On the Controllability of a Differential Equation with Delayed and Advanced Arguments

Abstract

A semigroup theory for a differential equation with delayed and advanced arguments is developed, with a detailed description of the infinitesimal generator. This in turn allows to study the exact controllability of the equation, by rewriting it as a classical Cauchy problem.

1. Introduction

In this paper, we will study the exact controllability of a functional differential equation with both delayed and advanced arguments. Such equations are often referred to in the literature as mixed-type functional differential equations (MTFDE) or forward-backward equations. The study of this type of equations is less developed compared with other classes of functional equations. Interest in MTDFEs is motivated by problems in optimal control [1] and applications, for example, in economic dynamics [2] and travelling waves in a spatial lattice [3]. See also [4]. In all these references, the reader can find interesting examples and applications.

In order to achieve our goal, first, we rewrite the equation as a classical Cauchy problem in a certain Banach space. Then we introduce the associated semigroup and its infinitesimal generator and prove some important properties of these operators (including some spectral properties). This will allow us to characterize the exact controllability, by applying a result of Bárcenas and Diestel (see [5]).

2. Preliminary Results

In [6], the following differential-difference equation is considered𝑥^′(𝑡)=𝑥(𝑡−1)+𝑥(𝑡+1),(2.1) where 𝑡≥0 and𝑥∶[−1,+∞)→ℂ^𝑛 is differentiable in [0,+∞).

Equation (2.1) may be written as 𝑥(𝑡+1)=𝑥^′(𝑡)−𝑥(𝑡−1)(2.2) or equivalently 𝑥(𝑡)=𝑥^′(𝑡−1)−𝑥(𝑡−2).(2.3) From this we have that in order to find the solution 𝑥(𝑡) on the interval [𝑚,𝑚+1], it is necessary to know its value on the interval [𝑚−2,𝑚], with 𝑚 being a positive integer. In particular, to determine the solution on the interval [1,2], it is necessary to know it on the interval [−1,1].

Accordingly, 𝑥(𝑡) is defined for 𝑡∈[−1,1] as𝜑𝑥(𝑡)=𝜑(𝑡)=¹𝜑(𝑡),𝑡∈[−1,0],₂(𝑡),𝑡∈[0,1],(2.4) where the function 𝜑 belongs to the space 𝐶^∞([−1,1],ℂ^𝑛). The solution of the initial value problem (2.1), (2.4) is constructed via an iterative process using the step derivation method.

It is𝑥(𝑡)=^𝑚_𝑖=0𝑐_1𝑖𝜑^(𝑖)(𝑡−𝑚)+^𝑚_𝑖=0𝑐_2𝑖𝜑^(𝑖)(𝑡−𝑚−1),(2.5) where 𝑐_1𝑖 and 𝑐_2𝑖 are constants not all necessarily different from zero.

This solution may be extended to the left by rewriting (2.1) as 𝑥(𝑡−1)=𝑥^′(𝑡)−𝑥(𝑡+1)(2.6) which allows to yield an expression for 𝑥(𝑡) analogous to (2.5).

In order to assure the existence, differentiability, and uniqueness of the solution 𝑥(𝑡), it is demanded that 𝑥(𝑡) must satisfy the relationship 𝜑^(𝑛)(0)=𝜑^(𝑛−1)(−1)+𝜑^(𝑛−1)(1)(2.7) for 𝑛=1,2,3,….

Further, if a differentiable solution 𝑥(𝑡) exists, then it belongs to the space 𝐶^∞([−1,+∞],ℂ^𝑛) (see [6, Theorems 3.1 and 3.2]).

Let 𝑀 be the following nonempty, closed subspace of the topological space 𝐶^∞([−1,1],ℂ^𝑛):𝑀={𝜑∈𝐶^∞([−1,1],ℂ^𝑛)∶𝜑^(𝑛)(0)=𝜑^(𝑛−1)(−1)+𝜑^(𝑛−1)(1),𝑛=1,2,3,…}.(2.8) The space 𝐶^∞([−1,1],ℂ^𝑛) is endowed with the topology induced by the following countable system of seminorms: 𝑃_𝑘(𝑓)=max_{𝑡∈[−1,1]}‖𝑓^(𝑘)(𝑥)‖_ℂ𝑛,for𝑘=0,1,2,….(2.9) The convergence in this topology means the uniform convergence of the function and each of its derivatives of any order. We denote‖𝑓‖_𝑘=^𝑘_𝑖=0𝑃_𝑖(𝑓),foreach𝑘=0,1,2,….(2.10) A sequence {𝑓_𝑛}^∞_𝑛=1 converges to 𝑓 if and only if ‖𝑓_𝑛−𝑓‖_𝑘 tends to 0, as 𝑛 tends to infinity, for each 𝑘.

For each 𝑡≥0, the operator 𝑇(𝑡) is defined on the solutions 𝑥(𝑡) of (2.1) as follows:𝑇(𝑡)𝑥(𝜃)=𝑥(𝑡+𝜃),𝜃∈[−1,1].(2.11) The following result originally appears on [7].

Theorem 2.1. The family {𝑇(𝑡)}_𝑡≥0 defines a strongly continuous semigroup on 𝐿(𝑀).

