# Exponential stability of C0-Semigroup via Lyapunov inequality in Banach
space

Belabbas Madani1 Zohra Bendaoud2

###### Abstract.

We give a relation between the exponential stability of $C_{0}-$semigroup
$\textbf{T}=\left\\{T(t)\right\\}_{t\geq 0}$ and the solutions of Lyapunov
inequality $\left\langle QAx,x\right\langle+\left\langle
Qx,Ax\right\langle\leq-||x||^{2},$ in $B^{+}(X,X^{*})$, with $X$ is a Banach
space. The solutions of this inequality characterizes, the boundedness of the
resolvent $R(\lambda,A)$ inside and outside of the left half-plane
$\Re\lambda\geq 0$, and also the left invertibility of the $C_{0}-$semigroup
T.

Key Words: semigroups, lyapunov, left-invertible, banach space, exponential
stability.

1University of Oran 1 Ahmed Ben Bella, Algeria. 1,2ENS of Laghouat, Laghouat,
Algeria
<EMAIL_ADDRESS><EMAIL_ADDRESS>

##  1\. Introduction

A classical result of R. Datko [3] stats that in Hilbert space H the
$C_{0}-$semigroup $T=\left\\{T(t)\right\\}_{t\geq 0}$ generated by A is
uniformly exponentially stable if and only if there exists a self-adjoint
positive operator $Q\in B(H)$ such that

$\left\langle QAx,x\right\rangle+\left\langle
Qx,Ax\right\rangle=-||x||^{2},\quad\forall x\in D(A).$

The authors in [6] and [5] studied the equivalence between the solvability of
Lyapunov operator equation (respectively inequality) and the exponential
stability in Banach space. However, the strongly positive taken in the two
proofs may not satisfy in arbitrary Banach space. So we give a necessary and
sufficient condition for a Banach space isomorphic to a Hilbert space and also
a sufficient inequality for Banach space which helps us to establish some new
results.

##  2\. Preliminaries

We denote by $X$ a Banach space and by $X^{*}$ its conjugate dual space i.e.
the set of all bounded and antilinear functionals on $X$, the duality pairing
between $X^{*}$ and $X$ denote by $\left\langle.,.\right\rangle$ and by
$B(X,X^{*})$ the space of all linear and bounded operators from $X$ to
$X^{*}$. An operator $Q\in B(X,X^{*})$ will be called symmetric if
$\left\langle Qx,y\right\rangle=\left\langle Qy,x\right\rangle$ for all
$x,y\in X$, and is called positive if $\left\langle Qx,x\right\rangle\geq 0$ ,
for all $x\in X$. If $Q$ is positive operator with $\left\langle
Qx,x\right\rangle\geq\theta\left\|x\right\|^{2}$, for some $\theta>0$ and all
$x\in X$, we called it strongly positive. Denote by $B^{+}(X,X^{*})$ the space
of all positive operator on $B(X,X^{*})$.
$\textbf{T}=\left\\{T(t)\right\\}_{t\geq 0}$ always $C_{0}-$semigroup on $X$
with generators $A$. The $C_{0}-$semigroup T is called uniformly exponentially
stable if there exist constants $\varepsilon>0,M\geq 1$ such that

$\left\|T(t)\right\|\leq Me^{-\varepsilon t},\>t\geq 0.$

It is well-known from [4] (see also[2]) that

######  Theorem 1.

A Banach space $X$ has the property that there exists a strongly positive
operator $P$ from $X$ onto $X^{*}$ such that $\left\langle
Px,x\right\rangle\geq\theta\left\|x\right\|^{2}$ for some $\theta>0$, if and
only if, the Banach space $X$ is isomorphic to a Hilbert space.

From the proof in [4], the theorem is still true if the operator $P$ it should
be not assumed onto.

##  3\. Exponential stability

######  Theorem 2.

Let $X$ be a Banach space isomorphic to a Hilbert space and $A$ the generator
of $C_{0}-$semigroup $\\{T(t)\\}_{t\geq 0}$. Then the following are
equivalent.

  1. (1)

$\\{T(t)\\}_{t\geq 0}$ is uniformly exponentially stable, with $||T(t)||\leq
Me^{\omega_{0}t}\qquad(M\geq 1,\omega_{0}<0)$.

