TITLE: Question about Lagrangian Multiplier
QUESTION [1 upvotes]: The question is $\max_{x,y} x^{2}+y^{2}$ subject to $x^{2}+xy+y^{2}-4 = 0$.
First, I set up the Lagrangian:
$L =x^{2}+y^{2}-\lambda (x^{2}+xy+y^{2}-4)$
Then, I calculated the first order condition:
$L_{x}^{'}=2x-(2x+y)\lambda=0 $
$L_{y}^{'}=2y-(2y+x)\lambda=0 $
$L_{\lambda}^{'}=x^{2}+xy+y^{2}-4=0 $
I don't know how to get $(x, y, \lambda)$ from these three conditions. Can someone teach me?
(When I check the solution, it says that from the first two conditions we can get $x^{2}=y^{2}$, I am confused by this statement.)

REPLY [1 votes]: Or rewrite first like $\frac{2-2\lambda }{\lambda}  \times x =y$
and substitute $y$ by left part in the second equation
you will get the quadratic equation for lambda
Solving it you will get two equation $x = y$, and $x = -y$
you can choose one trying each in the third equation.


text:
\begin{document}

\maketitle

\begin{abstract}
We consider switched systems on
Banach and Hilbert spaces
governed by
strongly continuous one-parameter semigroups of linear evolution operators.
We
provide necessary and sufficient conditions
for their global exponential stability, uniform with respect to
the switching signal,
in terms of the existence of a
Lyapunov function common to all modes.
\end{abstract}

\section{INTRODUCTION}

It is well known that the existence of a \emph{common Lyapunov function}
is necessary and sufficient for the global uniform
asymptotic
stability of
finite-dimensional continuous-time switched
dynamical systems
\cite{LinSontagWang1996}.
In the \emph{linear} finite-dimensional case,
the existence of a common Lyapunov function
is actually equivalent to
global uniform \emph{exponential}
stability \cite{MolchanovPyatnitskiy1986,MolchanovPyatnitskiy1989} and, provided that the system has finitely many  modes,
the Lyapunov function can be taken polyhedral or polynomial (see
\cite{Blanchini1994,BlanchiniMiani1999,DayawansaMartin1999} and also
\cite{BraytonTong1980}
for a discrete-time version).
A special role in the switched control literature has been played by common
quadratic Lyapunov functions, since their existence can be tested rather efficiently (see the surveys \cite{LinAntsaklis2009,ShortenEtAl2007} and the references therein). It is known, however, that

the
existence of a common quadratic Lyapunov function
is not necessary for the
global uniform exponential
stability of a linear switched system with finitely many modes. Moreover, there exists no uniform upper bound on the minimal
degree of a common polynomial Lyapunov function \cite{MasonBoscainChitour2006}.

The scope of this paper is to
prove that
the existence of a {common Lyapunov function}
is equivalent to the global uniform exponential
stability of
in\-fi\-ni\-te-di\-men\-sio\-nal switched systems of the type
\begin{equation}\label{eq:ssOpForm}
\left\{\begin{aligned}
 \frac{d}{dt}\,x(t)	&= A_{\sigma(t)} x(t),\quad t > 0\\
 	x(0)		&= x \in X
\end{aligned}\right.
\end{equation}
where each $A_j$ is a (possibly unbounded) operator generating a strongly continuous
semigroup $T_j(t)$ on a Banach space $X$ and $\sigma(\cdot)$ belongs to the class of
piecewise constant switching signals with values in an index set $Q$.

Such systems provide a convenient design paradigm for modeling a wide variety of complex
processes comprising distributed parameters, see \cite{HanteLeugeringSeidman2009} and the references therein
for examples in the context of networked transport systems.

Except for special cases,
that is,
when $X$ has a Hilbert structure and the infinitesimal generators
$A_j$ commute pairwise \cite{Sasane2005}, when the switching signals satisfy a
dwell-time constraint \cite{MichelSunMolchanov2005} or when $A_j$ is a linear convection-reaction operator with reflecting
boundary conditions \cite{AminHanteBayen2008}, global uniform exponential stability of systems such
as \eqref{eq:ssOpForm} has not been investigated (up to our knowledge).

The characterization of exponential stability for a single linear dynamical system on
Banach and Hilbert spaces dates back to Datko \cite{Datko1968} and Pazy \cite{Pazy1972} and has,
since then, seen a broad range of applications in control theory for partial differential equations (see, for instance, \cite{TucsnakWeiss2009}).
However, we recall that exponential stability of all subsystems (with $\sigma(t) \equiv j$
fixed in \eqref{eq:ssOpForm}) is of course necessary but not sufficient for the global uniform
exponential stability with respect to all possible
switching laws $\sigma(\cdot)$. This is a classical result for the finite dimensional case and we give an
infinite dimensional variant with interesting destabilizing properties
in Example~\ref{ex:notGUES} below.

Our starting point will be a switching system of the general form
\begin{equation}\label{eq:ss:intro}
\left\{\begin{aligned}
 x(t_{k+1}) &= T_{\sigma(t_k)}(t_{k+1}-t_k) x(t_k),\quad k\in\mathbb{N},\\
 x(0)&=x_0 \in X,
 \end{aligned}\right.
\end{equation}
where $\sigma\: [0,\infty) \to Q$ is a piecewise constant right-continuous
switching signal with switching times $0=t_0<t_1<\dots<t_k<\cdots$.
Each $t\mapsto T_j(t)$, $j\in Q$, is a strongly continuous semigroup on a Banach space $X$.
If $t\in(t_k,t_{k+1})$, then $x(t)=T_{\sigma(t_k)}(t-t_k) x(t_k)$.
This framework includes, in particular,
switched dynamical systems such as \eqref{eq:ssOpForm}, even
with the infinitesimal generators $A_j$ not sharing a common domain and also in the case of
infinitely many available modes $Q$. (The
semigroup formulation \eqref{eq:ss:intro}
corresponds to the choice of mild
solutions for the Cauchy problem \eqref{eq:ssOpForm}. The two formulations are clearly equivalent when $X$ has finite dimension, with $T_j(t)=e^{t A_j}$.)

The main result of this paper is

that
the following three conditions are equivalent:
\emph{
\begin{itemize}
 \item[(A)]
There exist two constants $K \geq 1$ and $\mu>0$
such that,
for every $\sigma(\cdot)$ and every $x_0$,  the solution $x(\cdot)$ to \eqref{eq:ss:intro} satisfies
\begin{equation}\label{thm0:GUES}
 \|x(t)\|_{X} \leq Ke^{-\mu t}\|x_0\|_{X}, \quad t \geq 0.
\end{equation}
\item[(B)] There exist two constants $M \geq 1$ and $\omega>0$
such that,
for every $\sigma(\cdot)$ and every $x_0$,  the solution $x(\cdot)$ to \eqref{eq:ss:intro} satisfies
\begin{equation}\label{thm0:UEBound}
 \|x(t)\|_{X} \leq Me^{\omega t}\|x_0\|_{X}, \quad t \geq 0,
\end{equation}
and there exists $V\: X \to [0,\infty)$
 such that
 $\sqrt{V(\cdot)}$ is a norm on $X$,

	\begin{equation}\label{thm0:VBound}
 		V(x) \leq C \|x\|_X^2, \quad x\in X
	\end{equation}
	for a constant $C>0$ and
	\begin{equation}\label{thm0:LjVest}
 		\liminf_{t \downarrow 0} \frac{V(T_j(t)x) - V(x)}{t} \leq -\|x\|_X^2,\quad
		j \in Q,~x\in X.
	\end{equation}

\item[(C)] There exists $V\: X \to [0,\infty)$
 such that
 $\sqrt{V(\cdot)}$ is a norm on $X$,

	\begin{equation}\label{thm00:VBound}
 		c\|x\|_X^2\leq V(x) \leq C \|x\|_X^2, \quad x\in X
	\end{equation}
	for some constants $c,C>0$ and
	\begin{equation}\label{thm00:LjVest}
 		\liminf_{t \downarrow 0} \frac{V(T_j(t)x) - V(x)}{t} \leq -\|x\|_X^2,\quad j \in Q,~x\in X.
	\end{equation}
		\end{itemize}
}

The equivalence between (A) and (C) extends to infinite-dimensional systems the well-known result obtained in \cite{MolchanovPyatnitskiy1986}
in the finite-dimensional setting.

Conditions \eqref{thm0:VBound} and \eqref{thm00:VBound} are redundant in the case of finite-dimensional systems, since
$\sqrt{V(\cdot)}$ and $\|\cdot\|_X$ are comparable, by compactness of the unit sphere.
Hence, condition \eqref{thm0:UEBound} in (B) could be dropped for finite-dimensional systems. This is not the case
for infinite-dimensional ones, as illustrated in
Remark~\ref{rem:Thm1sharp} by an example.

From the point of view of applications,
condition (B),
imposing less conditions on $V$ than (C), is better suited for
establishing that (A) holds (although the uniform exponential growth boundedness needs also be proved).
On the other hand, the implication (A)$\Rightarrow$(C)  can be used to
select a Lyapunov function with tighter requirements.

The construction of a common Lyapunov function
satisfying (B), under the assumption that (A) holds true, follows the same lines as in finite dimension. In particular, a possible choice of the Lyapunov function is
\begin{equation*}
V(x_0)=\sup\left\{ \int_0^\infty \|x(t)\|^2 dt : x(\cdot)~\text{solution to \eqref{eq:ss:intro} for some}~\sigma\right\}.
\end{equation*}
(Alternatively, one could take $V(x_0)=\int_0^\infty \sup_{\sigma(\cdot)}\|x(t)\|^2 dt$, as done in
\cite{HanteSigalotti2010}.)

