text:
\section{Introduction}

A notable result in the stability theory of ODEs lies in Datko's theorem.
In order to extend a classical theorem of A.M. Lyapunov to strongly continuous semigroups of operators in Hilbert spaces, R. Datko proved that if $\{T(t)\}_{t\geq 0}$ is a $C_0$-semigroup on a complex Hilbert space $X$ such that $\int_{0}^{\infty} \parallel T(t)x\parallel^2 dt<\infty$ for some $x\in X$, then $\lim\limits_{t\to\infty} T(t)x=0$ (Lemma 3 in \cite{Da.1970}). Recall that a family of bounded linear operators $\left\{  T(t)\right\}  _{t\geq0}$ acting on $X$ is  a
$C_{0}$-\emph{semigroup} if
\begin{enumerate}
\item[($s_1$)] $T(0)=\mathrm{Id}$;

\item[($s_2$)] $T(t)T(s)=T(t+s)$ for $t,s\geq0$;

\item[($s_3$)] $\lim\limits_{t\rightarrow0_{+}}T(t)x=x$ for every $x\in X$.
\end{enumerate}
Later, using distinct techniques, A. Pazy proved  that if for some  $p\in[1,\infty)$, the integral $\int_{0}^{\infty} \parallel T(t)x\parallel^p dt$ is finite for every $x\in X$,  then $\{T(t)\}_{t\geq 0}$ is uniform exponentially stable, i.e. there exist $M,\alpha>0$ such that $\parallel T(t)\parallel \leq M e^{-\alpha t}$, $t\geq 0$ (see \cite{Pa.1972} and \cite[Theorem 4.1, pp. 116]{Pa.1983}).
In 1972, R. Datko generalized  this result for evolution families on Banach spaces \cite{Da.1972}.
In fact, it is shown that a uniform exponentially bounded
evolution family $\left\{  U(t,s)\right\}_{t\geq s\geq0}$ on a
Banach space $X$ is uniform exponentially stable if and only if there exists
$p\in\lbrack1,\infty)$ such that
\[
\sup\limits_{t\geq0}\int_{t}^{\infty}\parallel U(\xi,t)x\parallel^{p
d\xi<\infty,\,x\in X.
\]
A. Ichikawa obtained in  \cite{Ic.1984} a Datko-type theorem for families of nonlinear operators $U(t,s):X_{s}\rightarrow X_{t}$, $t\geq s\geq0$, that satisfy:
\begin{enumerate}
\item[($e_1$)] $U(t,t)x=x$, $x\in X_{t}$;

\item[($e_2$)] $U(t,\tau)U(\tau,s)=U(t,s)$ on $X_{s}$, $t\geq\tau\geq s$;

\item[($e_3$)] the map $t\mapsto U(t,s)x$ is continuous on $[s,\infty)$ for every $x\in X_{s}$.
\end{enumerate}
In fact, Ichikawa proved that if there exists a continuous function
$g:\mathbb{R}_{+}\rightarrow(0,\infty)$ such that $\parallel U(t,s)x\parallel
\leq g(t-s)\parallel x\parallel$, $x\in X_{s}$, $t\geq s\geq 0$, then $\{U(t,s)\}_{t\geq s\geq 0}$
is uniform exponentially stable if and only if there exist constants $p,K>0$
such that
\[
\int_{t}^{\infty}\parallel U(\xi,t)x\parallel^{p}d\xi\leq K^{p}\parallel
x\parallel^{p},\,x\in X_{t},\,t\geq0.
\]
Using distinct techniques, this result has been generalized for weak
exponential stability in \cite{Lu.Me.Po}.   An interesting Datko-type theorem was obtained by P. Preda and M. Megan for uniform exponential dichotomy \cite{Pr.Me.1985}. Some nonlinear versions have been considered by B. Sasu and A.L. Sasu for both exponential dichotomy
\cite{Sa.2010-1} and exponential trichotomy  \cite{Sa.Sa.2011}.

It is worth to mention that the claim in Datko's theorem can be obtained as a consequence of
some results in the evolution semigroup theory (ex. \cite[Theorem
3.23, pp. 74-75]{Ch.La.1999}).

While in the Datko-type  theorems the integral is taken according to the first variable, in a similar result due to Barbashin the integral is taken with respect to the second one.  Recently, P.V. Hai unified the discrete-time versions  of the Datko's theorem and the Barbashin-type theorem \cite{Hai.2015}.

The theory of uniform behavior for evolution equations is too
restrictive; for instance one may consider examples of evolution families generated by differential equations which are not uniform exponentially bounded, but
having all the orbits asymptotically stable. In the last
decade the research in the field of evolution equations developed in a more general direction,
which is the nonuniform behavior. We need to mention here the consistent
contribution due to L. Barreira and C. Valls. A major motivation for considering a nonuniform behavior lies in the ergodic theory (we refer to  \cite{Ba.Va.2008-1} for a detailed exposition).
Notice that
C. Bu\c{s}e extended   Datko's theorem to nonuniform exponential
stability,  restricting his analysis to uniform exponentially bounded evolution
families \cite{Bu.1994}. Some of the most relevant generalizations of Datko's theorem in the case of nonuniform behavior can be
found in works as \cite{Be.Lu.Me.Si,Lu.Me.2014,Pr.Pr.Cr.2015,Pr.Pr.Cr.2012}.

The aim of the present paper is to give necessary and sufficient conditions for
the existence of nonuniform exponential stability of evolution families in the spirit of Datko's results.
In this regard, we introduce a class of  Banach function spaces for nonuniform exponentially bounded  evolution families. We
generalize the classical theorem of Datko on these spaces. Our results extend
conclusions in some previous works, also referring to new situations.

\section{Preliminaries}

\label{sec.preliminaries}

Let $X$ be a Banach space, and $\mathcal{L}(X)$ be the Banach algebra of all
bounded linear operators on $X$. In our paper we also make use of the
following notations: $C(\mathbb{R}_{+},X)$ denotes the space of all continuous
functions $u:\mathbb{R}_{+}\rightarrow X$, $C_{b}(\mathbb{R}_{+},X)$ is the
subspace of the bounded maps in $C(\mathbb{R}_{+},X)$, and $C_{c
(\mathbb{R}_{+},X)$ is the set of the maps in $C(\mathbb{R}_{+},X)$ with
compact support.

\begin{definition}
\label{d.ev} A family of operators $\mathcal{U}=\left\{
U(t,s)\right\}  _{t\geq s\geq0}$ in $\mathcal{L}(X)$ is called an
\emph{evolution family} if
\begin{enumerate}
\item[(i)] $U(t,t)=\mathrm{Id}$, $t\geq0$;

\item[(ii)] $U(t,\tau)U(\tau,s)=U(t,s)$, $t\geq\tau\geq s\geq0$;

\item[(iii)] the map $(t,s)\mapsto U(t,s)x$ is continuous for every $x\in X$.
\end{enumerate}

An evolution family $\mathcal{U}=\left\{  U(t,s)\right\}
_{t\geq s\geq0}$ is called \emph{reversible} if  $U(t,s)$ is invertible for all $t\geq s\geq 0$. In fact, in this case $\mathcal{U}$ can be extended to all of $\mathbb{R}_{+}^{2}$ by using
$U(s,t)=[U(t,s)]^{-1}$. Moreover, condition (ii) holds for all $t,\tau,s\geq 0$.
\end{definition}

The study of evolution families arose naturally from the theory of well-possed evolution
equations. Recall that the linear equation $dx/dt=A(t)x$, with bounded or
unbounded linear operator $A(t)$, is called \emph{well-posed }if the existence,
uniqueness and continuous dependence of solutions on the initial data is
guaranteed. In \cite{Na.Ni.2002} it is shown that well-posedness is
equivalent to the existence of an evolution family $\mathcal{U}=\left\{
U(t,s)\right\}  _{t\geq s\geq0}$ solving the equation, precisely $U\left(t,t_{0}\right)x_{0}=x\left(
t,t_{0},x_{0}\right)$, $t\geq t_{0}$, where
$x(\cdot,t_0,x_0)$ is the solution of the Cauchy problem
\begin{equation*}\label{eq.Cauchy}
\quad\left\{\begin{array}{ll}
\dot{x}(t)=A(t)x(t),\;t\geq t_0\geq 0,\\
x(t_0)=x_0.
\end{array}\right.
\end{equation*}

In this paper we will always assume that an  evolution family $\mathcal{U}=\left\{U(t,s)\right\}_{t\geq s\geq 0}$ is given.

\begin{definition}
\label{d.exp.bound}
For fixed $\alpha\in\mathbb{R}$, the evolution
family $\mathcal{U}$ is called \emph{$\alpha$-nonuniform exponentially
bounded} if there exists a continuous function $M:\mathbb{R}_{+
\rightarrow(0,\infty)$ such that
\begin{equation}
\parallel U(t,s)\parallel\leq M(s)e^{\alpha(t-s)}, \, t\geq s\geq
0.\label{eq.exp.bound
\end{equation}

If Eq. \eqref{eq.exp.bound} holds for some $\alpha<0$, then
$\mathcal{U}$ is called \emph{$\alpha$-nonuniform exponentially stable}. Each
$\alpha\in\mathbb{R}$ satisfying \eqref{eq.exp.bound} is called an \emph{admissible
exponent} for the evolution family $\mathcal{U}$, and $\mathcal{A
(\mathcal{U})$ denotes the set of all admissible exponents for $\mathcal{U}$. If $\mathcal{A
(\mathcal{U})\neq\emptyset$, then the evolution family $\mathcal{U}$ is said
to be \emph{nonuniform exponentially bounded}, and if $\mathcal{A
(\mathcal{U})$ contains negative admissible exponents, $\mathcal{U}$ is called
\emph{nonuniform exponentially stable}. In the above terminology, whenever
there exists a bounded function $M(s)$ satisfying
\eqref{eq.exp.bound}, which is equivalent to the existence of a constant one,
we just replace the term \textquotedblleft nonuniform\textquotedblright\ with
\textquotedblleft uniform\textquotedblright.
\end{definition}

