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Evolutionary Semigroups and

arXiv:math/9307201v2 [math.FA] 6 Dec 1999Evolutionary Semigroups andLyapunov Theorems in BanachSpacesYuri LatushkinUniversity of Missouri, Columbia, Missouri 65211andStephen Montgomery-Smith 1University of Missouri, Columbia, Missouri 652111. IntroductionLet us consider an autonomous differential equation v′ = Av in aBanach space E, where A is a generator of C0-semigroup {etA}t≥0.Denote, as usual, s(A) = sup{Re λ : λ ∈σ(A)} and ω(A) = inf{ω :∥etA∥≤Meωt}.A classical result of A. M. Lyapunov (see, e.g., [9]) shows that forany bounded operator A ∈B(E) the spectrum σ(A) of A is responsiblefor the asymptotic behavior of the solution y(t) = eAty(0) of the aboveequation.

For example, if σ(A) is contained in the left half-plane, thatis s(A) < 0, then the trivial solution is uniformly asymptotically stable,that is ω(A) < 0, and ∥etA∥→0 as t →∞. This fact follows from thespectral mapping theorem (see, e.g., [21]):σ(etA)\{0} = exp tσ(A),t > 0,(1)which always holds for bounded A.For unbounded A, equation (1) is not always true.

Moreover, thereare examples of generators A (see [21]) such that even s(A) < 0 doesnot guarantee ω(A) < 0 and ∥etA∥→0 as t →∞. Since σ(A) does notcharacterize the asymptotic behavior of the solution v(·), we would liketo find some other characterization that still does not involve solvingthe differential equation.1This author was supported by the National Science Foundation under the grantDMS-9201357.1

In this paper we solve precisely this problem in the following manner.Consider the space Lp(R; E) of E-valued functions for 1 ≤p < ∞, orthe space C0(R; E) of continuous vanishing at ±∞functions on R, andthe semigroup {etB}t≥0 of evolutionary operators(etBf)(x) = etAf(x −t), t ≥0,(2)generated by the operator B that is the closure of −ddx + A, x ∈R. Itturns out that it is σ(B) in Lp(R; E) (or in C0(R; E)) that is responsiblefor the asymptotic behavior of v(·) in E. For example, s(B) = ω(A),and s(B) < 0 in Lp(R; E) or C0(R; E) implies ∥etA∥→0 as t →∞onE.The order of the proofs is as follows.

First, we consider the evolu-tionary semigroup {etB} in the space Lp([0, 2π]; E) or C([0, 2π]; E) of2π-periodic functions. We prove that 1 /∈σ(e2πA) in E is equivalent to1 /∈σ(e2πB) or 0 /∈σ(B) in Lp([0, 2π]; E) or C([0, 2π]; E).

The mainpart of the proof uses a modification of an idea due to C. Chicone andR. Swanson [6].

Next, using this result, we give a variant of the spectralmapping theorem for a semigroup {etA} in a Banach space E. Thisspectral mapping theorem is a direct generalization of L. Gearhart’sspectral mapping theorem for Hilbert spaces (see, e.g., [21, p. 95]),and is related to the spectral mapping theorem of G. Greiner [21, p.94]. Finally, using a simple change of variables arguments, we provethat σ(etA) ∩T = ∅, t ̸= 0 in E, T = {z : |z| = 1} is equivalentto σ(etB) ∩T = ∅, t ̸= 0 which in turn is equivalent to 0 /∈σ(B) inLp(R; E) or C0(R; E).We will also consider the well-posed nonautonomous equation v′ =A(x)v in E, and its associated evolutionary family {U(x, s)}x≥s, whichcan be thought as a propagator of this equation, that is v(x) = U(x, s)v(s).We assume that U(· , ·) is strongly continuous and satisfies the usual([31, p. 89]) algebraic properties of the propagator.Instead of thesemigroup given by (2) we consider on Lp(R; E) or C0(R; E) the evo-lutionary semigroup(etGf)(x) = U(x, x −t)f(x −t),x ∈R, t ≥0.

(3)We will show that σ(G) characterizes the asymptotic behavior of v(·)and prove the spectral mapping theorem for the semigroup {etG}.2

We will be considering not only stability but also the exponentialdichotomy (hyperbolicity) for the solutions of the equation v′ = A(x)vor evolutionary family {U(x, s)}. We say, that an evolutionary fam-ily {U(x, s)}x≥s is (spectrally) hyperbolic if there exists a continuousin the strong sense, bounded, projection-valued function P : R →B(E) such that: a) The norm of the restrictions U(x, s) Im P(s) (resp.U(x, s) Ker P(s)) exponentially decreases (resp.

increases) with x −s,and b) Im U(x, s) Ker P(s) is dense in Ker P(x). Note, that b) au-tomatically follows from a) if the operators U(x, s) are invertible anddefined for all (x, s) ∈R2.

This happens, in particular, if U(· , ·) is anorm-continuous propagator for the differential equation v′ = A(x)vwith continuous and bounded A : R →B(E). For this case the hy-perbolicity of {U(x, s)} coincides with the exponential dichotomy (see,e.g., [9]) of the equation.

However, generally a) does not imply b) (see[27]).Exponential dichotomy in the theory of differential equations withbounded coefficients on E is a major tool used for proving instabil-ity theorems for nonlinear equations, and for showing existence anduniqueness of bounded solutions and Green’s functions, etc. (see, e.g.

[7, 9]). The spectral mapping theorem for the semigroup (3) which isgiven here allows one to extend these ideas to the case of unboundedcoefficients.It turns out that the spectrum σ(etG) for nonperiodic A(·) plays thesame role in the description of exponential dichotomy as the spectrumof the monodromy operator does in the usual Floquet theory for theperiodic case.

That is, the condition σ(etG) ∩T = ∅, t ̸= 0, or equiv-alently 0 /∈σ(G), is equivalent to the (spectral) hyperbolicity of theevolutionary family {U(x, s)}x≥s.Showing that the hyperbolicity σ(eG) ∩T = ∅of the operator eGimplies the hyperbolicity of {U(x, s)} is a delicate matter. It turnsout that the Riesz projection P for eG on Lp(R, E) or C0(R; E), thatcorresponds to σ(eG) ∩D, D = {z : |z| < 1}, has the form (Pf)(x) =P(x)f(x).

Here P(·) is a continuous in the strong sense, projection-valued function, that defines the hyperbolicity of {U(x, s)}. Note thateG = aR, where a is an operator of multiplication by the functiona(x) = U(x, x −1), that is (af)(x) = a(x)f(x), and R is a translation3

operator (Rf)(x) = f(x −1), x ∈R. Therefore, eG falls into the classof so-called weighted translation operators, which are well-understoodin the case that E is Hilbert space and p = 2 (see [1, 2, 18, 27], andalso [8, 23] and references therein).

If U(· , ·) is norm-continuous, thenP is an operator from a C∗-algebra generated by R and the C∗-algebraAnc of operators of multiplication by the norm-continuous, boundedfunctions from R to B(E). The techniques from the theory of weightedtranslation operators (see [1, 2, 18, 27]) allows one to conclude thatP ∈Anc.

