Infinite-time admissibility under compact perturbations

TL;DR

This paper explores how infinite-time admissibility of semigroups is affected by compact perturbations, demonstrating through examples that it is not necessarily preserved even under strong stability.

Contribution

It provides the first examples showing that infinite-time admissibility can be lost under compact perturbations of the generator.

Findings

Infinite-time admissibility is not preserved under compact perturbations.

Both original and perturbed semigroups can be strongly stable.

Examples illustrate the non-preservation of admissibility.

Abstract

We investigate the behavior of infinite-time admissibility under compact perturbations. We show, by means of two completely different examples, that infinite-time admissibility is not preserved under compact perturbations $Q$ of the underlying semigroup generator $A$, even if $A$ and $A+Q$ both generate strongly stable semigroups.

Equations

$$x^{\prime }=Ax+Bu(t)\text{and}x(0)=0$$

$${\mathbb{R}}_0^{+}:=[0,\infty ),{\mathbb{C}}^{+}:=\{z\in \mathbb{C}:\mathrm{Re}z>0\},{\mathbb{C}}^{-}:=\{z\in \mathbb{C}:\mathrm{Re}z<0\}.$$

$$x^{\prime }=Ax+Bu(t)\text{with}y(t)=Cx(t).$$

$$(0,\infty )\ni t\mapsto {\Phi }_t(u):={\int }_0^t{\mathrm{e}}^{A_{-1}s}Bu(s)\mathrm{d}s$$

$${\Phi }_t\in L(L^2({\mathbb{R}}_0^{+},U),X)(t\in (0,\infty )).$$

$$(0,\infty )\ni t\mapsto {\Phi }_t(u):={\int }_0^t{\mathrm{e}}^{A_{-1}s}Bu(s)\mathrm{d}s$$

$${\Phi }_t\in L(L^2({\mathbb{R}}_0^{+},U),X)(t\in (0,\infty ))\text{and}\underset{t\in (0,\infty )}{\sup }{\Vert {\Phi }_t\Vert }_{L^2({\mathbb{R}}_0^{+},U),X}<\infty .$$

$${\Vert {\int }_0^t{\mathrm{e}}^{(A_0-B{B}^{\ast })s}Bu(s)\mathrm{d}s\Vert }_X^2\le \frac{1}{2}{\Vert u\Vert }_2^2(u\in L^2({\mathbb{R}}_0^{+},U),t\in {\mathbb{R}}_0^{+}).$$

$$A_{0}x:=({\lambda }_{0k}{x}_k{)}_{k\in \mathbb{N}}(x\in D(A_0)),$$

$$\mathrm{Re}{\lambda }_{0k}=-{\alpha }_k:=-1/k(k\in \mathbb{N})\text{and}\mathrm{Im}{\lambda }_{0k}={\beta }_k⟶\infty (k\to \infty ).$$

$$Bu:=(u{b}_k{)}_{k\in \mathbb{N}}(u\in U),$$

$$b_k:=1/k^{3/8}(k\in I_1)$$

$$I_1:=\{l^2:l\in \mathbb{N}\}$$

$$A:=A_0-B{B}^{\ast }\text{and}{A}^{\prime }:=A_0$$

$${\lambda }_{0k}\in {\mathbb{C}}^{-}(k\in \mathbb{N})\text{and}\sup \{\mathrm{Re}{\lambda }_{0k}:k\in \mathbb{N}\}=0$$

$$\underset{{\Vert u\Vert }_2=1}{\sup }\underset{t\in (0,\infty )}{\sup }{\Vert {\int }_0^t{\mathrm{e}}^{A_{0}s}Bu(s)\mathrm{d}s\Vert }_X=\infty .$$

$$\underset{t\to \infty }{\mathrm{liminf}}{\Vert {\int }_0^t{\mathrm{e}}^{A_{0}s}Bu(s)\mathrm{d}s\Vert }_X^2$$

$$\displaystyle\geq\bigg{|}\int_{0}^{\infty}u(s)\mathrm{e}^{\lambda_{0n}s}\,\mathrm{d}s\bigg{|}^{2}|b_{n}|^{2}$$

