2 Exponential behavior and strong admissibility

Strong and weak admissibility of L ^∞ spaces

Luis Barreira

and Claudia Valls

Departamento de Matemática, Instituto Superior Técnico, Universidade de Lisboa 1049-001 Lisboa, Portugal

Received 27 June 2017, appeared 17 November 2017 Communicated by Christian Pötzsche

Abstract. For a dynamics with continuous time, we consider the notion of a strong exponential dichotomy with respect to a family of norms and we characterize it in terms of the admissibility of bounded solutions. Moreover, we consider both strong and weak admissibility, in the sense that the solutions are respectively of a nonautonomous linear equation defined by a strongly continuous function or of an integral equation obtained from perturbing a general evolution family. As a nontrivial application, we establish the robustness of the notions of a strong exponential dichotomy and of a strongnonuniform exponential dichotomy. We emphasize that the last notion is ubiquitous in the context of ergodic theory: for almost all trajectories with nonzero Lyapunov exponents of a measure-preserving flow, the linear variational equation admits a strong nonuniform exponential dichotomy..

Keywords: admissibility, continuous time, exponential dichotomies.

2010 Mathematics Subject Classification: 37D99.

1 Introduction

For a nonautonomous linear equation

x⁰ = A(t)x (1.1)

in a Banach space defined by a strongly continuous function A(t)and more generally for an evolution family T(t,s) in a Banach space, we introduce the notion of a strong exponential dichotomy with respect to a family of norms. This means that besides having the usual upper bounds in the stable direction for positive time and in the unstable direction for negative time, we have, in addition, lower bounds in the stable direction for positive time and in the unstable direction for negative time.

Moreover, at each time we consider a possibly different norm. The main motivation comes from ergodic theory. Indeed, for almost all trajectories with nonzero Lyapunov exponents of a measure-preserving flow, the linear variational equation admits a strong nonuniform expo- nential dichotomy (we refer to [2] for details and references). This last notion is a particular

BCorresponding author. Email: barreira@math.tecnico.ulisboa.pt

case of the notion of a strong exponential dichotomy with respect to a family of norms, more precisely a family of Lyapunov norms. Therefore, the type of exponential behavior considered in the paper, besides being very common in the context of ergodic theory, plays a unifying role. In particular, it includes as particular cases both the notions of uniform and nonuni- form exponential behavior, considering respectively families of constant norms and Lyapunov norms.

Our main aim is to characterize the notion of a strong exponential dichotomy in terms of the admissibility of bounded solutions. The latter corresponds to assume that there exists a unique bounded solution for each time-dependent bounded perturbation of the original dynamics. In addition to considering a nonautonomous linear equation and more generally an arbitrary evolution family, we also consider both strong and weak admissibility, which corresponds to the perturbations of each of those dynamics. More precisely, in the case of equation (1.1) we consider the perturbed equation

x⁰ = A(t)x+y(t) (1.2)

and its classical solutions, while in the case of an arbitrary evolution familyT(t,s)we consider the perturbed integral equation

x(t) =T(t,τ)x(τ) +

Z _t

τ

T(t,s)y(s)ds (1.3) and its mild solutions. We refer to the admissibility in the two perturbed equations, respec- tively, asstrong and weak admissibility. We emphasize that a priori none of them implies the other.

Our main results show that:

1. the evolution family defined by equation (1.1) admits a strong exponential dichotomy with respect to a family of norms if and only if it has bounded growth and there exists a unique bounded solution of equation (1.2) for each bounded perturbation y of the original dynamics (see Theorems2.1and2.3);

2. an arbitrary evolution familyT(t,s)admits astrongexponential dichotomy with respect to a family of norms if and only if it has bounded growth and there exists a unique bounded solution of equation (1.3) for each bounded perturbationy of the original dy- namics (see Theorems4.1and4.2).

Here, “bounded growth” and “bounded” are always with respect to the family of normsk·k_t under consideration. For example, a functiony: R → X with values in a Banach space X is said to bebounded(with respect to the normsk·k_t) if

sup

t∈_R

ky(t)k_t <+_∞.

For an evolution family with bounded growth defined by a differential equation as in (1.1), it follows from the latter results that there exists a unique bounded solution of equation (1.2) for each bounded perturbationyif and only if there exists a unique bounded solution of equation (1.3) for each bounded perturbation y. In other words, in our setting the notions of weak admissibility and strong admissibility are in fact equivalent. In fact, this can be considered the main contribution of our work.

The study of the admissibility property goes back to pioneering work of Perron in [8] who used it to deduce the stability or the conditional stability under sufficiently small perturbations

of a linear equation. For some of the most relevant early contributions in the area we refer to the books by Massera and Schäffer [6] and by Dalec⁰ki˘ı and Kre˘ın [4]. We also refer to [5] for some early results in infinite-dimensional spaces.

As a nontrivial application of these results, we establish the robustness of the notion of a strong exponential dichotomy with respect to a family of norms and of a strong nonuniform exponential dichotomy. This corresponds to show that any sufficiently small linear pertur- bation of the dynamics is still, respectively, a strong exponential dichotomy with respect to a family of norms and a strong nonuniform exponential dichotomy. We emphasize that the study of robustness has a long history; see in particular [3,7,9,10] and the references therein.

See also [1] for the study of robustness in the general setting of a nonuniform exponential behavior.

2 Exponential behavior and strong admissibility

2.1 Exponential dichotomies

Let X = (X,k·k) be a Banach space and let B(X)be the set of all bounded linear operators on X. A function A: R → B(X)is said to be strongly continuous if for each x ∈ X the map t 7→ A(t)x is continuous. We note that every continuous function A: R → B(X)is strongly continuous.

Let A:R→ B(X)be a strongly continuous function and consider the linear equation

x⁰ = A(t)x. (2.1)

Let also T(t,τ) be the associated evolution family. Moreover, we consider a family of norms k·k_t on Xfort∈_Rsuch that:

(i) there exist constantsCandε≥0 such that

kxk ≤ kxk_t ≤Ce^ε|t|kxk (2.2) forx ∈Xandt∈_R;

(ii) the mapt 7→ kxk_tis measurable for each x∈ X.