This technique is not applicable to the case where {U(·, ·)}is only strongly-continuous, nor also to the case when E is not Hilbertspace.In this paper we present some new approaches, which allows one toderive the above result for any Banach space E and is new even for theHilbert space case and when it is only known that U(·, ·) is stronglycontinuous. The main idea is to “discretize” the operator aR, that isto represent it by the family of operators πx(a)S, x ∈R, acting on the“discrete” space lp(Z; E).

Here S : (vn)n∈Z 7→(vn−1)n∈Z is the shiftoperator and πx(a) : (vn)n∈Z 7→(a(x + n)vn)n∈Z is a diagonal operatoron lp(Z; E). This idea goes back to the theory of regular representationsof C∗-algebras [26], and is related to works [1, 2, 13, 16, 17, 18].

As aresult we prove that σ(aR)∩T = ∅in Lp(R; E) implies σ(πx(a)S)∩T =∅in lp(Z; E) for each x ∈R, and derive from this fact that P ∈A,where A is the set of bounded functions a : R →B(E) which arecontinuous in strong operator topology on B(E).We point out that the investigation of evolutionary operators (2)–(3)has a long history, probably starting from [14] (see also [10, 11, 19, 22]).Recently significant progress has been made in the papers [3, 4, 25, 27].It is these papers that essentially motivated and influenced this presentwork.Finally, the results of this article can be generalized to the case of thevariational equation v′(t) = A(ϕtx)v(t) for a flow {ϕt} on a compactmetric space X, or to the linear skew-product flow ˆϕt : X × E →X × E : (x, v) 7→(ϕtx, Φ(x, t)v), t ≥0 (see [6, 12, 18, 29, 30] andreferences contained therein). Here Φ : X × R+ →L(E) is a cocycleover ϕt, that is, Φ(x, t+s) = Φ(ϕtx, s)Φ(x, t).

Let us recall (see [29, 30])that part of the purpose of the theory of linear skew-product flows was4

to be able to handle the equation v′ = A(t)v in the case when A(·) isalmost-periodic .To answer the question when ˆϕt is hyperbolic (or Anosov), insteadof (3) one considers the semigroup of so called weighted compositionoperators (see [6, 15, 18]) on Lp(X; µ; E):(T tf)(x) =dµ ◦ϕ−tdµ1/pΦ(ϕ−tx, t)f(ϕ−tx),x ∈X, t ≥0. (4)Here µ is a ϕt-quasi-invariant Borel measure on X.As above, thecondition σ(T t) ∩T = ∅is equivalent to the spectral hyperbolicity ofthe linear skew-product flow ˆϕt.

The spectral hyperbolicity coincideswith the usual hyperbolicity if Φ(x, t), x ∈X, t ≥0 are invertible orcompact operators. A detailed investigation of weighted compositionoperators and their connections with the spectral theory of linear skew-product flows and other questions of dynamical system theory may befound in [18] (see also [27]).We will use the following notations: D = {z : |z| < 1}; T = {z : |z| =1}; “” denotes the restriction of an operator; D(·) = DF(·) denotes thedomain of an operator in a space F; σ(·) = σ(· ; F) denotes the spec-trum; σap(·) = σap(· ; F) denotes the approximative point spectrum;σr(·) = σr(· ; F) denotes the residual spectrum; and ρ(·) = ρ(· ; F) de-notes the resolvent set of an operator on F. For an operator A in E wedenote by A the operator of multiplication by A in a space of E-valuedfunctions: (Af)(x) = Af(x), f : R →E.The authors would very much like to thank C. Chicone for help andsuggestions, and R. Rau for many illuminating discussions.The authors also would like to thank the referee for the suggestionto shorten the proof of Theorem 2.5 and Remark below.2.

Autonomous CaseLet A be a generator of a C0-semigroup {etA}t≥0 on a Banach spaceE. The semigroup is called hyperbolic if σ(etA) ∩T = ∅for t ̸= 0.In this section we will characterize the hyperbolicity of the semigroup{etA} in terms of evolutionary semigroup {etB}t≥0 and its generator B.This semigroup acts by the rule (etBf)(x) = etAf(x −t) on functionsf with values in E. In Subsection 2.1 we consider {etB} acting on the5

space Lp([0, 2π]; E) and C([0, 2π]; E). In Subsection 2.3 {etB} acts onLp(R; E) and C0(R; E).

Subsection 2.2 is devoted to a spectral map-ping theorem for {etA}t≥0 on E which generalizes the spectral mappingtheorem of L. Gearhart for Hilbert space.2.1. Periodic CaseLet F denote one of the spaces Lp([0, 2π], E), 1 ≤p < ∞or C([0, 2π], E)of 2π-periodic E-valued functions f, f(0) = f(2π).

Consider the evo-lutionary semigroup {etB}t≥0 acting on F, defined by the rule(etBf)(x) = etAf([x −t](mod 2π)), x ∈[0, 2π].Of course, [0, 2π] here was chosen for convenience, and for a semigroup(etBf)(x) = etAf ([x −t](mod t0)) the proofs below remain the samefor any t0 > 0.Note that etB in F is a product of two commuting semigroups (Utf)(x) =f([x −t](mod 2π)) and (etAf)(x) = etAf(x). Hence the generator B isthe closure of the operator(B0f)(x) = −ddxf(x) + Af(x),(5)where B0 is defined on the core D(B0) of B (see [21, p. 24]).

Moreover,DF(B0) = DF(−d/dx) ∩DF(A), where the derivative d/dx is takenin the strong sense in E, and DF(B0) = {f : [0, 2π] →D(A)f ∈F is absolutely continuous, ddxf ∈F, and Af ∈F}.Since Beik·f(·) = eik·(B−ik)f(·), k ∈Z, for the operator (Lkf)(x) =eikxf(x) one has BLk = Lk(B −ik). Therefore, the spectrum σ(B) inF is invariant under translations by i.We will need the following Lemma.Lemma 2.1.

If 1 ∈σap(e2πA) in E, then 0 ∈σap(B) in F.Proof. Fix m ∈N, m ≥2.

Since 1 ∈σap(e2πA), we can choose v ∈Esuch that ∥v∥E = 1 and ∥v −e2πAv∥E < 1m. Note also that ∥e2πAv∥E ≥1 −1m.Let α : [0, 2π] →[0, 1] be any smooth function with bounded deriv-ative such that α(x) = 0 for x ∈[0, 2π3 ] and α(x) = 1 for x ∈[ 4π3 , 2π].Define a function g : [0, 2π] →E by the the formulag(x) = [1 −α(x)]e(2π+x)Av + α(x)exAv, x ∈[0, 2π].

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Note that g(0) = g(2π) = e2πAv. Obviously, g ∈F.

Also,(etBg)(x) = [1 −α(x −t)]e(2π+x)Av + α(x −t)exAv,g ∈DF(B), and(Bg)(x) = α′(x)exA[e2πAv −v],x ∈[0, 2π]. (7)Let us denote a = max{|α′(x)| : x ∈[0, 2π]} and b = max{∥exA∥:x ∈[0, 2π]}.

Note, that ∥e2πAv∥= ∥e(2π−x)AexAv∥≤b∥exAv∥for anyx ∈[0, 2π].First let us suppose that F = Lp([0, 2π]; E). Then∥Bg∥Lp([0,2π];E) ≤(2π)1/pabm.On the other hand,∥g∥pLp([0,2π];E) ≥Z 2π4π3∥exAv∥pdx ≥2π3 b−p∥e2πAv∥p ≥2π3 b−p(1 −1m)p.Finally,∥Bg∥Lp([0,2π];E) ≤(2π)1/pabm ≤31/pab2∥g∥Lp([0,2π];E) ·1m −1.

