A New Estimation of the Growth Bound of a Periodic Evolution Family on Banach Spaces

Abstract

Let 𝒰={𝑈(𝑡,𝑠)}_{𝑡≥𝑠≥0} be a strongly continuous and 𝑞-periodic evolution family acting on a complex Banach space 𝑋. We prove that if 0<𝑝<∞,𝑥∈𝑋, and sup_𝑠≥0sup_||𝑥||≤1∫^∞₀‖𝑈(𝑡+𝑠,𝑠)𝑥‖^𝑝𝑑𝑡∶=𝐶_𝑝<∞, then the growth bound of the family 𝒰 is less than or equal to −1/(𝑝𝐶_𝑝).

1. Introduction

Knowing the rate of convergence of an iterative process, we can control the speed of its convergence. This helps us to obtain the limit of the process, with the desired accuracy, in a determined time. On the other hand, in the study of certain dynamical systems or differential equations, we meet many times 𝑋-valued functions 𝑓, defined on ℝ₊∶=[0,∞), having the property that the map 𝑡↦𝑒^−𝜔𝑡‖𝑓(𝑡)‖ is bounded on ℝ₊, for a suitable real number 𝜔. Here, and in the following, 𝑋 stands for a real or complex Banach space. Clearly, every 𝜔^′≥𝜔 has the same property. The infimum of all real numbers 𝜔 which verify the given boundedness condition is called exponential growth of the function 𝑓. In this note, we obtain an estimate of the exponential growth of a periodic evolution family that satisfies an integral condition, originally given by Datko. The theoretical result allows us to estimate the 𝐿²-norm of the solution of a periodic Cauchy problem with time-varying coefficients, which is naturally led by the one-dimensional heat equation.

The classical theorem of Datko [1] states that a strongly continuous evolution family 𝒰={𝑈(𝑡,𝑠)}_{𝑡≥𝑠≥0}, acting on a real or complex Banach space 𝑋, is uniformly exponentially stable (i.e., there are two positive constants 𝑁 and 𝜈) such that ‖𝑈(𝑡,𝑠)‖≤𝑁𝑒^{−𝜈(𝑡−𝑠)},∀𝑡≥𝑠≥0, (1) if, for some (and then for all) 1≤𝑝<∞, one has sup_𝑠≥0^∞₀‖𝑈(𝑡+𝑠,𝑠)𝑥‖^𝑝𝑑𝑡<∞,∀𝑥∈𝑋. (2) As usual, the norm of 𝑋 and of ℒ(𝑋) is denoted by ‖⋅‖. Here, ℒ(𝑋) denotes the Banach algebra of all the bounded linear operators acting on 𝑋. By 𝜎(𝑇), we denote the spectrum of the linear operator 𝑇, and when it is bounded, its spectral radius is defined by 𝑟(𝑇)∶=sup{|𝑧|∶𝑧∈𝜎(𝑇)}=lim_𝑛→∞𝑇^𝑛1/𝑛. (3) We use classical notations for the set of real numbers, complex numbers, and integer numbers. The set of all nonnegative integer numbers will be denoted by ℤ₊.

It is known that if 𝐓={𝑇(𝑡)}_𝑡≥0 is a strongly continuous semigroup on a complex Banach space 𝑋 and if 1≤𝑝<∞ and 𝐶_𝑝>0 are such that ^∞₀‖𝑇(𝑡)𝑥‖^𝑝𝑑𝑡≤𝐶_𝑝‖𝑥‖^𝑝,∀𝑥∈𝑋, (4) then 𝜔₀(𝐓)≤−1/(𝑝𝐶_𝑝). See [2], where the case of semigroups acting on Hilbert spaces is analyzed, and see [3, pages 81-82] for the general case.

In this note, we extend this result to periodic strongly continuous evolution families acting on real or complex Banach spaces. Moreover, we prove the result with 𝑝∈(0,∞) instead of 1≤𝑝<∞.

As an application of our theoretical results, we provide a simple example, which apparently cannot be treated using the results described above for semigroups.