We say that equation (2.1) admits a strong exponential dichotomywith respect to the family of normsk·k_tif:

(iii) there exist projectionsP(t)_fort ∈_{Rsuch that}

P(t)T(t,τ) =T(t,τ)P(τ), t,τ∈_R; (2.3) (iv) there exist constants

a ≤a<₀<b≤b and D>₀ such that

kT(t,τ)P(τ)xk_t ≤De^a(t−τ)kxk_τ,

kT(τ,t)Q(t)xk_τ ≤De^−b(t−τ)kxk_t ^(2.4) fort≥ τand

kT(t,τ)P(τ)xk_t ≤De^a(t−τ)kxk_τ,

kT(τ,t)Q(t)xk_τ ≤De^−b(t−τ)kxk_t ^(2.5) fort≤ _{τ, where}Q(τ) =_Id−P(τ)_.

2.2 From exponential behavior to admissibility

LetY be the set of all continuous functionsx: R→Xsuch that kxk_∞ :=sup

t∈_R

kx(t)k_t <+_∞.

One can easily verify that when equipped with the normk·k_∞ the setYis a Banach space.

We first show that for a strong exponential dichotomy the pair(Y,Y)is admissible in the strong sense, that is, considering classical solutions of equation (2.1).

Theorem 2.1. Assume that equation(2.1) admits a strong exponential dichotomy with respect to the family of normsk·k_t. Then:

1. for each y∈Y there exists a unique x∈Y such that

x⁰(t)−A(t)x(t) =y(t) (2.6) for t∈_R;

2. there exist K,a>₀such that

kT(t,τ)xk_t ≤Ke^a|t−τ|kxk_{τ (2.7)} for x∈ X and t,τ∈ _R.

Proof. For the first statement in the theorem, takey∈Y. For t∈_Rwe define x(t) =

Z _t

−_∞T(t,τ)P(τ)y(τ)dτ−

Z +_∞

t T(t,τ)Q(τ)y(τ)dτ. (2.8) It follows from (2.4) that

Z _t

−∞kT(t,τ)P(τ)y(τ)k_tdτ+

Z +_∞ t

kT(t,τ)Q(τ)y(τ)k_tdτ

≤Dkyk_{∞ Z t}

−_∞e^a(t−τ)dτ+

Z +_∞

t e^−b(τ−t)dτ

=D

−¹ a + ¹

b

kyk_∞

(2.9)

fort ∈_{Rand thus,}x(t)is well defined. Moreover, givent₀ ∈_{R, we have} x(t) =

Z _t

t0

T(t,τ)y(τ)dτ−

Z _t

t0

T(t,τ)P(τ)y(τ)dτ

−

Z _t

t₀ T(t,τ)Q(τ)y(τ)dτ+

Z _t

−_∞T(t,τ)P(τ)y(τ)dτ

−

Z _+∞

t

T(t,τ)Q(τ)y(τ)dτ

=

Z _t

t0

T(t,τ)y(τ)dτ+

Z _t0

−_∞T(t,τ)P(τ)y(τ)dτ

−

Z +_∞ t0

T(t,τ)Q(τ)y(τ)dτ

=

Z _t

t0

T(t,τ)y(τ)dτ+T(t,t0)x(t0)

(2.10)

and hence,

x(t) =T(t,t₀)x(t₀) +

Z _t

t0

T(t,τ)y(τ)dτ (2.11) for t ∈ _{R. Since} T(t,τ) is the evolution family of equation (2.1), it follows from (2.11) that the function x: R → X is differentiable and that identity (2.6) holds fort ∈ R. Moreover, it follows from (2.9) thatx∈Y.

Lemma 2.2. x is the unique function in Y satisfying(2.6).

Proof of the lemma. Since the map x 7→ y defined by identity (2.6) is linear, it is sufficient to show that if a function x∈Ysatisfiesx⁰(t) = A(t)x(t)fort ∈_{R, then}x =0. Let

x^s(t) =P(t)x(t) and x^u(t) =Q(t)x(t). Thenx(t) =x^s(t) +x^u(t)and it follows from (2.3) that

x^s(_t) =_T(_t,τ)_x^s(τ) _{and x}^u(_t) =_T(_t,τ)_x^u(τ) fort,τ∈_{R. Since}x^s(t) =T(t,t−τ)x^s(t−τ)forτ≥0, we have

kx^s(t)k_t= kT(t,t−τ)x^s(t−τ)k_t

= kT(t,t−τ)P(t−τ)x(t−τ)k_t

≤ De^aτkx(t−τ)k_t−τ

≤ De^aτkxk_∞

and lettingτ→+_∞yields thatx^s(t) =0 fort∈R. Similarly, sincex^u(t) =T(t,t+τ)x^u(t+τ) forτ≥0, we have

kx^u(t)k_t=kT(t,t+τ)x^u(t+τ)k_t

=kT(t,t+τ)Q(t+τ)x(t+τ)k_t

≤ De^−bτkx(t+τ)k_t+τ

≤ De^−bτkxk_∞

and hence,x^u(t) =0 fort∈_R. Therefore,x(t) =0 fort∈ _R.

It remains to establish the second statement in the theorem. It follows from (2.4) and (2.5) that

kT(t,τ)xk_t ≤ kT(t,τ)P(τ)xk_t+kT(t,τ)Q(τ)xk_t

≤De^a(t−τ)kxk_τ+De^b(t−τ)kxk_τ

≤2De^b(t−τ)kxk_τ fort ≥τand similarly

kT(t,τ)xk_t ≤2De^−a(τ−t)kxk_τ fort ≤τ. Therefore, (2.7) holds withK=2Danda=_max{b,−a}_.

2.3 From admissibility to exponential behavior

Now we establish the converse of Theorem 2.1, that is, we show that if the pair (Y,Y) is admissible, then equation (2.1) admits a strong exponential dichotomy.

