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Topological Methods in Nonlinear Analysis Journal of the Juliusz Schauder Center Volume 37, 2011, 357–376

ROBUSTNESS OF NONUNIFORM POLYNOMIAL DICHOTOMIES FOR DIFFERENCE EQUATIONS

Luis Barreira — Meng Fan — Claudia Valls — Jimin Zhang

Abstract. For a nonautonomous dynamics with discrete time defined by a sequence of linear operators in a Banach space, we establish the robustness of polynomial contractions and of polynomial dichotomies under sufficiently small linear perturbations. In addition, we consider the general case of nonuniform polynomial behavior.

1. Introduction We consider in this paper the robustness problem for difference equations defined by a sequence of linear operators in a Banach space, or equivalently for a nonautonomous dynamics with discrete time. In loose terms, the problem asks whether the behavior of a dichotomy does not change much under sufficiently small linear perturbations. Our main aim is to show that a relatively weak form of dichotomy, which we call polynomial dichotomy, persists under sufficiently small perturbations of the original dynamics. We also consider the general case of nonuniform polynomial behavior. The notion of exponential dichotomy, essentially introduced in seminal work of O. Perron [13], plays a central role in a substantial part of the theory of 2010 Mathematics Subject Classification. 34D09, 34D10, 37D25. Key words and phrases. Nonuniform polynomial dichotomy, robustness. L. Barreira and C. Valls were partially supported by FCT through CAMGSD, Lisbon. M. Fan and J. Zhang were supported by the NSFC and NCET-08-0755. c

2011 Juliusz Schauder Center for Nonlinear Studies

357

358

L. Barreira — M. Fan — C. Valls — J. Zhang

differential equations and dynamical systems, particularly in what concerns the study of stable and unstable invariant manifolds. Although strictly speaking the notion is not introduced in [13], together with Hadamard’s work on the geodesic flow on surfaces of negative curvature, the paper can be considered one of main original sources for the study of hyperbolicity. In particular, it may be considered the first source for the study of robustness, via the the notion of admissibility. Due to the role played by the notion of exponential dichotomy, it is not surprising that the study of robustness has a long history. In particular, the problem was discussed by J. Massera and J. Sch¨affer [9] (see also [10]), W. Coppel [7], and in the case of Banach spaces by Ju. Dalec’ki˘ı and M. Kre˘ın [8]. For more recent works we refer to [2], [6], [11], [15], [16] and the references therein. In this paper we consider a notion of polynomial dichotomy mimicking a corresponding notion of contraction introduced in [4], now with rates of expansion and contraction varying polynomially instead of exponentially. We note that it follows from results in that paper that the notion of nonuniform polynomial dichotomy occurs naturally, in fact being related to the nonvanishing of a certain Lyapunov exponent. To formulate a rigorous statement we first introduce the notion of polynomial dichotomy in a particular case. Let B(X) be the space of bounded linear operators in a Banach space X. For simplicity of the exposition, we assume that there is a decomposition X = E ⊕F . Moreover, given a sequence (Am )m∈N ⊂ B(X) of invertible operators we assume that Bm 0 Am = 0 Cm with respect to the above decomposition. We say that the sequence (Am )m∈N admits a nonuniform polynomial dichotomy if there exist constants a < 0, ε ≥ 0 and K > 0 such that (1.1)

kBm−1 . . . Bn k ≤ K(m/n)a nε , −1 kCm

−1 . . . Cn−1 k

a ε

≤ K(n/m) n ,

m > n, m < n.

We also introduce a natural notion of Lyapunov exponent in the present context. Namely, the polynomial Lyapunov exponent of a vector v ∈ X (with respect to the sequence (Am )m∈N ) is defined by λ(v) = lim sup n→∞

log kAn . . . A1 vk . log n

One can easily verify that if the sequence (Am )m∈N admits a nonuniform polynomial dichotomy, then (1.2)

λ|(E \ {0}) < 0

and λ|(F \ {0}) > 0.

On the other hand, it follows from results in [4] that if condition (1.2) holds, and the Lyapunov exponents of the sequences Am and (A∗m )−1 are finite for

Robustness of Nonuniform Polynomial Dichotomies

359

nonzero vectors, then the sequence (Am )m∈N admits a nonuniform polynomial dichotomy. In a certain sense, this shows that the notion of nonuniform polynomial dichotomy occurs naturally, in the sense that it can be deduced from the nonvanishing of an appropriate Lyapunov exponent. We emphasize that we also consider the general case of nonuniform polynomial behavior. Indeed, the constant ε in (1.1) may be positive. It turns out that the classical notion of (uniform) exponential dichotomy is very stringent for the dynamics and it is of interest to look for more general types of hyperbolic behavior. This is precisely what happens with the notion of nonuniform exponential behavior. We refer to [1] for a detailed exposition of the theory, which goes back to the landmark works of V. Oseledets [12] and Pesin [14]. In particular, the notion of nonuniform hyperbolicity plays an important role in the construction of stable and unstable invariant manifolds (see [14], [17], [18]). We refer to [1], [3] for related discussions. We note that a different notion of nonuniform polynomial dichotomy was introduced in [5]. It corresponds to replace the terms (m/n)a and (n/m)a in (1.1), respectively by (1 + m − n)a and (1 + n − m)a . However, since m/n = 1 + (m − n)/n ≤ 1 + m − n

