Brjuno Numbers and Complex Dynamics

Edgar Arturo Saenz Maldonado

Thesis submitted to the Faculty of theVirginia Polytechnic Institute and State University

in partial fulfillment of the requirements for the degree of

Master of Sciencein

Mathematics

William J. Floyd, ChairLeslie D. Kay

Peter E. Haskell

April 17th, 2008Blacksburg, Virginia

Keywords: Brjuno Numbers, Siegel Disks, Complex DynamicsCopyright 2008, Edgar Arturo Saenz Maldonado

Brjuno Numbers and Complex Dynamics

Edgar Arturo Saenz Maldonado

(ABSTRACT)

The Brjuno numbers play a fundamental role in the study of the 1-dimensional Com-plex Dynamics Theory. In this work we examine the proof of the Brjuno theorem by usingelements of Number Theory. We also examine the topological version of the proof given byJ. Yoccoz and his renormalization principle.

If α ∈ R \ Q, we also describe how the existence of a Siegel disk at the origin for thepolynomial P (z) = exp(2πiα) · (z − z2) implies the linearization of any germ of the formf(z) = exp(2πiα) · z +O(z2).

Dedication

To the memory of my mother: Gladys Felicita Maldonado Huapaya .

iii

Acknowledgments

I am very thankful to Dr. William Floyd for his support and patience while he was advisingme on this paper. Also, I would like to thank Dr. Thomson and Dr. Haskell for theircomments and suggestions.

I am very thankful to Dr. Leslie Kay, Dr. Peter Duren and Dr. Floyd for directing myattention to the proof of a certain inequality which was crucial in the steps needed to proveone of the main results of this paper.

I had the pleasure of meeting wonderful people in the Math department who have supportedme in very difficult moments in my life.

I have learned a lot of mathematics from my professors. I have also learned many things inmath from my classmates and friends. I wish to thank Bart, Moises, David, Eddie, Mami,Carlos, Daniel, Toufik etc.

Finally, I am very thankful to Cristina, Diana, Nohely and Dayana for their patience, supportand love. Words cannot even begin to express how much you mean to me.

iv

Contents

1 Introduction 1

2 From Siegel to Brjuno 4

2.1 Small Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 Siegel’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Diophantine Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4 Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.5 Brjuno’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Yoccoz’s Theorem 19

3.1 Notations and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 Germs of Analytic Diffeomorphism and Linearization . . . . . . . . . . . . . 20

3.3 The Renormalization Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.4 Construction of the n-th renormalization . . . . . . . . . . . . . . . . . . . . 29

3.5 Proof of Brjuno’s Theorem by Yoccoz . . . . . . . . . . . . . . . . . . . . . . 34

4 Linearization of Quadratic Polynomials 40

4.1 Douady - Hubbard Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.2 The Quadratic Polynomial Theorem . . . . . . . . . . . . . . . . . . . . . . . 50

A Arithmetical Lemmas 53

B e is diophantine 55

v

C Davie’s Lemma 58

D Proof of Theorem 2.5.5 and Theorem 2.5.6 65

E Conformal Capacity and Koebe Inequalities 68

F Controlling the height of renormalization 70

G Modulus of a Ring Domain and Quasiconformal Mappings 74

Bibliography 79

vi

Chapter 1

Introduction

The study of the dynamics of analytic germs in one complex variable in a neighborhoodof a fixed point had been the a motivation of several works during the twentieth century.The central problem is to describe the structure of the conjugacy classes in the group Gof the analytic germs that leaves the point z = 0 fixed. In other words, given an analyticgerm f(z) = λz +O(z2) with λ ∈ C∗, is there an analytic function hf defined in some openneighborhood of 0 ∈ C, satisfying hf (0) = 0, h′f (0) = 1 and f(hf (z)) = hf (λz)?

For the case |λ| 6= 1, the answer is affirmative. It was shown by G. Koenigs in 1884, who alsoproved that for 0 < |λ| < 1 the fixed point has to be an attracting point while for |λ| > 1 thefixed point is a repelling point (see L. Carleson-T.W.Gamelin[7], A.F. Beardon[3]). But, if λis a root of unity, i.e. λ = exp(2πip/q) with p/q ∈ Q, the answer for the question above canbe negative; however Ecalle[14] and Voronin[31] have done a complete classification whichguarantees whether hf exists or not.

The most interesting case happens when |λ| = 1 and λ is not a root of unity. This meansthat λ = exp(2πiα), where α ∈ R \ Q. In order to prove that hf can be an analyticfunction in some neighborhood at the origin, for the case λ = exp(2πiα), α ∈ R \Q, we canassume (formally) that hf has a power series representation at the origin. By the formulasin question, it is possible to determine the coefficients of the formal power series of hf ; thedenominators of these coefficients can be written as products of the form (λn−λ) for n ≥ 2,since α is an irrational number these products could be very small, which implies that hf

may not converge.

The first significant result, although negative, was given by H.Cremer[10] who consideringthe sequence of continued fractions (pn/qn)n≥0 associated to α, proved that the arithmeticalcondition

supn≥0

log(qn+1)

qn= +∞

causes the divergence of hf .

1

2

The first positive result was given by Carl Ludwig Siegel[29] in 1942, who proved in a histor-ical article that the formal series of hf around the origin has positive radius of convergence(which means that hf is analytic in some neighborhood at 0) when the irrational number αsatisfies a diophantine condition, i.e. if there exist c > 0 and µ > 2 such that∣∣∣∣α− p

q

∣∣∣∣ ≥ c

qµ

then the radius of convergence of hf is positive. In the original proof of Siegel[29], heprovided direct estimates for the coefficients of hf , but in this paper we present a differentapproach. Part of the organization of the proof given here has been taken from two books,A.F. Beardon[3] and L. Carleson-T.W.Gamelin[7].

In 1965, following the same ideas of the proof given by C.L. Siegel, A.D. Brjuno[4] improvedthe arithmetical condition using some properties of the continued fractions (see AppendixD). He proved that the arithmetical condition∑

n≥0

log(qn+1)

qn< +∞

implies the analyticity of hf in some open neighborhood of the origin.

In this paper we present two different proofs of the Brjuno theorem. In Chapter 2, ourattention is focused on the proof of the Siegel theorem and the first version of the Brjunotheorem. This first version has been taken from S. Marmi[24] who used two strong resultsthat significantly shorten the original proof given by Brjuno. These two of results are theDavie Lemma and the famous Bieberbach Theorem which was proven by L. De Branges[5]in 1984. Basically, this version is focused on direct estimates for the coefficients of the formalpower series of hf .

In Chapter 3, we are focused on the second version of the proof of the Brjuno theorem.We follow the same techniques used by J.Yoccoz in [30] and X.Buff and A.Cheritat in [6];however, we introduce some definitions given by S. Marmi in [24]. In Section 3.2, for anyf(z) = λz+O(z2) univalent on the open unit disk D, with |λ| ≤ 1, we introduce the definitionof stability at 0. The point 0 ∈ C is called stable for the map f if and only if 0 belongs to theinterior of the set Kf =

⋂n≥0 f

−n(D). In Section 3.2, we also prove that if f(z) = λz+O(z2)is univalent on the open unit disk D, with |λ| ≤ 1, then

0 is stable ⇔ hf is analytic at 0.

If f(z) = λz + O(z2) is univalent on the open unit disk D, where λ = exp(2πiα) andα ∈ (0, 1) \ Q, then f lifts to a map F : H → C by using the map E : H → D∗ defined byE(Z) = exp(2πiZ); such a map F is univalent, Z-periodic and satisfies lim

=(Z)→+∞(F (Z)−Z) =

α. For this reason, we define F(α) as the collection of the all mappings F as in the previoussentence. In Section 3.5, using the renormalization technique of Yoccoz (developed in Section

3

3.3 and Section 3.4) we prove that if F ∈ F(α) and α satisfies the Brjuno condition, thereexists C > 0 such that

F m(Z) ∈ H

for all Z ∈ H with =(Z) > C. As a consequence the open disk EZ ∈ H : =(Z) > C, whichcontains 0, is contained in Kf . This means that 0 is an interior point for Kf and therefore0 is stable. So, by the equivalence between stability of 0 and the analyticity of hf , given inSection 3.2, we conclude that the radius of convergence of hf is positive. The crucial fact inthis second version is that we prove the analyticity of hf without computing any estimateson the coefficients of the formal power series hf .

From the definition of the set Kf given above (see also Section 3.2), we define a Siegel diskat 0 as the connected component of the interior of Kf which contains zero. In Section 3.2,we also prove that stability of f at 0 is equivalent to the existence of a Siegel disk at 0.

In Chapter 4, our attention is focused on the Quadratic Polynomial Theorem which sates thatthe stability at 0 for the quadratic polynomial Pλ(z) = λ(z − z2), where λ = E(α) and α ∈R\Q, is enough to guarantee the stability at 0 for every germ of the form f(z) = λz+O(z2).This result was first given by Jean Yoccoz in 1985. In this paper we provide the proof of theQuadratic Polynomial Theorem with more specific details. In Section 4.1, for any univalentmap f(z) = λz +O(z2) defined on the open unit disk D with λ = E(α) and α ∈ R \Q, weapply the same technique used in the Straightening Theorem for polynomial-like mappings(of degree 2) given by A. Douady and J.Hubbard in [12] and prove that there exists a0 ∈ (0, 1]such that for all a ∈ (0, a0] and b ∈ C with |b| = 10, the map fa,b(z) = a−1f(az) + bz2 isquasiconformally conjugate to the quadratic polynomial λz + bz2. This means that thereexist a quasiconformal map φ ≡ φf,a,b such that

fa,b(φ(z)) = φ(λz + bz2)

for all a ∈ (0, a0] and b ∈ C with |b| = 10. As an immediate consequence, fa,b is quasiconfor-mally conjugate to the polynomial λ(z − z2). Since a quasiconformal map is in particular ahomeomorphism, the existence of a Siegel disk at 0 for the quadratic polynomial Pλ impliesthe existence of a Siegel disk at 0 for the mappings fa,b and therefore the analyticity of themappings hfa,b

. In Section 4.2 we describe how the analyticity of the mappings hfa,bfor

a ∈ (0, a0], b ∈ C with |b| = 10 implies the analyticity of the map hfa,0 near the origin, andtherefore the existence of a Siegel disk at 0 for the map f(z) = λz +O(z2).

Chapter 2

From Siegel to Brjuno

In this chapter we are going to prove one of the most interesting results in 1-dimensionalComplex Dynamics Theory, the Brjuno theorem. A weak version of this theorem was firstproven by Carl Ludwig Siegel in 1942 by using elements of Number Theory such as Diophan-tine numbers. The first three sections of this chapter are focused on this particular case. Inthe fourth section we give the definition and some properties of continued fractions whichare used in the proof of Davie’s Lemma[11]. Finally, in the fifth section we prove the Brjunotheorem by using Davie’s Lemma and the Bieberbach theorem[9].

2.1 Small Divisors

Consider the analytic map f given by f(z) = 3z − 4z3.Is there an analytic function h defined in some open neighborhood of 0 ∈ C , such thath(0) = 0, h′(0) = 1 and f(h(z)) = h(f ′(0)z) ? Well , in this case the answer is affirmativebecause h(z) = sin(z) satisfies all of the conditions required above. So given an analyticgerm f at the origin 0 ∈ C with f(0) = 0, f ′(0) = λ ∈ C∗, the next natural question wouldbe : Is there an analytic function h defined in some open neighborhood of 0 ∈ C, satisfying

h(0) = 0, h′(0) = 1 and f(h(z)) = h(λz) ? (2.1.1)

For the case |λ| 6= 1 (see [7]) the answer is affirmative. But, if λ is a root of unity theanswer can be negative. Details about this case may be found in Ecalle[14] and Voronin[31].However, here we are focused on the interesting case : |λ| = 1 and λ is not a root of theunity. This is the case

λ = exp(2πiα), (2.1.2)

where α ∈ R\Q. We will consider the power series associated to f

f(z) =∑i≥1

aizi, a1 = λ

4

5

and we assume that the formal power series of h is given by

h(z) =∑i≥1

hizi.

If h is the solution of the functional equation given by 2.1.1, the coefficients of the lastseries must satisfy ( formally ) the following recursive relation :

hn =

1 for n = 1,

1

λn − λ

n∑m=2

am

∑n1+...+nm=n

ni≥1

hn1hn2 . . . hnm for n ≥ 2 (2.1.3)

The terms of the form λn − λ may cause the divergence of h in any open neighborhood ofthe origin 0 ∈ C as we will see in the following result which was first given in 1917 by G.A.Pfeiffer.

Theorem 2.1.1. There is λ = exp(2πiθ) with θ ∈ R\Q for which the equation 2.1.1 has nosolution for any polynomial f of degree d ≥ 2.

Proof. Let f(z) = zd + · · · + λz and suppose there exists an analytic solution h of 2.1.1defined on ∆(0, δ) = z ∈ C : |z| < δ. Consider the dn fixed points of fn; that is , the rootsof the polynomial equation

fn(z)− z = zdn

+ . . .+ (λn − 1)z = 0

One of these roots is 0 ∈ C. Label the others µ1, µ2, . . . , µdn−1 and note that if the originis the unique zero of fn(z) = h(λnh−1(z)) in ∆(0, δ), then µj /∈ ∆(0, δ). Assuming δ small(δ 1), we have

δdn dn−1∏j=1

|µj| = |1− λn| (2.1.4)

We now construct a λ for which the last inequality is impossible, contradicting the exis-tence of h. Suppose q1 < q2 < q3 < . . . is an increasing sequence of integers such that

θ =+∞∑k=1

2−qk ∈ R\Q, and define λ = exp(2πiθ). It is not difficult to see that exp(2πi(θ2ql)) =

exp(2πiw) where w = 2ql−ql+1 +∑

k≥l+2

2ql−qk . By an elementary computation

|1− exp(2πiw)| = 2| sin(πw)| ≤ 2π|w|

and|w| = w ≤ 3 · 2ql−ql+1

6

Then|1− λ2ql | = |1− exp(2πiw)| ≤ 6π · 2ql−ql+1

If h exists, replacing n by 2ql in the inequality 2.1.4 we get

δd2ql

≤ |1− λ2ql | ≤ 6π · 2ql−ql+1

which implies thatql+1 ≤ ql + log2(6π) + [log2(

1δ)]d2ql

On the other hand, since d ∈ Z+, d ≥ 2 and ql ≥ 2 for all l ≥ 2, we obtain

ql < d2ql

6π < 2222 ≤ 2d2ql

Henceql + log2(6π) < 2 · d2ql for all l ≥ 2

Thenql+1 < cd2ql

where c = c(δ) = (2 + log2(1

δ)) > 0. But if we consider by induction (qk)k≥1 such that

log2(qk+1) > k2qk for all k ∈ Z+, we get a contradiction.

2.2 Siegel’s Theorem

Now we know that the equation 2.1.1 may not have a solution for the case λ = exp(2πiα),α ∈ R\Q. However, Siegel[29], assuming an arithmetical condition on α, proved the existenceof the solution of the equation 2.1.1 .

Theorem 2.2.1. Let f(z) =∑n≥1

anzn be an analytic function in a neighborhood of the origin

0 ∈ C with a1 = λ = exp(2πiα). If α ∈ R\Q and there exist constants c > 0, µ > 2 satisfying

|α − p

q| ≥ c

qµfor all

p

q∈ Q (q ∈ Z+), then the equation 2.1.1 has an analytic solution

in some open neighborhood of the origin 0 ∈ C.

In order to prove the Siegel theorem we have to do some preliminary analysis.

Proposition 2.2.2. The arithmetical condition implies that there exist c > 1 and k ∈ Z+

such that1

|λn − 1|≤( ck!

)nk for all n ∈ Z+

7

Proof. Let n be a natural number, and let m be the closest integer to nα. Since nα isirrational, |nα−m| < 1/2 . By an elementary computation

|λn − 1| = |e2πi(nα) − e2πi(m)| = 2| sin(πnα)| = 2| sin(π(nα−m))|

By the hypothesis and since2

π|y| ≤ | sin(y)| on [−π/2, π/2] , we have

c

nµ≤ 4n

c

nµ≤ 4n|α− m

n| =

4

π|π(nα−m)| ≤ 2| sin(π(nα−m))| ≤ |λn − 1|

Therefore, it is enough to consider c > maxk!

c, 1 and k = [[µ]] + 1, where [[ ]] denotes the

integer part.

On the other hand, solving the equation 2.1.1 is equivalent to finding an analytic functionϕ such that ϕ(0) = 0, ϕ′(0) = 1 and ϕ−1 f ϕ(z) = λz. Note that

f(z) = λz + F (z) (2.2.1)

ϕ(z) = z + Φ(z) (2.2.2)

whereF (z) =

∑i≥2

aizi and Φ(z) =

∑i≥2

bizi (2.2.3)

We have to find ϕ satisfying ϕ−1 f ϕ(z) = λz, which is equivalent to getting

f(ϕ(z)) = ϕ(λz)λϕ(z) + F (ϕ(z)) = λz + Φ(λz)

λz + λΦ(z) + F (ϕ(z)) = λz + Φ(λz)F (ϕ(z)) = Φ(λz)− λΦ(z)

Now the idea is to replace the last functional equation by the following simple equation:

F (z) = Φ(λz)− λΦ(z). (2.2.4)

Substituting 2.2.3 into 2.2.4, we obtain∑j≥2

ajzj =

∑j≥2

bj(λz)j − λ∑j≥2

bjzj

So

|bj| = |aj|1

|λj−1 − 1|for all j ≥ 2

By Proposition 2.2.2, |bj| ≤ |aj|( ck!

)(j − 1)k for all j ≥ 2 . Hence

lim sup j

√|bj| ≤ lim sup j

√|aj|.

8

From the last inequality, the radius of convergence of f is less than or equal to the radius ofconvergence of ϕ.

Consider A = 2c(20)k+2 and B = (30)k+2, and choose δ and θ such that0 < δ = δ0 <1

c

1

10k+2

1

22k+4.

1

ABθ = θ0 = 1

101

1+(2)0< 1

5

(2.2.5)

Note that cδ < θ2k+4 < θk+2 . By the continuity of F ′, there exists r > 0 such that if |z| ≤ rthen |F ′(z)| < δ. Hence, there exist positive real numbers δ, θ, and r satisfying :

(c1) f is analytic in ∆(0, r) := z ∈ C : |z| ≤ r.Therefore ϕ is analytic on ∆(0, r) := z ∈ C : |z| < r.

(c2) θ <1

5

(c3) cδ < θk+2

(c4) If |z| ≤ r then |F ′(z)| ≤ δ

Proposition 2.2.3. Using the preceding notation, for z ∈ ∆(0, r) we have

1. |Φ′(z)| ≤ c‖F ′‖[

r

r − |z|

]k+1

, where ‖F ′‖ = sup|F ′(z)| : |z| ≤ r .

2. If |z| < r(1− θ) then |Φ′(z)| < c‖F ′‖θk+1

≤ cδ

θk+1< θ .

Clearly the second part comes from the first part and the item (c3). So it is enough to provethe first part of this proposition.

Proof. From the Cauchy estimates for the power series coefficients of F ′ we have

|jaj| ≤‖F ′‖rj−1

Hence

|Φ′(z)| = |∑j≥2

jbjzj−1| ≤

∑j≥2

|jaj|c

k!(j − 1)k|z|j−1

≤ c‖F ′‖∑j≥2

(j − 1)k

k!(|z|r

)j−1 < c‖F ′‖∑j≥1

(j + k − 1)!

(j − 1)!k!(|z|r

)j−1

9

Therefore

|Φ′(z)| ≤ c‖F ′‖[

r

r − |z|

]k+1

Definition 2.2.4. For each m ∈ 0, 1, 2, 3, 4, 5, we define the open disks Dm by

Dm := z ∈ C : |z| < r(1−mθ) .