Theorem 2.3. Assume that for each y∈Y there exists a unique x∈Y such that:

1. identity(2.6)holds for t∈_R;

2. there exist K,a>0such that(2.7)holds for x ∈ X and t,τ∈ _R.

Then equation(2.1)admits a strong exponential dichotomy with respect to the family of normsk·k_t. Proof. Let Hbe the linear operator defined by

(Hx)(t) =x⁰(t)−A(t)x(t), t ∈_R (2.12) in the domainD(H)formed by allx ∈Ysuch thatHx∈Y.

Lemma 2.4. The operator H: D(H)→Y is closed.

Proof of the lemma. Let(x_k)_k∈Nbe a sequence inD(H)converging tox ∈Ysuch thaty_k = Hx_k converges toy∈Y. For eachτ∈_{R, we have}

x(t)−x(τ) = lim

k→_∞(x_k(t)−x_k(τ))

= lim

k→_∞ Z _t

τ

x⁰_k(s)ds

= lim

k→_∞ Z _t

τ

(y_k(s) +A(s)x_k(s))ds fort ≥τ. Moreover, it follows from (2.2) that

Z _t

τ

y_k(s)ds−

Z _t

τ

y(s)ds

≤

Z _t

τ

ky_k(s)−y(s)kds

≤

Z _t

τ

ky_k(s)−y(s)k_sds

≤(t−τ)ky_k−yk_∞. Sincey_k →yinY, we obtain

klim→_∞ Z _t

τ

y_k(s)ds=

Z _t

τ

y(s)ds.

Similarly,

Z _t

τ

A(s)x_k(s)ds−

Z _t

τ

A(s)x(s)ds

≤ M Z _t

τ

kx_k(s)−x(s)kds

≤ M(t−τ)kx_k−xk_∞, where

M =sup

kA(s)k:s∈[τ,t] . Since the functions 7→ A(s)x is continuous for eachx ∈X, we have

sup

τ≤s≤t

kA(s)xk<+_∞

and it follows from the Banach–Steinhaus theorem that M < +_{∞. Since} Hx_k → y in Y, we obtain

klim→_∞ Z _t

τ

A(s)x_k(s)ds=

Z _t

τ

A(s)x(s)ds.

Therefore,

x(_t)−x(τ) =

Z _t

τ

(_A(_s)_x(_s) +_y(_s))_ds, which implies that Hx=yandx ∈ D(H).

It follows from Lemma 2.4 and the closed graph theorem that the operator H has a bounded inverseG:Y→Y.

Forτ∈_{R, let}F_τ^sbe the set of allx∈ Xsuch that there exists a solutionuof equation (2.1) with u(τ) = xsatisfying

sup

ku(t)k_t :t∈[τ,+_∞) <+_∞. (2.13) Similarly, letF_τ^ube the set of all x∈Xsuch that there exists a solutionuof equation (2.1) with u(τ) =x satisfying

sup

ku(t)k_t :t∈(−_∞,τ] <+_∞. (2.14) One can easily verify that F_τ^sandF_τ^uare subspaces ofX.

Lemma 2.5. Forτ∈_{R, we have}

X= F_τ^s⊕F_τ^u. (2.15)

Proof of the lemma. Let φ: R → _R be a smooth function supported on[τ,+_∞)such that 0 ≤ φ ≤ 1, φ = 1 on [τ+1,+_∞) and sup_t∈R|φ⁰(t)| < +∞. Moreover, given x ∈ X let u be the solution of equation (2.1) withu(τ) = x. It follows from (2.13) that g := φ⁰u ∈Y. Since H is invertible, there exists v ∈ Y such that Hv = g. Letw = (1−φ)u+v. One can easily verify that Hw =0. Furthermore,

sup

kw(t)k_t:t ∈[τ,+_∞) ≤sup

ku(t)k_t:t ∈[τ,τ+1] +sup

kv(t)k_t :t∈ [τ,+_∞) <+_∞, and thusw(τ)∈F_τ^s. On the other hand,w−uis also a solution of equation (2.1) and

sup

k(w−u)(t)k_t:t ∈(−_∞,τ] =sup

kv(t)k_t:t ∈(−_∞,τ] <+_∞.

Hence,

w(τ)−x=w(τ)−u(τ) =v(τ)∈ F_τ^u andx ∈F_τ^s+F_τ^u.

It remains to show that F_τ^s∩F_τ^u = {0}. Take x ∈ F_τ^s∩ F_τ^u and let u be the solution of equation (2.1) withu(τ) =x. It follows from (2.13) and (2.14) thatu∈Y. SinceHis invertible, we must have u=0 and hencex =0.

Now letP(τ): X→ F_τ^s andQ(τ): X→ F_τ^u be the projections associated to the decomposi- tion in (2.15), withP(τ) +Q(τ) =Id. It follows readily from the definitions that property (2.3) holds.

Lemma 2.6. There exists M>0such that

kP(τ)xk_τ ≤ Mkxk_τ (2.16)

for x ∈X and τ∈_R.

Proof of the lemma. Using the same notation as in the proof of Lemma2.5, we have kP(τ)xk_τ = kw(τ)k_τ ≤ ku(τ)k_τ+kv(τ)k_τ

≤ kxk_τ+kvk_∞= kxk_τ+kGgk_∞. (2.17) Furthermore,

kgk_∞ =kφ⁰uk_∞ ≤Lsup

ku(t)k_t :t∈ [τ,τ+1] ,

where L = sup_t∈R|φ⁰(t)|. We note that the constant L is independent of τ. Using (2.7) we obtain

kgk_∞≤ LKe^akxk_τ and it follows from (2.17) that

kP(τ)xk_τ ≤ (1+kGkLKe^a)kxk_τ. This shows that (2.16) holds taking M=1+kGkLKe^a.

Lemma 2.7. There exist constantsλ,D>0such that

kT(t,τ)P(τ)xk_t≤ De^{−λ(t−τ)}kxk_{τ (2.18)} for x∈ X and t≥ τ.

