text:
\section{Introduction}

In the paper \cite{DOUBLEINT} we posed the following problem: consider the system
$\dot x = f(t,x) + u$ with $x\in\mathbb R^n$ and the stabilizing control $u=u^*$. Consider
now the system
\begin{equation}
\label{1000} \dot x = f(t,x) + \alpha(t)u \,,
\end{equation}
where rank$\{\alpha(t)\} \neq n$ for certain times $t$
({\it i.e.} $\alpha$ may be rank-deficient over possibly ``large'' intervals of time).
Under which conditions
imposed on $\alpha$ does the closed-loop system  \rref{1000} with
the {\em same} control $u^*$ is asymptotically stable?
It must be stressed that a complete knowledge of
$\alpha$ (and, in particular, precise information on the set of times
where it is rank deficient) would be a too restrictive condition to
impose on $\alpha$. We are rather looking for a condition valid for a
whole class $\cal {G}$ of functions $\alpha$ and, therefore, we
expect the closed-loop systems \rref{1000} with $u^*$ to be
asymptotically stable for every $\alpha\in {\cal {G}}$.

In order to characterize such a condition, let us consider a similar
problem stemming from identification and adaptive control. It concerns
the linear system $\dot x = -P(t)u$, where the matrix
$P(\cdot)$ is symmetric non-negative and now plays the role of $\alpha$. If $P\equiv I$
then $u^*=x$ stabilizes the system exponentially. But what if $P(t)$ is only
semidefinite for all $t$? \textcolor{black}{Under which conditions}
does $u^*=x$ still stabilize the system?  For this
particular case the answer to this question can be found in the literature:
from the seminal paper \cite{MORNAR} we know that for the system
\begin{equation}
\label{1001}
\dot x = -P(t)x
\end{equation}
with $x\in\mathbb R^n$, $P\geq0$ bounded and with bounded derivative, it is
{\em necessary and sufficient}, for global exponential stability, that $P$ be also
{\em persistently exciting} (PE), {\em i.e.}, that there exist $\mu>0$ and $T>0$ such that
\begin{equation}\label{1002}
\int_{t}^{t+T} \xi^\top P(\tau)\xi \geq \mu
\end{equation}
for all unitary vectors $\xi\in\mathbb R^n$ and all $t\geq 0$. Therefore, as regards the
stabilization of \rref{1000}, the notion of
{\em persistent excitation} seems to be a reasonable additional assumption to consider
for the signals $\alpha$.

In this paper, we focus on $n$-dimensional linear time-invariant systems
\begin{equation}\label{system}
\dot x = A x +\alpha(t)Bu\,,
\end{equation}
where the input perturbation $\alpha$ is a {\em scalar} PE signal,
i.e., $\alpha$ takes values in $[0,1]$ and
there exist
two positive constants $\mu,T$ such that, for every $t\geq 0$,
\begin{equation}
\label{Tmu}
\int_t^{t+T}\alpha(s)ds\geq \mu.
\end{equation}
Given two positive real numbers $\mu\leq T$, we denote by ${\cal
{G}}(T,\mu)$ the class of all PE signals verifying \rref{Tmu}. Note
that we do not consider here any extra assumption on the regularity
of the PE signal $\alpha$ (e.g. having a bounded derivative).

An interpretation of the stabilization mechanism can be given, in the case of scalar
systems of type \rref{1001}, in terms of ``average''. Roughly speaking, one can dare
say that, even though
it is not the control action $u^*$ that enters the system for each $t$, this
``ideal'' control does drive the system ``{\em in average}''.
Indeed, for $t\geq 0$, set
$\left\langle\alpha\right\rangle\!\!(t):=\frac1T\int_t^{t+T}\alpha(s)ds$ for any PE signal $\alpha$.
Then, along non-trivial trajectories of $\dot x(t)=-\alpha(t) x(t)$, one has
for every $t\geq 0$,
$$
\frac1T\ln\left(\frac{x(t+T)}{x(t)}\right)=\frac1T\int_t^{t+T}\frac{\dot x(s)}{x(s)}ds
=-\left\langle\alpha\right\rangle(t).
$$
The control action $u=-\alpha(t)x$ which, in
average, corresponds
to $u=-\left\langle\alpha\right\rangle\!\!(t)x$, is tantamount to applying $u^*=-x$, modulo a gain-scale that
only affects the rate of convergence but not the stabilization property of $u^*$.
Of course the previous naive thinking largely relies on the fact that we are dealing
with a scalar system. Indeed, persistency of excitation does not guarantee the existence
of an averaged system in the sense of {\it e.g.} \cite{TEPEAE99}. This observation makes
all the related techniques unapplicable.

Our main goal in this paper consists of stabilizing \rref{system} to the origin
with a linear feedback. Therefore, we will assume in the sequel that
$(A,B)$ is a stabilizable pair. It is obvious that if $\alpha\equiv 1$, for
a proper choice of $K$, we have that the control $u=u^*$ with $u^*=
Kx$ renders the closed-loop system globally exponentially stable. If
$\alpha$ is not constant, consider the following question:
\begin{quote}
\label{ques:01} (Q1-0)\ \ Does  $u=\alpha(t)u^*$, with $\alpha$ an
arbitrary PE signal, stabilize \rref{system}?
\end{quote}
For systems of the type \rref{system}, the answer to Question
($Q$1-0) is negative in general. Indeed, the scalar case
essentially corresponds to stabilizing
$\dot x(t)= \lambda x(t)+\alpha(t)u$. Using the linear feedback
$u^*=k x$ for some $k<0$, one gets, after simple computations, that
for $t\geq 0$,
\begin{equation}\label{triv}
\frac1T\ln\left(\frac{x(t+T)}{x(t)}\right)=\lambda+\left\langle\alpha\right\rangle\!(t)k\leq
\lambda+\frac{\mu}Tk,
\end{equation}
if $\alpha$ is a PE-signal verifying \rref{Tmu} for some fixed positive
constants $\mu\leq T$. One deduces from
\rref{triv} that global exponential stabilization occurs if the
negative constant $k$ is chosen so that $k<-\frac T{\mu}\lambda$. If
$\lambda>0$, then the choice of $u^*=k x$ depends on $\mu,T$, the parameters
of the PE-signal. Therefore, question ($Q$1-0) \textcolor{black}{may only receive a
positive answer} for systems \rref{system} with $A$ marginally stable, {\it i.e.} all the
eigenvalues of $A$ have non-positive real part:
\begin{quote}
\label{ques:1} (Q1)\ \ Given $A$ marginally stable, does  $u=\alpha(t)u^*$, with $\alpha$
an arbitrary PE signal, stabilize \rref{system}?
\end{quote}
Intuitively one may think that the global exponential stability of
the closed-loop system is guaranteed, at least, for a particular choice of
$K$ and a particular class of PE functions
$\alpha$. In that spirit, consider the following question:
\begin{quote}
\label{ques:2} (Q2)\ \ Given a class ${\cal {G}}(T,\mu)$ of PE signals, can one
determine a linear feedback $u^*= Kx$ such
that $u=\alpha(t)u^*$ stabilizes \rref{system}, \textcolor{black}{for all PE signal
$\alpha$ in} ${\cal {G}}(T,\mu)$?
\end{quote}
While, for the scalar case, the answer is clearly positive (as shown previously), the
general case $n>1$ is fundamentally different and, in view of the
available tools from the literature of adaptive control, a proof (or
disproof) of the conjecture above is far from evident. A first step
in the solution of $(Q2)$ for the case $n=2$ has been undertaken
in \cite{DOUBLEINT} where we showed that, for certain persistently
exciting functions $\alpha$ and certain values of $K
$,
exponential stability follows. \textcolor{black}{We underline the fact that,
in Question $(Q2)$, the gain $K$ is required to be valid \emph{for all} signals
in the considered class.}

A particular case of this problem may be re-casted in the context of
switched systems --{\it cf.}\, \cite{LIBBOOK}. Indeed, consider the particular case
in which the PE signal $\alpha$ takes only the values $0$ and $1$.
In this setting, the system \rref{system} after the choice
$u=u^*=Kx$, switches between the uncontrolled system $\dot x = Ax$
and the exponentially stable system
$\dot x= (A+BK)x$. At this point,
it is worth emphasizing that since switches occur between a possibly
unstable dynamics and a stable one, Lyapunov-based conditions for
switched system between stable dynamics are inapplicable; see for
instance \cite{SHOWIRMASWULKIN} for results using common quadratic
Lyapunov functions, \cite{BRANICKY98,COLGERAST} for theorems
relying on multiple Lyapunov functions and also \cite{BOSCAIN2002} for a more geometric
approach.

The first issue we address in this paper regards the
controllability of \rref{system} uniformly with respect to $\alpha\in {\cal {G}}(T,\mu)$.
If the pair $(A,B)$ is controllable, we prove that \rref{system} is
(completely) controllable in time $t$ if and only if $t>T-\mu$.

We next focus on the stabilization of \rref{system} by a linear
feedback $u=Kx$, that is, we address Question $(Q2)$ for system
\rref{system}. Namely, we look for the existence of a matrix $K$ of
size $m\times n$ such that, for every $\alpha\in {\cal {G}}(T,\mu)$, the origin of the
system
\[
\dot x=(A+\alpha(t) BK) x
\]
is globally asymptotically stable. It is
of course assumed that $(A,B)$ is stabilizable. We first treat the case where $A$ is
neutrally stable. We actually
determine a feedback $K$ as required, which in addition provides a
positive answer to Question $(Q1)$ posed for system \rref{system}.

Finally, we consider the case of the double integrator , \textcolor{black}{that is,
$$
A=\begin{pmatrix}
0&1\\
0&0
\end{pmatrix}, \ \ \
B=\begin{pmatrix} 0\\1
\end{pmatrix}\,.
$$
}

In \cite{DOUBLEINT} we
already studied such a system  under the assumption that
$\alpha$ is PE. The solutions given in that paper, however, do not
bring a complete answer to the questions posed above:
the first solution relies on backstepping and, therefore,
requires a bound on the derivative of $\alpha$ while the second is based on
a normalization of $\alpha$ \textcolor{black}{and imposes a relationship between the
parameters $T$ and $\mu$ involved in the persistency of excitation.} In the present
paper, we bring a positive
answer to Question $(Q2)$ for every class ${\cal {G}}(T,\mu)$ of PE signals
and a negative answer to Question \textcolor{black}{($Q$1)}.

The rest of the paper is organized as follows. In coming section we
provide the main notations and the result on the controllability of
multi-input linear systems subject to PE signals. We discuss
stabilization issues
in Section~\ref{neutr} for $A$
neutrally stable and in Section~\ref{sec:mainproof} for the double
integrator, \textcolor{black}{first in case of a scalar control
and then in a more general setup}. We
close the paper with an Appendix which
contains the proof of a crucial
technical
result.

