eqref{12.1} for t=t_2: \begin{eqnarray}|u(t_2)|&\leq & e^{(t_2-t_1)a}|u(t_1)|+\Lip (F)|v(t_2)|\int^{t_2}_{t_1} e^{(t_2-s)a}e^{(\Lip (G)-b)(t_2-s)}ds\nonumber\\ &=&e^{(t_2-t_1)a}|u(t_1)|+ \frac{\Lip (F)}{b-a-\Lip (G)}|v(t_2)|[1- e^{(b-a-\Lip (G))(t_1-t_2)}]. \label{12.3}\end{eqnarray} However, by the first assumption in \eqref{11.4} one has |v(t_2)|=|u(t_2)|. Denote, for brevity, \delta =b-a-\Lip (G), \ell =\Lip (F)/\delta. Solve \eqref{12.3} for |v(t_2)| to get:$$|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)|.$$Indeed, the expression in \{\quad \} is strictly less then 1 due to the gap condition \eqref{1.5}. But the inequality |v(t_1)|<|u(t_1)| contradicts \eqref{11.2}, and the claim is proved. To finish the proof of the lemma, we stress that \eqref{11.3} implies |z_1(t)-z_2(t)|=|v(t)| for all t\in [t_1,\infty ). Therefore, using \eqref{12.4} with t=t_1, we conclude that for an arbitrarily large t_2\geq t_1 one has: $$\label{13.1} |v(t_1)|\leq |v(t_2)|e^{(\Lip(G)-b) (t_2-t_1)}.$$ Recall, that z_1-z_2=f_{\xi_1}-f_{\xi_2}\in E_\sigma with \sigma1} \Lip (F_j), \Lip (G)=\Lip (F_1) and the gap condition \eqref{1.5} of Corollary~\ref{back} gives the existence of the desired \varphi _1=\psi :Y\to X. Similarly, for i=N Theorem \ref{main} gives \varphi _N=\varphi :X\to Y for X=X_N, Y=X_1+\ldots +X_{N-1}. Fix i=2,\ldots , N-1. We will show here only the existence of a map \varphi_i:X_i\to\hat{X}_i with the invariant graph. The estimates on the \Lip(\varphi_i) as in \eqref{bigestphilip} will require more work; they will follow from Theorem \ref{central} below. Define subspaces Y_-=X_1+\ldots +X_{i-1}, Y_+=X_{i+1}+\ldots +X_N, Y=Y_++Y_- and operators B_-=\mbox{diag} (A_j)_{j=1}^{i-1}, B_+=\mbox{diag} (A_j)^N_{j=i+1}. First, we apply Theorem \ref{main} for X=Y_++X_i, Y=Y_- and A=B_++A_i, B=B_- to get \varphi :Y_++X_i\to Y_- with (forward) invariant graph {\cal M}_\varphi =\{ y_++x_i+\varphi (y_++x_i)|y_++x_i\in Y_++X_i\}. Next, we apply Corollary~\ref{back} for X=Y_+, Y=X_i+Y_- and A=B_+, B=A_i+B_- to get \psi :X_i+Y_-\to Y_+ with (backward) invariant graph {\cal M}_\psi =\{\psi (x_i+y_-)+x_i+y_-|x_i+y_-\in X_i+Y_-\}. Note, that M={\cal M}_\varphi \cap {\cal M}_\psi is invariant in both directions, and$$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: \Lip (G_+)+\Lip (F)\sigma_- as follows: \begin{eqnarray}a+\Lip (F) &<&\sigma_+<\beta -\Lip (G_-),\nonumber\\ \alpha +\Lip (G_+)&<&\sigma _-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: $$\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.$$ As in Remark~\ref{justLip}, one already has that \Phi  is Lipschitz. We proceed to estimate \Lip (\Phi ). Fix \xi _1\neq \xi_2 in X, let$$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.$$The Gronwall's inequality implies, for t\geq 0: $$\label{24.2} |x_1(t)-x_2(t)|\leq e^{t(a+\Lip (F))}|\xi _1-\xi _2|,\quad t\geq 0.$$ Similarly, for t\leq 0:$$|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,$$and $$|x_1(t)-x_2(t)|\leq e^{t(b-\Lip (F))}|\xi_1-\xi_2|,\quad t\leq 0.\label{24.3}$$ We use \eqref{24.1} and Lemma \ref{centralxz} to estimate: \begin{eqnarray*}|\Phi (\xi _1)-\Phi (\xi_2)|&\leq & \max \{\int^0_{-\infty}e^{-s\alpha} \Lip (G_+)|z_1(s)-z_2(s)|ds,\\ &\quad & \int^\infty_0e^{-s\beta}\Lip (G_-)|z_1(s)-z_2(s)|ds\}=\\&=&\max \{\Lip (G_+)\int^0_{-\infty}e^{-s\alpha}| x_1(s)-x_2(s)|ds,\\ &\quad & \Lip (G_-)\int^\infty_0e^{-s\beta}|x_1(s)- x_2(s)|ds\}.