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Estimates of the Linearization of Circle Diffeomorphisms

Mostapha Benhenda

To cite this version:

Mostapha Benhenda. Estimates of the Linearization of Circle Diffeomorphisms. 2011. �hal-00628297�

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Estimates of the Linearization of Circle Di ff eomorphisms

Mostapha Benhenda

September 30, 2011

Abstract

A celebrated theorem by Herman and Yoccoz asserts that if the rotation number αof aC-diffeomorphism of the circle f satisfies a Diophantine condition, then fisC-conjugated to a rotation. In this paper, we establish explicit relationships between theCknorms of this conjugacy and the Diophantine condition onα. To obtain these estimates, we follow a suitably modified version of Yoccoz’s proof.

1 Introduction

In his seminal work, M. Herman [5] shows the existence of a set Aof Diophantine numbers of full Lebesgue measure such that for any rotation numberα ∈ Aof a cir- cle diffeomorphism f of classCω (resp. C), there is a Cω-diffeomorphism (resp.

C-diffeomorphism)hsuch thath f h−1 =Rα. In theCcase, J. C. Yoccoz [14] ex- tended this result to all Diophantine rotation numbers. Results in analytic class and in finite differentiability class subsequently enriched the global theory of circle dif- feomorphisms [11, 9, 8, 13, 7, 15, 4, 10]. In the perturbative theory, KAM theorems usually provide a bound on the norm of the conjugacy that involves the norm of the perturbation and the Diophantine constants of the numberα(see [5, 12, 3] for exam- ple). We place ourselves in the global setting, we compute a bound on the norms of this conjugacyhin function ofk,|D f|0,W(f),|S f|k−3,βandCd.

To obtain these estimates, we follow a suitably modified version of Yoccoz’s proof.

Indeed, Yoccoz’s proof needs to be modified because a priori, it does not exclude the fact that the set:

EX =n

|Dh|0/∃f ∈Diffk+1), f =h−1Rαh, α∈DC(β,Cd),max (k, β,Cd,|D f|0,W(f),|S f|k−3)≤Xo could be unbounded for any fixedX>0.

These estimates have natural applications to the global study of circle diffeomor- phisms with Liouville rotation number: in [2], they allow to show the following results:

1) Given a diffeomorphismf of rotation numberα, for a Baire-dense set ofα, it is pos- sible to accumulateRαwith a sequencehnf h−1n ,hn being a diffeomorphism. 2) Given two commuting diffeomorphisms f andg, with the rotation numberαof f belonging to a specified Baire-dense set, it is possible to approach each of them by commuting diffeomorphismsfnandgnthat are differentiably conjugated to rotations.

Laboratoire d’Analyse, Geometrie et Applications, Paris 13 University, 99 Avenue J.B. Clement, 93430 Villetaneuse, France. Contact: mostaphabenhenda@gmail.com

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1.1 Notations

We follow the notations of [14].

• The circle is noted”1. The group of orientation-preserving circle diffeomor- phisms of classCr is denoted Diffr+1). The group ofš-periodic diffeomor- phisms of classCr of the real line is notedCr1). We often work in the uni- versal coverDr1), which is the group of diffeomorphisms f of classCrof the real line such that f −Id ∈ Cr1). Note that if f ∈ Dr1) andr ≥ 1, then D f ∈Cr−11).

• The derivative of f ∈ D11) is notedD f. The Schwartzian derivativeS f of f ∈D31) is defined by:

S f =D2logD f −1

2(DlogD f)2

• The total variation of the logarithm of the first derivative of f is:

W(f)= sup

a0≤...≤an n

X

i=0

|logD f(ai+1)−logD f(ai)|

• For any continuous andš-periodic functionφ, let:

|φ|0 =kφk0 =sup

x∈’

|φ(x)|

• Let 0< γ0<1.φ∈C01) is Holder of orderγ0if:

|φ|γ0=sup

x,y

|φ(x)−φ(y)|

|x−y|γ0 <+∞

Letγ≥1 be a real number. In all the paper, we writeγ=r+γ0withr∈Žand 0≤γ0<1.

• A functionφ∈Cr1) is said to be of classCγifDrφ∈Cγ01). The space of these functions is notedCγ1) and is given the norm:

kφkγ=max max

0≤j≤rkDjφk0,|Drφ|γ0

!

Ifγ=0 orγ≥1, theCγ-norm ofφis indifferently denotedkφkγ or|φ|γ. Thus, when possible, we favor the simpler notation|φ|γ.