The construction of a Lyapunov function
satisfying (C) is similar, but one has to
augment \eqref{eq:ss:intro} with a further mode $T_{j^*}(t)=e^{-\mu t}I$ (where $\mu>0$ is the constant appearing in (A) and $I$ denotes the identity on $X$)
and to consider all the solutions to this augmented system in the definition of $V$.

In the case of an exponentially stable single mode ($Q=\{0\}$), it was observed by Pazy \cite{Pazy1972}
that $x\mapsto \int_0^\infty \|T_0(t)x\|^2\,dt$ defines a Lyapunov function that is comparable with the squared norm if and only if $T_0$ extends to an exponentially stable strongly continuous group. Notice that, as a consequence of the implication (A)$\Rightarrow$(C), even if $T_0$ does not admit an extension to a group, a Lyapunov function comparable with the squared norm can still be found (see Remark~\ref{one-mode-L}).

Concerning the regularity of the Lyapunov functions obtained through the construction described above, they are always convex and continuous (since $\sqrt{V(\cdot)}$ is a norm).
 In the special case in which $X$ is a separable Hilbert space, we
also prove the Fr\'echet directional differentiability of $V$ and we establish a characterization of the directional Fr\'echet derivatives.

The paper is organized as follows.
In Section~\ref{sec:notations} we introduce the main notations and discuss a motivation example.
Section~\ref{sec:first}
provides a first necessary and sufficient condition for global uniform exponential stability
in terms of the existence of a common Lyapunov function, namely, the equivalence of  (A) and (B)
(Theorem~\ref{thm:VnonComparable}).
We also discuss the possible redundancies of
condition (B),
showing that
\eqref{thm0:UEBound}
cannot be removed (Remark~\ref{rem:Thm1sharp}).
In Section~\ref{sec:second} a second converse Lyapunov theorem is proved, establishing that (A) and (C) are equivalent (Theorem~\ref{thm:VComparable}).
In Section~\ref{sec:Hilbert} we show the Fr\'echet differentiability of the common Lyapunov functions constructed in the previous sections when $X$ is a separable Hilbert space (Corollary~\ref{corl:Hilbert}).
In Section~\ref{sec:conclusions} we give some final remarks and point to open problems.

\section{
NOTATIONS AND PRELIMINARIES}\label{sec:notations}

By $\NN$, $\QQ$ and $\RR$ we denote the set of natural, rational and real numbers, respectively. Further, let $X$ be a Banach space, $\Lcal(X)$ be the space of bounded linear operators on $X$, $Q$ be a countable set
and, for all $j \in Q$, let $t\mapsto T_j(t)\in \Lcal(X)$, $t \geq 0$ be a strongly continuous semigroup.

We wish to investigate the qualitative behavior of
\begin{equation}\label{eq:ss}
 x(t) = T_{\sigma(\cdot)}(t)x
\end{equation}
for $x\in X$, where $\sigma\: [0,\infty) \to Q$ is a piecewise constant
switching signal and
\begin{equation}\label{eq:TsigmaDef}
T_{\sigma(\cdot)}(t)=T_{j_p}(t-\tau_p)T_{j_{p-1}}(\tau_p-\tau_{p-1}) \cdots T_{j_{1}}(\tau_{1})
\end{equation}
for $\sigma(\cdot)$ equal to $j_k$ on $(\tau_{k-1},\tau_k)$ for $k=1,\dots,p+1$ and
\begin{equation*}
0=\tau_0<\tau_1<\dots<\tau_{p+1}=t.
\end{equation*}
In particular, we wish to study the asymptotic behavior of $x(t)$ as $t$ tends
to $+\infty$, uniformly with respect to the switching law $\sigma(\cdot)$ in the set $\Sigma$ of all piecewise constant switching signals.
We note that, for any given $\sigma(\cdot)\in\Sigma$, the operator
$T_{\sigma(\cdot)}(t)\in\Lcal(X)$ is strongly continuous with respect to $t$
($\lim_{t\downarrow t_0}\|T_{\sigma(\cdot)}(t)-T_{\sigma(\cdot)}(t_0)\|=0$)
and satisfies
\begin{equation}\label{eq:SemigoupProperty}
 T_{\sigma(\cdot)}(t+s)=T_{\sigma_s(\cdot)}(t)T_{\sigma(\cdot)}(s)
\end{equation}
for some switching signal $\sigma_s(\cdot)\in\Sigma$ depending on $s$, but in general,
$T_{\sigma(\cdot)}(t)$ does not satisfy the semigroup property, i.\,e.,
equation \eqref{eq:SemigoupProperty} with $\sigma_s(\cdot)$ replaced by $\sigma(\cdot)$ (independently of $s$).

For a function $V\: X \to [0,\infty)$ we define the generalized derivative
\begin{equation}\label{eq:LinfjVdef}
 \Linf_j V(x)=\liminf_{t \downarrow 0} \frac{V(T_j(t)x) - V(x)}{t},
\end{equation}
noting the possibility that $|\Linf_j V(x)|=\infty$ for some $x\in X$ and $j \in Q$.
Further, we call a switched system \eqref{eq:ss} (completely determined
by $\{T_{j}\}_{j \in Q}$) \emph{globally uniformly exponentially stable} when there exist
constants $K \geq 1$ and $\mu>0$ such that
\begin{equation}\label{def:GUES}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq K e^{-\mu t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation}
It is clear that \eqref{def:GUES} implies
\begin{equation*}
\|x(t)\|_X \leq K e^{-\mu t} \|x\|_X, \quad t \geq 0
\end{equation*}
globally for all $x \in X$ and uniformly for all $\sigma(\cdot)\in\Sigma$
justifying the terminology. We point out that \eqref{def:GUES} implies
\emph{strong attractivity at the origin}, i.\,e.,
\begin{equation}\label{def:strongStab}
 \lim_{t\to\infty}\|T_{\sigma(\cdot)}(t)x\|_X = 0, \quad x \in X,~\sigma(\cdot)\in\Sigma,
\end{equation}
and \emph{uniform stability}, i.\,e.,
\begin{equation}\label{def:unifStab}
\left\{\begin{aligned}
&\text{for all}~\varepsilon>0~\text{there exists a}~\delta>0,\\
&\text{independent of}~\sigma(\cdot),~\text{such that}~\|x\|_X<\delta~\text{implies}\\
&\quad \|T_{\sigma(\cdot)}(t)x\|_{X}<\varepsilon,~t \geq 0,~\sigma(\cdot)\text{-uniformly}
\end{aligned}\right.
\end{equation}
but that the converse implication is
false in general, even for a single mode: As a counterexample it suffices to take the left translation semigroup
defined by
\begin{equation*}
(T(t)f)(s):=f(s+t)
\end{equation*}
on the Lebesgue space $X=L^1(\RR_+)$. This is in contrast to the equivalence of \eqref{def:strongStab} and \eqref{def:GUES} when $X$ is a $n$-dimensional real coordinate space, $Q$ is finite, and $T_{j}(t)$ is given by the matrix exponential $e^{A_jt}$ for some real $n \times n$-matrix
$A_j$, as a consequence of Fenichel's Uniformity Lemma (see, for instance, \cite[\S 5.2]{ColoniusKliemann2000}).

Before turning our attention to necessary and sufficient conditions for global
uniform exponential stability, we give an example of a switched system exhibiting
illustrative instability properties, though the subsystems are exponentially stable.

\begin{figure*}[t]
\centering
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\caption{Illustration of $T_j(t)$, $j=1,2$, (left) and the blow up of the operator norm of $T_{\sigma(\cdot)}(t)$ (right) in Example~\ref{ex:notGUES}.\label{f::1}}
\end{figure*}

\begin{example}\label{ex:notGUES} Consider the bimodal system $\{T_j(t)\}_{j=1,2}$ with
$T_j(t)$ defined on the Lebesgue space $X=L^1(-1,1)$ by
\begin{equation*}
 \left(T_1(t)f\right)(s)=\begin{cases}
                         	2 f(s+t),	& s \in [-1,1-t]\cap[-t,0]\\
                         	f(s+t),		& s \in [-1,1-t]\setminus[-t,0]\\
				0,		& s \in (1-t,1]
                         \end{cases}
\end{equation*}
and
\begin{equation*}
 \left(T_2(t)f\right)(s)=\begin{cases}
                         	2 f(s-t),	& s \in [-1+t,1]\cap[0,t]\\
                         	f(s-t),		& s \in [-1+t,1]\setminus[0,t]\\
				0,		& s \in [-1,-1+t).
                         \end{cases}
\end{equation*}
Notice that both $T_1(\cdot)$ and $T_2(\cdot)$ are nilpotent semigroups, since
$T_1(t)=T_2(t)=0$ for $t\geq 2$. In particular, each of them is exponentially stable.

