\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 218, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/218\hfil Weak Rolewicz's theorem]
{Weak Rolewicz's theorem in Hilbert spaces}

\author[C. Bu\c se, G. Rahmat \hfil EJDE-2012/218\hfilneg]
{Constantin Bu\c se, Gul Rahmat}  % in alphabetical order

\address{Constantin Bu\c se \newline
West University of Timisoara, Department of Mathematics,
Bd. V. Parvan No. 4, 300223-Timisoara, Rom\^ania}
\email{buse@math.uvt.ro}

\address{Gul Rahmat \newline
Government College University,
Abdus Salam School of Mathematical Sciences, Lahore, Pakistan}
\email{gulassms@gmail.com}

\thanks{Submitted October 3, 2012. Published November 29, 2012.}
\subjclass[2000]{47A30, 46A30}
\keywords{Uniform exponential stability; Rolewicz's type theorems;
\hfill\break\indent weak integral stability boundedness}

\begin{abstract}
 Let $\phi:\mathbb{R}_+:=[0, \infty)\to \mathbb{R}_+$ be a nondecreasing
 function which is positive on $(0, \infty)$ and let
 $\mathcal{U} =\{U(t, s)\}_{t\ge s\ge 0}$ be a positive strongly continuous
 periodic evolution family of bounded linear operators acting on a
 complex Hilbert space $H$. We prove that $\mathcal{U}$ is uniformly
 exponentially stable if for each unit vector $x\in H$, one has
 $$
 \int_0^\infty \phi(|\langle U(t, 0)x, x\rangle|)dt<\infty.
 $$
 The result seems to be new and it generalizes others of the same topic.
 Moreover, the proof is surprisingly  simple.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks


\section{Introduction}

The classical theorem of  Datko \cite{Da} states that a strongly
continuous semigroup $\mathbf{T}=\{T(t)\}_{t\ge 0}$,
acting on a real or complex Banach space $X$, is uniformly exponentially
 stable; i.e., there are two positive constants $N$ and $\nu$ such that
$$
\|T(t)\|\le Ne^{-\nu t}, \quad \forall t\ge 0,
$$
if and only if
\begin{equation}
\int_0^\infty \|T(t)x\|^2dt<\infty,\quad\forall x\in X.\label{e1}
\end{equation}
Obviously, for strongly continuous selfadjoint semigroups acting on a
Hilbert space $H$, the integral condition \eqref{e1} is equivalent to
\begin{equation}
\int_0^\infty |\langle T(t)x, x\rangle| dt<\infty,\quad\forall x\in H.\label{e2}
\end{equation}

In this article we prove a result of this type for positive periodic
evolution families, in a more general form, that is related to a theorem
of  Rolewicz.

The history of the Rolewicz theorem is well known among experts in the field.
However, we recall, in the following, a few facts to help readers compare
results.

The theorem of Datko has been extended by  Pazy, who  states that all
trajectories of a strongly continuous semigroup
$\mathbf{T}=\{T(t)\}_{t\geq 0}$, of bounded linear operators acting on
a Banach space $X$, belongs to the space
$L^p(\mathbb{R}_+, X)$ (for some, and then for all $p\geq 1$) if
and only if the semigroup $\mathbf{T}$ is uniformly exponentially
stable, or equivalently, its growth bound
$$
\omega_0(\mathbf{T})=\lim_{t\to\infty}\frac{\ln\|T(t)\|}{t},
$$ is
negative. See \cite{Pa1} and \cite{Pa} for further
details.

A real valued nondecreasing
function $\phi$, defined on $\mathbb{R}_+$, which is positive on
$(0, \infty)$, will be called (ad hoc) $\mathcal{R}$-function.

