\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2018 (2018), No. 188, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2018 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2018/188\hfil Global and local versions for a Ph\'ong V\~u theorem] {Global and local versions for a Ph\'ong V\~u theorem for periodic evolution families in Hilbert spaces} \author[C. Bu\c{s}e, L. T. Nguyen, D. O'Regan \hfil EJDE-2018/188\hfilneg] {Constantin Bu\c{s}e, Lan Thanh Nguyen, Donal O'Regan} \address{Constantin Bu\c{s}e \newline Politehnica University of Timisoara, Department of Mathematics, Piata Victoriei No. 2, 300006-Timisoara, Romania.\newline Western Kentucky University, Department of Mathematics, Bowling Green, KY 42101, USA} \email{constantin.buse@upt.ro, constantin.buse@wku.edu} \address{Lan Thanh Nguyen \newline Western Kentucky University, Department of Mathematics, Bowling Green, KY 42101, USA} \email{lan.nguyen@wku.edu} \address{Donal O'Regan \newline National University of Ireland, Galway, Ireland} \email{donal.oregan@nuigalway.ie} \dedicatory{Communicated by Jerome A. Goldstein} \thanks{Submitted November 7, 2017. Published November 20, 2018.} \subjclass[2010]{35B35 47A30, 46A30} \keywords{Uniform exponential stability; growth bounds; Fourier series; \hfill\break\indent exponentially bounded evolution families of operators} \begin{abstract} A theorem of Gearhart concerning strongly continuous semigroups in Hilbert spaces is extremely useful for stability analysis of concrete equations; see e.g.\ \cite{Hu}), and for control theory \cite{We} or \cite[page 475]{EN}. Ph\'ong V\~u introduced an equivalent condition in \cite{VP}. The aim of this article is to extend these results from the autonomous case to time dependent $1$-periodic evolution equations in Hilbert spaces. Both cases (continuous and discrete) are analyzed and global and local versions of the Ph\'ong V\~u theorem are provided. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} The following result is well known. \begin{theorem} For any strongly continuous semigroup $\mathbf{T}=\{T(t)\}_{t\ge 0}$ that acts on a complex Hilbert space $\mathcal{H}$ the following three statements are equivalent: \begin{itemize} \item[(1)] $\mathbf{T}$ is uniformly exponentially stable i.e. its growth bound \begin{equation}\label{ngb} \omega_0(\mathbf{T}):=\inf_{t\ge 0}\frac{\ln\|T(t)\|}{t}<0. \end{equation} \item[(2)] For every vector $b$ in $\mathcal{H}$ the solutions of the following Cauchy problems associated to the generator $A$ of $\mathbf{T}$ \begin{equation}\label{vucond1} \begin{gathered} u'(t)= Au(t)+e^{i\mu t}b, \quad t\ge 0,\; \mu\in\mathbb{R} \\ u(0)= 0 \end{gathered} \end{equation} are bounded on $\mathbb{R}_+$ (uniformly with respect to the parameter $\mu)$, or equivalently \begin{equation}\label{semest} \sup_{\mu\in\mathbb{R}}\sup_{t\ge 0}\|\int_0^te^{i\mu s}T(t-s)x ds\| :=M(x)<\infty\quad\forall x\in\mathcal{H}. \end{equation} \item[(3)] The resolvent operator $R(z, A)$ exists for $z\in\mathbb{C}_+:=\{\Re(z)>0\}$ and \begin{equation}\label{uniboundres} \sup_{z\in\mathbb{C}_+}\|R(z, A\|<\infty. \end{equation} \end{itemize} \end{theorem} The proof of $(2)\Rightarrow(3)$ was given by Ph\'ong V\~u in \cite{VP} and the implication $(3)\Rightarrow(1)$ is known as the Gearhart-Pr\"uss Theorem, \cite{Ge,Pr}. Different proofs of this result were given in the literature by many authors. For example, the proof in \cite{We} relies on the strong connection of this theorem and the half plane version of an old theorem