The following result is detailed in \cite{Lu.Po}:

\begin{remark}
\label{rem.Ly} Assume that the evolution family $\mathcal{U}$ is
reversible. If the Lyapunov exponent
\[
\label{eq.Ly}K_{L}=\inf\left\{  \alpha\in\mathbb{R}:\,\text{there exists
}M>0\text{ with }\Vert U(t,0)\Vert\leq Me^{\alpha t},\;t\geq0\right\}
\]
is finite, then
\[
(K_{L},\infty)\subset\mathcal{A}(\mathcal{U})\subset\lbrack K_{L},\infty).
\]
Furthermore, if the Lyapunov exponent is attained, then $\mathcal{A
(\mathcal{U})$ is closed at its left endpoint, that is $\mathcal{A}(\mathcal{U
)=[K_{L},\infty)$.
\end{remark}

Throughout our paper  assume that $\mathcal{A}(\mathcal{U})\neq\emptyset$, i.e.  $\mathcal{U}$ is
nonuniform exponentially bounded.
For  $\alpha\in\mathcal{A}(\mathcal{U})$, $t\geq0$ and $u\in{C
(\mathbb{R}_{+},X)$ we set
\begin{equation}
\varphi_{\alpha}(t,u)=\sup_{\tau\geq t}e^{-\alpha(\tau-t)}\parallel
U(\tau,t)u(t)\parallel. \label{eq.fi
\end{equation}

If in particular $u(t)\equiv x$ for some $x\in X$, we step over the norm on
$X$ defined in \cite[Eq. (5)]{Ba.Va.2010}, precisely
$\left\Vert x\right\Vert_{t}=\sup\limits_{\tau\geq t} e^{-\alpha(\tau-t)}\parallel U(\tau,t)x\parallel$. In this regard,  the map $\varphi_{\alpha}$  can be defined as $\varphi_{\alpha}(t,u)=\left\Vert u(t)\right\Vert_{t}$.

Let us notice that
\begin{equation}
\parallel u(t)\parallel\leq\varphi_{\alpha}(t,u)\leq M(t)\parallel
u(t)\parallel. \label{eq.init
\end{equation}

The following result
is partially proved in \cite{Lu.Po}, and
assures the existence
of the integrals in the next section.
For the sake of completeness and also for a better understanding, we prefer exposing its complete proof.

\begin{proposition}
\label{p2} The function $\mathbb{R}_{+}\ni t\mapsto\varphi_{\alpha
(t,u)\in\mathbb{R}_{+}$ is continuous for each $u\in{C}(\mathbb{R}_{+},X)$.
\end{proposition}

\begin{proof}
Set $V(t,s)=e^{-\alpha(t-s)}U(t,s)$. Clearly $\left\{
V(t,s)\right\}  _{t\geq s\geq0}$ is an evolution family, and $\parallel
V(t,s)\parallel\leq M(s)$. For any $u\in{C}(\mathbb{R}_{+},X)$, $t_{0}\geq0$
and $\varepsilon>0$ there exists $\delta=\delta(u,t_{0},\varepsilon)>0$ such
that
\begin{align*}
|t-t_{0}|<\delta &  \Rightarrow\,M(t)\parallel u(t)-u(t_{0})\parallel
<{\varepsilon}/{3},\\
t_{0}\leq t<t_{0}+\delta &  \Rightarrow\,M(t)\parallel
V(t,t_0)u(t_{0})-u(t_{0})\parallel<\varepsilon/{3},\\
t_{0}-\delta<t\leq t_{0}&  \Rightarrow\,M(t_0)\parallel
V(t_0,t)u(t_{0})-u(t_{0})\parallel<\varepsilon/{3},\\
t_{0}-\delta<t\leq\tau<t_{0}+\delta &  \Rightarrow\,\parallel V(\tau
,t)u(t_{0})-u(t_{0})\parallel<\varepsilon/{3}.
\end{align*}
Choose $t\geq0$ with $|t-t_{0}|<\delta$. We treat cases $t\geq t_{0}$ and
$t<t_{0}$ separately.
If $t\geq t_{0}$, then for any $\tau\geq t$ one ha
\begin{align*}
\left\Vert V(\tau,t)u\left(  t\right)  \right\Vert  &  \leq\left\Vert
V(\tau,t)\left(  u\left(  t\right)  -u\left(  t_{0}\right)  \right)
\right\Vert +\left\Vert V(\tau,t)u\left(  t_{0}\right)  -V(\tau,t_{0})u\left(
t_{0}\right)  \right\Vert \\
&  \quad+\left\Vert V(\tau,t_{0})u\left(  t_{0}\right)  \right\Vert \\
&  \leq \left\Vert
V(\tau,t)\right\Vert  \left\Vert u\left(  t\right)  -u\left(
t_{0}\right)  \right\Vert +\left\Vert V(\tau,t)\right\Vert \left\Vert u\left(
t_{0}\right)  -V\left(  t,t_{0}\right)  u\left(  t_{0}\right)  \right\Vert \\
&  \quad+\left\Vert V(\tau,t_{0})u\left(  t_{0}\right)  \right\Vert \\
&  \leq M\left(  t\right)  \left\Vert u\left(  t\right)  -u\left(
t_{0}\right)  \right\Vert +M\left(  t\right)  \left\Vert u\left(
t_{0}\right)  -V\left(  t,t_{0}\right)  u\left(  t_{0}\right)  \right\Vert +
\varphi_{\alpha}(t_{0},u)\\
&  <2\varepsilon/3+\varphi_{\alpha}(t_{0},u).
\end{align*}
Thus,  taking the supremum with respect to $\tau\geq t$ we get
\[
\varphi_{\alpha}(t,u)-\varphi_{\alpha}(t_{0},u)<\varepsilon.
\]
Now, if $t<t_{0}$, then for every $\tau\geq t_{0}$ we estimat
\begin{align*}
\left\Vert V(\tau,t)u\left(  t\right)  \right\Vert  &  \leq \left\Vert
V(\tau,t)\right\Vert \Vert
u(t)-u(t_{0})\Vert+\left\Vert V(\tau,t_{0})\right\Vert \left\Vert
V(t_{0},t)u(t_{0})-u(t_{0})\right\Vert \\
&  \quad+\left\Vert V(\tau,t_{0})u\left(  t_{0}\right)  \right\Vert \\
&  \leq M(t) \Vert u(t)-u(t_{0})\Vert+M(t_0) \left\Vert
V(t_{0},t)u(t_{0})-u(t_{0})\right\Vert
+\varphi_{\alpha}(t_{0},u) \\
&  <2\varepsilon/3+\varphi_{\alpha}(t_{0},u).
\end{align*}
One the other hand, for $\tau\in[t,t_0)$ one gets
\begin{align*}
\left\Vert V(\tau,t)u\left(  t\right)  \right\Vert  &  \leq M(t)\Vert
u(t)-u(t_{0})\Vert+\left\Vert V(\tau,t)u(t_{0})-u(t_{0})\right\Vert
+\left\Vert u(t_{0})\right\Vert \\
&  <2\varepsilon/3+\varphi_{\alpha}(t_{0},u).
\end{align*}
Consequently, $$\left\Vert V(\tau,t)u\left(  t\right)  \right\Vert <2\varepsilon/3+\varphi_{\alpha}(t_{0},u), \text{  for all $\tau\geq t$.}$$
This implies $\varphi_{\alpha}(t,u)-\varphi_{\alpha}(t_{0},u)<\varepsilon$. Proceeding as before, one can prove that $\varphi_{\alpha}(t_0,u)-\varphi_{\alpha}(t,u)<\varepsilon$. Thus, the function $t\mapsto\varphi_{\alpha}(t,u)$ is continuous at $t_0$ and
this completes the proof.
\end{proof}

For each $\alpha\in\mathcal{A}\left(  \mathcal{U}\right)  $ we set
\[
\mathcal{C}_{\alpha}=\left\{  u\in C(\mathbb{R}_{+},X):\,\sup\limits_{t\geq
0}\varphi_{\alpha}(t,u)<\infty\right\}  .
\]

It is easy to show that $\mathcal{C}_{\alpha}$ is a Banach
space equipped with the norm
\[
\parallel u\parallel_{\alpha}=\sup\limits_{t\geq0}\text{ }\varphi_{\alpha
}(t,u).
\]
From inequality
(\ref{eq.init}) we deduce inclusions $C_{c}(\mathbb{R}_{+},X)\subset
\mathcal{C}_{\alpha}\subset C_{b}(\mathbb{R}_{+},X)$.

\begin{remark}
\label{rem.incl} If $\alpha\in\mathcal{A}(\mathcal{U})$ and $\beta\geq\alpha$,
then $\beta\in\mathcal{A}(\mathcal{U})$ and $\mathcal{C}_{\alpha
\subset\mathcal{C}_{\beta}$.
\end{remark}

\begin{example}
\label{ex} For the evolution family $U(t,s)=e^{f(t)-f(s)}\,\mathrm{Id}$, $t\geq
s\geq0$, where $f(t)=-2t+t\sin^{2}t$, we have
\[
\mathcal{A}(\mathcal{U})=[-1,\infty) \text{ and } \mathcal{C}_{\alpha
}=\mathcal{C}_{-1} \text{ for every } \alpha\in\mathcal{A}(\mathcal{U}).
\]
\end{example}