  2. (2)

There exists an operator $Q\in B^{+}(X,X^{*})$ such that

(1) $\left\langle QAx,x\right\rangle+\left\langle
Qx,Ax\right\rangle\leq-\left\|x\right\|^{2},\>\forall x\in D(A).$

######  Proof..

$(1).\Rightarrow(2).$ By theorem 1 there exists an operator $P\in B(X,X^{*})$
such that $\left\langle Px,x\right\rangle\geq\theta\left\|x\right\|^{2}$ for
some $\theta>0$ and all $x\in X$. We consider the linear operator
$Q_{0}B(X,X^{*})$ defined by

$\left\langle
Q_{0}x,y\right\rangle:=\dfrac{1}{\theta}\int_{0}^{+\infty}\left\langle
PT(t)x,T(t)y\right\rangle dx,\>x,y\in X.$

Since T is uniformly exponentially stable, $Q_{0}$ is well-defined. We show
that $Q_{0}$ is positive bounded operator, this follows from the following,
for all $x\in X$ we have

$\left\|Q_{0}x\right\|=\sup_{||y||\leq 1}|\left\langle
Q_{0}x,y\right\rangle|\leq\int_{0}^{+\infty}\sup_{||y||\leq 1}|\left\langle
PT(t)x,T(t)y\right\rangle|dt\leq\dfrac{||P||M^{2}}{-2\omega_{0}}||x||^{2},$

also

$\left\langle
Q_{0}x,x\right\rangle=\frac{1}{\theta}\int_{0}^{+\infty}\left\langle
Q_{0}x,x\right\rangle\geq\int_{0}^{+\infty}||T(t)||^{2}dt\geq 0.$

Finally, since for all $x,y\in D(A),$

$\frac{d}{dt}\left\langle PT(t)x,T(t)x\right\rangle=\left\langle
PT(t)Ax,T(t)x\right\rangle+\left\langle PT(t)x,T(t)Ax\right\rangle,$

and

$\int_{0}^{+\infty}\frac{d}{dt}\left\langle
PT(t)x,T(t)y\right\rangle=-\left\langle Px,y\right\rangle,$

therefore

$\left\langle Q_{0}Ax,x\right\rangle\left\langle
Q_{0}x,Ax\right\rangle=-\frac{1}{\theta}\left\langle
Px,x\right\rangle\leq-||x||^{2},\>\forall x\in D(A).$

$2.\Rightarrow 1.$ Since 1 holds, by replacing $x$ by $T(t)x$ we have

$\left\langle QAT(t)x,T(t)x\right\rangle+\left\langle
QT(t)x,AT(t)x\right\rangle\leq-||T(t)x||^{2},$

$\forall x\in D(A),\ t\geq 0,$ this means that $\frac{d}{dt}\left\langle
QT(t)x,T(t)x\right\rangle\leq-\left\|T(t)x\right\|^{2}.$

By integration over $(0,s),$ we obtain for all $x\in D(A)$ and $s>0$

$-\left\langle QT(s)x,T(s)x\right\rangle+\left\langle
Qx,x\right\rangle\geq\int_{0}^{s}\left\|T(t)x\right\|^{2}dt.$

Since $Q$ is positive bounded operator and $A$ is a densely defined operator
on $X$, then

$\int_{0}^{s}\left\|T(t)x\right\|^{2}dt\leq\left\langle
Qx,x\right\rangle\leq||Q||||x||^{2},\forall x\in X,s>0,$

applying Datko-Pazy theorem (see Theorem 3.1.8, [7]) we obtain the exponential
stability of $\textbf{T}.$

∎

######  Remark 3.

From the proof above, it should be noted that the result $2.\Rightarrow 1.$ is
still true if X is arbitrary Banach space.