Proof of the lemma. Let ψ: R → _R be a smooth function supported on[τ,+_∞) such that 0 ≤ ψ ≤ _1, ψ = _{1 on} [τ+_1,+_∞) _{and supt∈R}|ψ⁰(t)| ≤ 2. Moreover, given x ∈ F_τ^s, let u be the solution of equation (2.1) withu(τ) =x. It follows from (2.13) that ψu∈Yand one can easily verify that H(ψu) =ψ⁰u. Moreover,

sup

ku(t)k_t :t∈[τ+1,+_∞) =sup

kψ(t)u(t)k_t :t∈[τ+1,+_∞)

≤ kψuk_∞ =kG(ψ⁰u)k_∞

≤ kGk · kψ⁰uk_∞

= kGksup

k(ψ⁰u)(t)k_t: t∈[τ,τ+1]

≤2kGksup

ku(t)k_t :t ∈[τ,τ+1]

=2kGksup

kT(t,τ)u(τ)k_t :t∈ [τ,τ+1]

≤2Ke^akGk · ku(τ)k_τ

=2Ke^akGk · kxk_τ,

using (2.7) in the last inequality. Hence, using again (2.7), we obtain

ku(t)k_t≤ Ckxk_τ for t ≥τ, (2.19) whereC=2Ke^amax{1,kGk}.

Now we show that there exists N∈_Nsuch that for everyτ∈_Randx∈ F_τ^s, ku(t)k_t≤ ¹

2kxk_τ for t−τ≥ N. (2.20)

In order to prove (2.20), take t₀ ∈ _R such that t₀ > τ and ku(t₀)k_t0 > kxk_τ/2. It follows from (2.19) that

1

2Ckxk_τ <ku(s)k_s ≤Ckxk_τ, τ≤s≤ t₀. (2.21)

Now take ε > 0 and let ψ: R → _R be a smooth function supported on [τ,t₀] such that 0≤ψ≤1 andψ=1 on [τ+ε,t0−ε]. Moreover, let

y(t) =ψ(t)u(t) and v(t) =u(t)

Z _t

−_∞ψ(s)ds

fort ∈_R. Clearly,yandvbelong toYand one can easily verify thatHv=y. Therefore, kGksup

ku(t)k_t: t∈[τ,t₀] ≥ kGk · kyk_∞ ≥ kvk_∞. Hence, it follows from (2.21) that

CkGk · kxk_τ ≥ kv(t₀)k_t0

≥ ku(t0)k_t0

Z _t0−ε

τ+ε

ψ(s)ds

≥ ¹

2C(t₀−τ−2ε)kxk_τ. Lettingε→0 yields the inequality

t₀−τ≤2C²kGk. Hence, property (2.20) holds takingN >2C²kGk.

In order to complete the proof, take t ≥ τ and write t−τ = kN+r, with k ∈ _{N and} 0≤r < N. By (2.16), (2.19) and (2.20), we obtain

kT(t,τ)P(τ)xk_t =kT(τ+kN+r,τ)P(τ)xk_τ+kN+r

≤ ¹

2^kkT(τ+r,τ)P(τ)xk_τ+r

≤ ^C

2^kkP(τ)xk_τ

≤2CMe^{−(t−τ)log 2/N}kxk_τ, forx ∈X. TakingD=2CMandλ=log 2/Kyields inequality (2.18).

Lemma 2.8. There exist constantsλ,D>0such that

kT(t,τ)Q(τ)xk_t≤ De^{−λ(τ−t)}kxk_τ (2.22) for x ∈X and t≤τ.

Proof of the lemma. Let ψ: R → _R be a smooth function supported on (−_∞,τ] such that 0 ≤ ψ ≤ 1, ψ = 1 on (−_∞,τ−1] and sup_t∈R|ψ⁰(t)| ≤ 2. Moreover, given x ∈ F_τ^u, let u be the solution of equation (2.1) withu(τ) =x. It follows from (2.14) thatψu∈Yand one can easily verify that H(ψu) =ψ⁰u. Moreover,

sup

ku(t)k_t:t ∈(−_∞,τ−1] =sup

kψ(t)u(t)k_t:t ∈(−_∞,τ−1]

≤ kψuk_∞ =kG(ψ⁰u)k_∞

≤ kGk · kψ⁰uk_∞

=kGksup

k(ψ⁰u)(t)k_t:t ∈[τ−1,τ]

≤2kGksup

ku(t)k_t :t∈ [τ−1,τ]

=₂kGk_supkT(t,τ)u(τ)k_t:t ∈[τ−_1,τ]

≤2Ke^akGk · ku(τ)k_τ =2Ke^akGk · kxk_τ,

using (2.7) in the last inequality. Hence, using again (2.7), we obtain

ku(t)k_t≤ Ckxk_τ for t ≤τ, (2.23) whereC=2Ke^amax{1,kGk}.

We also show that there exists N∈_Nsuch that for everyτ∈_Randx∈ F_τ^u, ku(t)k_t≤ ¹

2kxk_τ for τ−t≥ N. (2.24)

In order to prove (2.24), take t₀ ∈ _R such that t₀ < τ and ku(t₀)k_t0 > kxk_τ/2. It follows from (2.23) that

1

2Ckxk_τ <ku(s)k_s ≤Ckxk_τ, t₀≤s ≤τ. (2.25) Now take ε > _{0 and let} ψ: R → _R be a smooth function supported on [_t0,τ] _{such that} 0≤ψ≤1 andψ=1 on[t₀+ε,τ−ε]. Moreover, let

y(t) =−ψ(t)u(t) and v(t) =u(t)

Z _+∞

t

ψ(s)ds

fort ∈R. Clearly,yandvbelong toYand one can easily verify that Hv= y. Therefore, kGksup

ku(t)k_t :t∈ [t₀,τ] ≥ kGk · kyk_∞≥ kvk_∞. Hence, it follows from (2.25) that

CkGk · kxk_τ ≥ kv(t₀)k_t0

≥ ku(t₀)k_t0

Z _τ−ε

t₀+ε

ψ(s)ds

≥ ¹

2C(τ−t₀−2ε)kxk_τ. Lettingε →0 yields the inequality

τ−t₀ ≤2C²kGk. Hence, property (2.24) holds taking N>2C²kGk.