(8)Since this holds for all m, it follows that 0 ∈σap(B).Now suppose that F = C([0, 2π], E). Then∥Bg∥C([0,2π],E) ≤abm, ∥g∥C([0,2π],E) ≥∥g(0)∥E = ∥e2πAv∥≥1 −1m(9)and hence∥Bg∥C([0,2π];E) ≤abm −1∥g∥C([0,2π];E).

(10)Since this is true for all m, it follows that 0 ∈σap(B).Theorem 2.2. Let F be one of the spaces Lp([0, 2π], E), 1 ≤p < ∞or C([0, 2π], E).

Then the following are equivalent:1) 1 ∈ρ(e2πA) in E;2) 1 ∈ρ(e2πB) in F;3) 0 ∈ρ(B) in F.Proof. 1) ⇒2).

Note that (e2πBf)(x) = e2πAf(x). Hence σ(e2πA; E) =σ(e2πB; F).

Note also that σr(e2πA; E) = σr(e2πB; F).2) ⇒3) follows from the spectral inclusion theorem e2πσ(B) ⊂σ(e2πB)(see [24, p. 45]).7

3) ⇒1). Assume 0 ∈ρ(B; F) but 1 ∈σ(e2πA; E) = σap(e2πA; E) ∪σr(e2πA; E).

By Lemma 2.1 it follows that 1 ∈σr(e2πA; E), and hence1 ∈σr(e2πB; F). By the spectral mapping theorem for the residualspectrum ([24, Theorem 2.5 (ii)]) it follows that ik ∈σr(B; F) forsome k ∈Z.

Since σ(B; F) is invariant under translations by i, wehave that 0 ∈σ(B; F), contradicting 3).2.2. Spectral Mapping Theorem for Banach SpacesAs it is well-known (see, e.g., [21, p. 82–89]), that in general, theinclusion etσ(A) ⊂σ(eAt), t ̸= 0 for a C0-semigroup {etA}t≥0 on a Ba-nach space E is improper.

In particular, iZ ⊂ρ(A) is implied by butdoes not imply 1 ∈ρ(e2πA). For Hilbert space E, however, the follow-ing spectral mapping theorem of L. Gearhart (see [21, p. 95]) is true:1 ∈ρ(e2πA) if and only if iZ ⊂ρ(A) and supk∈Z ∥(A −ik)−1∥< ∞.We will now give a direct generalization of this result to any arbitraryBanach space E. This generalization is related (but independent) toG.

Greiner’s spectral mapping theorem [21, p. 94] that involves C´esarosummability of the seriesPk(A −ik)−1v, v ∈E.Theorem 2.3. Let {etA} be any C0-semigroup on a Banach space Eand let F be one of the spaces Lp([0, 2π], E), 1 ≤p < ∞or C([0, 2π], E).Then the following are equivalent:1) 1 ∈ρ(e2πA);2) iZ ⊂ρ(A) and there is a constant C > 0 such that∥Xk(A −ik)−1eikxvk∥F ≤C∥Xkeikxvk∥F(11)for any finite sequence {vk} ⊂E.Proof.

Consider the evolutionary semigroup {etB}t≥0 on F from theprevious subsection. Consider a finite sequence {vk} ⊂E.

Assumethat (A −ik)−1 exists for all k ∈Z. Define functions f, g ∈F by therulef(x) =Xk(A −ik)−1eikxvk, g(x) =Xkeikxvk,x ∈[0, 2π].

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Since (A −ik)−1 : E →D(A), one has Bf = g. Indeed(Bf)(x) = ddtetAf([x −t](mod 2π))t=0=Xk[A(A −ik)−1eikxvk −ik(A −ik)−1eikxvk] = g.1) ⇒2). If 1 ∈ρ(e2πA), then the inclusion iZ ⊂ρ(A) follows fromthe spectral inclusion theorem e2πσ(A) ⊂σ(e2πA).

In accordance withpart 1) ⇒3) of Theorem 2.2, the operator B has bounded inverse B−1on F provided that 1 ∈ρ(e2πA). Denote C = ∥B−1∥, and considerfunctions (12).

Then ∥f∥F = ∥B−1g∥F ≤C∥g∥F, and (11) is proved.2) ⇒1). First, we show that 2) implies 0 /∈σap(B).

Indeed, thefunctions of type g in (12) are dense in F. If we let uk = (A −ik)−1vk,then we note that the functions of type f are also dense in F. Now(11) implies ∥Bf∥F = ∥g∥F ≥C−1∥f∥F, and 0 /∈σap(B).Assume that 2) is fulfilled, but 1 ∈σ(e2πA) = σr(e2πA)∪σap(e2πA). If1 ∈σap(e2πA) in E then, by Lemma 2.1, 0 ∈σap(B), in contradiction tothe previous paragraph.

On the other hand, 1 ∈σr(e2πA) implies, bythe spectral mapping theorem for residual spectrum, that ik ∈σr(A)for some k ∈Z, contradicting iZ ⊂ρ(A).Remark. We note, that 1) ⇒2) can be also seen directly.

Indeed,assuming 1), let us denote φ(s) = (e2πA −I)−1esA, s ∈[0, 2π]. Thenthe convolution operator(Kf)(x) =2πZ0φ(s)f(x −s) dsis a bounded operator on F. But(A −ik)−1 =2πZ0e−iks(e2πA −I)−1esA ds,k ∈Z,are Fourier coefficients of φ : [0, 2π] →B(E).

Inequality (11) can beviewed as the condition of boundedness of K, which gives 2).Let us show now that Theorem 2.3 is really a direct generalizationof L. Gearhart’s Theorem, mentioned above. Indeed, for Hilbert space9

E and p = 2, Parseval’s identity implies:∥Xk(A −ik)−1eik·vk∥L2([0,2π];E) =2πXk∥(A −ik)−1vk∥2E1/2∥Xkeik·vk∥L2([0,2π];E) =2πXk∥vk∥2E1/2.Clearly, (11) is equivalent to the condition sup{∥(A−ik)−1∥: k ∈Z} <∞.We conclude this subsection by giving four more statements equiva-lent to 1) and 2) in Theorem 2.3:3) iZ ⊂ρ(A) and there is a constant C > 0 such that∥Bf∥L1([0,2π];E) ≥C−1∥f∥C([0,2π];E)for all f ∈C([0, 2π]; E) such that Bf ∈L1([0, 2π]; E);4) iZ ⊂ρ(A) and there is a constant C > 0 such that∥Bf∥C([0,2π];E) ≥C−1∥f∥L1([0,2π];E)for all f ∈L1([0, 2π]; E) such that Bf ∈C([0, 2π]; E);5) iZ ⊂ρ(A) and there is a constant C > 0 such that∥Xk(A −ik)−1eikxvk∥C([0,2π];E) ≤C∥Xkeikxv∥L1([0,2π];E)for any finite sequence {vk} ⊂E;6) iZ ⊂ρ(A) and there is a constant C > 0 such that∥Xk(A −ik)−1eikxvk∥L1([0,2π];E) ≤C∥Xkeikxv∥C([0,2π];E)for any finite sequence {vk} ⊂E.2.3. Real LineConsider now the evolutionary semigroup {etB}t≥0,(etBf)(x) = etAf(x −t)(13)acting on the space F = Lp(R; E), 1 ≤p < ∞, or F = C0(R; E).Formula (5) is valid and the identitiesetBeiξ·f(·) = eiξ·e−iξtetBf(·),Beiξ·f(·) = eiξ·(B −iξ)f(·),ξ ∈R,10

show that σ(etB) in F is invariant under rotations centered at the origin,and that σ(B), σap(B) and σr(B) in F are invariant under translationsparallel to iR.First we state a simple lemma. Let Fs be one of the spaces Fs =lp(Z; E), 1 ≤p < ∞or Fs = c0(Z; E) (of sequences (vn)n∈Z suchthat vn →0 as n →±∞).