of Paley and Wiener. The Gearhart-Pr\"uss Theorem is not valid for semigroups acting on arbitrary Banach spaces, see for example \cite{GVW} for a counterexample. In \cite{VP} the equivalence between \eqref{semest} and the negativeness of $\omega_0(\mathbf{T})$ was settled. We mention that the proof depends of the implication $(3)\Rightarrow(1)$. For information concerning strongly continuous semigroups we refer the reader to the monographs \cite{EN,Go,HP,Pa} and the references therein. Let $X$ be a complex Banach space and denote by $\mathcal{L}(X)$ the Banach algebra of all bounded linear operators acting on $X$. The norms in $X$ and in $\mathcal{L}(X)$ will be denoted by $\|\cdot\|$. Recall that a family $\mathcal{U}=\{U(t, s): (t, s)\in\mathbb{R}^2, t\ge s\}$ of bounded linear operators acting on $X$ is called a one periodic strongly continuous evolution family if $U(t, t)=I$- the identity operator on $X$, $U(t, s)U(s, r)=U(t, r)$ for all reals $t\ge s\ge r$, the map $(t, s)\mapsto U(t, s)$ is strongly continuous on the set $\Delta:=\{(t, s)\in \mathbb{R}^2: t\ge s\}$ and $U(t, s)=U(t+1, s+1)$ for every pair $(t, s)\in\Delta$. The evolution family $\mathcal{U}$ is called exponentially bounded if there exists a real number $\omega$ such that the map $(t, s)\mapsto e^{-\omega(t-s)}U(t, s)$ is bounded in $\mathcal{L}(X)$ and uniformly exponentially stable if there exists a negative $\omega$ with that property. Throughout the paper we assume that the evolution families are exponentially bounded. Clearly, if $\mathcal{U}$ verifies the convolution condition $U(t, s)=U(t-s, 0)$ for every $(t, s)\in\Delta$ then the family $\mathbf{T}=\{U(t, 0): t\ge 0\}$ is a strongly continuous semigroup that acts on $X$. Thus it is natural to ask if the family $\mathcal{U}$ is uniformly exponentially stable provided \eqref{semest} holds when $U(t, s)$ replaces $T(t-s)$. The proof of such a result cannot follow the ideas of V\~u because in the non-autonomous case an analogue for the implication $(3)\Rightarrow(1)$ does not exist. As expected, additional assumptions are required. The following result was obtained in \cite{BBBS}. \begin{theorem} A strongly continuous one periodic evolution family $\mathcal{U}$ of bounded linear operators acting on a Hilbert space $\mathcal{H}$ is uniformly exponentially stable provided that for every $x\in X$ the map $t\mapsto U(t, 0)x$ satisfies a Lipschitz condition on $(0, 1)$ and \eqref{semest} holds when $U(t, s)$ replaces $T(t-s)$. \end{theorem} Let $X$ be a complex Banach space and $\alpha\in(0, 1]$. Recall that a $X$-valued function $f$ defined on a real closed interval $[a, b]$ is called $\alpha$-H\"older continuous with the constant $L_f$ if \begin{equation} \|f(t)-f(s)\|\le L_f|t-s|^{\alpha}\quad\forall t, s\in[a, b]. \end{equation} The next Lemma collects some facts concerning $\alpha$-H\"older continuity. Its proof is obvious so we omit the details. As usual by $\|f\|_{\infty}$ we denote the uniform norm of $f$ in the space $C([a, b], X)$. \begin{lemma}\label{reshold} Let $h{:}[a, b]\to\mathbb{C}$ and $H{:}[a, b]\to X$ be $\alpha$-H\"older continuous functions on the closed interval $[a, b]$ with the constants $L_h$ and $L_H$, respectively. Then one has: \begin{itemize} \item[(i)] The map $hH$ is $\alpha$-H\"older continuous on $[a, b]$ with the constant $H_{\infty}L_h+h_{\infty}L_H$. \item[(ii)] $H$ is $\beta$-H\"older continuous for every $\beta\in(0, \alpha]$ (with the same constant). \item[(iii)] The function $t\mapsto \sin(\pi t): [0, 