This subsection will be devoted entirely to the notion of spectrum of bounded sequences. Almost all results of this subsection can be proved in the same way as in Chapter 1. We will discuss only some particular points which allow us to prove analogs of the results in Sections 2.1. and 2.2..
We will denote by l∞(X) the space of all two-sided sequences with sup-norm, i.e.,
l∞(X) :={(xn)n∈Z:xn∈X,sup
n∈Z
kxnk<∞}.
First we will make precise the definition of the spectrum of a bounded sequence g:={gn}n∈ZinXused in this section. Recall that the set of all bounded sequences in X forms a Banach spacel∞(X) with norm kgk := supnkgnkX. We will denote byS(k) thek-translation inl∞(X) , i.e. , (Sg)n=gn+k,∀g, n.
Definition 2.16 The subset of all λof the unit circle Γ := {z ∈C : |z| = 1} at which
ˆ g(λ) :=
P∞
n=0λ−n−1S(n)g, for|λ|>1,
−P∞
1 λn−1S(−n)g, for|λ|<1
has no holomorphic extension, is said to be the spectrum of the sequence g :=
{gn}n∈Z and will be denoted byσ(g)3.
We list below some properties of spectrum of{gn}.
Proposition 2.22 Let g:={gn} be a two-sided bounded sequence inX. Then the following assertions hold:
i) σ(g)is closed,
3In Definition 2.16 our notion of spectrum is somewhat stronger than that in [217].
ii) If gn is a sequence in l∞(X) converging to g such that σ(gn) ⊂ Λ for all n∈N, whereΛ is a closed subset of the unit circle, thenσ(g)⊂Λ,
iii) If g ∈ l∞(X) and A is a bounded linear operator on the Banach space X, thenσ(Ag)⊂σ(g), where the sequence (Ag)n :=Agn,∀n∈Z.
iv) Let the space Xnot contain any subspace which is isomorphic toc0 andx∈ l∞(X)be a sequence such that σ(x)is countable. Then xis almost periodic.
Proof. i) From the definition it is obvious that the set of regular points (at which ˆ
g(z) has analytic continuation) is open. Hence,σ(g) is closed.
ii) The proof can be taken from that of [185, Theorem 0.8, pp.21-22]. In fact, from the assumption, for every positiveε
n→∞lim
∞
X
k=−∞
e−ε|k|kgnk −gkk= 0. (2.181) This yields that gcn(λ) → g(λ) asˆ n → ∞ uniformly on every compact subsets of C\S1 (S1 denotes the unit circle). From the uniform convergence of gn to g, without loss of generality we can assume that supnkgn(k)−g(k)k ≤ kgk. Hence, we can assume thatkgnk ≤2kgk, ∀n. Thus
|cgn(λ)| ≤ {2kgk/(|λ| −1), ∀|λ|>1,2kgk/(1− |λ|), ∀|λ|<1 = 2kgk
|1− |λ||. (2.182) Now let ρ0 ∈ (C\Λ). Since Λ is closed, obviously that dist(ρ0,Λ) > 0. Thus we can choose r > 0 such that 0 < r < min(dist(ρ0,Λ),1/4). By assumption, ρ0 6∈
σ(gn), ∀n. This means thatcgn(λ) is analytic inBr(ρ0) for all n. Using exactly the argument of the proof of [185, Proposition 0.8, p. 21] for the sequence Fcn(ξ) :=
cgn(eξ), we can show that
|cgn(λ)| ≤M, ∀n∈N, λ∈ U,
where U is a neiborhood of ρ0. Thus, by Montel’s Theorem (see e.g. [48, p. 149]) the family gcn(λ)|U is normal, i.e. every subsequence of it contains a subsubse- quence which converges inC(U, l∞(X)). Obviously, since this sequence converges to ˆg(λ) pointwise, the limit function here should be ˆg(λ). SinceH(U) is closed in C(U, l∞(X)) and sincecgn(λ) is analytic inU this shows that ˆg(λ) is analytic inU. Henceρ06∈σ(g), so σ(g)⊂Λ.
iii) The assertion is obvious.
iv) The proof can be done in the same way as in [137, Chap. 6] or [8, Section 3].
In view of Proposition 2.22 if Λ is a closed subset of the unit circle, then the set of all bounded sequences ing∈l∞(X) such thatσ(g)⊂Λ forms a closed subspace ofl∞(X) which we will denote by Λ(Z,X).
The following lemma will be used in the next subsections. We will denote by Mg the closure of the subspace ofl∞(X) spanned by all elementsS(n)g, n∈Z.
Lemma 2.36 We can define the spectrum of a given sequence x = {xn} as σ(S|Mx).
Proof. First note thatMx is invariant under all translationsS(k). It is easy to see thatkS|Mxk=kS−1|Mxk= 1. Henceσ(S|Mx)⊂Γ . We will use the following indentity
(I−A)−1=
∞
X
n=0
An, (2.183)
for any bounded linear operator A such that kAk < 1. By assumptions and by definition for|z|>1 we have
ˆ x(z) =
∞
X
n=0
z−n−1S(n)x
= z−1
∞
X
n=0
z−nS(n)x
= z−1(I−z−1S)−1x
= (z−S)−1x. (2.184)
Similarly, for|z|<1, by using in addition the indentityI−(I−zS−1)−1 = (I− z−1S)−1we can show that
ˆ
x(z) = (z−S)−1x.