A family 𝒰∶={𝑈(𝑡,𝑠)}_{𝑡≥𝑠≥0}⊂ℒ(𝑋) is called a strongly continuous and  𝑞-periodic (for some 𝑞≥1) evolution family if it satisfies the following.(1)𝑈(𝑡,𝑡)=𝐼 for all 𝑡≥0.(2)𝑈(𝑡,𝑟)𝑈(𝑟,𝑠)=𝑈(𝑡,𝑠) for all 𝑡≥𝑟≥𝑠≥0.(3)𝑈(𝑡+𝑞,𝑠+𝑞)=𝑈(𝑡,𝑠) for all 𝑡≥𝑠≥0.(4)The map 𝑡→𝑈(𝑡,𝑠)𝑥∶[𝑠,∞)→𝑋 is continuous for every 𝑥∈𝑋 and is called exponentially bounded if there exist 𝜔∈ℝ and 𝑀_𝜔≥1, such that ‖𝑈(𝑡,𝑠)‖≤𝑀_𝜔𝑒^{𝜔(𝑡−𝑠)}for𝑡≥𝑠≥0. (5)The growth bound 𝜔₀(𝒰) of an exponentially bounded evolution family 𝒰 is the infimum of all 𝜔∈ℝ for which there exists 𝑀_𝜔≥1 such that (5) is fulfilled. It is known [4] that 𝜔₀(𝒰)=lim_𝑡→∞ln‖𝑈(𝑡,0)‖𝑡. (6) The family 𝒰 is uniformly exponentially stable if its growth bound is negative. A 𝑞-periodic and strongly continuous evolution family 𝒰={𝑈(𝑡,𝑠)∶𝑡≥𝑠≥0} is uniformly exponentially stable if and only if 𝜔₀(𝒰) is negative or, equivalently, if the spectral radius of the monodromy operator 𝑈(𝑞,0) is less than one. The next result from the abstract theory of operators, originally given by Müller [5], will be useful in what follows.

Lemma 1. Let 𝑋 be a complex Banach space, and let 𝑉∈ℒ(𝑋). If the spectral radius of  𝑉 is greater than or equal to 1, then, for all 0<𝜀<1 and any sequence (𝑎_𝑛) with 𝑎_𝑛→0 (as 𝑛→∞) and ‖(𝑎_𝑛)‖_∞≤1, there exists a unit vector 𝑢₀∈𝑋, such that 𝑉^𝑛𝑢₀≥(1−𝜀)⋅𝑎_𝑛,∀𝑛∈ℤ₊. (7)

Moreover, this result remains valid for real Banach spaces provided that the operator 𝑉 is power bounded; that is, sup{‖𝑉^𝑛‖∶𝑛∈ℤ₊} is finite. See [6] for further details.

2. Preliminary Results

The following two lemmas will be useful in the proof of Theorem 5 below. Its proof is essentially contained in [6, Theorem 1.2]. Because it has an interest in itself, we state it as a separate statement and infer its proof for the sake of completeness.

Lemma 2. Let 𝜉∶[0,∞)→[0,1] be a nonincreasing function such that lim_𝑡→∞𝜉(𝑡)=0. For each 𝛿>0, there exists a nonincreasing function 𝜌∶[0,∞)→[0,1], having the properties (1)𝜌(𝑡)≥𝜉(𝑡),  for all 𝑡≥0,(2)lim_𝑡→∞𝜌(𝑡)=0, (3)𝜌(𝑡+𝑠)≥(1+𝛿)⁻¹𝜌(𝑡)𝜌(𝑠),  for all 𝑡,𝑠≥0.

Proof. Let 0=𝐾₀<𝐾₁<𝐾₂… such that 0≤𝜉(𝑡)≤(1+𝛿)^−𝑛, when 𝑡≥𝐾_𝑛. Next, choose the integers 0=𝑅₀<𝑅₁<𝑅₂… in such a way that 𝑅_𝑛≥𝐾_𝑛 for all 𝑛 and 𝑅_𝑛+𝑅_𝑚≤𝑅_𝑛+𝑚 for all 𝑚,𝑛≥0. For example, we can take 𝑅_𝑛=𝐾₀+𝐾₁+𝐾₂+⋯+𝐾_𝑛.
Define the function 𝜌∶[0,∞)→[0,1] by 𝜌(𝑡)∶=(1+𝛿)^−𝑛when𝑅_𝑛≤𝑡<𝑅_𝑛+1. (8) Clearly, 𝜌 is nonincreasing, 0≤𝜉(𝑡)≤𝜌(𝑡)≤1, and lim_𝑡→∞𝜌(𝑡)=0.
We claim that 𝜌(𝑡+𝑠)≥(1+𝛿)⁻¹𝜌(𝑡)𝜌(𝑠) for all 𝑡,𝑠≥0. Indeed, choose the integers 𝑗_𝑡 and 𝑗_𝑠 such that 𝑅_𝑗𝑡≤𝑡<𝑅_𝑗𝑡+1, 𝑅_𝑗𝑠≤𝑠<𝑅_𝑗𝑠+1. Then, 𝜌(𝑡)=(1+𝛿)^−𝑗𝑡 and 𝜌(𝑠)=(1+𝛿)^−𝑗𝑠. Now, 𝑡+𝑠<𝑅_𝑗𝑡+1+𝑅_𝑗𝑠+1≤𝑅_{𝑗𝑡+𝑗𝑠+2} (9) yields 𝜌(𝑡+𝑠)≥(1+𝛿)^{−𝑗𝑡−𝑗𝑠−1}=(1+𝛿)⁻¹𝜌(𝑡)𝜌(𝑠). (10)

