chaotic dynamics of quadratic maps · 2004. 5. 7. · chaotic dynamics of quadratic maps 1...

Chaotic dynamics of quadratic maps

Fernando Jorge S. Moreira

1992

Faculdade de Ciências do PortoDepartamento de Matemática

Praça Gomes Teixeira4050 Porto - Portugal

e-mail: [email protected]

I wish to thank Maria Carvalho and Marcelo Vianafor the attention they gave me during the elabora-tion of this work. I also thank INIC for its partialfinantial support.

Chaotic dynamics of quadratic maps

1 Introduction

In this work we address ourselves to the study of the dynamical properties of thefamily of quadratic maps on the real line fa : R −→ R, x 7−→ 1− ax2, a > 0.

In spite of their simple mathematical formulation, these maps may exhibit verycomplicated dynamical properties, namely in what concerns the asymptotic be-haviour of orbits, i.e. the behaviour of iterates fna (x), when n → +∞. Moreover,such features may change in a dramatic way under variation of the parameter a.This is, of course, related to the fact that for large n, fna (x) being a high degreepolynomial, depends in a complicated way on x and a.

On the other hand, and this is a major motivation for the study of this particularfamily of transformations, the quadratic family can be used as a model for thedescription of such dynamics in much more general situations. See [V] and [MV]for an appropriate setting of this generalization.

A crucial role in the description of the (typical) behaviour of fa is played by theorbit of the unique critical point ξ0 = 0. In this description we also make use of thetwo fixed points −qa < 0 < pa, given by

−qa = −1−√

1 + 4a

2aand pa =

−1 +√1 + 4a2a

.

The situation for a > 2 is comparatively simple. One can easily check thatξn(a) = f

na (ξ0) → −∞ as n → +∞. Actually, the same is true for the iterates of

any “typical” point x ∈ R. More precisely, one haslim

n→∞ fna (x) = −∞ ⇐⇒ x 6∈ Ka

where Ka denotes the set of points x ∈ R whose iterates fna (x) belong to [−qa, qa]for every positive integer n. On the other hand Ka is a (nowhere dense) Cantor set

1

with zero Lebesgue measure. Moreover this exceptional set Ka is a hyperbolic setfor fa: there exist C > 0 and θ > 1 such that

∣∣∣(fna )′(x)

∣∣∣ ≥ Cθn for all x ∈ Ka and n ≥ 1 .

A remarkable consequence of this fact is the stability of the dynamics of fa|Kaunder small variations of the parameter: for any b, sufficiently close to a, fb|Kb isconjugate to fa|Ka , i.e. there exists a homeomorphism h = ha,b : Ka −→ Kb suchthat h ◦ fa|Ka = fb|Kb ◦ h.

For 0 < a ≤ 2 the interval [−qa, qa] is positively invariant under fa, that is,fa([−qa, qa]) ⊂ [−qa, qa]. It continues to be true that

x 6∈ [−qa, qa] =⇒ limn→∞ f

na (x) = −∞

but the asymptotic behaviour of points in [−qa, qa] is more interesting. A fairly easypicture of the dynamics is still possible if fa has an attracting periodic point, i.e. apoint p ∈ [−qa, qa] such that

fka (p) = p and∣∣∣(fka )

′(p)

∣∣∣ < 1 for some k ≥ 1 .

(This holds, for instance, if 0 < a < 34: just take p = pa and k = 1). We denote by

Ua the basin of attraction of the orbit of p

Ua ={x ∈ [−qa, qa] : dist

(fna (x), {p, fa(p), . . . , fk−1a (p)}

)→ 0 as n → +∞

}.

It is a classical result, going back to the works of Fatou and Julia (see also Singer[S]), that Ua must contain the critical orbit {ξn(a) : n ≥ 0}. In fact, Ua is an openand dense full Lebesgue measure subset of [−qa, qa]. Moreover Ka = [−qa, qa]\Ua isa hyperbolic (Cantor) set for fa with null Lebesgue measure.

Let, furthemore, H denote the set of values of a ∈ (0, 2] for which such an at-tracting periodic orbit exists. This is easily seen to be an open set and a famousconjecture asserts that H should be dense in (0, 2]. (Very recently has been an-nounced a positive solution of this conjecture by Swiatek). This means that, verylikely, the type of asymptotic behaviour described above, corresponding to the a-values in H (periodic or hyperbolic behaviour), is the “typical” one among quadraticmaps at least from a topological point of view.

Yet, the most recent developments in this area, mostly motivated by experimentaldata, have been putting increasing emphasis on a rather different point of view: thatof measure theory. More precisely, one is interested in the study of those phenomenawhich are measure-theoretical persistent in the sense that they occur for a positive

2

Lebesgue measure of parameter values. Measure-theoretical persistence means thatthe phenomena is statistically detectable in the physical observation of the dynamicalsystem and so this notion seems to be better fitted in the analysis of relevant featuresof a system.

Jakobson proved, in his celebrated paper [J], that the existence of an absolutelycontinuous invariant measure is a measure-theoretical persistence property in thequadratic family. In other words, there exists a positive Lebesgue measure setE ⊂ (0, 2] such that for every a ∈ E there exists a finite Borel measure µa on[−qa, qa] satisfying

µa(f−1a (B)) = µa(B) for every Borel set B ⊂ [−qa, qa]

(invariance with respect to fa) and

µa(B) = 0 if |B| = 0(absolute continuity with respect to Lebesgue measure on R, denoted by | |). Itis not difficult to check that E must be disjoint from H and so it follows fromJakobson’s theorem the nonexistence of attracting periodic orbits (nonperiodic orchaotic behaviour) is a measure-theoretical persistent phenomenon in the quadraticfamily. This is the all the more remarkable that the set E must be constructed ina very careful way and the dynamics of each fa, a ∈ E is highly unstable (if theprevious mentioned conjecture holds then any such fa is approximated by maps fbexhibiting attracting periodic orbits).

A closely related result was later proved by Collet-Eckmann in [CE]. In fact,they considered a slightly different family of transformations.

In [BC1] Benedicks and Carleson provided a new proof of Jakobson’s theorem.Their approach is based on a detailed study of the critical orbit, to prove that itexhibits a growing derivative. More precisely, they get, for a positive Lebesguemeasure set of parameter values, that

(1) |Dn(a)| =∣∣∣(fna )

′(1)∣∣∣ ≥ e

√n for every n ≥ 1 .

Clearly, this property prevents the existence of attracting periodic orbits for fa:by the previously mentioned result of Singer [S], such an orbit would contain inits basin the critical value ξ1(a) = 1, implying that |Dn(a)| → 0 as n → +∞.From the machinery to prove (1) they were able to construct absolutely continuousinvariant measures for the fa corresponding to such a-values (See [A] for a completeexposition).

