A zero-sqrt(5)/ 2 law for cosine families

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A zero-sqrt(5)/ 2 law for cosine families

Jean Esterle

To cite this version:

Jean Esterle. A zero-sqrt(5)/ 2 law for cosine families. 2015. �hal-01147792�

A zero-

law for cosine families

Jean Esterle

Abstract: Leta∈R, and letk(a) be the largest constant such thatsup|cos(na)− cos(nb)| <k(a) forb ∈R implies thatb ∈ ±a+2πZ. We show that if a cosine sequence (C(n))_n∈Z with values in a Banach algebraA satisfiessup_n≥1kC(n)− cos(na).1Ak <k(a), thenC(n)=cos(na) forn∈Z. Since ^p₂⁵ ≤k(a)≤ ₃^p8₃ for everya ∈R, this shows that if some cosine family (C(g))_g∈G over an abelian groupG in a Banach algebra satisfiessupg∈GkC(g)−c(g)k < ^p₂⁵ for some sca- lar cosine family (c(g))g∈G, thenC(g)=c(g) forg ∈G, and the constant

p5 2 is optimal. We also describe the set of all real numbersa∈[0,π] satisfyingk(a)≤³₂. Keywords : Cosine function, scalar cosine function, commutative local Ba- nach algebra, Kronecker’s theorem, cyclotomic polynomials

AMS classification : Primary 46J45, 47D09, Secondary 26A99

1 Introduction

LetG be an abelian group. Recall that aG-cosine family of elements of a unital normed algebraAwith unit element 1Ais a family (C(g))g∈G of elements ofAsatisfying the so-called d’Alembert equation

C0=1A,C(g+h)+C(g−h)=2C(g)C(h) (g∈G,h∈G). (1) AR-cosine family is called a cosine function, and aZ-cosine family is called a cosine sequence.

A cosine familyC=(C(g))g∈Gis said to be bounded if there existsM>0 such thatkC(g)k ≤Mfor everyg∈G. In this case we set

kCk∞=supg∈GkC(g)k,d i st(C1,C2)= kC1−C2k∞.

A cosine family is said to be scalar ifC(g)∈C.1Afor everyg∈G. It is easy to see and well-known that a bounded scalar cosine sequence satisfiesC(n)=cos(an) for somea∈R.

Strongly continuous operator valued cosine functions are a classical tool in the study of differential equations, see for example [2], [3], [15], [19], and a func- tional calculus approach to these objects was developped recently in [11].

Bobrowski and Chojnacki proved in [4] that if a strongly continuous opera- tor valued cosine function on a Banach space (C(t))_t∈Rsatisfiessup_t≥0kC(t)− c(t)k <1/2 for some scalar bounded continuous cosine functionc(t) thenC(t)= c(t) pourt ∈R, and Zwart and F. Schwenninger showed in [18] that this result remains valid under the conditionsup_t≥0kC(t)−c(t)k <1. The proofs were ba- sed on rather involved arguments from operator theory and semigroup theory.

Very recently, Bobrowski, Chojnacki and Gregosiewicz [5] showed more preci- sely that if a cosine functionC=C(t) satisfies sup_t∈RkC(t)−c(t)k <₃^p8₃for some scalar bounded continuous cosine functionc(t), thenC(t)=c(t) fort ∈R, wi- thout any continuity assumption onC, and the same result was obtained inde- pendently by the author in [10]. The constant ⁸

3p

3 is obviously optimal, since supt∈R|cos(at)−cos(3at)| =₃^p8₃for everya∈R\ {0}.

The author also proved in [10] that if a cosine sequence (C(t))t∈R satisfies sup_t∈RkC(t)−cos(at)1_Ak =m<2 for somea6=0, then the closed algebra ge- nerated by (C(t))t∈Ris isomorphic toC^k for somek ≥1, and that there exists a finite familyp1, . . . ,pk of pairwise orthogonal idempotents ofA and a family (b₁, . . . ,b_k) of distinct elements of the finite set∆(a,m) :={b≥0 : sup_t∈R|cos(bt)− cos(at)| ≤m} such that we have

C(t)= Xk j=1

cos(bjt)pj (j∈R).

Also Chojnacki developped in [7] an elementary argument to show that if (C(n))n∈Zis a cosine sequence in a unital normed algebraAsatisfyingsupn≥1kC(n)− c(n)k <1 for some scalar cosine sequence (c(n))_n∈Zthenc(n)=C(n) for everyn, which obviously implies the result of Zwart and F. Schwenninger. His approach is based on an elaborated adaptation of a very short elementary argument used by Wallen in [20] to prove an improvement of the classical Cox-Nakamura-Yoshida- Hirschfeld-Wallen theorem [8], [13], [16] which shows that if an elementaof a unital normed algebraAsatisfiessupn≥1kaⁿ−1k <1, thena=1.

Applying this result to the cosine sequencesC(ng) andc(ng) forg∈G, Cho- najcki observed in [7] that if a cosine familyC(g) satisfiessupg∈GkC(g)−c(g)k <

1 for some scalar cosine familyc(g) thenC(g)=c(g) for everyg∈G.

In the same direction Schwenninger and Zwart showed in [17] that if a cosine sequence (C(n))n∈Zin a Banach algebraAsatisfiessupn≥1kC(n)−1Ak <³₂, then C(n)=1Afor everyn.