Let $L_{A}$ the set of all operators $Q\in B^{+}(X,X^{*})$ such that, the
Lyapunov inequality $\left\langle QAx,x\right\rangle+\left\langle
Qx,Ax\right\rangle\leq-\left\|x\right\|^{2}$ hold for all $\forall x\in D(A)$.
With this notation, $L_{A}$ is non-empty implies the exponential stability of
T in arbitrary Banach space. It is clear that if $Q\in L_{A}$ then $cQ\in
L_{A}$ for all $c>0$. We can show easily that $L_{A}$ is closed subset of
$B^{+}(X,X^{*})$, using that if $\left\\{Q_{n}\right\\}_{n}\subset L_{A}$
converging to $Q$ then for $x\in D(A)$

$\left\langle QAx,x\right\rangle+\left\langle
Qx,Ax\right\rangle=\left\langle(Q-Q_{n})Ax,x\right\rangle+\left\langle(Q-Q_{n})x,Ax\right\rangle+\left\langle
Q_{n}Ax,x\right\rangle+\left\langle Q_{n}x,Ax\right\rangle.$

With

$\left\|\left\langle(Q-Q_{n})Ax,x\right\rangle+\left\langle(Q-Q_{n})x,Ax\right\rangle\right\|\leq
2\left\|Q-Q_{n}\right\|\left\|Ax\right\|\left\|x\right\|$

and by letting $n\rightarrow+\infty$ we obtain $Q\in D(A)$.

######  Corollary 4.

Let $A$ be the generator of $C_{0}-$semigroup T on Banach space isomorphic to
a Hilbert space. If T is uniformly exponentially stable and $B\in B(X)$ such
that $\left\|B\right\|\leq\dfrac{1}{2\alpha||Q||}$, for some $Q\in L_{A}$ and
$\alpha>1$. Then the $C_{0}-$semigroup generated by $A+B$ is uniformly
exponentially stable with $Q\in L_{A+B}$.

######  Proof..

If $B\in B(X)$ such that $\left\|B\right\|\leq\dfrac{1}{2\alpha||Q||}$ for
some $Q\in L_{A}$ and $\alpha>1$, then for all $x\in D(A)$ we have

$\begin{array}[]{ll}\left\langle Q(A+B)x,x\right\rangle+\left\langle
Qx,(A+B)x\right\rangle&\leq-\left\|x\right\|^{2}+\left\langle
QBx,x\right\rangle+\left\langle Qx,Bx\right\rangle\\\
&\leq-\left\|x\right\|^{2}+2\left\|Q\right\|\left\|B\right\|\left\|x\right\|^{2}\\\
&\leq-(\frac{\alpha-1}{\alpha})\left\|x\right\|^{2}.\end{array},$

then we obtain the desired result. ∎

In the following we establish a result of uniformly boundedness of the
resolvent inside and outside the left half-plane $Re\lambda\geq 0$, for this
we need some quantities; with $\sigma(A)$ the spectrum set of $A$, the
spectral bound of A, is defined by

$s(A)=\sup\left\\{Re\lambda\in\mathbb{R}:\lambda\in\sigma(A)\right\\}$

, the growth bound, is defined by

$\omega_{0}=\inf\left\\{\omega\in\mathbb{R}:\exists M\geq 1\ such\
that\left\|T(t)\right\|\leq Me^{\omega t}for\ all\ t\geq 0\right\\}$

.

If T is uniformly exponentially stable, then $s_{0}(A)\leq
s_{0}(A)\leq\omega_{0}$ (see ([1], Proposition 2.2) and [7]).

######  Theorem 5.

Let $A$ be the generator of $C_{0}-$semigroup T on Banach space $X$ suth that
$L_{A}$ non-empty. Then a boundedness of the resolvent of A is given by

(2) $\sup_{Re\lambda\geq 0}\left\|R(\lambda,A)\right\|\leq 2\inf_{Q\in
L_{A}}\left\|Q\right\|,$

and there exists $\delta_{0}>0$ such that

(3)
$\sup_{Re\lambda\leq-\delta_{0}}\left\|R(\lambda,A)\right\|\leq\dfrac{2\left\|Q\right\|}{1-2\delta_{0}\left\|Q\right\|},\quad\forall
Q\in L_{A}.$

######  Proof..

For the resolvent set $\rho(A)$, we recall from ([7], Corollary 2.2.2.) that,
T is uniformly exponentially stable then $\left\\{Re\lambda\geq
0\right\\}\subset\rho(A)$. Now let $\lambda\in\rho(A)$ and $Q\in L_{A}$ then
we have