It is clear that D5 ⊆ D4 ⊆ . . . ⊆ D1 ⊆ D0 = ∆(0, r) .

Proposition 2.2.5. The mappings ϕ, f, ϕ−1 satisfy the following diagram :

D4ϕ−→ D3

f−→ D2ϕ−1

−−→ D1.

Proof. This comes from the definition above and the items (c1) and (c3).

Since we changed the original functional equation for another simpler one, it implies thatϕ−1 f ϕ(z) = λz + “an analytic function” . Now we are focused on how “ big ” is thisanalytic function on D5.

Proposition 2.2.6. If we consider ϕ−1 f ϕ(z) = λz +G(z), then

|G′(z)| ≤ 2cδ2

θk+2 for all z ∈ D5 .

Proof. By the equation 2.2.4, it is easy to deduce that for all z ∈ D4,

G(z) = Φ(λz)− Φ(λz +G(z)) + F (z + Φ(z))− F (z)

Since D4 ⊆ D1 , we can apply the second part of Proposition 2.2.3 to D4 and obtain

|G(z)| ≤ θ|G(z)|+ δ sup|Φ′(τ)| : τ ∈ D4|z| ≤ θ|G(z)|+ δcδ

θk+1r

So

|G(z)| ≤ cδ2

θk+1

r

1− θ

Now, fix z ∈ D5 . Clearly w : |w − z| ≤ rθ ⊆ D4 . Then

|G′(z)| = | 1

2πi

∫γ

G(w)

(w − z)2dw| ≤ cδ2

θk+1

r

1− θ

θr

θr2,

where γ(t) = z + rθ exp(it), t ∈ [0, 2π]. Finally, since θ < 1/5 the assertion follows.

10

Proposition 2.2.7. Let k ∈ Z+ and c > 1, and consider A = 2c(20)k+2 and B = (30)k+2.

Choose δ0 with 0 < δ0 <1c

110k+2

12k+2

1AB

. Define by induction δn+1 = 2cδ2n

θk+2n

, where θn = 110

11+2n

for all n ≥ 0. Then

δn <1

c(θn)k+2 for all n ≥ 0.

Proof. See Appendix A.

Proof of Theorem 2.2.1. We are going to inductively define sequences (fn)n≥0, (Fn)n≥0,and (ϕn)n≥0 satisfying fn+1(z) = λz + Fn(z). For this, consider f0 = f and F0 = F asin the equation 2.2.1 and define ϕ0 by ϕ0(z) = z + Φ0(z) where Φ0 is the solution of thefunctional equation 2.2.4. This means

F0(z) = Φ0(λz)− λΦ0(z)

By Proposition 2.2.5 ϕ−10 f0 ϕ0 is well defined in some open neighborhood of zero. Now

definef1(z) = ϕ−1

0 f0 ϕ0(z)F1(z) = f1(z)− λz

It is not difficult to check that f1 and F1 have the form given in 2.2.1. Hence we can defineϕ1 from f1 and F1 in the same way as ϕ0 was defined from f0 and F0. Then we can define(formally)

f2(z) = ϕ−11 f1 ϕ1(z)

F2(z) = f2(z)− λz

Again, it is not difficult to check that f2 and F2 have the form given in 2.2.1. Hence we candefine ϕ2 from f2 and F2 in the same way as ϕ1 was defined from f1 and F1. Hence, weinductively define (formally)

fn+1(z) = ϕ−1n fn ϕn(z)

Fn(z) = fn(z)− λz

The well definition of these sequences depends on the right selection of the parameters δn,θn and rn ; for this, consider δ0 and δn as in proposition 2.2.7 and

θn =1

10

(1

1 + 2n

)and rn =

r

2

(1 +

1

2n

)where r is a fixed positive number such that

if |z| ≤ r then |F (z)| = |F0(z)| < δ0

On the other hand, we will have “new” D5, D4, . . . , D1, D0 in each step. That is why foreach n we define

Dn,m = z ∈ C such that |z| < rn(1−mθn)

11

where m ∈ 0, 1, 2, 3, 4, 5. By the definition of the sequence (rn), we have

rn+1 =r

2(1 +

1

2n+1) =

r

2(1 +

1

2n)(

2n

1 + 2n)(1 +

1

2n+1)

So

rn+1 = rn(2n

1 + 2n+

1

2

1

1 + 2n) = rn(1− 1

1 + 2n+

1

2

1

1 + 2n) = rn(1− 5θn)

which means that Dn+1,0 = Dn,5. This implies that

V ⊆ . . . ⊆ Dn+1,0 = Dn,5 ⊆ Dn,4 ⊆ . . .

where V = z ∈ C : |z| < r2. Therefore, by Proposition 2.2.5 fn+1 = ϕ−1

n fn ϕn are welldefined for each n. Using the condition (c4) for z ∈ Dn,0 we have |F ′

n(z)| ≤ δn, in particularthis holds on Dn,4. By Proposition 2.2.6,

|F ′n+1(z)| ≤ 2cδ2

n

θk+2n

for all z ∈ Dn,5

Hence |F ′n+1(z)| ≤ δn+1 for all z ∈ Dn,5 = Dn+1,0. By the selection of δ0, θn and Proposition

2.2.7, it is not difficult to check that δn → 0 as n → ∞. So F ′n+1(z) converges uniformly

to zero on V . Since each fn is analytic on V and F ′n+1(z) converges uniformly to zero on

V , Fn(z) must converge uniformly to zero on V . For each n, denote by ψn the functionϕ0 ϕ1 . . . ϕn. By the definition of fn+1 ,

ψ−1n f ψn(z) = fn+1(z) = λz + Fn+1(z) (2.2.6)

for all z ∈ V . By Proposition 2.2.5, we have

ψn := ϕ0 ϕ1 . . . ϕn : Dn,4 → D0,3

and since V ⊆ Dn,4 for all n ≥ 0, we conclude that ψnn≥0 is a normal family on V . So,by Montel’s Theorem (see [8], [7]) the sequence (ψn)n≥0 has a subsequence (ψnj

)j∈Z+ whichconverges. Let h := lim

j→∞ψnj

.

Since the ϕn’s are perturbations of the identity z 7→ z, the ψn’s are also, whence h(z) =z + O(z2) and h is injective. Applying Hurwitz’s Theorem [8], the sequence ψ−1

njconverges

uniformly to h−1. Then, by 2.2.6

ψ−1nj f ψnj

(z) = λz + Fnj+1(z) for all z ∈ V

Thereforeh−1 f h(z) = λz for all z ∈ V.

12

2.3 Diophantine Numbers

We do not know yet whether there exist irrational real numbers satisfying the hypothesisof Siegel’s Theorem. This means : Are there any irrational numbers α ∈ R for which ispossible to find constants c = c(α) > 0 and µ = µ(α) > 2 such that,∣∣∣∣α− p

q

∣∣∣∣ ≥ c

qµ(2.3.1)

for all pq∈ Q, q ∈ Z+ ?

The following theorem gives an affirmative answer for this question. For the details of theproof see [2], [17].

Theorem 2.3.1. Liouville’s Theorem. For every algebraic α ∈ R\Q of degree n > 1, thereexists a constant c = c(α) > 0 such that:∣∣∣∣α− p

q

∣∣∣∣ ≥ c

qn(2.3.2)

for all p ∈ Z and q ∈ Z+.

This result was the motivation of the following definition.

Definition 2.3.2. We say that α ∈ R\Q satisfies a diophantine condition (or α is diophan-tine) if there exist real numbers c > 0, µ > 2 for which the inequality 2.3.1 holds for allpq∈ Q, q ∈ Z+ .

Well, there are two natural questions.

• Is there any diophantine real number which is not algebraic?

• Is there any other way to prove that there exist irrational numbers which are notdiophantine without using Theorem 2.1.1 and Theorem 2.2.1 ?

We will answer these questions in an affirmative way after introducing the concept of con-tinued fraction.

2.4 Continued Fractions

Definition 2.4.1. Given α ∈ R\Q define the following sequencesa0 = [[α]] ; α0 = α− [[α]]

an =[[α−1

n−1

]]; αn = α−1

n−1 − an for all n ≥ 1(2.4.1)

13

The an are called partial coefficients associated to the number α. Using the preceding notation,we define (pn)n≥2, (qn)n≥−2 as follows : p−2 = q−1 = 0, p−1 = q−2 = 1, and

pn = anpn−1 + pn−2

qn = anqn−1 + qn−2

(2.4.2)

for all n ≥ 0. The fraction pn

qnis called the n-th continued fraction of α.

Proposition 2.4.2. Properties of Continued Fractions

1. For all n ≥ −1 : qnpn−1 − pnqn−1 = (−1)n.

2. For all n ≥ 0 :∣∣∣α− pn

qn

∣∣∣ < 1qnqn+1

.

3. q0 ≤ q1 and qn ≥ n.

4. 1qn+1+qn

< |qnα− pn| .

5. For all k ≥ 1 : qk ≥(√

5+12

)k−1

.

Proof. For the proof of these facts see [2], [17] or [22].

Definition 2.4.3. Given x ∈ R\Q, define ‖x‖ := min|n− x| : n ∈ Z.

Theorem 2.4.4. The following facts hold :

1. ‖qnα‖ = |qnα− pn|.

2. ‖q0α‖ ≥ ‖q1α‖ > ‖q2α‖ > . . . > ‖qnα‖ > . . . > 0.

3. ‖qnα‖ < |qα− p| for all p, q ∈ Z, 0 < q < qn and pq6= pn

qn.

4. ‖qnα‖ ≤ |qα− p| for all p, q ∈ Z and 0 < q < qn+1.

5. If∣∣∣α− p

q

∣∣∣ < 12q2 then p

qis a continued fraction of α.

Proof. For more details see [17],[22] and [25]

Theorem 2.4.5. The following facts hold :

(i) The number e is diophantine.

(ii) The number π is diophantine.

14

(iii) Let a0 = 0, an = 10n! for n ≥ 1, and consider α = limn→∞

pn

qnwhere (pn)n≥0 and (qn)n≥0

are constructed as in 2.4.2. Then α ∈ R\Q and α is not diophantine.

In this section, we just provide the proof of the third item of this theorem. The proof of thefirst item is given in Appendix B. For the proof of the second item see [23].

Proof. By the continued fractions algorithm, we can construct a bijection between the setof irrational numbers and the set

L = (a0, a1, . . . , an, . . .), such that a0 ∈ Z, ai ∈ Z+, for all i ≥ 1

The details of the construction of this bijection may be found in [17]. So we can concludethat α ∈ R\Q. By Definition 2.4.1, one sees that :

• q1 = a1 < a1 + 1.

• q2 = a2q1 + q0, but q0 = 1 < q1 = a1 = 10 so q2 < (a2 + 1)q1.

• q3 = a3q2 + q1, but q1 < q2 so q3 < (a3 + 1)q2

By induction qn < (a1 + 1)(a2 + 1) . . . (an + 1) for all n ≥ 1, so

qn <

(1 +

1

101!

)(1 +

1

102!

). . .

(1 +

1

10n!

)a1a2 . . . an for all n ≥ 1

An elementary calculation shows that1

10n!<

1

102n−1 for all n > 2, hence(1 +

1

101!

)(1 +

1

102!

). . .

(1 +

1

10n!

)≤(

1 +1

101

)(1 +

1

102

). . .

(1 +

1

102n−1

)Then

qn <

1− 1

102n

1− 1

10

a1 . . . an <

1

1− 1

10

a1 . . . an =

(10

9

)a1 . . . an

So qn < 10 · 101! · 102! . . . · 10n! and since 1 + 1! + 2! + ..... + (n − 1)! + n! ≤ 2(n!) for alln ≥ 1, we get qn < (10n!)2 = a2

n. By Proposition 2.4.2(2), we have |α − pn

qn| < 1

qnqn+1for all

n ≥ 0, hence ∣∣∣∣α− pn

qn

∣∣∣∣ < 1

an+1

=1

an+1n

<1

ann

<1

qn/2n

15

If α was diophantine, there must exist c > 0, µ > 2 such that

∣∣∣∣α− p

q

∣∣∣∣ ≥ c

qµfor all p ∈ Z

and q ∈ Z+. In particular, for p = pn and q = qn we get

c

qµn≤∣∣∣∣α− pn

qn

∣∣∣∣ < 1

qn/2n

So the inequalityc

qµn<

1

qn/2n

must hold. By Proposition 2.4.2(3), for n large we have

nn/2−µ < qn/2−µn <

1

c

which is a contradiction.

As a consequence, we can say that the last theorem has answered the two questions formu-lated at the end of Section 2.3 , because e is not an algebraic number and the number αfrom the item (iii) of this theorem is not diophantine.

2.5 Brjuno’s Theorem

We start this section with a new natural question. Suppose f(z) = λz + a2z2 + a3z

3 + . . .is analytic in some open neighborhood at the origin, and λ = exp(2πiα) where α is theirrational number given in Theorem 2.4.5(iii). Does there exist a solution for equation2.1.1 in this case? In 1965, following the same ideas of the proof given by Siegel, Brjuno[4] gave an answer to this question. Moreover, he gave a more general result for some subsetof the set of irrational numbers.

Theorem 2.5.1. Brjuno’s Theorem. Suppose f(z) = λz + a2z2 + a3z

3 + . . . , analyticin some open neighborhood at the origin 0 ∈ C with λ = exp(2πiα), where α ∈ R\Q. Let

(pn

qn)n≥0 be the sequence of continued fractions associated to α. If

∑k≥0

log(qk+1)

qk< +∞ then

the equation 2.1.1 has an analytic solution at the origin .

The sketch of the proof can be found in [24]. The proof is based on the ideas of Brjuno[4]and Davie[11]. In order to prove this theorem, we state an important result called Davie’sLemma [11].

Lemma 2.5.2. Davie’s Lemma. If α ∈ R\Q, α = limn→+∞

pn

qnand λ = exp(2πiα), then there

exists a map G : N → R which satisfies :

1. G(0) = 0

16

2. G(n− 1) ≤ G(n) for all n ≥ 1.

3. G(n1) +G(n2) ≤ G(n1 + n2) for all n1, n2.

4. − log |λn − 1| ≤ G(n)−G(n− 1) for all n ≥ 1.

5. There exists a universal constant γ3 > 0 such that for all n ≥ 0 the following inequalityholds

G(n) ≤ n

k(n)∑i≥0

log(qi+1)

qi+ γ3

where k(n) was chosen such that qk(n) ≤ n < qk(n)+1.

Proof. For the details see Appendix C.

Proposition 2.5.3. For all m ≥ 2, if n1 + n2 + . . .+ nm = n then

G(n1 − 1) +G(n2 − 1) + . . .+G(nm − 1) ≤ G(n− 1) + log |λn − λ|.

Proof. In fact, by Davie’s Lemma items (2) and (3) we have

G(n1 − 1) +G(n2 − 1) + . . .+G(nm − 1) ≤ G(n1 + . . .+ nm −m) ≤ G(n− 2)

By Davie’s Lemma item(4) we must have

G(n1 − 1) +G(n2 − 1) + . . .+G(nm − 1) ≤ G(n− 1) + log |λn−1 − 1|.

Since |λ| = 1, the assertion follows.

Proof of Brjuno’s Theorem. Since |f ′(0)| = 1, there exists r > 0 such that f is analytic andinjective on ∆(0, r). Now the map z → 1

rf(rz) is well defined in the open unit disk D, is

analytic and injective, leaves the point z = 0 fixed and its derivative at z = 0 is still λ. Sowithout loss of generality we may assume that f is analytic and injective on the open unitdisk D. We are going to consider the Koebe function given by

k(z) =z

(1− z)2= z + 2z2 + 3z3 + 4z4 + . . .

Clearly k(z) =∑n≥1

nzn and its radius of convergence is 1. Now, define L(z, t) = z+ k(t)−2t.

It is not difficult to see that Lt(z, t) |(0,0)= −1 6= 0; so by the Implicit Function Theorem[18], there exists σ analytic, defined in a open neighborhood of the origin, which satisfiesL(z, σ(z)) = L(0, 0) = 0 and σ(0) = 0. Then

z + 2(σ(z))2 + 3(σ(z))3 + 4(σ(z))4 + . . . = σ(z) =∑n≥1

σnzn

17

It is not difficult to check that the coefficients of σ satisfy

σn =

1 n = 1,

n∑m≥2

m ·∑

n1+...+nm=nni≥1

σn1σn2 . . . σnm

Furthermore, since σ is analytic, by the Cauchy estimate there exist positive constants γ1

and γ2 such that |σn| ≤ γ1γn2 for all n ≥ 1. On the other hand, we already know that there

exists a formal power series h =∑n≥1

hnzn verifying

hn =

1 for n = 1

1

λn − λ

n∑m=2

am

∑n1+...+nm=n

ni≥1

hn1 . . . hnm for n > 1

By the Bieberbach–De Brange’s Theorem (see [5],[9]), |an| ≤ n for all n ≥ 1, so

|hn| ≤1

|λn − λ|

n∑m=2

m∑

n1+...+nm=nni≥1

|hn1| . . . |hnm|. (2.5.1)

We are going to prove by induction that |hn| ≤ σneG(n−1) holds for all n ≥ 1, where hn,

σ(n) and G are defined as above. It is clear for the case n = 1. Suppose the assertion|hn′| ≤ σn′e

G(n′−1) is true for all n′ < n (Inductive Hypothesis). Using (2.5.1) and theinductive hypothesis we have

|hn| ≤1

|λn − λ|

n∑m,2

m∑

n1+...+nm=nni≥1

σn1 . . . σnmeG(n1−1) . . . eG(nm−1)

So, by Proposition 2.5.3 we obtain

|hn| ≤1

|λn − λ|

n∑m≥2

m∑

n1+...+nm=nni≥1

σn1 . . . σnmeG(n−1)+log |λn−λ|

Then

|hn| ≤ eG(n−1)

n∑m≥2

m∑

n1+...+nm=nni≥1

σn1 . . . σnm

Therefore, we conclude |hn| ≤ eG(n−1)σn. Finally, since there exist positive constants γ1 , γ2

such that |σn| ≤ γ1γn2 and since G(n− 1) ≤ G(n), we have

n√|hn| ≤ n

√γ1γ2e

1n

G(n)

18

By hypothesis, M = limn→∞

n∑i=0

log(qi+1)

qiexists. Then, by Davie’s Lemma(5)

1nG(n) ≤M + γ3 .

Therefore ( n√|hn|) is bounded, so lim

n→∞sup n

√|hn| ∈ R+, which proves the theorem.

The preceding theory was the motivation for the following definition.

Definition 2.5.4. We say that an irrational real number is a Brjuno Number if it satisfiesthe hypothesis of the Brjuno’s theorem. We will define and denote

B = x ∈ R\Q : x satisfies the arithmetical condition of Brjuno

Now we will state two results and we will give the answer to the question formulated atthe beginning of this section. For more details about this results, the interested reader isencouraged to check the Appendix D.

Theorem 2.5.5. Let α ∈ R\Q be diophantine and

(pn

qn

)be its sequence of continued

fractions. Then∞∑

k≥0

log(qk+1)

qk< +∞.

Theorem 2.5.6. Let α ∈ R\Q and (an) be its sequence of partial coefficients.