The purpose of this paper is to obtain optimal results of this type. We prove

a "zero-^p₂⁵" law : if a cosine family (C(g))_g∈Gsatisfiessup_g∈GkC(g)−c(g)k <

p5 2

for some scalar cosine family (c(g))g∈G thenC(g)=c(g) for everyg∈G. Since sup_n≥1¯¯cos¡_2nπ

5

¢−cos¡_4nπ

5

¢¯¯=cos¡_2π

5

¢−cos¡_π

5

¢=

p5

2 , the constant ^p₂⁵ is opti- mal.

In fact for everya∈Rthere exists a largest constantk(a) such that supn≥1|cos(nb)− cos(na)| <k(a) implies thatcos(nb)=cos(na) for n ≥1, and we prove that if a cosine sequence (C(n))n∈Z in a Banach algebra A satisfies supn≥1|C(n)− cos(na)1A| <k(a) thenC(n)=cos(na) forn≥1. This follows from the following result, proved by the author in [10].

Theorem 1.1. Let(C(n))n∈Zbe a bounded cosine sequence in a Banach algebra. If spec(C(1))is a singleton, then the sequence(C(n))_n∈Zis scalar, and so there exists a∈Rsuch that C(n)=cos(na)for n≥1.

The second part of the paper is devoted to a discussion of the values of the constantk(a). As mentioned above, it follows from [17] thatk(0)= ³₂, and it is obvious thatk(a)≤supn≥1|cos(na)−cos(3na)| ≤₃^p8₃ifa∉^π2Z. We observe that k(a)=₃^p8₃ if ^a_π is irrational, and we prove, using basic results about cyclotomic fields, thatk(a)<₃^p8₃if^a_π is rational.

We also show that the setΩ(m) :={a∈[0,π]|k(a)≤m} is finite for every m< ₃^p8₃. We describe in detail the setΩ¡₃

2

¢: it contains 43 elements, and the only values fork(a) for whichk(a)<³2are^p₅²=cos¡_π

5

¢+cos¡_2π

5

¢≈1.1180,p 2= cos¡_π

4

¢+cos¡_3π

4

¢≈1.4142, andcos¡_2π

11

¢+cos¡_3π

11

¢≈1.4961.

The zero-^p₂⁵law follows then from the fact thatk(a)≥cos¡_π

5

¢+cos¡_π

5

¢=

p5 2

for everya∈R.

We also show that given a ∈Randm <2 the set Γ(a,m) of scalar cosine sequences (c(n))n∈Zsatisfyingsupn∈Z|c(n)−cos(na)| ≤mis finite. This implies that if a cosine sequence (C(n))n∈Zsatisfies supn∈ZkC(n)−cos(an)1Ak ≤m, then there existsk≤c ar d(Γ(a,m)) such that the closed algebra generated by (C(n))n∈Z

is isomorphic to C^k, and there exists a finite family p₁, . . . ,p_k of pairwise or- thogonal idempotents of Aand a finite family c₁, . . . ,c_k of distinct elements of Γ(a,m) such that we have

C(n)= Xk j=1

c_j(n)p_j (n∈Z).

This last result does not extend to cosine families over general abelian group.

LetG=(Z/3Z)^N: we give an easy example of aG-cosine family (C(g))g∈G with values inl^∞such that the closed subalgebra generated by (C(g))_g∈G equalsl^∞, while supg∈Gk1l^∞−C(g)k =³₂.

The author warmly thanks Christine Bachoc and Pierre Parent for giving him the arguments from number theory which lead to a simple proof of the fact that k(a)<₃^p8₃ifa∉πQ.

2 Distance between bounded scalar cosine sequences

We introduce the following notation, to be used throughout the paper.

Definition 2.1. Let a∈πQ.The order of a, denoted by or d(a),is the smallest integer u≥1such that e^{i ua}=1.

Recall that a subsetSof the unit circleTis said to be independent ifz₁ⁿ¹. . .zⁿ_k^k 6=

1 for every finite family (z1, . . . ,zk) of distincts elements ofS and every family (n1, . . . ,n_k)∈Z^kof such thatzj6=0 forj≤j≤k. It follows from a classical theo- rem of Kronecker, see for example [14], page 21 that ifS={z1, . . . ,zk} is a finite independent set then the sequence (z₁ⁿ, . . . ,z_kⁿ)n≥1 is dense inT^k. We deduce from Kronecker’s theorem the following observation.

Proposition 2.2. Let a∈[0,π].For m≥0,set Γ(a,m)=©

b∈[0,π] :supn≥1|cos(na)−cos(nb)| ≤mª . ThenΓ(a,m)is finite for every m<2.

Proof : Fixm∈[1, 2). Notice that if b ∈R, and if the set ©

e^{i a},e^{i b}ª

is inde- pendent, then it follows from Kronecker’s theorem that the sequence¡¡

e^{i na},e^{i nb}¢¢

n≥1

is dense inT², and sosupn≥1|cos(na)−cos(nb)| =2, andb∉Γ(a,m).