$\left\langle Q(A-\lambda)x,x\right\rangle+\left\langle
Qx,(A-\lambda)x\right\rangle\leq-\left\|x\right\|^{2}-2Re\lambda\left\langle
Qx,x\right\rangle,\quad\forall x\in D(A).$

Let $y=(\lambda-A)x$, then the above inequality becomes

$\left\langle Qy,R(\lambda,A)y\right\rangle+\left\langle
QR(\lambda,A)y,y\right\rangle\leq-\left\|R(\lambda,A)y\right\|^{2}-2Re\lambda\left\langle
QR(\lambda,A)y,R(\lambda,A)y\right\rangle,\quad\forall y\in X.$

If $Re\lambda\geq 0$, then

$-\left\langle Qy,R(\lambda,A)y\right\rangle-\left\langle
QR(\lambda,A)y,y\right\rangle\leq-\left\|R(\lambda,A)y\right\|^{2},\quad\forall
y\in X,$

consequently, fot all $y\in X$ and all $Q\in L_{A}$,

$\left\|R(\lambda,A)y\right\|^{2}\leq\left\langle
Qy,R(\lambda,A)y\right\rangle+\left\langle QR(\lambda,A)y,y\right\rangle\leq
2\left\|Q\right\|\left\|R(\lambda,A)y\right\|\left\|y\right\|,$

thus

$\sup_{Re\lambda\geq 0}\left\|R(\lambda,A)\right\|\leq 2\inf_{Q\in
L_{A}}\left\|Q\right\|.$

Since the resolvent is uniformly bounded in the open right half plane then
$s_{0}(A)<0$, moreover, for $s(A)<Re\lambda<0$ we obtain that

$\left\langle Qy,R(\lambda,A)y\right\rangle+\left\langle
QR(\lambda,A)y,y\right\rangle\geq(1+2\left\|Q\right\|Re\lambda)\left\|R(\lambda,A)y\right\|^{2},$

then

$2\left\|Q\right\|\left\|R(\lambda,A)y\right\|\left\|y\right\|\geq(1+2\left\|Q\right\|Re\lambda)\left\|R(\lambda,A)y\right\|^{2},$

and therefore

$\left\|R(\lambda,A)y\right\|\leq\dfrac{2\left\|Q\right\|}{1+2\left\|Q\right\|Re\lambda}\left\|y\right\|,\quad\forall
y\in X.$

Since $\alpha Q\in L_{A}$ for all $\alpha\geq 1$, we can choose $\alpha\geq 1$
such that $\dfrac{-1}{2\alpha\left\|Q\right\|}\geq s(A)$. With $\delta_{0}$
fixed positive real numbre and $Q\in L_{A}$ such that
$-\delta_{0}>\dfrac{-1}{2\alpha\left\|Q\right\|}$, we have

$\dfrac{2\left\|Q\right\|}{1+2\left\|Q\right\|Re\lambda}\leq\dfrac{2\left\|Q\right\|}{1-2\delta_{0}\left\|Q\right\|},$

thus

$\sup_{Re\lambda\leq-\delta_{0}}\left\|R(\lambda,A)\right\|\leq\dfrac{2\left\|Q\right\|}{1+2\delta_{0}\left\|Q\right\|},\quad\forall
Q\in L_{A},$

with $\dfrac{2\left\|Q\right\|}{1-2\delta_{0}\left\|Q\right\|}$ depend only on
the value of $\left\|Q\right\|$. ∎

##  4\. Left invertibility

In following we study the relation between Lyapunov inequality and the left
invertibility of T. We begin with the definition of left invertibility.

######  Definition 6.

The $C_{0}-$semigroup T is left-invertible if there exists a function
$t\mapsto m(t)$ such that $m(t)>0$ and for all $x\in X$ there holds

$\left\|T(t)x\right\|\geq m(t)\left\|x\right\|,\quad\forall t\geq 0.$

We need the following results from [8] (see also [9]):

######  Proposition 7.

Let T be a $C_{0}-$semigroup on Banach space $X$. Then the following
statements are equivalent.

  1. (1)

T is left-invertible.