for m ≥ n,

and n/m ≤ 1 + (n − m)/m ≤ 1 + n − m for m ≤ n, our notion is less restrictive (recall that a < 0). Moreover, as explained above, the inequalities in (1.1) occur naturally, in the sense that they can be derived from the nonvanishing of a Lyapunov exponent. To the best of our understanding the corresponding notion of dichotomy in [5] has no corresponding motivation. 2. Robustness of polynomial contractions We consider in this section the simpler problem of the robustness of nonuniform polynomial contractions, asking whether a nonuniform polynomial contraction persists under sufficiently small linear perturbations. Let again B(X) be the space of bounded linear operators in a Banach space X. Given a sequence (Am )m∈N ⊂ B(X), we define ( Am−1 . . . An if m > n, A(m, n) = id if m = n. Following [4], the sequence (Am )m∈N is said to admit a nonuniform polynomial contraction if there exist constants a < 0, ε ≥ 0 and K > 0 such that (2.1)

kA(m, n)k ≤ K(m/n)a nε ,

m ≥ n.

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We also consider the perturbed dynamics (2.2)

xm+1 = (Am + Bm )xm

m ∈ N,

and we define ( e A(m, n) =

(Am−1 + Bm−1 ) . . . (An + Bn )

if m > n,

id,

if m = n.

The following is our robustness result for contractions. Theorem 2.1. Assume that (Am )m∈N admits a nonuniform polynomial contraction and that there exist constants η, ρ > 0 such that kBm k ≤ ηm−ρ for m ∈ N. If ρ > ε + 1 and η is sufficiently small, then (Am + Bm )m∈N admits a nonuniform polynomial contraction with (2.3)

e kA(m, n)k ≤

K (m/n)a nε , 1 − Kη2ε−a ζ(ρ − ε)

m ≥ n,

where ζ is the zeta function. Proof. For each n ∈ N we consider the space e e 0 < ∞}, Ω0 := {Ae = (A(m, n))m≥n : kAk with the norm e kA(m, n)k :m≥n . (m/n)a nε It is not difficult to verify that Ω0 is a Banach space. We define an operator T0 in Ω0 by m−1 X e e n). (T0 A)(m, n) = A(m, n) + A(m, k + 1)Bk A(k, e 0 := sup kAk

k=n

It follows from (2.1) that e k(T0 A)(m, n)k ≤ kA(m, n)k +

m−1 X

e n)k kA(m, k + 1)k · kBk k · kA(k,

k=n

≤ K(m/n)a nε + Kη

m−1 X

e 0 (m/(k + 1))a (k + 1)ε k −ρ (k/n)a nε kAk

k=n

≤ K(m/n)a nε + Kη(m/n)a nε

m−1 X

e 0 k a−ρ (k + 1)ε−a kAk

k=n

≤ K(m/n)a nε + Kη2ε−a (m/n)a nε

m−1 X

e 0, k ε−ρ kAk

k=n

and hence (2.4)

e 0 ≤ K + Kη2ε−a ζ(ρ − ε)kAk e 0 < ∞. kT0 Ak


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This shows that the operator T0 : Ω0 → Ω0 is well-defined. Moreover, for each Ae1 , Ae2 ∈ Ω0 and m ≥ n, we have k(T0 Ae1 )(m, n) − (T0 Ae2 )(m, n)k ≤

m−1 X

kA(m, k + 1)k · kBk k · kAe1 (k, n) − Ae2 (k, n)k

k=n

≤ Kη

m−1 X

(m/(k + 1))a (k + 1)ε k −ρ (k/n)a nε kAe1 − Ae2 k0

k=n

≤ Kη(m/n)a nε

m−1 X

(k + 1)ε−a k a−ρ

k=n

≤ Kη2

ε−a

ζ(ρ − ε)(m/n)a nε kAe1 − Ae2 k0 ,

and hence kT0 Ae1 − T0 Ae2 k0 ≤ Kη2ε−a ζ(ρ − ε)kAe1 − Ae2 k0 . Provided that η is sufficiently small, the operator T0 is a contraction. Therefore, e and one can easily verify that there exists a unique Ae ∈ Ω0 such that T0 Ae = A, it is a solution of (2.2). Identity (2.3) follows readily from (2.4). This completes the proof of the theorem. 3. Robustness of polynomial dichotomies We consider in this section the more general case of nonuniform polynomial dichotomies, and the related robustness problem. In particular, we establish the continuous dependence with the perturbation of the constants in the notion of nonuniform polynomial dichotomy. Given a sequence (Am )m∈N ⊂ B(X) of invertible operators, we define  A . . . An if m > n,   m−1 A(m, n) = id if m = n,   −1 −1 Am . . . An−1 if m < n. We say that the sequence (Am )m∈N admits a nonuniform polynomial dichotomy if there exist projections Pn : X → X for n ∈ N such that Pm A(m, n) = A(m, n)Pn ,