Finally, take t ≤τ and writeτ−t = kN+r, withk ∈ _{Nand 0}≤ r < N. By (2.16), (2.23) and (2.24), we obtain

kT(t,τ)Q(τ)xk_t=kT(τ−kN−r,τ)Q(τ)xk_τ−kN−r

≤ ¹

2^kkT(τ−r,τ)Q(τ)xk_τ−r

≤ ^C

2^kkQ(τ)xk_τ

≤2C(₁+M)e^{−(τ−t)log 2/N}kxk_τ, forx∈ X. TakingD=2C(1+M)andλ=log 2/Kyields inequality (2.22).

In order to complete the proof of the theorem, we note that it follows from (2.18) and (2.22) that (2.4) holds takinga= −λandb= λ. Moreover, it follows from (2.7) and (2.16) that (2.5) holds takingD=K(₁+M)_,a =−aandb= a.

3 Strong robustness

In this section we establish the robustness of the notion of a strong exponential dichotomy using its characterization in terms of admissibility of the pair (_Y,Y)in Theorems2.1 and2.3.

Theorem 3.1. Let A,B: R→ B(X)be strongly continuous functions such that:

1. equation (2.1) admits a strong exponential dichotomy with respect to a family of norms k·k_t satisfying(2.2)for some C>0andε ≥0;

2. there exists c≥0such that

kB(t)−A(t)k ≤ce^−ε|t|, t ∈_R. (3.1) If c is sufficiently small, then the equation x⁰ = B(t)x admits a strong exponential dichotomy with respect to the same family of norms.

Proof. Let Hbe the linear operator defined by (2.12) on the domainD(H). Forx ∈ D(H)we consider the graph norm

kxk⁰_∞ =kxk_∞+kHxk_∞. Clearly, the operator

H:(D(H),k·k⁰_∞)→(Y,k·k_∞)

is bounded. For simplicity, we denote it from now on simply byH. It follows from Lemma2.4 that (D(H),k·k⁰_∞)is a Banach space.

It follows from (2.7) and (3.1) that

k(B(t)−A(t))xk_t≤ cCkxk_t (3.2) forx ∈Xandt∈ R. We define a linear operatorL: D(H)→Yby

(Lx)(t) =x⁰(t)−B(t)x(t), t∈_R.

By (3.2) we have

k(H−L)xk_∞ ≤cCkxk⁰_{∞ (3.3)}

forx ∈ D(T). By Theorem2.1, the operator His invertible. Hence, it follows from (3.3) that if cis sufficiently small, then Lis also invertible. Furthermore, it follows from Theorem2.1that there exist constants K,a > 0 such that (2.7) holds forx∈ X andt,τ∈_R. Now letU(t,τ)be the evolution family associated to the linear equationx⁰ =B(t)x.

Lemma 3.2. There exist constants K⁰,a⁰ >0such that

kU(t,τ)xk_t≤ K⁰e^a0|t−τ|kxk_τ for x ∈X and t,τ∈_R.

Proof of the lemma. Letx(t)be a solution of the equationx⁰ = B(t)x. For eacht≥τwe have kx(t)k_t =

T(t,τ)x(τ) +

Z _t

τ

T(t,s)(B(s)−A(s))x(s)ds t

≤Ke^a(t−τ)kx(τ)k_τ+K Z _t

τ

e^a(t−s)k(B(s)−A(s))x(s)k_sds

≤Ke^a(t−τ)kx(τ)k_τ+cCK Z _t

τ

e^a(t−s)kx(s)k_sds.

This shows that the functionφ(t) =e^−atkx(t)k_t satisfies φ(t)≤Kφ(τ) +cCK

Z _t

τ

φ(s)ds and using Gronwall’s lemma we obtain

φ(t)≤Kφ(τ)e^cCK(t−τ). Hence,

kU(t,τ)xk_t≤ Ke^{(a+cCK)(t−τ)}kxk_τ fort ≥τ. One can argue in a similar manner for t≤τ.

SinceLis invertible, it follows from Theorem2.3together with Lemma3.2that the equation x⁰ = B(t)x admits a strong exponential dichotomy with respect to the family of norms k·k_t.

4 Exponential behavior and weak admissibility

In this section we consider a weak form of the admissibility property and we use it to give a characterization of the notion of a strong exponential dichotomy.

A family T(t,τ), for t,τ ∈ R, of bounded linear operators on Xis said to be an evolution familyif:

1. T(t,t) =Id fort∈_R;

2. T(t,s)T(s,τ) =T(t,τ)_fort,s,τ∈_R;

3. givent,τ∈_Randx∈ X, the mapss7→ T(t,s)x ands 7→T(s,τ)x are continuous.

We continue to consider a family of normsk·k_t satisfying conditions (i) and (ii). We say that an evolution familyT(t,s)admits a strong exponential dichotomywith respect to the family of normsk·k_t if conditions (iii) and (iv) hold.

We first show that the existence of a strong exponential dichotomy yields the weak admis- sibility of the pair(Y,Y).

Theorem 4.1. If the evolution family T(t,τ) admits a strong exponential dichotomy with respect to the family of normsk·k_t, then:

1. for each y∈Y there exists a unique x∈Y such that x(t) =T(t,τ)x(τ) +

Z _t

τ

T(t,s)y(s)ds for t≥ τ; (4.1) 2. there exist K,a>0such that(2.7)holds.

Proof. Takey∈ Y. Fort ∈_Rwe define x(t)as in (2.8). Then (2.9) holds and proceeding as in (2.10) we obtain

x(t) =

Z _t

τ

T(t,s)y(s)ds+T(t,τ)x(τ)

fort ≥τ. This shows that property (4.1) holds. It follows readily from (4.1) that the function x is continuous and thus x ∈ Y. The uniqueness of x follows from Lemma 2.2 (that can be obtained using the same proof). This establishes the first property of the theorem.