Proof. That 𝑇(0)=𝐼 and 𝑇(𝑡+𝑠)=𝑇(𝑡)𝑇(𝑠) for each 𝑡,𝑠≥0 are straightforward from the definition of 𝑇(𝑡). Since 𝑥(𝜃)|_{𝜃∈[−1,1]}=𝜑(𝜃)∈𝑀, the domain of 𝑇(𝑡) is 𝑀. On the other hand, the function 𝑦_𝑡(𝜃)∶=𝑇(𝑡)𝑥(𝜃)=𝑥(𝑡+𝜃)(2.12) belongs to 𝑀 for each 𝑡≥0, because if 𝑥(𝑡)∈𝐶^∞([−1,1],ℂ^𝑛), then 𝑦_𝑡(𝜃)∈𝐶^∞([−1,1],ℂ^𝑛) for each fixed 𝑡≥0. Additionally, for each 𝑡≥0, 𝑦_𝑡^(𝑛)(0)=𝑦_𝑡^(𝑛−1)(−1)+𝑦_𝑡^(𝑛−1)(1), for 𝑛=1,2,3,…, or equivalently 𝑥^(𝑛)(𝑡)=𝑥^(𝑛−1)(𝑡−1)+𝑥^(𝑛−1)(𝑡+1),𝑛=1,2,….(2.13) Hence 𝑦_𝑡(𝜃)∈𝑀.
In order to prove that 𝑇(𝜏) is continuous for each fixed 𝜏≥0, we will prove that there exists 𝑛_𝑘 for each 𝑘 and some constant 𝑐≥0 such that ‖𝑇(𝜏)𝑥(𝜃)‖_𝑘≤𝑐^𝑛𝑘_𝑖=0‖𝑥(𝜃)‖_𝑖.(2.14) In view that 𝑃_𝑘(𝑇(𝜏)𝑥(𝜃))=max_{𝜃∈[−1,1]}|𝑥^(𝑘)(𝜏+𝜃)|≤max_{𝑡∈[[𝜏]]−1,[𝜏]+2]}|𝑥^(𝑘)(𝑡)|≤max_{𝑡∈[𝜏]−1,[𝜏]]}|𝑥^(𝑘)(𝑡)|+max_{𝑡∈[[𝜏]],[𝜏]+1]}|𝑥^(𝑘)(𝑡)|+max_{𝑡∈[[𝜏]]+1,[𝜏]+2]}|𝑥^(𝑘)(𝑡)|(2.15) and using formula (2.5) with 𝑚=[𝜏], one obtains max_{𝜃∈[−1,1]}|𝑥^(𝑘)(𝜏+𝜃)|≤^𝑚−1_𝑖=0𝑐_1𝑖max_{𝑡∈[𝑚−1,𝑚]}|𝜑^(𝑖+𝑘)(𝑡−(𝑚−1))|+^𝑚−1_𝑖=0𝑐_2𝑖max_{𝑡∈[𝑚−1,𝑚]}|𝜑^(𝑖+𝑘)+(𝑡−(𝑚−1)−1)|_𝑚𝑖=0𝑐_3𝑖max_{𝑡∈[𝑚,𝑚+1]}|𝜑^(𝑖+𝑘)(𝑡−𝑚)|+^𝑚_𝑖=0𝑐_4𝑖max_{𝑡∈[𝑚,𝑚+1]}|𝜑^(𝑖+𝑘)+(𝑡−𝑚−1)|_𝑚+1𝑖=0𝑐_5𝑖max_{𝑡∈[𝑚+1,𝑚+2]}|𝜑^(𝑖+𝑘)+(𝑡−(𝑚+1))|_𝑚+1𝑖=0𝑐_6𝑖max_{𝑡∈[𝑚+1,𝑚+2]}|𝜑^(𝑖+𝑘)≤(𝑡−(𝑚+1)−1)|_{𝑚−1𝑖=0}(𝑐_1𝑖+𝑐_2𝑖)^𝑚−1_𝑖=0max_{𝜃∈[−1,1]}|𝜑^(𝑖+𝑘)(𝜃)|+^𝑚_𝑖=0(𝑐_3𝑖+𝑐_4𝑖)^𝑚_𝑖=0max_{𝜃∈[−1,1]}|𝜑^(𝑖+𝑘)+(𝜃)|_𝑚+1𝑖=0(𝑐_5𝑖+𝑐_6𝑖)^𝑚+1_𝑖=0max_{𝜃∈[−1,1]}|𝜑^(𝑖+𝑘)(𝜃)|≤𝑐₁‖𝜑‖_{𝑚+𝑘−1}+𝑐₂‖𝜑‖_𝑚+𝑘+𝑐₃‖𝜑‖_𝑚+𝑘+1≤(𝑐₁+𝑐₂+𝑐₃)(‖𝜑‖_{𝑚+𝑘−1}+‖𝜑‖_𝑚+𝑘+‖𝜑‖_𝑚+𝑘+1)=𝑐(‖𝜑‖_{𝑚+𝑘−1}+‖𝜑‖_𝑚+𝑘+‖𝜑‖_𝑚+𝑘+1).(2.16) It only remains to prove that 𝑇(𝑡)𝑥→𝑇(𝑡₀)𝑥, as 𝑡→𝑡_𝑜, for each 𝑥∈𝑀.
In fact, let 𝑡₀≥0. We have that lim_𝑡→𝑡0𝑇(𝑡)𝑥=𝑇(𝑡₀)𝑥⇔lim_𝜏→0𝑇(𝑡₀+𝜏)𝑥=𝑇(𝑡₀)𝑥⇔‖𝑇(𝑡₀+𝜏)𝑥(𝜃)−𝑇(𝑡₀)𝑥(𝜃)‖_𝑘→0 as 𝜏→0 for each 𝑘=0,1,…, ⇔max_{𝜃∈[−1,1]}|𝑥^(𝑘)(𝑡₀+𝜏+𝜃)−𝑥^(𝑘)(𝑡₀+𝜃)|→0 as 𝜏→0 for each 𝑘=0,1,….
Assuming that 0≤[𝜏]≤1 and taking into account the uniform convergence of 𝑥^(𝑘)(𝑡) in the closed interval [𝑎,𝑏]∶=[𝑡₀−2,𝑡₀+2] (𝑡,𝑡^′∈[𝑎,𝑏]), it follows that max_{𝜃∈[−1,1]}|𝑥^(𝑘)(𝑡₀+𝜏+𝜃)−𝑥^(𝑘)(𝑡₀+𝜃)|≤max_{|𝑡−𝑡′|≤[𝜏]}|𝑥^(𝑘)(𝑡)−𝑥^(𝑘)(𝑡^′)|→0(2.17) as 𝜏→0.

Some basic definitions and concepts on controllability are recalled below. Let 𝑈 and 𝑋 be Banach spaces. We consider the inhomogeneous differential linear system:𝑥̇(𝑡)=𝐴𝑥(𝑡)+𝐵𝑢(𝑡),𝑥(0)=𝑥₀∈𝐷(𝐴),(2.18) where 𝐴∶𝑋→𝑋 is the infinitesimal generator operator of a strongly continuous semigroup (𝑆_𝑡)_𝑡≥0; 𝐵∶𝑈→𝑈 is a bounded linear operator and 𝑢∶[0,+∞)→𝑈 is a strongly measurable essentially bounded function.

Let Ω be a nonempty separable weakly compact convex subset of 𝑈.

We recall (see [8]) that 𝑥₀∈𝑋 is controllable with respect to 𝑥₁ if there exist ^_𝑡≥0 and a control 𝑢∈𝐿_∞([0,^_𝑡];𝑈) such that 𝑥(^_𝑡)=𝑥₁ in (2.18).