Suppose that^a_π∈Q, and denote byuthe order ofa, so thate^{i ua}=1. If ^b_π∉Q, then the sequence¡

e^{i unb}¢

n≥1is dense inT, and so

2≥supn≥1|cos(na)−cos(nb)| ≥supn≥1|1−cos(nub)| =2, which shows thatb∉Γ(a,m).

The same argument shows that if_π^a∉Q, and if^b_π∈Q, thenb∉Γ(a,m). So we are left with two situations

1) ^a_π∉Q, and there existsp6=0,q6=0 andk∈Zsuch thatbq=ap+2kπ.

2) ^a_π∈Qand_π^b∈Q.

We consider the first case. Replacingb∈[0,π] by−b∈[−π, 0] if necessary we can assume thatp≥1 andq≥1, and we can assume that we have

qb=p a+2kπ r ,

withg cd(p,q)=1,r≥1,g cd(r,k)=1 ifk6=0.

We have, since^{r a}_π ∉Q,

supn≥1|cos(na)−cos(nb)| ≥supn≥1

¯¯cos(nr q a)−cos(nr qb)¯¯

=supn≥1¯¯cos(nr q a)−cos(nr p a)¯¯=supt∈R

¯¯cos(q t)−cos(p t)¯¯, Sinceg cd(p,q)=1, we havesupt∈R

¯¯cos(q t)−cos(p t)¯¯=2 ifporq is even, so we can assume thatpandqare odd. Sets=^q−₂¹.

It follows from Bezout’s identity that there existn≥1 such thate

2i npπ q =e^{2i sπq} and settingt=^2nπ_q , we obtain

sup_t∈R

¯¯cos(q t)−cos(p t)¯¯≥1−cos µ 2sπ

2s+1

¶

=1+cos µπ

q

¶ . The same argument shows that we have

supt∈R

¯¯cos(q t)−cos(p t)¯¯≥1+cos µπ

p

¶ . We obtain

p≤ π

ar ccos(m−1),q≤ π ar ccos(m−1). We also have

supn≥1|cos(na)−cos(nb)| ≥supn≥1

¯¯cos(nq a)−cos(nqb)¯¯

=sup_n≥1

¯¯

¯¯cos(nq a)−cos µ

np a+2nk qπ r

¶¯¯¯¯.

Assume thatk6=0, setd=g cd(r,q),r1= ^r_d,q1=_d^q. Theng cd(k q1,r1)=1, and so there existsu≥1 such that^{2uk q}_r ^1π

1 ∈^2π_r1 +2πZ. This gives sup_n≥1|cos(na)−cos(nb)| ≥sup_n≥1

¯¯

¯¯cos(nuq a)−cos µ

npua+2nπ r₁

¶¯¯¯¯.

Ifr1is even, setr2=^r2¹. We obtain supn≥1

¯¯

¯¯cos(nuq a)−cos µ

npua+2nπ r1

¶¯¯¯¯

≥supn≥0

¯¯cos((2n+1)r2uq a)−cos¡

(2n+1)r2up a)+π¢¯¯.

Since 2r2ua∉πQ, there exists a sequence (nj)j≥1of integers such that l i mj→+∞

¯¯

¯e^{i2njr2ua+i r2ua}¯¯¯=1, so that

l i mj→+∞

¯¯cos((2nj+1)r2uq a)−cos¡

(2nj+1)r2up a)+π¢¯¯=2, and in this situationsupn≥1|cos(na)−cos(nb)| =2.

So we can assume thatr₁is odd. Setr₂=^r−₂¹. The same calculation as above gives

sup_n≥1

¯¯

¯¯cos(nuq a)−cos µ

npua+2nπ r₁

¶¯¯¯¯

≥supn≥1

¯¯

¯¯cos((n(2r2+1)+r2)uq a)−cos µ

(n(2r2+1)+r2)up a+2(n(2r₂+1)+r₂) 2r2+1 π

¶¯¯¯¯

≥1+cos

µ π

2r2+1

¶ .

Hencer₁=2r₂+1≤_{ar ccos(m}^π ₋₁₎,r=r₁d≤r₁q≤³ _π

ar ccos(m−1)

´2

. This gives

2|k|π≤r|qb−p a| ≤2π

µ π

ar ccos(m−1)

¶3

,|k| ≤

µ π

ar ccos(m−1)

¶3

. We see thatΓ(a,m) is finite if_π^a∉Q, and that we have

c ar d(Γ(a,m))≤2

µ π

ar ccos(m−1)

¶7

.

Now consider the case where _π^a ∈Q, ^b_π∈Q. We first discuss the case where a=0,b6=0. We haveb=^pπ_q , where 1≤p≤q,g cd(p,q)=1

Ifp=q=1, thenb=π, andsupn≥1|1−cos(nπ)| =2. So we may assume that p≤q−1. Ifpis odd, then we have

sup_n≥1|1−cos(nb)| ≥ |1−cos(qb)| =1−cos(pπ)=2.

So we can assume thatpis even, so thatq is odd. Setr = ^q−₂¹. There exists n0≥1 andr∈Zsuch thatn0p−r∈qZ, and we have

supn∈Z|1−cos(nb)| ≥ |1−cos(2n0b)| =

¯¯

¯¯1−cos µ 2rπ

2r+1

¶¯¯¯¯=1+cos µπ

q

¶ . We obtain againq≤_{ar ccos(m}^π ₋₁₎, andc ar d(Γ(0,m))≤³

π ar ccos(m−1)

´2

.

