q\|x\|\mbox{ for all } T\ge\tau\ge 0. \label{e7} \end{equation} \end{lemma} \begin{lemma} \label{lm2.6} Under the hypotheses of Theorem \ref{thm2.4}, it follows that there is a positive constant $K$ such that \begin{equation} \sup_{s\ge 0}|V_s^x|_E\le K\|x\|\mbox{ for all } x\in X. \label{e8} \end{equation} \end{lemma} \begin{proof} For each $s\ge 0$ let us consider the linear and bounded operator $V_s:X\to E(\mathbb{R}_+, X)$ given by $$ (V_sx)(t):=U(s+t, s)x, \quad t\in\mathbb{R}_+,\; x\in X. $$ Then for each $x\in X$, we have $$ |V_sx|_{E(\mathbb{R}_+, X)}=| \|U(s+\cdot, s)\| |_E=|V_s^x|_E\le K(x). $$ The assertion of Lemma \ref{lm2.6} follows by the Uniform Boundedness Principle applied to the family $\mathcal{V}:=\{V_s: s\ge 0\}$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.4}] Suppose that $\mathcal{U}$ is not uniformly exponentially stable. Then from \eqref{e7} and \eqref{e8} follows that $$ K\ge q|\chi_{[0, T]}|_E $$ for all positive real number $T$, which is a contradiction. \end{proof} To the best of our knowledge the result in Theorem \ref{thm2.4} is new and generalizes to the non-autonomous case some recently obtained autonomous or periodic versions in literature; see (\cite[Theorem 4.2]{[Ne95]}) or (\cite[Theorem 4.5]{[BP01]}). Using the method developed by Schnaubelt (\cite{[Sch00]}), see also \cite{[CL99]}, we can prove the following generalization of the $L^1$-version of Datko theorem. \begin{theorem} \label{thm2.7} Let $\mathcal{U}:=\{U(t, s): t\ge s\ge 0\}$ be an evolution family with exponential growth on a Banach space $X$. We suppose that for each $x\in X$ the map $$ (t, s)\mapsto U(t, s)x:\{(t, s): t\ge s\ge 0\} $$ is measurable. Then $\mathcal{U}$ is uniformly exponentially stable if and only if \begin{equation} \sup_{s\ge 0}\int_s^\infty\|U(t, s)x\|dt<\infty \label{e9} \end{equation} for all $x\in X$. \end{theorem} \begin{proof} As in the proof of Lemma \ref{lm2.6}, there exists a positive constant $K$, (independent of $x$ and $s$), such that \begin{equation} \|U_s^x\|_{L^1(\mathbb{R}_+)}\le K\|x\|. \label{e10} \end{equation} Let us consider the evolution semigroup $\mathbf{T}=\{T(t)\}_{t\ge 0}$ associated with $\mathcal{U}$ on $L^1(\mathbb{R}_+, X)$. Recall that for each $t\ge 0$ and each $f\in L^1(\mathbb{R}_+, X)$ the map $T(t)f$ is given by $$ (T(t)f)(s)=\begin{cases} U(s, s-t)f(s-t),& s\ge t\\ 0,& 0\le sFrom the hypothesis on the measurability and using the fact that the evolution family $\mathcal{U}$ has exponential growth it follows that the map $T(t)f$ belongs to $L^1(\mathbb{R}_+, X)$ for all $t\ge 0$ and all $f\in L^1(\mathbb{R}_+, X)$. Moreover it is easy to see that the evolution semigroup $\mathbf{T}$ has exponential growth. Thus for each $f\in L^1(\mathbb{R}_+, X)$, the map $t\mapsto \|T(t)f\|_{L^1(\mathbb{R}_+, X)}$ is measurable, see e.g. (\cite[Remark 4.3]{[Ne95]}). From \eqref{e10} using the Fubini theorem follows \begin{align*} \int_0^{\infty} \|T(t)f\|_{L^1(\mathbb{R}_+, X)}dt &=\int_0^\infty\int_0^\infty\chi_{[t, \infty)}(s)\|U(s, s-t)f(s-t)\|\,ds\,dt\\ &=\int_0^\infty\int_0^s \|U(s, \xi)f(\xi)\|\,d\xi \,ds\\ &=\int_0^\infty\int_0^\infty\chi_{[0, s]}(\xi)\|U(s, \xi)f(\xi)\|\,ds\,d\xi\\ &=\int_0^\infty\int_\xi^\infty\|U(s, \xi)f(\xi)\|\,ds\,d\xi\\ &\le K\|f\|_{L^1(\mathbb{R}_+, X)}. \end{align*} Now we apply the Datko-Pazy theorem for $p=1$ (see the beginning of our paper) and