The controllability map on [0,^_𝑡] for some ^_𝑡≥0 is the linear map 𝐵^_𝑡∶𝐿_∞([0,^_𝑡];𝑈)→𝑋(2.19) defined by 𝐵^_𝑡𝑢=^_𝑡₀𝑆(^_𝑡−𝑠)𝐵𝑢(𝑠)d𝑠.(2.20) Now, one says that (2.18) is exactly controllable on [0,^_𝑡] if every point in 𝑋 can be reached from the origin at time ^_𝑡, that is, if ran(𝐵^_𝑡)=𝑋, which is equivalent to𝑥₁=^_𝑡₀𝑆(^_𝑡−𝑠)𝐵𝑢(𝑠)d𝑠,0=−𝑥₁+^_𝑡₀𝑆(^_𝑡−𝑠)𝐵𝑢(𝑠)d𝑠,or0=𝑆(^_𝑡)𝑥₀+^_𝑡₀𝑆(^_𝑡−𝑠)𝐵𝑢(𝑠)d𝑠.(2.21) In other words,ran(𝐵^_𝑡)=𝑋⇔0∈{𝑆(^_𝑡)𝑥₀+^_𝑡₀𝑆(^_𝑡−𝑠)𝐵𝑢(𝑠)d𝑠∶𝑢isacontrolin𝐿_∞([0,^_𝑡];𝑈)}.(2.22) The set Ω^__𝑡={𝑢∈𝐿_∞([0,𝑡];𝑈)∶𝑢∈Ωa.e.} is called the set of admissible controls of (2.18), while the set 𝐴^__𝑡(𝑥₀)={𝑆(^_𝑡)𝑥₀+∫^_𝑡₀𝑆(^_𝑡−𝑠)𝐵𝑢(𝑠)d𝑠∶𝑢∈Ω^__𝑡} is called the set of accessible points of (2.18). Therefore, the system (2.18) is controllable if 0∈𝐴^__𝑡(𝑥₀), for each ^_𝑡>0.

We will make use of the following theorem, which will be applied to problem (2.1),(2.4).

Theorem 2.2 (Bárcenas and Diestel [5]). Let X and U be Banach spaces. Let 𝐵∶𝑈→𝑋 be a bounded linear operator and 𝐴∶𝑋→𝑋 the infinitesimal generator of a 𝐶₀-semigroup (𝑆_𝑡)_𝑡≥0 on 𝑋 whose dual semigroup is strongly continuous on (0,+∞). Suppose Ω is nonempty separable weakly compact convex subset of 𝑈 containing 0. Then, for each ^_𝑡>0, 0∈𝐴^__𝑡(𝑥₀) if and only if for each 𝑥^∗∈𝑋^∗, 𝑥^∗(𝑆(^_𝑡)𝑥₀)+^_𝑡₀max_𝑣∈Ω𝑥^∗(𝑆(𝑡)𝐵𝑣)d𝑡≥0.(2.23)

The Bárcenas-Diestel Theorem is an important and recent achievement on exact controllability. Throughout the literature on optimal control in Banach Spaces, hypotheses like “separable and reflexive” are frequently encountered. Using techniques from Banach space theory and the theory of vector measures, the authors show how to remove the hypothesis of reflexivity (thus giving considerably greater generality to the resulting conclusions) and translate the question of accessibility of controls to a problem in semigroups of operators, namely, given a 𝑐₀-semigroup (𝑆(𝑡))_𝑡≥0 of operators on a Banach space 𝑋, under what conditions is the dual semigroup strongly continuous on (0,∞)?

It should be noted that, for each fixed 𝑡≥0 and each 𝑥^∗∈𝑋^∗, a bounded linear functional 𝑢^∗∈𝑈^∗ is defined by means of 𝑢^∗(𝑣)∶=𝑥^∗(𝑆(𝑡)𝐵𝑣). The maximum in Theorem 2.2 exists as a consequence of a now classical result of James [9], stating that a weakly closed subset 𝐶 of a Banach space 𝑍 is weakly compact if and only if each continuous linear functional on 𝑍 attains a maximum on 𝐶.

On the other hand, the following result is proven in [10]: for each 𝑥^∗∈𝑋^∗, the mapping of [0,𝜏] to [0,∞) that takes 𝑡∈[0,𝜏] to max_𝑣∈Ω𝑥^∗(𝑆(𝑡)𝐵𝑣) is continuous (see also [5]), and so the integral in Theorem 2.2 exists in the common Riemann sense.

3. The Cauchy Problem

We will formulate the problem (2.1),(2.4) in the form (2.18). One should observe that we are working on a topological space which is not a Banach space. Let us consider the space ℂ^𝑛×𝑀, endowed with the product topology, where 𝑀 is defined in (2.8), and let 𝑁 be the closed subspace of all pairs (𝑟,𝑓) in ℂ^𝑛×𝑀 such that 𝑓(0)=𝑟. On 𝑁 we define the following map:𝑆(𝑡)==𝜑(0)𝜑(⋅)𝑥(𝑡)𝑥(𝑡+𝜃)^𝑚𝑖=0𝑐_1𝑖𝜑^(𝑖)(𝑡−𝑚)+^𝑚_𝑖=0𝑐_2𝑖𝜑^(𝑖)(𝑡−𝑚−1)𝑥(𝑡+𝜃)(3.1) for each 𝑡≥0 and 𝜃∈[−1,1], where 𝑥(𝑡) is the solution (2.5) of the initial value problem (2.1),(2.4).

Now, one has the following result.

Theorem 3.1. The operator 𝑆 satisfies(i)𝑆(𝑡)∈𝐿(𝑁) for all 𝑡≥0;(ii)𝑆(𝑡) is a 𝐶₀-semigroup in 𝑁.