• Forα∈ ’(respectively,α∈ ”1), we denoteRα ∈ D1) (respectively,Rα ∈ Diff+1)). the mapx7→x+α.

• An irrational numberα ∈ DC(Cd, β) satisfies a Diophantine condition of order β≥0 and constantCd>0 if for any rational numberp/q, we have:

α− p q

≥ Cd q2+β Moreover, ifβ=0, thenαis of constant typeC .

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• Letα−2 = α, α−1 = 1. Forn ≥ 0, we define a real numberαn (theGauss sequenceofα) and an integeranby the relations 0< αn< αn−1and

αn−2=anαn−1n

• In the following statements,Ci[a,b, ...] denotes a positive numerical function of real variablesa,b, ..., with an explicit formula that we compute.

C[a,b, ...] denotes a numerical function ofa,b, ..., with an explicit formula that we do not compute.

• We use the notationsa∧b =ab,e(n)∧xthenth- iterate of x7→ expx,bxcfor the largest integer such thatbxc ≤ x, anddxefor the smallest integer such that dxe ≥x.

We recall Yoccoz’s theorem [14]:

Theorem 1.1. Let k ≥ 3an integer and f ∈ Dk1). We suppose that the rotation numberαof f is Diophantine of orderβ. If k>2β+1, there exists a diffeomorphism h∈D11)conjugating f to Rα. Moreover, for anyη >0, h is of class Ck−1−β−η.

1.2 Statement of the results

1.2.1 C1estimations

Theorem 1.2. Let f ∈ D31)of rotation numberα, such thatαis of constant type Cd. There exists a diffeomorphism h∈D11)conjugating f to Rα, which satisfies the estimation:

|Dh|0≤e∧ C1[W(f),|S f|0] Cd

!

The expression of C1is given page 10.

More generally, for a Diophantine rotation numberα∈DC(Cd, β), we have:

Theorem 1.3. Let k≥3be an integer and f ∈Dk1). Letα∈DC(Cd, β)be the rota- tion number of f . If k>2β+1, there exists a diffeomorphism h∈D11)conjugating

f to Rα, which satisfies the estimation:

|Dh|0≤C2[k, β,Cd,|D f|0,W(f),|S f|k−3] (1) The expression of C2is given page 23.

Moreover, if k≥3β+9/2, we have:

|Dh|0≤e(3)

C3[β]C4[Cd]C5[|D f|0,W(f),|S f|0]C6[|S f|d3β+3/2e]

(2) The expressions of C3,C4,C5,C6are given page 28.

Letδ=k−2β−1. Whenδ→0, we have:

|Dh|0≤e(3)∧ 1

δ2C7[k,Cd,|D f|0,W(f),|S f|0]+C[δ]

δ2 C[k,Cd,|D f|0,W(f),|S f|0,|S f|k−3]

!

(3) where C[δ]→δ→00. The expression of C7is given page 30.

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Remark 1.4. Katznelson and Ornstein [9] showed that the assumptionk >2β+1 in Yoccoz’s theorem is not optimal (instead it isk> β+2). Therefore, the divergence of the bound given by estimation (3) is because we compute the bound of the conjugacy by following the Herman-Yoccoz method.

Remark1.5. Letαnbe the Gauss sequence associated withα. Yoccoz’s proof already gives the following result: ifk≥3β+9/2 and if, for anyn≥0,

αn+1

αn ≥C8[n,k,W(f),|S f|k−3] (4) then:

|Dh|0 ≤exp

C9[k,W(f),|S f|k−3]C10(β)

|D f|20

The expressions ofC8,C9,C10are given page 30.

1.2.2 Cuestimations

Theorem 1.6. Let k≥3an integer,η >0and f ∈ Dk1). Letα∈DC(Cd, β)be the rotation number of f . If k >2β+1, there exists a diffeomorphism h∈ Dk−1−β−η1) conjugating f to Rα, which satisfies the estimation:

kDhkk−2−β−η ≤e(dlog((k−2−β)/η)/log(1+1/(2β+3))e)∧(C11[η,k, β,Cd,|D f|0,W(f),|S f|k−3]) (5) The expression of C11is given page 48.

Moreover, if k≥3β+9/2, we have:

kDhk k

2(β+2)12 ≤e∧

C12[k]e(2)∧(2+C3[β]C4[Cd]C5[|D f|0,W(f),|S f|0]C6[|S f|k−3]) (6) The expression of C12is given page 46.