use the well-known fact that if the semigoup $\mathbf{T}$ is exponentially stable then the evolution family $\mathcal{U}$ is uniformly exponentially stable as well, see \cite[Theorem 2.2]{[CLMR]}. \end{proof} \begin{remark} \label{rmk2.8} \rm (1) The result contained in the above theorem may be known. It follows, for example, from (\cite[Corollary 3.2]{[BD02]}), for $\phi(t)=t$, $t\ge 0$. However, the main hypothesis of this Corollary is the boundedness of the function $(s, x)\mapsto \int_s^\infty\phi(\|U(t, s)x\|)dt$ on $\mathbb{R}_+\times \overline{B}(0, 1)$, where $\overline{B}(0, 1)$ is the closed unit ball in $X$ and $\phi$ is a nondecreasing function such that $\phi(t)>0$ for every $t>0$, which seems to be a more strongly require than the similar one from Theorem \ref{thm2.7}. \noindent(2) The result stated in Theorem \ref{thm2.7} holds under the general hypothesis that for each $x\in X$ and some real-valued, strictly increasing (or nondecreasing and positive on $(0, \infty)$) and convex function $\Phi$ on $\mathbb{R}_+$, one has \begin{equation} \sup_{s\ge 0}\int_s^\infty \Phi(\|U(t, s)x\|)dt<\infty. \label{e10b} \end{equation} \end{remark} \begin{proof}[Proof of 2] For every $k=1, 2, 3, \cdots$ let us consider the set $$ X_k=\big\{ x\in X: \sup_{s\ge 0}\int_s^\infty \Phi(\|U(t, s)x\|)\le k. \big\} $$ By the assumption \eqref{e10} follows that $X=\cup_{k\ge 1}X_k$. Using the well-known Fatou Lemma it is easily to see that each $X_k$ is closed. Then there is a natural number $k_0$ such that $X_{k_0}$ has nonempty interior. Let $x_0\in X$ and $\delta>0$ such that $X_{k_0}$ contains the open ball with the centre in $x_0$ and radius $\delta$. We will prove that the open ball which the centre in origin and radius $\frac{\delta}{2}$ is also contained in $X_{k_0}$. Indeed for each positive $s$ and each $x\in X$ with $\|x\|\le \delta$, one has \begin{align*} \int_0^\infty \Phi(\|U(t, s)(\frac{1}{2}x)\|)dt &\le \int_s^\infty\Phi(\frac{\|U(t, s)(x+x_0)\|+\|U(t, s)x_0\|}{2})dt\\ &\le \frac{1}{2}(\int_s^\infty\Phi(\|U(t, s)(x+x_0)\|)dt +\int_s^\infty\Phi(\|U(t, s)x_0\|)dt)\\ &\le k_0. \end{align*} Now we can apply \cite[Corollary 3.2]{[BD02]}. We remark that in this proof only the strong measurability of the maps $t\mapsto U(t, s)$ $( s\ge 0, t\ge s)$ were used. The ``if" part can be obtained in the following way. Upon replacing $\Phi$ be a some multiple of itself we may assume that $\Phi(1)=1$. It is clear that $\Phi(0)=0$. Let $N$ and $\nu$ two positive constants such that $$ \|U(t, s)\|\le Ne^{-\nu(t-s)}\quad \mbox{for all } t\ge s\ge 0. $$ Then for a sufficiently large and positive $h$, (independent of s), we have $$ \int_s^\infty\Phi(\|U(t, s)x\|)\le \int_0^h\Phi(Ne^{-\nu u})du+\int_h^\infty Ne^{-\nu u}du<\infty. $$ Finally we remark that the result holds even if the set of all $x\in X$ for which \eqref{e10} holds is a second category in $X$. \end{proof} Another result of this type can be formulate as follows. \begin{theorem} \label{thm2.9} Let $E$ be a solid Banach function space over $\mathbb{R}_+$ which satisfies (H1) and $\mathcal{U}$ be an evolution family such that for each positive $s$ the map $t\mapsto U(t, s)$ is strongly measurable on $[s, \infty)$. If the norm of $E$ has the Fatou property \cite{[Ne02]} and if the set of all $x\in X$ for which \begin{equation} \sup_{s\ge 0}| \|U(\cdot+s, s)x\| |_E<\infty \label{e11} \end{equation} is of the second category then $\mathcal{U}$ is uniformly exponentially stable. \end{theorem} \begin{proof} As above, (see also \cite{[Ne02]} for the semigroup case), using the triangle inequality in the space $E$ instead of convexity it follows that \eqref{e11} holds for every $x\in X$. Then we apply Theorem \ref{thm2.4} above to complete the proof. \end{proof} The following result shows that the hypothesis on the convexity of $\Phi$ from Remark 2.8 may be removed. However the converse statement of the Theorem \ref{thm2.10} below does not hold without the convexity of $\Phi$, see \cite[Example 8.12.1]{[Rol87]}. \begin{theorem} \label{thm2.10} Let $\phi: \mathbb{R}_+\to \mathbb{R}_+$ be a nondecreasing function such that $\phi(t)>0$ for all $t>0$ and $\mathcal{U}=\{U(t, s)\}_{t\ge s}$ be an evolution family such that for each $s\ge 0$ the map $t\mapsto U(t, s)$ is strongly measurable. If the set of all $x\in X$ for which \begin{equation} M_{\phi}(x):=\sup_{s\ge 0}\int_s^\infty \phi(\|U(t, s)x\|)dt<\infty \label{e12} \end{equation} is of second category in $X$ then $\mathcal{U}$ is uniformly exponentially stable. \end{theorem} \begin{proof} First we prove that the family $\mathcal{U}$ is uniformly bounded. Indeed for each $x\in X$ satisfying \eqref{e12} there exists a real number $C(x)$ such that \begin{equation} \sup_{t\ge s\ge 0}\|U(t, s)x\|\le C(x), \label{e13} \end{equation} see \cite[Lemma1]{[BD01]}. It is clear that \eqref{e13} holds for every $x\in X$, because it holds for each $x$ in a set of second category in $X$. Then we apply the Uniform Boundedness Theorem to obtain the uniform boundedness of $\mathcal{U}$. On the other hand \eqref{e12} can be written as \begin{equation} M_{\phi}(x)=\sup_{s\ge 0}\int_0^\infty \phi(\|U(t+s, s)x\|)dt<\infty. \label{e14} \end{equation} From \cite[Lemma 3.2.1]{[Ne96]} follows that there exists an Orlicz's space $E$ which satisfies (H1) and such that for each $x$ which satisfies \eqref{e14}, the map $t\mapsto \|U(t+s, s)x\|$ belongs to $E$. Using \eqref{e14} we can derive \eqref{e11}. Now we apply Theorem \ref{thm2.9} to complete the proof. \end{proof} We conclude by stating another related result. \begin{proposition} \label{prop2.11} Let $\mathcal{U}=\{U(t, s): t\ge s\ge 0\}$ be an evolution family with exponential growth on a Banach space $X$ and $x\in X$ be fixed. If for each $s\ge 0$, the map $U_s^x$ (or the map $V_s^x$) belongs to a rearrangement invariant solid space $E$ which verifies the hypothesis (H1), then the trajectory $U(s+\cdot, s)x$ of the evolution family $\mathcal{U}$ is asymptotically stable, that is, for each $s\ge 0$, one has: $$ \lim_{t\to\infty} U(s+t, s)x=0.$$ \end{proposition} The proof of this proposition follows the arguments in \cite[Theorem 2.1]{[B96]}, and we omit it. \subsection*{Acknowlegment} The authors would like to thank Professor Yuri Latushkin for his idea in the proof of Lemma \ref{lm2.6}. \begin{thebibliography}{99} \bibitem{[B59]} I. Barb\u alat, Systems d'equations d'oscilations non-linearis, {\it Rev. Roumaine Math. Pures Appl.} iv, {\bf 2}(1959), 267--270. \bibitem{[BS88]} C. Bennett and R. Sharpley, {\it Interpolation of Operators}, Pure Appl. Math., Vol. {\bf 129}, (1988). \bibitem{[B96]} C. Bu\c se, Nonuniform exponential stability and Orlicz functions, {\it Comm. Math. 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