1. If an+1 < aρann for some 0 < ρ < 1, except for finitely many n, then α ∈ B.

2. If an+1 > aρann for some ρ > 1, except for finitely many n, then α /∈ B.

Recall that the partial coefficients of the number α ∈ R\Q in theorem 2.4.5(iii) are a0 = 0,an = 10n! for n ≥ 1. It is not difficult to see (n+1)! < (1/2) · (n!) · (10n!), for all n ≥ 1. Then

an+1 = 10(n+1)! < 10(1/2)·(n!)·(10n!) = (an)(1/2)(an)

for all n ≥ 1. So, by Theorem 2.5.6 we conclude that α ∈ B. Finally, by Theorem 2.5.1,if λ = exp(2πiα) where α is the irrational number as in Theorem 2.4.5(iii) and f(z) =

λz +∑n≥2

anzn is analytic in some open neighborhood of the origin, then the equation 2.1.1

can be solved.

Chapter 3

Yoccoz’s Theorem

The main goal in this chapter is provide a proof for the Brjuno theorem from a topologicalpoint of view by using the existence of Siegel disks. The viewpoint in the first two sectionsof this chapter is influenced by S.Marmi[24], and the last three sections follow the line ofreasoning from X.Buff, A.Cheritat[6] and Yoccoz[30]. In the second section we give thedefinition of linearization and stability and we prove that for some particular conditionsthese concepts are equivalent.

3.1 Notations and Definitions

S := the space of all germs of holomorphic diffeomorphisms f : D → C such that:

(s1) f is univalent on D.

(s2) f(0) = 0.

We will denote :

• Sλ the subspace of f such that f ′(0) = λ.

• ST the space of f such that |f ′(0)| = 1.

Given α ∈ R , we say that F ∈ F(α) if F is an analytic function on the upper half plane H,satisfying :

(c1) F is univalent on H.

(c2) T F = F T , where T (Z) = Z + 1.

19

20

(c3) lim=(Z)→∞

(F (Z)− Z) := α.

Note that this F can be written as : F (Z) = Z + α + ϕ(Z), where ϕ is a Z-periodicfunction and lim

=(Z)→∞ϕ(Z) = 0. We also define E : H → D by E(Z) = exp(2πiZ). Clearly

|E(Z)| = exp(−2π=(Z)).

3.2 Germs of Analytic Diffeomorphism and Lineariza-

tion

Let C[[z]] denote the ring of formal power series and Cz denote the ring of convergentpower series.Let G denote the group of germs of holomorphic diffeomorphisms of (C, 0) and let G denotethe group of formal germs of holomorphic diffeomorphisms of (C, 0). This means :

G = f ∈ zCz, f ′(0) 6= 0 , G = f ∈ zC[[z]], f ′(0) 6= 0

One has the trivial fibrations

π : G =⋃

λ∈C∗Gλ → C∗ , π : G =

⋃λ∈C∗

Gλ → C∗

defined byπ(f) = f ′(0) , π(f) = f ′(0)

where

Gλ = f(z) =∞∑

n=1

anzn ∈ C[[z]] , a1 = λ ,

Gλ = f(z) =∞∑

n=1

anzn ∈ Cz , a1 = λ .

Definition 3.2.1. Let f ∈ G. We say that a germ g ∈ G is equivalent or conjugate to f ifthere exists h ∈ G1 such that g = h−1 f h.

Definition 3.2.2. Let f ∈ G. We say that a germ g ∈ G is equivalent or conjugate to f ifthere exists h ∈ G1 such that g = h−1 f h.

Let Rλ denote the germ Rλ = λz. This is a simple element of Gλ.

21

Definition 3.2.3. A germ f ∈ Gλ is linearizable if there exists hf ∈ G1 (a linearization off) such that h−1

f f hf = Rλ, i.e. f is conjugate to (its linear part) Rλ.

Let us note that there is an obvious action of C∗ on G by homotheties :

(µ, f) ∈ C∗ ×G 7→ AdRµf = R−1µ f Rµ

Note that this action leaves the fibers Gλ invariant. Also, f ∈ Gλ is linearizable if and onlyif AdRµf is also linearizable for all µ ∈ C∗. Therefore, in order to study the problem ofthe existence of a linearization, it is enough to consider G/C∗, i.e. we identify two germs ofholomorphic diffeomorphisms which are conjugate by a homothety.

Definition 3.2.4. Let f ∈ S. 0 is called stable if there exists a neighborhood U of 0 suchthat fn is defined on U for all n ≥ 0 and for all z ∈ U and n ≥ 0 one has |fn(z)| < 1. Notethat U ⊂ D. To each germ f ∈ S with |f ′(0)| ≤ 1, one can associate a natural f -invariantset

0 ∈ Kf =⋂n≥0

f−n(D) (3.2.1)

Let Uf denote the connected component of the interior of Kf which contains 0. Then 0 isstable if and only if Uf 6= ∅ ; i.e. if and only if 0 belongs to the interior of Kf .

Theorem 3.2.5. Let f ∈ S with |f ′(0)| ≤ 1. 0 is stable if and only if f is linearizable.

Proof. The statement is non-trivial only if λ = f ′(0) has unit modulus. First, assume that0 is stable. Then Uf 6= ∅ and one can prove that it must also be simply connected (see [24],pg. 14). Now let ζ : D → Uf be the unique conformal and onto map satisfying ζ(0) = 0 andζ ′(0) > 0. Then the map g : D → D defined by g = ζ−1 f ζ leaves the point z = 0 fixedand verifies g′(0) = λ. By Schwarz Lemma one must have g(z) = λz. This implies that f isanalytically linearizable.

Conversely we assume hf is analytic. Then for r small (r > 0), hf maps a small disk ∆(0, r)around zero conformally into D. Since hf (0) = 0 and |fn(z)| < 1 for all z ∈ hf (∆(0, r)) onesees that 0 is stable.

Definition 3.2.6. Siegel Disk and Conformal Capacity. When λ = f ′(0) has modulus one,is not a root of unity and 0 is stable, then Uf is conformally equivalent to a disk and is calledthe Siegel disk of f (at 0).

Thus the Siegel disk of f is the maximal connected open set containing 0 on which f isconjugate to Rλ. The conformal representation hf : ∆(0, cf ) → Uf linearizes f . Here cf is

the conformal capacity of f with respect to 0, and hf satisfies hf (0) = 0 and h′f (0) = 1.

For more details of the Conformal Capacity, see Appendix E.

22

3.3 The Renormalization Principle

Definition 3.3.1. Given δ > 0, we denote by Sδ(α) the space of maps F ∈ F(α) such that

∀Z ∈ H, |F (Z)− Z − α| ≤ δα and |F ′(Z)− 1| ≤ δ (3.3.1)

Such a function F is uniformly continuous on H; this comes from the inequality |F ′(Z)| ≤1 + δ. Therefore F extends continuously to H

⋃R.

Assume F ∈ Sδ(α) and define l = iR⋂

H and l′ = [0, F (0)]. Using the inequalities in 3.3.1,for δ ∈ (0, 1/10), l

⋃l′⋃F (l) bounds an open strip U in C. Gluing the curves l and F (l)

in the boundary of U via F , we obtain a surface V , whose remaining boundary correspondsto the segment l′. Its interior is a Riemann surface for the complex structure inherited fromthe set U (the gluing is analytic). It is not difficult to see that this Riemann surface isbiholomorphic to the punctured disk D∗. Denote by ic : U → V the canonical map such thatic(l

′) = ∂V .

Let C be a Jordan curve ∈ C, such that the bounded connected component V of C\C contains0. Let y0 ∈ C; by the Caratheodory Theorem [7], there exists a unique homeomorphism

L : V → V \ 0, holomorphic on V \ ∂V such that L(ic(0)) = y0. For the particular case

C = ∂D, we have that L ic : U → D \ 0, so lifting via Z 7→ z = E(Z) we obtain acontinuous map L : U → H

⋃R whose restriction over U is an injective holomorphic map

into H. Furthermore, the following condition holds

∀Z ∈ l L(F (Z)) = L(Z) + 1 (3.3.2)

We normalize L by requiring L(0) = 0 .

Remark 3.3.2. Given two different points Z1, Z2 ∈ U satisfying L(Z1) − L(Z2) ∈ Z, thenZ1 ∈ l, Z2 = F (Z1) or Z2 ∈ l, Z1 = F (Z2). This comes immediately from the equality

1 = E(L(Z1)− L(Z2)) = L(ic(Z1))/L(ic(Z2)).

Theorem 3.3.3. For all δ ∈ (0, 1/10), all α ∈ (0, 1), all F ∈ Sδ(α), and all Z ∈ U ,

=(Z)− 2δ ≤ α=(L(Z)) ≤ =(Z) + 2δ . (3.3.3)

In order to prove the theorem above, we have to prove some extra lemmas. From now onwe will follow what Xavier Buff and Arnaud Cheritat did in [6]. The proofs of these lemmasand the proof of the theorem may be found in [6]; but for completeness we sketch them herewith more details, especially the first lemma.

Lemma 3.3.4. Assume ψ : (D, 0) → (D, 0) is a K - quasiconformal homeomorphism. Then,for all z ∈ D,

41−K |z|K ≤ |ψ(z)| ≤ 41−1/K |z|1/K .

23

Proof. For a given z ∈ D∗, let R be the ring formed by D \ [0, z]. The modulus of this ringis given by µ(|z|) , see Appendix G, because R can be mapped conformally onto the ringdomain D \ [0, |z|] and the modulus is invariant under conformal mappings. Since ψ fixes0, the image of R under ψ is a ring domain which separates the points 0 and ψ(z) fromw : |w| = 1. By Grotzsh’s Modulus Theorem, see Appendix G, we have that

M(ψ(R)) ≤ µ(|ψ(z)|)

Since ψ is a K-quasiconformal map, M(B) ≤ KM(ψ(B)) for any arbitrary ring domain Awith B ⊆ D (see the definition of quasiconformality in [20], pg. 13) and using the continuityof the modulus, see Appendix G, we have

M(R) ≤ KM(ψ(R))

where R is the ring domain D \ [0, z]. We denote the modulus of R by µ(|z|) so

µ(|z|)K

=µ(R)

K≤ µ(ψ(R)) ≤ µ(|ψ(z)|)

Since the map µ : (0, 1) → (0,+∞) is an onto and strictly decreasing map, see Appendix G,the inverse of this map is also strictly decreasing; therefore

|ψ(z)| ≤ µ−1(µ(|z|)K

)

On the other hand, consider s = µ−1(µ(|z|)K

). Since K ≥ 1 ,

µ(s) =µ(|z|)K

≤ µ(|z|)

Sos ≥ |z|

because µ is a decreasing function on (0, 1). Since the function µ(r)/ log(4/r)) is strictlydecreasing, see Appendix G, we have :

µ(s)/ log(4/s) ≤ µ(|z|)/ log(4/|z|)

ThenKµ(s)/ log(4/s) ≤ Kµ(|z|)/ log(4/|z|)

Soµ(|z|)/ log(4/s) ≤ Kµ(|z|)/ log(4/|z|)

Then1/ log(4/s) ≤ K/ log(4/|z|)

Solog(4/|z|) ≤ K · log(4/s)

24

Hence(4/|z|) ≤ (4/s)K

Therefore|ψ(z)| ≤ s ≤ 41−1/K |z|1/K

The lower bound is obtained by applying the upper bound to ψ−1 which is K-quasiconformal.

Lemma 3.3.5. If Ψ : H → H is a K - quasiconformal homeomorphism such that Ψ T =T Ψ, then

1

K=(Z)− K − 1

2πKlog(4) ≤ =(Ψ(Z)) ≤ K=(Z) +

K − 1

2πlog(4) .

Proof. Ψ is the lift, via Z → z = E(Z) , of a K quasiconformal homeomorphism ψ : (D, 0) →(D, 0) as in the previous lemma .

Lemma 3.3.6. Let ε and η be any two positive real numbers. Assume G : H → H is a(1+ε)-quasiconformal map such that G T = T G and

KG(X + iY ) ≤ 1 + εe−ηY

where KG =1 + |∂G/∂G|1− |∂G/∂G|

. Then

=(Z)− ε

η− ε

2π(1 + ε)log(4) ≤ =(G(Z)) ≤ =(Z) +

ε

η+

ε

2πlog(4) ,

which yields

|=(G(Z))−=(Z)| ≤ ε

η+

ε

2πlog(4) .

Proof. Let G1(X + iY ) = X + iτ(Y ), where τ(Y ) is defined by

τ(Y ) =1

1 + ε

(Y − ε

ηe−ηY +

ε

η

)Clearly (1 + ε) d

dYτ(Y ) = 1 + εe−ηY > 0, so τ(Y ) is a strictly monotone increasing function.

Since limY→+∞

τ(Y ) = +∞ and limY→0+

τ(Y ) = 0 we may conclude that τ : R+ → R+ is a

diffeomorphism. Combining these facts, it is easy to see that G1 is a diffeomorphism. Onthe other hand, it is not difficult to check that

KG1(X + iY ) =1 + ε

1 + εe−ηYand =(G1(Z)) ≤ 1

1 + ε

(=(Z) +

ε

η

)

Let G2 = G G−11 and let µG, µG1 and µG2 be the respective complex dilatations of G, G1

25

and G2 ([20], pg. 23). Using the standard formula for compositions with an inverse function([20], pg. 24), for ζ = G1(Z) we have

|µG2(ζ)| ≤

∣∣∣∣∣ µG(Z)− µG1(Z)

1− µG(Z)µG1(Z)

∣∣∣∣∣≤ |µG(Z)|+ |µG1(Z)|

1 + |µG(Z)||µG1(Z)|

Then,1 + |µG2(ζ)|1− |µG2(ζ)|

≤(

1 + |µG(Z)|1− |µG(Z)|

)(1 + |µG1(Z)|1− |µG1(Z)|

)By definition of complex dilatation ([20], pg. 23) we have

1 + |µG2(ζ)|1− |µG2(ζ)|

≤ KG(Z) ·KG1(Z)

Since KG(X + iY ) ≤ 1 + εe−ηY and KG1(X + iY ) =1 + ε

1 + εe−ηY, we obtain

1 + |µG2(ζ)|1− |µG2(ζ)|

≤ 1 + ε

Applying Lemma 3.3.5 to the K-quasiconformal map G2 with K = 1 + ε and using the factG = G2 G1, we get the upper bound for =(G(Z)).

To get the lower bound, repeat the same argument writing G = G4 G3 with

G3(X + iY ) = X + i(1 + ε)

(Y +

1

ηlog

1 + εe−ηY

1 + ε

).

By elementary computations we have

KG3(X + iY ) =1 + ε

1 + εe−ηYand =(G3(Z)) ≥ (1 + ε)

(=(Z)− ε

η

)

As we mentioned before, the next proof may be found in [6].

Proof of Theorem 3.3.3. Since F (Z)−Z − α is holomorphic on the upper half plane H andis periodic of period 1, we may apply Schwarz lemma and obtain

|F (Z)− Z − α| ≤ δα exp(−2π=(Z))

26

Since F ∈ F(α) and δ ∈ (0, 1/10), we may apply Theorem F.3 (see Appendix F). So thereexists Cδ > 0 such that for all Z ∈ H with =(Z) = Cδ + t ≥ Cδ we have

|F ′(Z)− 1| ≤ δ exp(−2πt) = δ exp(2πCδ) exp(−2π=(Z))

so lim=(Z)→+∞

(F ′(Z) − 1) = 0. Now, we may apply the Schwarz lemma to the periodic map

Z → F ′(Z)− 1 (with period 1) which by hypothesis satisfies |F ′(Z)− 1| ≤ δ. Then we have

|F ′(Z)− 1| ≤ δ exp(−2π=(Z))

Let B be the half strip Z ∈ H : 0 < <(Z) < 1 . Let H : B → U be the map defined by

H(X + iY ) = iαY +X[ F (iαY )− iαY ] (3.3.4)

By straightforward computations we have

∂H

∂X= F (iαY )− iαY

and∂H

∂Y= iα +X[iαF ′(iαY )− iα]

So

∂H =1

2

(∂H

∂X− i

∂H

∂Y

)=

1

2(F (iαY )− iαY + α + αX[F ′(iαY )− 1])

and

∂H =1

2

(∂H

∂X+ i

∂H

∂Y

)=

1

2(F (iαY )− iαY − α− αX[F ′(iαY )− 1])

Hence|∂H − α| ≤ δα exp(−2παY ) and |∂H| ≤ δα exp(−2παY )

Combining the last two inequalities, Z = X + iY ∈ H and δ ∈ (0, 1/10) we get

|∂H||∂H|

≤ δα exp(−2παY )

α− δα exp(−2παY )< 1

Moreover, the following inequalities hold:

1 +|∂H||∂H|

≤ 1

1− δ exp(−2παY )and 1− |∂H|

|∂H|≥ 1− 2δ exp(−2παY )

1− δ exp(−2παY )

27

Now, if we set KH =1 + |∂H/∂H|1− |∂H/∂H|

, we have :

KH(X + iY ) ≤ 1

1− 2δ exp(−2παY ).

Then using the fact 0 < δ < 1/10, we have the inequality

KH(X + iY ) ≤ 1 +5

2δ exp(−2παY ) ≤ 1 +

5

2δ

Using the definition of H and since U is an open strip in C we may conclude that H is a

homeomorphism. In particular, H is a (1 +5

2δ)-quasiconformal homeomorphism.

Moreover, by definition

H(X + iY )− α(X + iY ) = X[F (iαY )− iαY − α]

So

|H(X + iY )− α(X + iY )| = |X|| F (iαY )− iαY − α| ≤ 1 · δα exp(−2παY ) ≤ δα

Then=(H(Z))− δα ≤ α=(Z) ≤ =(H(Z)) + δα ,

and thus for all Z ∈ U , since 0 < α < 1,

=(Z)− δ ≤ α=(H−1(Z)) ≤ =(Z) + δ (3.3.5)

By the construction of L, given at the beginning of this section, we know that L is conformalon U , so the map G = L H is quasiconformal on U with the same dilatation as H. By thedefinition of H, equation 3.3.2 and since L is defined on U , we have

G(iY + 1) = L H(iY + 1) = L(F (iαY ))

andG(iY ) + 1 = L(iαY ) + 1 = L(F (iαY ))

ThereforeG(iY + 1) = G(iY ) + 1.

This implies that iR⋂

H is quasiconformally removable, so G extends to a quasiconformalhomeomorphism H → H. Now, we may apply Lemma 3.3.5 with ε = 5

2δ and η = 2πα. Using

the fact 0 < α < 1, we have

ε

η+

ε

2πlog(4) =

5δ

4πα(1 + α log(4)) ≤ δ

α

28

Then for all Z ∈ H|=(G(Z))−=(Z)| ≤ ε

η+

ε

2πlog(4) ≤ δ

α

So we have showed that for all Z ∈ H

α=(Z)− δ ≤ α=(G(Z)) ≤ α=(Z) + δ (3.3.6)

Finally, by 3.3.5 and 3.3.6 it follows that

=(Z)− 2δ ≤ α=(H−1(Z))− δ ≤ α=(L(Z)) ≤ α=(H−1(Z)) + δ ≤ =(Z) + 2δ

Theorem 3.3.7. Under the same assumptions as in Theorem 3.3.3, the map L extends toa univalent map on

W = U⋃Z ∈ C : −1 ≤ <(Z) ≤ 0 and =(Z) ≥ 4δ

and for all Z ∈ W=(Z)− 5δ ≤ α=(L(Z)) ≤ =(Z) + 5δ (3.3.7)

The definition of W is such that any point Z ∈ W is eventually mapped to U under iterationof F : F k(Z) = Z ′ ∈ U for some k ∈ N. In particular, L conjugates F to the translation T .

Proof. Let V = Z ∈ C : −1 ≤ <(Z) ≤ 0 and =(Z) ≥ 4δ and let V∗ = Z ∈ C : =(Z) ≥4δ|<(Z)|. It is easy to see that V ⊂ V∗− where V∗− = V∗

⋂Z ∈ C : −1 ≤ <(Z) ≤ 0. By

Lemma F.4 (see Appendix F), we have that F (V∗−) ⊂ V∗−⋃U .