Now assume thata6=0, and letu≥2 be the order ofa. We have supn≥1|1−cos(nub)| =supn≥1|cos(nua)−cos(nub)| ≤m,

and so there exists there existsc∈Γ(0,m) such thatcos(nc)=cos(nub) forn≥1.

In particularcos(c)=cos(ub), andb= ±_u^c +^2kπ_u , wherek∈Z. We obtain c ar d(Γ(a,m))≤2uc ar d(Γ(0,m))≤2uc ar d

µ π

ar ccos(m−1)

¶2

. ä

We do not know whether it is possible to obtain a majorant forc ar d(Γ(a,m)) which depends only onmwhena∈πQ.

Theorem 2.3. Let a∈R, let m<2,and let(C(n))n∈Z be a cosine sequence in a Banach algebra A such that sup_n≥1kC(n)−cos(na)k ≤m.Then there exists k≤c ar d(Γ(a,m))such that the closed algebra generated by(C(n))n∈Zis isomor- phic toC^k,and there exists a finite family p1, . . . ,pkof pairwise orthogonal idem- potents of A and a finite family b₁, . . . ,b_kof distinct elements ofΓ(a,m)such that we have

C(n)= Xk j=1

cos(nbj)pj (n∈Z).

Proof : Sincec_n=P_n(c₁), whereP_ndenotes then-th Tchebishev polynomial, A1is the closed unital subalgebra generated byc1 and the mapχ→χ(c1) is a bijection fromcA1ontospecA₁(c1). Now letχ∈Ac1. The sequence (χ(cn))n≥1is a scalar cosine sequence, and we have

supn≥1¯¯cos(na)−χ(cn)¯¯<2.

It follows then from proposition 2.2 thatspec_A1(c₁) :=©

λ=χ(c₁) :χ∈cA₁ª is finite. HencecA1is finite. Letχ1, . . . ,χmbe the elements ofAc1. It follows from the standard one-variable holomorphic functional calculus, se for example [9], that there exists for every j ≤man idempotent pj of A1 such thatχ_j(pj)=1 and χ_k(p_j)=0 fork6=j. Hencep_jp_k=0 for j6=k, and P^m

j=1

p_jis the unit element of A1.

Letx∈A₁. Then (p_jc_n)_n∈Zis a cosine sequence in the commutative unital Banach algebrapjA1, andspecpjA1(pjc1)={χj(c1)}.

Sincesupn≥1°°p_jcos(na)−pjcn°°≤2kpjk, the sequence (pjcn)n≥1is boun- ded, and it follows from theorem 2.3 that (p_jc_n)_n≥1 is a scalar sequence, and there existsβj∈[0,π] such thatpjcn=χj(cn)pj=cos(nβj)pjforn∈Z.

Hencecn = Pm j=1

χj(cn)pj = Pm j=1

cos(nβj)pj forn≥1. Since A1 is the closed subalgebra ofAgenerated byc1, we havex= ^mP

j=1

χj(x)pjfor everyx∈A1, which shows thatA1is isomorphic toC^m.ä

Corollary 2.4. Let a≥0∈R,and let k(a)be the largest positive real number m such thatΓ(a,m)={a}for every m<k(a).If(C(n))n∈Z is a cosine sequence in a Banach algebra A such that supn≥1kC(n)−cos(na)1Ak <k(a), then C(n)= cos(na)1_Afor n∈Z.

Theorem 2.3 does not extend to cosine families over general abelian groups, as shown by the following easy result.

Proposition 2.5. Let G:=(Z/3Z)^N.Then there exists a G-cosine family(C(g))g∈G

with values in l^∞which satisfies the two following conditions (i) sup_g∈Gk1_l∞−C(g)k =³₂,

(ii) The algebra generated by the family(C(g))g∈Gis dense in l^∞.

Proof : ElementsgofG can be written under the formg=(g_m)m≥1, where gm∈{0, 1, 2}. Set

C(g) := µ

cos µ2g_mπ

3

¶¶

m≥1

.

Then (C(g))_g∈Gis aG-cosine family with values inl^∞which obviously satis- fies (i) sincecos¡_2π

3

¢=cos¡_4π

3

¢= −¹₂.

Now letφ=(φm)m∈Zbe an idempotent ofl^∞, and letS:={m≥1|φm=1}.

Setg_m=1 ifm∈S,g_m=0 ifm≥1,m∉S, and setg=(g_m)_m≥1. We have C(0_G)−C(g)=1_l∞−C(g)=3

2φ,

and soφ∈A. We can identifyl^∞toC(βN), the algebra of continuous func- tions on the Stone-C˘ech compactification ofN, andβNis an extremely discon- nected compact set, which means that the closure of every open set is open, see for example [1], chap. 6, sec. 6. Since the characteristic function of every open and closed subset ofβNis an idempotent ofl^∞, the idempotents ofl^∞separate points ofβN, and it follows from the Stone-Weierstrass theorem thatAis dense inl^∞, which proves (ii).ä

3 The values of the constant k (a)

It was shown in [17] thatk(0)=³2. We also have the following result.

Proposition 3.1. We have k(a)=₃^p8₃ if ^a_π is irrrational, and k(a)< ₃^p8₃ if ^a_π is rational .