\section{Notations and basic result on controllability}
In this paper, we are concerned with linear systems subject to a
scalar persistently exciting signal, {\em i.e.}, systems of the type \rref{system}
where $\alpha$ is a PE signal. The latter is defined as follows.

\begin{deff}[$(T,\mu)$-signal]\label{Tm-signal}
Let $\mu\leq T$ be positive constants. A \emph{$(T,\mu)$-signal} is
a measurable function $\alpha:{\mathbb R_{\geq 0}}\to[0,1]$
satisfying
\begin{equation}\label{EP}
\int_t^{t+T}\alpha(s)ds \geq \mu\,,\quad \forall t\in\mathbb R_{\geq 0}\,.
\end{equation}
We use ${\cal {G}}(T,\mu)$ to denote the set of all $(T,\mu)$-signals.
\end{deff}

\textcolor{black}{Notice that, for any such signal $\alpha$, existence and uniqueness
of the solutions of \rref{system} is guaranteed.}

\begin{rmk}\label{notime}
If $\alpha(\cdot)$ is a $(T,\mu)$-signal, then, for every $t_0\geq
0$, $\alpha(t_0+\cdot)$ is again a $(T,\mu)$-signal.
\end{rmk}

\textcolor{black}{Our first result studies the following property for \rref{system}.}

\begin{deff}[Controllability in time $t$]
We say that system \rref{system} is controllable in time $t>0$ for
${\cal {G}}(T,\mu)$ if, for every $\alpha\in{\cal {G}}(T,\mu)$, the
time-varying linear controlled system defined by \rref{system} is
completely controllable in time $t$.
\end{deff}

More precisely, we establish the following result on the
controllability of system \rref{system}.

\begin{prop}\label{cont-t}
Let $\mu\leq T$ be two positive constants and $(A,B)$ a controllable
pair of matrices of size $n\times n$ and $n\times m$ respectively.
Then, system \rref{system} is controllable
in time $t$ for ${\cal {G}}(T,\mu)$ if and only if $t>T-\mu$.
\end{prop}

\underline{Proof of Proposition~\ref{cont-t}.} Following the
classical proof of the Kalman condition for the controllability of
autonomous linear systems \textcolor{black}{({\it cf.} {\it e.g} \cite{SONTAGBOOK})},
it is easy to see that the conclusion
does not hold if and only if there exists a non-zero vector $p\in
\R^n$ and a $(T,\mu)$-signal $\alpha$ such that the function
\begin{equation}\label{Kal}
\alpha(s)p^\top e^{A(t-s)}B= 0, \ \ \hbox{a.e. in } [0,t].
\end{equation}
If $t>T-\mu$, there exists a subset $J$ of $[0,T]$
of positive measure such that $\alpha(s)>0$ for $s\in J$. Therefore,
the real-analytic function $s\mapsto p^\top e^{A(t-s)}B$ is equal to
zero on $J\subset [0,T]$. It must then be identically equal to zero,
which implies that $(A,B)$ is not controllable. We reach a
contradiction and one part of the equivalence is proved.
If $t\leq T-\mu$, ${\cal {G}}(T,\mu)$
contains a PE-signal identically equal to zero on time intervals of
lengtht $t$ and any non-zero vector $p$ verifies \rref{Kal}.\ \hfill \rule{0.5em}{0.5em}

\textcolor{black}{The rest of the paper is concerned with the stabilization of
\rref{system}.
We address} the following problem. Given $T\geq \mu>0$, we want to find a matrix $K$
of size $m\times n$ which makes the origin of
\begin{equation}\label{feedback}
\dot x=(A+\alpha(t) BK) x
\end{equation}
globally asymptotically stable, uniformly with respect to every
$(T,\mu)$-signal $\alpha$ \textcolor{black}{({\it i.e.}, $K$ is required to depend only
on $A$, $B$, $T$ and $\mu$ and to be valid for all signals $\alpha$ in the class
$\mathcal G(T,\mu)$)}. Referring to
$$x(\cdot\,;t_0,x_0,K,\alpha)=
(x_1(\cdot\,;t_0,x_0,K,\alpha),\dots,x_n(\cdot\,;t_0,x_0,K,\alpha))^\top,
$$
as the solution of \rref{feedback} with initial condition
$x(t_0;t_0,x_0,K,\alpha)=x_0$, we introduce the following
definition.

\begin{deff}[$(T,\mu)$-stabilizer]\label{stab}
Let $\mu\leq T$ be positive constants. The gain $K$ is said
to be a \emph{$(T,\mu)$-stabilizer} for \rref{system} if there
exists a class $\mathcal{KL}$-function $\beta$ such that, for every
$\alpha\in{\cal {G}}(T,\mu)$, $x_0\in\mathbb R^n$, and $t_0\in\mathbb R_{\geq 0}$, the
solution of \rref{feedback} satisfies
$$
\norm{x(t;t_0,x_0,K,\alpha)}\leq \beta(\norm{x_0},t-t_0)\,,\quad
\forall t\geq t_0\,.
$$
\end{deff}
\begin{rmk}\label{exp0}
Since we are dealing with linear systems (in the state), it is a
standard fact that one can rephrase the above definition as follows:
the gain $K$ is said to be a \emph{$(T,\mu)$-stabilizer} for
\rref{system} if \rref{feedback} is exponentially stable, uniformly
with respect to every $(T,\mu)$-signal $\alpha$. Here, uniformity
means that the rate of (exponential) decrease only depends on
$(A,B)$ and $\mu,T$.
\end{rmk}

\section{The neutrally stable case}\label{neutr}
The purpose of this section consists of proving the following
theorem.
\begin{thm}\label{thm-neutrally}
Assume that the pair $(A,B)$ is stabilizable and that the matrix $A$
is neutrally stable.
Then there exists a matrix $K$ of size $m\times n$ such that, for
every $0<\mu\leq T$, the gain $K$ is a \emph{$(T,\mu)$-stabilizer}
for \rref{system}.
\end{thm}

Since the feedback $K$ determined above does not depend on the
particular class ${\cal {G}}(T,\mu)$, we bring a positive answer to
Question $(Q1)$ in the case where $A$ is neutrally stable.
\begin{rmk}
It can be seen along the proof below that
the gain $K=-rB^\top$, with an arbitrary $r>0$, does the job in
the case where $A$ is skew-symmetric and $(A,B)$ is controllable.
\end{rmk}

\underline{Proof of Theorem~\ref{thm-neutrally}.}  Recall that a
matrix is said to be neutrally stable if its eigenvalues have
non-positive real part and those with zero real part have trivial
corresponding Jordan blocks. The proof of
Theorem~\ref{thm-neutrally} is based on the following equivalence
result.

\begin{lemma}\label{reduction}
It is enough to prove Theorem~\ref{thm-neutrally} in the case where $A$ is skew-symmetric
and $(A,B)$ controllable.
\end{lemma}

\underline{Proof of Lemma~\ref{reduction}.} Let $(A,B)$ be a
stabilizable pair with $A$ neutrally stable. Since the non-controlled
part of the linear system $\dot x=Ax+Bu$ is already stable, it is
enough to focus on the controllable part of $(A,B)$. Hence, we
assume that $(A,B)$ is controllable. Up to a linear change of
variable, $A$ and $B$ can be written as
$$
A=\begin{pmatrix}
A_1&A_2\\
0&A_3
\end{pmatrix}, \ \ \
B=\begin{pmatrix} B_1\\B_3
\end{pmatrix},
$$
where $A_1$ is Hurwitz and all the eigenvalues of $A_3$ have zero
real part. From the neutral stability assumption, we deduce that
$A_3$ is similar to a skew-symmetric matrix. Up to a further linear change of
coordinates, we may assume that $A_3$ is indeed skew-symmetric. From the
controllability assumption, we deduce that $(A_3,B_3)$ is
controllable. Setting $x=(x_1^\top,x_3^\top)^\top$ according to the above decomposition,
the system \rref{system} can be written as
\begin{eqnarray}
\dot x_1&=&A_1x_1+A_2x_3+\alpha(t)B_1u,\label{h1}\\
\dot x_3&=&A_3x_3+\alpha(t)B_3u.\label{h2}
\end{eqnarray}
Assume that Theorem~\ref{thm-neutrally} holds for \r{h2}, {\em i.e.}, there
exists $K_3$ such that, for
every $0<\mu\leq T$, the gain $K_3$ is a \emph{$(T,\mu)$-stabilizer}
for \r{h2}. Take
$$K=\left(\begin{array}{c}{\bf 0}\\ K_3\end{array}\right),$$
where all entries of ${\bf 0}$ are null. Then the conclusion for
system \rref{system} follows, since an autonomous linear Hurwitz
system subject to a perturbation whose norm converges exponentially
(with respect to time) to zero is still asymptotically stable at the
origin. \ \hfill \rule{0.5em}{0.5em}

\vspace{0.5cm} Based on Lemma \ref{reduction}, we assume that $A$ is
skew-symmetric and $(A,B)$ is controllable for the rest of the argument.
We will prove that the
gain $K=-B^\top$ does the job.

We consider the time derivative of the Lyapunov function $V(x)=\norm x^2/2$
along non-trivial solutions of the closed-loop system
\begin{equation}\label{feedback2}
\dot x=\left(A-\alpha(t)BB^\top \right)x\,,
\end{equation}
and get $\dot V=-\alpha(t)\norm{B^\top x}^2$, which can be also written
$$
\frac{\dot V}{V}=-\alpha(t)\frac{
\norm{B^\top x}^2}{V}\,.
$$
Integrating both parts and defining $x(\cdot)$ as
$x(\cdot\,;t_0,x_0,K,\alpha)$ and $v(\cdot)=V(x(\cdot))$, where
$(t_0,x_0)$ and $\alpha$ denote, respectively, an arbitrary initial condition and an
arbitrary $(T,\mu)$-signal given arbitrary $T\geq\mu>0$, we get that
\begin{equation}\label{Vint}
\int_{t_0}^{t_0+T} \frac{\dot v(t)}{v(t)}dt=-\int_{t_0}^{t_0+T}
\alpha(t)\frac{\norm{B^\top x(t)}^2
}{v(t)}dt\,.
\end{equation}

\begin{lemma}\label{claim1}
For every $0<\mu\leq T$, there exists a positive constant $\eta$
such that, for any $(T,\mu)$-signal $\alpha$ and any initial state
$\norm{x_0}=1$, it holds that
$$
\int_{t_0}^{T+t_0} \alpha(t)\frac{\norm{B^\top x(t)}^2
}{v(t)}dt \geq
\eta\,.
$$
\end{lemma}

\underline{Proof of Lemma~\ref{claim1}.}  Because of
Remark~\ref{notime} we take, without loss of generality, $t_0=0$. We fix
$0<\mu\leq T$ and reason by contradiction, {\it i.e.}, we assume
that there exist a sequence $\{x_{0i}\}_{i\in\mathbb{N}}$ such that
$\norm{x_{0i}}=1$ for all $i\in\mathbb{N}$ and a sequence of
$(T,\mu)$-signals $\alpha_i$ such that
\begin{equation}\label{limint}
\lim_{i\to\infty} \int_{0}^{T} \alpha_i(t)\frac{\norm{B^\top
x_i(t)}^2}{v_i(t)}dt =0\,,
\end{equation}
where $x_i(\cdot)$ denotes $x(\cdot\,;0,x_{0i},K,\alpha_i)$ and
$v_i(\cdot)=V(x_i(\cdot))$. Since $\{x_{0i}\}_{i\in\mathbb{N}}$ belongs to a compact set,
there exists a
subsequence $\{x_{0i_j}\}_{j\in\mathbb{N}}$ such that
$$
\lim_{j\to\infty} x_{0i_j}=x_{0\star}\,,\quad \textrm{with}\quad
\norm{x_{0\star}}=1\,.
$$
On the other hand,
recall that the space $L^\infty(\mathbb R_{\geq 0},[0,1])$ is sequentially weakly-$\star$
compact (see, for instance, \cite[Chapter IV]{brezis83}), that is,
for every sequence $\{\beta_i\}_{i\in\mathbb N}\subset L^\infty(\mathbb R_{\geq 0},[0,1])$
there exist $\beta_\star\in L^\infty(\mathbb R_{\geq 0},[0,1])$ and a subsequence
$\{\beta_{i_j}\}_{j\in\mathbb N}$ such that
for every $\varphi\in L^1(\mathbb R_{\geq 0},\mathbb R)$ the following holds
\begin{equation}\label{weakstar}
\lim_{j\to\infty} \int_0^\infty \beta_{i_j}(s)\varphi(s)ds
=      \int_0^\infty \beta_\star(s) \varphi(s)ds\,.
\end{equation}

Therefore, we can extract a
subsequence $\{\alpha_{i'_j}\}_{j\in{\mathbb N}}$ of
$\{\alpha_{i_j}\}_{j\in{\mathbb N}}$ which
converges weakly-$\star$ to a measurable
function $\alpha_\star$. Note that $\alpha_\star$
is itself a
$(T,\mu)$-signal.