Ifαis of constant type, for any k>3, we have:

kDhkk

412 ≤e∧





C13[k]

"

C14[W(f),|S f|k−3]+C1[W(f),|S f|0] Cd

#4





 (7) The expressions of C13and C14are given page 47.

2 Preliminaries

Let f ∈ D01) be a homeomorphism and x ∈ ’. When n tends towards infinity, (fn(x)−x)/nadmits a limit independent of x, notedρ(f). We call it thetranslation numberof f. Two lifts of f ∈Diff0+1) only differ by a constant integer, so this is also the case for their translation numbers. We call the class ofρ(f) modštherotation numberof f. We still denote itρ(f). It is invariant by conjugacy. Let f ∈ D21).

Whenα=ρ(f) is irrational, Denjoy showed that f is topologically conjugated toRα. However, this conjugacy is not always differentiable (see [1, 5, 7, 15]). The regularity

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of this conjugacy depends on the Diophantine properties of the rotation numberα(see Yoccoz’s theorem 1.1).

Letαbe an irrational number. Let the distance ofαto the closest integer be:

||α||=inf

p∈š|α−p|

Forn≥1,an ≥1. Letα=a0+1/(a1+1/(a2+...)) be the development ofαin continued fraction. We denote itα=[a0,a1,a2, ...]. Letp−2=q−1=0,p−1=q−2=1.

Forn≥0, letpnandqnbe:

pn=anpn−1+pn−2

qn=anqn−1+qn−2

We haveq0 =1,qn ≥1 forn ≥1. The rationals pn/qnare called the convergents ofα. They satisfy the following properties:

1. αn =(−1)n(qnα−pn) 2. αn =||qnα||, forn≥1

3. 1/(qn+1+qn)< αn<1/qn+1forn≥0.

4. αn+2< 12αn,qn+2≥2qn, forn≥ −1

We recall thatDC(Cd, β) denotes the set of Diophantine numbers of constantsβand Cd. One of the following relations characterizesDC(Cd, β):

1. |α−pn/qn|>Cd/q2n+βfor anyn≥0 2. an+1<C1

dqβnfor anyn≥0 3. qn+1<C1

dq1n+βfor anyn≥0 4. αn+1>Cdα1n+βfor anyn≥0

In all the paper, we denoteCd0 =1/Cd.

• Letmn(x)=fqn(x)−x,n≥1,x∈”1, letMn=supx∈”1|fqn(x)−x|and mn=infx∈”1|fqn(x)−x|.

• For anyφ, ψ∈Cγ1), we have:

|φψ|γ≤ kφk0|ψ|γ+|φ|γkψk0 (8) kφψkγ≤ kφk0|ψ|γ+kφkγkψk0 (9)

• For any real numbersaandb,a∨bdenotes max(a,b).

In the rest of the paper, for any integeri,Cif denotes a constant depending only onW(f) and |S f|0 (i.e. Cif is a numerical function of these variables). Cif,kdenotes a constant depending only onk,W(f),|S f|0and|S f|k−3. Cidenotes a constant that might depend onk,W(f),|S f|0,|S f|k−3and alsoβandCd.

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3 C

1

estimations: constant type

3.1 A 2-parameters family of homographies

In this subsection, we show the existence of a lower bound on the norm of the conjugacy in function ofCdin the particular case of a 2-parameters family of homographies. We also establish an upper bound on theC1norm of the conjugacy for this family. These bounds are similar to what is given by the local KAM theory. However, these bounds are very specific to this setting. Our general bounds given in theorems 1.2 and 1.3 are much larger.

Proposition 3.1. Let f :{z∈ƒ/|z|=1} → {z∈ƒ/|z|=1}defined by f(z)=h−1Rθh(z), with Rθ(z)=ez and h is a homography defined by:

h(z)= z−a az−1

Let2>a>1, let Cdsuch that Cd−1 ≥6is a positive integer; and0< θ=2πCd ≤ π/3, (therefore,θ/(2π)=[0,Cd−1,1]is of constant type Cd). Let f˜:”1 →”1the circle diffeomorphism induced by f andh the conjugacy induced by h. We have the following˜ estimation:

2

πC15(|Df˜(0)|,|D2f˜(0)|)/Cd≤ |Dh|˜0≤9C15(|Df˜(0)|,|D2f˜(0)|)/Cd

Proof. For anyφ∈ ’/š, we can writeh(e)=eih(φ)˜ . By differentiating this expres- sion, we have:

Dh(φ)˜ =eDh(e) h(e) and

Dh(z)=(a−1)(a+1) (az−1)2 Therefore

Dh(φ)˜ =e a2−1 (ae−1)(e−a)

|Dh(φ)|˜ reaches its maximum forφ=0, and|Dh˜|0= aa−1+1. Moreover, we have:

D2f˜(φ)

Df˜(φ) =i+ieD2f(e)

D f(e) −ieD f(e) f(e) Since DD2f(φ)f˜˜(φ) ∈’andDf˜(φ)∈’, we have:

D2f(1) D f(1)

=





 D2f˜(0)

Df˜(0)

!2 +

Df˜(0)−12







1/2

=8C15(|Df˜(0)|,|D2f˜(0)|) Therefore, in order to get the proposition, it suffices to show:

1 4πC

D2f(1) D f(1)

≤ |Dh|0≤ 9 8C

D2f(1) D f(1)

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Let us write

f(z)= (e−a2)z−a(e−1)

a(e−1)z−(a2e−1) = bz−c cz+d We have

D f(z)= db+c2 (cz+d)2 and

D2f(z)=−2 D f(z) z+d/c Moreover,

D f(1)=(1+a)2e (ae+1)2 and

D2f(1)=−2D f(1) a(e−1) (1−a)(1+ae) We have

D2f(1) D f(1)

=2

a 1+ae

|e−1| a−1 Since|e−1| ≥sinθ≥2πθ(because 0≤θ≤π/2), then:

D2f(1) D f(1)

≥4 π

a 1+a

θ a−1 Therefore,

D2f(1) D f(1)

≥ 4 π

a

(1+a)2|Dh|0θ i.e.

9π 4θ

D2f(1) D f(1)

≥ |Dh|0

Hence the first part of the inequality.

On the other hand, sinceθ≤π/3, then|1+ae| ≥1+acosθ≥1+a/2≥1

2(a+1).

Furthermore,|e−1|2 =2−2 cosθ=2−2(cos2θ/2−sin2θ/2)=4 sin2θ/2≤θ2. Therefore,

D2f(1) D f(1)

≤ θ a−1

4a

a+1 ≤ 4θ

a−1 =4θ|Dh|0 a+1 i.e.

1 2

D2f(1) θD f(1)

≤ |Dh|0

Hence the second part of the inequality.

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3.2 Proof of theorem 1.2

The proof of theorem 1.2 is divided in three steps. The first step is based on the im- proved Denjoy inequality, which estimates theC0-norm of logD fql. In the second step, we extend this estimation to logD fN for any integerN. To do this, following Denjoy and Herman, we writeN=PS

s=0bsqs, withbsintegers satisfying 0≤bs≤qs+1/qsand we apply the chain rule. In the third step, we derive aC0-estimation of the derivative Dhof the conjugacyh.

The first step is based on the Denjoy inequality:

Proposition 3.2. Let f ∈Diff3+1)and x∈”1. We have:

|logD fql(x)| ≤W(f)

Proposition 3.2 is used to obtain an improved version of Denjoy inequality [14, p.342]:

Lemma 3.3. Let f ∈Diff3+1). We have:

|logD fql|0≤C16f Ml1/2

|D fql−1|0≤C17f Ml1/2 Moreover, we can take:

C16f =2

2(2eW(f)+1)eW(f)(|S f|0)1/2 and

C17f =6

2e3W(f)|S f|1/2

0

In the second step, we estimateDlogD fN independently ofN. This step is based on the following lemma:

Lemma 3.4. Let f ∈Diff3+1)and Ml=supx∈”1|fql(x)−x|. We have:

X

l≥0

pMl≤ 1 q

C19f −C19f with

C19f = 1 p

1+e−C22f

(10) and:

C22f =6

√ 2e2W(f)

|S f|1/2

0 ∨1

(11) Proof. To obtain this lemma, we need the claim:

(10)

Claim 3.5. Let f ∈Diff2+1)of rotation numberα, and let pn/qnbe the convergents ofα. Then for all x∈”1, we have:

[x,f2ql+2(x)]⊂[x,fql(x)]

Proof. By topological conjugation, it suffices to examine the case of a rotation of angle α. It is also sufficient to takex=0.

By absurd, if the lemma was false, then we would have the following cyclic order on”1: −ql+2α ≤(ql+2−ql)α≤ 0 ≤(ql−ql+2)α≤ql+2α. In particular, (ql+2−ql)α would be closer to 0 thanql+2α, which would contradict the fact that

kql+2αk=inf{kqαk/0<q≤ql+2}.