\begin{proof}
Let us notice that
\[
f(t)-f(s)=-(t-s)-t\cos^{2}t+s\cos^{2}s\leq s-(t-s),
\]
hence $-1\in\mathcal{A}(\mathcal{U})$. Remark \ref{rem.incl} yields
$[-1,\infty)\subset\mathcal{A}(\mathcal{U})$. For each $\varepsilon>0$ we
claim that $-1-\varepsilon\notin\mathcal{A}(\mathcal{U})$. Indeed, if we
consider
\[
E_{\varepsilon}(t,s)=f(t)-f(s)+(1+\varepsilon)(t-s),\;t\geq s\geq0,
\]
then $E_{\varepsilon}(t,s)=-t\cos^{2}t+s\cos^{2}s+\varepsilon(t-s).$ Since
$E_{\varepsilon}(n\pi+\pi/2,\pi/2)=n\pi\varepsilon\rightarrow\infty$ as
$n\rightarrow\infty$, it follows that $-1-\varepsilon\notin\mathcal{A
(\mathcal{U})$ and thus $\mathcal{A}(\mathcal{U})=[-1,\infty)$. For the second statement, according to Remark \ref{rem.incl}
it suffices to prove the inclusion $\mathcal{C}_{\alpha}\subset\mathcal{C
_{-1}$, if $\alpha\geq-1$. For $\alpha\geq-1$ and  $t\geq s\geq0$ let us consider
\[
E_{\alpha}(t,s)=f(t)-f(s)-\alpha(t-s)=-(\alpha+1)(t-s)-t\cos^{2}t+s\cos^{2}s.
\]
For any fixed $t\geq 0$ we set $f_{\alpha}(t)=\sup\limits_{\tau\geq t}E_{\alpha}(\tau,t)<\infty$.
Since $E_{-1}(\tau,t)\leq t\cos^{2}t$ for all $\tau\geq t$, and
$E_{-1}(n\pi+\pi/2,t)=t\cos^{2}t$ for sufficiently large $n$, we deduce that
$f_{-1}(t)=t\cos^{2}t$.
Let $u\in\mathcal{C}_{\alpha}$ and $t\geq0$. Then
\begin{align*}
\varphi_{-1}(t,u)  &  =\sup\limits_{\tau\geq t}e^{\tau-t}e^{f(\tau
)-f(t)}\|u(t)\|=e^{f_{-1}(t)}\|u(t)\|\\
&  =e^{f_{-1}(t)-f_{\alpha}(t)}\varphi_{\alpha}(t,u)\\
&  =e^{t\cos^{2}t-f_{\alpha}(t)}\varphi_{\alpha}(t,u)\\
&  \leq e^{(\alpha+1)(\tau_{t}-t)}\varphi_{\alpha}(t,u)\\
&  \leq e^{(\alpha+1)\pi}\parallel u\parallel_{\alpha},
\end{align*}
where $\tau_{t}\in\lbrack t,t+\pi)$ is such that $\cos^{2}(\tau_{t})=0$. This
proves that $u\in\mathcal{C}_{-1}$ and thus $\mathcal{C}_{\alpha
}=\mathcal{C}_{-1}$ for every $\alpha\in\mathcal{A}(\mathcal{U})=[-1,\infty)$.
\end{proof}

\section{The main results}

In this section we give necessary and sufficient conditions for the existence
of nonuniform exponential stability of evolution families.

For any fixed $t\geq0$ and $u\in C(\mathbb{R}_{+},X)$ we define a map
$\Phi(t)u:\mathbb{R}_{+}\to X$ by
\[
\left(  \Phi(t)u\right)  (\tau)=
\begin{cases}
U(\tau,t)u(t), & \tau\geq t,\\
u(\tau), & \tau<t.
\end{cases}
\]

Obviously $\Phi(t)u\in C(\mathbb{R}_{+},X)$, thus the operator
$\Phi(t):C(\mathbb{R}_{+},X)\rightarrow C(\mathbb{R}_{+},X)$ is well-defined.
Moreover, we have
\begin{equation}
\varphi_{\alpha}(t,\Phi(t)u)=\varphi_{\alpha}(t,u), \label{eq.1
\end{equation}
and
\begin{equation}
\Phi(t)\Phi(s)=\Phi(s)\Phi(t)=\Phi(s),\;t\geq s\geq0. \label{eq.pre.D
\end{equation}

Indeed, let $t\geq s\geq 0$ and $u\in C(\mathbb{R}_{+},X)$. For $\tau\geq t$ we get
\begin{align*}
\left(\Phi(t) (\Phi(s)u)\right)(\tau)&= U(\tau,t)(\Phi(s)u)(t)=U(\tau,t)U(t,s) u(s)\\
&=U(\tau,s)u(s)=(\Phi(s)u)(\tau).
\end{align*}
If $\tau<t$, it directly follows  that $\left(\Phi(t) (\Phi(s)u)\right)(\tau)=(\Phi(s)u)(\tau).$
The above identities imply $\Phi(t)\Phi(s)=\Phi(s).$ The second identity in \eqref{eq.pre.D} can be obtained following the same type of computations as before.

Setting  $t=s$ in \eqref{eq.pre.D}, we get $\Phi^{2}(t)=\Phi(t)$, that is $\Phi(t)$ is a
projection on $C(\mathbb{R}_{+},X)$.

\begin{lemma}
\label{lem.Datko} Let $\alpha\in\mathcal{A}\left(  \mathcal{U}\right)  $.  For each $u\in C(\mathbb{R}_{+},X)$ the following inequality
holds:
\begin{equation*}
\varphi_{\alpha}(t,\Phi(s)u)\leq e^{\alpha(t-\xi)}\varphi_{\alpha}(\xi
,\Phi(s)u),\;t\geq\xi\geq s\geq0. \label{eq.D
\end{equation*}

\end{lemma}

\begin{proof}
Indeed, for $t\geq\xi\geq s\geq0$ we have
\begin{align*}
\varphi_{\alpha}(t,\Phi(s)u)  &  =\sup\limits_{\tau\geq t} e^{-\alpha(\tau-t)}
\parallel U(\tau,t)(\Phi(s)u)(t)\parallel\\
&  =e^{\alpha(t-\xi)}\sup\limits_{\tau\geq t} e^{-\alpha(\tau-\xi)}\parallel
U(\tau,s)u(s)\parallel\\
&  \leq e^{\alpha(t-\xi)}\sup\limits_{\tau\geq\xi} e^{-\alpha(\tau-\xi
)}\parallel U(\tau,\xi)(\Phi(s)u)(\xi)\parallel\\
&  =e^{\alpha(t-\xi)} \varphi_{\alpha}(\xi,\Phi(s)u).
\end{align*}

\end{proof}

We present below some necessary and sufficient conditions for the existence of nonuniform exponential stability, specific to nonuniform behavior.
\begin{theorem}
\label{prop.admisible.nes} The evolution family $\mathcal{U}$ is nonuniform exponentially stable if and only if there exists $\alpha<0$ such that for every $t\geq 0$ one of the
following conditions holds:

\begin{enumerate}
\item[(i)] $\Phi(t)\mathcal{C}_{\alpha} \subset\mathcal{C}_{\alpha}$;

\item[(ii)] $\Phi(t)\mathcal{C}_{\alpha}$ is a closed subspace of
$\mathcal{C}_{\alpha}$;

\item[(iii)] $\Phi(t)\in\mathcal{L}\left(  \mathcal{C}_{\alpha}\right)$;

\item[(iv)] $\Phi(t)$ is a bounded projection on $\mathcal{C}_{\alpha}$.
\end{enumerate}
\end{theorem}

\begin{proof}
Since $\Phi(t)$ is a projection on
$C(\mathbb{R}_{+},X)$, it is obvious that (iv) is equivalent to (iii). Implications (iii) $\Rightarrow $ (ii) $\Rightarrow$ (i) are
trivial. Assume that (i) holds for some $\alpha<0$ and for all $t\geq 0$. Choose $s\geq0$ and $x\in X$ with $\Vert
x\Vert=1$, and pick $u\in\mathcal{C}_{\alpha}$ such that $u(s)=x$. We have
\begin{align*}
e^{-\alpha(t-s)}\left\Vert U(t,s)x\right\Vert  &  =e^{-\alpha(t-s)}\left\Vert
U(t,s)(\Phi(s)u)(s)\right\Vert \leq\varphi_{\alpha}(s,\Phi(s)u)\\
&  \leq\left\Vert \Phi(s)u\right\Vert _{\alpha}=: M(\alpha,s,x),
\end{align*}
for all $t\geq s$. Uniform boundedness
principle implies that there exists $M\left(\alpha,  s\right)  >0$ such that
\[
e^{-\alpha(t-s)}\left\Vert U\left(  t,s\right)  \right\Vert \leq M\left(\alpha,s\right)  ,\text{ for all }t\geq s,
\]
which proves that $\mathcal{U}$ is $\alpha$-nonuniform exponentially stable. Assume now that $\mathcal{U}$ is nonuniform exponentially stable, that is there exists $\alpha\in\mathcal{A}\left(  \mathcal{U}\right)  $ with  $\alpha<0$. Let $t\geq0$ and
$u\in\mathcal{C}_{\alpha}$. For every $\xi\in\lbrack0,t)$ one gets
\[
\varphi_{\alpha}(\xi,\Phi(t)u)=\sup\limits_{\tau\geq\xi}e^{-\alpha(\tau-\xi
)}\parallel U(\tau,\xi)u(\xi)\parallel=\varphi_{\alpha}(\xi,u)\leq\parallel
u\parallel_{\alpha}.
\]
On the other hand, for every $\xi\geq t$
we have
\begin{align*}
\varphi_{\alpha}(\xi,\Phi(t)u)  &  =\sup\limits_{\tau\geq\xi}e^{-\alpha
(\tau-\xi)}\parallel U(\tau,t)u(t)\parallel\\
&  \leq e^{-\alpha(t-\xi)}\sup\limits_{\tau\geq t}e^{-\alpha(\tau-t)}\parallel
U(\tau,t)u(t)\parallel\\
&  \leq e^{-\alpha t}\varphi_{\alpha}(t,u)\leq e^{-\alpha t}\parallel
u\parallel_{\alpha}.
\end{align*}
The above estimations imply $\left\Vert \Phi\left(  t\right)  u\right\Vert
_{\alpha}\leq e^{-\alpha t}\left\Vert u\right\Vert _{\alpha}$.  Thus, $\Phi(t)$ is a bounded linear operator on
$\mathcal{C}_{\alpha}$, which proves (iii). This completes the proof.
\end{proof}

\begin{proposition}
\label{th.Datko.nec} Let $p\in(0,\infty)$. If the evolution family
$\mathcal{U}$ is $\alpha$-nonuniform exponentially stable, then there exists
$K>0$ such that
\begin{equation}
\int_{t}^{\infty}\varphi_{\alpha}^{p}(\xi,u)d\xi\leq K\varphi_{\alpha
^{p}(t,u),\; u\in\Phi(t)\mathcal{C}_{\alpha}, \; t\geq0.
\label{eq.Datko
\end{equation}