The convergence of $\alpha_{i'_j}$ to $\alpha_\star$ implies a convergence
of the solutions
of their corresponding non-autonomous linear dynamical systems in the form
\r{feedback2}: this is a particular case of the classical Gihman
convergence with respect to
time-varying parameters for ordinary differential equations (see \cite{gihman}
and also \cite{liu-sus} for a general discussion in the framework of control
theory). For the sake of completeness, and because the situation faced here
can be handled  with a specific and simpler approach, we include in the
Appendix the proof of Proposition~\ref{technic}, whose statement allows to conclude that
$\{x_{i'_j}(\cdot)\}_{j\in\mathbb N}$ converges, uniformly on compact
time intervals, to
$x_{\star}(\cdot):=x(\cdot\,;0,x_{0\star},K,\alpha_\star)$ as $j$
tends to infinity. Letting $v_\star(\cdot)=V(x_\star(\cdot))$, one has,
for every $j\in \mathbb N$,
$$
\int_{0}^{T} \alpha_{i'_j}(t)\frac{\norm{B^\top x_{i'_j}(t)^2}}{v_{i'_j}(t)}dt
-\int_{0}^{T} \alpha_\star(t)\frac{\norm{B^\top
x_\star(t)}^2}{v_\star(t)}dt=
$$
$$
\int_{0}^{T} \big(\alpha_{i'_j}(t)-
\alpha_\star(t)\big)\frac{\norm{B^\top
x_\star(t)}^2}{v_\star(t)}dt+
\int_{0}^{T} \alpha_{i_j'}(t)\left(\frac{\norm{B^\top x_{i'_j}(t)^2}}{v_{i'_j}(t)}-
\frac{\norm{B^\top
x_\star(t)}^2}{v_\star(t)}\right) dt.
$$
Letting $j$ tend to infinity, taking into account the weak-$\star$ convergence of
$\alpha_{i'_j}$ to $\alpha_\star$,
 and applying \r{limint},
we get that
\begin{eqnarray*}
\int_{0}^{T} \alpha_\star(t)\frac{\norm{B^\top
x_\star(t)}^2}{v_\star(t)}dt &\leq& \liminf_{j\to\infty}\int_{0}^{T} \alpha_{i'_j}(t)\left|\frac{\norm{B^\top x_{i'_j}(t)^2}}{v_{i'_j}(t)}-
\frac{\norm{B^\top
x_\star(t)}^2}{v_\star(t)}\right|dt\\
&\leq& \liminf_{j\to\infty}\int_{0}^{T} \left|\frac{\norm{B^\top x_{i'_j}(t)^2}}{v_{i'_j}(t)}-
\frac{\norm{B^\top
x_\star(t)}^2}{v_\star(t)}\right|dt\\
&=&0\,,
\end{eqnarray*}
where the last equality follows from the uniform convergence of
$x_{i'_j}(\cdot)$ to $x_\star(\cdot)$ on the interval $[0,T]$.
Hence, we conclude that
\begin{equation}\label{zeroae}
\alpha_\star(t)B^\top x_\star(t)=0
\end{equation}
for almost every $t\in[0,T]$.
Furthermore, since $x_{\star}$ is a solution of \rref{feedback2}, we
get from \rref{zeroae} that $x_\star(t)=e^{At}x_{0\star}$ for
$t\in[0,T]$. Moreover, since $\alpha_{\star}$ is a $(T,\mu)$-signal,
it is strictly positive on a subset $J$ of $[0,T]$ of positive
measure, and then, according to \rref{zeroae}, $B^\top e^{At}x_{0\star}$ must be
equal to zero on $J$. Since the latter is real-analytic, it must be
identically equal to zero on $[0,T]$, and
therefore the pair $(A,B)$ is not controllable. This is a
contradiction and thus Lemma~\ref{claim1} is proved. \ \hfill \rule{0.5em}{0.5em}

By standard homogeneity arguments Lemma~\ref{claim1}
\textcolor{black}{together with \rref{Vint}} imply
uniform exponential convergence of $V$ to zero along every
trajectory corresponding to a $(T,\mu)$-signal.
Theorem~\ref{thm-neutrally} is therefore proved. \ \hfill \rule{0.5em}{0.5em}

\section{The double integrator}\label{sec:mainproof}

\subsection{Scalar control}

\textcolor{black}{This section addresses the same problem as
above for the double integrator. For this, let}
\begin{equation}\label{scandia}
A=\left(\begin{array}{cc}0&1\\0&0\end{array}\right),\ \ \ \textcolor{black}{B}=\left(\begin{array}{c}0\\1\end{array}\right),
\end{equation}
so that system \r{system} becomes
\begin{equation}\label{DI}
\left\{
\begin{array}{lcl}
\dot x_1&=&x_2,\\
\dot x_2&=&\alpha u,
\end{array}
\right.
\end{equation}
while, by considering $K=(-k_1,-k_2)$, the closed-loop system \r{feedback} reads
\begin{equation}\label{DIf}
\left\{
\begin{array}{lcl}
\dot x_1&=&x_2,\\
\dot x_2&=&-\alpha (k_1 x_1+k_2 x_2)\,. \end{array} \right. \end{equation}

\noindent Throughout the section we prove the following fact.
\begin{thm}\label{theo}
For every $0<\mu\leq T$ there exists a $(T,\mu)$-stabilizer for \r{DI}.
\end{thm}

\textcolor{black}{Again, we stress that, given any $T\geq \mu>0$,
the above result establishes the existence of a static linear feedback
that globally asymptotically stabilizes \rref{system} for the case that
$A$ and $B$ are given by \rref{scandia}, for \emph{all} $(T,\mu)$-signal $\alpha$.
Theorem \ref{theo} therefore gives a positive answer to Question $(Q2)$ for this particular case.}
\vspace{5mm}

\noindent \underline{Proof of Theorem~\ref{theo}.}  We first show how to
exploit the symmetries of system \r{DI} in order to identify a
one-parameter family of problems which are equivalent to the
research of a $(T,\mu)$-stabilizer.

\begin{lemma}\label{multi}
Let $\lambda$ be a positive real number. Then \r{DI} admits a
$(T,\mu)$-stabilizer if and only if it admits a
$(T/\lambda,\mu/\lambda)$-stabilizer. More precisely, $(-k_1,-k_2)$ is a
$(T,\mu)$-stabilizer if and only if $(-\lambda^2 k_1,-\lambda k_2)$ is a
$(T/\lambda, \mu/\lambda)$-stabilizer.
\end{lemma}
\underline{Proof of Lemma~\ref{multi}.} Let $K=(-k_1,-k_2)$ be a
$(T,\mu)$-stabilizer and fix
 an arbitrary $\lambda>0$. It suffices to prove that $(-\lambda^2 k_1,-\lambda k_2)$
is a $(T/\lambda, \mu/\lambda)$-stabilizer, the converse part of the statement being
equivalent (just making $1/\lambda$ play the role of $\lambda$).

Applying to $x(\cdot):=x(\cdot\,;t_0,x_0,K,\alpha)$ a time-rescaling and an anisotropic dilation, we define
\[
x_\lambda(t)=\left(\begin{array}{cc}1&0\\0&\lambda\end{array}\right) x(\lambda t)\,,\quad \forall
t\geq 0\,.
\]
Then
\begin{eqnarray*}
\frac{d}{dt}x_\lambda(t)&=&\lambda \left(\begin{array}{cc}1&0\\0&\lambda\end{array}\right) \dot x(\lambda t)=
\lambda \left(\begin{array}{cc}1&0\\0&\lambda\end{array}\right) \left(\begin{array}{cc}0&1\\ -k_1\alpha(\lambda
t)&-k_2\alpha(\lambda t)\end{array}\right) x(\lambda t)\\
&=&\left(\begin{array}{cc}0&1\\ -\lambda^2 k_1\alpha(\lambda t)&-\lambda k_2\alpha(\lambda t)\end{array}\right) x_\lambda(t)
=Ax_\lambda(t)-\alpha(\lambda t)b(\lambda^2 k_1,\lambda k_2) x_\lambda(t)\,,
\end{eqnarray*}
that is,
$$x_\lambda(\cdot)=x(\cdot\,;t_0,\mathrm{Diag}(1,\lambda)x_0,(-\lambda^2 k_1,-\lambda k_2),\alpha(\lambda \cdot)).$$
It is clear that $\alpha(\lambda\cdot)$ is a $\left( T/\lambda,
\mu/\lambda\right)$-signal if and only if $\alpha(\cdot)$ is a $(T,\mu)$-signal.
Therefore, if $\beta$ is a class $\mathcal{KL}$ function such that, for every
$\alpha\in{\cal {G}}(T,\mu)$, $x_0\in\mathbb R^2$ and $t_0\in\mathbb R_{\geq 0}$,
$$
\norm{x(t;t_0,x_0,K,\alpha)}\leq \beta(\norm{x_0},t-t_0)\,,\quad \forall t\geq t_0\,,
$$
then, for every  $\alpha_\lambda\in{\cal {G}}(T/\lambda,\mu/\lambda)$, $x_0\in\mathbb R^2$ and $t\geq t_0\geq0$,
$$
\norm{x(t;t_0,x_0,(-\lambda^2 k_1,-\lambda k_2),\alpha_\lambda(\cdot))}=
\norm{x(t;t_0,\mathrm{Diag}(1,1/\lambda)x_0,K,\alpha(\cdot/\lambda)}\leq
\beta(\max\{1,1/\lambda\}\norm{x_0},t-t_0)\,.
$$
Since $(s,t)\mapsto \beta(\max\{1,1/\lambda\}s,t)$ is a class $\mathcal{KL}$ function,
the lemma is proved. \ \hfill \rule{0.5em}{0.5em}

\vspace{5mm}
We prove below Theorem \ref{theo} by fixing a gain $K=(-k_1,-k_2)$
and showing that there exists $\lambda>0$  such that $(-\lambda^2
k_1,-\lambda k_2)$ is a $(T,\mu)$-stabilizer for \r{DI}. According
to Lemma~\ref{multi}, this is equivalent to proving that  there
exists $\lambda>0$  such that $K$ is a $(T/\lambda,\mu/\lambda)$-stabilizer.