For any intervalIof the circle, if|I|denotes the length ofI, lemma 3.3 implies the estimation:

|fql+2(I)|

|I| ≥e−C22fMl+21/2

Let x ∈ ”1 such that Ml+2 = fql+2(x)−xand let I = [x,fql+2(x)]. The former estimation implies

|f2ql+2(x)−fql+2(x)| ≥e−C22fM1/2l+2Ml+2 By applying claim 3.5, and sinceMn≤1, we obtain:

Mn+2+e−C22f Mn+2≤Mn+2+e−C22f Mn+21/2Mn+2≤Mn

Therefore, for anyl≥0,

Ml≤(C19f )l−1 (12)

with

C19f = 1 p

1+e−C22f Estimation (12) above gives:

X

l≥0

pMl≤ 1 q

C19f 1 1−

q C19f

≤ 1

q

C19f −C19f Hence lemma 3.4.

Now, letN be an integer. Following Denjoy, sinceαis of constant type, we can writeN =Ps

l=0blql, withblintegers satisfying 0 ≤bl ≤ql+1/ql≤C−1d . By the chain rule and by lemma 3.3, since for ally∈”1,D fN(y)>0, then :

|logD(fN)(y)|=|logD(fPl=0s blql)(y)|=|Ps l=0

Pbs

i=0logD fql◦ fiql(y)|

≤sup0≤l≤sblPs

l=0|log|D(fql)|0| ≤C−1d C22f P

l≥0M1/2l

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By taking the upper bound ony ∈ ”1 andN ≥ 0, we obtain an estimation of supN≥0|logD(fN)|.

We turn to the third step: we relate the norms ofDhandD fN. By [14],hisC1and conjugates f to a rotation. Therefore, we have:

logDh−logDh◦ f =logD f hence, for allninteger:

logDh−logDh◦ fn=logD(fn)

Since there is a pointzin the circle such thatDh(z)=1, we then have:

|logDh◦ fn(z)|=|logD(fn)(z)| ≤sup

i≥0

|logD(fi)|0

Moreover, since (fn(z))n≥0is dense in the circle, and sinceDhis continuous, then we obtain:

|logDh|0≤sup

i≥0

|logD(fi)|0 We conclude:

|Dh|0≤exp C−1d C22f q

eC22f max(M01/2,M1/21 )+1(p M0+p

M1)

!

(13) Finally, since max(M1/20 ,M1/21 )≤1, we obtain:

|Dh|0≤exp C1f/Cd

whereC1f =2C22f p

eC22f +1. We recall that:

C22f =6

√ 2e2W(f)

|S f|1/20 ∨1 Hence the theorem.

Corollary 3.6. Sincemin1

”1Dh ≤exp

supi≥0|logD(fi)|0

, the proof above also provides an estimation onmin1

”1Dh:

1

min”1Dh ≤exp C1f/Cd

4 C

1

estimations: non-constant type

We have maxn≥0|D fn|0≤maxn≥0Mn/mn, by [14, p. 348]. Therefore, in order to prove theorem 1.3, we can estimateMn/mn. To that end, we proceed in two steps: first, we establish some preliminary results. The most important result is corollary 4.6, which gives an estimation ofMn+1/Mn in function ofMnn+1nand a constantC28f,k. This estimation is already given in [14, p. 345], but we still recall the steps to reach it, because we need to estimate the constantC28f,kin function ofk,W(f),|S f|0and|S f|k−3. In the second step, we establish an estimation of theC1-conjugacy, based on a mod- ification of the proof given in [14]. The main idea is to establish an alternative between

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two possible situations for the sequencesMnandαn: the "favorable" situation (Rn) and the "unfavorable" situation (R0n) (proposition 4.10). The "unfavorable" situation only occurs a finite number of times, due to the Diophantine condition onα(propositions 4.12 and 4.14).

In the "favorable" situation (Rn), we can estimateMn+1n+1in function ofMnn (see estimation (26)) and likewise, we can estimateαn+1/mn+1 in function ofαn/mn. Therefore, we can estimate Mn/mn in function of Mn4/mn4, where n4 is the integer such that for anyn ≥ n4, the favorable case occurs (see proposition 4.19). We relate Mn4/mn4 toD f|

αn24

0 (proposition 4.17), and we compute a bound onαn4 (proposition 4.15). Yoccoz’s proof needs to be modified because in its original version, it does not allow to compute a bound onαn4.