Proof : Assume that^a_π∉Q. Then 3a∉ ±a+2πZ, and we have

k(a)≤supn≥1|cos(na)−cos(3na)| =supx∈R|cos(x)−cos(3x)| = 8 3p

3. We saw above that If ^b_π inQ, then supn≥1|cos(na)−cos(nb)| =2, and we also have supn≥1|cos(na)−cos(nb)| =2 ifp a−qb∉2πZfor (p,q)6=(0, 0). So if supn≥1|cos(na)−cos(nb)| <2, there existsp∈Z\ {0},q∈Z\ {0} andr ∈Zsuch thatp a−qb=2rπ.

Ifp6= ±qthen it follows from lemma 3.5 of [10] that we have

sup_n≥1|cos(na)−cos(nb)| ≥sup_n≥1|cos(nq a)−cos(nqb)| =sup_n≥1|cos(qna)−cos(pna)|

=sup_x∈R|cos(q x)−cos(p x)| =sup_x∈R

¯¯

¯¯cos µp

qx

¶

−cos(x)

¯¯

¯¯≥ 8 3p

3. We are left with the case whereb= ±a+^2sπ_r , wherer∈Z\ {−1, 0, 1}, and we can restrict attention to the case whereb=a+^2sπ_r wherer ≥2, 1≤s≤r−1, g cd(r,s)=1. It follows from Bezout’s identity that there exists for everyp ≥1 someu∈Zsuch thatub−ua−^2pπr ∈2πZ. Ifr is even, setp=^r2. We have, since the set©

e^i(2n+1)aª

n≥1is dense in the unit circle,

supn≥1|cos(nb)−cos(na)| =supn∈Z|cos(nb)−cos(na)|

≥supn≥1|cos((2n+1)ub)−cos((2n+1)ua)|

=2supn≥1|cos((2n+1)ua)| =2.

Now assume thatr is odd, and setp=^r−2¹. We have

supn≥1|cos(nb)−cos(na)| ≥supn≥1|cos((2n+1)ub)−cos((2n+1)ua)|

≥supn≥1

¯¯

¯cos((2nr+1)ua)−cos³

(2nr+1)ua+(2nr+1)³ π−π

r

´´¯¯¯

≥supx∈R

¯¯

¯cos(x)+cos³ x−π

r

´¯¯¯≥2cos³π 2r

´

≥p 3> 8

3p 3.

Now assume that_π^a is rational. If the order ofais equal to 1, thenk(a)=1.5, and we will see later that this is also true if the order ofaequals 2 or 4.

Otherwise we have

k(a)≤supn≥1|cos(na)−cos(3na)| =max1≤n≤u|cos(na)−cos(3na)|. We have|cos(nx)−cos(3nx)| <_π^p8₃ifx∉ ±ar ccos³

p1 3

´

+πZ. Ifna∈ ±ar ccos³

p1 3

´ + πZfor somen≥1, then ^{ar ccos}

³p1 3

´

π would be rational, andα:= ^p1₃+

p2i p3 would be a root of unity. Soβ=α²= −¹3+²

p2i

3 would have the formβ=e^{2i kπn} for some n≤1 and some positive integerk≥nsuch thatg cd(k,n)=1.

LetQ(β) be the smallest subfield of C containingQ∪β. Since 3β²+2β+ 3=0, the degree of Q(β) over Q is equal to 2. On the other hand the Galois groupG al(Q(β)/Q) is isomorphic to (Z/nZ)^×, the group of invertible elements ofZ/nZ, and we have, see [21], theorem 2.5

H(n)=d e g(Q(β)/Q)=2,

whereH(n)=c ar d((Z/nZ)^×) denotes the number of integersp∈{1, . . . ,n} such thatg cd(p,n)=1.

LetP(n) be the set of prime divisors ofn. It is weil-known that we have, wri- tingn=Πp∈P(n)p^αp, see for example [21], exercise 1.1,

H(n)=Πp∈P(n)p^αp−1(p−1).

It follows immediately from this identity that the only possibilities to get H(n)=2 aren=3,n=4, andn=6. Since β³6=1,β⁴6=1, andβ⁶6=1, we see that^β_π is irrational, and sok(a)<₃^p8₃if^a_π is rational.ä

ä

We know that if ^a_π is rational, and if ^b_π is irrational, thensup_n≥1|cos(na)− cos(nb)| =2. We discuss now the case where ^a_π and _π^b are both rational, with b∉ ±a+2πZ.

Lemma 3.2. Let a,b∈(0,π].

(i) If7a≤b≤^π₂,or if^π₂≤b≤^5π₆ ,with¯¯b−^2π₃ ¯¯≥7a,then sup_n≥1|cos(na)−cos(nb)| >1.55.

(ii) If ^5π₆ ≤b≤π,and if b≥4a,then

cos(a)−cos(b)>1.57.

Proof : (i) Assume that 7a≤b≤^π₂, letpbe the largest integer such thatpb<

3π

4 , and setq=p+1. We have^3π₄ ≤qb≤^5π₄ , 0≤q a≤^5π₂₈, and we obtain supn≥1|cos(na)−cos(nb)| ≥cos¡

q a¢

−cos¡ qb¢

≥cos µ5π

28

¶

+cos³π 4

´

>1.55.