By the first part of inequality 3.3.1, <(F k(Z)) ≥ <(Z) + k(1− δ)α for all Z ∈ H wheneverthe iterations are well defined. If we assume that there exists Z ∈ V such that F k(Z) /∈ Ufor any k, then F k(Z) ∈ V∗− for all k, so <(F k(Z)) ≤ 0 for all k. This implies that0 ≥ <(Z) + k(1− δ)α for all k, which is a contradiction. Hence, for each Z ∈ W there existsk ≡ k(Z) such that F k(Z) ∈ U . Now define

L(Z) = L(F k(Z))− k,

where k is the minimum nonnegative integer such that F k(Z) ∈ U . The continuity of thisextension follow easily (see [30], pg. 28). Let Z1, Z2 ∈ W such that L(Z1) = L(Z2), and letm1,m2 ∈ Z such that F m1(Z1), F

m2(Z2) ∈ U . Then

L(F m1(Z1))− L(F m2(Z2)) = m1 −m2,

so by Remark 3.3.2 we have three possibilities:

29

(1) m1 = m2, Fm1(Z1) = F m2(Z2);

(2) m1 = m2 + 1, F m2(Z2) ∈ l and F m1(Z1) = F (F m2(Z2));

(3) m2 = m1 + 1, F m1(Z1) ∈ l and F m2(Z2) = F (F m1(Z1)).

Since F is univalent, we conclude that Z1 = Z2. Therefore the extension L is injective.Now, for any Z ∈ W we know that Z = F 0(Z), F (Z) = F 1(Z), F 2(Z), F 3(Z), . . . andF k(F (Z)) = Z ′ are in H. Using the first inequality in 3.3.1, we have :

|=(F i+1(Z)− F i(Z))| ≤ δα =δ

1− δ(α− δα) ≤ δ

1− δ(<(F i+1(Z)− F i(Z)))

for all i = 0, 1, 2, . . . , k(Z)−1. Adding all of these terms and applying the triangle inequality,we get

|=(Z ′ − Z)| = |=(F k(Z)(Z)− F 0(Z))| ≤ δ

1− δ(<(Z ′ − Z))

Since Z ′ ∈ U and 0 < <(F (it)) ≤ α + δα for all t ∈ R, 0 < <(Z ′) ≤ α + δα must hold.Recall that −1 ≤ <(Z) ≤ 0. Therefore

|=(Z ′ − Z)| ≤ δ

1− δ(<(Z ′ − Z)) ≤ δ

1− δ(α + δα + 1)

Since 11−δ

(α + δα + 1) ≤ 3, the inequality 3.3.7 comes from a straightforward application ofTheorem 3.3.3 to Z ′. On the other hand, it is easy to see that k(F (Z)) = k(Z) − 1 whenk(Z) ≥ 1; thus

L(F (Z)) = L(F k(F (Z))(F (Z)))− k(F (Z))

Therefore, L(F (Z)) = L(F k(Z)(Z))− (k(Z)− 1) = L(Z) + 1. This means that L conjugatesF to the translation T , T (Z) = Z + 1.

Remark 3.3.8. From now on, L will refer to this extension.

3.4 Construction of the n-th renormalization

Given δ ∈ (0, 1/10) and F ∈ F(α), we will inductively define a sequence (Fn)n≥0 of univalentmaps such that Fn ∈ F(αn). Recall αn was defined in Chapter 2 (see 2.4.1). The con-struction depends on the choice at each step of some real number tn > 0 . We start withF0 = F − a0 ( where a0 = [[α]] ) and we assume Fn has been constructed. We choose tn suchthat the fundamental inequalities in 3.3.1 hold for =(Z) ≥ tn; this is always possible and theproof of this fact is given in Theorem F.3. For more details see Appendix F. It follows thatGn : Z 7→ Fn(Z + itn) − itn belongs to Sδ(αn). For this Gn, we construct Un , Wn and Ln

as we did before. Now let Jn be defined on Dn = Ln(Z ∈ Un : =(Z) > 4δ) by

Jn = Ln T−1 Ln−1 (3.4.1)

30

Proposition 3.4.1. If =(Z0) >6δ

αn

, there exists an integer k such that Z0 − k belongs to

Dn, the domain of Jn.

Proof. By 3.3.2 the equality Ln(Gn(Z)) = Ln(Z) + 1 holds for all Z ∈ iR⋂

H, since therestriction of Ln to U is a homeomorphism onto its image and Ln([0, Gn(0)]) = [0, 1]we haveH =

⋃k∈Z( Ln(Un)+k). Since Z0 ∈ H, there exists k ∈ Z such that Z0−k belongs to Ln(Un).

So there exists y0 ∈ Un such that Ln(y0) = Z0 − k. By Theorem 3.3.3

αn=(Ln(y0)) ≤ =(y0) + 2δ

Soαn=(Z0) ≤ =(y0) + 2δ

By 2.4.1, it is easy to verify that αn ∈ (0, 1). Since6δ

αn

< =(Z0), then 4δ < =(y0). Therefore

Z0 − k belongs to Dn .

Proposition 3.4.2. Let w = Ln(it) with t > 0. If =(w) >6δ

αn

, then

t > 4δ and (Jn T )(w) = (T Jn)(w) .

Proof. Since it ∈ Un, we can apply Theorem 3.3.3. Then

=(it)− 2δ ≤ α=(L(it)) ≤ =(it) + 2δ .

Now, by hypothesis 6δ < α=(w) ≤ =(it) + 2δ. Then 4δ < =(it) = t, so w = Ln(it) ∈ Dn

and Jn(w) makes sense. By an elementary computation we have

Jn(w) = Jn(Ln(it)) = Ln T−1 Ln−1(Ln(it)) = Ln(−1 + it)

HenceT Jn(w) = T Ln(−1 + it) = Ln(−1 + it) + 1

On the other hand, since Gn(it) ∈ Un and Ln(Gn(it)) = Ln(it) + 1, we can replace it byGn(it). Repeating the same argument as above, we get 4δ < =(Gn(it)). So, Ln(Gn(it)) ∈ Dn.Then

Jn T (w) = Jn( Ln(it) + 1 )

= Jn(Ln( Gn(it) )) by Theorem 3.3.7

= Ln T−1(Gn(it))

= Ln(Gn(−1 + it)) because Gn T = T Gn on H= T Ln(−1 + it ) by Theorem 3.3.7

= Ln(−1 + it) + 1

31

Lemma 3.4.3. Jn extends univalently to the upper half plane :

Z : =(Z) >6δ

αn

Proof. By Proposition 3.4.1, Dn + Z contains the half plane:

Z : =(Z) >6δ

αn

By Proposition 3.4.2, Jn commutes with the translation T on the set of the points inLn(i[0,+∞)) whose imaginary part is greater than 6δ/αn, so this set is analytically remov-able. Since =(it)− 2δ ≤ =(Ln(it)) for all t > 0, we have lim

t→+∞=(Ln(it)) = +∞. Combining

the facts above, the lemma follows.

Lemma 3.4.4. Jn(w)− w → −1

αn

= −an+1 − αn+1 , as =(w) → +∞.

Proof. Let us recall that by Theorem 3.3.7, for all Z ∈ Un we have

=(Z)− 5δ ≤ αn=(Ln(Z)) ≤ =(Z) + 5δ

Then=(Z) → +∞⇔ =(Ln(Z)) → +∞ .

By the last lemma, the set Ln(it) : =(Ln(it)) > 6δ/αn, t > 0 is analytically removable, so itis enough to consider w ∈ Ln(Un) with =(w) > 6δ/αn. By the definition of Ln, for w = Ln(Z)with Z ∈ Un and =(w) > 6δ/αn , we have Jn(Ln(Z))− Ln(Z) = Ln(Z − 1)− Ln(Z). So

lim=(Ln(Z))→+∞

( (Jn(Ln(Z)))− Ln(Z) ) = lim=(Z)→+∞

( Ln(Z − 1)− Ln(Z) )

We do not know yet whether Ln(Z − 1) makes sense. But, since w = Ln(Z) with Z ∈ Un

and =(w) > 6δ/αn we can repeat the same argument as in the proof of Proposition 3.4.2and get =(Z) > 4δ. Hence −1 + Z ∈ Wn and Ln(Z − 1) makes sense. On the other hand,note that

Jn(Ln(Z)))− Ln(Z) =1

αn

((αnLn(Z − 1)− (Z − 1))− (αnLn(Z)− Z)− 1)

Following the same arguments given by Yoccoz in [30], pg. 29-32, we can find a continuousfunction ϕ such that lim

=(Z)→+∞ϕ(Z) = 0 and

|αnL′n(Z)− 1| ≤ ϕ(Z)

32

Now consider the path γ(t) = Z + t with t ∈ [−1, 0]. Then

|(αnLn(Z − 1)− (Z − 1))− (αnLn(Z)− Z)| ≤∫

γ

|αnL′n(ξ)− 1||dξ|

≤∫ 0

−1

ϕ(γ(t))dt

≤ maxϕ(Z + t) : t ∈ [−1, 0]

Since lim=(Z)→+∞

ϕ(Z) = 0, we have

lim=(Z)→+∞

(αnLn(Z − 1)− (Z − 1))− (αnLn(Z)− Z) = 0

Therefore,

lim=(Z)→+∞

( Ln(Z − 1)− Ln(Z) ) =−1

αn

Definition 3.4.5. For each n ≥ 0 and n ∈ Z, we set

W ′n = Wn + itn (3.4.2)

and we define χn : W ′n → C

χn(Z) = s Ln(Z − itn)− i6δ

αn

(3.4.3)

where s(x+iy)=-x+iy .

Lemma 3.4.6. On W ′n

⋂F−1

n (W ′n), χn conjugates Fn to T−1.

Proof. Let Z + itn ∈ W ′n

⋂F−1

n (W ′n). Then

χn Fn(Z + itn) = s Ln(Fn(Z + itn)− itn)− i6δ

αn

Recall that Gn : Z 7→ Fn(Z + itn)− itn, so

χn Fn(Z + itn) = s Ln Gn(Z)− i6δ

αn

By the last part in the proof of Theorem 3.3.7 Ln Gn = T Ln, so

χn Fn(Z + itn) = s(Ln(Z) + 1)− i6δ

αn

Finally, by Definition 3.4.5, the lemma follows.

33

Definition 3.4.7. We define Fn+1 by Fn+1(Z) = χn T−1 χ−1n (Z)− an+1.

Theorem 3.4.8. The map Fn+1 is an element of F(αn+1).

Proof. First of all, it is easy to check that

(L−1n s)(Z + i

6δ

αn

) = χ−1n (Z)− itn

So

Fn+1(Z) = s Ln(χ−1n (Z)− 1− itn)− i

6δ

αn

− an+1

= s Ln((L−1n s)(Z + i

6δ

αn

)− 1)− i6δ

αn

− an+1

= s Ln((T−1 L−1n )(s(Z) + i

6δ

αn

))− i6δ

αn

− an+1

= s (Ln T−1 L−1n ) s(Z + i

6δ

αn

)− i6δ

αn

− an+1

Therefore Fn+1 is univalent in H. Moreover, s(Z + i6δ

αn

) is in the domain of definition of the

extension Jn for all Z ∈ H, so

Fn+1(Z) = s Jn s(Z + i6δ

αn

)− i6δ

αn

− an+1 (3.4.4)

Since s T = T−1 s and the extension Jn commutes with T , we have :

Fn+1(T (Z)) = (s Jn (s T ))(Z + i6δ

αn

)− i6δ

αn

− an+1

= (s Jn T−1 s)(Z + i6δ

αn

)− i6δ

αn

− an+1

= (T s Jn s)(Z + i6δ

αn

)− i6δ

αn

− an+1

= T (Fn+1(Z))

Therefore F T = T F on H. Furthermore :

Fn+1(Z)− Z = s(Jn(s(Z + i6δ

αn

))− s(Z + i6δ

αn))− an+1

By Lemma 3.4.4 and Definition 2.4.1 we have

lim=(Z)→+∞

Fn+1(Z)− Z = s(−1

αn

)− an+1 = α−1n − an+1 = αn+1

So the theorem follows.

34

3.5 Proof of Brjuno’s Theorem by Yoccoz

Definition 3.5.1. For each point Z ∈ H, we associate a sequence (Zn)n as follows. Wedefine Z0 = Z. If dn = =(Zn) ≥ 4δ + tn, we choose Z ′

n ∈ H such that Zn − Z ′n ∈ Z and

−1 ≤ <(Z ′n) < 0 and we define

Zn+1 = χn(Z ′n)

The sequence (Zn)n may be finite or infinite.

Proposition 3.5.2. If (Zn)n is defined for some n ≥ 0, then

=(Zn)− tn − 11δ ≤ αn=(Zn+1) ≤ =(Zn)− tn − δ

Proof. Applying the definition of (Zn) and Definition 3.4.5, one sees that

=(Zn+1) = =(χn(Z ′n))

= =(s Ln(Z ′n − itn)− i

6δ

αn

)

= =(Ln(Z ′n − itn))− 6δ

αn

Since =(Zn) = =(Z ′n) and Z ′

n − itn ∈ Wn, we can apply 3.3.7 and obtain

=(Z ′n − itn)− 5δ ≤ αn=(L(Z ′

n − itn)) ≤ =(Z ′n − itn) + 5δ

So=(Zn)− tn − 5δ ≤ αn=(L(Z ′

n − itn)) ≤ =(Zn)− tn + 5δ

Now, the proposition follows by an elementary replacement.

Let us recall that α ∈ (0, 1), α ∈ R \Q, and the sequence (αn)n was defined in Chapter 1(see definition 2.4.1).

Definition 3.5.3. (βn)n is the sequence defined by β−1 = 1 and βn = α0α1 . . . αn−1 for alln ≥ 0.

Proposition 3.5.4. Let (βn)n as above. The following facts hold:

(i) βn+2 ≤1

2βn for all n ≥ 0.

(ii) 1 + β0 + β1 + β2 + · · · ≤ 4.

35

Proof.

(i) It is enough to prove that αn+1αn+2 ≤1

2for all n ≥ 0 . In fact, fix n ≥ 0. Then there

exists a positive integer p such that1

p+ 1< αn+1 <

1

p. So we have 2 subcases:

(1) If p = 1, then 1/2 < αn+1 < 1. So 1 ≤[[α−1

n+1

]], which implies that

1/2 < αn+1 ≤ αn+1

[[α−1

n+1

]]= 1− α−1

n+1α−1n+1

Thereforeαn+1αn+2 = αn+1α−1

n+1 < 1/2

(2) If p > 1, then p < α−1n+1 < p+1, whence αn+1p = αn+1

[[α−1

n+1

]]. Since 1−αn+1 < pαn+1

we have that 1− αn+1 < αn+1

[[α−1

n+1

]]. But also

αn+1 <1

p≤ 1

2< 1

Therefore,1

2< 1− αn+1 < αn+1

[[α−1

n+1

]]= 1− αn+1αn+2

(ii) By the previous fact,

1 + β0 + β1 + . . . ≤ 1 + β0 + β1 +1

2β0 +

1

2β1 . . .

≤ 1 + β0(1 +1

2+

1

22+ . . . ) + β1(1 +

1

2+

1

22+ . . . )

≤ 1 + 2β0 + 2β1.

Since β0 = α0 ∈ (0, 1) and β1 = α0α1 ∈ (0, 1/2), the assertion follows.

Proposition 3.5.5. We use the same notation as in Definition 3.5.1. If Zn is defined forsome n ≥ 0, then :

βn−1dn +n−1∑i=0

βi−1(ti + δ) ≤ d0 ≤ βn−1dn +n−1∑i=0

βi−1(ti + 11δ)

where the β′is are as in Definition 3.5.3.

Proof. This follows by a direct application of Proposition 3.5.2

Proposition 3.5.6. Under the same assumptions as in the last proposition, the followinginequality holds :

n−1∑i=0

βi−1ti ≤ d0 − βn−1dn ≤ 44δ +n−1∑i=0

βi−1ti

36

Proof. This comes from a straightforward application of the last two propositions.

Proposition 3.5.7. If Z ∈ H and if there exists some nonnegative integer m such thatF m(Z) /∈ H, then the sequence (Zn)n is finite.

Proof. We define and denote by Hn the upper half plane Z ∈ C : =(Z) ≥ tn. If Zn isdefined, let 1 +kn be the rank of the first iterate of Zn under Fn : H → C that leaves Hn; i.e.kn is the smallest nonnegative integer such that: F 1

n (Zn), F 2n (Zn), . . . , F kn

n (Zn) ⊂ Hn

and F kn+1n (Zn) /∈ Hn. If =(Zn) ≥ 4δ + tn then kn cannot be zero. Otherwise Fn(Zn) /∈ Hn

which means that =(Fn(Zn)) < tn, so this implies that 4δ < =(Zn) − =(Fn(Zn)) which isimpossible since |Fn(W )−W−αn| < δαn < δ for all W ∈ Hn. As an immediate consequence,Zn+1 cannot be defined when kn = 0. Now assume that Zn+1 is defined and kn+1 > 0. SinceFn commutes with the translation Z → Z + 1 it is easy to see that =F j

n (Zn) = =F jn (Z ′

n)for all j ∈ 0, 1, . . . , kn (here Z ′

n is as in Definition 3.5.1). By using the definition of Gn atthe beginning of Section 3.4, one sees that

Gjn (Z ′

n − itn) = F jn (Z ′

n)− itn

for all j ∈ 0, 1, . . . , kn. Therefore,

=Gjn (Z ′

n − itn) = =F jn (Z ′

n)− tn

for all j ∈ 0, 1, . . . , kn. Since kn+1 > 0 and Zn+1 = χn(Z ′n), by equations 3.4.4 and 3.4.3

we have

=(Ln(Z ′n − itn)) >

6δ

αn

and

Fn+1(Zn+1) = s Jn(Ln(Z ′n − itn))− i

6δ

αn

− an+1

By Proposition 3.4.1 we may assume that there is a unique integer p and a unique y0 ∈ Un

with =(y0) ≥ 4δ such that Ln(Z ′n − itn) = Ln(y0) + p. Hence

Fn+1(Zn+1) = s Jn(Ln(Z ′n − itn))− i

6δ

αn

− an+1

= s Jn(Ln(y0))− i6δ

αn

− an+1 + p

= s Ln T−1(y0)− i6δ

αn

− an+1 + p

= s Ln(y0 − 1)− i6δ

αn

− an+1 + p

But, by definition of the extension Ln (see the proof of Theorem 3.3.7) we have

Ln(Z ′n − itn) = Ln(Gσ1

n (Z ′n − itn)) + σ1

37

where Gσ1n (Z ′

n − itn) ∈ Un. Hence

y0 = Gσ1n (Z ′

n − itn) and σ1 = p

So

Fn+1(Zn+1) = s Ln(Gσ1n (Z ′

n − itn)− 1)− i6δ

αn

− an+1 + σ1

with Gσ1n (Z ′

n− itn) ∈ Un and =Gσ1n (Z ′

n− itn) = =(y0) ≥ 4δ. Then Gσ1n (Z ′

n− itn)−1 ∈ Wn.Since =Fn+1(Zn+1) ≥ t1 > 0, it is easy to see that

=Ln(Gσ1n (Z ′

n − itn)− 1) >6δ

αn

Using again Proposition 3.4.1 and repeating the same idea as above, there exists a nonneg-ative integer σ2 such that

Ln(Gσ1n (Z ′

n − itn)− 1) = Ln(Gσ1+σ2n (Z ′

n − itn)− 1) + σ2

where Gσ1+σ2n (Z ′

n − itn)− 1 ∈ Un, whose imaginary part is at least 4δ. So we can write

Fn+1(Zn+1) = s Ln(Gτ1n (Z ′

n − itn)− 1)− i6δ

αn

− an+1 + ρ1

where τ1 = σ1 + σ2 and ρ1 = σ1 − σ2. Repeating this argument kn+1 times, we have

Fkn+1

n+1 (Zn+1) = s Ln(Gτ1+···+τkn+1n (Z ′

n − itn)− kn+1)− i6δ

αn

− an+1 + ρ1 + · · ·+ ρkn+1

where Gτ1+···+τkn+1n (Z ′

n− itn)− kn+1 ∈ Un and has imaginary part at least 4δ. By definitionof Gn, one sees that

Gτ1+···+τkn+1n (Z ′

n − itn)− kn+1 = Fτ1+···+τkn+1n (Z ′

n)− kn+1 − itn

So=(F

τ1+···+τkn+1n (Z ′

n)− kn+1 − itn) > 0

which implies that

Fτ1+···+τkn+1n (Z ′

n)− kn+1 ∈ Hn

Thereforeτ1 + τ2 + · · ·+ τkn+1 ≤ kn

Since |Fn(Z) − Z − αn| ≤ δαn <1

10αn on Hn, then |Fn(Z) − Z| < 11

10αn. Now we have

|F j+1n (Z ′

n − kn+1)− F jn (Z ′

n − kn+1)| <11

10αn for all j ∈ 1, 2, ...., l − 1 where l = τ1 + τ2 +

· · ·+ τkn+1 . Adding all of these terms and applying the triangle inequality, we obtain

|F ln (Z ′

n − kn+1)− (Z ′n − kn+1)| <

11

10αn · l ≤

11

10αn · kn

38

Since F ln (Z ′

n − kn+1) − itn = F ln (Z ′

n) − kn+1 − itn = Gln (Z ′

n − itn) − kn+1 ∈ Un, we have<(F l

n (Z ′n − kn+1)) ≥ 0. By definition of Z ′

n, <(Z ′n) < 0. So we conclude that

kn+1 <11

10αn · kn

By Proposition 3.5.4(i) αnαn+1 ≤ 12. This implies kn+2 <

121200· kn whenever defined.