Lemma 3. Let 𝒰={𝑈(𝑡,𝑠)}_{𝑡≥𝑠≥0} be a strongly continuous and 𝑞-periodic evolution family on a complex Banach space 𝑋 such that 𝑟(𝑈(𝑞,0))≥1. Then, for any 0<𝜖<1 and every nonincreasing function 𝜉∶[0,∞)→[0,1], with lim_𝑡→∞𝜉(𝑡)=0, there exist 𝑡₀≥0 and a unit vector 𝑥∈𝑋, such that 𝑈𝑡+𝑡₀,𝑡₀𝑥≥(1−𝜖)𝜉(𝑡)∀𝑡≥0. (11)

Proof.  
Step  1. Define 𝛽∶[0,∞)→[0,1] by 𝛽(𝑡)𝜉(0)𝜉∶=,if0≤𝑡<𝑞,𝑡−𝑞,if𝑡≥𝑞. (12) The map 𝛽(𝑡) is nonincreasing, and lim_𝑡→∞𝛽(𝑡)=0.
Since 𝑟(𝑈(𝑞,0))≥1, there exists a norm one vector 𝑥₀∈𝑋, such that 𝑈^𝑘𝑞,0𝑥₀≥12𝛽(𝑘)∀𝑘∈ℤ₊. (13) This follows by Lemma 1 with 𝑎_𝑛=𝛽(𝑛), 𝜖=1/2, and 𝑉=𝑈(𝑞,0). For 𝑡≥0, let 𝑁 be an integer number, such that 𝑁𝑞≤𝑡<𝑁𝑞+𝑞. We infer 𝑈𝑁𝑞+𝑞,0𝑥₀=𝑈𝑁𝑞+𝑞,𝑡𝑈(𝑡,0)𝑥₀≤𝑀𝑈(𝑡,0)𝑥₀, (14) where 𝑀=𝑀_𝜔𝑒^𝑞𝜔. Thus, 𝑈(𝑡,0)𝑥₀≥1𝑀𝑈𝑁𝑞+𝑞,0𝑥₀≥1𝛽≥12𝑀((𝑁+1))𝛽)𝑞=12𝑀(𝑁+1𝜉≥12𝑀𝑁𝑞𝜉(𝑡).2𝑀 (15)Step  2. Let us choose 𝛿>0 such that (1−𝛿)/(1+𝛿)≥1−𝜖. By Lemma 2, we can find a nonincreasing function 𝜌∶[0,∞)→[0,1], such that 𝜌(𝑡)≥𝜉(𝑡), lim_𝑡→∞𝜌(𝑡)=0, and 𝜌(𝑡+𝑠)≥(1+𝛿)⁻¹𝜌(𝑡)𝜌(𝑠), for all 𝑡,𝑠≥0.
In view of the first step, applied to the function 𝜌, we may choose a unit vector 𝑦₀∈𝑋, such that 𝑈(𝑡,0)𝑦₀≥1𝜌(𝑡)2𝑀∀𝑡≥0. (16) Denote 𝜂∶=inf_𝑡≥0𝑈(𝑡,0)𝑦₀𝜌(𝑡). (17) Then, 𝜂≥1/(2𝑀), and for all 𝑡≥0, we have that ‖𝑈(𝑡,0)𝑦₀‖≥𝜂𝜌(𝑡). Thus, sup_𝑡≥0𝜂𝜌(𝑡)𝑈(𝑡,0)𝑦₀=1. (18) Now, let us choose 𝑡₀≥0, such that 𝑡1−𝛿≤𝜂𝜌₀𝑈𝑡₀,0𝑦₀, (19) and let 𝑈𝑡𝑥∶=⁰,0𝑦₀𝑈𝑡₀,0𝑦₀. (20) We infer 𝑈𝑡+𝑡₀,𝑡₀𝑥=𝑈𝑡+𝑡₀,𝑡₀𝑈𝑡₀,0𝑦₀𝑈𝑡₀,0𝑦₀=𝑈𝑡+𝑡₀,0𝑦₀𝑈𝑡₀,0𝑦₀≥𝜂𝜌𝑡+𝑡₀𝑈𝑡₀,0𝑦₀≥𝜂𝜌(𝑡)𝜌𝑡₀(1+𝛿)⁻¹𝑈𝑡₀,0𝑦₀≥1−𝛿𝜉(𝑡)≥()𝜉(𝑡),1+𝛿1−𝜖 (21)
which completes the proof.