4.1 Preliminary results

First, we recall the following lemmas, which are in [14] (lemmas 3,4 and 5):

Lemma 4.1. For l≥1and x∈”1, we have:

qn+1−1

X

i=0

D fi(x)l

≤C23f Ml−1n mn(x)l with C23f (l)=elW(f).

Remark4.2. This lemma is obtained by applying Denjoy inequality.

Lemma 4.3. Let f ∈Diffk+1), k≥3. For any x∈”1, any n∈Ž, any0≤ p≤qn+1, we have:

|S fp|0≤C24f Mn

m2n

|S fp(x)| ≤C24f Mn

mn(x)2

|DlogD fp|0≤C25f M1/2n

mn

|DlogD fp(x)| ≤C26f Mn1/2 mn(x) with:

• C24f =|S f|0e2W(f)

• C25f = p

2|S f|0eW(f)

• C26f =9p

2|S f|0e4W(f)

(13)

Lemma 4.4. For1≤r≤k−1, n≥0,0≤p≤qn+1, x∈”1, we have:

|DrlogD fp(x)| ≤C27f (r)





 M1/2n

mn(x)





r

(14) with

C27f (1)=C26f , C27f (2)=82|S f|0e8W(f) and, for r≥3:

C27f (r)=h

82(2r)2r(1∨ |S f|r−2)2e(r+8)W(f)ir!

In particular,

C27f,k:=C27f (k−1)≤h

100(2k−2)2k−2(1∨ |S f|k−3)2e(k+7)W(f)i(k−1)!

Proof of lemma 4.4. The proof follows the line of [14], lemma 5: see appendix 6.1.

The important preliminary result, corollary 4.6, is obtained from the following proposition. It is obtained by computing the constants in proposition 2 of [14]:

Proposition 4.5. Let

C28f,k=(k+3)(k+3)!e(k+2)!W(f)(max(1,|S f|k−3))k! (15) For any x∈”1, we have:

mn+1(x)−αn+1

αn mn(x)

≤C28f,kh

Mn(k−1)/2mn(x)+Mn1/2mn+1(x)i

(16) Corollary 4.6.

Mn+1≤Mn αn+1

αn +C28f,kM(k−1)/2n

1−C28f,kMn1/2 (17)

mn+1≥mn

αn+1

αn −C28f,kMn(k−1)/2 1+C28f,kM1/2n

The proof of proposition 4.5 combines the following three lemmas [14, pp. 343- 344] (lemmas 6, 7 and 8):

Lemma 4.7. For any x∈”1, there exists y∈[x,fqn(x)],z∈[fqn+1(x),x]such that mn+1(y)= αn+1

αn

mn(z)

Lemma 4.8. Suppose that mn+1 is monotonous on an interval Iz =(z,fq(z)), z∈”1. Then, for any x∈”1, for any y∈Ix(Ix=(x,fq(x))), we have:

mn+1(y) mn+1(x)−1

≤C29f,kMn1/2 with

C29f,k=29(k+2)e(11+k/2)W(f)(C17f )2C26f

(14)

Lemma 4.9. If mn+1 is not monotonous on any interval of the form Iz = (z,fq(z)), z∈”1, then for any x∈”1, y∈Ix, we have:

|mn+1(y)−mn+1(x)| ≤C30f,kM(k−1)/2n mn(x) with

C30f,k=(C27f (k−1))eW(f) e(k/2+2)W(f)(1+eW(f))2e(k/2+2)W(f)−1 eW(f)−1

!k−1

Proof of proposition 4.5. Let us recall the proof of proposition 4.5 from these three lemmas. (see [14, p.344]). Letx∈”1andy∈Ix,z∈[fqn+1(x),x] the points given by lemma 4.7. By combining lemmas 4.8 and 4.9, we obtain:

|mn+1(y)−mn+1(x)| ≤ max

C29f,k,C30f,k M1/2n mn+1(x)+M(k−1)/2n mn(x) Moreover, by lemma 3.3, we have:

|mn(z)−mn(x)| ≤C17f Mn1/2|z−x| ≤C17f Mn1/2mn+1(x) By applying lemma 4.7, and sinceαn+1n≤1, we get:

mn+1(x)−αn+1

αn mn(x)

mn+1(x)−αn+1

αn mn(z) +αn+1

αn |mn(z)−mn(x)|

mn+1(x)−αn+1 αn

mn(x)

≤ |mn+1(y)−mn+1(x)|+|mn(z)−mn(x)|

Therefore, we have:

mn+1(x)−αn+1 αn

mn(x)

≤C31f,k

Mn1/2mn+1(x)+M(k−1)/2n mn(x) withC31f,k=max(C29f,k,C30f,k)+C17f .