m, n ∈ N,

and there exist constants a < 0 < b, ε ≥ 0 and K > 0 such that (3.1)

kA(m, n)Pn k ≤ K(m/n)a nε ,

m ≥ n,

kA(m, n)Qn k ≤ K(n/m)−b nε ,

m ≤ n,

where Qn = id − Pn is the complementary projection of Pn .

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We also consider the perturbed dynamics (2.2), and we  (A + Bm−1 ) . . . (An + Bn )   m−1 e A(m, n) = id   (Am + Bm )−1 . . . (An−1 + Bn−1 )−1

define if m > n, if m = n, if m < n,

whenever the inverses are well-defined. The following is our main robustness result. Theorem 3.1. Assume that (Am )m∈N admits a nonuniform polynomial dichotomy and that there exist constants η, ρ > 0 such that kBm k ≤ ηm−ρ for m ∈ N. If ρ > 2ε + 1, min{−a, b} > ε, and η is sufficiently small, then (Am + Bm )m∈N admits a nonuniform polynomial dichotomy. Namely, there exist projections Pem for m ∈ N such that (3.2)

e e Pem A(m, n) = A(m, n)Pen ,

m, n ∈ N,

and e1 2K K (m/n)a n2ε , b 1 − 2K e e e n k ≤ 2K K2 (n/m)−b n2ε , kA(m, n)Q b 1 − 2K e kA(m, n)Pen k ≤

m ≥ n, m ≤ n,

e n = id − Pen for each n ∈ N, and where where Q K , 1 − 2ε Kη(2−a + 1)ζ(ρ − ε) K e2 = K , 1 − 2ε Kηζ(ρ − ε) b = 2ε Kηζ(ρ − 2ε)(K e1 + K e 2 ). K e1 = K

(3.3)

Proof. We separate the proof into several steps. Step 1. Construction of bounded solutions of equation (2.2). For each n ∈ N we consider the spaces Ω1 = {U = U (m, n)m≥n ⊂ B(X) : kU k1 < ∞}, Ω2 = {V = V (m, n)m≤n ⊂ B(X) : kV k2 < ∞}, respectively with the norms kU (m, n)k kU k1 = sup :m≥n , (m/n)a nε

kV k2 = sup

kV (m, n)k :m≤n . (n/m)−b nε

One can easily verify that Ω1 and Ω2 are Banach spaces.


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Lemma 3.2. For each n ∈ N, equation (2.2) has a unique solution U ∈ Ω1 satisfying, for m ≥ n, (3.4)

U (m, n) = A(m, n)Pn +

m−1 X

A(m, k + 1)Pk+1 Bk U (k, n)

k=n

−

∞ X

A(m, k + 1)Qk+1 Bk U (k, n).

k=m

Proof. One can easily verify that any sequence (U (m, n))m≥n satisfying (3.4) is a solution of equation (2.2). We define an operator T1 in Ω1 by (T1 U )(m, n) = A(m, n)Pn +

m−1 X

A(m, k + 1)Pk+1 Bk U (k, n)

k=n

−

∞ X

A(m, k + 1)Qk+1 Bk U (k, n).

k=m

It follows from (3.1) that (3.5) kA(m, n)Pn k + +

m−1 X

kA(m, k + 1)Pk+1 k · kBk k · kU (k, n)k

k=n ∞ X

kA(m, k + 1)Qk+1 k · kBk k · kU (k, n)k

k=m a ε

≤ K(m/n) n + Kη

m−1 X

(m/(k + 1))a (k + 1)ε k −ρ (k/n)a nε

k=n

+

∞ X

((k + 1)/m)−b (k + 1)ε k −ρ (k/n)a nε kU k1

k=m a ε

≤ K(m/n) n + Kη

m−1 X

(k/(k + 1))a (k + 1)ε k −ρ

k=n

+

∞ X

((k + 1)/m)−b (k/m)a (k + 1)ε k −ρ (m/n)a nε kU k1

k=m a ε

≤ K(m/n) n + Kη

m−1 X

(k + 1)ε−a k a−ρ

k=n

+

∞ X

(k + 1)ε k −ρ (m/n)a nε kU k1

k=m

≤ K(m/n)a nε + 2ε Kη(2−a + 1)ζ(ρ − ε)(m/n)a nε kU k1 for each m ≥ n. This shows that T1 U is well-defined, and that (3.6)

kT1 U k1 ≤ K + 2ε Kη(2−a + 1)ζ(ρ − ε)kU k1 < ∞.