Theorem 4. Let 𝒰={𝑈(𝑡,𝑠)}_{𝑡≥𝑠≥0} be a strongly continuous and 𝑞-periodic evolution family on a complex Banach space 𝑋, and let 𝑝∈(0,∞). The following three statements are equivalent.(i)∫^∞₀‖𝑈(𝑡,0)𝑥‖^𝑝𝑑𝑡≤𝐶_𝑝<∞, for all unit vectors 𝑥∈𝑋. (ii)For every 𝑠≥0, one has sup_{‖𝑥‖≤1}^∞₀‖𝑈(𝑡+𝑠,𝑠)𝑥‖^𝑝𝑑𝑡∶=𝐶_𝑝(𝑠)<∞. (22)(iii)The growth bound, 𝜔₀(𝒰), is negative.

The equivalence of (i) and (iii) is given, for example, in [7, Theorem 3.3], and (ii) implying (i) is obvious. We have only to prove that (i) implies (ii). To do this, let 𝑠=𝑛𝑞+𝑟, with 𝑛∈ℤ₊ and 𝑟∈[0,𝑞). Let 𝜔<0 and 𝑀_𝜔≥1 such that ‖𝑈(𝑡,𝑠)‖≤𝑀_𝜔𝑒^{𝜔(𝑡−𝑠)}∀𝑡≥𝑠≥0. (23) Then, ^∞₀‖𝑈(𝑡+𝑠,𝑠)𝑥‖^𝑝=𝑑𝑡_∞0‖𝑈(𝑡+𝑟,𝑟)𝑥‖^𝑝=𝑑𝑡_0𝑞−𝑟‖𝑈(𝑡+𝑟,𝑟)𝑥‖^𝑝+𝑑𝑡_{∞𝑞−𝑟}‖𝑈(𝑡+𝑟,𝑟)𝑥‖^𝑝𝑑𝑡≤𝑞𝑀^𝑝_𝜔𝑒^𝑝𝑞𝜔+^∞_𝑞−𝑟𝑈𝑡+𝑟−𝑞,0𝑈𝑞,𝑟𝑥^𝑝𝑑𝑡=𝑞𝑀^𝑝_𝜔𝑒^𝑝𝑞𝜔+𝑈𝑞,𝑟𝑥^𝑝×^∞₀𝑈()𝑈𝑥𝑡,0𝑞,𝑟𝑈𝑞,𝑟𝑥^𝑝𝑑𝑡≤𝑞𝑀^𝑝_𝜔𝑒^𝑝𝑞𝜔+𝐶_𝑝𝑀^𝑝_𝜔𝑒^𝑝𝑞𝜔≤𝑀^𝑝_𝜔𝑒^𝑝𝑞𝜔𝑞+𝐶_𝑝∶=𝐾_𝑝. (24)

3. The Result and Its Consequences

Theorem 5. Let 𝒰={𝑈(𝑡,𝑠)}_{𝑡≥𝑠≥0} be a strongly continuous and 𝑞-periodic evolution family acting on a complex Banach space 𝑋. Suppose that the integral condition (22) is fulfilled for any 𝑠≥0 and some (and then for all) 0<𝑝<∞. Then, there exists 𝑡₀≥0 such that 𝜔₀(𝒰)≤−1/(𝑝𝐶_𝑝(𝑡₀)).