\end{proposition}

\begin{proof}
Let $t\geq0$ and $u\in\Phi(t)\mathcal{C}_{\alpha}$. According to Theorem
\ref{prop.admisible.nes}, Lemma \ref{lem.Datko} and Eq. \eqref{eq.1} we have
\begin{align*}
\int_{t}^{\infty}\varphi_{\alpha}^{p}(\xi,u)d\xi &  =\int_{t}^{\infty
\varphi_{\alpha}^{p}(\xi,\Phi(t)u)d\xi\leq\int_{t}^{\infty}e^{p\alpha(\xi
-t)}\varphi_{\alpha}^{p}(t,\Phi(t)u)d\xi\\
&  =1/({-p\alpha})\varphi_{\alpha}^{p}(t,u),
\end{align*}
which concludes the proof.
\end{proof}

\begin{corollary}
\label{th.Ly.nec} Let $p\in(0,\infty)$. If $\mathcal{U}$ is $\alpha
$-nonuniform exponentially stable, then there exist $W:\mathbb{R}_{+
\times\mathcal{C}_{\alpha}\to\mathbb{R}_{+}$ and $K>0$ such that
\begin{equation}
\label{eq.Ly.1}W(t_{0},u)\leq K\varphi_{\alpha}^{p}(t_{0},u),
\end{equation}
and
\begin{equation}
\label{eq.Ly.2}W(t,u)+\int_{t_{0}}^{t}\varphi_{\alpha}^{p}(\xi,u)d\xi
=W(t_{0},u),
\end{equation}
for all $t\geq t_{0}\geq0$ and $u\in\Phi(t_{0})\mathcal{C}_{\alpha}$.
\end{corollary}

\begin{proof}
Set $W(t,u)=\int_{t}^{\infty} \varphi_{\alpha}^{p}(\xi,\Phi(t)u)d\xi$, for
$t\geq0$ and $u\in\mathcal{C}_{\alpha}$. Inequality \eqref{eq.Ly.1} is a simple
consequence of Proposition \ref{th.Datko.nec}, and Eq. \eqref{eq.Ly.2} follows
from \eqref{eq.pre.D}.
\end{proof}

The following theorem can be considered a generalization of a theorem of Datko
\cite{Da.1972} for nonuniform exponential stability.

\begin{theorem}
\label{th.Datko.suf} Let $\alpha\in\mathcal{A}(\mathcal{U})$. If there exist
$p,K>0$ such that \eqref{eq.Datko} holds, then the evolution family
$\mathcal{U}$ is nonuniform exponentially stable.
\end{theorem}

\begin{proof}
It suffices to assume that $\alpha\geq0$. For convenience of the reader we divide the proof of the theorem into several steps.

\textit{Step 1}. We show that there exists a constant $N>0$ such that
\begin{equation}
\varphi_{\alpha}^{p}(t,\Phi(s)u)\leq N\varphi_{\alpha}^{p}(s,u), \;u\in\mathcal{C}_{\alpha}, \; t\geq s\geq 0. \label{eq.D1
\end{equation}

The arguments in this step closely follow a classical idea of Ichikawa (see
Theorem 3.1 in \cite{Pr.Pr.Cr.2012}). At
least for a better understanding, we prefer exposing all the details. Let
$t\geq s\geq0$. If $t\geq s+1$, then using Lemma \ref{lem.Datko},
\eqref{eq.Datko} and \eqref{eq.1} we have
\begin{align*}
\varphi_{\alpha}^{p}(t,\Phi(s)u)  &  =\int_{t-1}^{t}\varphi_{\alpha
^{p}(t,\Phi(s)u)d\xi\leq\int_{t-1}^{t}e^{p\alpha(t-\xi)}\varphi_{\alpha
^{p}(\xi,\Phi(s)u)d\xi\\
&  \leq e^{p\alpha}\int_{s}^{\infty}\varphi_{\alpha}^{p}(\xi,\Phi(s)u)d\xi\leq
e^{p\alpha}K\varphi_{\alpha}^{p}(s,\Phi(s)u)=e^{p\alpha}K\varphi_{\alpha
^{p}(s,u).
\end{align*}
On the other hand, if $s\leq t<s+1$, then  Lemma \ref{lem.Datko} and
Eq. \eqref{eq.1} yield
\[
\varphi_{\alpha}^{p}(t,\Phi(s)u)\leq e^{p\alpha(t-s)}\varphi_{\alpha
^{p}(s,\Phi(s)u)\leq e^{p\alpha}\varphi_{\alpha}^{p}(s,u).
\]
Setting now $N=\max\left\{  e^{p\alpha}K,e^{p\alpha}\right\}  $, clearly
\eqref{eq.D1} holds.

\textit{Step 2}. We show that for all non-negative integers $n\in\mathbb{N}$,
\begin{equation}
\varphi_{\alpha}^{p}(t,\Phi(s)u)\frac{(t-s)^{n}}{n!N^{n}}\leq N\varphi
_{\alpha}^{p}(s,u), \; u\in\mathcal{C}_{\alpha}, \; t\geq s\geq0.
\label{eq.D2
\end{equation}

We will prove  \eqref{eq.D2} by induction on $n$. \emph{Step 1} shows that it works
for $n=0$. Assume that \eqref{eq.D2} holds for some $n\in\mathbb{N}$. Let $u\in\mathcal{C}_{\alpha}$ and
$t\geq s\geq0$. According to \eqref{eq.D2}, for every $\xi\in[s,t]$ we have
\[
\varphi_{\alpha}^{p}(t,\Phi(\xi)u)\frac{(t-\xi)^{n}}{n!N^{n}}\leq
N\varphi_{\alpha}^{p}(\xi,u).
\]
Substituting above $u$ by $\Phi(s)u$, and using \eqref{eq.pre.D} one has
\begin{equation}
\varphi_{\alpha}^{p}(t,\Phi(s)u)\frac{(t-\xi)^{n}}{n!N^{n}}\leq N\varphi
_{\alpha}^{p}(\xi,\Phi(s)u). \label{eq.D3
\end{equation}
Now using \eqref{eq.D3}, \eqref{eq.Datko} and \eqref{eq.1} we get
\begin{align*}
\varphi_{\alpha}^{p}(t,\Phi(s)u)  &  \frac{(t-s)^{n+1}}{(n+1)!N^{n+1}
=\frac{1}{N}\int_{s}^{t}\varphi_{\alpha}^{p}(t,\Phi(s)u)\frac{(t-\xi)^{n
}{n!N^{n}}d\xi\\
&  \leq\int_{s}^{t}\varphi_{\alpha}^{p}(\xi,\Phi(s)u)d\xi\leq K\varphi
_{\alpha}^{p}(s,\Phi(s)u)\\
&  =K\varphi_{\alpha}^{p}(s,u)\leq N\varphi_{\alpha}^{p}(s,u),
\end{align*}
that is estimation (\ref{eq.D2}) holds for $n+1$. By induction argument it follows that
it holds for all $n\in\mathbb{N}$.

\textit{Step 3}. For any fixed $\delta\in(0,1)$, there exists a constant
$\widetilde{N}>0$ such that
\begin{equation}
\parallel U(t,s)u(s)\parallel\leq\widetilde{N}e^{-\frac{\delta}{pN
(t-s)}\varphi_{\alpha}(s,u), \; u\in\mathcal{C}_{\alpha}, \; t\geq s\geq0. \label{eq.D4
\end{equation}
Let $u\in\mathcal{C}_{\alpha}$,
$t\geq s\geq0$ and pick $\delta\in(0,1)$. Multiplying inequality \eqref{eq.D2} by
$\delta^{n}$ and summing over $n\in\mathbb{N}$, one gets
\[
\varphi_{\alpha}^{p}(t,\Phi(s)u)\sum\limits_{n=0}^{\infty}\frac{1}{n!}\left[
\frac{\delta(t-s)}{N}\right]  ^{n}\leq N\varphi_{\alpha}^{p}(s,u)\sum
\limits_{n=0}^{\infty}\delta^{n},
\]
which is equivalent to
\[
\varphi_{\alpha}(t,\Phi(s)u)\leq\left(  \frac{N}{1-\delta}\right)
^{1/p}e^{-\frac{\delta}{pN}(t-s)}\varphi_{\alpha}(s,u).
\]
Since
$
\parallel U(t,s)u(s)\parallel\leq\varphi_{\alpha}(t,\Phi(s)u),
$
we deduce that  (\ref{eq.D4}) works with $\widetilde{N}=\left(
\frac{N}{1-\delta}\right)  ^{1/p}$.

\textit{Step 4}. For any fixed $\delta\in(0,1)$, the following holds:
\begin{equation}
\parallel U(t,s)\parallel\leq\widetilde{N}M(s)e^{-\frac{\delta}{pN
(t-s)}\text{, }t\geq s\geq0. \label{eq.D5
\end{equation}

Picking $x\in X$, let us define the following map
\[
u_{x}(t)=\frac{1}{M(t)}\cdot x,\;t\geq0.
\]
Estimation \eqref{eq.init} readily implies $\varphi_{\alpha}(t,u_{x})\leq\Vert
x\Vert$, thus $u_{x}\in\mathcal{C}_{\alpha}$ and from \emph{Step 3} the proof is now complete.
\end{proof}

\begin{remark}
\label{rem2}From the above arguments we deduce that the condition
$u\in\Phi(t)\mathcal{C}_{\alpha}$ can be easily replaced by $u\in\mathcal{F}$,
where $\mathcal{F}$ is an arbitrary Banach space of continuous functions
$u:\mathbb{R}_{+}\rightarrow X$, containing the constant ones. The case when
$\mathcal{F}$ is the space of the constant functions (identified with $X$) was
proved in \cite{Pr.Pr.Cr.2012}, but using a quite different method. Moreover,
we emphasize that our results cannot be deduced or proved anyway using this
source.
\end{remark}

Our next result generalizes the conclusion from Theorem 3.3 in
\cite{Pr.Pr.Cr.2012}.