364


Therefore, we obtain an operator T1 : Ω1 → Ω1 . Moreover, for each U1 , U2 ∈ Ω1 and m ≥ n, we have k(T1 U1 )(m, n) − (T1 U2 )(m, n)k ≤

m−1 X

kA(m, k + 1)Pk+1 k · kBk k · kU1 (k, n) − U2 (k, n)k

k=n ∞ X

kA(m, k + 1)Qk+1 k · kBk k · kU1 (k, n) − U2 (k, n)k

+

k=m

≤ Kη

m−1 X

(m/(k + 1))a (k + 1)ε k −ρ (k/n)a nε

k=n

+

∞ X

−b

((k + 1)/m)

ε −ρ

(k + 1) k

a ε

(k/n) n

kU1 − U2 k1

k=m

≤ 2ε Kη(2−a + 1)ζ(ρ − ε)(m/n)a nε kU1 − U2 k1 , and hence (3.7)

kT1 U1 − T1 U2 k1 ≤ 2ε Kη(2−a + 1)ζ(ρ − ε)kU1 − U2 k1 .

Provided that η is sufficiently small, the operator T1 is a contraction, and there exists a unique U ∈ Ω1 such that T1 U = U . Lemma 3.3. For each n ∈ N, equation (2.2) has a unique solution V ∈ Ω2 satisfying, for m ≤ n,

(3.8)

V (m, n) = A(m, n)Qn +

m−1 X

A(m, k + 1)Pk+1 Bk V (k, n)

k=1

−

n−1 X

A(m, k + 1)Qk+1 Bk V (k, n).

k=m

Proof. Again one can easily verify that any sequence (V (m, n))m≤n satisfying (3.8) is a solution of equation (2.2). We define an operator T2 in Ω2 by

(T2 V )(m, n) = A(m, n)Qn +

m−1 X

A(m, k + 1)Pk+1 Bk V (k, n)

k=1

−

n−1 X k=m

A(m, k + 1)Qk+1 Bk V (k, n).


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By (3.1), for each m ≤ n, we have kA(m, n)Qn k +

m−1 X

kA(m, k + 1)Pk+1 k · kBk k · kV (k, n)k

k=1

+

n−1 X

kA(m, k + 1)Qk+1 k · kBk k · kV (k, n)k

k=m

≤ K(n/m)−b nε + Kη

m−1 X

(m/(k + 1))a (k + 1)ε k −ρ (n/k)−b nε

k=1

+

n−1 X

−b

((k + 1)/m)

ε −ρ

(k + 1) k

−b ε

(n/k)

n

kV k2

k=m −b ε

≤ K(n/m)

n + Kη

m−1 X

(m/(k + 1))a (m/k)−b (k + 1)ε k −ρ

k=1

+

n−1 X

((k + 1)/k)−b (k + 1)ε k −ρ (n/m)−b nε kV k2

k=m −b ε

≤ K(n/m)

n + Kη

m−1 X

(k + 1)ε k −ρ

k=1

+

n−1 X

(k + 1)ε k −ρ (n/m)−b nε kV k2

k=m

≤ K(n/m)−b nε + 2ε Kηζ(ρ − ε)(n/m)−b nε kV k2 . This shows that T2 V is well-defined, and that kT2 V k2 ≤ K + 2ε Kηζ(ρ − ε)kV k2 < ∞.

(3.9)

Hence, we obtain an operator T2 : Ω2 → Ω2 . For each V1 , V2 ∈ Ω2 and m ≤ n, we have k(T2 V1 )(m, n) − (T2 V2 )(m, n)k ≤

m−1 X

kA(m, k + 1)Pk+1 k · kBk k · kV1 (k, n) − V2 (k, n)k

k=1

+

n−1 X

kA(m, k + 1)Qk+1 k · kBk k · kV1 (k, n) − V2 (k, n)k

k=m

≤ Kη

m−1 X

(m/(k + 1))a (k + 1)ε k −ρ (n/k)−b nε

k=1

+

n−1 X

((k + 1)/m)−b (k + 1)ε k −ρ (n/k)−b nε kV1 − V2 k2

k=m

≤ 2 Kηζ(ρ − ε)(n/m)−b nε kV1 − V2 k2 , ε

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and hence, kT2 V1 − T2 V2 k2 ≤ 2ε Kηζ(ρ − ε)kV1 − V2 k2 .