1]\to\mathbb{C}$ is Lipschitz continuous (i.e. $1$-H\"older continuous) with the constant $\pi$. \item[(iv)] If, in addition, $H(a)=H(b)$ then its extension by periodicity to $\mathbb{R}$ is $\alpha$-H\"older continuous with the constant $L_H$. \end{itemize} \end{lemma} \begin{lemma}\label{replem} Let $\mathcal{U}=\{U(t, s), t\ge s\ge 0\}$ be a strongly continuous $1$-periodic evolution family of bounded linear operators acting on a complex Banach space $X$ and let $h(t):=\sin(\pi t)$, $ t\in[0, 1]$. For each $x\in X$ set \begin{equation}\label{themap} H_{xh}(t):=h(t)U(t, 0)x,\quad t\in[0, 1] \end{equation} and denote by $\tilde{H}_{xh}$ the extension by periodicity of $H_{xh}$ to the entire real axis. Then (clearly) $I(h, \mu):=\int_0^1 e^{i\mu t}h(t)dt\neq 0$ for every $\mu\in\mathbb{R}$. Let $T:=U(1, 0)$. Then for any positive integer $n$, one has \begin{equation}\label{intest} \sum_{j=1}^n e^{-i\mu j}T^jx=\frac{e^{-\mu jn}}{I(h, \mu)} \int_0^n e^{i\mu s}U(n, s)\tilde{H}_{xh}(s)ds. \end{equation} \end{lemma} Note that for every integer $k$ one has \begin{equation} I(h, 0)=\frac{2}{\pi}, I(h, (2k+1)\pi) =\pm\frac{1}{2}i \quad \text{and}\quad I(h, \mu)=\frac{\pi}{\pi^2-\mu^2}(1+e^{i\mu}), \end{equation} elsewhere. To obtain \eqref{intest} it is sufficient to write the integral in the right-hand side of \eqref{intest} as \begin{equation} \sum_{k=0}^{n-1}\int_k^{k+1} e^{i\mu s}U(n, s)\tilde{H}_{xh}(s)ds. \end{equation} and then to change the variable in the last integral. \begin{remark}\label{lipscond} \rm With the notation from Lemma \ref{replem} assume that the map $U(\cdot, 0)x$ is $1$-H\"older continuous on $[0, 1]$ with the constant $L(x)$. Then $\tilde{H}_{xh}$ is $1$-H\"older continuous on $\mathbb{R}$ with the constant \begin{equation}\label{consttil} L(\tilde{H}_{xh}):=\pi \|1_{[0, 1]}(\cdot) U(\cdot, 0)x\|_{\infty}+L(x), \end{equation} where $1_{[0, 1]}(\cdot)$ denotes the characteristic function of the interval $[0, 1]$. In the particular case when the family $\{U(t, 0): t\ge 0\}$ is a strongly continuous semigroup generated by $A$ and $x\in D(A)$, we can take $L(x):= \sup_{t\in[0, 1]}||U(t, 0)||\cdot||Ax||$ and then there exists a positive $c$ (independent of $x$) such that \begin{equation} L(\tilde{H}_{xh})\le c(||x||^2+||Ax||^2)^{\frac{1}{2}}. \end{equation} \end{remark} Denote by $CP_1(\mathbb{R}, X)$ the set of all $X$-valued continuous functions $f$ defined on $\mathbb{R}$ which are periodic of period $1$, i.e.\ $f(t+1)=f(t)$ for every $t\in\mathbb{R}$. The vectors in $X$ \begin{equation} c_n(f):=\int_0^1e^{2i n t\pi}f(t)dt, \quad n\in\mathbb{Z} \end{equation} are called the Fourier-Bohr coefficients associated to $f$. With $CP_1^1(\mathbb{R}, X)$ we will denote the subset of $CP_1(\mathbb{R}, X)$ consisting by all functions $f$ for which \begin{equation}\label{absconv} \|f\|_1:= \sum_{n\in\mathbb{Z}}\|c_n(f)\|<\infty. \end{equation} Further details concerning the space of all continuous functions satisfying \eqref{absconv} (not necessarily periodic) can be found in the monograph \cite{Co} by Corduneanu. For each $f\in CP_1^1(\mathbb{R}, X)$ consider its Fourier sum (in respect with the uniform norm) given by \begin{equation} s_f(t)=\sum_{n\in\mathbb{Z}}e^{-2int\pi}c_n(f), \quad t\in\mathbb{R}. \end{equation} Clearly $s_f\in CP_1(\mathbb{R}, X)$ and thus $f=s_f$. The next lemma highlights an important subset of $CP_1^1(\mathbb{R}, X)$ when $X$ is a complex Hilbert space. \begin{lemma} \label{Fourier} Let $\mathcal{H}$ be a complex Hilbert space and let $f\in C_1(\mathbb{R}, \mathcal{H})$. If $f$ is $\alpha$-H\"older continuous on $[0, 1]$ for some $\alpha>\frac{1}{2}$ with a constant $L=L(f)>0$, i.e.