\begin{corollary}
\label{th.Ly.suf} Let $\alpha\in\mathcal{A}(\mathcal{U})$. If there exist
$W:\mathbb{R}_{+}\times\mathcal{C}_{\alpha}\rightarrow\mathbb{R}_{+}$ and
$p,K>0$ such that \eqref{eq.Ly.1}--\eqref{eq.Ly.2} hold, then $\mathcal{U}$ is
nonuniform exponentially stable.
\end{corollary}

\begin{proof}
According to \eqref{eq.Ly.1} and \eqref{eq.Ly.2}, since $W(t,u)\geq0$ for all
$u\in\mathcal{C}_{\alpha}$, we get
\[
\int_{t_{0}}^{t}\varphi_{\alpha}^{p}(\xi,u)d\xi=W(t_{0},u)-W(t,u)\leq
W(t_{0},u)\leq K\varphi_{\alpha}^{p}(t_{0},u),
\]
for all $t\geq t_{0}\geq0$ and $u\in\Phi(t_{0})\mathcal{C}_{\alpha}$. Letting
above $t\rightarrow\infty$ and using Theorem \ref{th.Datko.suf}, we conclude
that $\mathcal{U}$ is nonuniform exponentially stable.
\end{proof}

\section*{Conflict of Interests}

The authors declare that there is no conflict of interests regarding the
publication of this paper.

redpajama_set_name: RedPajamaArXiv


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Where all them profits going?
I don't plan to devote many words to current events, but this one struck me as being somewhat interesting and has been a trend for the last 7 quarters, so perhaps there is something deeper and more interesting going on here.
Corporate profits have been climbing steadily and setting records recently, while unemployment has remained stagnant. This is sometimes used as an example of how strange the market is behaving these days.
I would contend that this behavior on the part of major corporations makes perfect sense given the economics shifts we have endured, and it points to a healthy, reasoned functioning of the free markets.
When the credit markets froze during the beginning of the recession, companies that relied on the free flow of credit found significant areas of their business in jeopardy. Without being able to freely borrow money and repay with quick turn around time (commercial paper) they found themselves unable to function efficiently. Companies that used to be able to borrow money to make payroll if their collections were low one month realized how close to the wire they were really running their company.
The response to this tightening of credit and the realization that those companies were so dependent on the credit market is cash hoarding. Companies are turning out record profits by keeping a tight control on expenses and continuing to do more with less in large part to shore up their balance sheets. They want enough cash in the bank to be able to survive the market swings we are experiencing.
You may have noticed the same thing in your own household. Keeping a little extra in the bank since you know it may not be quite so easy to go out and borrow more money. Or even if you could borrow that money, you don't like the idea of the bank having so much influence over your financial future.
We are the next depression era generation. I darned 3 pairs of socks last sunday night. I've never considered myself a crafty person, but the idea of saving 20 dollars by darning a pair of wool socks was pretty appealing. I'm paying down my debt as quickly as possible, while at the same time trying to put some money in the bank.
Large corporations are doing the same thing. The US economy was very highly leveraged 3 years ago, and now that leverage is unwinding. Where are those profits going if they are not going to increased hiring or wages? They are going to pay down debt, and to put some cash in the bank. The companies left are the ones that adjusted, that learned to survive in this new economy you need to be thrifty. You can't depend on freely flowing money anymore.
I expect to see the markets shift over the next 12 months. Already there is more talk of startups being funded. All this cash is good when hedging bets, but that cash in the bank doesn't grow until you start putting it to work.
So what can we learn from this? I think the first take-away is that corporations are not behaving much different than individuals in this regard. After all, companies are run by individuals. Their behavior typically doesn't stray far from our own patterns.
Secondly, this is a good opportunity to focus on the core 20%. Without a lot of extra money, companies and individuals should be seeking to return to their core principals. Cut out products and customers that aren't profitable. Drop expensive hobbies you rarely have time to practice. Darn a few socks, and make sure that next sweater you buy is going to last a few seasons.
Some are less so "if you create a document with these bizarre placeholders and hook me up to excel I will do a database driven mail merge for you".
All software makes promises. The best software delivers on it's promises. The worst software doesn't even understand the promises it's making.
If you are buying or using a piece of software, don't automatically assume that any struggles you encounter are due to your inability to understand "these damn things". It's probably not your fault.
If you are a software designer, take the time to explore your creation looking for broken promises. The more broken promises your users encounter, the more frustrated they are going to be with the experience, and frustrated users don't often translate into loyal users.
Finding hidden promises is difficult, but the easiest way to find them is through some kind of user testing. Sitting a novice user down in front of your software and asking them to perform common tasks will quickly reveal some promises you didn't know you were making.
You made something great, you broke through the roadblocks and finished what was thought impossible. You showed it at the staff meeting and it flopped. Turns out you were solving the wrong problem, and not only that, you solved it the wrong way. Full on blew it. What should we do in this situation? Take it personally. Sulk a little bit. They just told you your baby is ugly, not only ugly but smelly and probably is never going to grow into those ears. Jerks, all of them.
Taking criticism personally is the curse of the craftsman. You put a piece of yourself into everything you create. Those emotional twinges are not inappropriate.
Be careful though. There are real opportunities for failure here. By identifying too deeply with our creation, we rob ourselves of the opportunity to grow from the experience. An amateur will never get out of the emotional funk brought on by solid spot on criticism. Their next attempt will be at 90% then 80% then eventually they will simply coast along, having lost their willingness to emotionally connect with their creation. It is a defense mechanism. It spares them the pain of seeing their baby stumble.
Conversely, stellar craftmen are able to propose designs with passion, argue deeply for their position, and in the next breath, tear their ideas to shreds. They fully acknowledge every deficiency, knowing that a singular idea does not define them, that their worth is manifest in the process of creation, in the quality of the end product, and not in how many of their proposals were deemed brilliant during the brainstorming meeting.
We used to have a saying during a recent major project I worked on.
The Microsoft Office team went through a similar realization during the design phase of office 2007, when they introduced "the ribbon". They noticed that a lot of the features they had spent man years creating weren't being used.
Their work was not being realized by many users. You can argue whether the ribbon achieved it's stated goal, but the goal itself was sound. If a user can't find the feature, all the time and energy we spent creating it is wasted for that user.
Creating features that are discoverable can take significant design work and unfortunately, this design effort doesn't contribute directly to the "bullet list" of features we are trying to create when building software products. It is often difficult to justify the ROI of this design effort to the higher-ups. Walking them through the frustration users encounter (I'm sure they have all gone through similar experiences) can help build the case that spending the few extra days it takes to put a feature in the right place will pay dividends in user satisfaction and loyalty.
If this ethos can make it's way into your group's culture, you have taken one major step forward in the experience you give your users.


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deliver us, we pray, from this dark place.
as we, oh Lord, complete this two-horse race.
wickedness abounds and men are bold.
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¶ 4 Leave a comment on verse 4 0 Oh, guide us through the perils that we face.
dear Lord, as we complete this two-horse race.
and please deliver us from this dark place.
and let our forwards win promotion gold.
and deliver us, we pray, from this dark place.