Proof. (i) The linearity of 𝑆(𝑡) follows from the linearity of its components. In order to see the continuity for each 𝑡≥0, let us bear in mind that we are working with the product topology. The continuity and convergence for this topology are coordinatewise. We say that (𝑓(0),𝑓) is close or tends to (𝑔(0),𝑔) if ‖𝑓(0)−𝑔(0)‖_ℂ𝑛 is close to 0 and ‖𝑓−𝑔‖_𝑘 is close to 0, for each 𝑘, (𝑘=0,1,2,…), or, equivalently, if for each 𝑘‖𝑓(0)−𝑔(0)‖_ℂ𝑛+‖𝑓−𝑔‖_𝑘(3.2) is close to 0, where ‖⋅‖_𝑘 is defined as above. In the case of the linear map 𝑆(𝑡), the second coordinate is the continuous semigroup (2.11), and so, if 𝑐₁=max_{1≤𝑖≤𝑚}|𝑐_1𝑖| and 𝑐₂=max_{1≤𝑖≤𝑚}|𝑐_2𝑖|, we have, for each 𝑘‖𝑥(𝑡)‖_ℂ^𝑛+‖𝑥(𝑡+𝜃)‖_𝑘≤^𝑚_𝑖=0𝑐₁‖𝜑^(𝑖)(𝑡−𝑚)‖_ℂ𝑛+^𝑚_𝑖=0𝑐₂‖𝜑^(𝑖)(𝑡−𝑚−1)‖_ℂ𝑛+𝑐^𝑛𝑘_𝑖=0‖𝜑(⋅)‖_𝑖≤(𝑐₁+𝑐₂)‖𝜑(⋅)‖_𝑚+𝑐^𝑛𝑘_𝑖=0‖𝜑(⋅)‖_𝑖.(3.3) If (𝜑(0),𝜑(⋅)) is close to (0,0), then, in particular, ‖𝜑(⋅)‖_𝑘 is close to 0, for each 𝑘, and thus, by the previous estimate, 𝑆(𝑡)(^𝜑(0)_𝜑(⋅)) is close to (0,0). Being 𝑆(𝑡) linear, this is enough to prove that it is continuous.
(ii) Now we will check the semigroup properties. Obviously, 𝑆(0)=𝐼. To prove 𝑆(𝑡+𝑠)=𝑆(𝑡)𝑆(𝑠), one defines the function ℎ(𝑡)=𝑥(𝑡+𝑠), where 𝑥(⋅) is the solution of (2.1),(2.4). Therefore, ℎ(𝑡) satisfies ℎ̇(𝑡)=ℎ(𝑡−1)+ℎ(𝑡+1),0≤𝑡,ℎ(0)=𝑥(𝑠),ℎ(𝜃)=𝜑(𝑠+𝜃),𝜃∈[−1,1].(3.4) By the definition of 𝑆, one has ℎ(𝑡)ℎ(𝑡+⋅)=𝑆(𝑡)𝑥(𝑠)𝑥(𝑠+⋅)=𝑆(𝑡)𝑆(𝑠)𝜑(𝑠)𝜑(⋅).(3.5) On the other hand, 𝑆(𝑡+𝑠)==𝜑(0)𝜑(⋅)𝑥(𝑡+𝑠)𝑥(𝑡+𝑠+⋅)ℎ(𝑡)ℎ(𝑡+⋅)=𝑆(𝑡)𝑆(𝑠)𝜑(0)𝜑(⋅).(3.6) According to Theorem 2.1, we have that lim_𝑡→0⁺𝑥(𝑡+𝑠)=𝑥(𝑠); since 𝜑 is continuous, then lim_𝑡→0⁺𝑥(𝑡)=𝜑(0). Consequently, lim_𝑡→0⁺𝑆(𝑡)=𝐼 and 𝑆(𝑡) is a 𝐶₀-semigroup in 𝑁.

Lemma 3.2. Let 𝐴 be the infinitesimal generator associated to the semigroup 𝑆(𝑡). For 𝛼∈ℝ sufficiently large, the resolvent is given by (𝛼𝐼−𝐴)⁻¹=𝜑(0)𝜑(⋅)𝑔(0)𝑔(⋅),(3.7) where
(i)𝑔(𝜃)=𝑒^𝛼𝜃𝑔(0)−∫^𝜃₀𝑒^{𝛼(𝜃−𝑠)}𝜑(𝑠)d𝑠,𝜃∈[−1,1];(ii)𝑔(0)=(Δ(𝛼))⁻¹[𝜑(0)+∫⁰₋₁𝑒^{−𝛼(𝜃+1)}𝜑(𝜃)d𝜃−∫¹₀𝑒^{−𝛼(𝜃−1)}𝜑(𝜃)d𝜃] withΔ(𝜆)=(𝜆−𝑒^𝜆−𝑒^−𝜆)𝐼, for𝜆∈𝐶.Further, 𝑔 satisfies the following relation:(iii)𝛼𝑔(0)=𝜑(0)+𝑔(−1)+𝑔(1).

Proof. (i) From [8, Lemma 2.1.11], for 𝛼≥𝜔₀ ( 𝜔₀ is the growth bound of the semigroup), one has (𝛼𝐼−𝐴)⁻¹𝜑(0)𝜑(⋅)=₀^+∞𝑒^−𝛼𝑡𝑆(𝑡)𝜑(0)𝜑(⋅)d𝑡=₀^+∞𝑒^−𝛼𝑡𝑥(𝑡)𝑥(𝑡+⋅)d𝑡.(3.8) One defines for 𝜃∈[−1,1]𝑔(𝜃)=₀^+∞𝑒^−𝛼𝑡𝑥(𝑡+𝜃)d𝑡=_𝜃^+∞𝑒^{−𝛼(𝑠−𝜃)}𝑥(𝑠)d𝑠.(3.9) It is observed that 𝑔(𝜃) is a solution of the differential equation d𝑔d𝜃(𝜃)=𝛼𝑔(𝜃)−𝑥(𝜃).(3.10) The variation of constants formula for (3.10) shows that 𝑔(𝜃) is equal to (i).
(iii) Since, as it was defined above, 𝑔(𝜃)=₀^+∞𝑒^−𝛼𝑡𝑥(𝑡+𝜃)d𝑡,𝜃∈[−1,1],(3.11) one has after an integration by parts, 𝛼𝑔(0)=𝛼₀^+∞𝑒^−𝛼𝑡𝑥(𝑡)d𝑡=𝜑(0)+₀^+∞𝑒^−𝛼𝑡(𝑥(𝑡−1)+𝑥(𝑡+1))d𝑡=𝜑(0)+𝑔(−1)+𝑔(1)(3.12) which is (iii)
(ii) On the other hand, 𝛼𝑔(0)=𝜑(0)+₀⁻¹𝑒^−𝛼𝑡𝑥(𝑡+1)d𝑡+^+∞₋₁𝑒^−𝛼𝑡𝑥(𝑡+1)d𝑡+¹₀𝑒^−𝛼𝑡𝑥(𝑡−1)d𝑡+₁^+∞𝑒^−𝛼𝑡𝑥(𝑡−1)d𝑡=𝜑(0)+^+∞₋₁𝑒^−𝛼𝑡𝑥(𝑡+1)d𝑡+₁^+∞𝑒^−𝛼𝑡𝑥(𝑡−1)d𝑡+¹₀𝑒^−𝛼𝑡𝜑(𝑡−1)d𝑡−⁰₋₁𝑒^−𝛼𝑡𝜑(𝑡+1)d𝑡=𝜑(0)+𝑒^𝛼𝑔(0)+𝑒^−𝛼𝑔(0)+¹₀𝑒^−𝛼𝑡𝜑(𝑡−1)d𝑡−⁰₋₁𝑒^−𝛼𝑡𝜑(𝑡+1)d𝑡.(3.13) Therefore, [(𝛼−(𝑒^𝛼+𝑒^−𝛼))𝐼]𝑔(0)=𝜑(0)+⁰₋₁𝑒^{−𝛼(𝜃+1)}𝜑(𝜃)d𝜃−¹₀𝑒^{−𝛼(𝜃−1)}𝜑(𝜃)d𝜃(3.14) which is (ii) for 𝛼 sufficiently large.