$${\Vert u_n\Vert }_2$$

$$\displaystyle\bigg{|}\int_{0}^{\infty}u_{n}(s)\mathrm{e}^{\lambda_{0n}s}\,\mathrm{d}s\bigg{|}^{2}=\frac{1}{n}\bigg{|}\int_{0}^{n}\mathrm{e}^{-\alpha_{n}s}\,\mathrm{d}s\bigg{|}^{2}$$

$$\underset{{\Vert u\Vert }_2=1}{\sup }\underset{t\in (0,\infty )}{\sup }{\Vert {\int }_0^t{\mathrm{e}}^{A_{0}s}Bu(s)\mathrm{d}s\Vert }_X^2$$

$$\displaystyle\geq(1-\mathrm{e}^{-1})^{2}\sup_{n\in\mathbb{N}}\big{(}n|b_{n}|^{2}\big{)}.$$

$$\mathrm{Re}{\lambda }_{0k}\le 0(k\in \mathbb{N})\text{and}|{\lambda }_{0k}|⟶\infty (k\to \infty ),$$

$$\Vert (A-z{)}^{-1}\Vert \le \frac{M}{\mathrm{Re}z-\omega }(\mathrm{Re}z>\omega ).$$

$$\underset{n\to \infty }{\mathrm{limsup}}\Vert (A-{\lambda }_{0n}{)}^{-1}\Vert \le \underset{n\to \infty }{\mathrm{limsup}}\frac{M}{\mathrm{Re}{\lambda }_{0n}-\omega }=\frac{M}{|\omega |}.$$

$$(A-{\lambda }_{0n})e_n=-B{B}^{\ast }{e}_n=-b_n\cdot b⟶0(n\to \infty ).$$

$$1=\underset{n\to \infty }{\mathrm{limsup}}\Vert e_n\Vert =\underset{n\to \infty }{\mathrm{limsup}}\Vert (A-{\lambda }_{0n}{)}^{-1}{b}_n\cdot b\Vert \le \frac{M}{|\omega |}\underset{n\to \infty }{\mathrm{limsup}}\Vert b_n\cdot b\Vert =0.$$

$$A_{0}x:=({\lambda }_{0k}{x}_k{)}_{k\in I}(x\in D(A_0)),$$

$$Bu=(u{b}_k{)}_{k\in I}(u\in U)$$

$$\sum _{k\in I}\frac{|b_k{|}^2}{|z-{\lambda }_{0k}{|}^2}\le \frac{M}{\mathrm{Re}z}(z\in {\mathbb{C}}^{+}).$$

$${\Vert (z-A{)}^{-1}B\Vert }_{U,X}\le \frac{M}{\sqrt{\mathrm{Re}z}}(z\in {\mathbb{C}}^{+}).$$

$$Ax:=({\lambda }_k{x}_k{)}_{k\in \mathbb{Z}}(x\in D(A))\text{and}{A}^{\prime }x:=({\lambda }_k^{\prime }{x}_k{)}_{k\in \mathbb{Z}}(x\in D(A^{\prime })),$$

Full text

Index terms: infinite-time admissibility, compact perturbations, stabilization of collocated linear systems

1 Introduction

In this note, we investigate the behavior of infinite-time admissibility under compact perturbations of the underlying semigroup generator. So, we consider semigroup generators $A:D(A)\subset X\to X$ (with $X$ a Hilbert space) and possibly unbounded control operators $B$ (defined on another Hilbert space $U$) and we ask how the property of infinite-time admissibility of $B$ behaves under compact perturbations of the generator $A$. Infinite-time admissibility of $B$ for $A$ means that for every control input $u\in L^{2}([0,\infty),U)$ the mild solution of the initial value problem

[TABLE]

is a bounded function from $[0,\infty)$ with values in $X$. (A priori, the mild solution has values only in the extrapolation space $X_{-1}$ of $A$ and, a fortiori, need not be bounded in the norm of $X$, of course.)

It is well-known that (finite-time) admissibility is preserved under very general perturbations $Q$ of the generator $A$, in particular, under bounded perturbations. It is also clear that infinite-time admissibility, by contrast, is not preserved under bounded perturbations. Just think of a generator $A$ of an exponentially stable semigroup and a bounded perturbation $Q$ (for example, a sufficiently large multiple of the identity) such that $A+Q$ has spectral points in the right half-plane.