Now assume that ^π₂ ≤b≤^5π₆ , with|b−^2π₃ | ≥7a, and setc= |3b−2π|. Since

¯¯b−^2π₃ ¯¯≤^π₆, we have 21a≤c≤^π₂, and we obtain

supn≥1|cos(na)−cos(nb)| ≥supn≥1|cos(3na)−cos(3nb)|

=supn≥1|cos(3na)−cos(nc)| >1.55.

(ii) If^5π₆ ≤b≤π, and ifb≥4a, then 0<a≤^π4, and we have cos(a)−cos(b)≥cos³π

4

´

+cos³π 6

´

>1.57.

Lemma 3.3. Let p,q be two positive integers such that p<q.

(i) If q6=3p,then there exists u_p,q≥1such that, if or d(a)≥u_p,q we have sup_n≥1|cos(np a)−cos(nq a)| > 8

p3.

(ii) If q=3p,then for every m<₃^p8₃there exists up(m)≥1such that if or d(a)≥ u(m)we have

sup_n≥1|cos(np a)−cos(3np a)| >m.

Proof : Setλ=supx∈R|cos(p x)−cos(q x)| =supx≥0|cos(p x)−cos(q x)|. An elementary verification shows thatλ>₃^p8₃ ifq6=3p, andλ=₃^p8₃ ifq=3p, see for example [10]. Now letµ<λ, and let η<δbe two real numbers such that

|cos(p x)−cos(q x)| >µforη≤x≤δ. Since {e^{i an}}n≥1={e^2niπu }1≤n≤u, we see that supn≥1|cos(np a)−cos(nq a)| >µif^2π_u <δ−η, and the lemma follows.ä Lemma 3.4. Assume that^a_π and^b_πare rational, let u≥1be the order of a and let v be the order of b.

(i) If u 6=v,u6=3v,v6=3u then sup_n≥1|cos(na)−cos(nb)| ≥1+cos¡_π

5

¢>

1.8>₃^p8₃.

(ii) If u=v,and if b∉ ±a+2πZ,then there exists w∈Zsuch that2≤w≤^u2

and g cd(u,w)=1satisfying

supn≥1cos(na)−cos(nb)| =supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2nwπ u

¶¯¯¯¯. (2)

Conversely if a∈πQhas order u,then for every integer w such that g cd(w,u)= 1,there exists b∈πQof order u satisfying (2).

(iii) If v=3u,then there exists an integer w such that1≤w≤^u₂and g cd(u,w)= 1satisfying

supn≥1cos(na)−cos(nb)| =supn≥1

¯¯

¯¯cos µ2nπ

3u

¶

−cos

µ2nwπ u

¶¯¯¯¯. (3)

Conversely if a∈πQhas order u,then for every integer w such that g cd(w,u)= 1there exists b∈πQof order3u satisfying (3).

(iv) If u=3v,then there exists an integer w such that1≤w≤^u₆and g cd¡_u

3,w¢

= 1satisfying

supn≥1cos(na)−cos(nb)| =supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ6nwπ u

¶¯¯¯¯. (4)

Conversely if the order u of a∈πQis divisible by3,then for every integer w such that g cd¡_u

3,w¢

=1there exists b∈πQof order^u₃ satisfying (4).

Proof : (i) Assume thatu6=v, say,u<v, and letw 6=1 be the order ofub, which is a divisor ofv. We haveub=^2πα_w , withg cd(α,w)=1, and there exists γ≥1 such thatαγ−1∈wZ. We obtain

sup_n≥1|cos(na)−cos(nb)| ≥sup_n≥1|cos(nuγa)−cos(nuγb)| =sup_1≤n≤w1−cos µ2nπ

w

¶ . Ifwis even, thensupn≥1|cos(na)−cos(nb)| =2. Ifwis odd, sets=^w₂⁻¹. We

obtain

supn≥1|cos(na)−cos(nb)| ≥1−cos µ2sπ

w

¶

=1+cos³π w

´.

Ifw≥5, we obtain

supn≥1|cos(na)−cos(nb)| ≥1+cos³π 5

´

>1.8> 8 3p

3.

Ifw=3, letd=g cd(u,v), and setr=^u_d. Thenw=3=_d^v >r. So eitherr =1 orr=2.

If r =2, we have u =2d,v =3d, a = ^2pπ2d = ^pπd with p odd,b = ^2q3d^π with g cd(q, 3d)=1, and we obtain

supn≥1|cos(na)−cos(nb)| = |cos(3d a)−cos(3d b)|

≥ |cos(3pπ)−cos(2qπ)| =2.

Ifr=1 thenu=dandv=3d=u.

We thus see that ifv>uandv6=3u, thensup_n≥1|cos(na)−cos(nb)| ≥1+ cos¡_π

5

¢>1.8>^p3₃, which proves (i).