text:
Home / Review / Theatre: 1st Irish so far
Theatre: 1st Irish so far
Share:FacebookTwitterGoogle+PinterestTumblrLinkedInRedditEmail
By Gwen Orel
Pat Kinevane in Silent (@Ger Blanch)
How It's New York: The 1st Irish Festival is a New York event, conceived in New York, for a New York audience.
How It's Irish: It's the 1st IRISH Festival, with plays by Irish and Irish-American writers.
Here's my wrap-up of what I've seen so far in the Festival: these plays close this weekend, except for one that has already closed, but will be in the upcoming Solos Festival in October. New plays in the Festival have started: For Love, Jimmy Titanic, Brendan and House Strictly Private. The Festival ends this weekend, with the closing ceremony on October 1!
A version of this article was first published in Irish Examiner USA, Tuesday, Sept. 25.
The 1st Irish Festival, now in its fifth year, is halfway through.
Some of the shows have closed, others will be closing next week, and others will begin this week.
For details and to buy tickets visit www.1stirish.org.
Here's our take on what we've seen so far:
Running at Irish Arts Center, extended through September 30. Presented by Fishamble: The New Play Company in association with the Irish Arts Center
Written and directed by Pat Kinevane and directed by Fishamble's Artistic Director Jim Culleton, the play is a one man show about a homeless man, Corkonian Tino McGoldrig, who was named for Rudolph Valentino.He reflects on his closeted homosexual brother, who killed himself after some aborted (and hilarious) attempts on his own life, his failed marriage, and kindness.
Kinevane, as he did with The Forgotten in 2010, engagingly talks to the audience to pull them into the show, and uses his physicality engagingly.At times it felt a little gimmicky, however. I didn't really buy the device of a bag of exotic dance clothes as an excuse for Tino to dress up, for example, and the ending was a little disappointing.
But Kinevane's humor never lets you down: a comparison between the French and Cork accent is worth the price of admission alone.
Martin Lynch for Green Shoot Productions, Running at 59E59 through September 30.
Marie Jones, the author of November and Stones in His Pockets, wrote and directed this two-hander about careworkers in Belfast who panic when they find their elderly charge has died.
What begins as a slight panic soon turns into larceny and even arson, with the best intentions in the world.
Katie Tumelty,Tara Lynne O'Neill (Vinnie Loughran)
Frances, sharply portrayed by Katie Tumelty, is proud of her teenage son, who's been expelled from school but has entrepreneurial skills in selling bootlegged DVDs, and she's the mastermind of the two women. Fake whiplash? "That's how I got my new kitchen," she says without batting an eyelash.
Loretta, a sweet-faced, blonde and cuddly Tara Lynne O'Neill, has a stronger sense of ethics but with an unemployed husband and a Euro Disney-bound child, she's in sore need of funds.
The play is at its most effective when the reality of that situation comes through in O'Neill's sad voice.
The dead man was a Sinatra fan, hence the title.
There are a lot of laughs, but not a lot of plot development – the situation goes on a bit long to make its point.
But the women are so over the top ridiculous and fun to watch that you'll probably not mind very much.
Hard Times: An American Musical
Presented by and at The Cell Theatre through September 30, with an additional performance added on September 29.
Black 47's Larry Kirwan collaborated with Stephen Foster, the 19th-century composer, for this musical set against the draft Riots in Five Points, New York, in 1863.
There's energy to spare in Kirwan's play, which resets some of Foster's 19th-century ballads in a rock and roll vein, and also includes some Kirwan originals.
At times the characters' perspectives on minstrelsy, race relations and homosexuality (which was not conceived of as an identity until the 1890s) seem a little modern, but then again, we are singing Foster to a rock and roll beat, so it all makes sense.
Jed Peterson, Stephane Duret, Almeria Campbell
Almeria Campbell winningly plays Nelly Blyrh, the black owner of a bar much like McSorley's (there's even McSorley's ale for sale), which seems a kind of show bar where she'll occasionally let people drink if they sing. John Charles McLaughlin plays Owen Duignan, a hotheaded young Irishman who starts the play off vibrantly by singing in blackface about how much he hates singing in blackface.
Nelly is courted by Michael Jenkins, played by Philip Callen, a white man with conflicted feelings.
Other characters include Thomas Jefferson, played by Stephane Duret, Stephen Foster, played by Jed Peterson, and Jane, Foster's wife, played by Erin West.
There's a live band onstage and it's particularly clever the way Kirwan has woven in some reels throughout.
At times, lovely as the music was, and beautiful as the cast's voices were, the songs seemed an interruption to the scenes.
The songs are lovely, though, and include all those Foster songs that are suddenly all the rage again: "Beautiful Dreamer," "I Dream of Jeannie with the Light Brown Hair," "Oh, Susanna," and of course, "Hard Times."
Auditions, Zoe's Auditions, Part 2
Presented by the Be You, All Others Are Taken company, in association with the Drilling company. Closed September 23, but will have additional performances in United Solo Festival on October 15 and 23.
Suzanna Geraghty wrote and appeared in this one-hander about a sweet production assistant on a troubled production of A Christmas Carol who wants to be an actress.
Geraghty's piece is dear, funny and smarter than it seems – much like her character of Zoe Browne.
Andy Crook offered help with direction and movement.
With a Mary Poppins voice and a dizzy dame appeal, Zoe goes through a series of hilariously bad auditions, with a voice shouting "next!"
Suzanna Geraghty (@Ros Kavanagh)
The Riverdance audition ends with a bagpipe on Zoe's head, and her sincere wish that nobody was injured.
A voiceover at the top warns the audience that they will be expected to take part or risk ruining "the reality of the show and everybody's fun."
So people do take part, answering that no, we haven't heard her name called yet, or nominating the person sitting next to them to be "Olga," the director of The Three Sisters.
Geraghty has crafted a smart drama, not just a series of sketches: she shows Zoe interacting with her elderly, crotchety agent, with her dysfunctional cast, and, eventually, with a clever riff on "A Christmas Carol."
Geraghty has hedged her bets a little by showing us that Zoe doesn't just have a dream, she actually has talent, and some training – which doesn't work perfectly with someone who thinks The Three Sisters are people she should know.
But when the ghost of Christmas future answers her "to be or not to be" with "that isn't the question," the play goes somewhere surprisingly powerful.
The only New York journalist who writes for both the Forward and Irish Music Magazine.
McLean Avenue, on it's way to a screen or stream near you
Alice FarrellMay 6May 6
Last week, at the Irish Arts Center, I got the chance to get a sneak peak at the new TV pilot from New York writer Mike Farragher, McLean Avenue.
Michael FitzpatrickMay 5May 7
Johathan Irwin's animation in '99 Problems' How it's New York: '99 Problems' is part of the Tribeca Film Festival. How it's Irish: It's set in...
Craic Fest 2019 Launches Paddy's Week
Alice FarrellMarch 12March 15
How it's New York: The lead-up to St. Patrick's Day in New York City How it's Irish: Features Irish actors, musicians and film in the annual festival that...

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text:
Things to do during the Lockdown
Could you be a volunteer or do you want to know more about what volunteering services are doing? Follow the link: https://volunteering.royalvoluntaryservice.org.uk/volunteer-search
Cooking BSL
Making a smoothy in BSL - https://www.youtube.com/watch?v=UZU02FD-U4U
Marble cakes in BSL - https://www.youtube.com/watch?v=HDo5zqSOvtY
BBC Good food Video site - https://www.youtube.com/user/BBCFood
The Experiment (2010) - Adrian Broady and Forest Whitaker star. Twenty-six subjects are chosen to participate in a psychological experiment. While one group plays the role of prison guards, the others play inmates. Can the guards maintain order when mutiny erupts?
Free to watch here - https://www.youtube.com/watch?v=TgSYxWmgbmA
The Good, he Bad and the Ugly (1966) - The classic Italian epic Spaghetti Western film directed by Sergio Leone and starring Clint Eastwood as "the Good", Lee Van Cleef as "the Bad", and Eli Wallach as "the Ugly".
Free to watch here - https://www.youtube.com/watch?v=4oCdwxmeM2A
Big Jake (1971) - John Wayne is the avenging head of the McCandle clan returns to his estranged family and leads the search for his kidnapped grandson. Big Jake sets off, his sons in tow, to deliver the ransom to the kidnappers, but has little intention of handing it over without a fight.
Free to watch here - https://www.youtube.com/watch?v=X2fNV2jjlWw
I Want Him Dead (1968) Italian-Spanish Spaghetti Western film directed by Paolo Bianchiniand starring Craig Hill
Free to watch here - https://www.youtube.com/watch?v=vs_tr2K2EDs
Phenomenon (1996) - John Travolta stars as a simple mechanic who turns into a genius overnight after a strange incident, he decides to put his brains to good use. But everyone around him suddenly starts growing wary of him.
Free to watch here - https://www.youtube.com/watch?v=MEoT3d0GrfM
Over the Top (1987) - Sylvester Stallone is trucker who decides to mend his relationship with his son after his wife's death. In a turn of events, he also participates in an arm-wrestling championship in order to rebuild his life.
Free to watch here - https://www.youtube.com/watch?v=VFPBM-R5Riw
AWOL/Lion Heart (1990) - Jean Claude Van Damme is a foreign legion deserter who arrives in the US after learning that his brother has been murdered and his family are in financial distress. He participates in fights to support them. This is proper!!
Free to watch here - https://www.youtube.com/watch?v=CFxddFN7XEQ
No Reatret, No Surrender (1986) - A karate-kid wannabe seeks a master to bring his skills to the next level to avenge his father, and finds it unexpectedly in the spirit of Bruce Lee. An early film role for Jean Claude Van Damme.
Free to watch here - https://www.youtube.com/watch?v=gimBsaA3w5Y
Military/War
A Bridge Too Far (1977) - The epic war film features everyone. I mean everyone was in this. I was in it and I wasn't born yet. The film includes Dirk Bogarde, James Caan, Michael Caine, Sean Connery, Edward Fox, Elliott Gould, Gene Hackman, Anthony Hopkins, Hardy Krüger, Laurence Olivier, Ryan O'Neal, Robert Redford, Maximilian Schell and Liv Ullmann.
Free to watch here - https://www.youtube.com/watch?v=H21zi9hj9-A
Who Dares Wins (1982) - The British receive intelligence that an anti-nuclear militant group is planning to kill the U.S. ambassador at the Iranian embassy. The British Special Air Services enlists the help of Capt. Peter Skellen played by Lewis Collins (The Professionels).
Free to watch here - https://www.youtube.com/watch?v=2jGCKv1Tus8
Deadpool 2 - Deadpool protects a young mutant Russell from the authorities and gets thrown in prison. However, he escapes and forms a team of mutants to prevent a time-travelling mercenary from killing Russell.
Free to watch here - https://www.youtube.com/watch?v=jt2wXx_sUCI
Mr Bean TV Series - https://www.youtube.com/watch?v=2k9MpD1gM14
Live Action Kids
Bratz: The Movie - https://www.youtube.com/watch?v=3L63iDvQvW0
Matilda - https://www.youtube.com/watch?v=JbqtvounXs4
Pixar and Similar shorts
For te birds - https://www.youtube.com/watch?v=_nbVTUYVKxg&list=PLlbkyhAZrBl-XJQudaCfoMsGy_Jjau6HE&index=8
Snack Attach - https://www.youtube.com/watch?v=38y_1EWIE9I&list=PLlbkyhAZrBl-XJQudaCfoMsGy_Jjau6HE
Counting Sheep - https://www.youtube.com/watch?v=5oGyBXv6_eg&list=PLlbkyhAZrBl-XJQudaCfoMsGy_Jjau6HE&index=3
The Pyramids - https://www.youtube.com/watch?v=j6PbonHsqW0&list=PLlbkyhAZrBl-XJQudaCfoMsGy_Jjau6HE&index=4
The Present - https://www.youtube.com/watch?v=3XA0bB79oGc&list=PLlbkyhAZrBl-XJQudaCfoMsGy_Jjau6HE&index=5
Lifted - https://www.youtube.com/watch?v=i62znvPLlrw&list=PLlbkyhAZrBl-XJQudaCfoMsGy_Jjau6HE&index=7
The 30 day Lego challenge: https://mk0freehomescho2g81n.kinstacdn.com/wp-content/uploads/2016/01/Lego30daychallenge.pdf
North Wales Police have produced this colour in rainbow for you to print off and place in your window: Downlaod it here Colour Me Rainbow
Here is a link to the Catrin and Abi pages where there is activities, videos and colour in sheets: https://www.centreofsignsightsound.org.uk/catrin-and-abi
Edinburgh zoo have release live webcams of the animals: https://www.edinburghzoo.org.uk/webcams/penguin-cam/#penguincam
Tourism and Live WebCams
Learn about Conwy and its history. You will be able to watch videos here in BSL with Welsh and English Audio. The Password is Culture: https://vimeo.com/showcase/6626080
01492 530013 - Centre of Sign Sight Sound
info@signsightsound.org.uk