Finally, by definition F0(W ) = F (W ) − a0, so if F m(Z) /∈ H for some Z ∈ H and somepositive integer m, then k0 is finite, whence by the analysis above the proposition follows.

Proposition 3.5.8. Assume we can choose a sequence of nonnegative real numbers (tn)n sothat the n− th renormalization Fn verifies the fundamental inequalities given in 3.3.1 when

=(Z) ≥ tn and so that τ =∞∑

n=0

βn−1tn < +∞. Then

F m(Z) ∈ H

for all m ≥ 0 and for all Z ∈ Hτ , where

Hτ = Z ∈ C : =(Z) > τ + 44δ

Proof. Let Z ∈ Hτ and assume Zn is defined. We will prove that Zn+1 is also defined. ByProposition 3.5.5, we have :

d0 ≤ βn−1dn +n−1∑i=0

βi−1(ti + 11δ)

So

βn−1dn ≥ d0 −n−1∑i=0

βi−1(ti + 11δ)

= d0 −n−1∑i=0

βi−1ti − (1 + β0 + · · ·+ βn−2)(11δ)

≥ d0 − (τ −∞∑

i=n

βi−1ti) + (4− (1 + · · ·+ βn−1)− 4 + βn−1)11δ

≥ (d0 − 44δ − τ) +∞∑

i=n

βi−1ti + (4− (1 + · · ·+ βn−1) + βn−1)11δ

By Proposition 3.5.4, 4− (1 + · · ·+ βn−1) > 0 and since d0 = =(Z) > τ + 44δ, we have

βn−1dn ≥∞∑

i=n

βi−1ti + βn−1(11δ)

39

Soβn−1dn ≥ βn−1tn + βn−1(11δ)

Then=(Zn) = dn ≥ tn + 11δ > tn + 4δ

Thus, we can define Zn+1. This implies that the sequence (Zn)n is infinite, therefore theproposition follows from Proposition 3.5.7.

Theorem 3.5.9. Brjuno Theorem by Yoccoz. Let f ∈ Sλ where λ = exp(2πiα) and α ∈

(0, 1) \Q. If∞∑

n=0

βn−1 log(1

αn

) < +∞, then f is linearizable.

Proof. Fix δ ∈ (0, 1/10) and let F ∈ F(α) such that f(E(Z)) = E(F (Z)) for all Z ∈ H. ByTheorem F.3 (see Appendix F), there exists a universal constant C > 0, just depending onδ, such that

|Fn(Z)− Z − αn| ≤ δαn and |F ′n(Z)− 1| ≤ δ

for all Z ∈ H with =(Z) ≥ 1

2πlog

1

αn

+ C, where Fn is the n-th renormalization of F .

Putting tn =1

2πlog

1

αn

+ C, it is easy to see that τ =∞∑

n=0

βn−1tn < +∞. So by the last

proposition F m(Z) ∈ H for all m ≥ 0 and for all Z ∈ Hτ . Then

f m(E(Z)) = E(F m(Z)) ∈ H

for all Z ∈ Hτ and all m ≥ 0. Hence

f m(z) ∈ D

for all z ∈ E(Hτ ) = ∆(0, exp(−2π(τ + 44δ))). This means that 0 is stable. Therefore, byTheorem 3.2.5 we conclude that f is linearizable.

Remark 3.5.10. It is not difficult to see that∞∑

n=0

βn−1 log(1

αn

) converges if only of the series

∞∑n=0

qn+1

qnconverges. The idea of the proof of this fact is given in [30] pg. 13.

Chapter 4

Linearization of QuadraticPolynomials

The main goal in this chapter is to prove that the stability of the origin for the polynomialPλ(z) = λ(z−z2) is enough to guarantee the stability of the origin for any germ f ∈ Sλ whenλ = E(α) with α ∈ R \Q. The point of view in this chapter has been mainly influenced byYoccoz[30] and Lehto[20].

4.1 Douady - Hubbard Theorem

In order to get our main result, in this first section we are focused on prove that there exista0 ∈ (0, 1] such that for all a ∈ (0, a0] and b ∈ C with |b| = 10 the map a−1f(az) + bz2 isquasiconformally conjugate to the quadratic polynomial P (z) = λz + bz2, for any f ∈ Sλ

where λ = E(α) with α ∈ R \Q.

Definition 4.1.1. Let α ∈ R \Q. Let λ = E(α) and let f be an element of Sλ. This meansf is a holomorphic and injective map which is defined on the open unit disk D such thatf(0) = 0 and f ′(0) = λ.

Definition 4.1.2. Let f be an element of Sλ as above. For each b ∈ C, we denote by fb theholomorphic map on D defined by :

fb(z) = f(z) + bz2.

Proposition 4.1.3. Let Hb be the formal linearization of fb. Then Hb has the form

Hb(z) = z(1 +∑n≥1

Pn(b)zn)

where, for each n, the function Pn(b) is a polynomial of degree n in b .

40

41

Proof. This comes from a straightforward inspection of the coefficients of Hb following therecursive relation given in 2.1.1.

Definition 4.1.4. A triple (U,U ′, F ) is called polynomial-like of degree 2 if U , U ′ are opensimply connected domains of C, neither equal to C, U ′ is relatively compact in U and F :U ′ → U is holomorphic and proper of degree 2 .

Definition 4.1.5. W = z : |z| < 1336 , Wb = z : |z| ≤ 1/3 , fb(z) ∈ W

Lemma 4.1.6. For |b| ≥ 10, (W ,Wb, fb) is a polynomial-like of degree 2 .

Proof. Since f is univalent on D, we have :

|f(z)| ≤ |z|(1− |z|)2

for all z ∈ D

In particular, |f(z)| ≤ 3/4 for |z| = 1/3 . On the other hand, for |b| ≥ 10 :

|bz2| ≥ 10

9=

13

36+

3

4for all |z| =

1

3

Given any y ∈ W , the following inequality holds for all |z| = 1/3 :

|fb(z)− (y + bz2)| = |f(z)− y| ≤ |f(z)|+ |y| < 3

4+

13

36=

10

9≤ |bz2|

Thus, for all |z| = 1/3 we have |(fb(z)− y)− bz2| < |bz2|. Then by Rouche’s theorem thereare exactly two solutions for the equation fb(z) = y inside the set z : |z| < 1/3 . Thisproves that fb : ∆(0, 1/3)

⋂f−1

b (W) → W is surjective. Furthermore, since f is defined inD, the last inequality shows that z : |z| = 1/3

⋂f−1

b (W) is empty. Hence

Wb = ∆[0, 1/3]⋂

f−1b (W) = ∆(0, 1/3)

⋂f−1

b (W)

Now let K be a compact subset of W and let zn be a sequence of elements in f−1b (K).

Without loss of generality, since K is compact we may assume that fb(zn) converges to apoint y ∈ K. But since f−1

b (K) ⊂ Wb, zn is a bounded sequence, so there is a subsequenceznj

converging to a point z with |z| ≤ 1/3 ; since fb is defined in the whole unit diskby continuity fb(znj

) converges to fb(z) which implies that fb(z) = y. But by the lastinequality above |z| must be less than 1/3, thus z ∈ f−1

b (K). Therefore we conclude thatfb : Wb → W is a proper map of degree 2. Since fb is proper, it maps the boundary of Wb

into the boundary of W . This implies that the image under fb of each connected componentof Wb is the set W . In consequence, since the degree of fb is 2, if we assume that Wb is notconnected then the connected component containing zero will be holomorphically equivalentto the set W . Since the connected component C0 containing zero is contained in W and|λ| = 1, by Schwarz lemma fb(z) = λz for all z ∈ C0, which implies that W = C0 ⊂ Wb ⊂ W.But this is a contradiction because we assumed that Wb was not connected. Therefore, Wb

must be connected.

42

We denote by Γ the circle b ∈ C : |b| = 10 . For a ∈ (0, 1], b ∈ Γ, and f ∈ Sλ, we set

fa,b = a−1f(az) + bz2 (4.1.1)

The Straightening Theorem for polynomial-like maps (see [12]) guaranties that, for anyb ∈ Γ, the map fb is conjugated on Wb to a polynomial of degree 2 by a quasiconformalhomeomorphism which is defined on the whole complex plane. However, adapting the proofof the Straightening Theorem to the family fa,b : f ∈ Sλ, a ∈ (0, 1], b ∈ Γ we can show

that there exist an interval (0, a0] such that fa,b is linearizable for all a ∈ (0, a0]. Our goalin this section is to provide the complete proof of this version of the Straightening Theoremfollowing the sketch given by Yoccoz[30], pg. 61.

We start by choosing a C∞ function η defined on the whole real line R such that η(R) = [0, 1],η is 1 on (−∞, 1/3] and η is 0 on [13/36,+∞). Now we define, for a ∈ (0, 1], b ∈ Γ, and

f ∈ Sλ, a mapping f by the formula

fa,b(z) = η(|z|)fa,b(z) + (1− η(|z|))(λz + bz2).

Clearly these mappings are well defined over the whole complex plane C. Furthermore, eachis exactly fa,b on the disk z : |z| ≤ 1/3, is λz + bz2 outside the disk z : |z| < 13/36 andit is also a C∞ function. This means

fa,b(z) =

fa,b(z) = a−1f(az) + bz2 on ∆[0, 1/3],

η(|z|)fa,b(z) + (1− η(|z|))(λz + bz2) on z, 1/3 ≤ |z| ≤ 13/36,λz + bz2 on C \∆(0, 13/36)

(4.1.2)

Lemma 4.1.7. The net of mappings fa,b converges to the polynomial P (z) = λz + bz2 as atends to zero, uniformly in b ∈ Γ and f ∈ Sλ on every compact subset of the open unit diskD. In particular, it holds over the set z : 1/3 ≤ |z| ≤ 13/36.

Proof. Let K be a compact subset of the open unit disk D. Then there exists r ≡ r(K) suchthat r ∈ (0, 1) and |z| ≤ r for all z ∈ K.

Let f ∈ Sλ with power series∞∑

n=1

anzn. Then

|fa,b(z)− (λz + bz2)| = |a−1f(az) + bz2 − λz − bz2|

So|fa,b(z)− (λz + bz2)| = a−1|a2(az)2 + a3(az)3 + a4(az)4 + . . . |

By D’Branges theorem ([5], [9]),

|fa,b(z)− (λz + bz2)| ≤ a−1(2|az|2 + 3|az|3 + 4|az|4 + . . . )

43

So

|fa,b(z)− (λz + bz2)| ≤ a−1(|az|

(1− |az|)2− |az|)

Then

|fa,b(z)− (λz + bz2)| ≤ (|z|

(1− |az|)2− |z|)

Hence

|fa,b(z)− (λz + bz2)| ≤ r(1

(1− ar)2− 1) for all z ∈ K

Finally, taking the limit as a tends to zero, the lemma follows.

Corollary 4.1.8. The net of mappings fa,b converges to the polynomial P (z) = λz + bz2 asa tends to zero, uniformly in b ∈ Γ and f ∈ Sλ on every compact subset of the complex planeC.

Proof. This follows from the equation 4.1.2 and the lemma above.

Corollary 4.1.9. The net of mappings ∂∂zfa,b converges to the map zero as a tends to zero,

uniformly in b ∈ Γ and f ∈ Sλ on every compact subset of the complex plane C.

Proof. By the equality 4.1.2 and the corollary above, the assertion holds on every compactsubset of the complement of the set z : 1/3 ≤ |z| ≤ 13/36. Therefore, we just have toanalyze over z : 1/3 ≤ |z| ≤ 13/36. In fact, by an elementary computation

∂

∂zfa,b(z) = η′(|z|) · z

2|z|· fa,b(z)− η′(|z|) · z

2|z|· (λz + bz2)

Since η′ is continuous on the compact interval [1/3, 13/36] there must be a constant C > 0such that

| ∂∂zfa,b(z)| ≤ C · |fa,b(z)− (λz + bz2)|

Therefore, the corollary follows from Lemma 4.1.7.

Corollary 4.1.10. The net of mappings ∂∂zfa,b converges to the polynomial Q(z) = λ+ 2bz

as a tends to zero, uniformly in b ∈ Γ and f ∈ Sλ on every compact subset of the complexplane.

Proof. As in Corollary 4.1.9, it is enough to analyze over the closed annulusz : 1/3 ≤ |z| ≤ 13/36. In fact, by an elementary computation we have

∂

∂zfa,b(z)−Q(z) = η′(|z|) · z

2|z|· (fa,b(z)− (λz + bz2)) + η(|z|) · ( ∂

∂zfa,b(z)−Q(z))

44

Consider C1 > supη(r) : 1/3 ≤ r ≤ 13/36 ≥ 0. Since η′ is continuous on the compactinterval [1/3, 13/36] we can guarantee the existence of C1; and also consider the positiveconstant C given in the proof of Corollary 4.1.9. Then

| ∂∂zfa,b(z)−Q(z)| ≤ C · |fa,b(z)− (λz + bz2)|+ C1 · |

∂

∂zfa,b(z)−Q(z)|

Hence, we only have to prove that ∂∂zfa,b(z) converges to Q(z) uniformly on b and f . By

definition of fa,b, it is not difficult to see that

| ∂∂zfa,b(z)−Q(z)| = |f ′(az)− λ|

Now, if we set f(z) =∞∑

n=1

anzn with a1 = λ, by D’Branges theorem

|f ′(az)− λ| = |∞∑

n=2

nan(az)n−1| ≤∞∑

n=2

n2|az|n−1

Since |z| ≤ 13/36, we have

|f ′(az)− λ| ≤1 + 13a

36

(1− 13a36

)3− 1

for any f ∈ Sλ. Therefore, the corollary follows.

Lemma 4.1.11. There exist a0 and a continuous map k : [0, a0] → [0, 1) with k(0) = 0, such

that for all f ∈ Sλ, b ∈ Γ and a ∈ (0, a0], the map fa,b is a branched covering of degree 2 ofC satisfying

| ∂∂zfa,b(z)| ≤ k(a)| ∂

∂zfa,b(z)|

for all 1/3 ≤ |z| ≤ 13/36.

Proof. The fact that the map fa,b is a branched covering of degree 2 comes from the equation4.1.1, because a−1f(az) is also an element of Sλ and we already proved that for any |b| ≥ 10and any F ∈ Sλ, (W ,Wb, Fb) is polynomial-like of degree 2 (see Lemma 4.1.6). To getthe inequality in the assertion, we set L = z : 1/3 ≤ |z| ≤ 13/36, P (z) = λz + bz2

and ‖fa,b − P‖ = max|fa,b(z) − P (z)| : z ∈ L. The existence of the map k comes fromCorollaries 4.1.9 and 4.1.10 as follows:Given ε > 0 there exists a0 ∈ (0, 1] such that, if a ∈ (0, a0] then

| ∂∂zfa,b(z)− (λ+ 2bz)| < ε

and

| ∂∂zfa,b(z)| ≤ C‖fa,b − P‖ < ε

45

for all f ∈ Sλ and all b ∈ Γ and for all z ∈ L. Note that the constant C comes from theproof in Corollary 4.1.9. By the minimum modulus principle

−ε+17

3≤ −ε+ |λ+ 2bz| < | ∂

∂zfa,b(z)|

So, for very small ε we have1

| ∂∂zfa,b(z)|

<1

173− ε

Then| ∂∂zfa,b(z)|

| ∂∂zfa,b(z)|

≤ C173− ε

· ‖fa,b − P‖ < ε173− ε

for all f ∈ Sλ and for all b ∈ Γ. Therefore if we consider 0 < ε < 17/6 (fixed) we have

C173− ε

· ‖fa,b − P‖ < 1

Then, we can define

k(a) =

0 for a = 0

C173−ε· sup‖fa,b − P‖ : f ∈ Sλ, b ∈ Γ for a ∈ (0, a0]

(4.1.3)

By Lemma 4.1.7, this map is continuous at zero. The proof of the continuity of k(a) on(0, a0] is as follows. First of all, note that

| ‖fa,b − P‖ − ‖fs,b − P‖ | ≤ ‖fa,b − fs,b‖

where ‖fa,b − fs,b‖ = sup|fa,b(z) − fs,b(z)| : z ∈ L, b ∈ Γ. Now we prove that ‖fa,b − fs,b‖goes to zero when s is approaching a.

In fact, assume f(z) =∞∑i=1

aizi. Then

|fa,b(z)− fs,b(z)| = |1a

∞∑i=1

ai(az)i − 1

s

∞∑i=1

ai(sz)i| = |∞∑i=2

ai((a)i−1 − (s)i−1)zi|

Hence

|fa,b(z)− fs,b(z)| = |a− s||a2z2 + a3(a+ s)z3 + a4(a

2 + as+ s2)z4 + . . . |

By D’Branges theorem and since a, s ∈ (0, a0] ⊂ (0, 1], we get

|fa,b(z)− fs,b(z)| ≤ |a− s|(2 · 1|z|2 + 3 · 2|z|3 + 4 · 3|z|4 + 5 · 4|z|5 + ......)

46

Then|fa,b(z)− fs,b(z)| ≤ |a− s|(22|z|2 + 32|z|3 + 42|z|4 + 52|z|5 + ......)