\ \begin{equation} \|f(t)-f(s)\|\le L|t-s|^{\alpha}, \quad \forall t, s\in[0, 1]\,. \end{equation} Then $f\in CP_1^1(\mathbb{R}, X)$, i.e.\ \begin{equation}\label{develop} {f}(t)=\sum_{n\in\mathbb{Z}}e^{-2int\pi}c_n({f}), \quad t\in\mathbb{R}. \end{equation} In addition, there exists a positive constant $K$ depending only of $\alpha$ such that. \begin{equation}\label{n1est} \|f\|_1\le 3\|f\|_{\infty}+LK(\alpha). \end{equation} \end{lemma} \begin{proof} Let $\rho>0$ and set $g(t):={f}(t+\rho)-{f}(t-\rho)$. A simple calculation shows that \begin{gather}\label{concoef} c_n(g)=-2i\sin(2n\pi\rho)c_n({f}), \\ \label{normest} 4^{\alpha}\rho^{2\alpha}L^2\ge\int_0^1\|g(t)\|^2dt =\sum_{n\in\mathbb{Z}}\|c_n(g)\|^2. \end{gather} Let $p$ be any positive integer and $\mathcal{A}_p:=\{n\in\mathbb{Z}: 2^{p-1}<|n|\le 2^p\}$. For $\rho:=\frac{1}{2^{p+2}}$ and $n\in\mathcal{A}_p$ one has $\frac{\pi}{4}<2|n|\pi\rho\le\frac{\pi}{2}$ and then \begin{equation}\label{sineest} 2<4\sin^2(2n\pi\rho).\end{equation} Thus, since $\text{card}(\mathcal{A}_p)=2^{p-1}$, the usual H\"older inequality yields \begin{equation}\label{CSest} \Big(\sum_{n\in\mathcal{A}_p}\|c_n({f})\|\Big)^2 \le 2^{p-1}\sum_{n\in\mathcal{A}_p}\|c_n({f})\|^2. \end{equation} Now \eqref{concoef}, \eqref{normest}, \eqref{sineest} and \eqref{CSest} yield \begin{equation} 2\Big(\sum_{n\in\mathcal{A}_p}\|c_n({f})\|\Big)^2 \le 2^{p-1}\sum_{n\in\mathcal{A}_p}\|c_n({g})\|^2 \le 2^{p-1}\rho^{2\alpha}2^{2\alpha}L^2. \end{equation} Thus \begin{equation}\label{finest} \sum_{p=1}^\infty\sum_{n\in\mathcal{A}_p}\|c_n({f})\| \le 2^{-1}L\sum_{p=1}^\infty2^{(\frac{1}{2}-\alpha)p}=LK(\alpha). \end{equation} Finally we obtain the estimate $ \|f\|_1\le 3\|f\|_{\infty}+LK(\alpha)$. \end{proof} \begin{lemma}[\cite{BDL}]\label{lessrad} Let $T$ be a bounded linear operator acting on a complex Banach space $X$. If for every real number $\mu$ one has \begin{equation}\label{autonomest} \sup_{n\in\mathbb{Z}_+}\|\sum_{j=1}^n e^{-i\mu j}T^j\|:=M(\mu)<\infty, \end{equation} then the spectral radius of $T$ is less than $1$. \end{lemma} \section{Main results} The following result was stated in \cite[[Theorem 2.1]{BLNS}. \begin{theorem}\label{mr0} Let $\mathcal{U}=\{U(t, s): t\ge s\}$ be a strongly continuous and $1$- periodic evolution family of bounded linear operators acting on a complex Hilbert space $\mathcal{H}$. Assume that: \begin{itemize} \item[(1)] For every $x\in \mathcal{H}$ the trajectory $U(\cdot, 0)x$ is Lipschitz continuous on $(0, 1)$, and \item[(2)] \begin{equation}\label{mainest} \sup_{\mu\in\mathbb{R}}\sup_{t\ge s\ge 0} \|\int_\tau^te^{i\mu s}U(t, \tau)x d\tau\|:=M(x)<\infty. \end{equation} \end{itemize} Then the family $\mathcal{U}$ is uniformly exponentially stable. \end{theorem} In our first result (using a different approach) we show that both assumptions in Theorem \ref{mr0} can be relaxed. \begin{theorem}\label{mr1} Let $\mathcal{U}=\{U(t, s): t\ge s\}$ be a strongly continuous and $1$-periodic evolution family of bounded linear operators acting on a complex Hilbert space $\mathcal{H}$. Assume that for some $\alpha>\frac{1}{2}$ and every $x\in \mathcal{H}$ the trajectory $U(\cdot, 0)x$ is $\alpha$-H\"older continuous on $(0, 1)$ with the constant $L(x)$ and \begin{equation}\label{mainestb} \sup_{\mu\in\mathbb{R}}\sup_{t\ge 0}\|\int_0^te^{i\mu s}U(t, s)x ds\|:=M(x)<\infty. \end{equation} Then