The first, obliged, choice is on the sign of the entries of $K$: a
necessary (and sufficient) condition for the matrix $A+bK$ to be
Hurwitz is that $k_1,k_2>0$.

The choice of $K$ will be determined by the following request: we
ask that each matrix $A+\alpha b K$, with $\alpha\in[\mu/T,1]$ constant, has
real negative eigenvalues. Since the discriminant of
\[\det\left(\begin{array}{cc}-\sigma&1\\ -\alpha k_1&-\sigma-\alpha k_2\end{array}\right)=
\sigma^2+\alpha k_2 \sigma+\alpha k_1\] is given by $\alpha(\alpha k_2^2-4
k_1)$, this sums up to imposing that
\begin{equation}\label{k_int}
k_1<\frac{\mu }{4T}k_2^2.
\end{equation}

We fix for the rest of the argument a positive $\rho<\mu/2T$
and take $K=(-\rho k^2/2,-k)$ for $k>0$ to be fixed later. The choice of $\rho$ and $K$ is such that
inequality \r{k_int} is automatically verified. For $\bar{\alpha}\in\{\mu/T,1\}$, define
$\xi^{\bar{\alpha}}_{\pm}$ as the roots of
$$
\xi^2+\bar{\alpha}k\xi+\bar{\alpha}\frac{\rho k^2}2=0,
$$
with $\xi^{\bar{\alpha}}_+<\xi^{\bar{\alpha}}_-$. In addition, let
$\xi^s_+$ and $\xi^s_-$ be given by
$$\xi^s_+:=-\frac k2(1+\sqrt{1-\rho}),\ \ \
\xi^s_-:=-\frac k2\lp1-\sqrt{1-\lp2-\frac\rho 2\right)\rho}\right). $$ A simple
calculation shows that
\begin{equation}\label{xi0}
\xi^s_+<\xi^1_+<\xi^{\mu/T}_+<\xi^{\mu/T}_-<\xi^1_-<\xi^s_-<0.
\end{equation}

We next define a set of half-lines of the plane. Let $D_1$ and $D_2$
be the half-lines defined, respectively, by $x_2=0,x_1<0$ and
$x_2=0,x_1>0$. Let, moreover,
$D^s_{\pm}$
and
$D^{\bar{\alpha}}_{\pm}$ be the half-lines of the open
upper half-plane defined, respectively, by the equations
\begin{equation}\label{lines0}
D^s_{\pm}: \ \ x_2=\xi^s_{\pm} x_1, \
\ \ D^{\bar{\alpha}}_{\pm}: \ \ x_2=\xi^{\bar{\alpha}}_{\pm}x_1.
\end{equation}

Finally, we define
${\cal{C}}_1$, ${\cal{C}}^s$, and
${\cal{C}}_2$ as the closed cones contained in the upper half-plane and delimited, respectively, by $D_1$, $D^s_-$,
then $D^s_-$, $D^s_+$, and $D^s_+$, $D_2$: see Figure \ref{halfplane}.

\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.8]{quadrants2.eps}
\caption{The cones ${\cal{C}}_1$, ${\cal{C}}^s$, ${\cal{C}}_2$.}\label{halfplane}
\end{center}
\end{figure}

A key step in the proof of Theorem~\ref{theo} is to evaluate
for how much time a trajectory of  \r{DIf} stays in each of the three cones.
\begin{lemma}\label{finite}
There exists a positive constant $C_1=C_1(\rho,T,\mu)$ ({\it i.e.}, only
depending on $\rho$, $T$, and $\mu$) such that, for every
$\lambda\geq
k$, every $(T/\lambda,\mu/\lambda)$-signal $\alpha$, every $x_*\in\R^2\setminus\{0\}$, and every $t_0\geq 0$, if an interval $I\subset \R$ is such that the
trajectory $t\mapsto x(t;t_0,x_*,K,\alpha)$ stays in ${\cal{C}}_1\cup {\cal{C}}_2$ for every $t\in I$ then the  length of $I$ is smaller than
$C_1/k$.
\end{lemma}

\underline{Proof of Lemma~\ref{finite}.}
Consider $\alpha\in L^\infty(\mathbb R_{\geq 0},[0,1])$ and a trajectory $x(\cdot):=x(\cdot\,;t_0,x_*,K,\alpha)$ of \r{DIf} with $x_*\in\R^2\setminus\{0\}$ and $t_0\geq 0$. Let $I=[t_1,t_2]$
be a time-interval such that $x(t)$ belongs
to ${\cal{C}}_1\cup{\cal{C}}_2$ for every $t\in I$.

Using polar coordinates, $x(\cdot)$ can be represented
as $t\mapsto (r(t)\cos\theta(t),r(t)\sin\theta(t))$ and one has
$$
\dot\theta=\frac{\dot x_1\sin\theta-\dot x_2\cos\theta}{r}=-\sin^2\theta+\alpha\cos\theta\left(\frac{\rho k^2}2\cos\theta+k\sin\theta\right)
$$
almost everywhere in $I$.
Therefore,
$$
\dot\theta\leq -\alpha\left( \sin^2\theta+k\cos\theta\sin\theta+\frac{\rho
k^2}2\cos^2\theta\right).$$
We next show that there exists a positive
constant $c(\rho)$ such that
\begin{equation}\label{derPE}
\dot\theta\leq-\alpha c(\rho)(\sin^2\theta+k^2\cos^2\theta),
\end{equation}
almost everywhere in $I$. The claim can be proved by
taking $\tau:=\tan(\theta)$ and noticing that there exists $c(\rho)>0$ small enough
such that if
$\tau<\xi^s_+$ or $\tau>\xi^s_-$ then
$$
\tau^2+k\tau+\frac{\rho k^2}2\geq c(\rho)(\tau^2+k^2).
$$

Define now
$$
F(\theta)=\left\{\begin{array}{ll}\arctan\left( \frac{\tan\theta}{k}\right)&\mbox{if }0\leq \theta<\pi/2,\\ \pi/2&\mbox{if }\theta=\pi/2,\\ \arctan\left( \frac{\tan\theta}{k}\right)+\pi&\mbox{if }\pi/2< \theta\leq \pi.\end{array}\right.
$$
Notice that $F$ is a continuous reparameterization of the interval $[0,\pi]$.
From (\ref{derPE}), we get
\begin{equation}\label{new}
\frac{d}{d t} (F(\theta))\leq-c(\rho)k\alpha.
\end{equation}

Therefore, $F(\theta)$ is monotone non-increasing as long as $x(\cdot)$
stays in ${\cal{C}}_1\cup {\cal{C}}_2$.
If $\alpha$ is a
$(T/\lambda,\mu/\lambda)$-signal, then
\begin{equation}\label{gap}
F\left( \theta\left( t+\frac T{\lambda}\right)\rp -F(\theta(t))\leq -\frac{c(\rho)\mu k}{\lambda}
\end{equation}
for every $t$ with $[t,t+T/\lambda]\subset I$.
Let $l^{\max}$ be the largest
integer such that
$I$ contains $l^{\max}$ disjoint sub-intervals of length $T/\lambda$.
Then \r{gap} yields
\[ \pi \geq F(\theta(t_1))-F(\theta(t_2))\geq l^{\max}\frac{c(\rho)\mu k}{\lambda} \]
and thus
\[l^{\max}\leq \frac{\lambda\pi}{c(\rho)\mu k}.\]
Hence
\[t_2-t_1\leq (l^{\max}+1)\frac T\l
\leq \frac{ T\pi}{c(\rho)\mu k}+\frac T\lambda.\]
The lemma is concluded by recalling the hypothesis $\lambda\geq k$.
\ \hfill \rule{0.5em}{0.5em}

\vspace{5mm}
From now until the end of the proof, let $x_0\ne 0$ belong to  the cone
defined by $D_1$ and $D^s_+$, {\it i.e.}, to ${\cal{C}}_1\cup {\cal{C}}^s$.
Let, moreover,
$t_0\geq 0$, $\lambda\geq k$, $\alpha\in {\cal G}(T/\lambda,\mu/\lambda)$, and define $x(\cdot):=x(\cdot\,;t_0,x_0,K,\alpha)$.
Then  the following
alternative occurs for $x(\cdot)$:
\begin{description}
\item[$(i)$] for every $t\geq t_0$, $x(t)$ remains in ${\cal{C}}_1\cup {\cal{C}}^s$;
\item[$(ii)$] $x(\cdot)$ reaches $D^s_+$ in finite time.
\end{description}
In both cases, let $[t_0,t_1]$ be the time-interval  needed by
$x(\cdot)$ to reach ${\cal C}^s$. Recall that, by Lemma~\ref{finite}, one
has $0\leq t_1-t_0<C_1/k$.
We notice the following fact, which results from a trivial
computation.

\begin{lemma}\label{ff00}
The positive definite function $V(x):=x_1^2+2x_2^2/\rho k^2$,
evaluated along $x(\cdot)$, is non-increasing as long as $x(\cdot)$
remains in the fourth quadrant, {\it i.e.} $\{x\in\mathbb R^2 : x_1\leq 0, x_2\geq 0\}$.
\end{lemma}

\noindent The following lemma provides an exponential decay result for trajectories staying in ${\cal C}^s$ (in particular, for trajectories satisfying ($i$)).
\begin{lemma}\label{ff01}
There exist
$C_2=C_2(\textcolor{black}{\rho,k},T,\mu)>0$ and $\gamma=\gamma(\rho,T,\mu)>0$
such that,
for every $t\geq t_1$ such that $x(\cdot)$ stays in ${\cal C}^s$ along the interval $[t_1,t]$, it holds that
 \begin{equation}\label{ff11} |x(t)|\leq
C_2 e^{-k\gamma (t-t_1)}|x_0|.
\end{equation}
Moreover, $C_2=O(k)$ as a function of $k$ as $k$ tends to infinity.
\end{lemma}
\underline{Proof of Lemma~\ref{ff01}.}
Let $t$ be as in the statement of the lemma.
We deduce that, for $\tau\in (t_1,t)$,
$$
\dot x_2(\tau)=-k\alpha(\tau)w(\tau)x_2(\tau),
$$
where $w(\cdot)$ is a continuous function verifying
$$
0<1+\frac{\rho k}{2\xi^s_-}\leq w(\tau)\leq 1+\frac{\rho k}{2\xi^s_+}.
$$
Notice that the bounds on $w$ do not depend on $k$, due to the definition of $\xi^s_\pm$. We are back to the (PE) one-dimensional case studied in the
introduction. We deduce that there exist $C_*=C_*(\rho,T,\mu)>0$ and
$\gamma=\gamma(\rho,T,\mu)>0$ such that
\begin{equation}\label{inter00}
x_2(t)\leq C_* e^{-k\gamma(t-t_1)}x_2(t_1).
\end{equation}

Notice now that, for every $x\in {\cal C}^s$,
$$|x|\leq x_2\sqrt{1+\frac1{(\xi_+^s)^2}},\ \ \ \ \ \ x_2\leq k\sqrt{\frac\rho2}\sqrt{V(x)},\ \ \ \ \ \ \sqrt{V(x)}\leq \max\left\{1,\textcolor{black}{\frac 1 k\sqrt{\frac 2\rho}}\right\}|x|.$$

The proof of the lemma is concluded by
recalling  that $V(x(t_1))\leq V(x_0)$ and
plugging the above estimates in \r{inter00}.
\ \hfill \rule{0.5em}{0.5em}