On the other hand, it is not difficult to check that

∞∑i=1

i2ti =(1 + t)t

(1− t)3

So∞∑i=2

i2ti = ((1 + t)

(1− t)3− 1)t

Therefore

|fa,b(z)− fs,b(z)| ≤ |a− s| · ( 1 + |z|(1− |z|)3

− 1)|z|

Since 1/3 ≤ |z| ≤ 13/36, there must be a constant M > 0 such that

|fa,b(z)− fs,b(z)| ≤ |a− s| ·M

for all f ∈ Sλ, b ∈ Γ and z ∈ L. So

| ‖fa,b − P‖ − ‖fs,b − P‖ | ≤ ‖fa,b − fs,b‖ ≤M · |a− s|

for all f ∈ Sλ, and b ∈ Γ. Then

C173− ε

· |k(a)− k(s)| ≤M · |a− s|

for any a, s ∈ (0, a0]. Therefore k is continuous on [0, a0].

Remark 4.1.12. For each open set U ⊂ C, we identify the Beltrami forms on U as thefunctions µ ∈ L∞ with norm ‖µ‖L∞ < 1.

Definition 4.1.13. If f : U → V is a C1 mapping, and µ is a Beltrami form on V, the formf ∗µ is defined by :

(f ∗µ)(z) =∂∂zf(z)µ(f(z)) + ∂

∂zf(z)

∂∂zf(z)µ(f(z)) + ∂

∂zf(z)

Proposition 4.1.14. Let a ∈ (0, a0]. There exists a unique Beltrami form µ = µf,a,b on Cinvariant under fa,b , null on C \W and null on the Julia set

⋂n≥0

f−na,b (Wb) of fa,b. Further-

more, ‖µ‖L∞ ≤ k(a) holds.

47

Proof. Let ν the Beltrami form defined by

ν(z) =

∂∂zfa,b(z)

∂∂zfa,b(z)

, z ∈ C : 1/3 ≤ |z| ≤ 13/36

0 , otherwise

Now define the recursive sequence ξn = ϕ∗ξn−1 for all n ≥ 1, where ξ0 = ν and ϕ(z) = fa,b(z).By elementary computations, µ = lim

n→∞ξn verifies the conditions required, so the support of

µ is contained in W . By Lemma 4.1.6 fa,b is analytic on ∆(0, 1/3)⋂f−1

a,b (W). Then applying

Lemma 4.1.11 to the corresponding fa,b, the inequality ‖µ‖L∞ ≤ k(a) follows.

By the Ahlfors-Bers Theorem, there exists a unique quasiconformal homeomorphism φ ≡φf,a,b satisfying the following properties :

(i) For almost all z ∈ C : ∂∂zφ(z) = µ(z) ∂

∂zφ(z).

(ii) φ(0) = 0 .

(iii) The function φ(z)− z is bounded .

(iv) supp(µ) ⊂ W.

Lemma 4.1.15. Douady-Hubbard Theorem (Yoccoz’s Version). Assuming the same notationas above :

φ fa,b φ−1(z) = λz + bz2 .

Proof. The map φ fa,b φ−1 is quasi-regular and preserves the null Beltrami form of C.Hence, this is a holomorphic map. It is also a branched covering of degree 2 with 0 as fixedpoint, so

φ fa,b φ−1(z) = λ′z + b′z2

for some λ′ and b′ 6= 0 . By the equation 4.1.2, for |z| ≥ 13/36 we have

φ(λz + bz2) = λ′φ(z) + b′(φ(z))2

By item (iii) above, lim|z|→∞

φ(z)

z= 1, so

lim|z|→∞

φ(λz + bz2)

λz + bz2= λ′ lim

|z|→∞

φ(z)

z

1

λ+ bz+ b′ lim

|z|→∞(φ(z)

z)2 1

λz

+ b

48

Then

1 = λ′ · 1 · 0 + b′ · 1 · 1

bSo

b = b′

On the other hand, for each t ∈ (0, 1] define :

T = φ(t) and φt(z) =1

Tφ(tz)

Clearly φt(0) = 0, φt(1) = 1 and φt(∞) = ∞. Hence, φtt∈(0,1] is a normal family ofquasiconformal homeomorphisms (see [21], pg. 70). We set

ft(z) =1

tfa,b(tz)

It is not difficult to check that φ−1t (w) = 1

tφ−1(Tw), so

φt ft φ−1t (z) =

1

Tφ(t(ft φ−1

t )(z))

and

φt ft φ−1t (z) =

1

Tφ(fa,b(φ

−1(Tz)))

But we already know that φ fa,b φ−1(z) = λ′z + bz2. Then

φt ft φ−1t (z) =

1

T(λ′Tz + b(Tz)2) = λ′z + bTz2

Since φtt∈(0,1] is a normal family, let ψ be the limit of some sequence φtnn∈Z+ withlim

n→∞tn = 0 . So ψ is also a quasiconformal homeomorphism ([21], pg. 73-76), lim

n→∞φ(tn) = 0

and limn→∞

ftn(z) = λz. Thus

limn→∞

φtn ftn φ−1tn (z) = lim

n→∞λ′z + bφ(tn)z2 = λ′z

Thenψ(λψ−1(z)) = λ′z

soψ(λz) = λ′ψ(z)

From this equality one sees that ψ((λ)n) = (λ′)n for all n ∈ Z+. Since ψ is a homeomorphism,λ′ has unit modulus and is not a root of unity. Furthermore ψ : S1 → S1 is an orientation-preserving homeomorphism, because λn : n ∈ Z+ is dense in S1. Thus, F, F : S1 → S1 given

by F (z) = λz and F (z) = λ′z are conjugate by the orientation-preserving homeomorphismψ. So, by the theorem on the topological invariance of the rotation number ([7], pg. 47-48)we get λ = λ′.

49

Remark 4.1.16.Recall that each φ ≡ φf,a,b is K(a)-quasiconformal homeomorphism and satisfies:

(i) For almost all z ∈ C : ∂∂zφ(z) = µ(z) ∂

∂zφ(z).

(ii) φ(0) = 0 .

(iii) The function φ(z)− z is bounded .

(iv) supp(µ) ⊂ W.

From the last item each φ is univalent on C \W = z : |z| ≥ 13/36. Now, for each φ let gφ

be the map defined by

gφ(z) =1

φ(1

z).

It is easy to see that each gφ is well-defined and is univalent on the disk ∆(0, 36/13). Theneach gφ is univalent on D. By the items above, one sees that gφ(0) = 0 and g′φ(0) = 1. Sowe may apply Theorem E.6 (see Appendix E) to each gφ and obtain

|z|(1 + |z|)2

≤ |gφ(z)| ≤ |z|(1− |z|)2

.

In particular, for any z in the closed ring z : 1/2 ≤ |z| ≤ 3/4 we have

8

49≤ |gφ(z)| ≤ 12.

Hence, for any z3 in the closed ring 4/3 ≤ |z3| ≤ 2, the following inequality holds:

1

12≤ |φ(z3)| ≤

49

8.

Now, for each a ∈ (a, a0], we set

K(a) =1 + k(a)

1 + k(a)

Here a0 and k(a) are the same as in Lemma 4.1.11.

Lemma 4.1.17. Fix a ∈ (0, a0]. There exists a constant C(a), depending only on a, suchthat:

|φf,a,b(z1)− φf,a,b(z2)| ≤ C(a)|z1 − z2|1/K(a)

for all z1, z2 ∈ D.

50

Proof. First of all, fix z3 ∈ R with 4/3 ≤ z3 ≤ 2. Now, for each φf,a,b define

Λf,a,b(w) =φf,a,b(z3z)

φf,a,b(z3).

By an elementary computation (see [7], pg. 17), the dilatations µΛ ≡ µΛf,a,bµφ ≡ µφf,a,b

satisfyµΛ(w) = µφ(3w)

for almost all w. By Proposition 4.1.14 we conclude that ‖µΛ‖L∞ ≤ k(a). Therefore themaps Λf,a,b(w) are K(a)-quasiconformal homeomorphisms for all a ∈ (0, a0], b ∈ C with|b| = 10. Moreover, the maps Λf,a,b fix 0,1 and ∞. So Λf,a,b : f ∈ Sλ, b ∈ Γ is a normalfamily of K(a)-quasiconformal homeomorphisms (see[21], pg. 73). Since the equicontinuity,Holder continuity and normality of a family of K-quasiconformal mappings are equivalentconcepts ([21], pg. 72) and the closed disk w : |w| ≤ 3

4 is compact, there must be a

constant c(a) just depending on a such that :

|Λf,a,b(w1)− Λf,a,b(w2)| ≤ c(a)|w1 − w2|1/K(a)

for all w1, w2 ∈ w : |w| ≤ 34. Then

|φf,a,b(z3w1)− φf,a,b(z3w2)| ≤ c(a) · |φf,a,b(z3)| · |w1 − w2|1/K(a)

for all w1, w2 ∈ w : |w| ≤ 34. Since 4/3 ≤ z3 ≤ 2 we may apply the last inequality in

Remark 4.1.16. In particular we have

|φf,a,b(z3w1)− φf,a,b(z3w2)| ≤ c(a) · 49

8· |w1 − w2|1/K(a)

for all w1, w2 ∈ w : |w| ≤ 1

z3

. By the elementary change of variable z = z3w and using the

inequality 4/3 ≤ z3, the lemma follows.

4.2 The Quadratic Polynomial Theorem

Definition 4.2.1. For λ ∈ C with modulus 1 and not a root of unity, we denote by Pλ thequadratic polynomial

Pλ(z) = λ(z − z2)

Definition 4.2.2. For |λ| = 1 not a root of unity we set Hλ to be the unique formal seriesthat satisfies Hλ(0) = 0, H ′

λ(0) = 1 and

Hλ(λz) = Pλ(Hλ(z))

51

Remark 4.2.3. Assuming that the radius of convergence of Hλ is strictly positive, thenthe polynomial Pλ is linearizable. Thus it has a Siegel disk Uλ at 0 . Let c = c(Uλ, 0)be the conformal capacity of Uλ at 0. By Theorem E.4 (see Appendix E) ∆(0, c/4) ⊂ Uλ.Considering the map gb(z) = − b

λz, it is easy to check that gb(λz + bz2) = Pλ(gb(z)) i.e.,

Pλ is conjugated to the polynomial λz + bz2 for all b ∈ C∗, so the polynomial λz + bz2

has as Siegel disk Dλ = g−1b (Uλ). Therefore, we have ∆(0, c

4|b|) = g−1b (∆(0, c/4)) ⊂ Dλ.

In particular ∆(0, c40

) ⊂ Dλ for all b with |b| = 10. By Lemma 4.1.15, the map φf,a,b

conjugates the polynomial λz + bz2 to the map fa,b for all b with |b| = 10. So, since a

quasiconformal conjugacy is in particular a topological conjugacy, the map fa,b has as Siegel

disk D′λ = φ−1

f,a,b(Dλ) . Therefore the mappings fa,b are linearizable for all b with |b| = 10.

Lemma 4.2.4. (i) The Siegel disk D′λ is contained in Wb.

(ii) For all b with |b| = 10, the maps fa,b are linearizable and their Siegel disks are therespective D′

λ.

(iii) The disk z, |z| ≤ [ c40C(a)

]K(a) is contained in D′λ.

Proof. Using the equation 4.1.2, it is not difficult to see that if z ∈ C\Wb then limn→∞

fna,b(z) =

∞. So item (i) follows. Here fna,b is the n-th iteration of fa,b.

By the equation 4.1.2, fa,b = fa,b on ∆(0, 1/3) ; then item (ii) comes from item (i). Finally,let z such that |z| ≤ [ c

40C(a)]K(a). Then

C(a)|z − 0|1/K(a) <c

40

By Lemma 4.1.17

|φf,a,b(z)− 0| ≤ C(a)|z − 0|1/K(a) <c

40

Then φf,a,b(z) ∈ ∆(0, c40

) . So, by the remark above φf,a,b(z) ∈ Dλ. Then z ∈ D′λ = φ−1

f,a,b(Dλ)and item (iii) follows.

Theorem 4.2.5. The Quadratic Polynomial Theorem . Let f ∈ Sλ, λ ∈ C withmodulus 1 and not a root of unity. Assume the radius of convergence of Hλ is a positivenumber. Then f is linearizable.

Proof. By the previous remark and lemma, fa,b(z) = a−1f(az) + bz2 is linearizable . Notethat the conformal representation ha,b : ∆(0, c(D′

λ, 0)) → D′λ linearizes fa,b. Since the map

fa,b ∈ Sλ, its formal linearization Ha,b (see 4.1.3) is analytic on ∆(0, c(D′λ, 0)). Thus the

power series of ha,b and Ha,b coincide.

52

Let Ra = [ c40C(a)

]K(a). By the last lemma ∆(0, Ra) is strictly contained in D′λ. Then, by

Theorem E.3 (see Appendix E), we have Ra < c(D′λ, 0). Therefore the power series Ha,b is

well defined on the closed disk ∆[0, Ra]. Recall that for each b ∈ C , Ha,b has the form

Ha,b(z) = z(1 +∑n≥1

Pn(a, b)zn)

where the function Pn(a, b) is a polynomial of degree n in b. Since Ha,b is analytic on ∆(0, Ra)and continuous on ∆[0, Ra] for all |b| = 10, its coefficients satisfy the following inequality

|Pn(a, b)| ≤ max|Ha,b(z)| : |z| ≤ RaRn+1

a

Thus by the maximum principle one has |Pn(a, 0)| ≤ M(a)(1/Ra)n for all n ∈ Z+, for alla ∈ (0, a0], where M(a) = max|Ha,b(z)| : |z| ≤ Ra. Then Ha,0 is analytic which means thatfa,0 is linearizable for all a ∈ (0, a0] (recall that a0 comes from Lemma 4.1.11) . Therefore fis linearizable.

Appendix A

Arithmetical Lemmas

Lemma A.1. Let (δn)n≥0 be a sequence of positive real numbers such that δn+1 ≤ ABnδ2n

for all n where A > 1 and B > 1. Then ABn+1δn ≤ (ABδ0)2n

for all nonnegative integer n.

Proof. The case n = 0 is obvious. Suppose is true for n = l (inductive hypothesis). Sinceδl+1 ≤ ABlδ2

l , we have

ABl+2δl+1 ≤ A2B2(l+1)δ2l = (ABl+1δl)

2 ≤ ((ABδ0)2l

)2

Therefore AB(l+1)+1δl+1 ≤ (ABδ0)2l+1

Corollary A.2. Using the same notation as above, if ABδ0 < 1 then δn < (ABδ0)n+1 for

all n ≥ 0.

Proof. In fact, the case n = 0 follows from the hypothesis A > 1 and B > 1. Moreover, wehave 1 < ABn+1 when n ≥ 0. In particular

(ABδ0)n+1 < ABn+1(ABδ0)

n+1

for all n ≥ 1. By the last lemma, ABn+1δn ≤ (ABδ0)2n

for all n ≥ 1. Then ABn+1δn ≤(ABδ0)

2n ≤ (ABδ0)n+1 < (ABn+1)(ABδ0)

n+1. Therefore δn < (ABδ0)n+1 for all n ≥ 0.

Lemma A.3. Let k ∈ Z+ and c > 1, and consider A = 2c(20)k+2, B = (30)k+2. Choose

δ0 with 0 < δ0 <1c· 1

10k+2 · 12k+2 · 1

AB. Define by induction the sequence δn+1 = 2cδ2

n

θk+2n

, where

θn = 110∗ 1

1+2n for all n ≥ 0. Then

(i) δn+1 ≤ ABnδ2n

(ii) δn < (ABδ0)n+1

(iii) δn <1c(θn)k+2.

53

54

Proof of (i). By the choice of A, the case n = 0 follows. For the case n ≥ 1, since [10(1 +2n)]k+2 ≤ (30k+2)n, one sees that ( 1

θn)k+2 ≤ Bn. Then

2c δn2

θk+2n

≤ 2cBnδ2n

But 2c < A, so 2cδ2n

θk+2n

< ABnδ2n . Therefore, δn+1 ≤ ABnδ2

n for all n ≥ 0.

Proof of (ii). By the choice of δ0, ABδ0 < 1. Furthermore, it is easy to see that δn > 0for all nonnegative integers n. So the claim follows from a direct application of the lastcorollary.

Proof of (iii). By the choice of δ0, 2k+2ABδ0 <1

c

1

10k+2< 1. Then

(ABδ0)n+1 <

1

c

1

10k+2

(1

2k+2

)n+1

for all n ≥ 1. Moreover, since 1 + 2n ≤ 2n+1 for all n ≥ 1, one sees that

(ABδ0)n+1 <

1

c

(1

10

1

1 + 2n

)k+2

=1

c(θn)k+2

for all n ≥ 1. So, by (ii) we have δn <1

c(θn)k+2 for all n ≥ 1. Finally, for the case n = 0 it

is enough to see that δ0 <1

c(20)−(k+2) =

1

c(θ0)

k+2 .

Appendix B

e is diophantine

Proposition B.1. Let θ ∈ R\Q and let pn

qnbe its sequence of continued fractions. Then

1qn+1+qn

< |θqn − pn| < 1qn+1

for all n ≥ 1.

Proposition B.2. Let θ ∈ R\Q and let pn

qnbe its sequence of continued fractions. Then

qn ≥ (32)n for all n ≥ 5.

Proposition B.3. If |θ − a

b| < 1

2b2then

a

bis a continued fraction of θ.

Proposition B.4. The partial coefficients of the number e verify :

• a0 = 2.

• an = 1 for n = 3k and n = 3k − 2 with k ≥ 1.

• an = 2k for n = 3k − 1 , with k ≥ 1.

This means e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, . . .]

For details of the proofs of the facts above see [2], [17], [22]. For the rest of this section pn

qn

means the n-th continued fraction of e.

Lemma B.5. Given ε > 0 and d > 0, the set S =

n :

∣∣∣∣e− pn

qn

∣∣∣∣ < 1

dq2+εn

is finite.

Proof. By definition qn+1 = an+1qn + qn−1, and since qn−1 ≤ qn we have

qn+1 + qn = qn(an+1 + 1 + qn−1

qn) ≤ qn(an+1 + 2)

By Proposition B.1 and the last inequality, we have

1

(an+1 + 2)≤ qnqn+1 + qn

≤∣∣∣∣e− pn

qn

∣∣∣∣ ∗ qn2

55

56

By Proposition B.4, one sees that an+1 ≤ n− 2 holds for all n ≥ 8. Therefore

1

nq2n

<

∣∣∣∣e− pn

qn

∣∣∣∣for all n ≥ 8. On the other hand, it is clear that the inequality d1/n(3

2)ε > n1/n holds for n

large. So there exists N0 such that d1/n(32)ε > n1/n for all n > N0. Now, by Proposition B.2

we have dqεn > n for all n > N0. Finally, if n ∈ S and n ≥ 8

1

nq2n

<

∣∣∣∣e− pn

qn

∣∣∣∣ < 1

dq2+εn

.

Then dqεn < n, and thus n ∈ 1, 2, . . . , N0. Therefore S ⊂ 1, 2, . . . , N0.

Corollary B.6. For any given ε > 0 and d > 0, the set

S =

a

b∈ Q :

∣∣∣e− a

b

∣∣∣ < 1

db2+ε

is finite.

Proof. Assume S is an infinite set. In this case S contains elements which have large denom-inators. Now consider the partition S1, S2 of S given by S1 = a

b∈ S : dbε < 2 and S2 =

ab∈ S : dbε > 2. It is clear that S1 is finite and S2 6= ∅. For any a

b∈ S2, we have∣∣∣e− a

b

∣∣∣ < 1

db2+ε<

1

2b2

So, by Proposition B.3 the fraction ab

is a continued fraction of e. This means that ab

= pn

qn

for some n, hence∣∣∣e− a

b

∣∣∣ =

∣∣∣∣e− pn

qn

∣∣∣∣ < 1

dq2+εn

and thus n ∈ S. Now, by the last lemma

there must be a finite number of fractionsa

bfor which dbε > 2. This means that S2 is finite.