This approach has, since then, been applied with success to several other sit-uations of complicated dynamics ([BC2], [BY1], [MV], [V], [BY2]). Actually,

3

in order to extend their arguments to higher dimensions, more precisely to thequadratic - or Hénon - family of diffeomorphisms on the plane, Benedicks and Car-leson [BC2] were forced to prove a much stronger statement than (1), namely anexponential growth of the derivative on the critical orbit

(2) |Dn(a)| =∣∣∣(fna )

′(ξ1)∣∣∣ ≥ ecn for every n ≥ 1 and some c > 0 .

The proof of (2) is considerably complicated and it is spread between [BC1]and [BC2], combining ideas from the former with new ingredients introduced inSection 2 of the later paper. In view of the great importance of this property (mostespecially in applications in higher dimensions), it seems useful to try and give acomplete and coherent presentation of its proof. This is our main objective here andso, in the forthcoming sections we prove

Theorem A Given 0 < c < log 2 and 0 < ρ < 1 there exists a0 = a0(ρ, c) ∈ (1, 2)and a subset E(c, ρ) of [a0, 2], such that

(A1) |E(c, ρ)| ≥ (1− ρ)(2− a0)(A2) For a ∈ E(c, ρ) one has |Dn(a)| ≥ ecn for every n ≥ 1.

Our exposition of the proof of the result is based on (besides the aforementionedpapers of Benedicks-Carleson) a set of handwiritten seminar notes [CMV] of IMPA/ Rio de Janeiro.

We remark that the growth of the derivative on the critical orbit is directlyrelated to sensitivity of orbits of fa on their initial point (which is usually taken asthe most characteristic feature of chaotic dynamics). In fact, one can prove fromTheorem A that the following property (introduced by Guckenheimer in [G]) holdsfor every a ∈ E: there is ε > 0 such that given any x ∈ [−qa, qa] and a neighbourhoodV of x there exists y ∈ V and n ≥ 1 such that |fna (x)− fna (y)| ≥ ε.

2 How does the proof work

We fix, without loss of generality, 23

< c < log 2 and 0 < ρ < 1. We take c0 =12(c + log 2), and choose constants α, β, ², such that

0 < 80² < α < β <1

48

(1− c

c0

)

and an integer n0 such that

∞∏

j=n0

(1− e−²j) ≥ 1− ρ2

4

and ∞∑

j=n0

e−²j ≤ ρ2

.

We also will make use of two integers ∆ and N = N(∆) sufficiently large in orderto get validity on the estimates of the proof, and put δ = e−(∆+1).

In our proof, we will construct, inductively, a decreasing sequence of sets (Ωn)n∈N ,such that Ω1 = [a0, 2] and every a ∈ Ωn has the exponential growth

|Dj(a)| ≥ ecj for j = 1, . . . , n . (EGn)Besides this property, for every n ≥ 1, Ωn (every a ∈ Ωn) also satisfies the followingbasic assumption:

|ξn(a)| ≥ e−αn for i = 1, . . . , n . (BAn)Therefore, taking

Ω∞ =⋂

n∈NΩn ,

Ω∞ satisfies (A2) of Theorem A. In a first approach, the imposition of (BAn) willbe useful to avoid that ξn(a) = 0 and so, |Dn(a)| 6= 0, for a parameter a satisfying(EGn−1). However it is important to emphasize that (BAn) will give rise to stongerproperties over the set Ωn.

We will find, for N sufficiently large, a non degenerate interval Ω = [aρ, 2] suchthat every a ∈ Ω satisfies the condition (EGN−1) and (BAN−1). This initial setwill be obtained by the continuity of a 7−→ |Dj(a)|, at a = 2 and the fact that|Dj(2)| = 4j > ec0j. After this, we construct inductively the sequence of sets puttingΩj = Ω for the first N − 1 iterates and, for n ≥ N , assuming constructed a set Ωn−1satisfying (EGn−1) and (BAn−1), we remove from Ωn−1 parameter values a that donot satisfy |Dn(a)| ≥ ecn or |ξn(a)| ≥ e−αn in order to get (EGn) and (BAn). Thisremoving of parameters is made as follows. To each a ∈ Ωn−1 we will associate twofinite sequences (µi)i=0,...,s+1 and (pi)i=0,...,s with µ0 = 1, µs+1 = n, p0 = −1 and fori = 0, . . . s− 1, 1 ≤ µi + pi + 1 < µi+1 ≤ n. Putting

qi = µi+1 − (µi + pi + 1) for i = 0, . . . , s− 1

qs =

{0 if n ≤ µs + psµs+1 − (µs + ps + 1) if n ≥ µs + ps + 1

the two sequences have the following properties∣∣∣(f qia )

′(ξµi+pi+1(a))∣∣∣ ≥ ec0qi for i = 0, . . . , s− 1 (1)

∣∣∣(f qsa )′(ξµs+ps+1(a))

∣∣∣ ≥ δec0qs (2)

5

and∣∣∣(fpi+1a )

′(ξµi(a))

∣∣∣ ≥ 1 for i = 0, . . . , s (3)∣∣∣(fka )

′(ξµi+1(a))

∣∣∣ ≥ 1B

eck ∀k = 1, . . . , pi for i = 0, . . . , s, (4)

with B = B(α − β). Under these conditions, by the Chain Rule, we have that forn > µs + ps,

|Dn(a)| =s∏

i=0

∣∣∣(f qia )′(ξµi+pi+1(a))

∣∣∣ ·∣∣∣(fpi+1a )

′(ξµi(a))

∣∣∣

and for n ≤ µs + ps ,

|Dn(a)| = |(f ′a)(ξµs(a))|·∣∣∣(fn−µsa )

′(ξµs+1(a))

∣∣∣·s−1∏

i=0


∣∣∣ ·∣∣∣(f pi+1a )

′(ξµi(a))

∣∣∣ ;

it follows that if we define

Fn(a) = q0 + . . . + qs

we obtain

|Dn(a)| ≥ δec0Fn(a) if n ≥ µs + ps + 1 (5)

and

|Dn(a)| ≥ 2B|ξµs(a)|ec(n−µs−ps)ec0Fn(a) if n ≤ µs + ps . (6)

We form Ωn excluding parameters a ∈ Ωn−1 such that Fn(a) < (1 − α)n and|ξn(a)| < e−αn and thus, Ωn satisfies (BAn) and

Fn(a) ≥ (1− α)n . (FAn)

By the choice of the constants, (c− 3α)c0 ≥ c and then, from (5) and (6) we obtainfor every a ∈ Ωn that

|Dn(a)| ≥ δeαn · ecn (n > µs + ps)

|Dn(a)| ≥ 2B

eαn · ecn (n ≤ µs + ps) .