 An important generalization of Datko-Pazy's
theorem was given by  Rolewicz, \cite{Rol}. He  showed that
a strongly continuous semigroup $\mathbf{T}=\{T(t)\}_{t\ge 0}$ of bounded
linear operators acting on a Banach space $X$ is uniformly
exponentially stable, provided that for a given continuous,
$\mathcal{R}$-function $\phi$, one has
$$
\int_0^\infty \phi(\|T(t)x\|)dt<\infty,\quad\forall x\in X.
$$

Earlier, special cases of this result were obtained by  Zabczyk, \cite{Za}
and  Przyluski, \cite{Prz}.
Zheng and Littman subsequently provided new proofs for theorem of Rolewicz
 and, moreover, they removed the assumption of continuity of $\phi$.
See \cite{WL} and \cite{Zhe}.

Jan van Neerven emphasized a new method of demonstration for Rolewicz's
theorem using the theory of Orlicz spaces. See \cite{Nem}.


Recently, Storozhuk \cite{Sto1} has given a very short proof for
a Rolewicz's type theorem. See also \cite{BD1,BD2, BD3}
 for different approaches of the Rolewicz theorem.

Next, we introduce the ``weak" version (in the sense of the functional analysis)
of the Rolewicz theorem.

Let $X$ be a Banach space, $X^\ast$ be its dual,
and let $p\geq 1$ be a given real number. The semigroup
$\mathbf{T}=\{T(t)\}_{t\geq 0}$ is called \emph{weak-$L^p$-stable}
if
$$\int_0^\infty|\langle T(t)x, x^\ast\rangle |^pdt<\infty,
\quad  \text{for all } x\in X \text{ and all } x^\ast\in X^\ast.
$$
 The weak-$L^p$-stability of a semigroup
$\mathbf{T}$ does not imply the uniform exponential stability
of $\mathbf{T}$. See  \cite{GVW,NSW} for counterexamples.

In 1983,  Pritchard and  Zabczyk \cite{PZ} raised the following problem:
\begin{quote}
Is the uniform exponential stability of a semigroup
$\mathbf{T}=\{T(t)\}_{t\geq 0}$, acting on a Hilbert space,
a consequence of its weak-$L^p$-stability?
\end{quote}
The positive answer to the Pritchard and Zabczyk question was given
by Falun Huang. Further details could be found in
\cite{FH}. The general case has been treated independently by
 Weiss  \cite{GW}.

Also, it is known \cite{Nem} that a bounded strongly continuous
semigroup $\mathbf{T}=\{T(t)\}_{t\geq 0}$, of bounded linear operators
acting on a complex Hilbert space $H$, is uniformly exponentially stable,
if, for a given function $\varphi$ as above, one has
\begin{equation}
\int_0^\infty\varphi(|\langle T(t)x, y\rangle |)dt<\infty, \quad
\text{ for all } x, y\in H.\label{e3}
\end{equation}
 It is natural to ask whether the condition of boundedness of $\mathbf{T}$
 can be removed?
 A positive answer to this question was given recently by  Storozhuk.
In order to describe the Storozhuk theorem we briefly resume some
results targeting the same problem, but on  Banach spaces.


 We refer to \cite[Theorems 4.6.3(i), 4.6.4]{Nem}, for  results concerning
the exponential stability (rather than the uniform exponential stability)
of bounded linear semigroups acting on a Banach space.
 Recently  Storozhuk \cite{Sto} settled a more general problem.
He  stated that, if $s_0(\mathbf{T})\geq 0$,
then there exist an  $\mathcal{R}$-function $\phi$ and two vectors
$x_0, y_0\in H$, such that
$$
\int_0^\infty\phi(|\langle T(t)x_0, y_0\rangle |)dt=\infty.
$$
 More details on strongly continuous semigroups of operators,
including the precise definitions and characterizations of the
semigroups growth bounds $\omega_0(\mathbf{T})$ and $s_0(\mathbf{T})$,
can be found, for example, in the monographs
\cite{Pa,EN,ABHN}. In \cite{Ge} and \cite{Pr} it is shown
that $s_0(\mathbf{T})=\omega_0(\mathbf{T})$
for all strongly continuous semigroups $\mathbf{T}$ of bounded
linear operators acting on complex Hilbert spaces.