Theorem 3.3. Let 𝑆(𝑡) be the semigroup defined by (3.1). Then its infinitesimal generator is given by 𝐴=𝜑(0)=𝑟𝜑(⋅)𝜑(−1)+𝜑(1)d𝜑d𝜃(⋅),(3.15) where the domain of 𝐴 is 𝑟𝐷(𝐴)={𝜑(⋅)∈𝑁∶𝜑isabsolutelycontinuous,andd𝜑d𝜃∈𝑀}.(3.16)

Proof. Consider the operator 𝐴 defined by 𝐴𝑟=𝜑(⋅)𝜑(−1)+𝜑(1)d𝜑d𝜃(⋅)(3.17) with 𝑟𝐷(𝐴)={𝜑(⋅)∈𝑁∶𝜑isabsolutelycontinuous,andd𝜑d𝜃∈𝑀}.(3.18) We will prove that 𝐴=𝐴. Suppose 𝛼 is a sufficiently large number such that Lemma 3.2 is true. In this case, for (^𝑟_𝜑(⋅))∈𝑁, one can write (𝛼𝐼−𝐴)(𝛼𝐼−𝐴)⁻¹𝑟=𝜑(⋅)=(𝛼𝐼−𝐴)𝑔(0)𝑔(⋅)𝛼𝑔(0)−𝑔(1)−𝑔(−1)𝛼𝑔(⋅)−d𝑔=𝑟d𝜃(⋅)𝜑(⋅)(3.19) by differentiating (i) from Lemma 3.2.
Now, let us see that (𝛼𝐼−𝐴)(𝛼𝐼−𝐴)⁻¹𝑟=𝑟𝜑(⋅)𝜑(⋅).(3.20) To do this, one takes (^𝑟_𝜑(⋅))∈𝐷(𝐴) and defines 𝑟₁𝜑₁(⋅)∶=(𝛼𝐼−𝐴)⁻¹𝑟(𝛼𝐼−𝐴)𝜑(⋅).(3.21) We will show that (𝛼𝐼−𝐴) is injective.
Suppose that there exists (^𝑟0_𝜑0(⋅))∈𝐷(𝐴) such that 00𝑟=(𝛼𝐼−𝐴)⁰𝜑₀𝜑(⋅)=(𝛼𝐼−𝐴)⁰𝜑(0)₀=(⋅)𝛼𝜑₀(0)−𝜑₀(−1)−𝜑₀(1)𝛼𝜑₀(⋅)−d𝜑₀d𝜃(⋅)(3.22) which implies 𝛼𝜑₀(0)−𝜑₀(−1)−𝜑₀(1)=𝛼𝜑₀(0)−𝜑₀(0)𝑒^𝛼−𝜑₀(0)𝑒^−𝛼=(𝛼−𝑒^𝛼−𝑒^−𝛼)𝜑₀𝜑(0)=0,₀(𝜃)=𝜑₀(0)𝑒^𝛼𝜃.(3.23) As 𝛼−𝑒^𝛼−𝑒^−𝛼 is different from zero, then 𝜑₀(0)=𝑟₀=0 and so 𝜑₀(⋅)=𝜑₀(0)𝑒^𝛼(⋅)=0; thus (𝛼𝐼−𝐴) is injective. Consequently, 𝑟(𝛼𝐼−𝐴)¹𝜑₁(⋅)=(𝛼𝐼−𝐴)(𝛼𝐼−𝐴)⁻¹𝑟𝑟[(𝛼𝐼−𝐴)𝜑(⋅)]=(𝛼𝐼−𝐴)𝜑(⋅)(3.24) which implies that 𝑟₁𝜑₁=𝑟(⋅)𝜑(⋅).(3.25) In other words, for every (^𝑟_𝜑(⋅))∈𝐷(𝐴), one has (𝛼𝐼−𝐴)⁻¹𝑟=𝑟(𝛼𝐼−𝐴)𝜑(⋅)𝜑(⋅),(3.26) that it is equivalent to 𝑟𝑟(𝛼𝐼−𝐴)𝜑(⋅)=(𝛼𝐼−𝐴)𝜑(⋅).(3.27) Then 𝐴=𝐴, as we wanted to see.

For each 𝑚∈ℕ, let 𝑀_𝑚 be the space 𝑀 defined in (2.8) provided with the topology of 𝐶^𝑚([−1,1],ℂ^𝑛), as a closed subset of 𝐶^𝑚([−1,1],ℂ^𝑛). 𝑀_𝑚 is a Banach space. Therefore, ℂ^𝑛×𝑀_𝑚, for all 𝑚∈ℕ, is also a Banach space with the norm ‖𝑟𝑓‖_{ℂ𝑛×𝑀𝑚}=‖𝑟‖_ℂ^𝑛+‖𝑓‖_{𝐶𝑚([−1,1],ℂ𝑛)}.(3.28) Now, let 𝑁_𝑚 be the closed subspace of ℂ^𝑛×𝑀_𝑚 of all pairs (𝑟,𝑓) such that 𝑟=𝑓(0).

The semigroup 𝑆^𝑚(𝑡)∶𝑁_𝑚→𝑁_𝑚, defined in the same form as 𝑆(𝑡), is now a semigroup defined on a Banach space for all 𝑚∈ℕ.

Lemma 3.4. If 𝐴^𝑚 denote the infinitesimal generator of 𝑆^𝑚(𝑡), then 𝐴^𝑚=𝐴 for all 𝑚∈ℕ, where 𝐴 is the infinitesimal generator of 𝑆(𝑡).