the family $\mathcal{U}$ is uniformly exponentially stable. \end{theorem} \begin{proof} From the Uniform Boundedness Principle there exists a positive constant $M$ such that $M(x)\le M\|x\|<\infty$. We use again the notations in Lemma \ref{replem}. The map $\tilde{H}_{xh}$, given in \eqref{themap}, can be represented as in \eqref{develop} and in view of Lemma \ref{reshold} and then of Lemma \ref{Fourier} it belongs to $CP_1^1(\mathbb{R}, X)$. Now for any positive integer $n$, using \eqref{intest}, \eqref{develop}, \eqref{finest}, \eqref{consttil} and \eqref{mainest}, we obtain \begin{align*} \| \sum_{j=1}^n e^{-i\mu j}T^jx\| &\le\frac{1}{|I(h, \mu)|}\|\int_0^n e^{i\mu s}U(n, s)\tilde{H}_{xh}(s)ds\|\\ &\le\frac{1}{|I(h, \mu)|}\sum_{k\in\mathbb{Z}} \|\int_0^n e^{i(\mu-2k\pi) s}U(n, s)c_k(\tilde{H}_{xh})ds\| \\ &\le\frac{M}{|I(h, \mu)|}\|\tilde{H}_{xh}\|_1 \\ &\le\frac{M}{|I(h, \mu)|}(3\|1_{[0, 1]}(\cdot) U(\cdot, 0)x\|_{\infty} +K(\alpha)L(\tilde{H}_{xh})), \end{align*} where $L(\tilde{H}_{xh}))=\pi\|1_{[0, 1]}(\cdot)U(\cdot, 0)x\|_{\infty}+L(x)$. The assertion follows using Lemma \ref{lessrad}. \end{proof} The following particular case of Theorem \ref{mr1} is an extension of \cite[Theorem 3.2]{ABS}] where only $\mathbb{C}^n$-valued functions was considered. \begin{corollary}\label{difcor} Let $\mathcal{U}=\{U(t, s): t\ge s\}$ be a strongly continuous and $1$-periodic evolution family of bounded linear operators acting on a complex Hilbert space $\mathcal{H}$. Assume that for every $x\in\mathcal{H}$ the map $U(\cdot, 0)x$ is differentiable on $(0, 1)$ and there exist a family of bounded linear operators $\{A(t): t\in [0, 1]\}$ that acts on $\mathcal{H}$ such that the map $A(\cdot)U(\cdot, 0)x$ is bounded on $(0, 1)$ and \begin{equation}\label{difcond} \frac{d}{dt}[U(t, 0)x]=A(t)U(t, 0)x,\quad\forall t\in(0, 1),\quad \forall x\in\mathcal{H}. \end{equation} Then \eqref{mainest} holds true if and only if the evolution family $\mathcal{U}$ is uniformly exponentially stable. \end{corollary} \begin{proof} From \eqref{difcond} follows that the map $A(\cdot)U(\cdot, 0)x$ is measurable. Thus for $t, s\in(0, 1)$ one has \begin{align*} \|U(t, 0)x-U(s, 0)x\| &=\|\int_s^t A(\tau)U(\tau, 0)xd\tau\|\\ &\le |t-s|\|1_{[0, 1]}(\cdot)A(\cdot)U(\cdot, 0)x\|_{\infty}, \end{align*} which shows that $U(\cdot, 0)x$ is $1$-H\"older continuous on $[0, 1]$. Now we can apply Theorem \ref{mr1} to complete the proof. \end{proof} \begin{remark} \rm Assume that the map $t\mapsto A(t)$ is continuous in the uniform operator topology of $\mathcal{L}(\mathcal{H})$ and that $A(0)=A(1)$ and let $\tilde{A}(\cdot)$ be the extension by periodicity of $A(\cdot)$ to the entire real axis. We can associate with the operator family $\{\tilde{A}(t): t\in\mathbb{R}\}$ an evolution family $\mathcal{U}$ which satisfies all the assumptions in Corollary \ref{difcor}. \end{remark} We also refer the reader to \cite[Theorem 5.1]{Pa}]. The next example considers the well-posedness in Corollary \ref{difcor}. A nice text concerning well-posedness which addresses in particular the framework of this paper provided by Schnaubelt is \cite[pages 477-496]{EN}, see also \cite{Da}. \begin{example} \rm Let $(\mathcal{H}, \langle , \rangle)$ be a separable complex Hilbert space and let $B=\{b_n, n=1, 2,\dots\}$ be an orthonormal basis of it, i.e.