Proof. Let 𝜔=𝜔₀(𝒰). The family ℬ∶=𝐵(𝑡,𝑠)∶=𝑒^{−𝜔(𝑡−𝑠)}𝑈(𝑡,𝑠)_{𝑡≥𝑠≥0} (25) is a 𝑞-periodic evolution, family and 𝜔₀(ℬ)=0. Let 𝜉(𝑡)∶=𝑒^−𝛿𝑡 with 𝛿>0, and let 0<𝜖<1 be fixed. By applying Lemma 3 to the 𝑞-periodic evolution family ℬ, there exist 𝑡₀≥0 and a norm one vector 𝑥∈𝑋, such that 𝐵𝑡+𝑡₀,𝑡₀𝑥≥(1−𝜖)𝑒^−𝛿𝑡,∀𝑡≥0, (26) or equivalently 𝑈𝑡+𝑡₀,𝑡₀𝑥≥(1−𝜖)𝑒^{(𝜔−𝛿)𝑡},∀𝑡≥0. (27) Hence, via Theorem 4, ^∞₀(1−𝜖)^𝑝𝑒^{(𝜔−𝛿)𝑝𝑡}𝑑𝑡≤^∞₀𝑈𝑡+𝑡₀,𝑡₀𝑥^𝑝𝑑𝑡≤𝐶_𝑝𝑡₀. (28) That is, −(1−𝜖)^𝑝/(𝜔−𝛿)𝑝≤𝐶_𝑝(𝑡₀). Since 𝜖>0 and 𝛿>0 were arbitrary, the last inequality yields −1/𝑝𝜔≤𝐶_𝑝(𝑡₀) that is, 𝜔₀(𝒰)≤−1/(𝑝𝐶_𝑝(𝑡₀)).

Corollary 6. Let 𝒰={𝑈(𝑡,𝑠)}_{𝑡≥𝑠≥0} be a strongly continuous and 𝑞-periodic evolution family acting on a complex Banach space 𝑋. If 0<𝑝<∞ and 𝐶>0 are such that sup_𝑠≥0^∞₀‖𝑈(𝑡+𝑠,𝑠)𝑥‖^𝑝𝑑𝑡≤𝐶‖𝑥‖^𝑝,∀𝑥∈𝑋, (29) then 𝜔₀(𝒰)≤−1/(𝑝𝐶).

Corollary 7. Let 0<𝑝<∞ and 𝐓={𝑇(𝑡)}_𝑡≥0 be a 𝐶₀-semigroup on a Banach space X. If sup_{‖𝑥‖≤1}^∞₀‖𝑇(𝑡)𝑥‖^𝑝𝑑𝑡=𝐶_𝑝<∞, (30) then 𝜔₀(𝑇)1≤−𝑝𝐶_𝑝. (31)

Theorem 8. Let 𝒰={𝑈(𝑡,𝑠)}_{𝑡≥𝑠≥0} be a strongly continuous and 𝑞-periodic evolution family acting on a real or complex Banach space 𝑋. Assume that 𝜔₀∶=𝜔₀(𝒰) is negative, and, in addition, the family 𝒰₀∶=𝑒^{−𝜔0(𝑡−𝑠)}𝑈(𝑡,𝑠)∶𝑡≥𝑠≥0 (32) is uniformly bounded, and let 0<𝑝<∞. Then, there exists 𝑡₀≥0 such that 𝜔₀(𝒰)≤−1/(𝑝𝐶_𝑝(𝑡₀)), where 𝐶_𝑝(𝑡₀) is defined in (22).

Actually, under the addition boundedness assumption, the monodromy operator associated with the family 𝒰₀, that is, 𝑒^−𝑞𝑈(𝑞,0), is power bounded, and we can use [6, Theorem 2.1].

4. Some Examples and Remarks

Now we present two examples which illustrate the theoretical result.

Example 1. Let 𝑎∶ℝ→ℂ be a continuous and 𝑞-periodic function, and let us define 𝑈(𝑡,𝑠)∶=𝑒^{∫𝑡𝑠𝑎(𝜏)𝑑𝜏}. Clearly, the family 𝒰∶={𝑈(𝑡,𝑠)∶𝑡≥𝑠} is a norm continuous 𝑞-periodic evolution family on 𝑋∶=ℂ. If sup_𝑠∈ℝ^∞₀𝑒^{∫𝑠𝑠+𝑡Re[𝑎(𝜏)]𝑑𝜏𝑝}𝑑𝑡=𝐶_𝑝<∞, (33) Then, for every 𝜈>−1/(𝑝𝐶_𝑝) and every 𝑡≥𝑠, one has 𝑒^{∫𝑡𝑠Re[𝑎(𝜏)]𝑑𝜏}≤𝑀_𝜈𝑒^{𝜈(𝑡−𝑠)}. (34)