\section{Notation and preliminary results}

We denote by $\mathbb{R}$ the set of  real numbers and by
 $ \mathbb{C}$ the set of  complex numbers. Also we denote by
$\mathbb{Z}_+$ the set of  nonnegative integer numbers. As usual,
$\sigma(L)$ denotes the spectrum of the bounded linear operator acting
on a Banach space $X$. The spectral radius of $L$, denoted by $r(L)$, is given by
$$
r(L):=\sup\{|z|: z\in\sigma(L)\}=\lim_{n\to\infty}\|L^n\|^{1/n}.
$$
By $\mathcal{L}(X)$ we denote the Banach algebra of all bounded linear
operators acting on $X$. As usual, $\langle\cdot,\cdot\rangle$
denotes the scalar product on a Hilbert space $H$.
The norms in $X, H, \mathcal{L}(X),\mathcal{L}(H)$ will be denoted by
the same symbol, namely by $\|\cdot\|$.

We recall that a family
$\mathcal{U}:=\{U(t,s)\}_{t\geq s \geq 0}\subset\mathcal{L}(H)$
is called strongly continuous $q$-periodic evolution family
 (for some $q\ge 1$) if it satisfy the following conditions:
\begin{itemize}
\item[(i)] $U(t,t)=I$ for all $t\geq 0$.
\item[(ii)] $U(t,r)U(r,s)=U(t,s)$ for all $t\geq r \geq s\geq 0$.
\item[(iii)] $U(t+q,s+q)=U(t,s)$ for all $t\geq s\geq 0$.
\item[(iv)] The map $(t,s)\to U(t,s)x:\{(t,s): t\geq s\}\to H$
is continuous for all $t\geq s\geq 0$ and every $x\in H$.
\end{itemize}

 It is well known that any such evolution family $\mathcal{U}$ is
 exponentially bounded, that is, there exist $\omega \in \mathbb{R}$ and
$M_{\omega}\geq 0$ such that
\begin{equation}
\|U(t,s)\|\leq M_{\omega}e^{\omega(t-s)} \quad \text{for $t\geq s\geq 0$}.
\label{e4}
\end{equation}
See \cite{BP}.
Whenever the evolution family $\mathcal{U}$ is exponentially bounded
its growth bound is defined by
$$
\omega_0(\mathcal{U}):=\inf \{\omega \in \mathbb{R}:
\text{ there is $M_{\omega}\geq 0$ such that \eqref{e4} holds}\}.
$$
 A bounded linear operator $L$, acting on a Hilbert space $H$,
is positive if $\langle Lx, x\rangle\ge 0$ for every $x\in H$.

An evolution family $\{U(t, s): t\ge s\ge 0\}$ is called selfadjoint,
(positive), if each operator $U(t, s)$, with $t\ge s\ge 0$, is selfadjoint
and (respectively  positive).

The family $\mathcal{U}$ is uniformly exponentially stable if its
growth bound is negative.

In the following  we use a deep result from the theory of operators,
originally given by   M\"uller and Jan van Neerven. See \cite{VM} and
\cite{Ne1}, and also \cite{Be} for related results in the framework
of Hilbert spaces. For the reader convenience, we state this result here.

\begin{lemma}\label{lemma_muller_neerven}
Let $X$ be a complex Banach space and let $V\in\mathcal{L}(X)$. If the
spectral radius of $V$ is greater or equal to 1, then for all
$0<\varepsilon<1$ and any  sequence $(a_n)$ with $a_n\to 0$
(as $n\to\infty)$ and $\|(a_n)\|_\infty\le 1$,  there exists a
unit vector $u_0\in X$, such that
$$
\|V^nu_0\|\geq (1-\varepsilon)\cdot |a_{n}|, \quad\text{for all  }
n\in\mathbb{Z}_+.
$$
\end{lemma}


Throughout this article, $(t_n)$ will be a sequence of nonnegative real numbers,
such that $1\le q\le t_{n+1}-t_n\leq\alpha$ for every
 $n\in \mathbb{Z}_+$ and  some positive real number $\alpha$.