6

Since N is sufficiently large, we have that the first factor of this two inequalities isgreater or equal than 1, and so, |Dn(a)| ≥ ecn. Thus, taking into account the choiceof the initial interval Ω = [a0, 2], every a ∈ Ω∞ satisfies (EGn) ∀n ∈ N , and theassertion (A2) of Theorem A is proved taking E(c, ρ) = Ω∞. The assertion (A1) isalso verified by this choice, since, as we will see in Section 6, the sets Ωn satisfy therecurrence relation

|Ωn| ≥ (1− e−²n)|Ωn−1| − e−²n|Ω| (n > n0) ,

and thus, for every n > n0,

|Ωn| ≥

n∏

j=n0

(1− e−²j)−n∑

j=n0

e−²j · |Ω|

≥ (1− ρ)|Ω| by the choice of n0 .

So,

|Ω∞| ≥ (1− ρ)|Ω| .

The crucial argument in the proof is the split of the orbit of a parameter a,{ξk(a), k ≥ 1}, into the pieces corresponding to the periods of time:

{µi}⋃{µi + 1, . . . , µi + pi}

⋃{µi + pi + 1, . . . , µi+1 − 1} ,

that will be called respectively, returns, bound periods after the return µi and freeorbit before the return µi+1. Returns correspond, as the word says, to times when theorbit returns close to ξ0. The free orbit corresponds to times k such that ξk(a) staysaway (outside a small neighbourhood) from the critical point ξ0. This subdivision ismotivated by the knowledge that when ξk(a) stays away from a small neighbourhoodof ξ0 we have an expanding behaviour (exponential increasing on |Dk(a)|). However,we cannot avoid the orbit of the critical point from returning close to itself (the setof parameters for which we have not such returns has null Lebesgue measure). Ata return µ, since |Dµ(a)| = 2a|ξµ(a)||Dµ−1(a)|, we have a loss in the exponentialincreasing. But by imposing the condition (BAµ) we assure that the factor 2a|ξµ(a)|is not too small and it makes sense to try to compensate it with the forward iterates.Indeed, ξµ+k(a) and ξk(a) remain close to each other (we will define later this kindof shadowing in a precise way) by a period of time that is as long as ξµ(a) is near ξ0.Once more, because (BAµ) does not allow ξµ(a) to come too near ξ0, this shadowingperiod is not so long, and in fact, we will prove that it is less than n. This willcorrespond to the bound period. Since ξµ+k(a) is close to ξk(a) for k = 1, . . . , p,

7

with p < n, we can use, by an induction argument, the expanding behaviour of{ξk(a), k < µ} to obtain an analogous expanding for {ξµ+k(a), k ≤ p} and thenovercome the small factor introduced at the return µ at the iterate µ + p + 1.

The assumption (FAn) ensures that the time spent in bound periods is too smallcompared with the total time n and so we have the required expanding behaviour(EGn).

The formal definition of the sets (Ωn)n∈N , and the concepts of returns, boundperiods and free orbits that we have heuristically introduced will be given in Sec-tion 5. The properties for the bound period are developed in Section 4. To get aprecise meaning of “close to” and “away from” the critical point ξ0, we introduceneighbourhoods of ξ0 given by

Um = (−e−m, e−m) and U+m = (−e−(m−1), e−(m−1)), for m ≥ ∆− 1and consider also for m ≥ ∆− 1

Im = [e−(m+1), e−m) and I+m = [e

−(m+1), e−(m−1)) .

It will be useful to extend this definitions for m ≤ −(∆ − 1) by Um = U−mand Im = −I−m, and analogous equalities for U+m and I+m. Notice that we haveU∆\{0} = ⋃|m|≥∆ Im, and for |m| ≥ ∆, I+m = Im−1

⋃Im

⋃Im+1 ⊂ U+m.

Since we will not study separately the iterations of a parameter but the itera-tions of small parameter intervals, the notions to be introduced, like returns, boundperiods or free orbits, must be constant in small intervals. For example the notionof a return cannot be like “n is a return iff ξn(a) ∈ U∆”, because for a small intervalω we can have portions of ω inside and outside U∆.

3 Preliminary results

Throughout the proof we will use repeatedly information about the (expanding)behaviour of fa, for a close to 2, in pieces of orbit outside U∆, which will be obtainedby a pertubation estimate from the case a = 2. In order to do that, let us definethe homeomorphism h of the interval [−1, 1] given by

[−1, 1] −→ [−1, 1]θ 7−→ sin

(π2θ)

.

h−1 has derivative at every x 6= −1, 1 given by the formula:(h−1

)′(x) =

1√1− x2 .

8

Definingga = h

−1 ◦ fa ◦ hone has

g2(θ) =2

πarcsin

(1− 2sin2

(π

2θ))

=2

πarcsin(cos πθ)

=2

π

(π

2− π|θ|

)

= 1− 2|θ| .Thus, g2 is the tent map θ 7→ 1− 2|θ| and then, it has derivative 2 for every θ 6= 0.So, by the chain rule, |(gk2)′(θ)| = 2k, for every θ whose orbit for i = 0, . . . , k − 1does not hit 0. For a close to 2, the next Lemma gives a similar property for ga.

Lemma 3.1 Given δ < 1 the following holds:(a) If |fa(x)| ≤ 1− δ2 then 2− 3πδ3 (2− a) ≤ |g′a(h−1(x))| ≤ 2 + 3πδ3 (2− a).(b) If |x| ≤ 1− δ2 and |f ia(x)| ≥ δ ∀i = 0, . . . , k − 1 then

[2− 3πδ3

(2− a)]k

≤ |(gka)′(h−1)(x))| ≤ [2 + 3π

δ3(2− a)]

k

Proof: (a) By definition we have that

|g′a(θ)− g′2(θ)| =∣∣∣(h−1)′ (fa(x)) f ′a(x)h′(θ)− (h−1)′ (f2(x)) f ′2(x)h′(θ)

∣∣∣

=∣∣∣(h−1)′ (fa(x)) f ′a(x)− (h−1)′ (f2(x)) f ′a(x)++(h−1)′ (f2(x)) f ′a(x)− (h−1)′ (f2(x)) f ′2(x)

∣∣∣ · |h′(θ)|≤

(|f ′a(x)||(h−1)′ (fa(x))− (h−1)′ (f2(x)) |+

+|(h−1)′ (f2(x)) ||f ′a(x)− f ′2(x)|)|h′(θ)| .