(3.10)

For η sufficiently small the operator T2 is a contraction, and there exists a unique V ∈ Ω2 such that T2 V = V . Step 2. Properties of the bounded solutions. Lemma 3.4. We have U (m, l)U (l, n) = U (m, n) for each m ≥ l ≥ n. Proof. It follows from (3.4) that U (m, l)U (l, n) = A(m, n)Pn +

l−1 X

A(m, k + 1)Pk+1 Bk U (k, n)

k=n

+ −

m−1 X k=l ∞ X

A(m, k + 1)Pk+1 Bk U (k, l)U (l, n) A(m, k + 1)Qk+1 Bk U (k, l)U (l, n).

k=m

For a fixed l, writing h(m, l) = U (m, l)U (l, n) − U (m, n) for m ≥ l, we obtain L1 h = h, where (L1 H)(m, l) =

m−1 X

A(m, k + 1)Pk+1 Bk H(k, l) −

k=l

∞ X

A(m, k + 1)Qk+1 Bk H(k, l)

k=m

for each H ∈ Ωl1 and m ≥ l, where Ωl1 is obtained from Ω1 replacing n by l. It follows from (3.5) that L1 is well-defined. Moreover, k(L1 H)(m, l)k ≤

m−1 X

kA(m, k + 1)Pk+1 Bk H(k, l)k

k=l

+

∞ X

kA(m, k + 1)Qk+1 Bk H(k, l)k

k=m

≤ Kη

m−1 X

(m/(k + 1))a (k + 1)ε k −ρ (k/l)a lε

k=l

+

∞ X

((k + 1)/m)−b (k + 1)ε k −ρ (k/l)a lε kHk1

k=m

≤ 2ε Kη(2−a + 1)ζ(ρ − ε)(m/l)a lε kHk1 , that is, kL1 Hk1 ≤ 2ε Kη(2−a + 1)ζ(ρ − ε)kHk1 < ∞.


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We thus obtain an operator L1 : Ωl1 → Ωl1 . For each H1 , H2 ∈ Ωl1 and m ≥ l, we have k(L1 H1 )(m, l) − (L1 H2 )(m, l)k ≤

m−1 X

kA(m, k + 1)Pk+1 k · kBk k · kH1 (k, l) − H2 (k, l)k

k=l

+

∞ X

kA(m, k + 1)Qk+1 k · kBk k · kH1 (k, l) − H2 (k, l)k

k=m

≤ Kη

m−1 X

(m/(k + 1))a (k + 1)ε k −ρ (k/l)a lε

k=l

+

∞ X

((k + 1)/m)−b (k + 1)ε k −ρ (k/l)a lε kH1 − H2 k1

k=m

≤ 2ε Kη(2−a + 1)ζ(ρ − ε)(m/l)a lε kH1 − H2 k1 . Therefore, kL1 H1 − L1 H2 k1 ≤ 2ε Kη(2−a + 1)ζ(ρ − ε)kH1 − H2 k1 , and for η sufficiently small there exists a unique H ∈ Ωl1 such that L1 H = H. Since 0 ∈ Ωl1 also satisfies this identity, we have H = 0. Moreover, since h ∈ Ωl1 we conclude that h = 0. Lemma 3.5. We have V (m, l)V (l, n) = V (m, n) for each m ≤ l ≤ n. Proof. The argument is analogous to that in the proof of Lemma 3.4. By (3.8), we have V (m, l)V (l, n) = A(m, n)Qn −

n−1 X

A(m, k + 1)Qk+1 Bk V (k, n)

k=l

+

m−1 X

A(m, k + 1)Pk+1 Bk V (k, l)V (l, n)

k=1

−

l−1 X

A(m, k + 1)Qk+1 Bk V (k, l)V (l, n).

k=m

Now set h(m, l) = V (m, l)V (l, n) − V (m, n) for each m ≤ l. Then L2 h = h, where (L2 H)(m, l) =

m−1 X k=1

A(m, k + 1)Pk+1 Bk H(k, l) −

l−1 X

A(m, k + 1)Qk+1 H(k, l)

k=m

for each H ∈ Ωl2 and m ≤ l, where Ωl2 is obtained from Ω2 replacing n by l. Proceeding in a similar manner to that in the proof of Lemma 3.4, one can show

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that 0 is the unique fixed point of L2 in Ωl2 , and since h ∈ Ωl2 we conclude that h = 0. Step 3. Construction of the projections Pem in (3.2). Given p ∈ N we define (3.11)

e e m), P m = A(m, p)U (p, p)A(p,

e e m) Qm = A(m, p)V (p, p)A(p,

for each m ∈ N. We emphasize that the operators P m and Qm may depend on p. It follows from Lemmas 3.4 and 3.5 that: (a) P m and Qm are projections for each m ∈ N; e e e e (b) P m A(m, n) = A(m, n)Qn for each m, n n) = A(m, n)P n and Qm A(m, in N. Moreover, since (3.12)

P p = U (p, p) = Pp −

∞ X

A(p, k + 1)Qk+1 Bk U (k, p)

k=p

(3.13)

Qp = V (p, p) = Qp +

p−1 X

A(p, k + 1)Pk+1 Bk V (k, p),

k=1

we obtain: (c) Pp P p = Pp , Qp Qp = Qp , Qp (id − P p ) = id − P p , Pp (id − Qp ) = id − Qp . e (m, p) = U (m, p)Pp satisfies identity (3.4) with n = p. Since We also note that U e U ∈ Ω1 , it follows from the uniqueness in Lemma 3.2 that U (m, p)Pp = U (m, p). Similarly, Ve (m, p) = V (m, p)Qp satisfies identity (3.8) and the uniqueness in Lemma 3.3 implies that V (m, p)Qp = V (m, p). Setting m = p we obtain: (d) P p Pp = P p and Qp Qp = Qp . Lemma 3.6. If η is sufficiently small, then the operator Sp = P p + Qp is invertible. Proof. It follows from (c) that (3.14)

P p + Qp − id = Qp P p + Pp Qp .