It is easy to see that for
suitable
switching signals $\sigma(\cdot)\in\Sigma$, e.\,g., switching at $\tau_k=k\delta$, $k\in\NN$,
for a fixed $\delta<1$,
\begin{equation*}
\|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)}\to+\infty \quad\text{as}~t \to +\infty.
\end{equation*}
In fact, the speed of blow-up is not uniformly exponentially bounded over the set of all possible $\sigma(\cdot)$, i.\,e.,
for any fixed $t>0$, we have
\begin{equation*}
\|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)}\geq 2^{\lceil\frac{t}{\delta}\rceil}\to+\infty \quad\text{as}~\delta \to 0,
\end{equation*}
with $\lceil\tau\rceil=\min\{k\in \mathbb{N} : \tau\leq k \}$ for $\tau>0$ (see Figure~\ref{f::1}). This can be seen by taking $L^1(-1,1)$-functions $f$ of norm one, identically constant near $x=0$ on progressively smaller intervals and zero elsewhere. \hfill$\square$
\end{example}

\section{FIRST CONVERSE LYAPUNOV THEOREM}\label{sec:first}
In this section we establish a first equivalence result for the global uniform exponential stability of a switched of the form~\eqref{eq:ss}. The crucial step is given by the following lemma, related to the blow-up phenomenon illustrated in Example~\ref{ex:notGUES} in the previous section. It
is a variant of a result obtained in \cite{Triggiani1994} in the framework of
strongly continuous semigroups. While extending the property to
switched system of the form \eqref{eq:ss}, the proof given in \cite{Triggiani1994}
should be modified in order to replace
the semigroup property by \eqref{eq:SemigoupProperty}.
We include the modified proof for the sake of completeness.
\begin{lemma}\label{lem:IntUBound} Assume that
\begin{itemize}
\item[i)] there exist constants $M \geq 1$ and $\omega>0$
such that
\begin{equation*}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Me^{\omega t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly};
\end{equation*}
\item[ii)] there exists a constant $C>0$
and some $p\in[1,\infty)$
such that
\begin{equation*}
 \int_0^\infty \|T_{\sigma(\cdot)}(t)x\|^p_X\,dt \leq C\|x\|^p_X, \quad x \in X,~\sigma(\cdot)\text{-uniformly}.
\end{equation*}
\end{itemize}
Then, there exist constants $K \geq 1$ and $\mu>0$
such that
\begin{equation*}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Ke^{-\mu t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation*}
\mbox{}
\end{lemma}
\begin{proof}
First, we show that under the assumptions i) and ii), for every $x \in X$, there exists a constant $C_x>0$
such that
\begin{equation}\label{eq:XNormUBound}
 \|T_{\sigma(\cdot)}(t)x\|_{X} \leq C_x,~t \geq 0, \quad\sigma(\cdot)\text{-uniformly}
\end{equation}
and that, for all $\sigma(\cdot)$ and for all $x \in X$,
\begin{equation}\label{eq:StrongStab}
 \lim_{t\to+\infty}\|T_{\sigma(\cdot)}(t)x\|_{X}=0.
\end{equation}
To this end, let
$t>\frac{1}{\omega}$ and set $\Delta(t)=[t-\frac{1}{\omega},t]$.
Observe that, for every $\sigma(\cdot)$ and every $\tau\in\Delta(t)$, there exists a
$\sigma_{\tau}(\cdot)$ such that
\begin{equation}\label{eq:DeltaiEst}
\begin{split}
 \|T_{\sigma(\cdot)}(t)x\|_X	&=\|T_{\sigma_{\tau}(\cdot)}(t-\tau)T_{\sigma(\cdot)}(\tau)x\|_X\\
				&\leq \|T_{\sigma_{\tau}(\cdot)}(t-\tau)\|_{\Lcal(X)}\|T_{\sigma(\cdot)}(\tau)x\|_X.
\end{split}
\end{equation}
Moreover, by assumption i)
and by definition of $\Delta(t)$, we have
\begin{equation}\label{eq:OpNormOnDeltaiEst}
 \|T_{\sigma_{\tau}(\cdot)}(t-\tau)\|_{\Lcal(X)} \leq M e^{\omega(t-\tau)}\leq M e^{\omega\frac{1}{\omega}}=Me,
\end{equation}
yielding
\begin{equation}\label{eq:XY}
 \|T_{\sigma(\cdot)}(\tau)x\|_X \geq \frac{\|T_{\sigma(\cdot)}(t)x\|_X}{Me}, \quad\tau\in\Delta(t).
\end{equation}

Now suppose \eqref{eq:XNormUBound} does not hold.
Then, there exist $x \in X$, a sequence of switching signals $(\sigma_i(\cdot))_{i\in\mathbb{N}}$ in $\Sigma$ and a sequence of times $(t_i)_{i\in\mathbb{N}}$
such that
\begin{equation}\label{eq:XNormDiv}
 \delta_i=\|T_{\sigma_i(\cdot)}(t_i)x\|_X \to +\infty \quad\text{as}~i\to+\infty.
\end{equation}
Assumption i) guarantees that $t_i$ is diverging. Without loss of generality, $t_i>\frac1\omega$ for every $i\in\mathbb{N}$.

For $\tau\in\Delta(t_i)$, \eqref{eq:XY} yields
\begin{equation}\label{eq:XNormOnDeltaiEst}
 \|T_{\sigma_i(\cdot)}(\tau)x\|_X \geq \frac{\delta_i}{Me}, \quad\tau\in\Delta(t_i).
\end{equation}
Hence, using \eqref{eq:XNormOnDeltaiEst} and again the size of $\Delta(t_i)$, we obtain
from \eqref{eq:XNormDiv}
\begin{equation*}
\begin{split}
 \int_0^\infty\|T_{\sigma_i(\cdot)}(\tau)x\|_X^p\,d\tau	&\geq \int_{\Delta(t_i)} \|T_{\sigma_i(\cdot)}(\tau)x\|_X^p\,d\tau\\
						&\geq \left(\frac{\delta_i}{Me}\right)^p\frac{1}{\omega} \to \infty
\end{split}
\end{equation*}
as $i \to \infty$. This contradicts assumption ii). Hence, \eqref{eq:XNormUBound} holds true.

Next, suppose \eqref{eq:StrongStab} does not hold. Then, there exist $x\in X$,
$\sigma(\cdot)\in\Sigma$, $\delta>0$ and a diverging sequence of times $(t_i)_{i\in\NN}$ such that
\begin{equation}\label{eq:NoStrongStab}
 \|T_{\sigma(\cdot)}(t_i)x\|_X \geq \delta \quad\text{for all}~i.
\end{equation}
Without loss of generality $t_i>t_{i-1}+\frac1\omega$ for every $i\in\mathbb{N}$
with $t_0=0$.
For $\tau\in\Delta(t_i)$,
\eqref{eq:XY} yields
\begin{equation}\label{eq:XNormOnDeltaiEst2}
 \|T_{\sigma(\cdot)}(\tau)x\|_X \geq \frac{\delta}{Me}, \quad\tau\in\Delta(t_i).
\end{equation}
Hence, using \eqref{eq:XNormOnDeltaiEst} and again the size of $\Delta(t_i)$, we obtain
\begin{equation*}
\begin{split}
 \int_0^\infty\|T_{\sigma(\cdot)}(\tau)x\|_X^p\,d\tau	&\geq \sum_{i=1}^\infty \int_{\Delta(t_i)} \|T_{\sigma(\cdot)}(\tau)x\|_X^p\,d\tau\\
						&\geq  \left(\frac{\delta}{Me}\right)^p\sum_{i=1}^\infty\frac{1}{\omega} =  \infty.
\end{split}
\end{equation*}
This again contradicts assumption ii) and hence \eqref{eq:StrongStab} holds true.

Let
\begin{equation*}
 t_{x,\sigma(\cdot)}(\rho)=\max\{ t : \|T_{\sigma(\cdot)}(t)x\|_X \geq \rho \|x\|_X,~0 \leq s \leq t\}.
\end{equation*}
By \eqref{eq:StrongStab}, $t_{x,\sigma(\cdot)}(\rho)$ is finite (and positive) for every $\sigma(\cdot)$ and $x \in X\setminus\{0\}$.
By strong continuity,
\begin{equation*}
 \|T_{\sigma(\cdot)}(t_{x,\sigma(\cdot)}(\rho))x\|_X = \rho \|x\|_X.
\end{equation*}
Using assumption ii),
\begin{equation*}
\begin{split}
 t_{x,\sigma(\cdot)}(\rho)\rho^p\|x\|_X^p 	&\leq \int_0^{t_{x,\sigma(\cdot)}(\rho)}\|T_{\sigma(\cdot)}(t)x\|_X^p\,dt\\
						&\leq \int_0^{\infty}\|T_{\sigma(\cdot)}(t)x\|_X^p\,dt\leq C\|x\|_X^p
\end{split}
\end{equation*}
whereby,
\begin{equation*}
 t_{x,\sigma(\cdot)}(\rho) \leq \frac{C}{\rho^p}=:t_0,~\text{independent of}~\sigma(\cdot).
\end{equation*}

By the principle of uniform boundedness, \eqref{eq:XNormUBound} implies that there exists
a constant $k>0$
such that
\begin{equation}\label{eq:kdef}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq k, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation}
Hence, for $t>t_0$, we have
\begin{align*}
 \|T_{\sigma(\cdot)}(t)x\|_X &\leq \sup_{\tilde\sigma(\cdot)}\|T_{\tilde\sigma(\cdot)}(t-t_{x,\sigma(\cdot)}(\rho))\|_{\Lcal(X)}\rho\|x\|_X\\
 &\leq k\rho\|x\|_X, \quad \sigma(\cdot)\text{-uniformly}.
\end{align*}
Choose $\rho>0$ such that $\beta:=k\rho<1$, so that
\begin{equation*}
 \|T_{\sigma(\cdot)}(t)x\|_X \leq \beta \|x\|_X, \quad t \geq t_0,~\sigma(\cdot)\text{-uniformly}.
\end{equation*}
Finally, let $t_1>t_0$ be fixed and let $t=nt_1+s$, $0 \leq s < t_1$. Then,
 \begin{align*}
  \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} &\leq \sup_{\tilde\sigma(\cdot),\;\hat \sigma(\cdot)} \|T_{\tilde\sigma(\cdot)}(s)\|_{\Lcal(X)} \|T_{\hat\sigma(\cdot)}(nt_1)\|_{\Lcal(X)}\\
& \leq k \left(\sup_{\hat\sigma(\cdot)} \|T_{\hat\sigma(\cdot)}(t_1)\|_{\Lcal(X)}\right)^n\\
& \leq k \beta^n \leq K e^{-\mu t},~t \geq 0,~\sigma(\cdot)\text{-uniformly}
 \end{align*}
with $K=\frac{C}{\beta}$ and $\mu=-\left(\frac{1}{t_1}\right)\ln\beta>0$.
\end{proof}