Attending to |f ′a(x)| ≤ 4, |h′(θ)| ≤ π2 and |f ′a(x)− f ′2(x)| ≤ 2|a− 2|, it follows that|g′a(θ)− g′2(θ)| ≤ 2π

∣∣∣(h−1)′ (fa(x))− (h−1)′ (f2(x))∣∣∣

+π∣∣∣(h−1)′ (f2(x))

∣∣∣ · |a− 2| . (7)Taking into account that, for |y| ≤ 1− δ2,

∣∣∣(h−1)′(y)∣∣∣ ≤ 1√

1− (1− δ2)2=

1√2δ2 − δ4

≤ 1δ

9

An immediate consequence of the Lemma 3.1 is the

Corollary 3.2 Given 0 < c1 < log 2 and δ < 1 there exists a1 = a1(δ, c1) such thatfor a ≥ a1 the following holds:(a) If |fa(x)| ≤ 1− δ2 then |g′a(h−1(x))| ≥ ec1.(b) If |x| ≤ 1− δ2 and |f ia(x)| ≥ δ for i = 0, . . . , k − 1 then |(gka)′(h−1)(x))| ≥ ec1k.Proof: It is sufficient to take a1 ≥ 2− δ33π (2− ec1). 2

Now, we are in condition to state the Proposition that establishes in a precise waythat fa has an expanding behaviour outside a small neighbourhood of the criticalpoint.

Proposition 3.3 For every 0 < c0 < log 2 and ∆ sufficiently large there exists1 < a1 < 2 such that for every x ∈ [−1, 1] and a ∈ [a1, 2] one has:(a) If x, fa(x), f

2a (x), . . . , f

k−1a (x) 6∈ U∆+1 then

∣∣∣∣(fka

)′(x)

∣∣∣∣ ≥ δec0k (δ = e−(∆+1)).(b) If x, fa(x), f

2a (x), . . . , f

k−1a (x) 6∈ U∆+1 and fka (x) ∈ U+∆ then

∣∣∣∣(fka

)′(x)

∣∣∣∣ ≥ ec0k.(c) If x, fa(x), f

2a (x), . . . , f

k−1a (x) 6∈ U∆+1 and fka (x) ∈ U1 then

∣∣∣∣(fka

)′(x)

∣∣∣∣ ≥ 45ec0k.

Proof: Take a1 = a1(δ, c1) given in Corollary 3.2, where c0 < c1 < log 2, and let pbe the first integer 0 ≤ p ≤ k − 1 such that |fpa (x)| < 1 − δ2 (if such integer doesnot exist take p = k). Since

∣∣∣(fka )′(x)

∣∣∣ =∣∣∣(fk−pa )

′(fpa (x))

∣∣∣ ·∣∣∣(fpa )

′(x)∣∣∣ ,

if we conclude that for a > a1∣∣∣(f pa )

′(x)∣∣∣ ≥ 2p (I)

and ∣∣∣(fk−pa )′(x)

∣∣∣ ≥ Lec0(k−p) (II)where we can choose L = δ (respectively L = 1, L = 4

5) under the hypothesis of (a)

(respectively under the hypothesis of (b), (c)), then the Proposition is proved. So,the rest of the proof will be reduced to get (I) and (II) above.

11

proof of (I): If p = 0 the inequality is obvious. So, assume that p ≥ 1. Then∣∣∣(f pa )

′(x)∣∣∣ =

p−1∏

i=0

∣∣∣2a(f ia(x))∣∣∣

≥ 3p(1− δ2)p since a ≥ 32

≥ 2p since ∆ ≥ 1

proof of (II): (a) As before it is enough to prove this inequality for p < k.Taking by convenience of notation q = k − p and y = fpa (x), we have that

|(f qa)′(y)| =∣∣∣h′(gqa(h−1(y))

∣∣∣ ·∣∣∣(gqa)′(h−1(y))

∣∣∣ ·∣∣∣(h−1)′(y)

∣∣∣

=|(h−1)′(y)|

|(h−1)′(f qa(y))| ·∣∣∣(gqa)′(h−1(y))

∣∣∣ .

By the choice of p, setting

L =|(h−1)′(y)|

|(h−1)′(f qa(y))| ,

Corollary 3.2 leads to∣∣∣(f qa)

′(y)∣∣∣ ≥ Lec1q . (11)

Now the previous Claim says us that |f qa(y)| ≤ 1− δ2, and so, as y 6∈ U∆+1

L =

√√√√1− (f qa(y))21− y2 ≥

√√√√1− (1− δ2)21− δ2 ≥ δ .

(b) If fka (x) ∈ U∆−1, then |f qa(y)| = |fka (x)| ≤ e−∆+1 and

L ≥√√√√1− (e−∆+1)2

1− (e−∆−1)2 = L1(∆) .

Sincelim

∆→∞L1(∆) = 1

we can find ∆0 such that ∆ ≥ ∆0 implies L1(∆) ≥ ec0−c1 ≥ e(c0−c1)q. Therefore,from (11) we conclude that ∣∣∣(f qa)

′(y)∣∣∣ ≥ ec0q .

12

(c) If fka (x) ∈ U1, then

L ≥√

1− 1e2≥

√3

4≥ 4

5. 2

The definition that will be given in Section 5 for free orbits up to n, {µi + pi +1, . . . , µi+1 − 1}, leads us to conclude that ξk(a) is outside U∆+1 during this period.This together with the previous Proposition leads to


∣∣∣ ≥ δec0qi .

Moreover, the definition of returns will imply that for i < s, ξµi+1(a) ∈ U∆−1 andconsequentely

f qia (ξµi+pi+1(a)) ≥ ec0qi (i < s) .The next proposition shows that under strong growth of |Dn(a)|, the parameter

and space derivatives are comparable.

Proposition 3.4 Given 23

< c < log 2 there exists N0 such that for every n ≥ N0:

(a) |Dj(a)| ≥ 3j for 1 ≤ j ≤ N0(b) |Dj(a)| ≥ ecj for 1 ≤ j ≤ n− 1

}⇒ 1

A≤ |ξ

′n(a)|

|Dn−1(a)| ≤ A

where A = 8.

Proof: From the Chain Rule, for k ≥ 1, setting f(a, x) = fa(x),

Dk(a) =∂f

∂x

∣∣∣∣∣(a,ξk(a))

·Dk−1(a)

= 2aξk(a)Dk−1(a)

ξ′k+1(a) =∂f

∂x

∣∣∣∣∣(a,ξk(a))

· ξ′k(a) +∂f

∂a

∣∣∣∣∣(a,ξk(a))

= 2aξk(a)ξ′k(a)− (ξk(a))2 .