\begin{lemma}\label{lem2}
Let $\mathcal{U}=\{U(t,s)\}_{t\geq s \geq 0}$ be a strongly continuous
$q$-periodic ($q\geq 1$) evolution family of bounded linear operators
acting on a Banach space $X$ and let $(t_n)$ be a sequence as given before.
If the evolution family is not uniformly exponentially stable, then there
exists a positive constant $C$, having the properties: for every
$\mathbb{C}$-valued sequence $(b_n)$ with $b_n\to 0$ (as $n\to\infty$) and
 $\|(b_n)\|_\infty\leq 1$, there exists a unit vector $u_0\in X$, such that
\begin{equation}
\|U(t_n,0)u_0\|\geq C\cdot |b_{n+1}|, \quad\text{for all } n\in\mathbb{Z}_+.
\label{e5}
\end{equation}
\end{lemma}

\begin{proof}
Let $V:=U(q,0)$. Since $\mathcal{U}$ is not uniformly exponentially stable,
$r(U(q,0))\geq 1$. From Lemma \ref{lemma_muller_neerven} follows that
for every $\varepsilon\in(0, 1)$ and every sequence $(a_n)$ with
$a_n\to 0$ (as $n\to\infty$) and
$\|(a_n)\|_\infty = 1$ there
exists a unit vector $y_0\in X$ such that
$\|U(nq,0)y_0\|\geq(1-\varepsilon)\cdot |a_n|$, for any natural number $n$.

Let $k:\mathbb{Z_{+}}\to \mathbb{Z_{+}}$ be the function defined by,
 $k(n)=[\frac{t_n}{q}]$. Here $[\frac{t_n}{q}]$ denotes the integer
part of the real number $\frac{t_n}{q}$. In view of
the properties of the sequence $(t_n)$, the function $k(\cdot)$ is increasing.
Let $(b_n)$ be a sequence having the assumed properties, and
let us consider a sequence $(a_n)$ defined by
$$
a_{n}=\begin{cases}
   b_{k^{-1}(n)},&\text{for } n\in k(\mathbb{Z}_+) \\
    0,& \text{otherwise}
\end{cases}
$$
Then, we have
\begin{equation}
\|U(qk_n,0)y_0\| \geq(1-\varepsilon)\cdot |a_{k(n)}|
\geq(1-\varepsilon)\cdot |b_n|, \quad n\in\mathbb{Z}_+.\label{e6}
\end{equation}
Choose $u_0:=y_0$. Using \eqref{e4} and \eqref{e6}, we obtain
\begin{align*}
(1-\varepsilon)\cdot |b_{n+1}|
&\leq\|U(qk_{n+1},0)u_0\|\\
& = \|U(qk_{n+1},t_n)U(t_n,0)u_0\| \\
& \leq  Me^{\omega\cdot\alpha}\cdot\|U(t_n,0)u_0\|.
\end{align*}
Hence, \eqref{e5} holds  with
$C:=\frac{(1-\varepsilon)}{Me^{\omega\cdot\alpha}}$.
\end{proof}

A periodic evolution family
$\mathcal{U}=\{U(t,s)\}_{t\geq s \geq 0}$ satisfies the \emph{strong discrete
Rolewicz condition} related to the $\mathcal{R}$-function
$\varphi$ and  the sequence $(t_n)$, if
\begin{equation}
\sum_{n=0}^\infty\phi(\|U(t_n,0)x\|)<\infty, \quad
\forall x\in X,\; \|x\|=1.\label{e7}
\end{equation}

Whenever the evolution family $\mathcal{U}$ satisfies
the condition \eqref{e7} we can highlight new qualities of the
function $\varphi$. Such qualities are listed in the following.
\begin{itemize}
\item The inequality  \eqref{e7} holds for all $x\in X$ with $\|x\|\le 1$.

\item $\varphi(0)=0$. To justify, putting $x=0$ in \eqref{e7}.

\item  We may suppose that $\phi(1)=1$. To justify, putting a suitable
multiple ($\alpha\varphi$, with $\alpha>0$),
instead of $\varphi$.