Proof. lim_𝑡→01𝑡‖𝑆^𝑚(𝑡)−‖𝜑(0)𝜑(⋅)𝜑(0)𝜑(⋅)−𝑡𝐴𝜑(0)𝜑(⋅)_{ℂ𝑛×𝑀𝑚}=lim_𝑡→01𝑡‖−−𝑡𝑥(𝑡)𝑥(𝑡+𝜃)𝜑(0)𝜑(⋅)𝑡(𝜑(−1)+𝜑(1))d𝜑‖d𝜃(⋅)_{ℂ𝑛×𝑀𝑚}=lim_𝑡→0[1𝑡‖𝑥(𝑡)−𝜑(0)−𝑡(𝜑(−1)+𝜑(1))‖_ℂ𝑛+1𝑡‖𝑥(𝑡+𝜃)−𝜑(⋅)−𝑡d𝜑d𝜃(⋅)‖_𝑀𝑚]≤lim_𝑡→0[1𝑡‖𝑥(𝑡)−𝜑(0)−𝑡(𝜑(−1)+𝜑(1))‖_ℂ𝑛+1𝑡^𝑚_𝑘=0𝑃_𝑘(𝑥(𝑡+𝜃)−𝜑(⋅)−𝑡d𝜑d𝜃(⋅))]=0.(3.29)

Theorem 3.5. Let 𝐴 be the infinitesimal generator of 𝑆(𝑡).
The spectrum of 𝐴 is discrete and it is defined by 𝜎(𝐴)={𝜆∈ℂ∶det(Δ𝜆)=0},(3.30) where Δ𝜆=(𝜆−𝑒^𝜆−𝑒^−𝜆)𝐼 for all 𝜆∈ℂ, and the multiplicity of each eigenvalue is finite. For every 𝛿∈ℝ, there exist only a finite number of eigenvalues in 𝐶⁺_𝛿={𝑠∈ℂ∶Re(𝑠)>𝛿}. If 𝜆∈𝜎(𝐴), then (^𝑟_𝑒^𝜆⋅_𝑟), where 𝑟≠0 satisfies Δ𝜆𝑟=0, is an eigenvector of 𝐴 with eigenvalue 𝜆. On the other hand, if Φ is an eigenvector of 𝐴 with eigenvalue 𝜆, then Φ=(^𝑟_𝑒^𝜆⋅_𝑟) with Δ𝜆𝑟=0.

Proof. According to the previous Lemma 3.2, for 𝛼∈ℝ sufficiently large, one has (𝛼𝐼−𝐴)⁻¹𝑟=𝜑(⋅)𝑔(0)𝑔(⋅),(3.31) where 𝑔(𝜃) and 𝑔(0) are as (i) and (ii), respectively in Lemma 3.2. Denote by 𝑄_𝜆 the extension of (3.31) to ℂ, that is, 𝑄_𝜆𝑟𝜑(⋅)∶=𝑔(0)𝑔(⋅).(3.32) A simple calculation shows that if 𝜆∈ℂ satisfies det(Δ𝜆)≠0, then 𝑄_𝜆 is a bonded linear operator from 𝑁 to 𝑁. Furthermore, for these 𝜆 we have (𝜆𝐼−𝐴)𝑄_𝜆=𝐼 and (𝜆𝐼−𝐴) is injective. Therefore, 𝑄_𝜆=(𝜆𝐼−𝐴)⁻¹=𝜚(𝐴), the resolvent operator of 𝐴.
Since {𝜆∈ℂ∶det(Δ𝜆)≠0}⊆𝜚(𝐴), then 𝜎(𝐴)⊂{𝜆∈ℂ∶det(Δ𝜆)=0}.
On the other hand, if det(Δ𝜆)=0, then there exists a 𝜉∈ℂ^𝑛 such that (𝜆𝐼−𝑒^−𝜆−𝑒^𝜆)(𝜉)=0.
The following element of 𝑁, 𝑧₀=(^𝜉_{𝑒𝜆⋅𝜉}) belongs to 𝐷(𝐴) and (𝜆𝐼−𝐴)𝑧₀=𝜆𝜉−𝑒^−𝜆𝜉−𝑒^𝜆𝜉𝜆𝑒^𝜆𝜃d𝜉−𝑒d𝜃^𝜆𝜃𝜉=00.(3.33) Then {𝜆∈ℂ∶det(Δ𝜆)=0}⊂𝜎(𝐴).
Let 𝜆 be an element in 𝐶⁺_𝛿 with 𝑒^𝛿+𝑒^−𝛿<|𝜆|.
For this 𝜆, one has |𝑒^𝜆+𝑒^−𝜆|≤𝑒^𝛿+𝑒^−𝛿<|𝜆|, and from [8, Corollary A.4.10], one concludes that 𝜆𝐼−𝑒^𝜆−𝑒^−𝜆 is invertible in 𝑁_𝑚, where 𝑁_𝑚 is defined as before. Thus det(Δ𝜆)≠0 and 𝜆∈𝜌(𝐴). Since det(𝜆𝐼−𝑒^𝜆−𝑒^−𝜆) is an entire function, it has finitely many zeros in the compact set 𝐶⁺_𝛿∩{𝜆∈ℂ∶|𝜆|≤𝑒^𝛿+𝑒^−𝛿} (see in [8, Theorem A.1.4.6b]) and we have showed that in the rest of 𝐶⁺_𝛿 there are none. Therefore, there are only a finite number of eigenvalues in 𝐶⁺_𝛿.
Let Φ=(^𝑟_𝜑(⋅)) be an eigenvector of 𝐴 with eigenvalue 𝜆. From the definition of 𝐴, one obtains (d𝜑/d𝜃)(𝜃)=𝜆𝜑(𝜃) for 𝜃∈[−1,1] which gives 𝜑(𝜃)=𝑒^𝜆𝜃𝜑(0).
Since Φ∈𝐷(𝐴), one has 𝜑(0)=𝑟. Using the first equation of the definition of 𝐴, 𝑒^𝜆+𝑒^−𝜆𝑟=𝜆𝑟 which shows that Δ𝜆𝑟=0. The other implication is obvious.
Finally, we will show that the multiplicity of each eigenvalue is finite. From Lemma 3.2, one has (𝛼𝐼−𝐴)⁻¹=𝑒𝜑(0)𝜑(⋅)𝑔(0)_𝛼+0𝑔(0)−^⋅₀𝑒^{𝛼(⋅−𝑠)}𝜑(𝑠)d𝑠,(3.34) where 𝑔(0) is given in (ii) of Lemma 3.2. We deduce from this expression that the resolvent operator, as an operator from 𝑁_𝑚 to 𝑁_𝑚, is the sum of an operator with finite range and an integral operator. The first operator is compact (see [8, Lemma A.3.22a]), and so is the second one, as we will see. Therefore, (𝛼𝐼−𝐴)⁻¹ is compact as an operator of 𝐿(𝑁_𝑚), for all𝑚∈ℕ. From [8, Theorem A.4.18 and Lemma A.4.19], one obtains that the eigenvalue's multiplicity is finite for (𝛼𝐼−𝐴) and 𝐴.
In order to prove the compactness of the second operator, let us observe that in [11, Example 1, page 277], it is seen that the operator 𝑇𝜑(𝑎)=^𝑎₀𝑒^{𝛼(𝑎−𝑠)}𝜑(𝑠)d𝑠(3.35) is compact in 𝐶[0,𝑎]. Let us suppose that 𝑎≤1 and let ℬ be the closed unit ball in 𝐶^𝑚[−1,1]; then ℬ∩𝑀 is the unit ball of 𝑀_𝑚 and it is valid that 𝑇(ℬ∩𝑀)⊆𝑇(ℬ) in the topology of 𝐶^𝑚[−1,1].
Therefore, it is enough to prove that 𝑇(ℬ) is compact in 𝐶^𝑚[−1,1]. Let ℬ^′ be the closed unit ball of 𝐶[0,𝑎]; then ℬ⊆ℬ^′ which implies 𝑇(ℬ)⊆𝑇(ℬ^′). If 𝒲 is the closure of 𝑇(ℬ) in 𝐶[0,𝑎], then, as 𝑇(ℬ)⊆𝑇(ℬ^′), 𝒲 is compact in 𝐶[0,𝑎].
Now, let {𝑉_𝛼}_𝛼∈𝐽 be a covering consisting of open balls in 𝐶^𝑚[−1,1] of 𝒲. Each 𝑉_𝛼 is contained in an open ball ℬ_𝛼 of 𝐶[0,𝑎]. Therefore, {ℬ_𝛼}_𝛼∈𝐽 is an open covering of 𝒲 in 𝐶[0,𝑎], hence there is a finite open covering {ℬ_𝛼𝑖}^𝑘_𝑖=1. At last, {𝑉_𝛼𝑖∶ℬ_𝛼𝑖⊆𝑉_𝛼𝑖}^𝑘_𝑖=1 is a covering of 𝒲 and so 𝒲 is compact in 𝐶^𝑚[−1,1].
Since 𝑇(ℬ)^{𝐶𝑚[−1,1]}⊆𝑇(ℬ)^𝐶[0,𝑎]=𝒲, 𝑇(ℬ) is compact in 𝐶^𝑚[−1,1].