Lemma~\ref{lem:IntUBound} allows us to prove the first of the converse Lyapunov theorems that are the object of this paper.
\begin{theorem}\label{thm:VnonComparable}
The following conditions:
\begin{itemize}
\item[i)] there exist constants $M \geq 1$ and $\omega>0$
such that
\begin{equation}\label{thm1:UEBound}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Me^{\omega t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly},
\end{equation}
 \item[ii)]
 there exists $V\: X \to [0,\infty)$
 such that
 $\sqrt{V(\cdot)}$ is a norm on $X$,

	\begin{equation}\label{thm1:VBound}
 		V(x) \leq C \|x\|_X^2, \quad x\in X,
	\end{equation}
	for a constant $C>0$ and
	\begin{equation}\label{thm1:LjVest}
 		\Linf_j V(x) \leq -\|x\|_X^2,\quad j \in Q,~x\in X,
	\end{equation}
	with $\Linf_j V(x)$ defined as in \eqref{eq:LinfjVdef}
\end{itemize}
are necessary and sufficient for the existence of constants $K \geq 1$ and $\mu>0$
such that
\begin{equation}\label{thm1:GUES}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Ke^{-\mu t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation}
\mbox{}
\end{theorem}
\begin{proof}
Assume that the conditions i) and ii) hold. For all $\sigma(\cdot)\in\Sigma$, $x \in X$
and for $t \geq 0$ small enough so that the restriction of $\sigma(\cdot)$ to the interval $[0,t]$ is constant, we have
\begin{equation*}
 0 \leq V(T_{\sigma(\cdot)}(t)x) \leq V(x) - \int_0^t \|T_{\sigma(\cdot)}(\tau)x\|_X^2\,d\tau
\end{equation*}
as it follows from \eqref{thm1:LjVest} and \cite[\S VI.7]{Saks1964} (see also \cite{HagoodThomson2006}). Thus, for all $\sigma(\cdot)$ and $x \in X$,
\begin{equation}\label{eq:IntUBound}
\int_0^\infty \|T_{\sigma(\cdot)}(\tau)x\|_X^2\,d\tau \leq V(x)\leq C\|x\|_X^2.
\end{equation}
The uniform exponential decay \eqref{thm1:GUES} now follows from \eqref{thm1:UEBound} and \eqref{eq:IntUBound}, thanks to
Lemma~\ref{lem:IntUBound} with $p=2$.

Conversely, assume that \eqref{thm1:GUES} holds for some constants $K \geq 1$ and $\mu>0$.
Then, \eqref{thm1:UEBound} holds for $M=K$ and arbitrary $\omega>0$. Define $V\: X \to [0,\infty)$ by
\begin{equation}\label{eq:Vdef}
V(x) = \sup_{\sigma(\cdot)\in\Sigma}\int_0^\infty \|T_{\sigma(\cdot)}(t)x\|_X^2\,dt.
\end{equation}
Then, by assumption, $V(x)$ satisfies
\begin{equation*}
 V(x) \leq \sup_{\sigma(\cdot)}\int_0^\infty K^2e^{-2\mu t}\|x\|_X^2\,dt=\frac{K^2}{2\mu}\|x\|_X^2,
\end{equation*}
establishing \eqref{thm1:VBound} with $C=\frac{K^2}{2\mu}$. In particular, $V$ is well-posed.

Notice that, by definition, $V$ is positive definite and homogenous of degree $2$.
In order to show that it is the square of a norm, we are left to prove that it is convex and continuous.

The convexity of $V$ follows from the fact that each
$$x\mapsto \int_0^\infty \|T_{\sigma(\cdot)}(t)x\|_X^2\,dt$$
is convex.

In order to verify the continuity of $V$, let $(x_n)_{n\in\NN}$ be a sequence in $X$ converging to $x$ in $X$. By definition of $V$,
\begin{equation}\label{eq:Vnest}
\int_0^\infty \|T_{\sigma(\cdot)}(t)x_n\|_X^2\,dt \leq V(x_n),
\end{equation}
for all $\sigma(\cdot)\in\Sigma$. So taking the $\liminf$ over $n\in\NN$ in \eqref{eq:Vnest} on both sides
and using the continuity of $T_{\sigma(\cdot)}(t)$ for all $t\geq 0$, we have
\begin{equation}\label{eq:Vcont}
 \int_0^\infty \|T_{\sigma(\cdot)}(t)x\|_X^2\,dt \leq \liminf_{n\to\infty} \int_0^\infty \|T_{\sigma(\cdot)}(t)x_n\|_X^2\,dt \leq \liminf_{n\to\infty} V(x_n).
\end{equation}
Taking the $\sup$ over $\sigma(\cdot)$ in \eqref{eq:Vcont} then yields
\begin{equation*}
 V(x) = \sup_{\sigma(\cdot)}\int_0^\infty \|T_{\sigma(\cdot)}(t)x\|_X^2\,dt \leq \liminf_{n\to\infty} V(x_n),
\end{equation*}
proving that $V$ is lower semi-continuous. On the other hand, for a fixed $\varepsilon>0$, there exist $\sigma_{\varepsilon}(\cdot)\in\Sigma$ such that
\begin{align*}
 V(x_n) - \frac{\varepsilon}{2} &< \int_0^\infty\kern-0.4em\|T_{\sigma_{\varepsilon}(\cdot)}(t)x_n\|^2\,dt\\
&\leq (1+m)\int_0^\infty\kern-0.4em \|T_{\sigma_{\varepsilon}(\cdot)}(t)(x_n-x)\|^2\,dt+ \left(1+\frac1m\right)\int_0^\infty\kern-0.4em \|T_{\sigma_{\varepsilon}(\cdot)}(t)x\|^2\,dt,
\end{align*}
for any $m>0$.
Notice that, by definition of $V$,
\begin{align*}
 \int_0^\infty \|T_{\sigma_{\varepsilon}(\cdot)}(t)(x_n-x)\|^2\,dt&\leq
 V(x_n-x)
 \leq C\|x_n-x\|_X^2,\\
 \int_0^\infty \|T_{\sigma_{\varepsilon}(\cdot)}(t)x\|^2\,dt &\leq V(x).
\end{align*}
Thus, for any $m>0$, we have
\begin{equation}\label{eq:uscestLast}
V(x_n) - \frac{\varepsilon}{2} < C (1+m)\|x_n-x\|^2_X + \left(1+\frac1m\right)V(x).
\end{equation}
In particular, choosing $m$ such that $(1+\frac1m)V(x)<V(x)+\frac\varepsilon4$ and taking $n$ sufficiently
large, so that $(1+m) C\|x_n-x\|_X^2 \leq \frac\varepsilon4$, we have from \eqref{eq:uscestLast}
\begin{equation*}
 V(x_n) < V(x) + \varepsilon, \quad n~\text{sufficienty large}.
\end{equation*}
This implies the upper semi-continuity of $V$.
Resuming, we proved the
continuity of $V$.

To complete the proof of the theorem, we are left to show that $V$ satisfies \eqref{thm1:LjVest}.
Fixing $t > 0$, $j \in Q$ and letting
\begin{equation*}
\Sigma_{t,j}=\{\sigma(\cdot)\in\Sigma : \sigma|_{[0,t]}\equiv j\}
\end{equation*}
be the set of switching signals whose restriction to the interval $[0,t]$ is constantly equal to $j$, we have, since $\Sigma_{t,j}\subseteq\Sigma$,
\begin{align}
 V(x)	&\geq\sup_{\sigma(\cdot)\in\Sigma_{t,j}} \int_0^\infty \|T_{\sigma(\cdot)}(\tau)x\|_X^2\,d\tau\nonumber\\
 	&=\int_0^t \|T_{j}(\tau)x\|_X^2\,d\tau + \sup_{\sigma(\cdot)\in\Sigma_{t,j}} \int_t^\infty \|T_{\sigma(\cdot)}(\tau)x\|_X^2\,d\tau.\label{half}
\end{align}
Moreover, thanks to \eqref{eq:SemigoupProperty} and the invariance of $\Sigma$ by time-shift,
\begin{align*}
V(T_j(t)x)	&=\sup_{\sigma(\cdot)\in\Sigma} \int_0^\infty \|T_{\sigma(\cdot)}(\tau)T_j(t)x\|_X^2\,d\tau\\
		&=\sup_{\sigma(\cdot)\in\Sigma} \int_t^\infty \|T_{\sigma(\cdot)}(\tau-t)T_j(t)x\|_X^2\,d\tau\\
 		&=\sup_{\sigma(\cdot)\in\Sigma_{t,j}} \int_t^\infty \|T_{\sigma(\cdot)}(\tau)x\|_X^2\,d\tau.
\end{align*}
This and \eqref{half} yield
\begin{equation*}
 V(T_j(t)x)-V(x) \leq - \int_0^t \|T_j(\tau)\|_X^2\,d\tau,
\end{equation*}
for all $j\in Q$ and $t > 0$. Therefore
\begin{equation*}
\begin{split}
 &\Linf_j V(x) = \liminf_{t \downarrow 0} \frac{V(T_j(t)x) - V(x)}{t}\\
 &\qquad\leq - \limsup_{t \downarrow 0} \frac{1}{t}\int_0^t \|T_j(\tau)x\|_X^2\,d\tau=-\|x\|_X^2
\end{split}
\end{equation*}
for all $j\in Q$, establishing \eqref{thm1:LjVest}.
\end{proof}