  2. (2)

There exists $t_{0}>0$ and $m_{0}>0$, such that $\left\|T(t_{0})\right\|\geq
m_{0}\left\|x\right\|$, for all $x\in X$.

######  Proposition 8.

Let T be a $C_{0}-$semigroup on Banach space $X$. Then the following
statements are equivalent.

  1. (1)

T is left-invertible.

  2. (2)

There exist two constants and $\alpha,c>0$ such that

$\left\|T(t)x\right\|\geq ce^{-\alpha t}\left\|x\right\|\quad\forall x\in X,\
t\geq 0.$

######  Theorem 9.

Let $A$ be the generator of $C_{0}-$semigroup T on Banach space $X$ isomorphic
to a Hilbert space. If T is uniformly exponentially stable, then the following
are equivalent.

  1. (1)

T is left-invertible.

  2. (2)

All of the elements in $L_{A}$ are strongly positive.

######  Proof..

$1.\Rightarrow 2.$ Let $Q\in L_{A}$, so that $\left\langle
QAx,x\right\rangle+\left\langle Qx,Ax\right\rangle\leq-\left\|x\right\|^{2}$
for all $x\in D(A)$. This implies that

$\dfrac{d}{dt}\left\langle
QT(t)x,T(t)x\right\rangle\leq-\left\|T(t)x\right\|^{2},\quad\forall x\in
D(A).$

As in the proof of theorem 2 use that $A$ is a densely defined operator on
$X$, we obtain that

$\left\langle
Qx,x\right\rangle\geq\int_{0}^{+\infty}\left\|T(t)x\right\|^{2}dt\quad\forall
x\in X.$

Thus, by the left invertibility of T, and Proposition 8, there exists
$\alpha,c>0$ such that

$\left\langle Qx,x\right\rangle\geq c^{2}\int_{0}^{+\infty}e^{-2\alpha
s}\left\|x\right\|^{2}ds=\dfrac{c^{2}}{2\alpha}\left\|x\right\|^{2},\ \forall
x\in X$

$2.\Rightarrow 1.$ We show in the proof of theorem 2 that the operator
$Q_{0}\in B(X,X^{*})$ defined by

$\left\langle
Q_{0}x,y\right\rangle:=\dfrac{1}{\theta}\int_{0}^{+\infty}\left\langle
PT(t)x,T(t)y\right\rangle dt,\quad x,y\in X,$

is a solution of the Lyapunov inequality (1) i.e. $Q_{0}\in L_{A}$. If $Q$ is
strongly positive then for all $x\in X$,

$\dfrac{1}{\theta}\int_{0}^{+\infty}\left\langle PT(t)x,T(t)x\right\rangle
ds\geq\theta\left\|x\right\|^{2},$

for some $\theta>0$.

Since $\dfrac{1}{\theta}\int_{0}^{+\infty}\left\langle
PT(t)x,T(t)x\right\rangle
ds\leq\left\|P\right\|\int_{0}^{+\infty}\left\|T(t)x\right\|^{2}dt$ and

$\dfrac{\theta}{\left\|P\right\|}\left\|x\right\|^{2}=\int_{0}^{+\infty}e^{-\frac{||P||}{\theta}s}\left\|x\right\|^{2}ds,$

then for all $x\in X$,

$\int_{0}^{+\infty}\left\|T(s)x\right\|^{2}ds\geq\int_{0}^{+\infty}e^{-\frac{||P||}{\theta}s}\left\|x\right\|^{2}ds.$

Clealy, for two positive real functions $f(t):=\left\|T(t)x\right\|^{2}$ and
$g(t):=e^{-\frac{||P||}{\theta}s}\left\|x\right\|^{2}$, such that
$\int_{0}^{+\infty}f(t)dt\geq\int_{0}^{+\infty}g(t)dt$, there exists a set
$\Omega_{0}\subset\mathbb{R}_{+}$ with non-zero measure such that $f(f)\geq
g(t)$ for all $t\in\Omega_{0}$, then there exists $t_{0}\in\Omega_{0}$,

$\left\|T(t_{0})x\right\|\geq
e^{-\frac{||P||}{2\theta}t_{0}}\left\|x\right\|,$

for all $x\in X$. By the proposition 7, T is left-invertible.

∎

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