Finally, let us estimateC31f,k. Sincek≥3, then:

[4(k/2+1)(200k)]2≤(k+3)(k+3)(k+2)k/2 and therefore,

22(k−1)(k/2+1)k−1(200k)2(k+1)(k−1)!≤(k+3)(k+3)!/2 Therefore, we have:

C30f,k+C17f ≤(k+3)(k+3)!e(k+2)!W(f)(max(1,|S f|k−3))k!

Sincek≥3, we also have:

C29f,k+C17f ≤(k+3)(k+3)!e(k+2)!W(f)(max(1,|S f|k−3))k!

Therefore,C31f,k ≤C28f,k=(k+3)(k+3)!e(k+2)!W(f)(max(1,|S f|k−3))k!. Hence proposi- tion 4.5.

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4.2 Estimation of the C

1

-conjugacy. Proof of estimation (1).

We choose an integern1such that for anyn≥n1, we have:

C28f,kMn1/2 ≤C28f,k(C19f )n−12 <1/2 (18) We take:

n1=











−log

2C28f,k/(C19f )1/2 log

(C19f )1/2











We choose a parameterθsuch that (k+1)/2−θ > (1+β+θ)(1+θ) (for the interpretation of this parameterθ, see the remark after proposition 4.10). We take:

θ=min





1/2, 3+β 4

! 





−1+ 1+2(k−2β−1) (3+β)2

!1/2











 (19) (in the proof of estimation (2), we takeθ=1/2 instead).

We recall that for x ≥ 0, 1+x ≤ ex and for 0 ≤ x ≤ 1/2, log (1/(1−x)) ≤ x/(1−x)≤2x. We apply estimation (18), we use the definition ofn1and the fact that θ≤1/2. We get:

+

Y

n=n1

1+Mθn

≤exp







+

X

n=n1

Mθn







≤exp







1 2C28f,k(1−(C19f )θ)







+

Y

n=n1





 1 1−C28f,kMn1/2







≤exp







+

X

n=n1

2C28f,kMn1/2







≤exp





 1 1−(C19f )1/2







Therefore,

+

Y

n=n1







1+Mnθ 1−C28f,kMn1/2







≤exp





 2 1−(C19f )θ





=C32 (20) Let:

C33=max

(4C28f,k)(1+β+θ)(1+θ)−11 ,C32

(21) For any

n≥

−log 2(C33)2

logC19f +1=C34 (22)

we have:

Mn≤(C19f )n−1≤ 1

2C233 (23)

We use this estimation in the second step of the proof, to which we come now:

Let

(16)

n2 =max(n1,n˜2) (24) where ˜n2is the integer defined by

C34+ 4

log 2log(1/Cd)+1≤n˜2<C34+ 4

log 2log(1/Cd)+2 (25) Having defined the integern2, we can present the alternative between the "favor- able" case (Rn) and the "unfavorable" case (R0n).

Proposition 4.10. Let an2 = 1/((C33)2). Let1 ≥ ηn ≥ 0 be a sequence such that αn1−ηn+1n. For any n≥n2, we can define a sequence an,1/((C33)2)≤an≤1/C33and a sequenceρn<1such that Mn=anαρnn. The sequence anis defined by:

if

(Rn) C28f,kM(kn+1)/2−θ≤Mn

αn+1 αn

then an+1=an

1+Mθn 1−C28f,kM1/2n and if

(R0n) C28f,kMn(k+1)/2−θ>Mn

αn+1

αn

then an+1=an

Moreover, if(Rn)holds, thenρn+1≥ρnn(1−ρn);

and if(R0n)holds, thenρn+1≥((k+1)/2−θ)(1−ηnn. In particular, the sequence (ρn)n≥n2is increasing.

The threshold between the alternatives (Rn) and (R0n) is controlled with a parameter θ, which could be freely chosen such thatθ >0 and (k+1)/2−θ≥(1+β+θ)(1+θ).

Whenθincreases, the numbern3of occurrences of (R0n) increases. Whenn3increases, all other quantities being equal, the bound on the norm of the conjugacy increases.

Moreover, ifθgets too large, we can no longer show thatn3 is finite (see proposition 4.14), and therefore, we can no longer estimate the norm of the conjugacy.