\end{eqnarray*} Now apply \eqref{24.2}-\eqref{24.3} to get: \begin{eqnarray*}\Lip (\Phi )&\leq & \max \{ \Lip (G_+)\int^0_{-\infty}e^{-s\alpha }e^{s(b-\Lip (F))}ds,\\ &\quad \qquad & \Lip (G_-)\int^\infty_0e^{-s\beta }e^{s(a+\Lip (F))}ds\}=\\ &=&\max\left\{\frac{\Lip (G_+)}{b-\alpha -\Lip (F)},\frac{\Lip (G_-)}{\beta -a-\Lip (F)}\right\}.\quad \Box\end{eqnarray*} \noindent{\bf Proof of Lemma \ref{centralxz}.} Some preparations are in order. Denote z(t)=z_1(t)-z_2(t), x(t)=x_1(t)-x_2(t), y^\pm (t)=y^\pm _1(t)-y^\pm _2(t). We want to prove that $$\label{25.1} |x(t)|\geq \max \{|y^+(t)|, |y^-(t)|\},\quad t\in {\Bbb R}.$$ \begin{prop}\label{contrprop} For each t_0\in {\Bbb R} {\bf neither} one of the following inequalities is possible: \begin{enumerate}\item[(a)] \max\{|x(t_0)|,|y^-(t_0)|\}<|y^+ (t_0)|; \item[(b)] \max \{|x(t_0)|,|y^+(t_0)|\}<|y^-(t_0)|; \item[(c)] |x(t_0)|<|y^+(t_0)|=|y^-(t_0)|. \end{enumerate}\end{prop} To derive \eqref{25.1} from the proposition, let us suppose that, say, |x(t_0)|<|y^+(t_0)| for some t_0\in {\Bbb R}. Clearly, |y^-(t_0)|\neq |y^+(t_0)|, otherwise we get (c). By |y^-(t_0)|>|y^+(t_0)| yields (b), while |y^- (t_0)|<|y^+(t_0)| yields (a), a contradiction that proves the lemma.\quad \Box \noindent{\bf Proof of Proposition~\ref{contrprop}.} Let us suppose that (a) holds. We claim, that (a) implies $$\label{26.1} \max \{ |x(t)|, |y^-(t)|\}<|y^+(t)|\mbox{ for all }t\leq t_0.$$ to prove the claim \eqref{26.1}, let us denote: u(t)=x(t)+y^-(t), v(t)=y^+(t). Thus, (a) means that |u(t_0)|<|v(t_0)|. Suppose, there exists a point t_10 by the choice of \sigma_- in \eqref{21.1}. So, we conclude that 0=|v(t_0)|=|y^+(t_0)|, which contradicts to the strict inequality in (a). Let us suppose that (b) holds. We claim that (b) implies $$\max\{|x(t)|,|y^+(t)| \}<|y^-(t)|\mbox{ for all }t\geq t_0.\label{29.1}$$ Indeed, denote v(t)=x(t)+y^+(t), v(t)=|y^-(t)|, f(t)=F(z_1(t))-F(z_2(t))+G_+(z_1 (t))-G_+(z_2(t)), g(t)=G_-(z_1(t))-G_-(z_2(t)). Suppose, there exists t_2>t_0 such that |u(t)|<|v(t)| for all t\in [t_0,t_2), but |u(t_2)|=|v(t_2)|. Now, for t\in [t_0,t_2], \begin{eqnarray} u(t)&=&e^{(t-t_0)(A+B_+)}u(t_0)+ \int^t_{t_0}e^{(t-s)(A+B_+)}f(s)ds \label{29.2}\\ v(t)&=&e^{(t-t_2)B_-}v(t_2)- \int^{t_2}_te^{(t-s)B_-}g(s)ds. \label{29.3}\end{eqnarray} Gronwall's inequality and the estimate |g(s)|\leq \Lip (G_-)|v(s)| applied in \eqref{29.3} give $$|v(t)|\leq e^{(t-t_2)(\beta -\Lip (G_-))}|v(t_2)|.\label{30.1}$$ In particular, for t=t_0 one gets: $$|v(t_0)|\leq e^{(t_0-t_2)(\beta -\Lip (G_-))}|v(t_2)|.\label{30.2}$$ As before, apply \eqref{30.1} in u-equation \eqref{29.2} at t=t_2. For the x-component one has: $$|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}$$ By our assumption |v(t_2)|=|u(t_2)|=\max \{|x(t_2)|,\quad |y^+(t_2)|\}. If |v(t_2)|=|x(t_2)|, similarly to the case (a), \eqref{30.3} leads to a contradiction. Analogous argument works for the y^+-component in the u-equation \eqref{29.2}, and the claim \eqref{29.1} is proved. Now, since z(\cdot )=z_1(\cdot )-z_2(\cdot )\in E_{\sigma _\pm } and |z(t_2)|=|v(t_2)| for any t_2>\max \{0,t_0\}, we have |v(t_2)|\leq \|z \|_{\sigma_\pm }e^{\sigma _+t_2}. Thus, by \eqref{30.2},$$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. 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