\ $\langle b_n, b_m\rangle=0$ when $m\neq n$, $\|b_n\|^2:=\langle b_n, b_n\rangle=1$ and the linear span of $B$ is dense in $\mathcal{H}$. Thus any $x\in\mathcal{H}$ can be represented uniquely as $x=\sum_{n=1}^{\infty}\langle x, b_n\rangle b_n$. Let $(\lambda_n)$ be a sequence of real numbers with $\lambda_n\le -1$ for every positive integer $n$. For every $t\ge 0$ and every $x\in\mathcal{H}$ let \begin{equation} T(t)x:=\sum_{n=1}^{\infty}e^{\lambda_n t}\langle x, b_n\rangle b_n. \end{equation} It is well-known, \cite{Za}, and easy to prove that: \begin{itemize} \item[(1)] The family $\mathbf{T}:=\{T(t): t\ge 0\}$ is a strongly continuous semigroup that acts on $\mathcal{H}$. \item[(2)] $\omega_0(\mathbf{T})=\sup_{n\ge 1}\lambda_n\le -1$. \item[(3)] $D(A)=\{x\in\mathcal{H}: \sum_{n=1}^{\infty}|\lambda_n|^2|\langle x, b_n\rangle|^2<\infty\}$ \item[(4)] $Ax:=\sum_{n=1}^{\infty}\lambda_n\langle x, b_n\rangle b_n$ for all $x\in D(A)$. \item[(5)] The resolvent set of $A$, i. e. $\rho(A)$ consist by all complex numbers $z$ for which $\inf_{n\ge 1}|z-\lambda_n|>0$. In particular, $\{\lambda\in\mathbb{C}: \Re(\lambda)\ge 0\}\subset \rho(A)$. \item[(6)] $R(\lambda, A)=\sum_{n=1}^{\infty}\frac{1}{\lambda-\lambda_n} \langle x, b_n\rangle b_n$, for all $\lambda\in\rho(A)$. Clearly, \begin{equation}\|R(\lambda, A)\|\le\frac{2}{|\lambda|+1}\quad \forall \lambda\in\mathbb{C},\text{ with } \Re(\lambda)\ge 0. \end{equation} \end{itemize} Let $a:\mathbb{R}\to [1, \infty)$ be a $1$-periodic function having the property that there exists two constants $c>0$ and $\alpha\in(0, 1]$ such that \begin{equation}\label{hca} |a(t)-a(s)|\le c|t-s|^{\alpha}, \quad \forall t, s\in[0, 1]. \end{equation} Set $A(t):=a(t)A$ for all $t\in[0, 1]$. Clearly $D(A(t))=D(A)$ for all $t\in[0, 1]$ and \begin{equation} \|R(\lambda, A(t))\|\le\frac{2}{|\lambda|+1}\quad \forall \lambda\in\mathbb{C},\text{ with } \Re(\lambda)\ge 0. \end{equation} Further, in view of \eqref{hca} one has \begin{equation} \|(A(t)-A(s))A^{-1}(\tau)\| =\frac{1}{a(\tau)}|a(t)-a(s)|\le c|t-s|^{\alpha}, \quad \forall t, s\in[0, 1]. \end{equation} Thus, the family \begin{equation}\label{famin} U(t, s)x:=T\Big(\int_s^ta(\tau)d\tau\Big)x,\quad 0\le s\le t\le 1, x\in\mathcal{H} \end{equation} solves the non-autonomous Abstract Cauchy Problem \begin{equation} u'(t)=A(t)u,\quad 0\le s0$ be such that $N(\mu, f)\le N\|f\|_{\infty}$. For any $t\ge 0$, \eqref{estevol} yields \begin{equation} \big\|\Big(\int_0^te^{i\mu s}\mathcal{T}(t-s)fds\Big)(t)\big\| =\big\|\int_0^te^{i\mu s}U(t, s)f(s)ds\big\| \le N\|f\|_{\infty}. \end{equation} Now, writing the latter inequality for $t=n\in\mathbb{Z}_+$ and with $f$ replaced by $\tilde{H}_{xh}$ (the extension by periodicity to the entire axis of the map $H_{xh}$ defined in Lemma \ref{replem}) and taking into account \eqref{intest} we obtain \begin{align*} \| \sum_{j=1}^n e^{-i\mu j}T^jx\| &\le\frac{1}{|I(h, \mu)|}\|\int_0^n e^{i\mu s}U(n, s)\tilde{H}_{xh}(s)ds\| \\ &\le\frac{N}{|I(h, \mu)|}\|1_{[0, 1]}(\cdot)U(\cdot, 0)x\|_{\infty}. \end{align*} Since the above estimates hold for all $x\in X$ we can apply Lemma \ref{lessrad} to finish the proof. \end{proof} \section{Discrete case} Let $q\ge 2$ be a integer number, $X$ be a complex Banach space and let $\mathcal{U}=\{U(n, m): n\ge m\ge 0\}\subset\mathcal{L}(X)$ be a $q$-periodic discrete evolution family on $X$, that is $U(n, n)x=x, U(m, n)U(n, r)=U(m, r)$ and $U(n+q, m+q)=U(n, m)$ hold for all nonnegative integers $m, n, r$ with $m\ge n\ge r$ and all $x\in X$. We denote by $T_q:=U(q, 0)$ the monodromy operator associated to the evolution family $\mathcal{U}$. As is well known the family $\mathcal{U}$ is uniformly exponentially stable, that