\begin{figure*}[t]
\centering
\scalebox{1}[1]{
\begin{pspicture}(5,5)
\pspolygon[linestyle=none,fillstyle=crosshatch,hatchangle=45,hatchwidth=0.1pt](1,4)(1,1.5)(1.5,1)(1.5,3.5)(1,4)
\pspolygon[linestyle=none,fillstyle=vlines,hatchangle=45,hatchwidth=0.1pt](1,1)(1.5,1)(1,1.5)(1,1)
\pspolygon[linestyle=none,fillstyle=vlines,hatchangle=45,hatchwidth=0.1pt](1.5,1)(4,1)(1.5,3.5)(1.5,1)
\psline{->}(1,1)(4.5,1)
\psline{->}(1,1)(1,4.5)
\psline[linestyle=dotted]{-}(4,1)(4,4.5)
\psline[arrowsize=0.2,linewidth=0.5pt]{->}(4,1)(1,4)
\psline[arrowsize=0.2,linewidth=0.5pt]{->}(1.5,1)(1,1.5)
\psline[linestyle=dotted]{-}(1.5,1)(1.5,4.5)
\rput(0.5,2.5){$t$}
\rput(5,0.75){$s$}
\psline[linestyle=dotted]{-}(0.95,4)(4,4)
\psline{-}(0.95,4)(1.05,4)
\rput(0.75,4){$1$}
\psline{-}(1,0.95)(1,1.05)
\rput(0.9,0.75){$0$}
\psline{-}(1.5,0.95)(1.5,1.05)
\rput(1.5,0.775){$4^{-j}$}
\psline{-}(4,0.95)(4,1.05)
\rput(4,0.75){$1$}
\end{pspicture}
}
\caption{Illustration of $T_j(t)$, $j \in Q$, of the example in Remark~\ref{rem:Thm1sharp}.\label{f::2}}
\end{figure*}

\begin{remark}\label{rem:Thm1sharp} We show here, through an example, that  condition i) appearing in the statement of Theorem~\ref{thm:VnonComparable}
cannot be removed.

Consider the family of semigroups $\{T_j(t)\}_{j \in Q}$ with $Q=\NN$ and $T_j(t)$ defined on the Lebesgue space
$X=L^p(0,1)$, $p\in[1,\infty)$, by
\begin{equation}\label{eq:TcounterexDef}
 \left(T_j(t)f\right)(s)=\begin{cases}
                         	2^{\frac1p} f(s+t),	& s \in [0,1-t]\cap[0,{4^{-j}})\\
                         	f(s+t),			& s \in [0,1-t]\setminus[0,4^{-j})\\
				0,			& s \in (1-t,1],
                         \end{cases}
\end{equation}
cf. also Figure~\ref{f::2}. Notice that, for all $j \in Q$, $T_j(\cdot)$ is a nilpotent semigroup, since $\|T_j(t) f\|_X=0$ for $t > 1$. In particular, each of them is exponentially stable. Moreover, for all $\sigma(\cdot)\in\Sigma$, we have
\begin{equation*}
\left|\left(T_{\sigma(\cdot)}(t)f\right)(s)\right|\leq 2^{\frac k p}|f(s+t)|,\quad\mbox{with }k=\#\{l\in\NN :
 s<4^{-l}\leq s+t\}.
\end{equation*}
 In particular,
 \begin{equation*}
\left|\left(T_{\sigma(\cdot)}(t)f\right)(s)\right|\leq 2^{\frac k p}|f(s+t)|,\quad\mbox{if }  4^{-k-1}\leq s< 4^{-k},
\end{equation*}
 yielding
\begin{align*}
 \int_0^\infty \|T_{\sigma(\cdot)}(t)f\|_X^p
 &= \int_0^1 \int_0^{1-t}\left|\left(T_{\sigma(\cdot)}(t)f\right)(s)\right|^p\,ds\,dt\\
&= \int_0^1 \int_0^{1-s}\left|\left(T_{\sigma(\cdot)}(t)f\right)(s)\right|^p\,dt\,ds\\
&\leq \sum_{k=0}^\infty \int_{4^{-k-1}}^{4^{-k}} 2^{k}\left( \int_0^{1-s}| f(s+t)|^p\,dt\right)ds\\
&\leq
\sum_{k=0}^\infty \left(\frac{1}{{4^{k}}}-\frac{1}{{4^{k+1}}}\right)2^{k}\int_0^1|f(t)|^p\,dt = \frac{3}{2}\|f\|_X^p.
\end{align*}

Hence, defining $V(x)$ as in \eqref{eq:Vdef}, we have
\begin{equation*}
 V(x)=\sup_{\sigma(\cdot)} \int_0^\infty \|T_{\sigma(\cdot)}(t)x\|_X^2\,dt \leq \frac{3}{2}\|x\|_X^2
\end{equation*}
and, by the same arguments as in the proof of Theorem~\ref{thm:VnonComparable},
\begin{equation*}
\Linf_j V(x) \leq -\|x\|_X^2, \quad\text{for all}~j \in Q,
\end{equation*}
so condition ii) in Theorem~\ref{thm:VnonComparable} holds with $C=\frac{3}{2}$.

Nevertheless, for a sequence of switching signals $(\sigma_n(\cdot))_{n\in\NN}\subset\Sigma$ with switching times $\tau_k=\frac{1}{4^k}$ and modes $j_k=k+1$, $0\leq k\leq n$, we have for functions $\One_{[s,1]}$ of $L^p$ norm one
concentrated on the interval $[s,1]$,
\begin{equation*}
 T_{\sigma_n(\cdot)}(1-\epsilon) \One_{[s,1]} = 2^{\frac{n}{p}}
  \One_{[s-1+\epsilon,\epsilon]}, \quad\mbox{if }1\geq s\geq 1-\epsilon>1-4^{-n}.
\end{equation*}
Therefore, for $\epsilon<4^{-n}$,
\begin{align*}
\|T_{\sigma_n(\cdot)}(1-\epsilon)\|_{\Lcal(X)} 	&= \sup_{\|f\|_X=1} \|T_{\sigma_n(\cdot)}(1-\epsilon)f\|_X\\
						&\geq \lim_{s \uparrow 1} \|T_{\sigma_n(\cdot)}(1-\epsilon) \One_{[s,1]}\|_X
						= 2^{\frac n p}.
\end{align*}

Hence,
\begin{equation}\label{eq:exampleblowup}
\sup\{\|T_{\sigma_n(\cdot)}(1-\epsilon)\|_{\Lcal(X)} : \epsilon\in[0,1],~n\in\NN\}=+\infty
\end{equation}
violating any uniform bound of the form \eqref{thm1:GUES}.

This example also shows that assumption i) appearing in Lemma~\ref{lem:IntUBound} is
necessary for the validity of the lemma.
\hfill$\square$
\end{remark}

\section{SECOND CONVERSE LYAPUNOV THEOREM}\label{sec:second}

If one wishes to conclude that a switched system is globally uniformly exponentially stable, Theorem~\ref{thm:VnonComparable}
requires the knowledge of a squared norm $V(\cdot)$ satisfying \eqref{thm1:VBound},
\eqref{thm1:LjVest} for all modes $j \in Q$ and the knowledge of a global uniform exponential bound \eqref{thm1:UEBound}.
As an alternative, we will, in Theorem~\ref{thm:VComparable}, show that the
existence of a Lyapunov norm $\sqrt{V(\cdot)}$ that is comparable with the norm $\|\cdot\|_X$ allows to
conclude that the system is globally uniformly exponentially stable, without knowledge of a global uniform exponential
bound of the type \eqref{thm1:UEBound}.

Notice that
the norm $\sqrt{V(\cdot)}$ constructed in the proof of Theorem~\ref{thm:VnonComparable} (see definition \eqref{eq:Vdef}) is in general not comparable with
$\|\cdot\|_{X}$, i.\,e., in general
it
does not satisfy a lower bound of the form
\begin{equation}\label{lower_est}
 c\|x\|_X \leq \sqrt{V({x})}, \quad x \in X
\end{equation}
for a constant $c>0$.
Such a lower bound always holds, on the contrary,
when $X$ has finite dimension,
as it is exploited in \cite{MolchanovPyatnitskiy1986,DayawansaMartin1999}.
The bound \eqref{lower_est} may fail to hold even in the case of a single strongly continuous semigroup, as it is the case, for instance, of the semigroups $T_1(\cdot)$ and $T_2(\cdot)$ introduced in Example~\ref{ex:notGUES}.   (For a characterization of
exponentially stable strongly continuous semigroups
whose Lyapunov function defined as in \eqref{eq:Vdef}

is comparable with the squared norm, see \cite{Pazy1972}.)

In order to obtain a Lyapunov norm
comparable with
$\|\cdot\|_X$
for infinite dimensional switched
systems, we make use of the following lemma imposing a stronger assumption on the
family of semigroups $T_j(\cdot)$.

\begin{lemma}\label{lem:VComparable}
Assume that there exists $j^\ast \in Q$ such that $T_{j^\ast}(\cdot)$ can be extended to a group of bounded
linear operators on $X$. Moreover, assume that there exist constants $K \geq 1$ and $\mu>0$
such that
\begin{equation}\label{lem2:GUES}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Ke^{-\mu t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation}
Then there exists
a

function
$V\: X \to [0,\infty)$

such that
$\sqrt{V(\cdot)}$ is a norm on $X$,
\begin{equation}\label{lem2:VBounds}
	c\|x\|_X^2 \leq V(x) \leq C \|x\|_X^2,\quad x\in X,
\end{equation}
for constants $c,C>0$ and
\begin{equation}\label{lem2:LjVest}
	\Linf_j V(x) \leq -\|x\|_X^2, \quad j \in Q,~x\in X,
\end{equation}
with $\Linf_j V(x)$ defined as in \eqref{eq:LinfjVdef}.
\end{lemma}
\begin{proof}
Assume that \eqref{lem2:GUES} holds for some constants $K \geq 1$ and $\mu>0$ independent of $\sigma(\cdot)$.
Define $V(\cdot)$ by \eqref{eq:Vdef}. As seen in
the proof of Theorem~\ref{thm:VnonComparable},
\eqref{lem2:GUES} guarantees that
$V$ is

the square of a norm,
satisfies \eqref{lem2:LjVest} and that there exist $C>0$ such that
$V(x) \leq C \|x\|_X^2$.