The second property follows exactly as in the proof of Theorem2.1.

Now we establish the converse of Theorem4.1.

Theorem 4.2. Assume that for each y∈Y there exists a unique x ∈Y such that(4.1)holds and that there exist constants K,a >0such that(2.7)holds for x∈ X and t,τ∈R. Then the evolution family T(t,τ)admits a strong exponential dichotomy with respect to the family of normsk·k_t.

Proof. Let H be the linear operator defined by Hx = y in the domain D(H) formed by all x∈Yfor which there existsy∈Ysatisfying (4.1). In order to show thatHis well defined, let y₁,y₂∈Y be such that

x(t) =T(t,τ)x(τ) +

Z _t

τ

T(t,s)y₁(s)ds and

x(t) =T(t,τ)x(τ) +

Z _t

τ

T(t,s)y2(s)ds fort ≥τ. Then

1 t−τ

Z _t

τ

T(t,s)y₁(s)ds= ¹ t−τ

Z _t

τ

T(t,s)y2(s)ds

and since the map s7→ T(t,s)y_i(s)is continuous for i= 1, 2, lettingτ→ tyields thaty₁(t) = y₂(t)_fort∈ _R.

Lemma 4.3. The operator H: D(H)→Y is closed.

Proof of the lemma. Let (x_n)_n∈N be a sequence in D(H) converging to x ∈ Y such that Hx_n converges to y∈Y. For eachτ∈_{R, we have}

x(t)−T(t,τ)x(τ) = lim

n→_∞(x_n(t)−T(t,τ)x_n(τ))

= lim

n→_∞ Z _t

τ

T(t,s)y_n(s)ds fort ≥τ. Furthermore,

Z _t

τ

T(t,s)yn(s)ds−

Z _t

τ

T(t,s)y(s)ds

≤ M Z _t

τ

kyn(s)−y(s)kds

≤ M Z _t

τ

kyn(_s)−y(_s)k_sds

≤ Mky_n−yk_∞(t−τ)_, where

M =sup

kT(t,s)k:s∈ [τ,t] . Since the maps7→ T(t,s)x is continuous for eachx∈X, we have

sup

τ≤s≤t

kT(t,s)xk<+_∞

and it follows from the Banach–Steinhaus theorem that M < +_{∞. Since}y_n converges toyin Y, we conclude that

nlim→_∞ Z _t

τ

T(t,s)y_n(s)ds=

Z _t

τ

T(t,s)y(s)ds and

x(t)−T(t,τ)x(τ) =

Z _t

τ

T(t,s)y(s)ds fort ≥τ. This shows that (4.1) holds. Hence,Hx =yandx∈ D(H)_.

It follows from the closed graph theorem that H has a bounded inverse G: Y → Y. For eachτ∈_Rwe define

F_τ^s=

x∈ X: sup

t≥τ

kT(t,τ)xk_t< +_∞

and

F_τ^u=

x∈ X: sup

t≤τ

kT(t,τ)xk_t< +_∞

. One can easily verify thatF_τ^sandF_τ^uare subspaces of X.

Lemma 4.4. Forτ∈_R, we have

X=F_τ^s⊕F_τ^u. (4.2)

Proof of the lemma. Let φ: R → _R be a continuous function supported on [τ,τ+1]such that Rτ+1

τ φ(_s)_ds=_{1. Given x}∈ X, we define a functiong: R→Xby g(t) =φ(t)T(t,τ)x.

Clearly,g∈Y. Since His invertible, there existsv∈Ysuch that Hv=g. Moreover, it follows from (4.1) that

v(t) =T(t,τ)(v(τ) +x)

fort≥ τ+1 and thus v(τ) +x ∈ F_τ^s. Furthermore, again by (4.1), we havev(t) =T(t,τ)v(τ) fort ≤τand thusv(τ)∈F_τ^u. This shows thatx ∈F_τ^s+F_τ^u.

Now take x ∈ F_τ^s∩F_τ^u. We define a function u: R → X by u(t) = T(t,τ)x. It follows from the definitions of F_τ^s and F_τ^u that u ∈ Y. Moreover, Hu = 0 and u ∈ D(H). Since H is invertible, we obtainu=0 and hencex =0.

Now let P(τ): X →F_τ^sandQ(τ): X →F_τ^ube the projections associated to the decomposi- tion in (4.2), withP(τ) +Q(τ) =Id.

Lemma 4.5. There exists M>0such that

kP(τ)xk_τ ≤ Mkxk_τ (4.3)

for x∈ X andτ∈_R.

Proof of the lemma. Using the same notation as in Lemma4.4, we have kP(τ)xk_τ =kv(τ) +xk_τ

≤ kv(τ)k_τ+kxk_τ ≤ kvk_∞+kxk_τ

=kGgk_∞+kxk_τ ≤ kGk · kgk_∞+kxk_τ. On the other hand, it follows from (2.7) thatkgk_∞ ≤CKe^akxk_τ, where

C=_sup|φ(t)|_:t ∈[_τ,τ+₁] _. This shows that (4.3) holds taking M=CKe^akGk+1.

Lemma 4.6. There exist constantsλ,D>0such that

kT(t,τ)P(τ)xk_t≤ De^{−λ(t−τ)}kxk_τ (4.4) for x∈ X and t≥ τ.

Proof of the lemma. Takex ∈ F_τ^s and letu(t) = T(t,τ)x. Moreover, letψ: R →_R be a smooth function supported on[τ,+_∞)such that 0≤ ψ≤1,ψ=1 on[τ+1,+_∞)and sup_t∈R|ψ⁰(t)| ≤ 2. Clearly,ψu ∈Yand one can easily verify thatH(ψu) =ψ⁰u. Moreover,

sup

ku(t)k_t:t ∈[τ+1,+_∞) =sup

kψ(t)u(t)k_t:t ∈[τ+1,+_∞)

≤ kψuk_∞ =kG(ψ⁰u)k_∞

≤ kGk · kψ⁰uk_∞

=kGksup

k(ψ⁰u)(t)k_t:t ∈[τ,τ+1]

≤2kGksup

ku(t)k_t :t∈ [τ,τ+1]

=₂kGksup

kT(t,τ)u(τ)k_t:t ∈[_τ,τ+₁]

≤2Ke^akGk · ku(τ)k_τ =2Ke^akGk · kxk_τ, using (2.7) in the last inequality. Hence, again using (2.7), we obtain

ku(t)k_t≤Ckxk_τ for t≥ τ, (4.5) whereC=2Ke^amax{1,kGk}.