In this note, we will show by way of two completely different kinds of examples that infinite-time is also not preserved under compact perturbations $Q$ which are such that both $A$ and $A+Q$ generate stronly stable (but not exponentially stable) semigroups. So, in other words, we show that there exist semigroup generators $A$ and $A+Q$ with $Q$ being compact and a control operator $B$ such that

• •

the semigroups $\mathrm{e}^{A\cdot}$ and $\mathrm{e}^{(A+Q)\cdot}$ are strongly stable but not exponentially stable

• •

$B$ is infinite-time admissible for $A$ but not infinite-time admissible for $A+Q$.

In our first – more elementary – example, we will use an old and well-known result from the 1970s, namely a stabilization result for collocated linear systems. In that example, the compact perturbation $Q$ will be of rank $1$ and the control operator $B$ will be bounded. In particular, none of the technicalities coming along with unbounded control operators will bother us there. In our second – less elementary – example, we will use a more advanced result from the 1990s, namely a characterization of infinite-time admissibility for diagonal semigroup generators. In that example, the control operator $B$ will be unbounded and the compact perturbation $Q$ will be of rank $\infty$.

In the entire note, we will use the following notation.

[TABLE]

As usual, $L(X,Y)$ denotes the Banach space of bounded linear operators between two Banach spaces $X$ and $Y$ and $\left\|\cdot\right\|_{X,Y}$ stands for the operator norm on $L(X,Y)$. Also, $\left\|u\right\|_{2}$ denotes the norm of a square-integrable function $u\in L^{2}(\mathbb{R}^{+}_{0},U)$ with values in the Banach space $U$. And finally, for a semigroup generator $A$ and bounded operators $B,C$ between appropriate spaces, the symbol $\mathfrak{S}(A,B,C)$ will stand for the state-linear system [3]

[TABLE]

2 Some basic facts about admissibility and infinite-time admissibility

In this section, we briefly recall the definition of and some basic facts about admissibility and infinite-time admissibility. If $A:D(A)\subset X\to X$ is a semigroup generator on the Hilbert space $X$ and $X_{-1}$ is the corresponding extrapolation space, then an operator $B\in L(U,X_{-1})$ (with $U$ another Hilbert space) is called control operator for $A$. Also, $B$ is called a bounded control operator iff $B\in L(U,X)$ and an unbounded control operator iff $B\in L(U,X_{-1})\setminus L(U,X)$. See  [9] (Section 2.10) or [4] (Section II.5) for basic facts about extrapolation spaces.

Definition 2.1**.**

Suppose $A:D(A)\subset X\to X$ is a semigroup generator on $X$ and $B\in L(U,X_{-1})$, where $X,U$ are both Hilbert spaces. Then $B$ is called admissible for $A$ iff for every $u\in L^{2}(\mathbb{R}^{+}_{0},U)$

[TABLE]

is a function with values in $X$, where $A_{-1}$ is the generator of the continuous extension of the semigroup $\mathrm{e}^{A\cdot}$ to $X_{-1}$.

Clearly, for a given semigroup generator $A$ every bounded control operator $B\in L(U,X)$ is admissible (because $\mathrm{e}^{A_{-1}s}|_{X}=\mathrm{e}^{As}$ for $s\in\mathbb{R}^{+}_{0}$). It should also be noted that if $B\in L(U,X_{-1})$ is admissible for $A$, then for every $t\in(0,\infty)$ the linear operator $L^{2}(\mathbb{R}^{+}_{0},U)\ni u\mapsto\Phi_{t}(u)\in X$ defined in (2.1) is closed and thus continuous by the closed graph theorem. Consequently, $B\in L(U,X_{-1})$ is admissible for $A$ if and only if

[TABLE]

Definition 2.2**.**

Suppose $A:D(A)\subset X\to X$ is a semigroup generator on $X$ and $B\in L(U,X_{-1})$, where $X,U$ are both Hilbert spaces. Then $B$ is called infinite-time admissible for $A$ iff for every $u\in L^{2}(\mathbb{R}^{+}_{0},U)$

[TABLE]

is a function with values in $X$ that is bounded (in the norm of $X$), where $A_{-1}$ is the generator of the continuous extension of the semigroup $\mathrm{e}^{A\cdot}$ to $X_{-1}$.