Let S be a shift operator on Fs, that isS : (vn) 7→(vn−1). For an operator a on E we will denote by Da thediagonal operator on Fs, acting by the rule Da : (vn)n∈Z 7→(avn)n∈Z.Lemma 2.4.

The following are equivalent:1) σ(a) ∩T = ∅in E;2) σ(DaS) ∩T = ∅in Fs.Proof. We will give the proof for the case when Fs = lp = lp(Z; E).The case Fs = c0(Z; E) can be considered similarly.1) ⇒2).

Since σ(a) ∩T = ∅, there exists a Riesz projection ˆp fora in E that corresponds to the part of the spectrum σ(a) ∩D. Defineˆq = I −ˆp, and consider in F complimentary projections Dˆp and Dˆq.Since DaSDˆp = DaDˆpS, the decomposition Fs = Im Dˆp ⊕Im Dˆq isDaS-invariant.

For the spectral radiusr(·) = limn→∞∥(·)n∥1none has r(ˆpaˆp) < 1 and r([ˆqaˆq]−1) < 1 in E. Hence, r(DaSDˆp) < 1,r([DaSDˆq]−1) < 1, and σ(DaS) ∩T = ∅in Fs.2) ⇒1). Assume that I −DaS is invertible in lp, but for any ǫ > 0there is a vector v ∈E such that ∥v∥E = 1 and ∥v −av∥E < ǫ. Fixq > 0 such that1 −e±q < ǫ,and define a sequence (vn) ∈lp by vn = e−q|n|v, n ∈Z.

ThenI −DaS : (vn)n∈Z 7→e−q|n|(v −av) + (e−q|n| −e−q|n−1|)avn∈Z.A direct calculation shows that∥(I −DaS)(vn)∥lp ≤(1 + ∥a∥) · ǫ · ∥(vn)∥lp,contradicting the invertibility of I −DaS in lp.Let us show now that I −a has a dense range in E, provided I −DaSis an operator onto lp.Indeed, for any u ∈E consider a sequence(un) ∈lp defined by u0 = u and un = 0 for n ̸= 0. Find a sequence11

(vn) ∈lp such that (I −DaS)(vn) = (un), that is vn −avn−1 = un forn ∈Z. But then for k ∈N one hasu =kXn=−k(vn −avn−1)= vk −av−k−1 + (yk −ayk),whereyk =k−1Xn=−kvn.Therefore, Im(I −a) ∋yk −ayk →u, since vk →0 and av−k−1 →0 ask →∞.Theorem 2.5.

Let F be one of the spaces Lp(R; E), 1 ≤p < ∞orC0(R; E), and let t > 0. Then the following are equivalent:1) σ(etA) ∩T = ∅in E;2) σ(etB) ∩T = ∅in F;3) 0 ∈ρ(B) in F.Proof.

2) ⇒3) follows from the spectral inclusion theorem for {etB}.3) ⇒2) we will prove for F = Lp(R; E); the arguments for F =C0(R; E) are similar.Since σ(etB) is invariant under the rotations with the center at origin,it suffices to prove that 3) implies 1 ∈ρ(etB). Also, to confirm ourprevious notations, we will consider only the case t = 2π.

The proofstays the same for any t.The idea is to apply Theorem 2.2, to show that 1 ∈ρ(e2πB) is impliedby 0 ∈ρ(B′). Here the operator B′ = −dds −ddx +A acts on Lp([0, 2π]×R; E), s ∈[0, 2π], x ∈R.

Indeed, by formula (5) one has B = −ddx +A.Hence, B′ on Lp([0, 2π]; Lp(R; E)) is the generator of the evolutionarysemigroup for the semigroup {etB} on Lp(R; E). But the change ofvariables u = [s −x](mod 2π), v = x shows that ρ(B′) = ρ(−ddv + A) =ρ(B).

Let us now make this argument more formal.Consider the semigroups(etB′h)(s, x) = etAh([s −t](mod 2π), x −t), t > 0, s ∈[0, 2π], x ∈R,(etBh)(s, x) = etAh(s, x −t), t > 0, s ∈[0, 2π], x ∈R,and an invertible isometry J,(Jh)(s, x) = h([s + x](mod 2π), x),12

acting on the spaceLp([0, 2π] × R; E) = Lp([0, 2π]; Lp(R; E)).Since etB acting on Lp([0, 2π]; Lp(R; E)) is actually the operator of mul-tiplication by the operator etB in Lp(R; E), one has:σ(etB) = σ(etB) and σ(B; Lp([0, 2π]; Lp(R; E))) = σ(B; Lp(R; E)).Also,JetB′h(s, x) = etAh([s + x −t](mod 2π), x −t) =etBJh(s, x),and hence one has JetB′ = etBJ and JB′ = BJ. Therefore,σ(etB) = σ(etB′) and σ(B) = σ(B′) in Lp([0, 2π]; Lp(R; E)).Thus 3) implies 0 ∈ρ(B) and 0 ∈ρ(B′).The semigroup {etB′} acts on Lp([0, 2π]; Lp(R; E)) by the rule(etB′f)(s) = etBf([s −t](mod 2π)),where f(s) = h(s, ·) ∈Lp(R; E) for almost all s ∈[0, 2π].

Hence, {etB′}on the space Lp([0, 2π]; Lp(R; E)) is the evolutionary semigroup for thesemigroup {etB} on Lp(R; E). Now one can apply the part 3) ⇒1) ofthe Theorem 2.2 and conclude that 1 ∈ρ(e2πB) on Lp(R; E).1) ⇔2) we will prove for F = Lp(R; E); the arguments for F =C0(R; E) are similar.Let us denote, for brevity, a = e2πA and (Rf)(x) = f(x −2π) onLp(R; E).

Thus e2πB = aR. Consider the invertible isometryj : Lp(R; E) →lp(Z; Lp([0, 2π]; E)) :f 7→(fn), fn(s) = f(s + 2πn), n ∈Z, s ∈[0, 2π).Let S : (fn) 7→(fn−1) be a shift operator on lp(Z; Lp([0, 2π]; E)).

ThenjaR = DaSj and σ(aR) = σ(DaS).Therefore, 2) is equivalent toσ(DaS)∩T = ∅. By Lemma 2.4 this in turn is equivalent to σ(a)∩T = ∅in Lp(R; E).Note, that 3) ⇒1) in the above theorem can also be derived directlyby constructing a function g in a similar manner as in the proof ofLemma 2.1.