4. Controllability

Let 𝑈 be a Banach space. One will consider the following linear system:𝑥^′(𝑡)=𝐴𝑥(𝑡)+𝐵𝑢(𝑡),𝑥(0)=𝑥₀,𝑥₀∈𝐷(𝐴),𝑥(𝜃)=𝜑(𝜃),𝜃∈[−1,1],(4.1) where 𝐴∶𝐷(𝐴)⊆𝑁→𝑁 is the infinitesimal generator of the semigroup (𝑆(𝑡))_𝑡≥0, 𝐵∶𝑈→𝑁 is a bonded linear operator, 𝑢∶[0,∞)→𝑈 is a strongly measurable, essentially bounded function.

In this section, we will study the controllability of the system (4.1). The mild solution of (4.1) is given by 𝑥(𝑡)=𝑆(𝑡)𝑥₀+^𝑡₀𝑆(𝑡−𝑠)𝐵𝑢(𝑠)d𝑠.(4.2) Further, we will suppose that Ω is a separable weakly compact convex subset of 𝑈.

For 𝜏>0, the set Ω_𝜏={𝑢∈𝐿^∞_𝑈[0,𝜏]∶𝑢∈Ωa.e.}(4.3) is called the set of admissible controls of (4.1), while the set 𝐴_𝜏(𝑥₀)={𝑆(𝜏)𝑥₀+^𝜏₀𝑆(𝜏−𝑠)𝐵𝑢(𝑠)d𝑠∶𝑢∈Ω_𝜏}(4.4) is called the set of accessible points of (4.1).

The system (4.1) is controllable if 0∈𝐴_𝜏(𝑥₀), for each 𝜏>0.

Let 𝑥₀,𝜑,𝑈,𝑢,𝐵 be as (4.1). Consider for each 𝑚∈ℕ, the system𝑥^′(𝑡)=𝐴^𝑚𝑥(𝑡)+𝐵𝑢(𝑡),𝑥(0)=𝑥₀,𝑥₀∈𝐷(𝐴),𝑥(𝜃)=𝜑(𝜃),𝜃∈[−1,1],(4.5) where 𝐵∶𝑈→𝑁_𝑚.

Note that, in this case, 𝐵 is also a bounded operator. In fact, since for each 𝑢∈𝑈, 𝐵𝑢∈𝑁_𝑚, we can write 𝐵𝑢=(𝐵₁𝑢,𝐵₂𝑢), where 𝐵₁∶𝑈→ℂ^𝑛 and 𝐵₂∶𝑈→𝑀_𝑚 are bounded linear operators. If {𝑢_𝑗}^∞_𝑗=0⊆𝑈 is a convergent sequence to 𝑢, then ‖𝐵𝑢_𝑗−𝐵𝑢‖_{ℂ𝑛×𝑀𝑚}𝐵=‖¹𝑢_𝑗𝐵₂𝑢_𝑗−𝐵₁𝑢𝐵₂𝑢‖_{ℂ𝑛×𝑀𝑚}𝐵=‖¹𝑢_𝑗−𝐵₁𝑢𝐵₂𝑢_𝑗−𝐵₂𝑢‖_{ℂ𝑛×𝑀𝑚}=‖𝐵₁𝑢_𝑗−𝐵₁𝑢‖_ℂ𝑛+‖𝐵₂𝑢_𝑗−𝐵₂𝑢‖_𝑀𝑚≤‖𝐵₁𝑢_𝑗−𝐵₁𝑢‖_ℂ𝑛+𝑃_𝑚(𝐵₂𝑢_𝑗−𝐵₂𝑢)(4.6) which converges to zero as 𝑗 tends to +∞.

Let us consider Ω and Ω_𝜏 as before.

The system (4.5) is controllable if 0∈𝐴^𝑚_𝜏(𝑥₀), for each 𝜏>0, where 𝐴^𝑚_𝜏(𝑥₀)={𝑆^𝑚(𝜏)𝑥₀+^𝜏₀𝑆^𝑚(𝜏−𝑠)𝐵𝑢(𝑠)d𝑠∶𝑢∈Ω_𝜏}.(4.7)

Lemma 4.1. 𝐴_𝜏𝑥₀=⋂^∞_𝑚=1𝐴^𝑚_𝜏𝑥₀.