The remaining bound $c\|x\|^2 \leq V(x)$ for some constant $c>0$ is a consequence of the assumption that $T_{j^\ast}(\cdot)$  can be extended to a group. Indeed, $T_{j^\ast}(t)$ is then invertible for every $t \geq 0$ and satisfies $\|T_{j^\ast}(t)x\|_X \geq (\|T_{j^\ast}(-t)\|_{\Lcal(X)})^{-1}\|x\|_X$  (cf.~\cite{Pazy1972}). Hence
\begin{equation*}
\begin{split}
 V(x)	&= \sup_{\sigma(\cdot)} \int_0^\infty \|T_{\sigma(\cdot)}(t)x\|_X^2\,dt \geq \int_0^\infty \|T_{j^\ast}(t)x\|_X^2\\
	&\geq \int_0^\infty (\|T_{j^\ast}(-t)\|_{\Lcal(X)})^{-2}\,dt \|x\|_X^2 = c\|x\|_X^2
\end{split}
\end{equation*}
with $c=\int_0^\infty (\|T_{j^\ast}(-t)\|_{\Lcal(X)})^{-2}\,dt<\infty$.
\end{proof}

We can now state and prove our second converse Lyapunov theorem.

\begin{theorem}\label{thm:VComparable} The existence of
a
function
$V\: X \to [0,\infty)$

such that
$\sqrt{V(\cdot)}$ is a norm on $X$,
\begin{equation}\label{thm2:VBounds}
	c\|x\|_X^2 \leq V(x) \leq C \|x\|_X^2, \quad x\in X,
\end{equation}
for constants $c,C>0$ and
\begin{equation}\label{thm2:LjVest}
	\Linf_j V(x) \leq -\|x\|_X^2, \quad j \in Q,~x\in X,
\end{equation}
with $\Linf_j V(x)$ defined as in \eqref{eq:LinfjVdef} is necessary and sufficient for
the existence of constants $K \geq 1$ and $\mu>0$
such that
\begin{equation}\label{thm2:GUES}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Ke^{-\mu t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation}
\mbox{}
\end{theorem}
\begin{proof}
Assume that there exists a function $V\: X \to [0,\infty)$ such that \eqref{thm2:VBounds}
and \eqref{thm2:LjVest} hold for
$c,C>0$. Then, for all $x \in X$, $\sigma(\cdot)\in\Sigma$, and $t \geq 0$,
\begin{equation*}
 c\|T_{\sigma(\cdot)}(t)x\|_X^2 \leq V(T_{\sigma(\cdot)}(t)x) \leq V(x) - \int_0^t \|T_{\sigma(\cdot)}(\tau)x\|_X^2\,d\tau,
\end{equation*}
so, using again \eqref{thm2:VBounds} and dividing by $c$,
\begin{equation*}
 \|T_{\sigma(\cdot)}(t)x\|_X^2 \leq \frac{C}{c}\|x\|_X^2 - \frac{1}{c}\int_0^t \|T_{\sigma(\cdot)}(\tau)x\|_X^2\,d\tau.
\end{equation*}
From Gronwall's Lemma, we obtain
\begin{equation*}
 \|T_{\sigma(\cdot)}(t)x\|_X^2 \leq \frac{C}{c} e^{-\frac{1}{c}t} \|x\|_X^2, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation*}
Hence,
\begin{equation*}
\|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Ke^{-\mu t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly},
\end{equation*}
for the constants $K=\sqrt{\frac{C}{c}} \geq 1$ and $\mu=\frac{1}{2c}>0$, establishing \eqref{thm2:GUES}.

Conversely, assume that \eqref{thm2:GUES} holds for
$K \geq 1$ and $\mu>0$ and consider the
switched system with $\sigma^\ast(\cdot)$ taking values in $Q^\ast=Q\cup\{j^*\}$, where
\begin{equation}\label{def:SSextended}
 T_{j^\ast}(t)=e^{-\mu t} I
\end{equation}
and $I\: X \to X$ denotes the identity.
Then,
\begin{equation*}
 T_{j^\ast}(t) T_j(s) = T_j(s) T_{j^\ast}(t), \quad t,s \geq 0,~j \in Q.
\end{equation*}
Moreover, for $t^\ast=|\{\tau \in [0,t] : \sigma^\ast(\tau)=j^\ast\}|$ and
some $\sigma(\cdot)$ just taking values in $Q$,
\begin{align}\label{eq:SSextendedGUES}
\|T_{\sigma^\ast(\cdot)}\|_{\Lcal(X)}
&= \|T_{\sigma(\cdot)}(t-t^\ast)e^{-\mu t^\ast}\|_{\Lcal(X)} \leq K e^{-\mu(t-t^\ast)} e^{-\mu t^\ast}\nonumber\\
					&=K e^{-\mu t}, \quad\quad t \geq 0,~\sigma(\cdot)\text{-uniformly}
\end{align}
where we have used \eqref{thm2:GUES}.
The existence of a squared norm
$V\: X \to [0,\infty)$
and constants $c,C>0$
such that \eqref{thm2:VBounds}
and \eqref{thm2:LjVest} hold
now follows from \eqref{eq:SSextendedGUES}
and Lemma~\ref{lem:VComparable}, noting that
\eqref{def:SSextended}
actually defines
 a group.
\end{proof}

\begin{remark}\label{one-mode-L}
In the case of a single strongly continuous semigroup $T(\cdot)$,
Theorem~\ref{thm:VComparable} shows
that its global exponential stability is equivalent to
the existence of a Lyapunov norm comparable with the norm $\|\cdot\|_{X}$.

When $T(\cdot)$ is globally exponentially stable, such a Lyapunov norm can be obtained
by
the construction suggested in the proof of Theorem~\ref{thm:VComparable}.

An alternative construction of a common Lyapunov function has been proposed in \cite{HanteSigalotti2010}, where

\eqref{eq:Vdef}
is replaced by
\begin{equation}\label{defV:CDC}
\tilde V(x) = \int_0^\infty \sup_{\sigma(\cdot)\in\Sigma}\|T_{\sigma(\cdot)}(t)x\|_X^2\,dt.
\end{equation}
In the case of a single mode and following the strategy of augmenting $Q$ by adding a group corresponding to  a diagonal operator, this construction leads to the explicit expression
\begin{equation*}
\tilde V(x)=\int_0^\infty \left(\max_{s\in[0,\tau]}e^{2\mu(s-\tau)}\|T(s)x\|_X^2\right)d\tau,
\end{equation*}
for any fixed $\mu>0$.

It should be noticed that, although not stated in \cite{HanteSigalotti2010}, the definition of $\tilde V$ given in \eqref{defV:CDC}
identifies a function which is positive definite, homogenous of degree 2,
continuous and convex, i.e., a squared norm. The proof
of this fact can be rather easily obtained by
adapting the proof of Theorem~\ref{thm:VnonComparable}.
\hfill$\square$
\end{remark}

\section{COMMON LYAPUNOV FUNCTIONS ON
HILBERT SPACES}\label{sec:Hilbert}
The goal of this section is to prove further regulatity properties

of the common Lyapunov functions constructed in the previous sections
in the case in which $X$ is a separable Hilbert space.
The proof of the following lemma adapts
arguments presented in \cite[\S4.3.1]{BonnansShapiro2000}.

Recall that a function $V\: X\to \RR$ is said to be {\it directionally differentiable in the sense of Fr\'{e}chet}
at $x\in X$ if for every $\psi\in X$ there exists
$$V'(x,\psi)=\lim_{t\downarrow 0}\frac{V(x+t\psi)-V(x)}{t}$$
and, moreover,
$$\lim_{\psi\to 0}\frac{V(x+\psi)-V(x)-V'(x,\psi)}{\|\psi\|_X}=0.$$
Notice that, in general, $V'(x,\cdot)$ needs not be linear.

\begin{lemma} \label{lem:Hilbert}
Let $X$ be a separable Hilbert space with scalar product $\langle\cdot,\cdot\rangle$ and assume that there exist constants $K \geq 1$ and $\mu>0$
such that
\begin{equation}\label{prop1:GUES}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Ke^{-\mu t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation}
Then, there exists a
subset $\B$ of $\Lcal(X)$, compact
for the weak operator topology and
made
of self-adjoint operators, such  that
\begin{equation}\label{defVviaB}
 \sup_{\sigma(\cdot)\in\Sigma}\int_0^\infty \|T_{\sigma(\cdot)}(t)x\|_X^2\,dt =\max_{B\in\B} \la x, B x \ra=:V(x).
\end{equation}

In particular, $V$ is
directionally differentiable in the sense of Fr\'{e}chet and its derivative in the direction $\psi \in X$ is given by
\begin{equation}\label{eq:VDdef}
  V'(x,\psi) = \max_{\hat{B}\in\Scal(x)} 2\la \psi,\hat{B} x\ra,
\end{equation}
where
\begin{equation}\label{eq:Sdef}
 \Scal(x)=\argmax_{B\in\B} \la x, B x \ra.
\end{equation}
\end{lemma}
\begin{proof}
For a fixed $\sigma(\cdot)\in\Sigma$ and all $t\geq 0$, let $T_{\sigma(\cdot)}^*(t)\in\Lcal(X)$ be the adjoint operator of $T_{\sigma(\cdot)}(t)\in\Lcal(X)$, uniquely defined by
\begin{equation*}
 \langle T_{\sigma(\cdot)}(t)x,x'\rangle = \langle x,T_{\sigma(\cdot)}^*(t)x'\rangle, \quad x,x'\in X.
\end{equation*}
If $T_{\sigma(\cdot)}(t)$ has the expression given in \eqref{eq:TsigmaDef},
then
\begin{equation*}
 T_{\sigma(\cdot)}^*(t)=T_{j_{1}}^*(\tau_{1})\cdots T_{j_{p-1}}^*(\tau_p-\tau_{p-1})T_{j_p}^*(t-\tau_p),
\end{equation*}
where $T_{j}^*(t)$ denotes the adjoint semigroup of $T_{j}(t)$, $t \geq 0$, $j \in Q$.