\item  Also, we can put a suitable multiple of the function
$\overline{\varphi}$, instead of $\varphi$. Here, by
$\overline{\varphi}$ we understand the function defined by:
\[
\overline{\varphi}(t)= \begin{cases}
\int_{0}^t \varphi(s)ds, & \text{for } t\in [0, 1)\\
 \frac{at}{at+1-a}, &\text{for } t\geq1,
\end{cases}
\]
where $a=\int_0^1\varphi(s)ds$.
\end{itemize}
We remark that $\overline{\varphi}$ is continuous and increasing
function on $\mathbb{R}_+$, and, moreover, it is convex on the
interval $[0, 1]$. In addition, the evolution family
$\mathcal{U}$ satisfies the condition \eqref{e7} in respect to
$\overline{\varphi}$, because $\overline{\varphi}(t)\leq\varphi(t)$,
for all $t\in\mathbb{R}_+$.

Given the above, the following lemma proof becomes clear.
 We insert it for the sake of completeness.

\begin{lemma}\label{lem2.3}
Let $\varphi$ be an $\mathcal{R}$-function and let
 $\mathcal{U}=\{U(t,s)\}_{t\geq s \geq0}$
be a strongly continuous $q$-periodic ($q\geq 1$) evolution family
acting on a Banach space $X$. If the family $\mathcal{U}$
satisfies \eqref{e7} then it is uniformly
exponentially stable.
\end{lemma}

\begin{proof}
Assume that $r(U(q,0))\geq 1$. We replace $\varphi$
by $\overline{\varphi}$. Let $C$ be as in Lemma \ref{lem2} and
let $|b_n|=\frac{1}{C}\cdot\overline{\varphi}^{-1}(\frac{1}{n}),
n\geq 1$. Obviously the sequence $(b_n)$ satisfies the
requirements in Lemma \ref{lem2}. As a consequence, there
exists a unit vector $u_0\in X$ such that $\|U(t_n,0)u_0\|\geq
C\cdot |b_{n+1}|$ for any natural number $n$.
Hence,
$$
\sum_{n=0}^\infty\overline{\varphi}(\|U(t_n,0)u_0\|)
\geq \sum_{n=0}^\infty\overline{\varphi}(C\cdot |b_{n+1}|)
=\sum_{n=0}^\infty\frac{1}{n+1}=\infty.
$$
\end{proof}

\section{Weak integral conditions and exponential stability}

Let $\mathbf{T}=\{T(t)\}_{t\geq 0}$ be a strongly continuous
semigroup acting on a complex Hilbert space $H$. When $\mathbf{T}$ is
selfadjoint (i.e. $T(t)=T(t)^\ast$, for every $t\geq
0$), then $\langle T(t)x, x\rangle \geq 0$, for each $t\geq 0$ and every $x\in H$.
Indeed, for any $t\geq 0$, we have
$$
\langle T(t)x, x\rangle =\langle T(t/2)T(t/2) x, x\rangle
=\|T(t/2)x\|^2\geq 0.
$$
 In the proof of the next theorem, we use the following
inequality of the Cauchy-Buniakovski-Schwartz type. \emph{Let $A$
and $B$ be two selfadjoint operators acting on the complex
Hilbert space $H$. Then,}
\begin{equation}
|\langle ABx, y\rangle |^2\leq \langle A^2y, y\rangle \langle B^2 x, x\rangle ,
\quad \text{for all } x, y\in     H.\label{e8}
\end{equation}
In fact,
$$
|\langle AB x, y\rangle|^2=|\langle Bx, Ay\rangle|^2
\leq \|Bx\|^2\|Ay\|^2=\langle A^2y, y\rangle\langle B^2 x, x\rangle.
$$

\begin{theorem}\label{evo}
Let $\phi$ be an $\mathcal{R}$-function and let
$\mathbf{T}=\{T(t)\}_{t\geq 0}$ be a selfadjoint, strongly continuous
semigroup of bounded linear operators acting on a complex Hilbert space $H$.
If
\begin{equation}
I(x):=\int_{0}^\infty\phi(\langle T(t)x, x\rangle )dt<\infty, \quad
\text{ for all }x\in H \text{ with } \|x\|=1,\label{e9}
\end{equation}
then the semigroup $\mathbf{T}$ is uniformly exponentially stable.
\end{theorem}