(ii) Assume thatu=v, and thatb∉ ±a+2πZ. There existsα,β∈{1, . . . ,u− 1}, withα6=β,α6=u−βsuch that a ∈ ±^2απ_u +2πZand b ∈ ±^2βπ_u +2πZ, and g cd(α,u)=g cd(β,u)=1. It follows from Bezout’s identity that there existsγ∈Z such thatαγ−1∈uZ. Ifβγ±1∈uZthen we would haveαβγ±α∈αuZ⊂uZ, and β±α∈uZ, which is impossible. Henceγβ−w∈uZfor somew∈{2, . . . ,u−2}, g cd(w,u)=1 sinceg cd(γ,u)=g cd(β,u)=1, and we have

supn≥1|cos(na)−cos(nb)| ≥supn≥1|cos(nγa)−cos(nγb)|

=supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2nwπ u

¶¯¯¯¯≥supn≥1

¯¯

¯¯cos µ2nαπ

u

¶

−cos

µ2nαwπ u

¶¯¯¯¯

=sup_n≥1

¯¯

¯¯cos µ2nαπ

u

¶

−cos µ2nβπ

u

¶¯¯¯¯=sup_n≥1|cos(na)−cos(nb)|.

By replacingwbyu−wif necessary, we can assume that 2≤w≤^u₂.

Now letw∈Zsuch thatg cd(u,w)=1. We havea=^2απ_u , withg cd(α,u)=1.

The same argument as above shows that we have sup_n≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2nwπ u

¶¯¯¯¯=sup_n≥1cos(na)−cos(nb)|,

withb=^2wαπ_u , which has orderu.

(iii) Now assume thatv=3u. There existsα∈{1, . . . ,u−1} andβ∈{1, . . . , 3u− 1} such thata∈ ±^2απ_u +2πZandb∈ ±^2βπ_3u +2πZ, andg cd(α,u)=g cd(β, 3u)=1.

Letγ∈Zsuch thatβγ−1∈3uZ. Theng cd(γ, 3u)=1, and a fortiorig cd(γ,u)=1.

There existsw∈Zsuch thatαγ∈ ±w+uZ, and we see as above that we have

supn≥1|cos(na)−cos(nb)| =supn≥1

¯¯

¯¯cos µ2nαπ

u

¶

−cos µ2nβπ

3u

¶¯¯¯¯

=supn≥1

¯¯

¯¯cos

µ2nαγπ u

¶

−cos

µ2nβγπ 3u

¶¯¯¯¯=supn≥1

¯¯

¯¯cos

µ2nwπ u

¶

−cos µ2nπ

3u

¶¯¯¯¯.

Conversely leta=^2απu ∈πQhave orderu, and letw∈Zbe such thatg cd(u,w)= 1. Ifαis not divisible by 3, theng cd(α, 3u)=1. Ifαis divisible by 3, thenuis not divisible by 3, and soα+u∈α+uZis not divisible by 3. So we can assume wi- thout loss of generality thatαis not divisible by 3, and there existsβ≥1 such thatαβ−1∈3uπZ. Similarly we can assume without loss of generality thatwis not divisible by 3, and there existsγ≥1 such thatwγ−1∈3uπZ. Setb=^2αγπ_3u . Thenbhas order 3u, and we see as above that we have

supn≥1

¯¯

¯¯cos

µ2nwπ u

¶

−cos µ2nπ

3u

¶¯¯¯¯≥supn≥1

¯¯

¯¯cos

µ2nαγwπ u

¶

−cos

µ2nαγπ 3u

¶¯¯¯¯

=supn≥1|cos(na)−cos(nb)| ≥supn≥1

¯¯

¯¯cos

µ2nαγwβwπ u

¶

−cos

µ2nαγβwπ 3u

¶¯¯

¯¯

=supn≥1

¯¯

¯¯cos

µ2nwπ u

¶

−cos µ2nπ

3u

¶¯¯¯¯,

which concludes the proof of (iii).

(iv) Clearly, the first assertion of (iv) is a reformulation of the first assertion of (iii). Now assume that the orderuofa∈πQis divisible by 3, setv=^u3, write a = ^2απ_u , and let w∈Zsuch thatg cd(w,v)=1. We see as above that we can assume without loss of generality thatg cd(u,w)=1.

Sinceg cd(α,u)=1, we have a fortiorig cd(α,v)=1, so thatg cd(αw,v)=1, so thatb:=^6αw_u has ordervand we see as above thata,b,uandwsatisfy (4).ä

In order to use lemma 3.4, we introduce the following notions.

Definition 3.5. Let u≥2,and denote by∆(u)the set of all integers s satisfying 1≤s≤^u₂,g cd(u,s)=1,and let∆1(u)=∆(u) \ {1}.We set

σ(u)=i n f_w∈∆(u)

· sup_n≥1

¯¯

¯¯cos µ2π

3u

¶

−cos µ2wπ

u

¶¯¯¯¯

¸ , θ(u)=i n fw∈∆1(u)

· supn≥1

¯¯

¯¯cos µ2π

u

¶

−cos µ2wπ

u

¶¯¯¯¯

¸ . with the conventionθ(u)=2if∆1(u)= ;.

Notice that∆1(u)= ;ifu=2, 3, 4 or 6, and that∆1(u)6= ;otherwise since as we observed aboveH(n)=c ar d((Z/nZ)^×)≥3 ifn∉{1, 2, 3, 4, 6}.

We obtain the following corollary, which shows in particular that the value ofk(a) depends only on the order ofa.

Corollary 3.6. Let a∈πQ,and let u≥1be the order of a.