This proof will be given elsewhere.13

3. Non-autonomous CaseConsider a non-autonomous differential equation v′(x) = A(x)v(x),x ∈R in E. We will assume that this equation is well-posed.

Thismeans that there exists an evolutionary family {U(x, s)}x≥s (propaga-tor) for the equation, that is v(x) = U(x, s)v(s), x ≥s. Recall thedefinition of evolutionary family (see, e.g., [27, 31]).Definition 3.1.

A family {U(x, s)}x≥s of bounded in E operators U(x, s)is called an evolutionary family if the following conditions are fulfilled:(i) for each v ∈E the function (x, s) 7→U(x, s)v is continuous forx ≥s;(ii) U(x, s) = U(x, r)U(r, s), U(x, x) = I, x ≥r ≥s;(iii) ∥U(x, s)∥≤Ceβ(x−s), x ≥s for some constants C, β > 0.The evolutionary family {U(x, s)} generates an evolutionary semi-group {etG}t≥0 acting on the space F = Lp(R; E), 1 ≤p < ∞orF = C0(R; E) by the rule(etGf)(x) = U(x, x −t)f(x −t), x ∈R. (14)In Subsection 3.1 we will prove the spectral mapping theorem σ(etG)\{0} =etσ(G), t ̸= 0 for {etG}.

We will achieve this by applying a simple changeof variables argument (cf. the proof of Theorem 2.5) to deduce this fromTheorem 2.5.

In Subsection 3.2 we will prove that the hyperbolicityσ(etG) ∩T = ∅of the semigroup in F is equivalent to the so-calledspectral hyperbolicity of the family {U(x, s)}. Spectral hyperbolicityis a generalization of the notion of exponential dichotomy (see, e.g.,[9]) for the equation v′(x) = A(x)v(x) with bounded A : R →B(E).3.1.

The Spectral Mapping Theorem for EvolutionarySemigroupLet G be the generator of the evolutionary semigroup {etG}t≥0 actingon the space F = Lp(R; E), 1 ≤p < ∞or F = C0(R; E) by equation(14).Theorem 3.1. The spectrum σ(G) is invariant under translations alongthe imaginary axis, and the following are equivalent:1) 0 ∈ρ(G) on F;2) σ(etG) ∩T = ∅on F, t > 0.14

Proof. For any ξ ∈R it is true that etGeiξ·f(·) = eiξ(·−t)etGf(·) andGeiξ· = eiξ·(G −iξ).

Hence σ(etG) is invariant under rotations cen-tered at the origin, and σ(G) is invariant under translations along theimaginary axis.2) ⇒1) follows from the spectral inclusion theorem for {etE}.1) ⇒2). We will first consider the case when F = Lp(R; E), 1 ≤p < ∞.The idea of the proof is almost identical to the proof of 3) ⇒2) fromTheorem 2.5.

If U(·, ·) is a smooth propagator for the equation v′ =A(x)v, then G = −ddx + A(x). Consider the evolutionary semigroupfor {etG}, that is the semigroup with the generator B = −dds + G onLp(R; Lp(R; E)).

Theorem 2.5 shows that 1 ∈ρ(etG) is implied by 0 ∈ρ(B), where B = −dds −ddx +A(x), s, x ∈R by formula (5). The changeof variables u = s−x, v = x shows that ρ(B) = ρ(−ddv +A(v)) = ρ(G).Let us now make this argument more formal.Consider the semigroups {etB}t≥0 and {etG}t≥0 acting on the spaceLp(R × R; E) = Lp(R; Lp(R; E)) by(etBh)(s, x) = U(x, x −t)h(s −t, x −t), (s, x) ∈R2, t > 0,(etGh)(s, x) = U(x, x −t)h(s, x −t).Note that etG in Lp(R; Lp(R; E)) is the operator of multiplication byetG, that is (etGf)(s) = etGf(s), where f(s) = h(s, ·) ∈Lp(R; E).Similarly, (Gf)(s) = Gf(s) = Gh(s, ·) for f(s) = h(s, ·) ∈DLp(R;E)(G)for almost all s ∈R.Consider an isometry J on Lp(R × R; E) defined by (Jh)(s, x) =h(s + x, x).

Then for h ∈Lp(R × R; E) one has:(etGJh)(s, x) = U(x, x −t)h(s + x −t, x −t) = (JetBh)(s, x). (15)Also (15) impliesGJh = JBh, h ∈D(B) and J−1Gh = BJ−1h, h ∈D(G).Therefore, σ(G) = σ(B) on Lp(R × R; E).Note that G on Lp(R × R; E) has bounded inverse (G−1f)(s) =G−1f(s), s ∈R, provided G has bounded inverse G−1 on Lp(R; E).Hence 1) implies 0 ∈ρ(B).Let us apply the part 3) ⇒1) of Theorem 2.5 to the semigroups{etG} and {etB}.

To this end we note that (etBf)(s) = etGf(s −t) for15

f : R →Lp(R; E) : s 7→h(s, ·). Hence 0 ∈ρ(B) on Lp(R; Lp(R; E))implies σ(etG) ∩T = ∅, t ̸= 0 on Lp(R; E).The proof for the case F = C0(R; E) is identical, and uses exactlythe same semigroups and isometries on C0(R; F) = C0(R × R; E).3.2.

HyperbolicityLet {U(x, s)}x≥s be an evolutionary family on a Banach space E. Inthis subsection we relate the (spectral) hyperbolicity of the evolutionaryfamily and the hyperbolicity of the evolutionary semigroup {etG} on thespace Lp = Lp(R; E) in the case when the Banach space E is separable.The case F = C0(R; E) (without the assumption of separability) andthe case of a Hilbert space E and p = 2 was considered in [27, 28].Definition 3.2. An evolutionary family {U(x, s)}x≥s is called (spec-trally) hyperbolic if there exists a projection-valued, bounded functionP : R →B(E) such that the function R ∋x 7→P(x)v ∈E is contin-uous for every v ∈E and for some constants M, λ > 0 and all x ≥sthe following conditions are fulfilled:a) P(x)U(x, s) = U(x, s)P(s):b) ∥U(x, s)v∥≤Me−λ(x−s)∥v∥if v ∈Im P(s),∥U(x, s)v∥≥M−1eλ(x−s)∥v∥if v ∈Ker P(s);c) Im(U(x, s)| Ker P(s)) is dense in Ker P(x).This notion generalizes the notion of exponential dichotomy (see,e.g., [9]) for the solutions of differential equation v′(x) = A(x)v(x),x ∈R, with bounded and continuous A : R →B(E).

In this case theevolutionary family (propagator) {U(x, s)}(x,s)∈R2 consists of invertibleoperators, the function (x, s) 7→U(x, s) is norm-continuous, and P(·)from Definition 3.2 is also a bounded, norm-continuous function P :R →B(E).The second inequality in b) implies that the restriction U(x, s)| Ker P(s)is uniformly injective as an operator from Ker P(s) to Ker P(x) (thatis ∥U(x, s)v∥≥c∥v∥for some c > 0 and all v ∈Ker P(s)). Thuscondition c) implies that U(x, s)| Ker P(s) is invertible as an opera-tor from Ker P(s) to Ker P(x).