Therefore, whenever Dk(a) > 0

ξ′k(a)Dk−1(a)

− ξ′k+1(a)

Dk(a)=

ξk(a)

2aDk−1(a). (12)

13

After summation of both sides of (12) over k = 1, . . . , n− 1 we get

ξ′1(a)D0(a)︸︷︷︸

=1

− ξ′n(a)

Dn−1(a)=

n−1∑

k=1

ξk(a)

2aDk−1(a).

Taking modulos we obtain:

∣∣∣∣∣ξ′n(a)

Dn−1(a)− 1

∣∣∣∣∣ ≤n−1∑

k=1

|ξk(a)|2a|Dk−1(a)| ≤

n−1∑

k=1

1

2|Dk−1(a)|hence

1− 12

n−1∑

k=1

1

|Dk−1(a)| ≤∣∣∣∣∣

ξ′n(a)Dn−1(a)

∣∣∣∣∣ ≤ 1 +1

2

n−1∑

k=1

1

|Dk−1(a)| . (13)

Let N0 be an integer ≥ 2 such that ∑∞k=N0+1 e−23k ≤ 1

4. Then, if |Dk(a)| ≥ 3k,

for every k = 0, . . . , N0 and |Dk(a)| ≥ e 23k for every k = N0 + 1, . . . , n− 2, we have:n−1∑

k=1

1

|Dk−1(a)| ≤N0∑

k=0

1

3k+

n−2∑

k=N0+1

1

e23k

≤∞∑

k=0

1

3k+

∞∑

k=N0+1

1

e23k

≤ 32

+1

4

≤ 74

From (13) it follows immediately that:

1

8≤

∣∣∣∣∣ξ′n(a)

Dn−1(a)

∣∣∣∣∣ ≤15

8.2

It will be useful to emphasize that if we have an interval ω such that everya ∈ ω satisfies (a) and (b) of the previous Proposition then ξ′k does not vanish inω for k = N0, . . . n, since |ξ′k(a)| ≥ 1A |Dk−1(a)|. In particular ξk restricted to ω is

14

a homeomorphism. The same holds for the closure ω of ω. This simple assertionwill play a fundamental role in the inductive construction of the sets Ωn, n ≥ N .Moreover, for such ω and i, j between N0 and n we can define the function

ψ : ξi(ω) −→ ξj(ω)x 7−→ ξj ◦ ξ−1i (x) . (14)

whose derivative is given by

ψ′(ξi(t)) =ξ′j(t)

ξ′i(t), t ∈ ω . (15)

Lemma 3.5 Suppose that ω ⊂ [1, 2] is an interval satisfying (a) and (b) of Propo-sition 3.4. Then, for every integers i, j such that N0 ≤ i ≤ j ≤ n we have that:

(i) ∀a, b ∈ ω 1A2

·∣∣∣(f j−it )

′(ξi(t))

∣∣∣ ≤ |ξj(a)− ξj(b)||ξi(a)− ξi(b)| ≤ A2

∣∣∣(f j−it )′(ξi(t))

∣∣∣ , t ∈ ω

(ii)1

A2·∣∣∣(f j−it )

′(ξi(t))

∣∣∣ ≤ |ξj(ω)||ξi(ω)| ≤ A2

∣∣∣(f j−it )′(ξi(t))

∣∣∣ , t ∈ ω .

Proof: (i) By the Mean Value Theorem defining ψ as in (14) we have that, for somet between a and b (it can be taken distinct from a and b),

|ξj(a)− ξj(b)| = |ψ′(ξi(t))| | · |ξi(a)− ξi(b)| .

Now, from equality (15) and the previous Proposition, it follows that

1

A2· |Dj−1(t)||Di−1(t)| ≤ |ψ

′(ξi(t))| ≤ A2 · |Dj−1(t)||Di−1(t)| .

And so, we have the assertion, since

∣∣∣(f j−it )′(ξi(t))

∣∣∣ = |Dj−1(t)||Di−1(t)| .

(ii) The result follows immediately from (i) since for k ≥ N0, ξk|ω is a homeomor-phism and so, |ξk(ω)| = |ξk(a)− ξk(b)|, where a and b are the extreme points of theclosure of ω (notice that in (i), t can be choosen different from a and b). 2

Next Proposition will give us the parameter interval to begin later our construc-tion of the sets (Ωn)n∈N .

15

Proposition 3.6 Given a1 such that ec0 < a1 < 2 and N1 ≥ N0, there exists an

integer N ≥ N1 and an interval Ω ⊂ [a1, 2] such that:

(a) ξj(Ω) ∩ U1 = ∅ for every 1 ≤ j ≤ N − 1(b) ξN(Ω) ⊃ U1(c) |Dj(a)| ≥ 3j for every a ∈ Ω and 1 ≤ j ≤ N0(d) |Dj(a)| ≥ ec0j for every a ∈ Ω and 1 ≤ j ≤ N − 1 .

Proof: For 1 ≤ n ≤ N1, setΦn : [1, 2] −→ [−1, 1]× [0, +∞)

a 7−→ (ξn+1(a), |Dn(a)|)We have Φn(2) = (−1, 4n), so attending to the continuity of Φn, it follows that thereexists a neighbourhood Vn of 2 in [1, 2] of the form [tn, 2] such that

Φn(Vn) ⊂ [−1,−12 ]× [3n, +∞) .We can take VN1 ⊂ VN1−1 ⊂ . . . ⊂ V2 ⊂ V1 ⊂ [a1, 2]. So for every 1 ≤ n ≤ N1 andevery a ∈ VN1

ξn+1(a) ≤ −12

|Dn(a)| ≥ 3n

At this time VN1 satisfies statement (c). Now we are going to shrink VN1 to obtainthe other properties. We first note that while ξj(a) ≤ −12 for every 1 ≤ j ≤ k, wehave:

|Dk(a)| =k∏

j=1

2a|ξj(a)|

≥ ak≥ ec0k

since a ≥ ec0 . Thus, if ξj(VN1) is contained in [−1,−12 ] for every j < n then anya ∈ VN1 satisfies the hypothesis of Proposition 3.4 and then

|ξn(VN1)| = |ξ′n(t)| · |VN1|, t ∈ VN1≥ 1

A· |Dn−1(t)| · |VN1|

≥ ec0(n−1) · |VN1|A

.