Assuming that there exist constants $K \geq 1$ and $\mu>0$, independent of $\sigma(\cdot)$, such that
\eqref{prop1:GUES} holds, the operator $B_{\sigma(\cdot)}\: X \to X$, given by
\begin{equation*}
 B_{\sigma(\cdot)}x = \int_0^\infty T_{\sigma(\cdot)}^*(t)T_{\sigma(\cdot)}(t)x\,dt,
\end{equation*}
is linear, self-adjoint, and satisfies
\begin{align*}
 \|B_{\sigma(\cdot)}x\|_X	&\leq \int_0^\infty \|T_{\sigma(\cdot)}^*(t)\|_{\Lcal(X)}\|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)}\|x\|_X\,dt\\
				&\leq \int_0^\infty K^2e^{-2\mu t}\|x\|_X\,dt=\frac{K^2}{2\mu}\|x\|_X,
\end{align*}
where we have used \eqref{prop1:GUES}. In particular, $B_{\sigma(\cdot)}\in\Lcal(X)$ for all $\sigma(\cdot)\in\Sigma$ and
\begin{equation}\label{eq:BsigmanBound}
 \|B_{\sigma(\cdot)}\|_{\Lcal(X)} \leq \frac{K^2}{2\mu}, \quad\sigma(\cdot)\text{-uniformly}.
\end{equation}
Therefore, the set
\begin{equation*}
 \B=\{ B \in \Lcal(X) :~\text{there exists a sequence}~(\sigma_n(\cdot))_{n\in\NN}\subset \Sigma~\text{such that}~ B_{\sigma_n(\cdot)} \wot B \}
\end{equation*}
is compact for the weak operator topology. We recall that
$B_{\sigma_n(\cdot)}\wot B$ (i.\,e., $B_{\sigma_n(\cdot)}$ converges to $B$ for the \emph{weak operator topology}) if, for every sequence
$(x_n,y_n)$ converging to $(x,y)$ in $X\times X$,
we have
 \begin{equation}\label{wstard}
\lim_{n\to\infty}\langle x_n,B_{\sigma_n(\cdot)}y_n\rangle=\langle x,B y\rangle,
\end{equation}
and that every bounded closed subset of $L(X)$ is sequentially compact for the weak operator topology (see, for instance, \cite[Theorem III.4]{Fabec}).
Notice that \eqref{wstard} guarantees that $\B$
consists of self-adjoint operators.

Define $V$ as in \eqref{defVviaB}. The maximization makes sense because of
\eqref{wstard} (with $x_n=y_n=x$) and of the compactness of $\B$.
Moreover,
\begin{align*}
 V(x) &= \max_{B\in\B}\la x,Bx \ra = \sup_{\sigma(\cdot)\in\Sigma}\la x,B_{\sigma(\cdot)}x \ra = \sup_{\sigma(\cdot)\in\Sigma}\la x, \int_0^\infty T_{\sigma(\cdot)}^*(t)T_{\sigma(\cdot)}(t) x\,dt \ra\\
 &=\sup_{\sigma(\cdot)\in\Sigma}\int_0^\infty \la x, T_{\sigma(\cdot)}^*(t)T_{\sigma(\cdot)}(t) x \ra dt
= \sup_{\sigma(\cdot)}\int_0^\infty \| T_{\sigma(\cdot)}(t) x\|^2\,dt,
\end{align*}
for all $x \in X$.
Hence $V$ coincides with the Lyapunov function defined in  \eqref{eq:Vdef} and we recover that $V$ satisfies the condition ii) of Theorem~\ref{thm:VnonComparable}. In particular, $V$ is
continuous.

In order to verify the directional differentiability in the sense of Fr\'{e}chet and
to prove \eqref{eq:VDdef}, we show below that
\begin{equation}\label{Frechet_D}
\lim_{\psi\to 0}\frac{V(x+\psi)-V(x)-\max_{\hat{B}\in\Scal(x)} 2\la \psi,\hat{B} x\ra}{\|\psi\|_X}=0.
\end{equation}

First observe that the map
$x \mapsto \la x,Bx \ra$ is differentiable on $X$ in the sense of Fr\'{e}chet for all $B\in\Lcal(X)$. For $B$ self-adjoint, the derivative is given by $2\la \cdot, Bx\ra$. Fix now some $x_0 \in X$ and define
\begin{equation*}
\Phi(B,x) = 2 \|Bx - B x_0\|_{X}.
\end{equation*}
We claim that
\begin{equation}\label{eq:PhiunifToZero}
 \lim_{x \to x_0} \Phi(B,x)=0, \quad\text{uniformly with respect to}~B\in\B.
\end{equation}
Indeed, suppose by contradiction that there exists some $\varepsilon>0$, a sequence $(x_n)_{n\in\NN}$ converging to $x_0$ in $X$ and a sequence $(B_n)_{n\in\NN}$ in $\B$ such that
\begin{equation}\label{eq:proofcontra1}
\Phi(B_n,x_n)>\varepsilon \quad\text{for all}~n\in\NN.
\end{equation}
Thanks to the compactness of $\B$, there exist $B\in\B$ and  a subsequence $n(k)$ such that
\begin{equation*}
B_{n(k)}\wot B\quad\mbox{as $k\to\infty$}.
\end{equation*}
Then, it follows from  \eqref{wstard} that
\begin{equation*}
 \Phi(B_{n(k)},x_{n(k)}) =  0,
\end{equation*}
contradicting \eqref{eq:proofcontra1}.

The Mean Value Theorem then gives
\begin{equation}\label{eq:MVTest1}
\begin{split}
&|\la x_0 + \psi, B (x_0 + \psi)\ra - \la x_0, B x_0 \ra - 2 \la \psi,B x_0 \ra| \\
&\qquad\qquad\qquad\qquad\leq \|\psi\|_X \int_0^1 \Phi(B,x_0+\xi\psi)\,d\xi \leq \varepsilon \|\psi\|_X
\end{split}
\end{equation}
for $\varepsilon>0$, $\psi$ sufficiently close to zero, and uniformly with respect to
$B\in\B$, as it follows from \eqref{eq:PhiunifToZero}.

Let $\Scal(\cdot)$ be defined as in \eqref{eq:Sdef}.
Notice that, for every $x\in X$, $\Scal(x)$ is close and therefore compact for the weak operator topology.
For any $\hat{B}\in\Scal(x_0)$,
$V(x_0)=\la x_0,\hat{B}x_0\ra$ and $V(x_0+\psi) \geq \la x_0+\psi,\hat{B}(x_0+\psi)\ra$. Thus, using \eqref{eq:MVTest1},
\begin{equation*}
 V(x_0+\psi) - V(x_0) \geq \max_{\hat{B}\in\Scal(x_0)} 2\la \psi,\hat{B} x_0\ra - \varepsilon \|\psi\|_X.
\end{equation*}
In order to prove \eqref{Frechet_D}, it therefore remains to show that for all $\varepsilon>0$ and $\psi$ close to zero,
\begin{equation}\label{eq:proofcontra2}
 V(x_0+\psi) - V(x_0) \leq \max_{\hat{B}\in\Scal(x_0)} 2\la \psi,\hat{B} x_0\ra + \varepsilon \|\psi\|_X.
\end{equation}
So, suppose \eqref{eq:proofcontra2} not to hold. Then there exist $\varepsilon>0$ and a sequence $(\psi_n)_{n\in\NN}$ in $X$ converging to zero such that
\begin{equation}\label{eq:proofcontra3}
 V(x_0+\psi_n) - V(x_0) > \max_{\hat{B}\in\Scal(x_0)} 2\la \psi_n,\hat{B} x_0\ra + \varepsilon \|\psi_n\|_X.
\end{equation}
By definition of $V$,
there exist $B_0,\hat{B}_n\in\B$
such that
\begin{align*}
 V(x_0+\psi_n)-V(x_0)& = \la x_0 + \psi_n,\hat{B}_n (x_0+\psi_n)\ra - \la x_0,{B_0} x_0 \ra \\
 &\leq \la x_0 + \psi_n,\hat{B}_n (x_0+\psi_n)\ra - \la x_0,\hat{B}_n x_0 \ra.
\end{align*}
Again by the Mean Value Theorem,
\begin{equation*}
\begin{split}
& |\la x_0 + \psi_n,\hat{B}_n (x_0+\psi_n)\ra - \la x_0,\hat{B}_n x_0 \ra - 2 \la\psi_n,\hat{B}_n x_0\ra|\\
& \qquad\qquad\qquad\qquad\qquad\qquad\leq \|\psi_n\|_X \int_0^1 \Phi(\hat{B}_n,x_0+\psi_n)\,dt \leq \frac\varepsilon2 \|\psi_n\|_X
\end{split}
\end{equation*}
for all $n$ large enough. Thus,
\begin{equation*}
 V(x_0+\psi_n)-V(x_0) \leq 2 \la\psi_n,\hat{B}_n x_0\ra + \frac\varepsilon2 \|\psi_n\|_X.
\end{equation*}
Using one more time the compactness of $\B$, there exist a subsequence $n(k)$
such that $\hat{B}_{n(k)}\wot\hat{B}$ for some $\hat{B}\in\B$.
Moreover,
by continuity of $V$ and because of \eqref{wstard}, $\hat{B}\in \Scal(x_0)$.
Hence,
\begin{align*}
 V(x_0+\psi_{n(k)})-V(x_0) &\leq 2 \la\psi_{n(k)},\hat{B} x_0\ra + 2 \la\psi_{n(k)},(\hat{B}_{n(k)}-\hat{B}) x_0\ra + \frac\varepsilon2 \|\psi_{n(k)}\|_X\\
&\leq \max_{B\in\Scal(x_0)} 2 \la \psi_{n(k)},Bx_0 \ra + \frac{3}{4}\varepsilon\|\psi_{n(k)}\|
\end{align*}
where we used that $2\la\psi_{n(k)},(\hat{B}_{n(k)}-\hat{B}) x_0\ra \leq \frac\varepsilon4\|\psi_{n(k)}\|_X$ for $k$ sufficiently large. This contradicts \eqref{eq:proofcontra3} and completes the proof.
\end{proof}