Obviously, if U(x, s) is invertible inE or dim Ker P(x) ≤d < ∞, condition c) in Definition 3.2 is redun-dant. The inequality dim Ker P(x) ≤d < ∞holds, for example, if the16

U(x, s) are compact operators in E ([R. Rau, private communication]).Generally, of course, b) does not imply c).¿From now on we will assume that the Banach space E is separable.As we will see below, the spectral hyperbolicity of the evolution-ary family {U(x, s)} is equivalent to the hyperbolicity σ(etG) ∩T = ∅,t > 0 of the evolutionary semigroup {etG}t≥0 in Lp(R; E). Therefore,by Theorem 3.1 the spectral hyperbolicity of the evolutionary family{U(x, s)}x≥s is also equivalent to the condition σ(G) ∩iR = ∅.

Thatis why we used the term spectral hyperbolicity in Definition 3.2. Aremarkable observation by R. Rau [27] shows that generally the condi-tion c) in Definition 3.2 cannot be dropped: there exists an evolutionaryfamily that satisfies conditions a) and b) but σ(etG) ∩T ̸= ∅for theassociated evolutionary semigroup.If the operator T = eG is hyperbolic in Lp(R; E), that is σ(T)∩T = ∅,we let P denote the Riesz projection for T, corresponding to the partσ(T) lying inside the unit disk D, and set Q = I −P.Lemma 3.2.

If σ(etG) ∩T = ∅, t > 0, then P has a form (Pf)(x) =P(x)f(x), where P : R →B(E) is a bounded projection-valued functionsuch that the function R ∋x 7→P(x)v ∈E is (strongly) measurablefor each v ∈E.Proof. We will show first thatχP = Pχ(16)for any scalar function χ ∈L∞(R; R).Then we will derive that(Pf)(x) = P(x)f(x) from (16).Note that the decomposition Lp(R; E) = Im P ⊕Im Q is T-invariant.Denote TP = PTP = T| Im P, TQ = QTQ = T| Im Q.Note thatσ(TP) ⊂D, and TQ is invertible with σ(T −1Q ) ⊂D in Im Q.

Hence forsome λ, M > 0 and all n ∈N, the following inequalities hold:∥T nP f∥Lp ≤Me−λn∥f∥Lp,f ∈Im P,(17)∥T nQf∥Lp ≥M−1eλn∥f∥Lp,f ∈Im Q. (18)We show first that Im P = {f ∈Lp(R; E) : T nf →0 as n →∞}.Indeed, f ∈Im P implies that T nf →0 by (17).

Conversely, if T nf →0, then for f = Pf + Qf, the inequality (18) implies∥Qf∥≤Me−λn∥T nQQf∥≤Me−λn{∥T nf∥+ ∥T nP f∥} →0,17

and hence f ∈Im P.Consider on Lp = Lp(R; E) the operator χ of multiplication by χ(·) ∈L∞(R; R). Note that (T nχf)(x) = χ(x −n)(T nf)(x).

Hence for f ∈Im P∥T nχf∥Lp ≤∥χ∥L∞∥T nf∥Lp →0 as n →∞,and so χf ∈Im P. Thus, to prove (16), it suffices to show that f ∈Im Q implies χf ∈Im Q.Fix f ∈Im Q. Recall that TQ is invertible on Im Q.

Let fn = T −nQ f,and define functions gn(x) = χ(x + n)fn(x), n = 0, 1, . .

. .

Decomposegn = Pgn + Qgn. Since the decomposition Lp(R; E) = Im P ⊕Im Q isT-invariant, one has:χf = T ngn, Pχf = T nP Pgn, Qχf = T nQQgn, n = 0, 1, .

. .

.Now (17)–(18) imply:∥Pχf∥≤Me−λn∥Pgn∥≤Me−λn{∥gn∥+ ∥Qgn∥}≤Me−λn{∥χ∥L∞∥fn∥+ Me−λn∥Qχf∥}≤Me−λn{∥χ∥L∞Me−λn∥f∥+ Me−λn∥Qχf∥} →0,and hence χf ∈Im Q. Thus (16) is proved.In order to define P(·) such that (Pf)(x) = P(x)f(x), fix m ∈Zand let χm(x) = 1 if x ∈[m, m + 1), and χm(x) = 0 otherwise.

Let{en}n∈Z be a linearly independent set with dense span E0.Consider the function f ∈Lp(R; E), defined by f(x) = χm(x)en.Since P is bounded on Lp(R; E), it is true that Pf ∈Lp(R; E). Forx ∈[m, m + 1) define a vector P(x)en ∈E as P(x)en = (Pf)(x).

Forv = Pkn=1 dnen ∈E0 set P(x)v = Pkn=1 dnP(x)en.Let ∆be a measurable subset in [m, m + 1), and let χ∆be its char-acteristic function. Now (16) implies thatZ∆∥P(x)v∥pE dx =ZR ∥χ∆(Pχmv)(x)∥pE dx =ZR ∥(Pχ∆v)(x)∥pE dx≤∥P∥pB(Lp(R;E))Z∆∥v∥pE dx.Therefore, ∥P(x)v∥E ≤∥P∥B(Lp)∥v∥E for a.e.

x ∈R and all v ∈E0.Hence, P(x) can be extended to a bounded operator on E, such that∥P(x)∥≤∥P∥for a.e. x ∈R.

That the function x 7→P(x)v is ameasurable function for all v ∈E0 (and, hence, for all v ∈E) followsfrom the fact that the function x 7→(Pf)(x) is measurable.18

To show that (Pf)(x) = P(x)f(x) we can assume that f is a simplefunction, f = P χ∆kvk, where ∆k ⊂[mk, mk+1), mk ∈Z, and vk ∈E0.Then (16) implies:(Pf)(x) =Xχmk(x)(Pχ∆kvk)(x) =Xχ∆k(x)P(x)vk = P(x)f(x).Let us stress that the function P(·) above is only defined on a setR0 ⊂R such that mes(R \ R0) = 0. In Theorem 3.4 below we willestablish that, in fact, this function P(·) can be extended to all of R asa continuous function (in the strong operator topology in B(E)).

Toprove this fact we will need a few definitions and Lemma.Let A be the set of all operators a in Lp(R; E) of the form (af)(x) =a(x)f(x), where the function a : R →B(E) is bounded and the func-tion R ∋x 7→a(x)v ∈E is continuous for each v ∈E. For a ∈A andx ∈R let us define an operator πx(a) on lp(Z; E) by the ruleπx(a) = diag{a(x + n)}n∈Z : (vn)n∈Z 7→(a(x + n)vn)n∈Z.

(19)Finally, let S : (vn)n∈Z 7→(vn−1)n∈Z be a shift operator on lp(Z; E).Let us denote: T = eG, a(x) = U(x, x −1), (Rf)(x) = f(x −1).Then T = aR.For λ ∈T set b = λI −aR, and for x ∈R setπx(b) = λI −πx(a)S.Lemma 3.3. If σ(T) ∩T = ∅in Lp(R; E) then σ(πx(a)S) ∩T = ∅inlp(Z; E) for all x ∈R.