By (3.12) and (3.13), we obtain Pp Qp = Pp V (p, p) =

p−1 X

A(p, k + 1)Pk+1 Bk V (k, p),

k=1 ∞ X

Qp P p = Qp U (p, p) = −

A(p, k + 1)Qk+1 Bk U (k, p).

k=p

On the other hand, by (3.6) and (3.3), for each m ≥ n we have (3.15)

e 1 (m/n)a nε , kU (m, n)k ≤ K


369

and by (3.9) and (3.3), for each m ≤ n we have e 2 (n/m)−b nε . kV (m, n)k ≤ K

(3.16)

It follows from (3.14)–(3.16) that kP p + Qp − idk ≤

∞ X

kA(p, k + 1)Qk+1 k · kBk k · kU (k, p)k

k=p

+

p−1 X

kA(p, k + 1)Pk+1 k · kBk k · kV (k, p)k

k=1

e 1 Kη ≤K

∞ X

((k + 1)/p)−b (k + 1)ε k −ρ (k/p)a pε

k=p

e 2 Kη +K

p−1 X

(p/(k + 1))a (k + 1)ε k −ρ (p/k)−b pε

k=1

e 1 Kη ≤K

∞ X

((k + 1)/p)−b (k + 1)ε k −ρ (k/p)a k ε

k=p

e 2 Kη +K

p−1 X

(p/(k + 1))a (k + 1)ε k −ρ (p/k)−b+ε (p/k)−ε pε

k=1

e 1 Kη ≤K

∞ X

e 2 Kη (k + 1)ε k ε−ρ + K

k=p

p−1 X

(k + 1)ε k ε−ρ

k=1

ε

e1 + K e 2 ) = K. b ≤ 2 Kηζ(ρ − 2ε)(K This implies that for η sufficiently small, the operator Sp is invertible. Now we set (3.17)

e e m), Pem = A(m, p)Sp Pp Sp−1 A(p,

e m = A(m, e e m) Q p)Sp Qp Sp−1 A(p,

e m are projections for each for each m ∈ N. It is easy to show that Pem and Q e e m = id for each m ∈ N. fixed m ∈ N, and that (3.2) holds. We note that Pm + Q Step 4. Norm bounds for the evolution operators. e 1 (m/n)a nε for m ≥ n. e Lemma 3.7. We have kA(m, n)|ImP n k ≤ K Proof. We first show that if (zm )m≥n is a bounded solution of equation (2.2), then (3.18)

zm = A(m, n)Pn zn +

m−1 X

A(m, k + 1)Pk+1 Bk zk

k=n

−

∞ X k=m

A(m, k + 1)Qk+1 Bk zk

370


for each m ≥ n. Note that zm = Pm zm + Qm zm , where Pm zm = A(m, n)Pn zn +

m−1 X

A(m, k + 1)Pk+1 Bk zk ,

k=n

Qm zm = A(m, n)Qn zn +

m−1 X

A(m, k + 1)Qk+1 Bk zk .

k=n

We rewrite the last identity in the form Qn zn = A(n, m)Qm zm −

(3.19)

m−1 X

A(n, k + 1)Qk+1 Bk zk .

k=n

On the other hand, we have ∞ X

kA(n, k + 1)Qk+1 Bk zk k ≤ Kη

k=n

∞ X

((k + 1)/n)−b (k + 1)ε k −ρ kzk k

k=n ε

≤ Kη2 ζ(ρ − ε) sup kzk k, k≥n

and kA(n, m)Qm zm k ≤ K(m/n)−b mε kzm k = K(m/n)−b+ε nε kzm k. Therefore, letting m → ∞ in (3.19) yields Qn zn = −

∞ X

A(n, k + 1)Qk+1 Bk zk .

k=n

Consequently, Qm zm = − =−

∞ X

A(m, k + 1)Qk+1 Bk zk +

k=n ∞ X

m−1 X

A(m, k + 1)Qk+1 Bk zk

k=n

A(m, k + 1)Qk+1 Bk zk ,

k=m

which yields (3.18). e Now given ξ ∈ X we consider the solution zm = A(m, n)P n ξ of equation (2.2) for m ≥ n. By (3.11) we have e e n)ξ = U (m, p)A(p, e n)ξ. zm = A(m, p)U (p, p)A(p, e The last identity follows from the fact that both A(m, p)U (p, p) and U (m, p) are solutions of equation (2.2), which coincide for m = p. Since m 7→ U (m, p) is