\vspace{5mm}
Let us now
establish a lower bound on the time needed to go across
 the cone
${\cal C}^s$.
\begin{lemma}\label{final0}
There exists $\lambda_0>0$ such that
if $\lambda\geq\lambda_0$ and $x_0\in D^s_-$ then $x(\cdot)$ stays in ${\cal C}^s$ for all $t\in [t_0,t_0+1]$.
\end{lemma}
\underline{Proof of Lemma~\ref{final0}.}
Fix $x_0\in D^s_-$.
Reasoning by contradiction (and exploiting the homogeneity of \r{DIf}) we assume that there exist a strictly
increasing unbounded sequence $\{\lambda_i\}_{i\in \mathbb N}$
and a sequence $\{\alpha_i\}_{i\in \mathbb N}\subset L^\infty(\mathbb R_{\geq 0},[0,1])$ with $\alpha_i\in
{\cal {G}}(T/\lambda_i,\mu/\lambda_i)$ such that each
$x_i(\cdot):=x(\cdot\,;t_0,x_{0},K,\alpha_i)$, $i\in \mathbb N$,
reaches $D^s_+$ in time smaller than one. Because of the sequential
weak-$\star$  compactness of $L^\infty(\mathbb R_{\geq 0},[0,1])$, there
exists $\alpha_\star\in L^\infty(\mathbb R_{\geq 0},[0,1])$ and a
subsequence of $\{\lambda_i\}_{i\in \mathbb N}$ (still denoted by
$\{\lambda_i\}_{i\in \mathbb N}$) such that $\{\alpha_i\}_{i\in \mathbb N}$ converges
weakly-$\star$ to $\alpha_\star$.

By Proposition~\ref{technic},
we deduce that the sequence $\{x_i\}_{i\in \mathbb N}$ converges, uniformly
on compact time-intervals, and in particular on $[t_0,t_0+1]$,
to $x_{\star}(\cdot):=x(\cdot\,;t_0,x_0,K,\alpha_\star)$. Since, for every
$i\in \mathbb N$, $x_i(\cdot)$ reaches $D^s_+$ in time smaller than
one, we deduce that $x_{\star}(t_0+t_{\star})\in D^s_+$ for some
$t_{\star}\in (0,1]$.

Let us show that $\alpha_{\star}\geq \mu/T$ almost everywhere on
$\mathbb R_{\geq 0}$. For every interval $J\subset\mathbb R_{\geq 0}$ of finite length $\ell>0$,
apply  \r{weakstar} to the characteristic function
of $J$.
Since each
$\alpha_{\lambda_i}$ is a $(T/{\lambda_{i}},\mu/{\lambda_{i}})$-signal, it
follows that
\[
\frac 1 \ell \int_J \alpha_\star(s)ds=\lim_{i\to\infty}
\frac 1 \ell \int_J \alpha_{i}(s)ds \geq \liminf_{i\to\infty} \frac{\mu}{\ell \lambda_{i}}
{\cal I}\left(\frac{\ell\lambda_{i}}{T}\right)=\frac{\mu}{T}\,,
\]
where ${\cal I}(\cdot)$ denotes the integer part.
Recall that since $\alpha_\star$ is measurable and bounded (actually, $L^1$ would be enough),
almost every $t>0$ is a Lebesgue point for $\alpha_\star$, {\em i.e.}, the limit
\[\lim_{\varepsilon\to0+}\frac 1 {2\varepsilon} \int_{t-\varepsilon}^{t+\varepsilon} \alpha_\star(s)ds\]
exists and is equal to $\alpha_\star(t)$ (see, for instance,
\cite{leb_points}). We conclude that, as claimed,
$\alpha_\star(t)\geq \mu/T$ almost everywhere.

Therefore,
$x_{\star}(\cdot)$ is actually a trajectory of
the switching system
\begin{equation}\label{swi0}
\dot x =\left[u(A+bK^\top)+(1-u)\left( A+\frac{\mu}TbK^\top\right)\right]x,
\end{equation}
where $u$ is a measurable
function defined on $\mathbb R_{\geq 0}$ and taking values in $[0,1]$ (see
\cite{BOSCAIN2002} and references therein for more on switching
systems). According to the taxonomy and the results in \cite[page 93]{BOSCAIN2002}, \r{swi0}
is a switching system of type
$(RR.2.2.A)$ and, as a consequence, the curve
$x_{\star}$ stays below the trajectory $\bar{x}$ of \r{swi0} starting from
$x_0$ at time $t_0$ and corresponding to the input $u\equiv 0$.
Since $\bar x$ converges to zero in the cone delimited by $D^s_-$ and $D^{\mu/T}_-$, $x_{\star}$ must stay in the same
cone, contradicting the fact that $x_{\star}$ reaches
$D^s_+$ in finite time. Lemma~\ref{final0} is proved. \ \hfill \rule{0.5em}{0.5em}

\vspace{5mm}
Let us now focus on the behavior of trajectories exiting ${\cal C}^s$ or, equivalently,
such that $x_0\in D^s_+$.
\begin{lemma}\label{c2}
There exists $\lambda_1\geq k$ such that
if $\lambda\geq \lambda_1$ and $x_0\in D^s_+$ then \textcolor{black}{there exists a finite time
$t_f>0$ such that} $x(\cdot)$ satisfies
\begin{equation}\label{c22}
|x(t_f)|\leq|x_0|,
\end{equation}
with $x(t_f)\in D_2$ and $x(t)\in{\cal C}_2$ for all $t\in [t_0,t_f]$.
\end{lemma}

\underline{Proof of Lemma~\ref{c2}.} Fix $x_0\in D^s_+$. We reason again by
contradiction and follow the same procedure as in Lemma~\ref{final0}.
This is possible since, according to
Lemma~\ref{ff01}, the time needed by any trajectory of \r{DIf} to
go across in ${\cal {C}}_2$ is bounded (uniformly with respect to
$\lambda\geq k$ and $\alpha\in{\cal {G}}(T/\lambda,\mu/\lambda)$) by $C_1/k$.
We obtain that,
for some $t_0<t'_f\leq C_1/k$, the limit trajectory $x_\star$
is contained in ${\cal C}_2$ on $[t_0,t_f']$, reaches $D_2$ at time $t_f'$, and satisfies  $|x_{\star}(t'_f)|\geq|x_0|$.
 According
to \cite{BOSCAIN2002}, the trajectory $x_{\star}$ is, inside ${\cal
{C}}_2$, below the integral curve $x_{\mu/T}$ of $\dot x=
(A+\frac{\mu}TbK^\top)x$ with initial
condition $x_{\mu/T}(t_0)=x_0$. That means in particular that
$|x_{\star}(t'_f)|\leq |x_{\mu/T}(t''_f)|$, where $t''_f$ is the first time larger than $t_0$ such that $x_{\mu/T}(t''_f)\in D_2$. However, a lengthy but straightforward computation
shows that $|x_{\mu/T}(t''_f)|<|x_0|$ and we reach a
contradiction. \ \hfill \rule{0.5em}{0.5em}

\vspace{5mm}
Define $\lambda_*=\max\{\lambda_0,\lambda_1\}$ where $\lambda_0$ and $\lambda_1$ are the quantities appearing in the statements of Lemmas~\ref{final0} and \ref{c2}.
Whenever $x(\cdot)$ satisfies ($ii$) we let $(t_1\leq)t_2<t_f$ be such that
$x(t)\in{\cal C}^s$ for $t\in[t_1,t_2]$, $x(t)\in{\cal C}_2$ for $t\in[t_2,t_f]$, and $x(t_f)\in D_2$.

As a final technical result for the completion of the argument,
we need the following lemma.
\begin{lemma}\label{ouf0}
There exist $k_*=k_*(\rho,T,\mu)>0$ and
$\gamma_*=\gamma_*(\rho,T,\mu)>0$
such that
if $x(\cdot)$ satisfies $(ii)$,
$t_f-t_0\geq 1$, $k\geq k_*$, and $\lambda\geq \lambda_*$,
then
\begin{equation}\label{ouf}
|x(t_f)|\leq
\frac12 e^{-k\gamma_* (t_f-t_0)}|x_0|.
\end{equation}
\end{lemma}

\underline{Proof of Lemma~\ref{ouf0}.}
Applying estimates (\ref{ff11}) and (\ref{c22}) we get that
\[
|x(t_f)|\leq|x(t_2)|\leq C_2 e^{-k\gamma (t_2-t_1)}|x_0|\,.
\]
Moreover, according to Lemma~\ref{finite} and the hypothesis $t_f-t_0\geq 1$,
\[
|x(t_f)|\leq C_2 e^{2\gamma C_1} e^{-k\gamma (t_f-t_0)}|x_0|\leq C_2 e^{2\gamma C_1-k\gamma/2} e^{-k\gamma (t_f-t_0)/2}|x_0|\,.
\]
The lemma is proved by taking $\gamma_*=\gamma/2$ and
$k_*$ large enough in order to have  $C_2 e^{2\gamma C_1- k\gamma/2}\leq 1/2$. Such a $k_*$ does exist because, in view of Lemmas \ref{finite} and \ref{ff01}, neither $C_1$ nor $\gamma$  depend on $k$, while $C_2=O(k)$ as $k$ tends to infinity.
 \ \hfill \rule{0.5em}{0.5em}

\vspace{5mm}
We have developed enough tools to conclude the proof of Theorem~\ref{theo}.
Take $k\geq k_*$ and $\lambda\geq \lambda_*$.
Let $\xi_0\in\R^2\setminus\{0\}$, $t_0\geq 0$, $\alpha\in{\cal G}(T/\lambda,\mu/\lambda)$, and denote by ${\cal T}$ the set of
times such that $\xi(\cdot)=x(\cdot\,;t_0,\xi_0,\alpha)$ belongs to the $x_1$-axis.
Choose one representative for every connected component of ${\cal T}$.
Denote by ${\cal T}'$ the set of such representatives and by $j\in\mathbb{N}\cup\{+\infty\}$ its cardinality. By monotonically enumerating the elements of ${\cal T}'$, we have
${\cal T}'=\{\tau_i\mid 0\leq i<j\}$ with $\tau_{i-1}< \tau_{i}$ for $0<i<j$.
Let, moreover, $\tau_{-1}=t_0$ and $\tau_{j}=+\infty$.
Since,  for every $-1\leq i<j$, either $\xi(\cdot)$ or $-\xi(\cdot)$, restricted to $(\tau_{i},\tau_{i+1})$, is contained in the upper half-plane, then the previously established estimates apply to it.

Take
$-1\leq i<j$ and $t\in[\tau_i,\tau_{i+1})$.
Then there exists $C_3=C_3(K,T,\mu)>1$ such that
\begin{equation}\label{kind}
|\xi(t)|\leq C_3 e^{-k\gamma_* (t-\tau_{i})}|\xi(\tau_{i})|.
\end{equation}
Indeed, let
\begin{eqnarray*}
s_1&=&\sup\{s\in [\tau_i,t]\mid \xi([\tau_i,s])\subset {\cal C}_1\mbox{ or }-\xi([\tau_i,s])\subset {\cal C}_1\},\\
s_2&=&\inf\{s\in [\tau_i,t]\mid \xi([s,t])\subset {\cal C}_2\mbox{ or }-\xi([s,t])\subset {\cal C}_2\},
\end{eqnarray*}
and notice that $\max\{s_1-\tau_i,t-s_2\}< C_1/k$. Then \r{kind} immediately follows from
Lemma~\ref{ff01}.