Moreover, for all λ ∈T the following estimateholds:∥[πx(b)]−1∥B(lp(Z;E)) ≤∥b−1∥B(Lp(R;E)),x ∈R.(20)Proof. First, for any ξ ∈R one has:πx(a)SL = e−iξLπx(a)S,where L is the operator (vn) 7→(eiξnvn).

Hence σ(πx(a)S) is invariantunder rotations centered at the origin. Thus it suffices to prove theLemma for the special case λ = 1, that is, to show that if b = I −aR isinvertible in Lp(R; E) then for each x0 ∈R that the operator πx0(b) =I −πx0(a)S is invertible in lp(Z; E), and that estimate (22) is valid forthis b and x = x0.19

Further, it suffices to prove the Lemma only for x0 = 0. Indeed, letus denote ˆa(x) = a(x + x0), x ∈R for any fixed x0 ∈R.

Obviously,πx0(I −aR) = I −πx0(a)S = I −π0(ˆa)S = π0(I −ˆaR).Consider the invertible isometry Jx0 on Lp(R; E), defined as (Jx0f)(x) =f(x+x0). Clearly, I−ˆaR = Jx0(I−aR)J−1x0 .

Hence, the operator I−ˆaRis invertible if and only if the operator b = I −aR is invertible, and∥(I −ˆaR)−1∥= ∥b−1∥. Therefore, the estimate (20) for x = x0 followsfrom the estimate (20) for x = 0.Thus our purpose is to prove if b = I −aR is invertible in Lp(R; E),then the operator π0(b) = I −π0(a)S is invertible in lp(Z; E), and∥[π0(b)]−1∥B(lp(Z;E)) ≤∥b−1∥B(Lp(R;E)).

(21)We first show that for any (vn) ∈lp(Z; E) the following estimateholds:∥(vn)∥lp(Z;E) ≤∥b−1∥B(Lp(R;E))∥π0(b)(vn)∥lp(Z;E). (22)Let us fix a sequence (vn)n∈Z ∈lp(Z; E), a natural number N > 1, andǫ > 0.Recall that the function R ∋x 7→a(x)v ∈E is continuous for eachv ∈E.

Choose δ < 1 such that∥[a(x + n) −a(n)]vn−1∥E < ǫ,∀x ∈[0, δ], n = −N, . .

. , N.(23)Define f ∈Lp(R; E) by f(x) = vn for x ∈[n, n + δ], |n| ≤N, andf(x) = 0 otherwise.

Since b is an invertible operator in Lp(R; E), itfollows that:∥b−1∥pB(Lp)∥bf∥pLp ≥∥f∥pLp =NXn=−NZ n+δn∥vn∥p dx = δNXn=−N∥vn∥p. (24)20

On the other hand, using (23), one has:∥bf∥pLp =ZR ∥f(x) −a(x)f(x −1)∥pE dx=NXn=−N+1Z n+δn∥vn −a(x)vn−1∥pE dx +Z −N+δ−N∥v−N∥pE dx+Z N+1+δN+1∥a(x)vN∥pE dx≤NXn=−N+1Z δ0 ∥vn −a(n)vn−1 −[a(x + n) −a(n)]vn−1∥pE dx+ δ∥v−N∥pE + δ maxx∈R ∥a(x)∥p · ∥vN∥pE≤δNXn=−N(∥vn −a(n)vn−1∥E + ǫ)p + δ∥v−N∥pE + δ maxx∈R ∥a(x)∥p∥vN∥pE.Combining this inequality with (24), one has:NXn=−N∥vn∥p ≤∥b−1∥pB(Lp)NXn=−N(∥vn −a(n)vn−1∥E + ǫ)p+ ∥v−N∥pE + maxx∈R ∥a(x)∥p∥vN∥pE.If ǫ →0 and N →∞, then∥(vn)∥lp(Z;E) ≤∥b−1∥B(Lp)∞Xn=−∞∥vn −a(n)vn−1∥pE1/pand (22) is proved.Note, that (22) is sufficient to show (21) provided the operator π0(b)is invertible, and so it only remains to show that π0(b) is an operatoronto lp(Z; E).Fix any (vn) ∈lp(Z; E) and let f(x) = U(x, n −1)vn−1, x ∈[n −1/2, n + 1/2), n ∈Z.Property (iii) from the Definition 3.1 of theevolutionary family {U(x, s)} implies that∥U(x, n −1)∥≤Ceβ(x−n+1) ≤Ce32 β for x ∈[n −1/2, n + 1/2).Hence f ∈Lp(R; E). Since the operator b, defined by (bg)(x) = g(x) −U(x, x−1)g(x−1), is invertible in Lp(R; E), it follows that there existsa unique function g ∈Lp(R; E) such thatg(x) −U(x, x −1)g(x −1) = f(x)(25)21

for almost all x ∈R. Since∥g∥pLp =Xn∈ZZ −n+1/2−n−1/2 ∥g(s)∥p ds =Z 1/2−1/2 Xn∈Z∥g(s + n)∥pds < ∞,the sequence (g(s + n))n∈Z belongs to lp(Z; E) for all s ∈Ωfor somesubset Ω⊂(−1/2, 1/2) of full measure.

For each s ∈Ω, let us define afunction hs by the rule:hs(x) =g(x),if n −12 ≤x ≤n + s,U(x, n + s)g(n + s),if n + s ≤x < n + 12,n ∈Z.Clearly, hs ∈Lp(R; E) for each s ∈Ωbecause (g(s+n))n∈Z ∈lp(Z; E),and∥U(x, n + s)∥≤Ceβ(1/2−s) for x ∈[n + s, n + 1/2).We note that hs is a solution of equation (25).Indeed, for x ∈[n−1/2, n+s], equation (25) implies hs(x)−U(x, x−1)hs(x−1) = f(x).For x ∈[n + s, n + 1/2), one has:hs(x) −U(x, x −1)hs(x −1) =U(x, n + s)[g(n + s) −U(n + s, n −1 + s)g(n −1 + s)]= U(x, n + s)f(n + s) = U(x, n + s)U(n + s, n −1)vn−1 = f(x).But equation (25) has only one solution g in Lp(R; E). Hence g = hsfor all s ∈Ω.Fix s < 0, s ∈Ω.

The function hs(·) is defined for x = n, n ∈Z.Moreover, the sequence (hs(n))n∈Z belongs to lp(Z; E).Indeed,hs(n) = U(n, n+s)g(n+s), the sequence (g(n+s))n∈Z ∈lp(Z; E), and∥U(n, n + s)∥≤Ce−βs by (iii) from Definition 3.1.Set un = hs(n)+vn. Since hs(·) satisfies the equation (25) for x = n,n ∈Z, we have:π0(b)(un) =un −U(n, n −1)un−1 = hs(n) + vn −U(n, n −1)hs(n −1) −U(n, n −1)vn−1 =vn + hs(n) −U(n, n −1)hs(n −1) −f(n) = (vn).This identity proves that π0(b) is an operator onto Lp(Z; E).22

Theorem 3.4. Let {U(x, s)}x≥s be an evolutionary family in a separa-ble Banach space E, and let {etG}t≥0 be the evolutionary semigroup act-ing on Lp(R; E), 1 ≤p < ∞by the rule (etGf)(x) = U(x, x−t)f(x−t).The following conditions are equivalent:1) {U(x, s)}x≥s is (spectrally) hyperbolic in E;2) σ(etG) ∩T = ∅, t ̸= 0, in Lp(R; E);3) 0 ∈ρ(G) in Lp(R; E).Moreover, the Riesz projection P that corresponds to the part σ(eG)∩Dof the spectrum of the hyperbolic operator eG is related to a strongly con-tinuous, projection-valued function P : R →B(E) that satisfies Defi-nition 3.2 by the formula (Pf)(x) = P(x)f(x), x ∈R, f ∈Lp(R; E).Proof.