Therefore S is finite, which is a contradiction.

Even though e is not an algebraic number, we have proved that (taking d=1 ) Se =a

b∈ Q :

∣∣∣e− a

b

∣∣∣ < 1

b2+ε

is finite .

Recall that θ ∈ R\Q satisfies a diophantine condition if there exist c > 0, µ > 0 such that∣∣θ − mn

∣∣ ≥ cn2+µ for all m

n∈ Q and n ∈ Z+.

Theorem B.7. e satisfies a diophantine condition.

57

Proof. Fix ε > 0. Since Se is finite, Se =

a1

b1, . . . ,

as

bs

, so we can choose a large ν such

that1

b2+νi

< |e− ai

bi| for i = 1, 2, . . . , s. On the other hand, if

a

b/∈ Se we have

1

b2+ε≤∣∣∣e− a

b

∣∣∣.Therefore, taking µ = maxε, ν ∣∣∣e− a

b

∣∣∣ ≥ 1

b2+µ

for all ab∈ Q, b ∈ Z+.

Appendix C

Davie’s Lemma

Let α ∈ R\Q. Through this Appendix we denote by pn

qnthe n-th continued fraction of α and

we consider ‖nα‖ = min|nα−m| : m ∈ Z.Lemma C.1. If r ∈ Z and 0 < r < qk then ‖rα‖ ≥ 1

2qk.

Proof. In fact, if ‖rα‖ < 12qk

, by definition there exists p ∈ Z such that |rα − p| = ‖rα‖ <1

2qk< 1

2r. This implies that ∣∣∣α− p

r

∣∣∣ < 1

2r2

So, by Proposition B.3, pr

is a continued fraction of α. This means that p = pj and r = qj forsome j. Since (qn)n is an increasing sequence and qj < qk, j < k must hold. By PropositionB.1 we have

1

2qj+1

≤ 1

qj+1 + qj< |qjα− pj| <

1

qj+1

So 12qj+1

< ‖qjα‖ = ‖rα‖ < 12qk

. Then qj < qk < qj+1 which is a contradiction.

Corollary C.2. Let k be a non-negative integer . If n ∈ Z+ and ‖nα‖ ≤ 14qk

, then n ≥ qkand either qk divides n or n ≥ qk+1

4.

Proof. Clearly if n satisfies the hypothesis of the lemma above we get an obvious contra-diction, so n ≥ qk. Now, if qk does not divide n and n < qk+1

4then n = mqk + r where

0 < r < qk. It is easy to check that m < qk+1

4qkholds. By Proposition B.1 we know that

‖qkα‖ < 1qk+1

, so ‖mqkα‖ < mqk+1

< 14qk

. Now let τ1, τ2 ∈ Z satisfying ‖nα‖ = |(mq+ r)α− τ1|and ‖mqkα‖ = |mqkα− τ2|. Then ‖nα‖ = |(mq + r)α− τ1| ≥ |rα− (τ1 − τ2)| − |mqα− τ2|,so we get ‖nα‖ ≥ ‖rα‖ − ‖mqα‖ > 1

2qk− 1

4qk= 1

4qk, which is a contradiction.

Definition C.3. Let q be a positive integer and E a real number with E ≥ q. Suppose A isa set of non-negative integers such that 0 ∈ A and such that whenever j1 , j2 ∈ A and qdoes not divide j1 − j2 we have |j1 − j2| ≥ E .

58

59

Definition C.4. Let A∗ be a set of non-negative integers, namely the set of j such thateither j ∈ A or j1 < j < j2 and q divides j− j1, for some j1 , j2 ∈ A with j2− j1 < E. Notethat if the second condition holds, then q also divides j2 − j1 which implies that q dividesj2 − j.

Claim C.4.1. If j, j′ ∈ A∗ with q < j′− j < E there must exist j′′ ∈ A∗ with j < j′′ < j′.

Proof. We have 4 cases :

1. If j, j′ ∈ A , since q < j′ − j < E , it is enough to consider j′′ = j + q .

2. If j ∈ A and j′ ∈ A∗\A , there exist j1, j2 ∈ A such that j1 < j′ < j2 and q dividesj′ − j1 with j2 − j1 < E . If j1 ≤ j , take j′′ = j′ − q ; this j′′ verifies the second partof the definition of A∗. Otherwise, if j1 > j consider j′′ = j1 .

3. If j ∈ A∗\A and j′ ∈ A , there exist j1, j2 ∈ A such that j1 < j < j2 and q dividesj − j1 with j2 − j1 < E . If j2 ≥ j′ , take j′′ = j + q ; this j′′ verifies the second partof the definition of A∗. Otherwise, if j2 < j′ consider j′′ = j2 .

4. If j ∈ A∗\A and j′ ∈ A∗\A there exist j1, j2, τ1, τ2 ∈ A such that j1 < j < j2,τ1 < j′ < τ2, q divides j−j1 with j2−j1 < E and q divides j′−τ1 with τ2−τ1 < E.If j < j2 < j′ or j < τ1 < j′, it is enough to consider j′′ = j2 or j′′ = τ1. Otherwise,since τ1 < j′ < τ2, q divides j′ − τ1, τ2 − τ1 < E and q < j′ − j < E, it is easy to seethat j′′ = j′ − q ∈ A∗.

Claim C.4.2. Let mn = max j : 0 ≤ j ≤ n , j ∈ A∗ .

1. If n ∈ A∗ and mn−1 + q ∈ A∗, then mn−1 = n− q.

2. If n ∈ A∗ and mn−1 + q /∈ A∗, then n−mn−1 ≥ E .

Proof of 1. The assumption mn−1 + q ≤ n − 1 contradicts the definition of mn−1, whencemn−1 + q ≥ n. Assuming that mn−1 + q > n = mn, we will have 4 subcases :

1. Case I : mn−1 ∈ A and n ∈ A .Since 0 < n−mn−1 < q, q does not divide n−mn−1, so n−mn−1 ≥ E. Then q > E,which is a contradiction .

2. Case II : mn−1 ∈ A∗\A and n ∈ A.By definition of A∗, there exist j1, j2 ∈ A such that j1 < mn−1 < j2 and q dividesmn−1−j1 with j2−j1 < E . Since there is not any element of A∗ between mn−1 and n,j2 ≥ n must hold.

60

If q does not divide n − j1, by definition of A we have n − j1 ≥ E, which impliesE > j2 − j1 ≥ n − j1 ≥ E. Hence q divides n − j1 and since q divides mn−1 − j1, qdivides also n−mn−1 . But this is a contradiction because 0 < n−mn−1 < q.

3. Case III : mn−1 ∈ A and n ∈ A∗\A .By definition of A∗, there exist j1, j2 ∈ A such that j1 < n < j2 and q dividesn− j1 with j2 − j1 < E. Since there is not any element of A∗ between mn−1 and n,j1 ≤ mn−1 must hold.If q does not divide j2−mn−1, by definition of A we have j2−mn−1 ≥ E, which impliesE > j2 − j1 ≥ j2 −mn−1 ≥ E. Hence q divides j2 −mn−1 and since q divides j2 − n, qalso divides n−mn−1. But this is a contradiction again.

4. Case IV : mn−1 ∈ A∗\A and n ∈ A∗\A.By definition, there exist j1, j2, τ1, τ2 ∈ A such that j1 < n < j2, τ1 < mn−1 < τ2,q divides n − j1 with j2 − j1 < E and q divides mn−1 − τ1 with τ2 − τ1 < E.In particular j1 < τ2. Since q divides τ2 − mn−1 then mn−1 + q ≤ τ2 . By the sameargument j1 ≤ n− q.Suppose q does not divide τ2 − j1 . By definition of A, τ2 − j1 ≥ E holds. Thisimplies that j1 < τ1 and j2 < τ2 . If q does not divide τ1 − j1, then E ≥ j2 − j1 >τ1−j1 > E, which is a contradiction . Hence, q must divide τ1−j1. Therefore q dividesn−mn−1 which is a contradiction again .Suppose q divides τ2 − j1 . Since q divides τ2 −mn−1 and n− j1, we conclude thatq divides n−mn−1 which is impossible. That is why this case never happens.

Since all of these cases imply contradictions then our initial assumption mn−1 + q > n failsto be true. Therefore mn−1 + q = n holds.

Proof of 2. By hypothesis n ∈ A∗ and mn−1 + q /∈ A∗, hence mn−1 6= n − q. We have 4subcases :

1. Case I : mn−1 ∈ A and n ∈ A.If 0 < n−mn−1 < q, q does not divide n−mn−1, so n−mn−1 ≥ E.If q < n − mn−1 , then n − mn−1 ≥ E. Otherwise by Claim C.4.1 there must existj′′ ∈ A∗ such that mn−1 < j′′ < n. But this contradicts the definition of mn−1.

2. Case II : mn−1 ∈ A∗\A and n ∈ A.By definition of A∗, there exist j1, j2 ∈ A such that j1 < mn−1 < j2 and q dividesmn−1− j1 with j2− j1 < E. The inequality n−mn−1 < q does not hold. Otherwise itimplies that q does not divides n−mn−1, which implies that q does not divide n− j1;but as in item 1, Case II, this is a contradiction . Hence q < n−mn−1. Following thesame argument as in Case I above, n−mn−1 ≥ E must hold.

3. Case III : mn−1 ∈ A and n ∈ A∗\A.By definition of A∗, there exist j1, j2 ∈ A such that j1 < n < j2 and q divides

61

n − j1 with j2 − j1 < E. If n − mn−1 < q, then q does not divide n − mn−1, andhence q does not divide mn−1 − j1. By definition of A, we have mn−1 − j1 ≥ E. ThusE > j2 − j1 ≥ j2 −mn−1 ≥ E, hence n−mn−1 ≥ q. Following the same argument asin Case I above, n−mn−1 ≥ E must hold.

4. Case IV : mn−1 ∈ A∗\A and n ∈ A∗\A .If q < n − mn−1 < E, then by Claim C.4.1 there must exist j′′ ∈ A∗ such thatmn−1 < j′′ < n But this contradicts the definition of mn−1. So either n −mn−1 ≥ Eor n−mn−1 < q. If n < mn−1 + q, repeating exactly the same steps as in item 1, CaseIV, we can prove that this case never happens. Therefore n−mn−1 ≥ E holds.

Definition C.5. For each n ≥ 0 , n ∈ Z define

l(n) = max

(1 + τ)

n

q− 2,

τmn + n

q− 1

where mn = maxj : 0 ≤ j ≤ n, j ∈ A∗ and τ = 2

q

E.

Now we define

h(n) =

q−1(mn + τn)− 1, mn + q ∈ A∗

l(n), mn + q /∈ A∗

Claim C.5.1. By Claim C.4.2 and Definition C.5, one sees that:

1. (1 + τ)n

q− 2 ≤ h(n) ≤ (1 + τ)

n

q− 1 for all n.

2. If n ∈ A∗ and n > 0, then h(n) ≥ h(n− 1) + 1.

3. h(n) ≥ h(n− 1) for all n > 0.

4. h(n+ q) ≥ h(n) + 1 for all n.

Definition C.6. For each k, let Ak be the set defined by

Ak =

n ≥ 0 : ‖nα‖ ≤ 1

8qk

By Corollary C.2 these sets satisfy the definition C.3 where q = qk, E = maxqk, qk+1

4,

τk = 2 qk

E≤ 8qk

qk+1. So, for each k we can define hk as in Definition C.5. Now, for each k we

define gk(n) := maxhk(n) ,[[

nqk

]] .

Remark C.7.

62

1. By C.5.1, hk(0) ∈ [−2,−1]. So we have gk(0) = 0.

2. gk(1) ≥[[

1q

]]≥ 0

Proposition C.8. The function gk is non-negative, and using Claim C.5.1 it is easy tocheck that gk satisfies :

1. gk(n) ≤ (1 + τ)q−1n.

2. gk(n1) + gk(n2) ≤ gk(n1 + n2).

3. If n ∈ A and n > 0, then gk(n) ≥ gk(n− 1) + 1

Definition C.9. For each n ≥ 0, n ∈ Z, define

G(n) = 2n log(2) +

k(n)∑k=0

gk(n) log(2qk+1) ,

where k(n) is defined by the condition

qk(n) ≤ n < qk(n)+1.

Since qn ∈ Z+ and 1 = q0 ≤ q1 < q2 < q3 . . . , k(n) is well defined.

Proposition C.10. The function G(n) satisfies :

1. G(0) = 0

2. G(n− 1) ≤ G(n), for all n ≥ 1.

3. G(n1) +G(n2) ≤ G(n1 + n2), for all n1, n2 ∈ N.

4. There exists a universal constant γ3 > 0 such that for all n ∈ Z+0 the following inequality

holds

G(n) ≤ n

k(n)∑i=0

log qi+1

qi+ γ3

5. − log |λn−1| ≤ G(n)−G(n−1) , for all n ≥ 1, where λ = exp(2πiα) and α = lim

n→∞

pn

qn.

Proof of 1. This follows by definition.

Proof of 2. Use the fact k(n− 1) ≤ k(n) and the definition of G(n).

63

Proof of 3. G(ni) = 2ni log(2) +

k(ni)∑k=0

gk(n) log(2qk+1) where i ∈ 1, 2. Consider r =

k(n1 +n2). By definition of k(n) it is not difficult to see that k(n1) < r and k(n2) < r. Then

G(n1) +G(n2) ≤ 2(n1 + n2) log(2) +r∑

k=0

(gk(n1) + gk(n2)) log(2qk+1)

So, by Proposition C.8 the assertion follows.

Proof of 4. By the first item of proposition C.8, gk(n) ≤ (1+τk)q−1k n where τk = 2 qk

Ek≤ 8 qk

qk+1.

So

G(n) ≤ 2n log 2 +

k(n)∑k=0

[n

qk+

8n

qk+1

]log(2qk+1).

Then

G(n) ≤ n

2 log 2 +

k(n)∑k=0

log(2) + log(qk+1)

qk+ 8

k(n)∑k=0

log 2qk+1

qk+1

By induction, it is not difficult to check that qk+2 ≥ 2qk for all k ≥ 0. So

∞∑k=0

1

qk≤ 4

On the other hand, by induction is also easy to check that qk ≥

(√5 + 1

2

)k−1

for all k ≥ 1.

Since log(t)/t is decreasing on (e,+∞) and

(√5 + 1

2

)3

> e, we have

log qk+1

qk+1

≤ k ·log(

√5+12

)(√5+12

)k

for all k ≥ 3. Since q1, q2 and q3 are positive integers we have log(qi)/qi < 1 for i ∈ 1, 2, 3.Now by elementary computations is easy to see that

∞∑k=0

log qk+1

qk+1

< 3 + log(

√5 + 1

2) ·

∞∑k=3

k

(2√

5 + 1

)k

Therefore, there exists a universal constant γ3 > 0 such that

G(n) ≤ n

k(n)∑k=0

log qk+1

qk+ γ3

64

Proof of 5. By an elementary computation | sin(t)| ≥ 2π|t| on [−π

2, π

2] and ‖nα‖ ≤ 1

2for all

n ≥ 0, where ‖nα‖ = min|nα −m| : m ∈ Z. Let m ∈ Z such that ‖nα‖ = |nα −m|. So2‖nα‖ = 2

π|π(nα−m)| ≤ | sin(π(nα−m))|. Hence

2‖nα‖ ≤ | sin(nπα)|

for all n ≥ 0. Then |λn − 1|−2 = (2 sin(nπα))−2 ≤ (4‖nα‖)−2.

Fix n ∈ Z+. We have two possibilities :

(i) If ‖nα‖ ≥ 1

8=

1

8q0then |λn−1|−2 ≤ 4, and hence − log |λn−1| ≤ log 2. By Definition

C.9, one sees that 2 log 2 ≤ G(n)−G(n− 1). Then

− log |λn − 1| ≤ G(n)−G(n− 1) .

(ii) If1

8qk0+1

≤ ‖nα‖ < 1

8qk0

, then n ∈ Ak0 and ‖nα‖ < 1

4qk0

. By Corollary C.2 n ≥ qk0 .

Hence k(n) ≥ k0.

By Proposition C.8(3), we have that gk0(n−1)+1 ≤ gk0(n). Thus gk0(n−1) log(2qk0+1)+log(2qk0+1) ≤ qk0(n) log(2qk0+1). Since k0 ∈ 1, 2, . . . , k(n), we obtain

log(2qk0+1) ≤k(n)∑k=0

[gk(n)− gk(n− 1)] log(2qk+1)

Therefore − log |λn − 1| ≤ log(2qk0+1) ≤ G(n)−G(n− 1)

Appendix D

Proof of Theorem 2.5.5 and Theorem2.5.6

Proof of Theorem 2.5.5. By hypothesis, there exist constants c > 0 and µ > 2 such that∣∣∣∣α− p

q

∣∣∣∣ ≥ c

qµ

for all p ∈ Z, q ∈ Z+. By Proposition B.1,

∣∣∣∣α− pk

qk

∣∣∣∣ < 1

qkqk+1

for all k. Then

log qk+1

qk<

log c−1

qk+ (µ− 1)

log qkqk

By Proposition B.2, qk ≥(

3

2

)k

for all k ≥ 5, so we have∞∑

k=0

log c−1

qk< +∞. On the other

hand, sincelog t

tis decreasing on (e,+∞), we obtain

log qkqk

<

k log

(3

2

)(

3

2

)k

So∞∑

k=0

log qkqk

< +∞. Therefore∞∑

k=0

log qk+1

qk< +∞.

Lemma D.1. Let (ak)k≥1 be a sequence of positive integers. Then

log

(ak +

1

ak−1

)< log(ak) + 1.

65

66

Proof. See also [19]. An elementary computation shows that x > 1e−1

implies log(x + 1) <

log x+ 1. Since ak ≥ 1 > 1e−1

, we have log(ak + 1) < log(ak) + 1.

Lemma D.2. Let α ∈ R\Q, (ak)k≥1 its sequence of partial coefficients and (pk

qk) its sequence

of continued fractions. Then

∞∑k≥0

log qk+1

qk< +∞ if and only if

∞∑k≥0

log ak+1

qk< +∞.

Proof. See also [19].

[⇒ ]This follows from the fact ak+1 ≤ qk+1.

[⇐ ]By equation (2.4.2), qk+1 = ak+1qk + qk−1 and qk = akqk−1 + qk−2 > akqk−1 hold for allk ≥ 2, so

qk+1 =

(ak+1 +

qk−1

qk

)qk <

(ak+1 +

1

ak

)qk.

By Lemma D.1, we havelog qk+1

qk<

log ak+1

qk+

1

qk+

log qkqk

for all k ≥ 2. By hypothe-

sis∑ log ak+1

qk< +∞, and since the series

∑ 1

qk,∑ log qk

qkconverge (see the proof of

Theorem 2.5.5), we obtain∞∑

k≥0

log qk+1

qk< +∞ .

Proof of Theorem 2.5.6. See also [19].

(i) We will apply the ratio test to the series∑n≥0

log an+1

qn. It is clear that

log an+1

qn

qn−1

log an

<

ρan log an

qn

qn−1

log an

, solog an+1

qn

qn−1

log an

< ρanqn−1

qn. Since qn = anqn−1 + qn−2 > anqn−1, we have

log an+1

qn

qn−1

log an

< ρ < 1.

So∞∑

n≥0

log an+1

qn< +∞. Applying Lemma D.2, we have

∑n≥0

log qn+1

qn< +∞. Therefore

α ∈ B.(ii) Note that qn = anqn−1 + qn−2 = an(qn−2an−1 + qn−3) + qn−2 for all n ≥ 2, so

1

anan−1

>qn−2

qn

67

By hypothesis an+1 > aρann for some ρ > 1, except for finitely many n. Therefore the

coefficients an are growing quickly, so

qn−2

qn→ 0 .