The following Corollary is now a direct consequence of Lemma~\ref{lem:Hilbert} and the choice
of the Lyapunov function $V(\cdot)$ in the proof of Theorem~\ref{thm:VnonComparable}.

\begin{corollary}\label{corl:Hilbert}
 Let $X$ be a separable Hilbert space
and assume that there exist constants $K \geq 1$ and $\mu>0$ such that
\begin{equation}\label{corl:GUES}
 \|T_{\sigma(\cdot)}(t)\|_{\Lcal(X)} \leq Ke^{-\mu t}, \quad t \geq 0,~\sigma(\cdot)\text{-uniformly}.
\end{equation}
Then, there exists a
function
$V\: X \to [0,\infty)$

such that
$\sqrt{V(\cdot)}$ is a norm on $X$, $V(\cdot)$ is directionally Fr\'{e}chet differentiable,
\begin{equation}\label{corl:VBounds}
	c\|x\|_X^2 \leq V(x) \leq C \|x\|_X^2, \quad x\in X,
\end{equation}
for constants $c,C>0$ and
\begin{equation}\label{corl:LjVest}
	\Linf_j V(x) \leq -\|x\|_X^2, \quad j \in Q,~x\in X,
\end{equation}
with $\Linf_j V(x)$ defined as in \eqref{eq:LinfjVdef}.
\end{corollary}

\section{FINAL REMARKS AND OPEN PROBLEMS}\label{sec:conclusions}

We presented necessary and sufficient conditions for a (possibly infinite) family of semigroups to be
globally uniformly exponentially stable for arbitrary switching signals $\sigma(\cdot)$, in terms of the
existence of a common Lyapunov function. In particular, our results apply to switched  dynamical
systems such as \eqref{eq:ssOpForm}, involving (possibly unbounded) operators on a Banach space $X$.

We have shown that the existence of a  norm,
decaying uniformly along trajectories
and either
bounded from above by a multiple of the Banach norm
in presence of
a uniform
exponential growth bound for the switched system or
comparable with the Banach norm
is equivalent to the switched system being globally uniformly exponentially stable. The latter
generalizes a
well-known
result for switched linear
dynamical systems in
$\RR^n$, $n\in\NN$ \cite{MolchanovPyatnitskiy1986}.
In the case in which $X$ is a separable Hilbert space the common Lyapunov function is shown to be Fr\'echet directionally differentiable.

As an application, our results answer for example the question of existence of a common Lyapunov
function for the switched linear hyperbolic system with reflecting boundaries considered in \cite{AminHanteBayen2008}.

Concerning the differences between Theorems~\ref{thm:VnonComparable} and~\ref{thm:VComparable}, we already noticed that a Lyapunov function $V(\cdot)$ satisfying condition ii) of the
statement of Theorem~\ref{thm:VnonComparable} does not necessarily satisfy the stronger conditions appearing in the statement of Theorem~\ref{thm:VComparable}. Hence, Theorem~\ref{thm:VnonComparable} is better suited for proving
the global uniform exponential stability of a switched system (although the uniform exponential growth boundedness should also be proved), while Theorem~\ref{thm:VComparable} provides more information on a switched system that is known to be globally uniformly exponentially stable, by tightening the properties satisfied by $V(\cdot)$.

In the same spirit one could characterize
the global uniform exponential stability of a switched system by
(a priori) loosening  the hypotheses on $V(\cdot)$,
replacing inequalities \eqref{thm1:LjVest} and \eqref{thm2:LjVest}
by
$$\Linf_j V(x)\leq -\kappa \|x\|_X^2, \quad\text{for all}~j \in Q,~\text{for some}~\kappa>0.$$
One could also remove the hypothesis that
$V$ is the square of a norm.
Conversely,  one could (a priori) tighten the same hypotheses replacing $\Linf_j V(x)$ by
$$\Lsup_j V(x)=\limsup_{t \downarrow 0} \frac{V(T_j(t)x) - V(x)}{t}.$$

As an open problem it remains
to understand if smoothing can improve the
regularity properties of the Lyapunov function $V(\cdot)$. For the
finite-dimensional case, it is for instance known that $V(\cdot)$ can be taken polyhedral or polynomial
\cite{Blanchini1994,BlanchiniMiani1999,DayawansaMartin1999}.
It would be interesting to recover results in the same direction for
infinite dimensional switched systems.

\bibliographystyle{plain}
\bibliography{convTHMs}

\end{document}

config: arxiv


TITLE: How to evaluate $\lim_{n\to\infty} \dfrac{1^p+2^p+...+n^p}{n^p}-\frac{n}{p+1}$?
QUESTION [3 upvotes]: $$\lim_{n\to\infty} \dfrac{1^p+2^p+...+n^p}{n^p}-\frac{n}{p+1}\text{ with } p\in\mathbb N$$
I don't really know how to start, I mean... I could try substituting by  $\Big(\dfrac{n(n+1)}{2}\Big)^p$ but that would do me anything? Any hints?

REPLY [2 votes]: If you know generalized harmonic numbers
$$\sum_{k=1}^n k^p=H_n^{(-p)}$$ and their asymptotics
$$H_n^{(-p)}=n^p \left(\frac{n}{p+1}+\frac{1}{2}+\frac{p}{12
   n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-p)$$ making
$$\frac{\sum_{k=1}^n k^p }{n^p }-\frac n {p+1}=\frac{1}{2}+\frac{p}{12
   n}+\cdots$$ which shows the limit and how it is approached.


TITLE: Can particles become entangled by every of the three basic forces, or even by gravity?
QUESTION [0 upvotes]: A pair of particles can become entangled after having had an e.m interaction. Almost every example of entanglement is that of electrons whose spins become entangled having had an e.m. interaction, or the entanglement of photons, the carriers of the e.m. force. There is such a thing as "entanglement swapping" (see the Wikipedia article) but that's begging the question.
So is the same (becoming entangled) true for particles that have interacted by means of one of the other two basic forces or maybe even gravity, which, I suppose, is overwhelmed by the other forces, but the particles that interact gravitationally are at least indistinguishable, and the same before and after the interaction and this is not the case in most of the weak and strong force interactions (correct me if I am wrong)?

REPLY [2 votes]: Entanglement is not something magical. It is  a consequence of the underlying quantum mechanical framework of all nature. It really means that there exists a quantum mechanical solution which describes the system, where all participating particles have a mathematically defined role. So mathematically, in principle, the answer is yes. BUT
The strong force and the weak force are very short ranged, the strong because of asymptotic  freedom  and color neutrality,and the weak because of the very large mass of the W and Z. One cannot set up experiments with color entanglement, for example, the way one can do spin entanglements. One can describe the wavefunction of a system but there is no handle to manipulate it in the way one can  the long range electromagnetic interaction.
Gravity is long range, but so very weak that its effects cannot be measured between elementary particles. Only in large massive ensembles,
 where the quantum mechanical nature is statistically overwhelmed, gravity plays a role.


TITLE: Does the inequality still hold if I invert both sides with cyclic sums?
QUESTION [0 upvotes]: For example if I invert both sides of
$$ \sum_{\sigma} \frac{x(x+2)}{2x^2 + 1} \geq 0$$
So that it becomes
$$\sum_{\sigma}\frac{2x^2+1}{x(x+2)} \leq 0$$
Will the inequality hold?

REPLY [2 votes]: Hint:
Does $x\geq 0$ imply $\dfrac{1}{x}\leq 0$?


TITLE: Green's theorem vector form interal
QUESTION [0 upvotes]: I use the fact that the integral is equal to $divF = sint-cost$ . Now I'm supposed to integrate that w.r.t. to the area, which is $dA=dydx$ . I then try to replace $\frac{dy}{dt}=-sint$ . I do the same for dx and get $dx=-cosdt$ The integral is now a double integral with 2 dt's. I don't think it's right. Also how would I compute the limits?

REPLY [0 votes]: Area calculations using Green's theorem can go as follows,
\begin{eqnarray}
\text{Area} &=& \int \int dA \\
            &=& \frac{1}{2} \int_{C} x dy - y dx
\end{eqnarray}
And take it from there.