In particular, if $j>0$
then
$$ |\xi(\tau_{0})|\leq
C_3 e^{-k\gamma_* (\tau_{0}-t_{0})}|\xi_0|.
$$
Moreover, for every $0<i<j$, Lemma~\ref{ouf0} yields
$$ |\xi(\tau_{i})|\leq
\frac12 e^{-k\gamma_* (\tau_{i}-\tau_{i-1})}|\xi(\tau_{i-1})|,
$$
which implies, by recurrence, that
$$ |\xi(\tau_{i})|\leq
\frac12 e^{-k\gamma_* (\tau_{i}-\tau_{0})}|\xi(\tau_{0})|.
$$
Therefore, independently of $j$, for every $-1\leq i<j$ we have
$$ |\xi(\tau_{i})|\leq C_3 e^{-k\gamma_* (\tau_{i}-t_{0})}|\xi_{0}|. $$
Applying again \r{kind} we obtain that for every $t\in[t_0,+\infty)=\bigcup_{i=1}^{j-1}[\tau_{i-1},\tau_i)$,
$$ |\xi(t)|\leq C_3^2 e^{-k\gamma_* (t-t_{0})}|\xi_{0}|\,, $$
which proves Theorem~\ref{theo}.
\ \hfill \rule{0.5em}{0.5em}

\vspace{5mm}
Let us go back to the natural question posed in $(Q1)$, that is, whether the choice of a stabilizer
$K$ can be made independently of $T$ and $\mu$.
Unlike the case when $A$
is neutrally stable, we prove below that the answer is negative when
the double integrator is considered.

\begin{prop}\label{prop1}
For every $K\in\R^2$, there exist $T\geq\mu>0$ such that $K$ is not
a $(T,\mu)$-stabilizer for system \r{DI}.
\end{prop}

\underline{Proof of Proposition~\ref{prop1}.} Let
$K=(-k_1,-k_2)\in\R^2$. If $k_1\leq 0$ or $k_2\leq0$ then $A+b K$ is
not Hurwitz and thus $K$ is not  a $(T,\mu)$-stabilizer for system
\r{DI}, whatever $\mu$ and $T$. Let now $k_1,k_2>0$. Among the possible configurations of the right-hand side of
\r{DIf}, we focus our attention on the linear vector fields $L_0$ and
$L_1$, corresponding to $\alpha(t)=0$ and to $\alpha(t)=1$
respectively, that is,
$$
L_0(x):=A\,x\,,\quad L_1(x):=\left( \begin{array}{cc}0&1\\ -k_1& -k_2\end{array} \right) x\,.
$$
Consider the set where $L_0$ and $L_1$ are collinear, {\em i.e.},
the union of the axis $\{x_2=0\}$ and the line $D$ defined by
$$
D:=\{(x_1,x_2)\in\mathbb R^2\mid x_2=-(k_1/k_2)x_1\}\,.
$$
We denote by $Q_1,Q_2,Q_3,$ and $Q_4$ the four regions of the plane
delimited by these two lines and defined as follows
\begin{eqnarray*}
Q_1&:=& \{(x_1,x_2)\in\mathbb R^2\mid x_2\geq -(k_1/k_2)x_1\,, x_2> 0\},\\
Q_2&:=& \{(x_1,x_2)\in\mathbb R^2\mid x_2> -(k_1/k_2)x_1\,, x_2\leq 0\},\\
Q_3&:=& \{(x_1,x_2)\in\mathbb R^2\mid x_2\leq -(k_1/k_2)x_1\,, x_2< 0\},\\
Q_4&:=& \{(x_1,x_2)\in\mathbb R^2\mid x_2< -(k_1/k_2)x_1\,, x_2\geq 0\}.
\end{eqnarray*}
Roughly speaking, on $Q_2\cup Q_4$ the vector field $L_1$ points
``more outwards'' than $L_0$  (with respect to the origin). Therefore,
on $Q_2\cup Q_4$ the excitation ($\alpha=1$) is not helping the stability
towards the origin (see Figure~\ref{quadrants}).

\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.4]{quadrants.eps}
\end{center}
\caption{Partition of the state-space into the four regions
$Q_1,Q_2,Q_3,Q_4$.}\label{quadrants}
\end{figure}

We look for a ``destabilizing" $\alpha$ in the form $\zeta(x(\cdot))$, where
$x(\cdot)$ is an unbounded solution corresponding to the feedback
\[
\zeta(x)=\left\{\begin{array}{ll}
1&\mbox{if }x\in Q_2\cup Q_4,\\
\mu/T&\mbox{if }x\in Q_1\cup Q_3,
\end{array}\right.
\]
$\mu$ and $T$ to be chosen. Clearly, any $\alpha$ obtained in such
a way is a $(T,\mu)$-signal. (It should be stressed that
although $\zeta$ is a discontinuous feedback, its solutions are anyway well-defined
because all trajectories of \r{DI} are rotating around the origin).

Let $\bar x$ be the first intersection with $D$ of the trajectory of
\r{DIf} starting from $(-1,0)$ and corresponding to $\alpha\equiv 1$.
Notice that there exists
$\nu>0$ small enough
such that
 the trajectory of \r{DIf} starting from
$\bar x$ and corresponding to $\alpha\equiv \nu$ crosses the $x_1$-axis
at a point $(\xi,0)$ such that $\xi>1$.  This is because
as
$\nu\to 0$ such trajectories converge uniformly on compact intervals to the horizontal curve $t\mapsto e^{t A}\bar x$.

Fix $T\geq \mu>0$ such that
$\mu/T\geq \nu$. Then the trajectory $x(\cdot)$ of
\r{DIf} starting from $(-1,0)$ and corresponding to the feedback $\zeta$ first leaves the upper half-plane through $[\xi,+\infty)\times\{0\}$.
Then, by symmetry and
homogeneity of the system, $|x(t)|$ goes to infinity as $t\to\infty$.
Thus, $\alpha(t)=\zeta(x(t))$ is the required destabilizing signal.
\ \hfill \rule{0.5em}{0.5em}

\subsection{Non-scalar control}
It makes sense to consider the stability properties of the system whose linear dynamics is the same as that of the double integrator, but which has a
different controlled part. That is, we study system
 \begin{equation}\label{multiDI}
\dot x=\left(\begin{array}{cc}0&1\\0&0\end{array}\right) x+\alpha B u,\ \ \ \ \ x\in\R^2,\ \ \ \ \ u\in\R^m,
\end{equation}
where $B$ is a general $2\times m$ matrix such that the pair $(A,B)$ is controllable
($A$ denotes the nilpotent $2\times2$ matrix appearing in \r{multiDI}).
Since the image of $A$ is one-dimensional,
the controllability of $(A,B)$ implies the existence of  a column $b$ of $B$ such that $(A,b)$ is controllable.
Moreover, by a linear change of coordinates,
we can transform $(A,b)$ into its Brunovsky normal form, that is, into the
matrix $A$ and the vector $b=(0 , 1)^\top$ as in \r{scandia}.
Therefore Theorem~\ref{theo} guarantees that for every pair of positive constants $\mu\leq T$ there exists a $(T,\mu)$-stabilizer for \r{system}.

The difference with respect to the scalar case is that
Proposition~\ref{prop1}
is not valid anymore when the rank of $B$ equals two, that is, the answer to question $(Q1)$ becomes positive.

\begin{prop}\label{q1yes}
If $\mathrm{rank}(B)=2$, then
there exists $K$ of size $m\times 2$ such that
for every  $T\geq\mu>0$
the gain $K$ is a
$(T,\mu)$-stabilizer for system \r{multiDI}.
\end{prop}

\underline{Proof of Proposition~\ref{prop1}.} Let us remark that, up to a reparameterization of $\R^m$ of the type $u'=M u$, $M$ invertible, we can rewrite $B$ as $(\mathrm{Id}\ {\bf 0})$, where $\mathrm{Id}$ is the $2\times 2$ identity matrix and all the entries of the $2\times(m-2)$ matrix ${\bf 0}$ are null.
That is, \r{multiDI} is equivalent to
 \begin{equation}\label{multiDI2}
\dot x=A x+\alpha u,\ \ \ \ \ x\in\R^2,\ \ \ \ \ u\in\R^2.
\end{equation}
Fix $k>0$ and take $u=-k x$. If $t\mapsto x(t)$ is a solution of
\[\dot x=(A -\alpha k\mathrm{Id})x,\]
then $t\mapsto y(t):=e^{-At}x(t)$ satisfies $\dot y=-k\alpha y$. Therefore
$$
y(t)= \exp\left(-k\int_0^t\alpha(\tau)d\tau\right) y(0)\,,
$$
which implies that
$$
\norm{x(t)}\leq \norm{e^{A t}}\exp\left(-k\int_0^t\alpha(\tau)d\tau\right)\norm{x(0)}.
$$
Since $e^{A t}=\mathrm{Id}+tA$ is linear in $t$,
we have proved that for every $T\geq\mu>0$ and for every $k>0$ the gain $K=-k\mathrm{Id}$ is a
$(T,\mu)$-stabilizer for system \r{multiDI2}.
\ \hfill \rule{0.5em}{0.5em}

\section{Appendix}

We first provide a simple result used several times in the
paper.

\begin{prop}\label{technic}
Consider system \rref{feedback}, where $A,B,K$ are matrices of size
$n\times n$, $n\times m$ and $m\times n$ respectively, and $\alpha$
is a $(T, \mu)$-signal. Take a sequence $x_{0i}$ of norm-one
vectors converging to $x_{0*}$ and a bounded sequence $\alpha_i$ in
$L^\infty({\mathbb R})$ which converges weakly-$\star$ (in
$L^\infty({\mathbb R})$) to a measurable function $\alpha_\star$.
Then the sequence
$\{x_{i}(\cdot):=x(\cdot,0,x_{0i},K,\alpha_i)\}_{i\in\mathbb N}$ converges,
uniformly on compact time intervals, to
$x_{\star}(\cdot):=x(\cdot,0,x_{0\star},K,\alpha_\star)$ as $i$
tends to infinity.
\end{prop}

\underline{Proof of Proposition~\ref{technic}.} Recall that the
weakly-$\star$ convergence of $\alpha_i$ to $\alpha_\star$ means that, for
every $\varphi\in L^1(\mathbb R_{\geq 0},\mathbb R)$, it holds that
\[
\lim_{i\to\infty} \int_0^\infty \alpha_{i}(s)\varphi(s)ds =
\int_0^\infty \alpha_\star(s)\varphi(s)ds\,.
\]
Taking as $\varphi$ the characteristic function of an arbitrary interval
of length $T$
shows that $\alpha_\star$ is a
$(T,\mu)$-signal.
Moreover, the norm of $x_{0*}$ is equal to one. For $i\in \mathbb N$, set
$$
e_i(\cdot)=x_{i}(\cdot)-x_{\star}(\cdot), \ \
e_{0i}=x_{0i}-x_{0*},$$
$$
A_i(\cdot)=A+\alpha_i(\cdot)BK,
$$
and let $\phi_i(\cdot,\cdot)$ be the fundamental solution of $\dot
x=A_i(t)x$. Integrating the differential equation verified by $e_i$,
one gets, for $t\geq 0$,
\[
e_i(t)=\int_0^tA_i(s)e_i(s)ds+h_i(t),
\]
where
$h_i(t)=\phi_i(t,0)e_{0i}+\int_0^t(\alpha_i(s)-\alpha_\star(s))BKx_{\star}(s)ds$.
Note that the functions $h_i$ are uniformly bounded over compact
time-intervals and the sequence they define converges point-wise to
zero as $i$ tends to infinity. Therefore, by combining Gronwall
Lemma and the bounded convergence theorem, it follows that the
sequence $\{e_i(\cdot)\}_{i\in\mathbb N}$ converges point-wise to
zero as $i$ tends to infinity. The uniform convergence on compact
time intervals results from the above combined with Ascoli theorem.
\ \hfill \rule{0.5em}{0.5em}