16

Thus while |ξj(VN1)| remains outside [−1,−12 ] for N1 ≤ j ≤ n, we have an expo-nential growth of the length of ξn(VN1) and then there must exist an integer k suchthat ξk(VN1) is not contained in [−1,−12 ]. Let N2 be the first integer in these condi-tions. Since ξn(2) = −1, by the continuity of ξN2 there exists t∗ ∈ [tN1 , 2] such thatξN2(t∗) = −12 . Let Ω = [t∗, 2]. Now, since ξN2+1(2) = −1 and ξN2+1(t∗) = 1− t∗4 ≥ 12we obtain, by the continuity of ξN2+1, that ξN2+1(Ω) ⊃ [−12 , 12 ] ⊃ U1. So our propo-sition is proved choosing N = N2 + 1. 2

4 The bound period

In Section 2 we have seen the important role of the subsequent periods associatedto returns, the so called bound periods. We have said that after a return µ, theorbits {ξµ+k(a), 1 ≤ k ≤ p} and {ξk(a), 1 ≤ k ≤ p} are close to each other, but wedid not establish in what sense this happens. In this section we are going to definethis concept in a precise way, and we will obtain the corresponding results used inSection 2 to get the exponential growth of (EG) under (BA) and (FA) assumptions.

For every x ∈ U+m, we have that|fa(x)− ξ1(a)| = ax2

≤ ae−2(∆−1)≤ e−β since ∆ ≥ 2 .

So, it makes sense to define p(a,m) as the maximal p ∈ N = N ⋃{∞} satisfyingthe following binding condition:

∀x ∈ U+m |f ja(x)− ξj(a)| < e−βj for 1 ≤ j ≤ p . (BC)An immediate consequence of this definition is that

{ |f ja(U+m)| ≤ 2e−βj for 1 ≤ j ≤ p(a,m)∣∣∣fp(a,m)+1a (U+m)∣∣∣ ≥ e−βp(a,m)+1 if p(a,m) ∈ N . .

With this definition, we establish the following Lemma which states that ifwe choose as bound period for a return µ with ξµ(a) ∈ I+m, the time interval{µ + 1, . . . , µ + p}, it satisfies the required properties (3) and (4) concerning boundperiods in Section 2. Here we mean by Ωk a set satisfying (EGk) and (BAk) (indeed,this sets will be explicitly constructed in the next section satisfying these proper-ties). To simplify the notation we put Dk(a, x) = (f

ka )′(fa(x)) (notice that then

Dk(a, 0) = Dk(a)).

17

Lemma 4.1 Suppose that a ∈ Ωn−1 and ξn(a) ∈ I+m with ∆ ≤ |m| ≤ [αn] − 1 andput

p = p(a,m) .

Then:

(a) There is a constant B1 = B1(α − β) such that for every x ∈ U+m andk = 1, . . . , p

1

B1≤ |Dk(a, x)||Dk(a)| ≤ B1

(b) p < 3|m|

(c)∣∣∣(fp+1a )

′(x)

∣∣∣ ≥ e(1−4β)|m| for every x ∈ I+m.

Proof:(a) For k = 1, . . . , min{p, n},

|Dk(a, x)||Dk(a)| =

k∏

j=1

2a|f ja(x)|2a|ξj(a)|

=k∏

j=1

[1 +

|f ja(x)| − |ξj(a)||ξj(a)|

](16)

≤k∏

j=1

[1 +

|f ja(x)− ξj(a)||ξj(a)|

]

≤ exp

k∑

j=1

|f ja(x)− ξj(a)||ξj(a)|

a satisfies (BAn) and (BC), and so, since j ≤ n and j ≤ p we have that|f ja(x)− ξj(a)|

|ξj(a)| ≤e−βj

e−αj

and|Dk(a, x)||Dk(a)| ≤ exp

∞∑

j=1

e(α−β)j = B

′1

Now, each fraction of the product of (16) is less than one, so we also have

|Dk(a, x)||Dk(a)| ≥

k∏

j=1

[1− |f

ja(x)− ξj(a)||ξj(a)|

]

18

≥k∏

j=1

(1− e(α−β)j

)

≥∞∏

j=1

(1− e(α−β)j

)= B′′1

Setting

B1 = max

{B′1,

1

B′′1

}

the proof of part (a) is finished provided that we garantee that p < n. See part (b).

(b) Choose x = e−|m| ∈ U+m and j = min{p, n} − 1.∣∣∣f j+1a (x)− ξj+1(a)

∣∣∣ =∣∣∣f ja(fa(x))− f ja(fa(0))

∣∣∣

=∣∣∣(f ja)

′(y))

∣∣∣ |fa(x)− fa(0)| , y ∈ fa(Um)≥ B′1|Dj(a)|ax2 .

The binding condition says us that |f j+1a (x) − ξj+1(a)| < e−β(j+1) so, from the lastinequality, it follows that

B′1|Dj(a)|e−2|m| ≤ e−β(j+1)B′1e

cj−2|m| ≤ e−β(j+1) (a ∈ Ωn−1)and therefore,

j ≤ 2|m| − log B′1

c + β

≤ 3|m| − 1 since ∆ is sufficiently large and c > 23

.

So, j < 3[αn] − 1 < n (α < 14). Thus j = p − 1 and attending to (17) we have, as

required,p < 3|m| .

(c) From the binding condition,

|fp+1a (U+m)| ≥ e−β(p+1)

and, by the Mean Value Theorem, for some y ∈ U+m|fp+1a (U+m)| = |(f p+1a )′(y)||U+m| ,

19

thus

∣∣∣(f pa )′(fa(y))

∣∣∣ |(f ′a(y))| >e−β(p+1)

|U+m|·

So, for every x ∈ I+m,∣∣∣(fp+1a )

′(x)

∣∣∣ =∣∣∣(f pa )

′(fa(x))∣∣∣ |(f ′a(x))|

≥ 1B21

∣∣∣(fpa )′(fa(y))

∣∣∣ |(f ′a(x))|, by (a)

≥ 1B21

· e−β(p+1)

|U+m|· |x||y| ·

Since |x| ≥ e−(|m|+1), |y| < e|m|−1, p < 3|m| and |U+m| < 2e|m|−1, we get∣∣∣(fp+1a )

′(x)

∣∣∣ ≥ e(1−4β)|m| ,

providing that ∆ is sufficiently large so that |m| ≥ ∆ ⇒ eβ|m| ≥ 12e3B21

. 2

Notice that under the hypothesis of this Lemma p is finite (by statement (b)),for k = 1, . . . , p

∣∣∣(fka )′(ξn+1(a))

∣∣∣ ≥ 1B1|Dk(a)| by part (a)

≥ 1B1

eck since a satisfies (BAk)

and, by part (c)

∣∣∣(f p+1a )′(ξn(a))

∣∣∣ ≥ e(1−4β)|m| ≥ 1 .