We show that there existsN∈_Nsuch that for everyτ∈_Randx∈ F_τ^s, ku(t)k_t ≤ ¹

2kxk_τ for t−τ≥ N. (4.6)

In order to prove (4.4), take t₀ ∈ _R such that t₀ > τ and ku(t₀)k_t0 > kxk_τ/2. It follows from (4.5) that

1

2Ckxk_τ <ku(s)k_s≤Ckxk_τ, τ≤s≤t₀. (4.7) Now take ε > 0 and let ψ: R → _R be a smooth function supported on [τ,t₀] such that 0≤ψ≤_{1 and}ψ=_{1 on} [τ+_ε,t₀−ε]. Moreover, let

y(t) =ψ(t)u(t) _and v(t) =u(t)

Z _t

−_∞ψ(s)ds

fort ∈R. Clearly,yandvbelong toYand one can easily verify thatHv=y. Therefore, kGksup

ku(t)k_t: t∈[τ,t₀] ≥ kGk · kyk_∞ ≥ kvk_∞. Hence, it follows from (4.7) that

CkGk · kxk_τ ≥ kv(t₀)k_t0

≥ ku(t₀)k_t0

Z _t0−ε

τ+ε

ψ(s)ds

≥ ¹

2C(t0−τ−2ε)kxk_τ. Lettingε→0 yields the inequality

t0−τ≤2C²kGk. Hence, property (4.6) holds takingN>2C²kGk_.

Now take t ≥ τ and write t−τ = kN+r, with k ∈ _N and 0 ≤ r < N. By (4.3), (4.5) and (4.6), we obtain

kT(t,τ)P(τ)xk_t=kT(τ+kN+r,τ)P(τ)xk_τ+kN+r

≤ ¹

2^kkT(τ+r,τ)P(τ)xk_τ+r

≤ ^C

2^kkP(τ)xk_τ

≤2CMe^{−(t−τ)log 2/N}kxk_τ, forx∈ X. TakingD=2CMandλ=log 2/Kyields property (4.4).

Lemma 4.7. There exist constantsλ,D>0such that

kT(t,τ)Q(τ)xk_t ≤De^{−λ(τ−t)}kxk_τ (4.8) for x∈ X and t≤ τ.

Proof of the lemma. Take x ∈ F_τ^u and let u(t) = T(t,τ)x. Moreover, let ψ: R → _R be a smooth function supported on (−_∞,τ] such that 0 ≤ ψ ≤ 1, ψ = 1 on (−_∞,τ−1] and sup_t∈R|ψ⁰(t)| ≤2. Clearly,ψu∈Yand one can easily verify that H(ψu) =ψ⁰u. Moreover,

sup

ku(t)k_t :t∈(−_∞,τ−1] =sup

kψ(t)u(t)k_t :t∈(−_∞,τ−1]

≤ kψuk_∞ =kG(ψ⁰u)k_∞

≤ kGk · kψ⁰uk_∞

= kGksup

k(ψ⁰u)(t)k_t: t∈[τ−1,τ]

≤2kGksup

ku(t)k_t :t ∈[τ−1,τ]

=₂kGk_supkT(t,τ)u(τ)k_{t :}t∈ [τ−_1,τ]

≤2Ke^akGk · ku(τ)k_τ

=2Ke^akGk · kxk_τ,

using (2.7) in the last inequality. Hence, again using (2.7), we obtain

ku(t)k_t≤ Ckxk_{τ for} t ≤_{τ, (4.9)} whereC=2Ke^amax{1,kGk}.

Now we show that there exists N∈_Nsuch that for everyτ∈_Randx∈ F_τ^u, ku(t)k_t≤ ¹

2kxk_τ for τ−t≥ N. (4.10)

In order to prove (4.10), take t0 ∈ _R such that t0 < τ and ku(_t0)k_t0 > kxk_τ/2. It follows from (4.9) that

1

2Ckxk_τ <ku(s)k_s ≤Ckxk_τ, t₀≤s ≤τ. (4.11) Now take ε > 0 and let ψ: R → _R be a smooth function supported on [t0,τ] such that 0≤ψ≤1 andψ=1 on[t₀+ε,τ−ε]. Moreover, let

y(t) =−ψ(t)u(t) and v(t) =u(t)

Z _+∞

t ψ(s)ds

fort ∈R. Clearly,yandvbelong toYand one can easily verify thatHv=y. Therefore, kGksup

ku(t)k_t: t∈[t₀,τ] ≥ kGk · kyk_∞ ≥ kvk_∞. Hence, it follows from (4.11) that

CkGk · kxk_τ ≥ kv(t₀)k_t0

≥ ku(t₀)k_t0

Z _τ−ε

t₀+ε

ψ(s)ds

≥ ¹

2C(τ−t₀−2ε)kxk_τ. Lettingε→0 yields the inequality

τ−t₀≤2C²kGk_. Hence, property (4.8) holds takingN>2C²kGk_.

Finally, taket ≤ τ and write τ−t = kN+r, with k ∈ _N and 0≤ r < N. By (4.3), (4.9) and (4.10), we obtain

kT(t,τ)Q(τ)xk_t =kT(τ−kN−r,τ)Q(τ)xk_τ−kN−r

≤ ¹

2^kkT(τ−r,τ)Q(τ)xk_τ−r

≤ ^C

2^kkQ(τ)xk_τ

≤2C(1+M)e^{−(τ−t)log 2/N}kxk_τ, forx ∈X. TakingD=2C(1+M)andλ=log 2/Kyields property (4.8).