371

bounded, this shows that (zm )m≥n is a bounded solution of equation (2.2) with zn = P n ξ. By (3.18), for each m ≥ n we have e P m A(m, n)ξ = A(m, n)Pn P n ξ +

m−1 X

e n)ξ A(m, k + 1)Pk+1 Bk P k A(k,

k=n

−

∞ X

e n)ξ. A(m, k + 1)Qk+1 Bk P k A(k,

k=m

e Then, writing B = (A(m, n)|ImP n )m≥n we obtain e kP m A(m, n)ξk ≤ K(m/n)a nε kP n ξk + Kη + Kη

m−1 X k=n ∞ X

e n)ξk (m/(k + 1))a (k + 1)ε k −ρ kP k A(k, e n)ξk ((k + 1)/m)−b (k + 1)ε k −ρ kP k A(k,

k=m

= K(m/n)a nε kP n ξk + Kη + Kη

m−1 X k=n ∞ X

e n)P n ξk (m/(k + 1))a (k + 1)ε k −ρ kP k A(k, e n)P n ξk ((k + 1)/m)−b (k + 1)ε k −ρ kP k A(k,

k=m

≤ K(m/n)a nε kP n ξk + Kη + Kη

m−1 X k=n ∞ X

(m/(k + 1))a (k + 1)ε k −ρ (k/n)a nε kBk1 kP n ξk ((k + 1)/m)−b (k + 1)ε k −ρ (k/n)a nε kBk1 kP n ξk

k=m

≤ K(m/n)a nε kP n ξk + 2ε Kη(2−a + 1)ζ(ρ − ε)(m/n)a nε kBk1 kP n ξk. Therefore, kBk1 ≤ K + 2ε Kη(2−a + 1)ζ(ρ − ε)kBk1 , and since η is sufficiently small (see (3.7)), kBk1 ≤

K e1. =K 1 − 2ε Kη(2−a + 1)ζ(ρ − ε)

This yields the desired inequality.

e e 2 (n/m)−b nε for m ≤ n. Lemma 3.8. We have kA(m, n)|ImQn k ≤ K Proof. Since −a > ε, proceeding in a similar manner to the proof of Lemma 3.7, we conclude that if (zm )m≤n is a bounded solution of equation (2.2),

372


then (3.20)

zm = A(m, n)Qn zn +

m−1 X

A(m, k + 1)Pk+1 Bk zk

k=1

−

n−1 X

A(m, k + 1)Qk+1 Bk zk .

k=m

Now given ξ ∈ X we have e n)ξ e zm := A(m, n)Qn ξ = V (m, p)A(p,

for m ≤ n.

Therefore, (zm )m≤n is a bounded solution of equation (2.2) with zn = Qn ξ, and it follows from (3.20) that e Qm A(m, n)ξ = A(m, n)Qn Qn ξ +

m−1 X

e n)ξ A(m, k + 1)Pk+1 Bk Qk A(k,

k=1

−

n−1 X

e n)ξ. A(m, k + 1)Qk+1 Bk Qk A(k,

k=m

e Therefore, writing C = (A(m, n)|ImQn )m≤n we obtain e kQm A(m, n)ξk ≤ K(n/m)−b nε kQn ξk + Kη

m−1 X

e n)ξk (m/(k + 1))a (k + 1)ε k −ρ kQk A(k,

k=1

+ Kη

n−1 X

e n)ξk ((k + 1)/m)−b (k + 1)ε k −ρ kQk A(k,

k=m −b ε

n kQn ξk

= K(n/m) + Kη

m−1 X

e n)Qn ξk (m/(k + 1))a (k + 1)ε k −ρ kQk A(k,

k=1

+ Kη

n−1 X

e n)Qn ξk ((k + 1)/m)−b (k + 1)ε k −ρ kQk A(k,

k=m

≤ K(n/m)−b nε kQn ξk + Kη

m−1 X

(m/(k + 1))a (k + 1)ε k −ρ (n/k)−b nε kCk2 kQn ξk

k=1

+ Kη

n−1 X

((k + 1)/m)−b (k + 1)ε k −ρ (n/k)−b nε kCk2 kQn ξk

k=m −b ε

≤ K(n/m)

n kQn ξk + 2ε Kηζ(ρ − ε)(n/m)−b nε kCk2 kQn ξk,

and hence, kCk2 ≤ K + 2ε Kηζ(ρ − ε)kCk2 .


373

Since η is sufficiently small (see (3.10)), it follows that kCk2 ≤

K e2. =K 1 − 2ε Kηζ(ρ − ε)

This yields the desired inequality.