However our intention is to obtain similar relations when p is constant in smallparameter intervals (in order to obtain bound distorsion of Dn in such intervalswhich will play a fundamental role in the estimation of the excluded set). With thispurpose, for a parameter interval ω such that

ξn(ω) ⊂ I+m , |m| ≥ ∆

definep(ω, m) = min

a∈ω p(a,m) .

20

With this definition, the statements (a) and (b) of Lemma 4.1 follow immediately(p(ω, m) ≤ p(a,m) ∀a ∈ ω). But (c) is more difficult to establish, and this is whatwe are going to do from now on. In the remaining of this section we write ω ∈ Pn−1to mean that ω is an interval contained in Ωn−1, with n sufficiently large. First weprove two Lemmas that are consequence of the exponential growth of |ξk(ω)|, k ≥ 1,under the hypothesis of ω ∈ Pn−1.

Lemma 4.2 If ω ∈ Pn−1 then for every a, b ∈ ω:

(a) |a− b| ≤ 4Ae− 23n

(b)(

a

b

)k≤ e2A ∀k = 1, . . . , n .

Proof:

2 ≥ |ξn(ω)| = |ξ′n(t)||ω|, for some t ∈ ω≥ |ξ′n(t)||a− b|≥ 1

A|Dn−1(t)||a− b| (by Proposition 3.4)

≥ 1A

e−23(n−1)|a− b| (t ∈ Ωn−1)

so|a− b| ≤ 2Ae− 23 (n−1) ≤ 4Ae− 23n.

Finally

(a

b

)k≤

(1 +

|a− b|b

)k

≤ exp{2kAe−

23n}

≤ e2A . 2

Lemma 4.3 If ω ∈ Pn−1 and ξn(ω) ⊂ I+m with ∆ ≤ |m| ≤ [αn]− 1, then for everya, b ∈ ω:

|ξj(a)− ξj(b)| ≤ e−βj for j = 1, . . . , p(ω,m)

Proof: We have seen in (13) that

∣∣∣∣∣ξ′j+1(a)

Dj(a)

∣∣∣∣∣ ≤ 1 +1

2

j∑

i=1

1

|Di−1(a)| .

21

Since ω ∈ Pn−1 and j ≤ p(ω, m) ≤ n− 1 it follows that:∣∣∣∣∣ξ′j+1(a)

Dj(a)

∣∣∣∣∣ ≤ 1 +2

3

∞∑

i=0

1

e23i

= L1 .

Therefore

|ξj(a)− ξj(b)| = |ξ′j(t)||a− b| , t ∈ (a, b)≤ L1|Dj−1(t)||a− b| . (17)

By the Mean Value Theorem,∣∣∣f jt (U+m)

∣∣∣ = |Dj−1(t, y)|∣∣∣ft(U+m)

∣∣∣ , y ∈ U+m≥ 1

B21|Dj−1(t)|

∣∣∣ft(U+m)∣∣∣ .

So, from (17) we have that

|ξj(a)− ξj(b)| ≤ L1B21|a− b||ft(U+m)|

∣∣∣f jt (U+m)∣∣∣ .

Applying to this estimate the fact that∣∣∣ft(U+m)

∣∣∣ ≥∣∣∣ft(e−|m|)− ft(0)

∣∣∣ ≥ te−2m ≥ e−2αn

we obtain finally

|ξj(a)− ξj(b)| ≤ L2∣∣∣f jt (U+m)

∣∣∣ e(2α− 23 )n , L2 = 4AL1B21 . (18)

Attending to the binding condition we get∣∣∣f jt (U+m)

∣∣∣ ≤ 2e−βj

and since n is sufficiently large,

2L2e(2α− 2

3)n ≤ 1 ,

so the result follows immediately from (18). 2

22

Lemma 4.4 If ω ∈ Pn−1 and ξn(ω) ⊂ I+m with ∆ ≤ |m| ≤ [αn]− 1, then settingp = p(ω, m)

we have for every a, b ∈ ω and x, y ∈ U+m|Dj(a, x)||Dj(b, y)| ≤ B3 = B3(α− β) ∀j = 1, . . . , p

Proof: We have

|Dj(a, x)||Dj(b, y)| =

|Dj(a, x)||Dj(a)| ·

|Dj(a)||Dj(b)| ·

|Dj(b)||Dj(b, y)|

≤ B21|Dj(a)||Dj(b)| by Lemma 4.1 .

Therefore it is enough to find a good upper bound for

|Dj(a)||Dj(b)| =

(a

b

)j·

j∏

i=1

|ξi(a)||ξi(b)| .

Since |ξi(a)||ξi(b)| ≤ 1 +

|ξi(a)− ξi(b)||ξi(b)| ,

then

|Dj(a)||Dj(b)| ≤

(a

b

)j j∏

i=1

[1 +

|ξi(a)− ξi(b)||ξi(b)|

]

≤(

a

b

)jexp

j∑

i=1


≤ e2Ak∏

i=1

[1 +


]. (19)

Now, by Lemma 4.3, for every i = 1, . . . , p

|ξi(a)− ξi(b)| ≤ e−βi

and also,

|ξi(b)| ≥ |ξi(a)| − e−βi≥ e−αi − e−βi≥ e−αi(1− e(α−β)j)≥ L1e−αj , L1 = (1− eα−β)−1 .

23

Therefore,

j∑

i=1

|ξi(a)− ξi(b)||ξi(b)| ≤ L1

∞∑

i=1

e(α−β)i < ∞ .

So, applying the last inequality to (19) the result follows immediately. 2

Proposition 4.5 Suppose that ω ∈ Pn−1 and ξn(ω) ⊂ I+m with ∆ ≤ |m| ≤ [αn]. Ifwe put

p = p(ω, m)

then

(a) There is a constant B = B(α−β) such that for every a ∈ ω and x ∈ U+m:1

B≤ |Dk(a, x)||Dk(a)| ≤ B ∀k = 1, . . . , p.

(b) p < 3|m|.

(c)∣∣∣(fp+1a )

′(x)

∣∣∣ ≥ e(1−5β)|m| for every a ∈ ω and x ∈ U+m.

Proof: Let a∗ ∈ ω such that p(ω, m) = p(a∗,m). From (c) of Lemma 4.1,∣∣∣(f p+1a∗ )

′(x)

∣∣∣ ≥ e(1−4β)|m|

and by the previous Lemma,

|Dp(a∗, x)||Dp(a, x)| ≤ B2 .