\begin{proof}
Having in mind that $\phi$ may be considered a continuous
function and by applying the Mean-Value Theorem to the function
$t\mapsto\phi(\langle T(2t)x, x\rangle) $ on the interval $[n, n+1]$,
 we find a real number
$t_n(x)\in[n, n+1]$, such that
\begin{align*}
\frac{1}{2}I(x)
&=\int_0^\infty\phi(\langle T(2t)x, x\rangle )dt\\
&=\sum_{n=0}^\infty\phi(\langle T(2t_n(x)x, x\rangle )\\
&\ge\sum_{n=0}^\infty\phi(\langle T(2t_{4n}(x)x, x\rangle ).
\end{align*}
Set $s_n\in[2n+1, 2n+2]$ and let $y$ be a unit vector in $H$.
In view of \eqref{e8}, successively one has
\begin{align*}
  |\langle T(2s_n)x, y\rangle |^2
& =  |\langle T(2s_n-t_{4n}(x))T(t_{4n}(x))x, y\rangle |^2\\
& \leq  \langle T(4s_n-2t_{4n}(x)) y, y\rangle
 \langle T(2t_{4n}(x)) x, x\rangle  \\
& \leq  Me^{8\omega}\langle T(2t_{4n}(x)) x, x\rangle .
\end{align*}
Hence, for any unit vector $x\in H$, one has
\begin{equation}
\sum_{n=0}^{\infty}\phi\Big(\frac{1}{Me^{8\omega}}\|T(s_n)x\|^4\Big)
\le\sum_{n=0}^\infty \phi(\langle T(2t_{4n}(x))x, x\rangle ).\label{e10}
\end{equation}
Clearly
$1\le s_{n+1}-s_n\leq 3$. By assumption and relation \eqref{e10}, we obtain
$$
\sum_{n=0}^\infty \phi\Big(\frac{1}{Me^{8\omega}}\| T(s_n)x\|^4\Big)<\infty.
$$
Now replacing $ \phi(\frac{1}{Me^{8\omega}}t^4)$ by $\varphi(t)$.
Obviously, the map $\varphi$ is an $\mathcal{R}$-function.
The assertion follows by Lemma \ref{lem2.3}.
\end{proof}

In the following we recall a concrete example of semigroup verifying \eqref{e9}.
That the Dirichlet semigroup is exponentially stable is more quite standard.
 There are several possible ways of showing this.
The following is yet another demonstration in addition to this well known fact.

 The state space is $H:=L^2([0, \pi], \mathbb{C})$,
endowed with the usual inner product and norm,  becomes a complex Hilbert
space. In addition, the one parameter family $\{T(t)\}_{t\ge 0}$,
given by
$$\left(T(t)x\right)(\xi)=\frac{2}{\pi}\sum_{n=1}^\infty 
e^{-tn^2}\sin n\xi\Big(\int_0^\pi x(s)\sin nsds\Big),\quad 
\xi\in[0,\pi],\; t\geq 0,
$$ 
is a strongly continuous semigroup on $H$. Moreover, this semigroup 
solves the following Cauchy Problem with boundary conditions
\begin{gather*}
  \frac{\partial{u(t, \xi)}}{\partial {t}}
= \frac{\partial^2{u(t, \xi)}}{\partial^2 {\xi}},\quad t>0,\; \xi\in[0, \pi] \\
    u(t, 0)= u(t, \pi)=0, \quad t\ge 0 \\
    u(0, \xi)= x(\xi),
  \end{gather*}
where $x(\cdot)$ is a given function in $H$.