On the other hand, whenθis smaller,C32 increases. It increases the numbern2

above which we consider the alternatives (Rn) and (R0n).C35 increases too (see propo- sition 4.19). WhenC32 andC35 increase, all other quantities being equal, the bound on the norm of the conjugacy increases. Moreover, whenθ → 0,C32 → +∞, which makes this bound on the conjugacy diverge.

Thus, the variation ofθhas contradictory influences on the bound of the norm of the conjugacy, and there is a choice ofθthat optimizes this bound. However, in this paper, we do not seek this optimalθ, since it would complicate further the expression of the final estimate. Instead, in estimation (2), we fixθ=1/2, which allows simplifying the expression of the estimate. In estimation (3), we takeθ → 0, which also allows simplifying the estimate.

Proof of proposition 4.10: For anyn≥n2, sincen2≥n1, an2= 1

C233 ≤an≤an2 +

Y

n=n1







1+Mθn 1−C28f,kM1/2n







≤C32

C332 ≤ 1 C33 and since

(17)

αρnn >anαρnn=Mn≥αn thenρn<1.

Second, if (Rn) holds, then by applying corollary 4.6, we have:

Mn+1≤ 1+Mnθ 1−C28f,kM1/2n Mn

αn+1 αn

(26) Therefore,

Mn+1=an+1αρnn+1+1 ≤an+1αn+1αρnn−1=an+1αn+1α(1−ηn+1n)(ρn−1) and then:

ρn+1−1≥(1−ηn)(ρn−1) hence the estimation:

ρn+1≥ρnn(1−ρn)

If (R0n) holds, sinceC28f,kM1/2n ≤1/2, then by applying corollary 4.6, we obtain:

Mn+1≤4C28f,kM(kn+1)/2−θ Moreover, sincean≤1/C33<1, then:

a(kn+1)/2−θ≤a(1n+β+θ)(1+θ)=ana(1n+β+θ)(1+θ)−1≤ an

C(133+β+θ)(1+θ)−1

≤ an

4C28f,k Therefore, by combining these two estimations, we obtain:

an+1αρnn+1+1 =Mn+1≤4C28f,kMn(k+1)/2−θ≤4C28f,ka(kn+1)/2−θαρnn((k+1)/2−θ)≤anαρnn((k+1)/2−θ) Moreover, sincean+1=an, then

1≤αn((k+1)/2−θ))(1−ηn)−ρn+1 n+1

hence the estimation:

ρn+1≥(ρn((k+1)/2−θ))(1−ηn)

The reader can notice that until now, we have not used the Diophantine condition onαyet. Now, we introduce this condition in order to estimateρn2from below (propo- sition 4.11), and in order to determine a bound ρabove which (Rn) always occurs (proposition 4.12).

(18)

Proposition 4.11. Ifβ >0, we have the estimation:

ρn2≥ log 2

((1+β)n2+1−1) log(1/Cd)/β Ifβ=0, we have the estimation:

ρn2≥ log 2 (n2+1) log(1/Cd)

Proof. Sinceαis Diophantine, we have:αn+1≥Cdα1n+β. Therefore, forβ >0, log 1

αn+1

!

+log(1/Cd)

β ≤(1+β) log (1/αn)+log(1/Cd) β

!

and sinceα−1=1, then by iteration, for anyn≥0, log (1/αn)≤

(1+β)n+1−1log(1/Cd) β Ifβ=0, we have:

log (1/αn)≤(n+1) log(1/Cd)

Moreover, sinceρn2 =−log(Mn2/an2)/log(1/αn2) andMn2/an2 ≤1/2, then we get proposition 4.11.

Proposition 4.12. Letβ1=β+(n2 log(1/C2−1) log 2d). If

ρn≥ β1

(k−1)/2−θ =ρ (27)

then(Rn)occurs.

Remark4.13. Note thatρ <1, because (k+1)/2−θ≥(1+β+θ)(1+θ) andβ1 ≤β+1/2.

Proof. Sinceαn≤(1/2)n−12 , then

0< logCd

logαn ≤ −logCd

n−1

2 log 2 (28)

Furthermore, sinceαn+1n1−ηn1 ≥Cdα1n+β, then 1

1−ηn

logαn≥logCd+(1+β) logαn

and since logαn<1 forn≥0, then by (28), 1

1−ηn −1≤β+logCd

logαn ≤β+log(1/Cd)

n−1 2 log 2 Therefore, if estimation (27) holds, then

k−1 2 −θ!

ρn+1− 1 1−ηn

≥0

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