Example 2. Let us choose 𝐻∶=𝐿²([0,𝜋],ℂ) to be the state space. Endowed with the usual inner product and norm, it becomes a complex Hilbert space, and, in addition, the one parameter family {𝑇(𝑡)}_𝑡≥0, given by (𝑇(𝑡)𝑥𝜉)=2)(𝜋^∞_𝑛=1𝑒^−𝑡𝑛2𝑐_𝑛(𝑥)sin𝑛𝜉,𝜉∈[0,𝜋],𝑡≥0, (35) where 𝑐_𝑛(𝑥)∶=∫^𝜋₀𝑥(𝑠)sin(𝑛𝑠)𝑑𝑠, is a strongly continuous semigroup on 𝐻. The semigroup 𝐓={𝑇(𝑡)}_𝑡≥0 is generated by the linear operator 𝐴 given by 𝐴𝑥=𝑥̈, and the maximal domain of 𝐴 is the set 𝐷(𝐴) of all 𝑥∈𝐻 such that 𝑥 and 𝑥̇ are absolutely continuous, 𝑥̈∈𝐻, and 𝑥(0)=𝑥(𝜋)=0. Moreover, 𝑇(𝑡) is a self-adjoint operator for every 𝑡≥0 [8, Example 1.3, pages 178, 198]. Consider the nonautonomous Cauchy problem 𝜕𝑧(𝑡,𝜉)(𝑡)𝜕𝜕𝑡=𝑓²𝑧(𝑡,𝜉)𝜕²𝜉[]𝑧()()𝑧()(𝜉),,𝑡>0,𝜉∈0,𝜋𝑡,0=𝑧𝑡,𝜋=0,𝑡≥00,𝜉=𝑥 (36) where 𝑥(⋅) is a given function in 𝐻, and 𝑓∶ℝ₊→[1,∞) is a function having the following properties.(1)𝑓(⋅) is 𝑞-periodic for some 𝑞≥1.(2)There exist 𝛼∈(0,1] and 𝑐>0 such that 𝑓(𝑡)−𝑓(𝑠)≤𝑐|𝑡−𝑠|^𝛼∀𝑡,𝑠≥0. (37)
Let 𝐹(𝑡)=∫^𝑡₀𝑓(𝑡)𝑑𝑡. As is well known, the solution 𝑧(⋅) of the above Cauchy problem satisfies the evolution condition 𝑧(𝑡)=𝑈(𝑡,𝑠)𝑧(𝑠), (38) where 𝑈(𝑡,𝑠)=𝑇(𝐹(𝑡)−𝐹(𝑠)). See [9, Example 2.9b].
We can estimate a positive constant 𝜈 having the property that the map 𝑡↦𝑒^𝑣𝑡‖𝑧(𝑡)‖ is bounded on ℝ₊.
Indeed, ^∞₀‖𝑈(𝑡,0)𝑥‖²=2𝑑𝑡𝜋^∞₀^∞_𝑛=1𝑐²_𝑛(𝑥)𝑒^{−2𝑛2𝐹(𝑡)}=2𝑑𝑡𝜋^∞_𝑛=1𝑐²_𝑛(𝑥)^∞₀𝑒^{−2𝑛2𝐹(𝑡)}=‖𝑥‖𝑑𝑡_22∞0𝑒^{−2𝑛2𝐹(𝑡)}≤‖𝑥‖𝑑𝑡_22∞0𝑒^{−2𝐹(𝑡)}𝑑𝑡. (39) On the other hand ^∞₀𝑒^{−2𝐹(𝑡)}𝑑𝑡=^∞_𝑘=0^(𝑘+1)𝑞_𝑘𝑞𝑒^{−2𝐹(𝑡)}=𝑑𝑡_{∞𝑘=0𝑞0}𝑒^{−2𝐹(𝑘𝑞+𝑟)}=𝑑𝑟_∞𝑘=0𝑒^{−2𝑘𝐹(𝑞)𝑞}₀𝑒^{−2𝐹(𝑟)}𝑑𝑟≤𝑞^∞_𝑘=0𝑒^{−2𝑘𝐹(𝑞)}=𝑞𝑒^2𝐹(𝑞)𝑒^2𝐹(𝑞)−1∶=𝐶₂. (40)
Hence, ^∞₀‖𝑈(𝑡,0)𝑥‖²𝑑𝑡≤𝐶₂‖𝑥‖²₂. (41)
Now by Theorem 8, we get 𝜔₀(𝒰)≤−1/(2𝐾₂), with 𝐾₂ being given in (24).