(i) If u is not divisible by3,then k(a)=i n f(σ(u),θ(u)).

(ii) If u is divisible by3,then k(a)=i n f(σ¡_u

3

¢,σ(u),θ(u)).

Proof : Set

— Λ1(a)=©

b∈πQ|b∉ ±a+2πZ,or d(b)=or d(a)ª ,

— Λ2(a)=©

b∈πQ|or d(b)=3or d(a)ª ,

— Λ3(a)=©

b∈πQ|3or d(b)=or d(a)ª ,

— Λ4(a)=©

b∈πQ|or d(b)6=or d(a)6=3or d(b)ª , and for 1≤i≤4, set

λ_i(a)=i n f_b∈Λi(a)sup

n≥1|cos(na)−cos(nb)|, with the conventionλi(a)=2 ifΛi(a)= ;.

Sinceb∉ ±a+2πZifor d(b)6=or d(a), we haveλ2(a)≤₃^p8₃, and it follows from lemma 3.4(i) that we have

k(a)=i n f1≤i≤4λi(a)=i n f1≤i≤3λi(a),

and it follows from lemma 3.4 (ii), (iii) and (iv) thatλ1(a)=θ(u) if∆1(u)6= ;, thatλ₂(a)=σ(u), and thatλ₃(a)=σ¡_u

3

¢ifuis divisible by 3.ä We have the following result.

Theorem 3.7. Let m<₃^p8₃.Then the setΩ(m) :={a∈[0,π] :k(a)≤m}is finite.

Proof : It follows from lemma 3.3 applied to^2π_u and^6π_u that there existsu0≥1 such that we have, foru≥u0,

(i)supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2w nπ u

¶¯¯¯¯>m if 2≤w≤i n f ³u 2, 6´

,

(i i)supn≥1

¯¯

¯¯cos µ6nπ

u

¶

−cos

µ2(3w+1)nπ u

¶¯¯¯

¯>m if 0≤w≤6, (i i i)supn≥1

¯¯

¯¯cos µ6nπ

u

¶

−cos

µ2(3w+2)nπ u

¶¯¯¯¯>m if 0≤w≤6.

Letu≥u0, and letw be an integer such that 2≤w≤ ^u2. Il ^2wπ_u ≤π/2, or if

2wπ

u ≥^5π₆ , it follows from lemma 3.2 and property (i) that we have sup_n≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2w nπ u

¶¯¯¯¯>m.

Now assume that^π₂≤^2wπ_u ≤^5π₆. If¯¯w−^u₃¯¯≥7, it follows from lemma 3.2 that we have

supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2w nπ u

¶¯¯¯¯>1.55>m.

If¯¯w−^u3

¯¯<7, setr= |3w−u|. Then 0≤r≤20, and we have

supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2w nπ u

¶¯¯¯¯≥supn≥1

¯¯

¯¯cos µ6nπ

u

¶

−cos µ2nrπ

u

¶¯¯¯¯.

Ifuis not divisible by 3, then eitherr =3s+1 orr =3s+2, with 0≤w≤6, and it follows from (ii) and (iii) that we have

supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2w nπ u

¶¯¯¯¯>m.

Ifu is divisible by 3 thenr is also divisible by 3. Setv= ^u₃ ands= ^r₃. Then 0≤s≤6, and we have

supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2w nπ u

¶¯¯¯¯≥supn≥1

¯¯

¯¯cos µ2nπ

v

¶

−cos µ2nsπ

v

¶¯¯¯¯.

Ifs∈{2, 3, 4, 5, 6} it follows from (i) that we have, ifu≥3u0, sup_n≥1

¯¯

¯¯cos µ2nπ

v

¶

−cos µ2snπ

u

¶¯¯¯¯>m.

Now assume thats=0. Ifu≥15, thenv≥5, and we have

sup_n≥1

¯¯

¯¯cos µ2nπ

v

¶

−cos µ2snπ

u

¶¯¯¯¯=sup_n≥1

¯¯

¯¯cos µ2nπ

v

¶

−1

¯¯

¯¯≥1+cos³π 5

´

>1.8>m.

Now assume thats=1. We have, withǫ= ±1,

supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2w nπ u

¶¯¯¯¯=supn≥1

¯¯

¯¯cos µ2nπ

3v

¶

−cos µ2nπ

3v +2nǫπ 3

¶¯¯¯¯

≥supn≥1

¯¯

¯¯cos

µ2(3n+1)π 3v

¶

−cos

µ2(3n+1)π

3v +2ǫπ

3

¶¯¯¯¯=p 3

¯¯

¯¯si n µ2nπ

v +2π 3v+ǫπ

3

¶¯¯¯¯.

There existsp≥1 andq∈Zsuch that ^π₂−^πv ≤ ^2pπv +^2π3v +^ǫπ3 +2qπ≤^π2+^πv, and we obtain, foru≥21,w=v±1,

supn≥1

¯¯

¯¯cos µ2nπ

u

¶

−cos

µ2w nπ u

¶¯¯¯¯≥p

3cos³π v

´

≥p

3cos³π 7

´

≥1.56>m.

We thus see that ifu ≥u₀ is not divisible par 3, or ifu ≥max(21, 3u₀) is divisible by 3, we have, for 2≤w≤^u₂,