Lemma 3.9. We have e e 1 (m/n)a nε kPen k, m ≥ n, kA(m, n)Pen k ≤ K e en k ≤ K e 2 (n/m)−b nε kQ e n k, m ≤ n. kA(m, n)Q

(3.21) (3.22)

Proof. By property (d), we have Sp Pp = (P p + Qp )Pp = P p ,

Sp Qp = (P p + Qp )Qp = Qp .

e e m) for m ∈ N, we can show that Setting Sm = A(m, p)Sp A(p, Pem Sm = P m

e m Sm = Qm . and Q

Indeed, by (3.17) we have e e m) = A(m, e e m) = P m , Pem Sm = A(m, p)Sp Pp A(p, p)P p A(p, since P p = U (p, p). The other identity can be obtained in a similar manner. Since Sm is invertible, we conclude that ImPem = ImP m

and

e m = ImQm . ImQ

Thus, by Lemmas 3.7 and 3.8, we obtain e e kA(m, n)Pen k ≤ kA(m, n)|ImPen k · kPen k e 1 (m/n)a nε kPen k e = kA(m, n)|ImP n k · kPen k ≤ K for m ≥ n, and e e n k ≤ kA(m, e e n k · kQ en k kA(m, n)Q n)|ImQ e en k ≤ K e 2 (n/m)−b nε kQ en k = kA(m, n)|ImQn k · kQ for m ≤ n.

Lemma 3.10. Provided that η is sufficiently small, for each m ∈ N we have kPem k ≤

2K b 1 − 2K

mε

and

em k ≤ kQ

2K b 1 − 2K

Proof. Given ξ ∈ X, we set 1 e zm = A(m, n)Pen ξ 2 e en ξ zm = A(m, n)Q

for m ≥ n, for m ≤ n.

mε .

374


1 2 By Lemma 3.9, (zm )m≥n and (zm )m≤n are bounded solutions of equation (2.2). It thus follows from (3.18) and (3.20) that

e Pem A(m, n)ξ = A(m, n)Pn Pen ξ +

m−1 X

e n)ξ A(m, k + 1)Pk+1 Bk Pek A(k,

k=n

−

∞ X

e n)ξ, A(m, k + 1)Qk+1 Bk Pek A(k,

k=m

and e m A(m, e en ξ + Q n)ξ = A(m, n)Qn Q

m−1 X

e k A(k, e n)ξ A(m, k + 1)Pk+1 Bk Q

k=1

−

n−1 X

e k A(k, e n)ξ. A(m, k + 1)Qk+1 Bk Q

k=m

Taking m = n, we obtain Qm Pem ξ = −

∞ X

e m)ξ, A(m, k + 1)Qk+1 Bk Pek A(k,

k=m

em ξ = Pm Q

m−1 X

e k A(k, e m)ξ. A(m, k + 1)Pk+1 Bk Q

k=1

By (3.21) and (3.22), we have e 1 Kη kQm Pem k ≤ K

∞ X

((k + 1)/m)−b (k + 1)ε k −ρ (k/m)a mε kPem k

k=m

e 1 Kη2ε ζ(ρ − 2ε)kPem k ≤K and em k ≤ K e 2 Kη kPm Q

m−1 X

em k (m/(k + 1))a (k + 1)ε k −ρ (m/k)−b mε kQ

k=1

e 2 Kη2ε ζ(ρ − 2ε)kQ e m k. ≤K Moreover, since kPm k ≤ Kmε and kQm k ≤ Kmε , we obtain kPem k ≤ kPem − Pm k + kPm k = kPem − Pm Pem − Pm + Pm Pem k + kPm k e m k + kPm k = kQm Pem − Pm Q e m k + kPm k ≤ kQm Pem k + kPm Q e 1 Kη2ε ζ(ρ − 2ε)kPem k + K e 2 Kη2ε ζ(ρ − 2ε)kQ e m k + Kmε ≤K b Pem k + kQ e m k) + Kmε , ≤ K(k


375

and e m k ≤ kQ e m − Qm k + kQm k kQ b Pem k + kQ e m k) + Kmε . = kPem − Pm k + kQm k ≤ K(k Summing the two yields e m k ≤ 2K(k b Pem k + kQ e m k) + 2Kmε . kPem k + kQ This completes the proof of the lemma.

The statement of Theorem 3.1 follows now readily from the above lemmas.

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Manuscript received July 2, 2010

Luis Barreira and Claudia Valls Departamento de Matem´ atica Instituto Superior T´ ecnico 1049-001 Lisboa, PORTUGAL E-mail address: [email protected], [email protected] Meng Fan School of Mathematics and Statistics Northeast Normal University 5268 Renmin Street Changchun, Jilin, 130024, P. R. CHINA E-mail address: [email protected] Jimin Zhang School of Mathematics and Statistics Northeast Normal University 5268 Renmin Street, Changchun Jilin, 130024, P. R. CHINA and School of Mathematical Sciences Heilongjiang University 74 Xuefu Street Harbin, Heilongjiang, 150080, P. R. CHINA E-mail address: [email protected]

TMNA : Volume 37 – 2011 – No 2