More exactly, the unique solution of this Cauchy Problem is given by
\begin{equation}
u(t, \xi, x(\cdot))=(T(t)x)(\xi).\label{e11}
\end{equation}
The semigroup $\mathbf{T}=\{T(t)\}_{t\ge 0}$ is generated by the
linear operator $A$ given by $Ax=\ddot{x}$. The maximal domain of $A$
is the set $D(A)$ of all $x\in H$ such that $x$ and $\dot{x}$
are absolutely continuous, $ \ddot{x}\in H$  and $ x(0)=x(\pi)=0$.
Moreover, $T(t)$ is selfadjoint for all $t\ge 0$.
\cite[Example 1.3, pp. 178, 198]{Zabczyk}.
Based on  Theorem 3.1, we prove that the solution given in
\eqref{e11} is exponentially stable; i.e., there exist the positive
constants $N$ and $\nu$ such that for each $x(\cdot)\in H$, one has
$$
\|u(t, \cdot, x(\cdot))\|_2\le Ne^{-\nu t}\|x(\cdot)\|_2\quad\forall t\ge 0.
$$
For this to be completed, we remark that for each $x(\cdot)\in H$, one has
\begin{align*}
\int_0^\infty \langle T(t)x(\cdot), x(\cdot)\rangle dt
&= \frac{2}{\pi}\sum_{n=1}^\infty\frac{1}{n^2}\|\int_0^\pi x(s)\sin(ns)ds\|^2\\
&\le \sum_{n=1}^\infty\|\int_0^\pi x(s)\sin(ns)ds\|^2
<\infty
\end{align*}
The latter estimate is obtained in the view of Bessel inequality.

Next, we comment on the relationship between the integral conditions 
\eqref{e3} and \eqref{e9}. Clearly, \eqref{e3} implies \eqref{e9},
 but it is not clear whether \eqref{e9} implies \eqref{e3}.
However, this happens if we admit that \eqref{e9} holds for all $x\in H$ 
and the map $\phi$ is subadditive. This latter fact is a consequence 
of the well-known formula of polarization.

Our approach in the semigroup case allow us to extend this result 
to the nonautonomous case of positive and periodic evolution families 
acting on complex Hilbert spaces. The next result could be known. 
See for example \cite{Bu,BBCD} for its different counterparts.
Here we present a new proof.

\begin{theorem}
Let $\phi$ be an $\mathcal{R}$-function and let
$\mathcal{U}=\{U(t, s)\}_{t\geq s\ge 0}$
be a selfadjoint strongly continuous and $q$-periodic $(q\geq 1)$ 
evolution family acting on a complex Hilbert space $H$.
If
$$
J(x):=\int_{0}^\infty\phi(\|U(t,0)x\|)dt<\infty, \quad
\text{for all }x\in H \text{ with } \|x\|=1,
$$
then $\mathcal{U}$ is uniformly exponentially stable.
\end{theorem}

\begin{proof}
Adopting the technique used in Theorem \ref{evo}, we find some
$t_n^x\in[nq, (n+1)q]$ such that
\begin{align*}
           J(x)&= q \sum_{n=0}^\infty\phi(\|U(t_{n}^{x},0)x\|) \\
               & \geq  \sum_{n=0}^\infty\phi(\|U(t_{2n}^{x},0)x\|).
\end{align*}
Set $s_n=(2n+2)q$ and let $x$ and $y$ be two unit vectors. In view of \eqref{e8},
successively one has
\begin{align*}
  |\langle U(s_n,0)x, y\rangle |^2 
&= |\langle U(s_n,t_{2n}^{x})U(t_{2n}^{x},0)x, y\rangle |^2\\
& \leq  \langle U^{2}(s_n,t_{2n}^x) y, y\rangle \langle U^{2}(t_{2n}^x,0) x, x\rangle  \\
& \leq  M^{2}e^{4q \omega}\|U(t_{2n}^x,0) x\|^{2} .
\end{align*}
Hence, for any unit vector $x\in H$, one has
\begin{equation}
\begin{aligned}
  \sum_{n=0}^\infty \phi(\|U(t_{2n}^x,0) x\|)
& \geq  \sum_{n=0}^\infty\phi(\frac{1}{Me^{2q \omega}}|
 \langle U(s_n,0)x, x\rangle |)\\
&= \sum_{n=0}^\infty\phi(\frac{1}{Me^{2q \omega}}\| U((n+1)q,0)x\|^{2}).
\end{aligned} \label{e12}
\end{equation}