text:
فارسی عربي Shqip Saturday, Jan 16, 2021
habilian
Iran slams designation of Cuba as state sponsor of terrorism
The Champion of Fight Against Terrorism
MKO claims responsibility for assassination of Basij's head in Ramsar
Meet the Terrorism Supporter vying for House foreign affairs leadership
Iran has a clear track record against al-Qaeda terrorism: FM
A Brief History of the MeK's Activities between 1971 and 1975 (Part Two)
MKO claims responsibility for eleven terrorist attacks in a single week
Iranian Foreign Ministry vows bringing assassins of Gen. Soleimani to justice
Who is the link between the Mek and the Zionists in Canada?
Fight against terrorism requires setting aside dual behaviors
Lobbyists of Terror
Terror Victims
A List of 17000 terror victims in Iran
MKO terrorists must stand trial in Iran: Minister
Written by Press TV
Iran's Justice Minister Mostafa Pour-Mohammadi says the terrorist Mujahedin-e Khalq Organization (MKO) members should be handed over to Iran to be put on trial for their crimes against the Iranian nation.
"These individuals are criminals and guilty and they must be handed over to our courts so that we would deal with their offenses," Pour-Mohammadi said on Sunday. He said that the MKO agents must stand trial for killing "thousands of our citizens", damaging the Iranians' public properties and "helping our enemies." "In order to restore the rights of the Iranian nation and enforce justice, these traitors must be punished," Pour-Mohammadi said. Pour-Mohammadi further pointed to remarks by sIraqi Minister of Justice Hassan al-Shammari on March 3 that Tehran can ask Baghdad to extradite MKO members. "The handover of the criminal traitors (MKO agents) to Iran does not need a request, but if the Iraqi justice minister has made such a suggestion, we will again lodge an official request to turn them over to our tribunals," Pour-Mohammadi said. The UN High Commissioner for Refugees (UNHCR) said last December that the agency and the United Nations Assistance Mission for Iraq (UNAMI) had relocated 311 out of 3,200 residents of the Camp Hurriya (Liberty) to third countries. In December 2011, the UN and Baghdad agreed to relocate some 3,000 MKO members from Camp New Iraq, formerly known as Camp Ashraf, to the former US-held Camp Hurriya. The last group of the MKO terrorists was evicted by the Iraqi government in September 2013 and relocated to Camp Hurriya to await potential relocation to other countries. The MKO is listed as a terrorist organization by much of the international community and has committed numerous terrorist acts against Iranians and Iraqis. The group fled Iran in 1986 for Iraq, where it received now-executed Iraqi dictator Saddam Hussein's support and set up Camp Ashraf near the Iranian border.
http://www.presstv.ir/detail/2014/03/09/353955/mko-terrorists-must-stand-trial-in-iran/
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Trump must be punished for Soleimani assassination: Judiciary chief
IRNA
US terrorist adventurism lead to Ukrainian plane incident: IRGC
Iran's speaker: US assassination of Gen. Soleimani organized state terrorism
Gen. Soleimani flag-bearer of fight against terrorism: envoy
Afghan political party leader decries General Soleimani's assassination "blatant violation of international law"
Albanian Journalists interrogated by MEK
A Brief History of the MeK's Activities between 1971 and 1975 (Part One)
Iran Accuses U.K.'s G4S of Role in Gen. Soleimani Assassination
Medicine embargo on Iran example of terrorism
Gen. Soleimani assassination flagrant violation of int'l law
Copyright © 2005-2016 Habilian Association
Americans describe the MeK


Blue Sky Bookshelf: End of the Year 2013 Reading Cram Read-a-thon!
End of Year 2013 Read-a-Thon #ReadingCram is hosted by Jenny from Juliababyjen's Reading Room and Dana from Dana Square and will run from the 9th of December (today!) to the 22nd of December.
1. To chisel off some of your TBR pile.
2. Do all the fun challenges and win some prizes.
3. Encourage others to read like a speed demon (and yourself) + reading with others is loads of fun.
So let's get in our last books of 2013 together in the #ReadingCram!
Embarassingly, most some of these books are ones I've been saying all year that I'll be reading. I posted about them in the summertime, in the spring, every season. But now I have this challenge to really push me. I already completed 3 books these past couple of days so I won't add those.
And if I can actually achieve the above titles, I'll dig into Edelweiss.
There are other Edelweiss copies that I won't download until the last possible second because I really think I would like them better as an actual book in my hands. I'm still wary about ebooks, but they're so tempting.
My goals: If I really wanted to complete my goal, it would be to finish every one of these. But life gets in the way so I would like to finish at least four books, which means that if I finish a trilogy, that's three right there. A mini-challenge may be fun, but that's secondary to me.
This is my first read-a-thon or challenge of any sort so hopefully I don't completely fail. If anyone else is dong this read-a-thon (or any reading goal for December), I'd love to hear what your goals are and what books you'll be reading. Did I miss a certain book that is a NECESSITY?
Do you have a December goal? Any personal reading challenges before 2013 ends?
I don't know why I haven't read it yet! And I hope you like Everneath!
I am terrible at read-a-thons. I always start out with such lofty goals, but life gets in the way and I get *maybe* three books read. I'm not a speed-reader...I like to comprehend what I'm reading, and if I read as fast as I can, I won't. That's not to say I'm a slow reader, but compared to a lot of other people I know, I guess I am. :P I hope you get all of these read, Sunny. I've read all but Scarlet and Code Name Verity, and I can attest that all are awesome reads, worthy of your read-a-thon goals. :) Good luck!
I've always been nervous because I don't like commitments and usually when I say I'll do something, I completely fail at it and would've been better off just not trying because that's when I usually DO succeed, oddly enough. But maybe I'll do good with a goal! Thanks, Jen! I'm in the midst of Scarlet and loving it!
How are you liking Graceling so far?
I have never done a read-a-thon because I feel like I'd be terrible and finish 1 book total. lol I'm such a slow reader. I wish you lots of luck with this! Graceling is amazing! And I'm about halfway through Evertrue but it's going really slow for me. :( Good luck!
Next time, do it with me! We could fail together ;) Aw no. I've heard such great (mostly shocked) things about that one!
Thanks, Megan! So far, so good, thankfully.
I KNOW. It's truly horrible. Thanks, Alise!
I AM! Man, I'm so happy with my picks.
I read it and really liked it! The beginning was slow, but it definitely picked up. I hope you get to it soon!
I keep hearing code name verity is great but I keep putting off reading it myself!
This sounds fun! I've never done a read-a-thon because I actually tend to enjoy books more when I read them slowly - and this one is at a bad time for me, what with finals and such - but I hope you achieve your goals. Let me know when you finish Code Name Verity and Everbound!
So far...kinda good. I did great in the beginning, but now in a slump of sorts. And not enough time. Finished Code Name Verity!
Graceling, Throne of Glass, and Code Name Verity are all amazing 5 star reads!! Great list!!
2 out of the 3 you mentioned I've read and they were really good, so you're right!
Everbound is an amazing sequel, definitely one of the best I've read! Dark Triumph is also awesome :) Maybe One Day sounds really good, too. I hope you complete all your goals!
I have Scarlet in my pile. I really hope to get to it soon.


If you think about what's happened the last 4-5 years in smartphones, it's truly remarkable. Apple has created an entirely new, multibillion-dollar franchise. Nokia and RIM's dominance have been severly damaged. And entirely new app and OS systems have been developed.
What's it all mean? As we surmised back in January, it probably means more pain for the legacy market leaders — people such as Nokia, Ericsson, and RIM. Clearly their market-share is eroding, but more importantly they are losing their grips on OS and apps markets. Apps willd drive this market forward, and Apple and Google's Android are leading here.
With Verizon rumored to be close to picking up the iPhone, look for them to sell upwards of 12 million (Apple Insider). Boy, Nokia and RIM are totally screwed now.
Credit where credit is due: Bloomberg had the original iPhone-on-Verizon story (Bloomberg). The big Bloomie can still rock out!
RIM CEO puts on the gameface, even though quarterly subscribers were weak and clearly the iPhone-on-Verizon story is scary for RIM (WSJ). Rearranging deck chairs, anybody?
RIM is trying to lure developers to its App store (WSJ). Here, kitty kitty.
How to troubleshoot the iPhone 4 antenna problems (BGR). 1) Get out of car 2) Put on tin-foil hat.
Cisco introduced a tablet device! (ZDnet). Picture John Chambers speaking in his soothing tones: "This will really enable collaboration and innovation in the workplace.. " yada yada yada.
Cool international espionage news of the day: A Russian spy worked in the Barclays office in London (Guardian). I love the fact that she had a LinkedIn profile. Think of all your Facebook friends that could be spies!
Market looks pretty snoozy today (MarketWatch). Time for the hedgies to spruce of their end-of-Q marketups and markdowns and then grab a facial. Don't be lured into complacency, becuase the next market facial may be coming.
Wall Street psychology cheat sheet (The Big Picture).
Profiles of "Open Carry" advocates (Time). In Montana, this is not news. My friend pointed out that he walked by a guy carrying a shotgun down the street the other day.


Liane is a Portuguese singer, songwriter, composer and multi-instrumentalist, who has been studying the voice for over 12 years, and performing and singing for about 8 of those years.
Her live events performances are acoustic, classy, warm and energetic sessions, in which she reinterprets cover songs and originals alike. Specialising in live looping as well, Liane can create either more "chilled" sets that fit elegant, relaxed atmospheres, or more upbeat, funky sets with guitar percussion to create that great, danceable mood.