Proof. If 𝑧∈𝐴_𝜏𝑥₀, there exist 𝑢∈Ω_𝜏 such that 𝑧=𝑆(𝜏)𝑥₀+∫^𝜏₀𝑆(𝜏−𝑠)𝐵𝑢(𝑠)d𝑠. Since 𝑆(𝜏)𝑥₀=𝑆^𝑚(𝜏)𝑥₀ and 𝑆(𝜏−𝑠)𝐵𝑢(𝑠)=𝑆^𝑚(𝜏−𝑠)𝐵𝑢(𝑠), for all𝑚∈ℕ, then 𝑧∈⋂^∞_𝑚=1𝐴^𝑚_𝜏𝑥₀. If 𝑧∈⋂^∞_𝑚=1𝐴^𝑚_𝜏𝑥₀, then there exists, for every 𝑚∈ℕ, 𝑢_𝑚∈Ω_𝜏, such that 𝑧=𝑆^𝑚(𝜏)𝑥₀+∫^𝜏₀𝑆^𝑚(𝜏−𝑠)𝐵𝑢_𝑚(𝑠)d𝑠. In particular, for 𝑚=1, 𝑧=𝑆¹(𝜏)𝑥₀+∫^𝜏₀𝑆¹(𝜏−𝑠)𝐵𝑢₁(𝑠)d𝑠 and as 𝑆(𝜏)𝑥₀=𝑆¹(𝜏)𝑥₀ and 𝑆(𝜏−𝑠)𝐵𝑢₁(𝑠)=𝑆¹(𝜏−𝑠)𝐵𝑢₁(𝑠), then 𝑧=𝑆¹(𝜏)𝑥₀+∫^𝜏₀𝑆¹(𝜏−𝑠)𝐵𝑢₁(𝑠)d𝑠, that is, 𝑧∈𝐴_𝜏𝑥₀.

Before stating the main result on controllability, we need to prove the following lemma.

Lemma 4.2. The map 𝑇_𝑡∶𝑀_𝑚→𝑀_𝑚, defined by 𝑇_𝑡(𝑢)(𝜃)=𝑢(𝑡+𝜃), is compact.

Proof. To prove this lemma, we will follow the next five steps.
(1) The Kondrasov's Theorem [12, 13] gives that the following canonical injection is compact 𝐻^𝑚(−1,1)↪^𝑐𝐶⁰([−1,1])if𝑚∈ℕ,(4.8) where 𝐻^𝑚(−1,1)={𝑓∈𝐿²(−1,1)∶𝑓^(𝑘)∈𝐿²(−1,1),𝑘=1,…,𝑚}(4.9) with 𝑓^(𝑘) being the derivative of order 𝑘 in the sense of distributions. 𝐻^𝑚(−1,1) is endowed with the norm ‖𝑓‖²_𝑚=^𝑚_𝑘=0¹₋₁|𝑓^(𝑘)(𝑡)|²d𝑡.(4.10) This implies that 𝐻^𝑚[−1,1] is a Banach space.
(2) Using (1), for 𝑚=1, 𝐻¹[−1,1]↪^𝑐𝐶⁰([−1,1]). Thus, for 𝑚≥1, if 𝑓∈𝐻^𝑚+1(−1,1), then 𝑓^(𝑚)∈𝐻¹(−1,1). Therefore, 𝐻^𝑚+1(−1,1)↪^𝑐𝐶^𝑚([−1,1]).
(3) By (2) and the definition of 𝑀_𝑚+1, it follows that 𝑀_𝑚+1⊆𝐶^𝑚+1([−1,1])⊆𝐻^𝑚+1(−1,1)↪^𝑐𝐶^𝑚([−1,1]).(4.11) Hence, for every bounded sequence {𝑓_𝑘}_𝑘∈ℕ in 𝑀_𝑚+1 with the topology of 𝐶^𝑚([−1,1]), it can be found a convergent subsequence {𝑓_𝑘𝑙}_𝑙∈ℕ⊆{𝑓_𝑘}_𝑘∈ℕ that converges to 𝑓 in 𝐶^𝑚([−1,1]); but 𝑓^(𝑚) is continuous and 𝑓^(𝑚+1)∈𝐿²(−1,1). Thus {𝑓_𝑘^(𝑚+1)_𝑙}_𝑙∈ℕ converges to 𝑓^(𝑚+1)a.e, and so this subsequence converges to 𝑓 in 𝑀_𝑚+1.
(4) To prove that 𝑇_𝑡 is compact, it must be proven that for every bounded sequence {𝑢_𝑖}_𝑖∈ℕ in 𝑀_𝑚, a subsequence {𝑢_𝑖𝑗}_𝑗∈ℕ can be found such that the sequence {𝑇_𝑡(𝑢_𝑖𝑗)}_𝑗∈ℕ converges in 𝑀_𝑚. As 𝑇_𝑡(𝑢_𝑖)(𝜃)=𝑢_𝑖(𝑡+𝜃), then the previous fact is equivalent to obtain a convergent subsequence {𝑢_𝑖𝑗}_𝑗∈ℕ from {𝑢_𝑖}_𝑖∈ℕ in 𝑀_𝑚.
(5) Using (3) and the conclusion of (4), it is obtained that for all 𝑡∈ℝ, 𝑇_𝑡 is compact as a bounded linear operator from 𝑀_𝑚 to 𝑀_𝑚.

Theorem 4.3. 0∈𝐴_𝜏𝑥₀ for each 𝜏>0 (i.e., the system (4.1) is controllable) if and only if 𝑥^∗(𝑆^𝑚(𝜏)𝑥₀)+^𝜏₀max_𝑢∈Ω𝑥^∗(𝑆^𝑚(𝑡)𝐵𝑢)d𝑡≥0(4.12) for every 𝑚∈ℕ and each 𝑥^∗∈(𝑁_𝑚)^∗.

Proof. First, we will prove that𝑆^𝑚(𝑡), for all 𝑡≥0, is compact, for all 𝑚∈ℕ; in fact, 𝑆^𝑚𝑟==0+0(𝑡)𝜑(⋅)𝑥(𝑡)𝑥(𝑡+𝜃)𝑥(𝑡)𝑥(𝑡+𝜃).(4.13) The first term has finite rank, therefore, it is compact. According to Lemma 4.2, the second term is compact in 𝑀_𝑚. Consequently, 𝑆^𝑚(𝑡) is compact, as an operator in 𝐿(ℂ^𝑛×𝑀_𝑚). In particular, it is compact as an operator in 𝐿(𝑁_𝑚). Hence, the adjoint (𝑆^𝑚(𝑡))^∗ is strongly continuous in (0,+∞) (see, e.g., [14]). On the other hand, in view of Lemma 4.1, 0∈𝐴_𝜏𝑥₀ if and only if 0∈𝐴^𝑚_𝜏𝑥₀, for all𝑚∈ℕ and this is true, after Theorem 1.1 of Bárcenas-Diestel [5], if and only if 𝑥^∗(𝑆^𝑚(𝜏)𝑥₀)+∫^𝜏₀max_𝑢∈Ω𝑥^∗(𝑆^𝑚(𝑡)𝐵𝑢)d𝑡≥0for all𝑚∈ℕ and each 𝑥^∗∈(𝑁_𝑚)^∗.