Thus it is obvious that ifz0∈ρ(S|Mx), thenz0is a regular point ofx. Conversely, suppose thatz0 is a regular point of ˆx. We will prove that the mappingz0−S|Mx : Mx → Mx is one-to-one and onto to establish z0 ∈ ρ(S|Mx). From the above calculation it follows that for anyy∈ Mx,
ˆ
y(z) = (z−S|Mx)−1y or (z−S|Mx)ˆy(z) =y
whenever |z| 6= 1. Hence, for every y ∈ Mx we get (z−S|Mx)ˆy(z) = y on U (because of the analyticity of the function ˆy(z) on U). In particular, the mapping z0−S|Mx : Mx → Mx is ”onto”. Furthermore, we show that this mapping is one-to-one. Indeed, if (z0−S|Mx)a= 0 for ana={a(n)} ∈ Mx, then a(n+ 1) = z0a(n) ∀n∈Z, and hence,a(n) =z0na(0) ∀n∈Z. Then
ˆ
a(z) = a z−z0
(∀|z| 6= 1).
Since ˆais analytic inU because ofa∈ Mx, we havea= 0, as required. This shows thatz0∈ρ(S|Mx), completing the proof of the lemma.
Corollary 2.22 Let x={xn}be an element of l∞(X)such thatxn=xn+1=c6=
0,∀n∈Zif and only if σ(x) ={1}. Similarly, x∈l∞(X)such that xn=−xn+16=
0,∀n∈Zif and only if σ(x) ={−1}.
Proof. Ifxn =c6= 0,∀n∈Z, then it is easy to computeσ(x) ={1}. Conversely, letσ(x) ={1}. Then by Lemma 2.36,σ(S1) ={1}, where S1 is the restriction of S toMx. In view of Gelfand’s TheoremS1=IMx which completes the proof. For the second assertion note that in this casexn=xn+2. Using the previous argument forS(2) we get the assertion.
Corollary 2.23 Let Λ(Z,X) denote the subspace of l∞(X) consisting of all se- quences x such that σ(x) ⊂ Λ for given closed subset Λ of the unit circle. Then the translation S leaves Λ(Z,X) invariant and its restriction to Λ(Z,X) which is denoted bySΛ has the property that
σ(SΛ) = Λ. (2.185)
Proof. For x0 6= 0 put xn = λnx0, λ ∈ Λ. It is easy to see that σ(x) = {λ}
and λ ∈ σ(SΛ). Now we prove the converse, i.e., C\Λ ⊂ C\σ(SΛ) = ρ(SΛ). To this end, suppose thatλ0∈C\Λ. By definition of Λ(Z,X), we have that for every y ∈ Λ(Z,X), λ0 ∈/ σ(y). To show that λ0 ∈ ρ(SΛ) we will prove that for every y∈Λ(Z,X) there exists a unique solutionx∈Λ(Z,X) such that
λ0x−Sx=y . (2.186)
In fact, first the existence of such a solutionxis obvious in view of Lemma 2.36. Now we show the uniqueness of such a solutionx. Equivalently, we show that equation λ0w−Sw = 0 has only the trivial solution in Λ(Z,X). In fact, since w,0 belong toMy in view of Lemma 2.36 w= 0 is the unique solution to the above equation.
Hence,λ0∈ρ(SΛ).
The following result will relate the Bohr spectrum of an almost periodic function f(t) and the spectrum of the sequence f(n), n∈ Z. Before stating this result we recall the Approximation Theorem saying that for everyX-valued almost periodic functionf(ã) there exists a sequence of trigonometric polynomials
Pn(t) =Pn:=
N(n)
X
j=1
ajeiλjt, t∈R
which converges uniformly on the real line to the function f (for more informa- tion see [137]). Obviously, every almost periodic function is bounded and uniformly continuous. Moreover, for everyλ∈Rthe following limit
a(λ) := lim
T→∞
1 2T
Z T
−T
e−λξf(ξ)dξ
exists and there are at most countably many realsλsuch thata(λ)6= 0. We define Bohr spectrum off as the setσb(f) :={λ:a(λ)6= 0}and use the following notation σ(f) := eiσb(f). Note that [137] the exponents of the approximate trigonometric polynomials of an almost periodic function can be chosen from its Bohr spectrum.
Proposition 2.23 Let f(ã) be an almost periodic function in Bohr’s sense on the real line with Bohr spectrum σb(f). Then the spectrum of the sequence x :=
{f(n), n∈Z} satisfies
σ(x)⊂σ(f) :=eiσb(f). (2.187) Proof. Let Pn := PN(n)
j=1 ajeiλjt be a sequence of trigonometric polynomials which approximates the almost periodic function f(ã) with λj ∈ σb(f). Then as in the proof of Corollary 2.23 we have σ(Qn) = {eiλj, j = 1,ã ã ã, N(n)}, where Qn(k) :=Pn(k),∀k∈Z. Henceσ(Qn)⊂eiσb(f):=σ(f) and in view of Proposition 2.22σ(x)⊂σ(f).