2) ⇔3) was proved in Theorem 3.1.1) ⇒2).Without loss of generality assume t = 1.¿From theprojection-valued function P(·) from Definition 3.2, let us define aprojection P in Lp(R; E) by the rule (Pf)(x) = P(x)f(x). DenoteQ = I −P.For T = eG, condition a) of Definition 3.2 impliesPT = TP.Set TP = PTP and TQ = QTQ.Then b) impliesσ(TP) ⊂D in Im P = {f ∈Lp(R; E) : f(x) ∈Im P(x)}.

Also b)and c) imply that the operator TQ, which can be written as (TQf)(x) =Q(x)U(x, x−1)Q(x−1)f(x−1), is an invertible operator, and σ(T −1Q ) ⊂D in Im Q = Ker P. Hence, σ(T) ∩T = ∅.2) ⇒1). Let B be a Banach algebra with a norm ∥· ∥1 consistingof the operators d on Lp(R; E) of the formd =∞Xk=−∞akRk, ak ∈A, ∥d∥1 :=∞Xk=−∞∥ak∥B(Lp(R;E)) < ∞.We first show that if b = λ−T is invertible in Lp(R; E) for all λ ∈T,then (λ −T)−1 ∈B.

This fact will be proved in several steps.First, without loss of generality let λ = 1. Since σ(T) ∩T = ∅, byLemma 3.2 the Riesz projection P has a form (Pf)(x) = P(x)f(x),where the function R0 ∋x 7→P(x)v ∈E is a bounded, measurable(in the strong sense in E) function for each v ∈E, where R0 is a setof full measure in R. Recall also that Q(x) = I −P(x).

Decomposeb = I −T = (P −TP) ⊕(Q −TQ). Since σ(TP) ⊂D and σ(T −1Q ) ⊂D,23

one has that b−1 = (P −TP)−1 ⊕(Q −TQ)−1, where(P −TP)−1 =∞Xk=0T kP;(Q −TQ)−1 = [−TQ(Q −T −1Q )]−1 = −−1Xk=−∞T kQ. (26)Notice that T −1Q= (QaRQ)−1 = (QaQ(· −1)R)−1 = R−1(QaQ(· −1))−1 = [Q(· + 1)a(· + 1)Q(·)]−1R−1, and that TP = aRP = aP(· −1)R.Hence both operators T kP and T kQ can be written as akRk forsome multiplication operators ak.

The Neumann series in (26) convergeabsolutely. Therefore,b−1 =∞Xk=−∞akRk,∞Xk=−∞∥ak∥B(Lp(R;E)) < ∞,(27)for each k ∈Z the function ak : R0 →B(E) is bounded, and thefunction R0 ∋x 7→ak(x)v ∈E is measurable for each v ∈E.Our next aim is to show that the ak from (27) belong to A, that is,the function x 7→ak(x)v ∈E extends to a continuous function fromR for each v ∈E.

To this end let us define for ak from (27) and allx ∈R0 the operator πx(ak) in lp(Z; E) as in (19). Denote:πx(b−1) =Xπx(ak)Sk,πx(ak) = diag{ak(x + n)}n∈Zfor x ∈R0.

(28)Identities bb−1 = b−1b = I in Lp(R; E) imply that πx(b) · πx(b−1) =πx(b−1) · πx(b) = I in lp(Z; E) for x ∈R0.Since the operator b isinvertible in Lp(R; E) by assumption, for each x ∈R the operatorπx(b) is invertible in lp(Z; E) by Lemma 3.3. Henceπx(b−1) = [πx(b)]−1 for x ∈R0.

(29)Recall that the function R ∋x 7→a(x)v ∈E is continuous for eachv ∈E. Also, the function R ∋x 7→∥a(x)∥∈R+ is bounded.

Hencefor each (vn) ∈lp(Z; E) the function R ∋x 7→πx(b)(vn) ∈lp(Z; E) iscontinuous. By Lemma 3.3 ∥[πx(b)]−1∥B(lp) are uniformly bounded forx ∈R.

This implies that the function R ∋x 7→[πx(b)]−1(vn) ∈lp(Z; E)is continuous for each (vn) ∈lp(Z; E). Indeed,[πx(b)]−1 −[πx0(b)]−1(vn)lp(Z;E) =[πx(b)]−1 · [πx(b) −πx0(b)] · [πx0(b)]−1(vn)lp(Z;E)for any x, x0 ∈R, and the function R ∋x 7→[πx(b)]−1(vn) ∈lp(Z; E)is continuous at x = x0.24

Fix k0 ∈Z, x0 ∈R, and v ∈E. Define (˜vn) ∈lp(Z; E) as ˜v−k0 = vand ˜vn = 0 for n ̸= −k0.

Consider a sequence xm →x0, xm ∈R0. Wewill show that {ak0(xm)v}m∈N is a Cauchy sequence in E and will defineak0(x0)v = limm→∞ak0(xm)v. Then the function R ∋x 7→ak0(x)v ∈Ebecomes a continuous function, and ak0 ∈A.Note, that the πxm(b−1) are defined by the formula (28) since xm ∈R0.

For the sequence (˜vn) one has the following estimate:∥[πxm′(b−1) −πxm′′(b−1)](˜vn)∥plp=Xn∈ZXk∈Z[ak(xm′ + n) −ak(xm′′ + n)]˜vn−kpE≥Xk[ak(xm′) −a(xm′′)]˜v−kpE= ∥[ak0(xm′) −ak0(xm′′)]v∥pE. (30)Since xm ∈R0, formula (29) is applicable.

Then the sequenceπxm(b−1)(˜vn) = [πxm(b)]−1(˜vn)is a Cauchy sequence in lp(Z; E) since the function R ∋x 7→[πx(b)]−1(˜vn) ∈lp(Z; E) is continuous. In accordance with (30) the sequence {ak0(xm)v}m∈Nis a Cauchy sequence in E, and ak0 ∈A.Since the ak from (27) are continuous, we have proved that (λI −T)−1 ∈B for all λ ∈T.The rest of the proof is standard (cf.

[2, 18, 27]). Indeed, considerthe absolutely convergent Fourier series f : λ 7→λI −aRλ0 with thecoefficients from B.

For each λ ∈T, the operator f(λ) = b is invertiblein B. Hence the function T ∋λ 7→[f(λ)]−1 ∈B is expandable (see,e.g., [5]) into an absolutely convergent Fourier series[f(λ)]−1 = (λI −aR)−1 =∞Xk=−∞dkλk,Xk∥dk∥< ∞, dk ∈B.By the integral formula (see, e.g., [9, p. 20]) for the Riesz projection P,we conclude that P = d−1 ∈B.

Hence for some ak ∈A one has thatP =∞Xk=−∞akRk, where∞Xk=−∞∥ak∥< ∞.25

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