On the other hand one sees thatlog an+1

qn

qn−1

log an

> ρan log an

qn

qn−1

log an

, so

log an+1

qn

qn−1

log an

> ρanqn−1

qn= ρ

(1− qn−2

qn

).

Now consider ε ∈ (0, 1− 1ρ). So there is n0 such that if n ≥ n0 then

qn−2

qn< ε. Then,

log an+1

qn

qn+1

log an

> ρ(1− ε) > 1 for n ≥ n0

thus the series∑n≥0

log an+1

qndiverges. Therefore, by Lemma D.2 we conclude that α /∈ B.

Appendix E

Conformal Capacity and KoebeInequalities

Theorem E.1. Riemann Mapping Theorem. Given any simply connected region Ωwhich is not the whole plane, and a point z0 ∈ Ω, there exists a unique conformal map (theRiemann map) ζ : D → Ω such that ζ is onto, ζ(0) = z0 and ζ ′(0) > 0.

The Riemann Mapping Theorem is a classical result in the one-dimensional complex theory.For more details see [8].

Definition E.2. The number ζ ′(0) is called the conformal capacity of Ω with respect to z0

and will be denoted by c(Ω, z0).

Considering the map h : ∆(0, c(Ω, z0)) → Ω defined by h(z) = ζ( zζ′(0))

), we have h(0) = z0

and h′(0) = 1. This map h is called the conformal representation of Ω at z0.

Theorem E.3. Let U be an open simply connected set and let x0 ∈ U . If V is an opensimply connected subset of U with x0 ∈ V then c(V , x0) ≤ c(U , x0) with equality if and onlyif U = V.

Proof. Let ζ : D → U and η : D → V be as in the theorem above . Then ζ−1 η : D → D isanalytic and has 0 as a fixed point. So, the theorem follows from a simple application of theSchwarz lemma.

Theorem E.4. Koebe One-Quarter Theorem. If f ∈ S1, then the image of f coversthe open disk centered at 0 of radius one-quarter, that is f(D) ⊃ ∆(0, 1/4) .

Proof. If we consider f(z) =∞∑

n=1

anzn, by Bieberbach’s Theorem (see [13], pg.30), we have

68

69

|a2| ≤ 2. Now fix a point τ and suppose τ ∈ C \ f(D) . Then

τf(z)

τ − f(z)= z + (a2 +

1

τ)z2 + . . .

belongs to S1. Applying the Bieberbach’s Theorem twice, we obtain

1

|τ |≤ |a2|+ |a2 +

1

τ| ≤ 2 + 2 = 4

Theorem E.5. Let ζ : D → Ω and z0 ∈ Ω as in the Riemann Mapping Theorem. ThenΩ ⊃ ∆(z0,

c(Ω,z0)4

) .

Proof. In fact, define f(z) = 1ζ′(0)

(ζ(z) − z0). Clearly f belongs to S1 and by the theoremabove

1

ζ ′(0)(Ω− z0) ⊃ ∆(0, 1/4)

Then,Ω− z0 ⊃ ∆(z0, ζ

′(0)/4)

Therefore

Ω ⊃ ∆(z0,c(Ω, z0)

4)

The next result is the well-known Koebe Distortion Theorem.

Theorem E.6. If f ∈ S1 and |z| < 1, then

1− |z|(1 + |z|)3

≤ |f ′(z)| ≤ 1 + |z|(1− |z|)3

and|z|

(1 + |z|)2≤ |f(z)| ≤ |z|

(1− |z|)2.

Proof. For the details of the proof see [9], pg. 65 .

Appendix F

Controlling the height ofrenormalization

Lemma F.1. For all f ∈ S1,

|z f′(z)

f(z)− 1| ≤ 2|z|

1− |z|

Proof. For all f ∈ S1 and for each fixed z in the open unit disk, the following inequalityholds

| log(zf ′(z)

f(z))| ≤ log(

1 + |z|1− |z|

)

The proof of this fact comes from the Goluzin inequalities. This was brought to my attentionby Dr. Peter Duren and Dr. Leslie Kay. The interested reader may find the Goluzininequalities in [13], pg. 126. Now let w = log( zf ′(z)

f(z)). Then

e|w| ≤ 1 + |z|1− |z|

On the other hand it is not difficult to check that

|zf′(z)

f(z)− 1| = |ew − 1| = |

∑n≥0

wn

n!− 1| ≤

∑n≥1

|w|n

n!= e|w| − 1

Therefore

|zf′(z)

f(z)− 1| ≤ e|w| − 1 ≤ 2|z|

1− |z|

70

71

Theorem F.2. Let F (Z) = Z + α + ϕ(Z) be a element of F(α). For all Z ∈ H :

|ϕ′(Z)| ≤ 2 exp(−2π=(Z))

1− exp(−2π=(Z))

|ϕ(Z)| ≤ −1

πlog(1− exp(−2π=(Z)))

Proof. Let f ∈ S1 such that f(E(Z)) = E(Z+ϕ(Z)) for all Z ∈ H, where E(Z) = exp(2πiZ).Then

ϕ′(Z) = E(Z)f ′(E(Z))

f(E(Z))− 1

Therefore, the first inequality follows from the last lemma. Now, in order to prove the secondinequality we can apply the first inequality and obtain

|ϕ(Z)− ϕ(Z + ni)| ≤∫ n

0

|ϕ′(Z + si)|ds ≤∫ n

0

2 exp(−2π(=Z + s))

1− exp(−2π(=Z + s))ds

Then

|ϕ(Z)− ϕ(Z + ni)| ≤ 1

πlog(1− exp(−2π=(Z)) exp(−2πs))]n0

So

|ϕ(Z)− ϕ(Z + ni)| ≤ 1

πlog

(1− exp(−2π=(Z)) exp(−2πn)

1− exp(−2π=(Z))

)But, since exp(−2πn) → 0 as n tends to infinity and lim

=(Z)→∞ϕ(Z) = 0, we obtain the second

inequality .

Theorem F.3. For all δ ∈ (0, 1/10), there exists a constant Cδ such that for all F ∈ F(α),

=(Z) ≥ Cδ ⇒ |F ′(Z)− 1| ≤ δ

and

=(Z) ≥ 1

2πlog(

1

α) + Cδ ⇒ |F (z)− z − α| ≤ δα

Proof. Since F (Z) = Z + α + ϕ(Z), we can use the last theorem and obtain

|F ′(Z)− 1| ≤ 2 exp(−2π=Z)

1− exp(−2π=(Z))

|F (Z)− Z − α| ≤ −1

πlog(1− exp(−2π=(Z)))

Let Cδ be the positive constant given by Cδ =−1

2πlog(

δ

2 + δ). Then for all Z ∈ H such that

=(Z) = Cδ + t ≥ Cδ we have

|F ′(Z)− 1| ≤ δ exp(−2πt)

72

Since Cδ =−1

2πlog(

δ

2 + δ) >

−1

2πlog(1− 1

eπδ), then

−1

πlog(1− kδ) ≤ δ where kδ =

δ

2 + δ.

On the other hand, note that for any t > 0

log(1− exp(−2πt) · kδ) ≥ log(1− kδ) ≥ log(1− kδ) exp(−2πt)

Hence, for any t > 0

−1

πlog(1− exp(−2πt) · kδ) ≤

−1

πlog(1− kδ) · exp(−2πt) ≤ δ exp(−2πt)

If =(Z) = t + Cδ ≥ Cδ, by the second inequality of the last theorem and the previousinequality, we have

|F (Z)− Z − α| ≤ δ exp(−2πt)

Therefore, if =(Z) ≥ Cδ then |F ′(Z)− 1| ≤ δ.

Furthermore, if =(Z) ≥ 1

2πlog(

1

α) + Cδ then |F (Z)− Z − α| ≤ δα.

Remark. In particular, F can not have fixed points above1

2πlog(

1

α) plus some universal

constant.

Lemma F.4. Suppose F ∈ Sδ(α) with δ ∈ (0, 1/10) and consider the corresponding sets V,V∗, V∗− as in the proof of Theorem 3.3.6. Then

F (V∗−) ⊂ V∗−⋃U .

Proof. By the definition of Sδ(α), |F (Z)− Z − α| ≤ δα holds. Then =(F (Z)) ≥ =(Z)− δαfor all Z ∈ H. In particular =(F (Z)) ≥ =(Z)− δα ≥ 4δ(−<(Z))− δα for all Z ∈ V∗−.Since F ∈ Sδ(α) we also have <(F (Z)− Z)− α ≥ −δα. Hence, if <(F (Z)) < 0 we have,

=(F (Z)) ≥ 4δ(−<(Z))− δα ≥ 4δ(−<(F (Z))) + 4δ(α− δα)− δα ≥ 4δ(−<(F (Z)))

This means that for all Z ∈ V∗− with <(F (Z)) < 0, F (Z) has to be in V∗−.Since F is univalent on H and periodic with period 1, the sets V , F (V) are homeomorphic,and the following union

F (−1 + it : t ≥ 4δ)⋃

F (t+ i4δ : −1 ≤ t ≤ 0)⋃

F (it : t ≥ 4δ)

bounds the open strip F (V). By definition of Sδ(α), it is easy to see that limt→+∞

(F (−1 + it)−it) = −1 + α < 0, so <(F (−1 + it)) < 0 for t large. Therefore, we have two possibilities.If <(F (−1 + it)) < 0 for some t ≥ 4δ, by the first conclusion F (−1 + it) ∈ V∗− because−1 + it ∈ V∗− when t ≥ 4δ.

73

If <(F (−1 + it)) ≥ 0 then 0 ≤ <(F (−1 + it)) = −1 + <(F (it)) < <(F (it)) which impliesthat the line segment [F (−1 + it), F (it)] is contained in U because F (−1 + it) = F (it) − 1for all t ≥ 4δ; therefore the inclusion F (−1 + it : t ≥ 4δ) ⊂ V∗−

⋃U holds. On the other

hand, it is obvious that F (it : t ≥ 4δ) ⊂ V∗−⋃U and using the definition of Sδ(α) one sees

that =(F (s+ i4δ)) ≥ =(s+ i4δ)− δα ≥ 4δ − δα. Then we conclude that F (V) ⊂ V∗−⋃U .

Since −1 + i4δ ∈ V∗−, from one of the inequalities above =F (−1 + i4δ) ≥ 4δ − δα. On theother hand it is easy to see =(F (0)) ≤ δα. Now consider the map ϕ(t) = =(F (t − i4δt))where t ∈ [−1, 0]. By the previous statements ϕ(−1) > ϕ(0). Clearly ϕ is continuouson [−1, 0], and differentiable on (−1, 0). Moreover ϕ′(t) does not change sign on (−1, 0);otherwise there exists some t∗ ∈ (−1, 0) such that 0 = ϕ′(t∗) = =(F ′(t∗ − i4δt∗) · (1− i4δ)).Since F ∈ Sδ(α), we have |F ′(t∗ − i4δt∗)− 1| ≤ δ. Then

|F ′(t∗ − 4δt∗i)(1− i4δ)− (1− i4δ)| ≤ |1 + i4δ|δ

So4δ = =(F ′(t∗ − i4δt∗)(1− i4δ)− (1− i4δ)) ≤ (1 + 4δ)δ

which is a contradiction because δ ∈ (0, 1/10). Since ϕ(−1) > ϕ(0) and ϕ′(t) 6= 0 we cansay that ϕ is strictly decreasing with minimum at 0, so =(F (t − i4δt)) ≥ =(F (0)) for allt ∈ [−1, 0]. Thus the curve F (t− i4δt) cannot intersect the segment line [0, F (0)].Now, if <(F (t−i4δt)) < 0 then F (t−i4δt) ∈ V∗−, because t−i4δt ∈ V∗−. If <(F (t−i4δt)) > 0,the point F (t − i4δt) cannot be at the right side of the image of the imaginary axis underF ; otherwise the curve ϕ intersects the image of the imaginary axis under F in at least onepoint. But that is impossible because F is injective, so the point F (t− i4δt) ∈ U . Therefore,the image of the curve ϕ is also in V∗−

⋃U .

Finally we will analyze the remainder of the set V∗−. This means the closure of the set V∗− \Vwhich is Z : 4δ ≥ =(Z) ≥ −<(Z) ; −1 ≤ <(Z) ≤ 0. Since the boundary of this set ismapped into V∗−

⋃U and F is univalent, in particular a homeomorphism, we may conclude

that F (V∗− \ V) is also a subset of V∗−⋃U .

Appendix G

Modulus of a Ring Domain andQuasiconformal Mappings

Definition G.1. A plane domain G is called simply connected if its complement is con-nected. If the complement of G consists of n components (1 < n < ∞), then G is calledn-tuply connected. A doubly connected domain is called a ring domain or ring.

The definition of the modulus of a ring domain is based on the following mapping theorem:Every ring domain B can be mapped conformally onto an annulus z : 0 ≤ r1 < |z| <r2 ≤ ∞. Such a mapping is called a canonical mapping and the corresponding annulus acanonical image of B. If r1 > 0 and r2 < ∞ for one canonical image of B, then the ratior2/r1 is the same for all canonical images of B, see ([27], Pg. 292). The number

M(B) = log(r2r1

),

which then determines the conformal equivalence class of B, is called the modulus of B. Onthe other hand, if r1 = 0 or r2 = ∞ for any canonical image, then we define M(B) = ∞.This happens if and only if at least one boundary component of B consists of a single point.

Definition G.2. Continuity of the Modulus. A sequence of sets En is said to convergeuniformly to a set E if, for each ε > 0, there exists an N such that n > N implies each pointof En lies within distance ε of E and each point of E lies within distance ε of En.

The next result is know as the continuity property for the modulus of a ring. For moredetails see [16], pg. 367 .

Lemma G.3. Let Rn be a sequence of rings and let R be a bounded ring. If each of thecomponents of ∂Rn converges uniformly to the corresponding component of ∂R, then

M(R) = limn→∞

M(Rn)

74

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Definition G.4. Quasiconformal Mappings. A sense preserving homeomorphism f ofa domain A is a K-quasiconformal ( K ≥ 1 ) if and only if M(B) ≤ KM(f(B)) for anyarbitrary ring domain B, B ⊆ A.

For more details about this definition see [20], pg. 13 . For these kinds of mappings we havethe following generalized extension theorem. For the details of the proof see [21], pg. 42.

Theorem G.5. Extension Theorem for Quasiconformal Mappings. A quasicon-formal mapping of a Jordan domain onto another Jordan domain can be extended to ahomeomorphism between the closures of the domains.

Lemma G.6. Assume ψ : (D, 0) → (D, 0) is a K-quasiconformal homeomorphism. For eachr ∈ (0, 1) set R = D \ [0, r]. Then

M(R) ≤ KM(ψ(R))

Proof. Fix r ∈ (0, 1). There exists n0 such that n ≥ n0 implies 1n< 1− r. For n ≥ n0 let Rn

be the bounded ring domain whose boundary is the disjoint union of the line segment [0, r]and the circle z : |z| = 1− 1

n. Clearly Rn converges uniformly to the bounded ring domain

R. Moreover by the definition of quasiconformal map, we have

M(Rn) ≤ KM(ψ(Rn))

From this inequality and the continuity of the modulus it is enough to prove that the bound-ary of ψ(Rn) converges uniformly to the boundary of ψ(R). But from the definition of thesequence Rnn≥n0 it is enough to show that ψ(z : |z| = 1− 1

n) converges uniformly to ∂D.

Since D is a Jordan domain and ψ is quasiconformal we can extend ψ homeomorphically tothe boundary; we also call this extension ψ. On the other hand, let En be the set defined byEn = ψ(z : |z| = 1− 1

n).

Given ε > 0 there exists N such that n > N implies dist(x, ∂D) < ε for all x ∈ En. Otherwisethere exist ε0 > 0 and a sequence (znk

) satisfying

|znk| = 1− 1

nk

and dist(ψ(znk), ∂D) > ε0;

we may assume (znk) converges to a point a ∈ ∂D. So using the extension of ψ we get

dist(ψ(a), ∂D) = limk→∞

dist(ψ(znk), ∂D) ≥ ε0, which is a contradiction since ψ(a) ∈ ∂D. Now,

given y ∈ ∂D by the extension there exists a unique x ∈ ∂D such that ψ(x) = y. Since thesequence of circles z : |z| = 1− 1

n converges uniformly to ∂D there exists a sequence (zn)

which converges to x and satisfies |zn| = 1− 1n. Then lim

n→∞dist(y, En) ≤ lim

n→∞|ψ(x)−ψ(zn)| =

0. Define gn(y) = dist(y, En); clearly this is a sequence of continuous functions on ∂D andfor each y ∈ ∂D the sequence (gn(y)) converges monotonically (decreasing) to zero. Since∂D is compact we can apply Dini’s Theorem. Thus dist(y, En) converges uniformly in y tozero. Therefore En converges uniformly to ∂D and the lemma follows.

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The following result is known as Grotzsch’s Modulus Theorem. For the proof see [21], pg.54.

Theorem G.7. Grotzsch’s Modulus Theorem Suppose r ∈ R with 0 < r < 1. If thering domain B separates the points 0 and r from the circle |z| = 1 then

M(B) ≤M(B(r))

where B(r) is the ring domain whose boundary consists of the circle |z| = 1 and the segment0 ≤ x ≤ r of the real axis.

This ring domain is called the Grotzsch extremal domain. From now on, we denote M(B(r))by µ(r). For any r ∈ (0, 1), the following equality holds :

µ(r) =π

2

K(√

1− r2)

K(r),

where K(r) is the Legendre normal integral given by

K(r) =

∫ 1

0

dt√

1− t2√

1− r2t2

This means that µ(r) can be expressed in terms of elliptic integrals. For the proof of the lastequality above see [21], pg. 59-60. Now, it is easy to check that µ is a continuous functionon (0, 1) satisfying µ(0+) = +∞ and µ(1−) = 0. By an elementary computation (see [1],pg. 50)

d

dr(K(√

1− r2)

K(r)) =

−π2r(1− r2)(K(r))2

Therefore, we conclude that µ is a strictly decreasing homeomorphism from (0, 1) onto(0,+∞).

Lemma G.8. The function h(r) = µ(r) + log( r1+r′

), where r′ =√

1− r2, is strictly decreas-ing.

Proof. . By a computation

d

dr(h(r)) =

π2

4rr′K2(4r′K2

π2− 1) < 0

where K(r) is the Legendre normal integral .

Lemma G.9. The function f(r) = µ(r) + log(r) is strictly decreasing .

Proof. Clearly f(r) = h(r) + log(1 + r′) where r′ =√

1 + r2 . Since log(1 + r′) is strictlydecreasing, the monotonicity of f follows .

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Lemma G.10. The function µ(r)/ log(4/r) is strictly decreasing on (0,1).

Proof. This follows from the lemma above, since

log(4/r)

µ(r)= 1 +

log(4)− (µ(r) + log(r))

µ(r)

78

List of symbolsC∗ = C \ 0.D = ∆(0, 1) := z ∈ C : |z| < 1.H = UpperHalfP lane := Z ∈ C : =(Z) > 0.∆(z0, r) := z ∈ C : |z − z0| < r.∆[z0, r] := z ∈ C : |z − z0| ≤ r.(a, b):= the open interval with extremes a and b[0, z]:= the closed line segment with extremes 0 and z .l = iR

⋂H := it ∈ C : t ≥ 0.

e=exp(1)f : (C, 0) → (C, 0) , means that f leaves the point z=0 fixed.[[α]]:=integer part of α.α:=fractional part of α||α||:=min|n− α| : n ∈ Z.dist(y, V ):=inf|y − v| : v ∈ V

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