I will fully admit that despite the fact that I spend a big part of my week showing women how to make more time for themselves — that it's one of the hardest parts for me to get through to women.
It's really hard for busy moms to choose to stop and take a break for themselves.
There is a real resistance at play here.
That just goes to show you how hard it is, in fact, to make time for what matters where there is little time to be found.
In my work as a family wellness coach with women around the world, I am constantly amazed by the resistance that comes into play in our everyday lives around taking better care of ourselves.
I deliciously DO make time for myself in way more ways than most — even if it's not always ALL that I'd like to make time for — but it's still really hard for me to do as well.
Making time for yourself is something that can — and should be — intentional and deliberate and something that you consider a top priority in motherhood.
For instance, I'd like to make more time for longer walks, longer yoga sessions and doing more creative writing and art. I'd also like to bake more.
But the truth is that I take very good care of myself. I take the time I need when I can … and I communicate my needs to my family when I need my time to breathe and think and just be.
And I follow a few rules as well and I thought I'd share them with you. These are simple ways for me to make sure I'm working in some peace and quiet – and time for doing what makes me happy each day.
Say what you want about forcing yourself to get out of bed early … but it's worth its weight in gold when you do it. So it's Rule No. 1. Always for me. Weekends, included. Vacations included. Join my Rise and Shine Challenge here. It's free.
Hence the first rule … and this is also a great time for me to relax and be more attentive to my body. My bedtime includes a gratitude practice, journaling and reading — with the rare TV show to cap it all off.
OK, this is my dog's rule but it's working in my favor. This rule is one I cannot break. I walk the dog every afternoon and we take a few walks in different directions so that I'm moving my body, noticing the beauty of the neighborhood and enjoying fresh air.
Taking are of myself and getting time for myself doesn't have to be complicated. I am not going to sit around waiting for a chance to escape the house or homework battles or bedtime duty. I work in my favorite ways to enjoy life WHILE doing motherhood at the same time. I call this soulcare in my program. When you know what makes your soul happy, it makes it easy to work into your day!
What are your tips for making time for yourself?


text:
That is, they who are the children of the flesh, these are not the children of God; rather, the children of the promise are counted as the seed.
That is, it is not the children of the flesh that are children of God; but the children of the promise are reckoned for a seed.
That is, it is not the children of the body [Abraham's natural descendants] who are God's children, but it is the children of the promise who are counted as [Abraham's true] descendants.
That is to say, it is not the children of the body [of Abraham] who are made God's children, but it is the offspring to whom the promise applies that shall be counted [as Abraham's true] descendants.
That is, They which are the children of the flesh, these are not the children of God: but the children of the promise are counted for the seed.
That is, it is not the children by physical descent who are God's children, but the children of the promise are considered to be the offspring.
That means it isn't the natural children who are God's children, but it is the children from the promise who are counted as descendants.
In other words, it is not the physical children who are children of God, but the children the promise refers to who are considered seed.
In fact, when God made the promise to Abraham, he meant only Abraham's descendants by his son Isaac. God was talking only about Isaac when he promised
That is, [they that are] the children of the flesh, these [are] not the children of God; but the children of the promise are reckoned as seed.
DLNT
That is, these children of the flesh are not children of God, but the children of the promise are counted for seed.
That is to say, not they that are the children of the flesh, are the children of God; but they, that are the children of the promise, are accounted for the seed.
This means that not all of Abraham's descendants are God's true children. Abraham's true children are those who become God's children because of the promise he made to Abraham.
EHV
This means that it is not the children of the flesh who are God's children, but it is the children of the promise who are counted as his descendants.
This means that it is not the children of the flesh who are the children of God, but the children of the promise are counted as offspring.
ESVUK
EXB
This means that ·not all of Abraham's descendants [L it is not the children of the flesh who] are God's true children. [L But; Rather] ·Abraham's true children [L Those counted as Abraham's descendants/seed] are ·those who become God's children because of the promise God made to Abraham [L the children of the promise].
That is, they which are the children of the flesh, are not the children of God: but the children of the promise, are counted for the seed.
This means that children born by natural descent from Abraham are not necessarily God's children. Instead, children born by the promise are considered Abraham's descendants.
This means that the children born in the usual way are not the children of God; instead, the children born as a result of God's promise are regarded as the true descendants.
This means that not all of Abraham's descendants are God's true children. Abraham's true children are those who become God's children because of the promise God made to Abraham.
That is, it is not merely the children born through natural descent who were regarded as God's children, but it is the children born through the promise who were regarded as descendants.
That means that it is not the natural descendants who automatically inherit the promise, but, on the contrary, that the children of the promise (i.e. sons of God) are to be considered truly Abraham's children. For it was a promise when God said: 'At this time I will come and Sarah shall have a son'. (Everybody, remember, thought it quite impossible for Sarah to have a child.) And then, again, a word of promise came to Rebecca, at the time when she was pregnant with two children by the one man, Isaac our forefather. It came before the children were born or had done anything good or bad, plainly showing that God's act of choice has nothing to do with achievements, good or bad, but is entirely a matter of his will. The promise was: 'The older shall serve the younger'.
That is, Those who are sons of the flesh, these are not the sons of God; but those who are sons of the promise are counted in the generation.
That is, it is not the children by human descent who are children of God, but the children of the promise are counted as descendants.
This means that not all of Abraham's children are children of God, but only those who believe the promise of salvation which he made to Abraham.
Don't suppose for a moment, though, that God's Word has malfunctioned in some way or other. The problem goes back a long way. From the outset, not all Israelites of the flesh were Israelites of the spirit. It wasn't Abraham's sperm that gave identity here, but God's promise. Remember how it was put: "Your family will be defined by Isaac"? That means that Israelite identity was never racially determined by sexual transmission, but it was God-determined by promise. Remember that promise, "When I come back next year at this time, Sarah will have a son"?
So those who are the children of the flesh are not the children of God, but the children of the promise are counted as descendants.
MOUNCE
This means it is not the children of the flesh who are children of God, but the children of the promise are counted as descendants.
This means that it is not the children of the flesh who are the children of God, but the children of the promise are counted as descendants.
That is, it is not the children of the flesh who are children of God, but the children of the promise are regarded as descendants.
NCV
This means it is not the children of the flesh who are the children of God; rather, the children of promise are counted as descendants.
In other words, God's children are not just in the family line of Abraham. Instead, they are the children God promised to him. They are the ones considered to be Abraham's children.
In other words, it is not the children by physical descent who are God's children, but it is the children of the promise who are regarded as Abraham's offspring.
That is, those who are the children of the flesh, these are not the children of God; but the children of the promise are counted as the seed.
This means that children born to Abraham are not all children of God. Only those that are born because of God's promise to Abraham are His children.
This means that Abraham's physical descendants are not necessarily children of God. Only the children of the promise are considered to be Abraham's children.
That is to say, the children of the flesh are not the children of God, but the children of promise are counted the seed.
NRSVA
NRSVACE
NRSVCE
That means that it isn't the flesh-and-blood children who are God's children; rather, it is the children of the promise who will be calculated as 'seed'.
That is, it is not the b'nei habasar (old humanity without hitkhadshut) who are the b'nei HaElohim (children of G-d) but the b'nei HaHavtachah (children of the promise) who are reckoned as ZERA (seed, children, including the right of the heir in relation to the father).
This confirms that it is not merely the natural offspring of Abraham who are considered the children of God; rather, the children born because of God's promise are counted as descendants.
That is, those who are the children of the flesh are not the children of God. But the children of the promise are counted as the seed.
This means that it is not the children of the flesh who are the children of God, but the children of the promise are reckoned as descendants.
That is, it is not the children of the flesh who are children of God; rather, the children of the promise are counted as seed.
The proper interpretation is this: Abraham's children by natural descent are not necessarily God's covenant people; what matters is that His children receive and live the promise.
That is, it is not the children of the flesh who are children of God, but the children of the promise are counted as heirs.
For this is what God promised, `About this time next year I will come and Sarah shall have a son.'
WYC
that is to say, not they that be sons of the flesh, be sons of God, but they that be sons of [the] promise be deemed in the seed.
that is, the children of the flesh -- these [are] not children of God; but the children of the promise are reckoned for seed;
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The Oliva Verde Ristorante is located on Old Town Square, in the historic centre of Prague, opposite its most famous monument, the historic Prague Astronomical Clock or Orloj.
The environment of our gothic building, "At the Blue Star", with its Romanesque foundations from the 13th Century will evoke for you the atmosphere of sunny Italy.
On our variegated menu created by our chef Jaroslav Zahálka you will discover fresh fish and succulent meats dishes together with a wide range of pasta dishes. Of course also not to be missed are our traditional Italian pizzas. Lastly, you can enjoy the real Italian tiramisu that is prepared in our own patisserie at the Grand café Praha.
May we invite you for a great Italian breakfast, which is served between 9.00 a.m. and 11.00 a.m.?


After breaking ground in the international calling space, Ringo, the low-cost calling app, has introduced local calling in India. With the app, Ringo users can call any landline or mobile in the country at a flat rate of 19 p/min, without any additional charges like roaming, STD or top-up cards. Unlike other VoIP apps in the market, Ringo uses telephone networks instead of phone data or the Internet for a phone call, and thus offers superior calling experience, without any call drops.
Ringo is a one of its kind app in India that eliminates all the hassles of communication today, including switching between apps for different calling options, confusing tariff plans, high costs, and low quality calls. It can even make calls if the recipient of the call does not have the app installed. With Ringo's standard rate of 19 paise, it is 90% cheaper than most carriers and 25% cheaper than the most popular calling apps that are Internet based. Additionally, the app offers 50 minutes of free talk time to every user who installs the app for the first time, and 50 additional minutes for every friend a user refers to Ringo.
Intending to revolutionize the domestic calling space in the Indian telecom market with this launch, Bhavin Turakhia, CEO and Founder of Ringo, said, "Our vision has been to make calling across the globe simple, seamless and low-cost for the masses. Our local calling service is not only the cheapest in India right now, but it is also free of typical telecom-related hassles such as STD charges and differential pricing. In a country where 3G penetration is not fairly high throughout, we also have a strong edge over Internet telephony apps, as Ringo calls do not use data, and are comparable in price."
Ringo is able to offer the unprecedented price of 19 p/min by purchasing minutes in bulk at extremely low prices, and passing on the accrued savings to its customers by leveraging top-end technology. Ringo does not use your talktime, by using a unique call flow that connects both the initiator and the recipient of a call via a conference bridge.
Additionally, Ringo also comes with transparent and easy-to-read itemized bills, and makes earning free credits possible by downloading other apps from its Offers section.
Commenting on future plans for Ringo, Bhavin added, "Our aim is to become a smart, one-stop calling solution for people across the globe, and local calling in India is certainly our biggest step in that direction."
Go ahead and make your first Ringo call now.
Ringo is a calling app that uses a unique call flow to provide high-quality calls at costs that are typically significantly lower than other popular calling options. Ringo calls do not require Internet or data connectivity, and instead make use of traditional phone lines. Ringo provides its users with popular calling options in a single, intuitive app, giving users the power to choose the one that suits them the best. As an app, Ringo is unparalleled in cost, convenience, simplicity and reliability. Ringo is a Directi company founded by Bhavin Turakhia.