is, there exists the positive constants $N$ and $\nu$ such that $$ \|U(n, m)\|_{\mathcal{L}(X)}\le Ne^{-\nu(n-m)} \quad \text{for all } n\ge m $$ if and only if the spectral radius of $T_q$ $$ r(T_q):=\lim_{k\to\infty}\|T_q^kx\|^{\frac{1}{k}}, $$ is less than $1$. \begin{theorem} Let $T$ be a bounded linear operator acting on a Banach space $X$. If for a given $x\in X$ one has \begin{equation}\label{boundsing} \sup_{\mu\in\mathbb{R}}\sup_{n\in\mathbb{Z}_+} \|\sum_{k=0}^ne^{i\mu k}T^{n-k}x\|=:K(x)<\infty, \end{equation} then \begin{equation}\label{indasym} T^nx\to 0\quad \text{as } n\to\infty. \end{equation} \end{theorem} \begin{proof} Consider the sequence of holomorphic functions $$ h_n^x(z):=\sum_{j=1}^n \frac{T^jx}{z^{j+1}}, \quad |z|>\frac{1}{2}. $$ From the classical Abel-Dirichlet Theorem in Banach spaces, the sequence $(h_n^x(z))$ converges for each $z\in\mathbb{C}$ with $|z|>1$. Moreover, when $|z|>\|A\|+1$ we have that $h_n^x(z)\to (zI-T)^{-1}x$ as $n\to\infty$ and $\|h_n^x(z)\|\le \|T\|\|x\|$ for every $n\in\mathbb{Z}_+$. Then from the Phragmen-Lindel\"of Theorem the sequence $(h_n^x(z))$ is uniformly bounded on $\{|z|\ge 1\}$ with the bound $\max\{K(x), \|A\|\|x\|\}$. The Vitali theorem (\cite[Theorem 3.1]{Ar}, \cite[Theorem 3.14.1]{HP}) assures us that $(h_n^x(z))_n$ converges for each $z$ with $|z|\ge 1$. Thus $T^jx\to 0$ as $j\to\infty$ as well. \end{proof} \begin{remark} \rm (1) If in addition $T$ is power bounded, i.e., $\sup_{n\in\mathbb{Z}_+}\|T^n\|<\infty$, and \eqref{boundsing} is satisfied for $x$ in a dense linear subspace $D$ of $X$ then $T$ is strongly stable, i.e.\ \eqref{indasym} holds for all $x$ in $X$. (2) Let $b\in X$ be a given vector and let $\mathcal{T}(b)$ be the trajectory of $T$ generated of $b$, i.e.\ the set $\{T^nb: n\in\mathbb{Z}_+\}$. With $X_b$ we denote the smallest closed subspace of $X$ which contains $\mathcal{T}(b)$. If for each $\tilde{b}\in X_b$ one has \begin{equation}\label{boundsing1} \sup_{n\in\mathbb{Z}_+}\|\sum_{k=0}^ne^{i\mu k}T^{n-k}\tilde{b}\| :=K(\tilde{b})<\infty, \end{equation} then every trajectory of $T$ starting from $X_b$ is exponentially stable, i.e.\ for every $x\in X_b$ there exist $\nu>0$ and $N>0$ such that $$ \|T^nx\|\le Ne^{-\nu n}\|x\|. $$ Indeed $X_b$ is an invariant subspace of $T$ (i.e.\ $T(X_b)\subseteq X_b)$ and the assertion follows by applying Lemma \ref{lessrad} to the restriction of the operator $T$ to $X_b$. \end{remark} \begin{theorem} Let $q\ge 2$ be an integer, $\mathcal{H}$ be a complex Hilbert space and let $\mathcal{U}=\{U(n, m): n\ge m\ge 0\}$ be a $q$-periodic evolution family of bounded linear operators acting on $\mathcal{H}$. For a given nonzero vector $b\in\mathcal{H}$, let \begin{equation} \mathcal{H}_b:=\overline{\operatorname{span}}\{ b, U(1, 0)b, U(2, 0)b, \dots, U(q, 0)b\}. \end{equation} If $\mathcal{H}_b$ is an invariant subspace for all operators $U(1, 0), U(2, 0), \dots, U(q, 0)$ and if there exists an absolute positive constant $K$ such that \begin{equation}\label{uniformcond} \sup_{\mu\in\mathbb{R}}\sup_{n\in\mathbb{Z}_+} \|\sum_{k=0}^ne^{i\mu k}U(n, k)v\|\le K\|v\|\quad\forall v\in \mathcal{H}_b, \end{equation} then for every $\tilde{b}\in \mathcal{H}_b$, the trajectory $U(\cdot, 0)\tilde{b}$ is uniformly exponentially stable, that is there exist the positive constants $N$ and $\nu$ such that $\|U(n, 0)\tilde{b}\|\le N^{-\nu n}\|\tilde{b}\|$ for all integers $n$. \end{theorem} \begin{proof} \textbf{Step 1.