Clearly, if $B\in L(U,X_{-1})$ is infinite-time admissible for a given semigroup generator $A$, then it is also admissible for $A$. It should also be noted that, by the uniform boundedness principle, $B\in L(U,X_{-1})$ is infinite-time admissible for $A$ if and only if

[TABLE]

Some authors [8], [2], [10] use the term input-stability for the system $\mathfrak{S}(A,B)$ instead of infinite-time admissibility.

It is well-known that admissibility is preserved under bounded perturbations.

Proposition 2.3**.**

Suppose $A:D(A)\subset X\to X$ is a semigroup generator on $X$ and $B\in L(U,X_{-1})$, where $X,U$ are both Hilbert spaces. Also, let $Q\in L(X)$. Then $B$ is admissible for $A$ if and only if $B$ is admissible for $A+Q$.

In fact, the conclusion of this proposition remains true for much more general perturbations $Q$, namely for perturbations of the (feedback) form $Q=B_{0}C_{0}$, where $B_{0}\in L(U_{0},X_{-1})$ is an admissible control operator for $A$ and $C_{0}\in L(X,U_{0})$ with $U_{0}$ an arbitrary Hilbert space. See Corollary .5.5.1 from [9], for instance.

Proposition 2.4**.**

Suppose $A:D(A)\subset X\to X$ is the generator of an exponentially stable semigroup on $X$ and $B\in L(U,X_{-1})$ is admissible for $A$. Then $B$ is even infinite-time admissible for $A$.

See Proposition 4.4.5 in [9], for instance, and notice that for bounded control operators $B$ the above proposition is trivial. In view of that proposition, it is clear that infinite-time admissibility – unlike admissibility – is not preserved under bounded perturbations. Choose, for example, a bounded generator $A$ of an exponentially stable semigroup and let $Q:=-A\in L(X)$ and $B:=I\in L(X,X)$ (identity operator on $X$).

3 An example using a stabilization result for collocated linear systems

3.1 Stabilization of collocated linear systems

We will use the following well-known stabilization result for collocated systems, that is, systems of the form $\mathfrak{S}(A,B,B^{*})$ with a bounded control operator $B$. It essentially goes back to [1] (Corollary 3.1) and, in the form below, can be found in [8] (Lemma 2.2.6), for instance. (Actually, for the more general version with the countability assumption on $\sigma(A_{0})\cap\mathrm{i}\mathbb{R}$ we have to refer to [10], but this more general version will not be used in the sequel.)

Theorem 3.1**.**

Suppose $A_{0}$ is a contraction semigroup generator on a Hilbert space $X$ with compact resolvent (or, more generally, with $\sigma(A_{0})\cap\mathrm{i}\mathbb{R}$ being countable). Suppose further that $B\in L(U,X)$ with another Hilbert space $U$ and that $\mathfrak{S}(A_{0},B,B^{*})$ is approximately controllable in infinite time or approximately observable in infinite time. Then

• (i)

$B$* is infinite-time admissible for $A_{0}-BB^{*}$, more precisely,*

[TABLE]

• (ii)

$\mathrm{e}^{(A_{0}-BB^{*})\cdot}$* is a strongly stable contraction semigroup on $X$.*

A far-reaching generalization of this result to the case of unbounded control operators was obtained by Curtain and Weiss [10]. See Theorem 5.1 and 5.2 in conjunction with Proposition 1.5 from [10]. We also refer to [2] for a parallel result on exponential stabilization.

3.2 Infinite-time admissibility under compact perturbations

Example 3.2**.**

Set $X:=\ell^{2}(\mathbb{N},\mathbb{C})$ and let $A_{0}:D(A_{0})\subset X\to X$ be defined by

[TABLE]

where $D(A_{0}):=\{(x_{k})\in X:(\lambda_{0k}x_{k})\in X\}$ and $\lambda_{0k}:=-\alpha_{k}+\mathrm{i}\beta_{k}$ with

[TABLE]

Set $U:=\mathbb{C}$ and let $B:U\to\mathbb{C}^{\mathbb{N}}$ be defined by

[TABLE]

where

[TABLE]

Clearly, $(b_{k})\in X$ and therefore $B\in L(U,X)$. We now define

[TABLE]

and show, in various steps, that $A$ and $A^{\prime}$ are generators of strongly but not exponentially stable contraction semigroups on $X$, that $A^{\prime}=A+Q$ for a compact perturbation $Q$ of rank one, and that $B$ is infinite-time admissible for $A$ but not infinite-time admissible for $A^{\prime}$.