By the assumption and relation \eqref{e12}, we obtain
$$
\sum_{n=0}^\infty\phi\Big(\frac{1}{Me^{2q \omega}}\| U((n+1)q,0)x\|^{2}\Big)
<\infty,
$$
for all unit vectors $ x\in H$.
Now replace $\phi(\frac{1}{Me^{2q \omega}}t^{2})$ by $\varphi(t)$.
The assertion follows by Lemma \ref{lem2.3}.
\end{proof}

For positive evolution families we can give even a weak version 
of the Rolewicz's theorem.


\begin{theorem} \label{thm3.3}
Let $\phi$ be an $\mathcal{R}$-function and let
$\mathcal{U}=\{U(t,s)\}_{t\geq s\ge 0}$
be a positive strongly continuous and $q$-periodic $(q\geq 1)$ evolution 
family acting on a complex Hilbert space $H$.
If
$$
J(x):=\int_{0}^\infty\phi(\langle U(t,0)x,x\rangle)dt<\infty,\quad
 \text{for all }x\in H \text{ with } \|x\|=1,
$$
then $\mathcal{U}$ is uniformly exponentially stable.
\end{theorem}

\begin{proof}
Again using the technique used in Theorem \ref{evo}, we find some
$t_n^x\in[nq, (n+1)q]$ such that
$$
J(x)= q \sum_{n=0}^\infty\phi(\langle U(t_{n}^{x},0)x,x\rangle) 
 \geq  \sum_{n=0}^\infty\phi(\langle U(t_{2n}^{x},0)x,x\rangle)).
$$
Set $s_n=(2n+2)q$. In view of \eqref{e8},
successively one has
\begin{align*}
  |\langle U^{1/2}(s_n,0)x, y\rangle |^2
&= |\langle U^{1/2}(s_n,t_{2n}^{x})U^{1/2}(t_{2n}^{x},0)x, y\rangle |^2\\
&\leq  \langle U(s_n,t_{2n}^x) y, y\rangle \langle U(t_{2n}^x,0) x, x\rangle  \\
&\leq  Me^{2q \omega}\langle U(t_{2n}^x,0) x, x\rangle .
\end{align*}
Hence, for any unit vector $x\in H$, one has
\begin{equation}
\begin{aligned}
  \sum_{n=0}^\infty \phi(\langle U(t_{2n}^x,0) x, x\rangle)
& \geq  \sum_{n=0}^\infty\phi(\frac{1}{Me^{2q \omega}}|
 \langle U^{1/2}(s_n,0)x, x\rangle |^{2})\\
&= \sum_{n=0}^\infty\phi\Big(\frac{1}{Me^{2q \omega}}|
\langle U((n+1)q,0)x, x\rangle |^{2}\Big).
\end{aligned}\label{e13}
\end{equation}
By the assumption and relation \eqref{e13}, we obtain
$$
\sum_{n=0}^\infty\phi\Big(\frac{1}{Me^{2q \omega}}|\langle U((n+1)q,0)x,
 x\rangle |^{2}\Big)<\infty,
$$
for all unit vectors $ x\in H$.
In particular for $n+1=2m$, we have
$$
\sum_{m=0}^\infty\phi \Big(\frac{1}{Me^{2q \omega}}\|U(mq,0)x\|^{4}\Big)<\infty,
$$
Now replace $\phi(\frac{1}{M^{2}e^{4q \omega}}t^{4})$ by $\varphi(t)$
and apply Lemma \ref{lem2.3}.
\end{proof}

We leave open the question whether Theorem 3.3 remains valid
 for periodic (selfadjoint) evolution families.

\subsection*{Acknowledgements}
The authors want to thank the anonymous referees for their careful
 reading and helpful comments.


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\end{document}