} We prove that for every $(\mathcal{H}_b)$-valued, $q$-periodic sequence $w=(w_j)$, with $w_0=0$, one has \begin{equation}\label{seqcond} \sup_{\mu\in\mathbb{R}}\sup_{n\in\mathbb{Z}_+} \|\sum_{k=0}^ne^{i\mu k}U(n, k)w_k\|:=M(w)<\infty. \end{equation} Let $f$ be the continuous function verifying $f(j)=w_j$ for every $j=0, 1, \dots, q-1, q$ and such that $f$ is linear on each of the intervals $[0, 1], [1, 2],\dots, [q-1, q]$. The function is $(\mathcal{H}_b)$-valued since it is piecewise linear and satisfies a Lipschitz condition on $[0, q]$. We denote by $\tilde{f}$ its extension by periodicity to the entire real axis which satisfies a Lipschtiz condition on $\mathbb{R}$, as well. Taking into account that $\mathcal{H}_b$ is a complex Hilbert space and arguing as in Lemma \ref{Fourier}, we obtain \begin{equation}\label{fqdevelop} w_j=\sum_{n\in\mathbb{Z}}e^{2\pi in\frac{j}{q}}c_n(f), \end{equation} where the Fourier coefficients associated to $f$, given by \begin{equation} c_n(f):=\frac{1}{q}\int_0^qe^{-2i\pi n\frac{t}{q}}f(t)dt, \end{equation} belong to $\mathcal{H}_b$. In addition, one has $$ \sum_{n\in\mathbb{Z}}\|c_n(f)\|:=c(w)<\infty. $$ Now, \eqref{uniformcond} and \eqref{fqdevelop} yield \begin{align*} &\sup_{\mu\in\mathbb{R}}\sup_{n\in\mathbb{Z}_+} \|\sum_{k=0}^ne^{i\mu k}U(n, k)w_k\| \\ &=\sup_{\mu\in\mathbb{R}}\sup_{n\in\mathbb{Z}_+} \|\sum_{k=0}^ne^{i\mu k}U(n, k)\sum_{j\in\mathbb{Z}}e^{2\pi ij\frac{k}{q}}c_j(f)\| \le Kc(w). \end{align*} Hence \eqref{seqcond} is satisfied with $M(w):=Kc(w)$. \smallskip \noindent\textbf{Step 2.} We show that all the hypothesis in Lemma \ref{lessrad} are fulfilled when $\mathcal{H}_b$ replaces $X$. Indeed, let $v\in \mathcal{H}_b$ be randomly chosen and let $w$ be the $q$-periodic sequence defined by $$ w_k:=k(q-k)U(k, 0)v, \quad k=0, 1, \dots, q-1. $$ By assumption $(w_k)$ is an $(\mathcal{H}_b)$-valued sequence. As in \cite[Theorem 2]{BCDS} we can show that \eqref{autonomest} holds with $(\mathcal{H}_b)$ instead of $X$ and $T_q$ instead of $T$ (The fact that $(\mathcal{H}_b)$ is an invariant subspace of $T_q$ was used as well). Thus from Lemma \ref{lessrad}, the spectral radius of the restriction of $T_q$ to $(\mathcal{H}_b)$ is less than $1$ and we obtain the conclusion. \end{proof} The following example shows that the uniformity of boundedness in respect to the parameter $\mu$ in \eqref{uniformcond} cannot be removed. \begin{example} \rm Let $\mathcal{U}$ be a $2$-periodic evolution family on $X=\mathbb{C}^2$ (endowed with the usual inner product) given by \begin{equation} U(1, 0)= \begin{pmatrix} 1/2 & 0 \\ 0 & -1 \end{pmatrix}, \quad U(2, 0)= \begin{pmatrix} 1/2 & 0 \\ 0 & 1 \end{pmatrix}, \quad U(2, 1)= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \end{equation} \end{example} Set $b=(0, 1)^T$. Thus $U(1, 0)b=-b$ and $U(2, 0)b=b$ and $\mathcal{H}_b=\{\lambda b, \lambda\in\mathbb{C}\}$ is an invariant subspace for the operators $U(1, 0)$ and $T_2:=U(2, 0)$. Moreover, for each real number $\mu$ one has \begin{equation}\label{evenest} \sup_{n\in\mathbb{Z}_+}\|\sum_{k=0}^{2n}e^{i\mu k}U(2n, k)b\| =\sup_{n\in\mathbb{Z}_+}\|\sum_{k=0}^{n}e^{2i\mu k}T_q^{n-k}(U(2, 1) +e^{i\mu}b)\|<\infty. \end{equation} See \cite[Example 1]{ABNZ} for further details. On the other hand \begin{equation} \sum_{k=0}^{2n+1}e^{i\mu k}U(2n+1, k)b =e^{i\mu(2n+1)}+U(1, 0)\sum_{k=0}^{2n}e^{i\mu k}U(2n, k)\cdot b. \end{equation} Hence, in view of \eqref{evenest} the sequence \begin{equation} \Big(\sum_{k=0}^{2n+1}e^{i\mu k}U(2n+1, k)b\Big)_n \end{equation} is bounded as well. Finally for a certain $0