Remark 9. Even in the case of Hilbert spaces, the inequality 𝜔₀(𝐓)1≤−𝑝𝐶_𝑝 (42) could be sharp.
Indeed, let 𝐻∶=𝑙²(ℤ₊,ℂ) be the Hilbert space of all ℂ-valued sequences 𝑥=(𝑥_𝑛) verifying ‖𝑥‖₂∶=^∞_𝑛=1𝑥_𝑛^21/2<∞. (43)
Let 𝜀>0 be fixed, and set (𝑇(𝑡)𝑥)_𝑛∶=𝑒^−𝜀𝑡𝑥_𝑛,𝑛∈ℤ₊,𝑡≥0. (44) Obviously, the family 𝐓={𝑇(𝑡)}_𝑡≥0 is a strongly continuous semigroup on 𝑙²(ℤ₊,ℂ), and 2𝜔₀(𝐓)⋅sup_{‖𝑥‖2≤1}^∞₀‖𝑇(𝑡)𝑥‖²₂𝑑𝑡=−1. (45)

Remark 10. It seems that the theoretical result of this paper cannot be applied to wave equations. We try to justify this statement in the following.
Let 𝐴 be a closed operator on a complex Banach space 𝑋, and let 𝑥∈𝑋. Consider the Cauchy Problem 𝑃²()𝑢𝑥,0^′′(𝑡)(𝑡)𝑢(0)𝑢=𝐴𝑢)(𝑡≥0=𝑥_′(0)=0. (46)
Recall that a classical solution of 𝑃²(𝑥,0) is a function 𝑢∈𝐶²(ℝ₊,𝑋) such that 𝑢(𝑡)∈𝐷(𝐴) for all 𝑡≥0, and 𝑃²(𝑥,0) holds. A function 𝑢∈𝐶(ℝ₊,𝑋) is called a mild solution for 𝑃²(𝑥,0) if ^𝑡₀()𝑢(𝑠)(𝐴),𝑢(𝑡)𝑡−𝑠𝑑𝑠∈𝐷=𝑥+𝐴^𝑡₀(𝑡−𝑠)𝑢(𝑠)𝑑𝑠 (47) for all 𝑡≥0. Any 𝐶²(ℝ₊,𝑋)-valued mild solution of 𝑃²(𝑥,0) is a classical solution. The mild solution of 𝑃²(𝑥,0) leads to the notion of cosine function. See [10] or [11, pages 206–221], for further details.
Recall that a strongly continuous function cos∶ℝ₊→ℒ(𝑋) is called cosine function if(1)cos(0)=𝐼, where 𝐼 is the identity operator on 𝑋,(2)2cos(𝑡)cos(𝑠)=cos(𝑡+𝑠)+cos(𝑡−𝑠) for all 𝑡≥0 and 𝑠≥0.It is well known that any cosine function is exponentially bounded; that is, there are 𝜔∈ℝ and 𝑀>0 such that ‖cos(𝑡)‖≤𝑀𝑒^𝜔𝑡 for all 𝑡≥0. Moreover, the uniform growth bound of any cosine function is nonnegative [11, Proposition 3.14.6]. Any cosine function has a generator. More exactly, there exist a positive 𝜔 and a linear operator 𝐴 such that (𝜔²,∞)⊂𝜌(𝐴) and 𝜆𝑅𝜆²,𝐴=^∞₀𝑒^−𝜆𝑡cos(𝑡)𝑑𝑡(𝜆>𝜔). (48) See [11, Proposition 3.14.4]. On the other hand, if 𝐴 is a generator of a cosine function, the mild solution of the Cauchy Problem 𝑃²(𝑥,0) is given by 𝑢(𝑡)=cos(𝑡)𝑥, 𝑡≥0. Finally, we remind the reader that the result of this paper refers to exponentially stable evolution families, and then it cannot be applied to cosine functions.

Acknowledgment

The authors thank the anonymous referees for their useful suggestions on the preliminary version of this paper. In particular, the authors added the final remarks at the suggestion of the referees.