\label{sec:concl}

\newcommand{\textsc}[1]{{\rm #1}}
\newcommand{\textit}[1]{{``#1''}}
\newcommand{\titl}[1]{{\em #1}}
\newcommand{\textrm}[1]{{\em #1}}
\newcommand{\vol}[1]{vol.~{#1}}
\newcommand{\pp}[1]{pp.~#1}

\newcommand{\SortNoop}[1]{}
  \def\nesic{Nesi\'{c}\,} %
  \def\astrom{{\SortNoop{As}\AA}str{\"{o}}m\,}\let\c=\cedille

redpajama_set_name: RedPajamaArXiv


text:
CampCounts 2021 Report
CampCounts 2021 is the latest iteration of ACA's annual business and industry research survey, which aims to support camp professionals in their operational decision-making and to support ACA's advocacy efforts.
CampCounts 2021 is the latest iteration of ACA's annual business and industry research survey, which aims to support camp professionals in their operational decision-making and to support ACA's advocacy efforts. Responses are anonymous and reported in aggregate by the following segments: program types offered (day, overnight, or other alternative programming), primary program type (50% or more day camp programming, overnight camp programming, or other alternative programming), and camp owner/sponsor (independent nonprofit, for-profit, municipal/government, nonprofit affiliated with another entity).
In addition to the standard survey questions carried forward from previous business and industry surveys, CampCounts 2021 included unique questions intended to capture some of the ongoing effects of the COVID-19 pandemic on camps in the US, such as staffing impacts and supply chain challenges. Additionally, some questions asked camps to report information from 2021 as well as a more "typical" (pre-pandemic) year, with the hopes of gleaning more representative benchmark data. Results are compared to the 2017 ACA Business Survey (now incorporated into CampCounts) and the CampCounts 2020 survey where possible.
The American Camp Association would like to discourage journalists from using the CampCounts 2021 Report to cover the average price of camp. The 2021 data is based on a sample of less than 300 respondents and that sample doesn't represent the full breadth of camp pricing across the field in our country. Average day camp prices range from completely free to $200+ per day and overnight camp prices range from free to $500+ per day. We do expect to see increases in camp prices for 2023 due to increasing costs across the board this year (It is important to note that many camps set prices for the following summer in late August or fall after camp concludes). In turn, there is a tremendous case for additional philanthropic support for low-income families given the tremendous need. ACA would also highlight the 93 percent of all camps who provide need-based financial aid available for low-income families. To learn more, please email the ACA PR Team at pr@acacamps.org.
Download the 2021 CampCounts Report
Download the 2021 CampCounts Infographic
Blog: Key Takeaways from the 2021 CampCounts Report
Infographic: 2020 CampCounts Report
Download the Comprehensive Report
Download the Parent Perspectives Report

redpajama_set_name: RedPajamaCommonCrawl


Thank you all for coming to an opening of my first solo exhibition. It was great to have you there!
Contemporary teapots by Radka Passianova, recipient of the 2009 Jewellery and Object Design Award, Design Centre Enmore.
The series is inspired by human evolution and specifically personal evolution; the evolution of skills, aesthetics and knowledge.
This 'Evolution Tea Pot' series considers the aesthetics and processes of the traditional raising techniques of silversmithing and responds to these in a contemporary manner.
This also coincides with the 2 Danks Street gala opening of Sculpture 2010. All galleries within the complex will be open between 6-8pm showcasing new sculpture work for the year..
Featured on the Jewellery & Objects Website.


The Union Springs Public Library is one of two remaining Carnegie Libraries operating as a library, in the state of Alabama, we are over a hundred years old. We house approximately 10,000 volumes of Adult, Young Adult and Children's books. We subscribe to about 22 periodicals. We have 16 computers available for public use, divided into two separate work areas for Adults and Students age 12-18. Wireless internet use is available. The Union Springs Public Library strives to serve and better the community we all live in with annual Summer Reading Programs, monthly Career Resource Training, and a number of other community outreach events and programs.


text:
The Former Bartender Behind Fishers Island Lemonade Makes a Mean Mignonette Sauce
Recipes, Tastemakers May 27, 2022
STORY BY Micki Wagner
Just two miles off the coast of Connecticut, right at the edge of the Long Island Sound, is a small sliver of land called Fishers Island. Home to just 125 people, the island is said to host up to 2,000 people throughout the summer. Which feels fitting for a place that has a canned spiked lemonade named after it.
Bronya Shillo, founder of that canned spiked lemonade company, Fishers Island Lemonade, grew up on the island and began bartending at her parents bar, The Pequot Inn, at just 18 years old. Soon after that, she saw the opportunity to turn her parents' recipe into a canned cocktail people could enjoy anywhere—even outside of Fishers Island. As the weather begins to heat up in New York, the brand has launched its first variety pack—an eight-pack with the original spiked lemonade (a combination of vodka, whiskey, lemon and honey), along with the spiked tea (vodka, whiskey, lemon and black tea) and pink flamingo (vodka, whiskey, lemon and cranberry juice) varieties. Ahead of this launch, Edible Manhattan sat down with Shillo to discuss all things Fishers Island Lemonade, including the surprising recipe she makes with it.
What sparked the idea for Fishers Island Lemonade?
I grew up in a funky little place called Fishers Island. My parents owned the only bar out there, which is called The Pequot Inn, and Fishers Island Lemonade was actually the house recipe of that really fun place that I had the opportunity to grow up at and get my bartending skills from. Pretty much as soon as I could start legally working, I was making pizzas in the back kitchen or helping out my parents with inventory. I have an older sister and cousins, so I just tagged along. By the time I was legally able to step behind the bar, I had seen quite a bit at a young age and so I felt like I was able to start pretty quickly. And I started listening to consumers and understanding what they wanted. I was also brewing beer and making a lot of different cocktails outside of bar hours and decided to can our cocktail because I felt like that was something that hadn't been done yet.
Tell me a little bit about the packaging for Fishers Island Lemonade. It's so cute!
We took a lot of pride in creating a really fun package that spoke to the roots of Fishers Island. So, the yellow and white striping has a lot to do with the 1950s-style bathhouses.
What do you hope consumers feel when drinking Fishers Island Lemonade?
I love the discovery mode, where they're just like, "Oh my gosh, I can't believe that all of this is in the can. It tastes so good; it tastes just like the bartender made it." We really do take a lot of pride in everything from the exterior of the cans to the interior of the cans to how we promote our story and our messaging. So, our team has a very high standard for that quality throughout. And I think that that's something that consumers continue to come back for with our brand.
How do you do that? How do you ensure that it tastes like the bartender just made it?
This was sort of the "aha moment" for me too, as I was building a business. As a bartender, I was like, we're going to use vodka, we're going to use whiskey, we're going to use freshly squeezed lemons and real honey. And when it came down to creating that business not only from all the costs of the components, but also from the manufacturing side, there were a lot of areas that I just was uncovering, and I didn't want to cut corners. I was really proud of what we were creating, and I didn't want to slip something in there just because it costs us less. You can see it right on our cans, we're talking about a canned cocktail—it's truly a cocktail made with real spirits. We use a whiskey that's been barrel-aged, and that's something that you could pour out of a bottle off the shelf that you'd want to drink neat. It's not something that has a masked flavor in it. So I think when it comes down to why consumers continue to come back to Fishers and become lifelong customers is because they know that we're putting the real deal on the can, and it's not just about the marketing.
I know lemonade is not a drink you traditionally pair with food, but what would you want to eat with Fishers Island Lemonade?
Growing up on Fishers Island, I've always been a pretty avid fisherwoman, so I love fish tacos. My husband and I met spearfishing, so we love to kind of go get our groceries, so to speak. I'm also a really big fan of our Fishers Island oysters. And our lemonade makes a really fun mignonette sauce. So if you were to take a champagne vinegar, a little black peppercorn, some Fishers Island Lemonade, you get that nice salty brightness, but a little touch of the Fishers Island flare with both Fisher Island oysters and Fishers Island Lemonade.
RELATED: The Billion Oyster Project: Reversing History to Move Forward
Fishers Island Lemonade Mignonette Sauce
1 can Fishers Island Lemonade
2/3 cup champagne vinegar
2 medium shallots minced, about 1/2 cup
2 Tablespoons coarse black pepper, butcher crack will do
Stir all ingredients together with a fork and chill.
Feature photo courtesy of Maaike Bernstrom.
Tags: Bronya Shillo, Canned Cocktails, drinks, Fishers Island Lemonade, Summer
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When does traditional statistical modelling (TSM) become machine learning (ML)?[i] "Machine learning" has truly become a buzzword that is applied rather liberally to a wide range of modelling applications. But, the difference is far from a question of semantics: there are fundamental differences between ML and TSM that data practitioners should keep in mind.
But, let's start off with some commonalities between ML and TSM. In both disciplines our aim is to build a (statistical) model (to use TSM terminology) that minimises loss, that is, that achieves the smallest possible difference between observed values and the values estimated by the model. In so doing, we have to achieve a successful balance between model complexity and generalisability: pick too complex a model and you'll achieve a great fit on the data that you used to develop it; but your predictive power will be limited on unseen data.
Some of the techniques that we use to achieve our goal – of minimising loss – are similar between ML and TSM too. For example, both rely on a cost function (such as the mean squared error, MSE), and on a way of optimising that cost function.
ML and TSM have fundamentally different philosophical roots. As a branch of artificial intelligence[ii] and a sub-field of computer science, ML's main purpose is to "learn" from data, that is, to optimise the weights on features (or: parameters), and to apply that information to generate new predictions. Its approach is inductive: we let the data tell give us the answers, and do not rely on ex ante expectations of how feature x may affect phenomenon y. This requires fewer distributional assumptions about our predictor variables, and fewer assumptions about which features are most predictive of our outcome of interest. (Note however that this is not to say that ML is not interested in selecting appropriate features to include! In fact, this is a crucial part of any successful ML application).
By contrast, traditional statistical modelling attempts to use mathematical formulas to formalise the relationship between two or more variables. As such, it is a subfield of mathematics, and is usually applied in a deductive context, that is, a research design where hypotheses are formulated and subsequently put to the test.
In addition, a linear model in a machine learning setup would use a different optimisation algorithms such as gradient descent (GD). While TSM applications of linear regression can (and do) use GD, it becomes particularly relevant in an ML context. When we have large numbers of predictors (which usually is the case in ML problems, where we may have thousands of features), we need to use an optimiser that is computationally cheap enough to process large amounts of data. Gradient descent saves a lot of time on calculations compared to calculating parameters analytically (note that some optimisation problems may not even have a closed-form solution due to their complexity!). We can significantly reduce our computation time in ML by using GD in mini-batch form, or in particular, in the stochastic variant, where we sample from data and change the model parameters just a little bit after each sampling.
In sum, there are important differences between machine learning and traditional statistical modeling. And, your choice between either should really be informed by the data problem that your facing. For example, TSM is preferable by far if you have a relatively small dataset that consists of structured data, and if your purpose is to identify if and to what extent a (limited) set of variables affect a phenomenon of interest. In other words: use the TSM toolbox when you are interested in taking a deductiveapproach to your research problem, i.e. when you start from theory and hypotheses, and use empirical data to confirm or reject your theoretical propositions.
This article was originally published on the Official Inspera Blog.
[i] I use terms from either discipline rather liberally in this post. For example, whereas ML speaks of "weights", TSM usually refers to "coefficients". The same goes for "features" vs. "independent variables". I also do not venture into a discussion of supervised vs. unsupervised learning, and/or deep learning.
[ii] In turn, a key difference between AI and ML is that the latter does not try to imitate "intelligent" behaviour. Rather, ML is intended to complement and outperform human tasks using the strengths of computers.
[iii] ML is of course not altogether assumption-free. For one, in ML we assume the training samples the we draw from our distribution are i.i.d. (independently and identically distributed).