If follows from (4.4) and (4.8) that (2.4) holds witha=−λandb= λ. Moreover, it follows from (2.7) and (4.3) that (2.5) holds with D = (1+M)K, a = −a andb = a. This completes the proof of the theorem.

5 Weak robustness

In a similar manner to that in Section3 we establish, once more, the robustness of the notion of a strong exponential dichotomy but now using its characterization in terms of the weak admissibility of the pair (Y,Y)in Theorems4.1 and4.2.

Theorem 5.1. Assume that the evolution family T(t,τ)admits a strong exponential dichotomy with respect to the family of normsk·k_tand that B: R→B(X)is a strongly continuous function such that kB(t)k ≤ce^−ε|t|, t ∈_R. (5.1) If c is sufficiently small, then the evolution family U(t,τ)defined by

U(t,τ)x= T(t,τ)x+

Z _t

τ

T(t,s)B(s)U(s,τ)x ds, t,τ∈_R admits a strong exponential dichotomy with respect to the same family of norms.

Proof. LetLbe the linear operator associated to the evolution familyU(t,τ), defined byLx= y on the domainD(L)formed by allx ∈Yfor which there existsy∈Ysuch that

x(t) =U(t,τ)x(τ) +

Z _t

τ

U(t,s)y(s)ds for t ≥τ.

For eachx,y ∈Ysuch thatLx=y, we have x(t) =U(t,τ)x(τ) +

Z _t

τ

U(t,s)y(s)ds

= T(t,τ)x(τ) +

Z _t

τ

T(t,s)B(s)U(s,τ)x(τ)ds +

Z _t

τ

T(t,s)y(s)ds+

Z _t

τ

Z _t

s T(t,w)B(w)U(w,s)y(s)dw ds

= T(t,τ)x(τ) +

Z _t

τ

T(t,w)B(w)U(w,τ)x(τ)dw +

Z _t

τ

T(t,s)y(s)ds+

Z _t

τ

Z _w

τ

T(t,w)B(w)U(w,s)y(s)ds dw

= T(t,τ)x(τ) +

Z _t

τ

T(t,s)y(s)ds +

Z _t

τ

T(t,w)B(w)

U(w,τ)x(τ) +

Z _w

τ

U(w,s)y(s)ds

dw

= T(t,τ)x(τ) +

Z _t

τ

T(t,w)(y(w) +B(w)x(w))dw

(5.2)

for t ≥ τ. Now we introduce an operator P: Y → Y by (Px)(t) = B(t)x(t). It follows from (2.2) and (5.1) that

kB(t)x(t)k_t≤Ce^ε|t|kB(t)x(t)k

≤cCkx(t)k ≤cCkx(t)k_t ^(5.3) for t ∈ _R and hence, P is a well defined bounded linear operator. Furthermore, it follows from (5.2) thatD(H) =D(L)and thatH=L+P. For x∈ D(H)we consider the graph norm

kxk⁰_∞ =kxk_∞+kHxk_∞. Clearly, the operator

H: (D(H),k·k⁰_∞)→(Y,k·k_∞)

is bounded and for simplicity we denote it simply by H. Moreover, since H is closed, (D(H),k·k⁰_∞)is a Banach space. By (5.3) we have

k(H−L)xk_∞ ≤ cCkxk_∞ ≤cCkxk⁰_∞ (5.4) forx ∈ X. On the other hand, by Theorem4.1, the operator His invertible. Hence, it follows from (5.4) that ifcis sufficiently small, thenLis also invertible.

It remains to show that there existK⁰,a⁰ >0 such that

kU(t,τ)xk_t ≤K⁰e^a0|t−τ|kxk_{τ for} t,τ∈_{R. (5.5)}

By (2.7), we have

kU(t,τ)xk_t =

T(t,τ)x+

Z _t

τ

T(t,s)B(s)U(s,τ)x ds t

≤Ke^a(t−τ)kxk_τ+K Z _t

τ

e^a(t−s)kB(s)U(s,τ)xk_sds

≤Ke^a(t−τ)kxk_τ+cCK Z _t

τ

e^a(t−s)kU(s,τ)xk_sds fort ≥τ. Hence, the functionφ(t) =e^−atkU(t,τ)xk_tsatisfies

φ(t)≤Kφ(τ) +cCK Z _t

τ

φ(s)ds and it follows from Gronwall’s lemma that

φ(t)≤Kφ(τ)e^cCK(t−τ).

This shows that property (5.5) holds for t ≥ τ taking K⁰ = Kφ(τ) _and a⁰ = cCK. A similar argument can be used for t ≤ τ. One can now apply Theorem 4.2 to conclude that the evolution familyU(t,τ)admits a strong exponential dichotomy.

6 Strong nonuniform exponential dichotomies

In this section we consider briefly the notion of a strong nonuniform exponential dichotomy and we obtain a corresponding robustness result.

We say that an evolution familyT(t,τ), fort,τ∈R, admits astrong nonuniform exponential dichotomy if there exists:

(i) projections P(t)fort∈_Rsatisfying (2.3);

(ii) constants

λ≤λ<₀<µ≤_µ, ε≥_{0 and} D>₀ such that

kT(t,τ)P(τ)xk ≤De^{λ(t−τ)+ε|τ|}kxk, kT(_τ,t)Q(t)xk ≤De^{−µ(t−τ)+ε|t|}kxk fort≥ τand

kT(t,τ)P(τ)xk ≤De^{λ(t−τ)+ε|τ|}kxk, kT(τ,t)Q(t)xk ≤De^{−µ(t−τ)+ε|t|}kxk fort≤ τ, whereQ(τ) =Id−P(τ).

We first relate this notion to the notion of a strong exponential dichotomy with respect to a family of norms.

Proposition 6.1. The following properties are equivalent:

1. T(t,τ)admits a strong nonuniform exponential dichotomy;

2. T(t,τ)admits a strong exponential dichotomy with respect to a family of normsk·k_t satisfying conditions (i) and (ii).