As a first step, we observe that $A^{\prime}=A+Q$ with $Q:=BB^{*}$ and that $Q$ has rank one (because the same is true for $B$), whence $Q$ is compact.

As a second step, we observe from

[TABLE]

that $A^{\prime}$ is the generator of a strongly stable but not exponentially stable contraction semigroup on $X$.

As a third step, we show that $B$ is not infinite-time admissible for $A^{\prime}$. In view of (2.4) we have to show that

[TABLE]

We first observe by Fatou’s lemma that

[TABLE]

for every $u\in L^{2}(\mathbb{R}^{+}_{0},U)$ and $n\in\mathbb{N}$. Setting $u_{n}(s):=n^{-1/2}\chi_{[0,n]}(s)\cdot\mathrm{e}^{-\mathrm{i}\beta_{n}s}$ for $s\in\mathbb{R}^{+}_{0}$ and $n\in\mathbb{N}$, we see that

[TABLE]

for every $n\in\mathbb{N}$. Combining now (3.2), (3.3) and (3.4) we get

[TABLE]

Since \sup_{n\in\mathbb{N}}\big{(}n|b_{n}|^{2}\big{)}\geq\sup_{n\in I_{1}}\big{(}n|b_{n}|^{2}\big{)}=\infty, the desired relation (3.1) follows.

As a fourth step, we show that $B$ is infinite-time admissible for $A$ and that $A$ is the generator of a strongly stable contraction semigroup on $X$. In order to do so, we apply the stablization theorem above (Theorem 3.1). Since

[TABLE]

we see that $A_{0}$ is a contraction semigroup generator on $X$ with compact resolvent, and since the eigenvalues $\lambda_{0k}$ of $A_{0}$ are pairwise distinct and $b_{k}\neq 0$ for every $k\in\mathbb{N}$, we see that the collocated linear system $\mathfrak{S}(A,B,B^{*})$ is approximately controllable and approximately observable in infinite time (Theorem 4.2.3 of [3]). So, by the stablization theorem above (Theorem 3.1), $B$ is infinite-time admissible for $A_{0}-BB^{*}=A$ and $\mathrm{e}^{A\cdot}$ is a strongly stable contraction semigroup on $X$.

As a fifth and last step, we convince ourselves that the semigroup generated by $A$ is not exponentially stable. Assume the contrary. Then there exist $M\geq 1$ and $\omega<0$ such that $\{z\in\mathbb{C}:\operatorname{Re}z>\omega\}\subset\rho(A)$ and

[TABLE]

So, since $\operatorname{Re}\lambda_{0n}\longrightarrow 0$ as $n\to\infty$, we conclude that

[TABLE]

We now observe that

[TABLE]

Combining (3.5) and (3.6) we arrive at

[TABLE]

Contradiction! $\blacktriangleleft$

4 An example using an admissibility result for diagonal linear systems

4.1 Characterization of infinite-time admissibility

We will use the following well-known characterization of infinite-time admissibility for diagonal semigroup generators $A_{0}$. It essentially goes back to [5] (Proposition 2.2) and can also be found in [9] (Theorem 5.3.9 in conjunction with Remark 4.6.5), for instance.

Theorem 4.1**.**

Suppose $X=\ell^{2}(I,\mathbb{C})$ with a countable infinite index set $I$ and let $A_{0}:D(A_{0})\subset X\to X$ be the diagonal operator given by

[TABLE]

where $D(A_{0}):=\{(x_{k})\in X:(\lambda_{0k}x_{k})\in X\}$ and $\lambda_{0k}\in\mathbb{C}^{-}$ for every $k\in I$. Suppose further that $B\in L(U,X_{-1})$ with $U:=\mathbb{C}$, that is,

[TABLE]

for a uniqe sequence $(b_{k})\in X_{-1}=\{(c_{k})\in\mathbb{C}^{I}:\sum_{k\in I}|c_{k}|^{2}/(1+|\lambda_{k}|^{2})<\infty\}$. Then the following statements are equivalent:

• (i)

$B$* is infinite-time admissible for $A_{0}$*

• (ii)

there exists a constant $M\in\mathbb{R}_{0}^{+}$ such that

[TABLE]

Clearly, in the situation of the above theorem the condition (ii) is equivalent to the existence of a constant $M\in\mathbb{R}_{0}^{+}$ such that

[TABLE]

A far-reaching generalization of the above theorem to the case of general contraction semigroup generators $A_{0}$ on a separable Hilbert space $X$ was obtained by Jacob and Partington [6]. See Theorem 1.3 from [6]. It states that for a contraction semigroup generator $A_{0}$ on a separable Hilbert space $X$ a control operator $B\in L(U,X_{-1})$ with $U:=\mathbb{C}$ is infinite-time admissible if and only if there is a constant $M\in\mathbb{R}_{0}^{+}$ such that the resolvent estimate (4.1) is satisfied. We also refer to [7] and [9] (Section 5.6) for an overview of many more admissibility results, for example, for infinite-dimensional input-value spaces $U$.

4.2 Infinite-time admissibility under compact perturbations

Example 4.2**.**

Set $X:=\ell^{2}(\mathbb{Z},\mathbb{C})$ and let $A:D(A)\subset X\to X$ and $A^{\prime}:D(A^{\prime})\subset X\to X$ be defined by

[TABLE]

where $D(A):=\{(x_{k})\in X:(\lambda_{k}x_{k})\in X\}$ and $D(A^{\prime}):=\{(x_{k})\in X:(\lambda_{k}^{\prime}x_{k})\in X\}$ with

[TABLE]

Set $U:=\mathbb{C}$ and let $B:U\to\mathbb{C}^{\mathbb{Z}}$ be defined by

[TABLE]

where

[TABLE]

Clearly, $\sum_{k\in\mathbb{Z}}|b_{k}|^{2}/(1+|\lambda_{k}|^{2})<\infty$ and $\sum_{k\in\mathbb{Z}}|b_{k}|^{2}=\infty$ whence $(b_{k})\in X_{-1}\setminus X$. And therefore

[TABLE]

We now show, in various steps, that $A$ and $A^{\prime}$ are generators of strongly but not exponentially stable contraction semigroups on $X$, that $A^{\prime}=A+Q$ for a compact perturbation $Q$ of infinite rank, and that $B$ is infinite-time admissible for $A$ but not infinite-time admissible for $A^{\prime}$.

As a first step, we observe from

[TABLE]

that $A$ and $A^{\prime}$ are generators of strongly stable but not exponentially stable contraction semigroups on $X$.

As a second step, we observe that $A^{\prime}=A+Q$ for a compact operator $Q$ of infinite rank. Indeed, the operator $Q:X\to X$ defined by

[TABLE]

is a bounded operator on $X$ because $(\lambda_{k}^{\prime}-\lambda_{k})_{k\in\mathbb{Z}}$ is a bounded sequence. Also, $Q$ is the limit in norm operator topology of the finite-rank operators $Q_{N}:X\to X$ defined by

[TABLE]

and therefore $Q$ is compact, as desired.

As a third step, we show that $B$ is infinite-time admissible for $A$. We have that

[TABLE]

for every $z\in\mathbb{C}^{+}$ and that

[TABLE]

So, by the admissibility theorem above (Theorem 4.1), the claimed infinite-time admissibility of $B$ for $A$ follows from (4.2) and (4.3).

As a fourth and last step, we show that $B$ is not infinite-time admissible for $A^{\prime}$. We have that

[TABLE]

for every $z\in\mathbb{C}^{+}$ and $n\in\mathbb{N}$. Choosing $z_{n}:=\mathrm{e}^{-n}+\mathrm{i}n\in\mathbb{C}^{+}$ for $n\in\mathbb{N}$, we see from (4.4) that

[TABLE]

So, by the admissibility theorem above (Theorem 4.1), $B$ is not infinite-time admissible for $A^{\prime}$, as desired. $\blacktriangleleft$

Acknowledgements

I would like to thank the German Research Foundation (DFG) financial support through the grant “Input-to-state stability and stabilization of distributed-parameter systems” (DA 767/7-1).