Up until May 2015, I had always worked in some form of the client-service model, offering design-related services to anyone from sports teams to the largest corporations in the world. I had spent the first 12 years working, for the most part, in a freelance/contract capacity in the Advertising and Digital Marketing (Creative) industry. But in 2013, I decided to take a full-time offering as I wanted to be more immersed in projects, take ownership and form a closer bond with a team.
After 14 years of client-service, I wanted something different, something that got me closer to the users of products. In client-service, the client is usually a proxy to the user, which made it difficult to understand their needs at the level required to design great products. I was also bored and fairly sure that I wanted to start a product company of my own at some point.
Wanting to learn the inner workings of startups, I took the position of Product Design Lead at a Fintech startup with the additional goal of doing something I hadn't seen very much of at the time - make a financial services product an enjoyable experience to use.
Given my background and the support I had (i.e., Account Managers) or having worked solo, I assumed that everyone's knowledge of the "Design Process" was a given and that this would be like any other project I'd worked on before, just longer. Immerse myself in the business and get to know it inside and out, review any data that was available, audit the current offering to see what was working and what wasn't, and start producing a story map, some interaction models, and wireframes.
In hindsight, it sounds so obviously wrong as to be absurd to think this, but it was at least six months before I became aware that I even had these assumptions — something about a forest and some trees should probably go here. It was at this point that I was trying to understand why I was so frustrated working there. I'd never even considered what I'd been doing for so many years as something that required much explanation to others, and that I even had a "Design Process" (let alone formally document it). So, of course, it makes sense now that it shouldn't be a given.
It was at this point that I was explaining my dilemma to a friend I had previously worked with and he summed it up perfectly - I was a fish out of water. My way of working had been so engrained that trying to explain a "Design Process" was like trying to explain how to breathe.
As with many startups (from what I've read), this one was developer-led, and the development team had their problems trying to form a process that worked for them. Requirements would usually go directly from the business to the development team. Not recognising that I even had a Design Process until it was too late to "rock the boat" enough to get it into the mix, it was tough to make a positive impact on the product, from a UX perspective. To be honest, I had taken up a defeatist attitude at that point and didn't particularly want to push as hard as I needed to have a Design Process be part of a holistic product development process (more on that concept in another post).
From a Product Design perspective, that just left me ensuring there was some thinking that went into anything put in front of users that didn't violate basic usability rules. Not the least bit challenged, or interested, I was ready to leave when we decided to pivot. Oooooooh, a Pivot! Seriously, though, I was here to learn about startups, so I saw this as a great learning opportunity.
Not being able to influence the UX in a way that I wanted to, I took this as an opportunity to influence the CX (Customer Experience) side of the business. The differences being subtle (in my opinion), this was a chance to add the tooling necessary for me to create an enjoyable experience for users, outside of what the development team built. Mostly in the form of chat, email, and analytics.
For the last nine months, my role was more of a Product Manager/Customer Success Manager. I went on an event tracking rampage, adding Segment tracking calls for almost everything in the new app. Completely obsessed with all of this, I integrated Intercom for customer engagement and support, Mixpanel for funnel tracking and segmentation, and a plethora of other tools that gave us great insights into what users were doing – a level of understanding we never had before. Adding all of this gave life to an entirely new way of approaching how we handled the Customer Experience and gave us the data to support suggested changes to the product.
This change and the abilities it afforded us is what I'm most proud of during my time there. I put us in a position to be able to ditch a very expensive CRM, use data to make decisions and answer intelligent questions, be more informed about what was happening and when with regards to the product, and empower others to uncover new insights. I can now do the same things for many other startups as a new service offering.
Reflecting on my experience, I learned a lot. My design skills were not improved at all (which I expected), but I learned how (and how not) to do my work in an entirely new setting. The importance of explaining to others your process and how you work is critical if you want them to see value in what you do. I also realise the need to improve my ability to recognise problems earlier and find ways to get unstuck.
If there isn't an existing Design Process, be prepared to create, rationalise, fight for, and enforce one.
Be prepared to back up your process and follow it.
Educate those that have not worked with designers on how you work.
Measure everything, it's cheaper than calling everyone.
Use "The Window" in the first few weeks to take your position and stand your ground.
Insist that the designer explains their process and why they do what they do.
Question whether or not this process is something that you see value in and are willing to incorporate into any existing processes, given what stage your company is at right now. There's no point in hiring them if you're not willing or able to.
The outcome of all this has been great for me. I've been able to crystalise much more clearly what it is I want and what it isn't, which I'll probably write about at some point. I've also discovered a new offering that I'd never even thought of before this. Product analytics – strategy and implementation. If you're interested in learning more about this, get in contact.


text:
The Quest for a Better Protein Bar
Sally Arsenault
Contributor - Brazilian Jiu Jitsu, Gear Reviews
Brazilian Jiu Jitsu, MMA
Breaking Muscle receives no compensation in exchange for reviews. We received this product for free and did not experience typical customer service. The opinions expressed belong solely to the writer.
If you've ever tried to eat a protein bar, you know that they can be hard to swallow. Dry, chewy, awful tasting, and I've seen bars that cost over five dollars. Ridiculous!
Quest Nutrition has changed the game. Delicious and affordable, their products leave you feeling satisfied.
In 2010, a group of wellness enthusiasts on a quest gathered in a kitchen. Their goal was to produce a protein bar that people actually wanted to eat, but was free of sugar and other junk fillers. A bar so delicious, it felt like a cheat meal.
When the company was founded, they started with two flavors: Vanilla Almond Crunch and Peanut Butter Supreme. Now, Quest offers seventeen different flavors. For this review, I was sent a 12-Flavor Sample Pack.
"What I like about this company is that throughout the development of their products, they refuse to compromise."
I'm not a huge fan of eating protein bars, so it took me about a year to get through a box of twelve. I kept notes as I ate each one. The qualities consistent with each bar were the texture (chewy and grainy, with kind of a brownie texture) and how filling they were. I could rarely eat a whole bar, but I'm about 105lb and I generally halve a lot of my supplements, so it didn't surprise me.
I can't honestly say that eating these bars felt like a cheat. When I ate them, I knew I was eating a protein bar, and in some cases didn't really enjoy them. However, there were some exceptions. In my tasting notes, I stated the following:
"The Cookies & Cream bar tastes like an Oreo blizzard. It appears to have chunks of cookies."
"The Lemon Cream Pie bar has a nice, light lemon flavor. Chewy texture, a little gritty. My favorite so far."
"Vanilla Almond Crunch is delicious. It tastes like that green pistachio ice cream. It's not as chewy or gritty as the other bars. There are pieces of almond in it. I ate the whole thing at once."
"Chocolate Chip Cookie Dough is delicious, but still obviously a protein bar. It's not gritty and the chocolate chips appear to be real. One of my favorites."
"White Chocolate Raspberry seems to have real raspberry seeds. It is delicious, one of my favorites. The raspberry flavor is mild and there are chunks of what appears to be white chocolate."
The remaining bars were okay, but I probably wouldn't buy them in the future if the flavors above were available. Compared to what some of the other companies are offering, though, even an okay Quest bar is preferable.
Quality of Ingredients
In my notes, I commented that some bars had products that "appeared to be" cookies or chocolate chips or white chocolate. It turns out that Quest Nutrition's commitment to using high quality, cutting-edge ingredients is so strong, they've manufactured those, too:
We even manufacture the inclusions that we put in our bars. The chocolate crumbles in Cookies and Cream and the white chocolate chunks in the White Chocolate Raspberry… we make those from scratch. It's the only way to get the perfect taste and nutritional profile you come to expect from Quest.
What I like about this company is that throughout the development of their products, they refuse to compromise. The "About" section of their website explains:
Every consultant and industry expert recommends we change our formulation, cut some corners or add some sugar. We know what's down that road: compromise. A failure to uphold our stated mission. This wasn't just about starting a business. It was about revolutionizing the way people eat. Everyone else is adding sugar and junk. For us, it's personal. We have friends and family struggling with obesity and diabetes in desperate need of an alternative to what's available. So we forge ahead. We fly to different production labs to meet with engineers that promise a solution. Time-and-time again Quest Bars proves too much for any pre-made equipment.
Quest bars became so popular, customers developed a community, sharing their experience, and even baking with the bars as an ingredient. It turns out some of my favorite bars can be turned into other delicious treats, as explained on the company website:
"Customers and fans are baking with Quest Bars. Quest-inspired recipes pour in from kitchens around the country. And we're not talking about one or two. It's hundreds.
Suddenly it's not us huddled around the kitchen island trying to invent something new, it's our community. To them, it's not only a protein bar. It's an ingredient. They use our bars to bake cookies, cake, and crepes. We pick a few out and give them a try. Our reaction: "Whoa! This is delicious!"
The company owners loved readers' recipes so much, they developed a series of cooking videos, with host Cassey Ho. The video series is called Cooking Clean with Quest, and there's also an online Cheat Clean Cookbook. Be sure to check them out if you want to experiment with clean eating in the kitchen.
Beyond Protein Bars
Over the past couple of years, Quest Nutrition has been expanding beyond the protein bar market to offer protein powder, protein chips, peanut butter cups, and guilt-free pasta. I was also sent the spaghetti, spinach fettuccini, and peanut butter cups. I don't eat a lot of pasta so unfortunately the fettuccini expired before I could use it, but I took notes on the spaghetti and the peanut butter cups.
"I had the Quest Spaghetti with meat sauce I made at home. It seems to retain a bit more water than regular pasta, so I should have let drain a bit longer than normal. It doesn't have a distinct taste, it just tastes like the sauce. Not as filling as regular pasta but still good. Easy to eat, faster to prepare than other pasta."
"The Quest Cravings Peanut Butter Cups are similar to Reese's only not as creamy. It's a lot drier and it takes a while to eat just one. I found it stuck to my teeth and the roof of my mouth. I had to use my finger to scrape it off after licking like a dog for some time. I couldn't eat two, one was filling enough."
Quest Nutrition products are available for $2.39 and up at QuestNutrition.com.
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