Therefore∣∣∣(fp+1a∗ )

′(x)

∣∣∣∣∣∣(fp+1a )

′(x)

∣∣∣=

a∗a

|Dp(a∗, x)||Dp(a, x)|

≤ 2B2 .So,

∣∣∣(fp+1a )′(x)

∣∣∣ ≥ 12B2

∣∣∣(fp+1a∗ )′(x)

∣∣∣

≥ 12B2

e(1−4β)|m|

≥ e(1−5β)|m| if ∆ is sufficiently large . 2

24

5 Basic construction

In this section, we are going to define precisely the sets (Ωn)n and for a ∈ Ωn thesequences (µi)i, (pi)i refered at Section 2. First we make a partition of the intervals

Im in the following way. For m ≥ ∆ consider the ratio rm = |Im|m2 and then, for1 ≤ k ≤ m2, we take the intervals

Im,k =[e−m − krm , e−m − (k − 1)rm

).

For m = ∆− 1 and k ≥ 1,

I∆−1,k =[e−∆, e−∆ + r∆−1

), r∆−1 =

|I∆−1|(∆− 1)2 .

We extend these definitions for m ≤ −(∆ − 1) setting Im,k = I−m,k. Therefore for|m| ≥ ∆ we have a partition of Im into intervals of equal length

Im = Im,m2⋃

. . .⋃

Im,1

and Im,k has two adjacent intervals: Im,k−1 and Im,k+1 for Im,k with 1 < k < m2,Im−1,(m−1)2 and Im,2 for Im,1, Im+1,1 and Im,m2−1 for Im,m2 . We set

I+m,k = Im1,k1⋃

Im,k⋃

Im2,k2 ,

where Im1,k1 and Im2,k2 are the adjacent intervals to Im,k. Note that Im,k ⊂ Im,I+m,k ⊂ I+m and |I+m,k| ≤ 3|Im|m2 if k 6= 1 and |I+m,k| ≤ 5|Im|m2 if k = 1, providing ∆ is largeenough. It is useful to consider also the sets I+∆−1,k = (0, 1] and I

+1−∆,k = [−1, 0).

Related to these splitting of U∆ we will define inductively partitions Pn of theparameter intervals in order to have bound distorsion of ξn and Dn−1 on ω ∈ Pn−1.The sets Ωn are then defined as

Ωn =⋃{ω : ω ∈ Pn} .

The bound distorsion on the components of Pn will have most importance to esti-mate the excluded sets Ωn\Ωn−1. For every ω ∈ Pn, we will introduce some concepts,like returns and bound periods, and for a ∈ Ωn we set the same concept to a as theone we associated to the component ω(a) to which a belongs. For example, if wesay that µ is a return for ω then µ is a return for every a ∈ ω.

To begin the induction we need ∆ and N1 = N1(∆) sufficiently large such thatfor all n ≥ N ≥ N1 and |m| ≥ ∆ we get validity on our estimates (in particular

25

N1 ≥ n0). Then we take a parameter a1 close to 2 such that Proposition 3.3 worksand

[2 + 3π

δ3(2− a)

2− 3πδ3

(2− a)

]4∆≤ 2 (20)

2− a ≤ 8e−4∆ (21)

holds for a ≥ a1 Now Proposition 3.6 gives us an integer N ≥ N1 and the startinginterval Ω = [a0, 2] of our induction. We put for i = 1, . . . , N − 1, Ωi = Ω andPi = {Ω}. Now by induction on n ≥ N assume the following assertions are true forevery ω ∈ Pn−1:

1. There is a sequence of parameter intervals Ω = ω1 ⊃ . . . ⊃ ωn−1 = ω suchthat, for k = 1, . . . , n− 1, ωk ∈ Pk.

2. There is a set Rn−1 = Rn−1(ω) = {µ0, . . . , µs}, s = s(n − 1), such thatµ0 = 1. For i = 1, . . . , s, we say that µi is a return for ω and for k < n − 1,Rk(ωk) = Rn−1(ω)

⋂{1, . . . , k}. Notice that when Rn−1 = {1}, ω has noreturns.

3. For any return µi ∈ Rn−1 (i.e. i ≥ 1) we have associated an interval I+mi,ki ,|mi| ≥ ∆, called the host interval of ω at the return µi, such that ξµi(ωµi) iscontained in I+mi,ki . Then, defining pi = p(ωµi ,mi), we call {µi +1, . . . , µi +pi}the bound period associated to the return µi. By convenience of notation wealso put p0 = −1.

4. The periods {µi + pi + 1, . . . , µi+1 − 1} (i < s) and {µs + ps + 1, . . . , n− 1} ifn > µs + ps are called the free period after µi. During these free times j, wehave

ξµi+pi+j(ω)⋂

U∆+1 = ∅ ∀j = 1, . . . , qi ,and so, by Proposition 3.3 we have, for every a ∈ ω, assertions (1) and (2) ofSection 2.

5. For every k = 1, . . . , n − 1, ωk satisfies (BAk) and (EGk). Thus, for everyreturn µi ∈ Rn−1, ωµi satisfies (BAµi) and (EGµi−1), and so, we can applyProposition 4.5 to obtain that pi < 3|mi|. In particular the bound period isfinite. Moreover, since ω ⊂ ωµi , for every a ∈ ω,

∣∣∣(fpi+1a )′(ξµi(a))

∣∣∣ ≥ e(1−5β)|m| ≥ 1

26

so we have (3) of Section 2. By the same Proposition, for j = 1, . . . , pi,

∣∣∣(f ja)′(ξµi+1(a))

∣∣∣ ≥ 1B· |Dj(a)| ,

and then, since j ≤ pi ≤ 3|mi| < 3αµi < µi and a satisfies (EGµi−1),∣∣∣(f ja)

′(ξµi+1(a))

∣∣∣ ≥ 1B· ecj ,

which is assertion (4) of Section 2.

We remark that all these properties are trivially verified for n ≤ N takingRn−1(Ω) = {1}, i.e. there is no returns up to N − 1. Let us see now how theinductive step works. We will make use of an auxiliary partition P ′n, that willcontain portions of ω ∈ P ′n−1, satisfying (BAn). So take ω ∈ Pn−1.(a) If n belongs to a bound period associated to a previous return, we put ω ∈ P ′n

and Rn(ω) = Rn−1(ω). (During bound periods we can have the orbit insideU∆+1).

(b) If ξn(ω)⋂

U∆ ⊂ I∆,1 ⋃ I−∆,1 then we put also ω ∈ P ′n and Rn(ω) = Rn−1(ω).Note that in this situation we have ξn(ω) ⊂ UC∆+1, and so n is a free time.

(c) If the two above conditions do not hold we say that n is a return situation forω. We have to consider two cases:

i. ξn(ω) does not cover completely an interval Im,k. We say that the returningsituation is inessential. Since (b) does not hold, we have x1 ∈ ξn(ω) ⋂ U∆such that x1 does not belong to the extreme intervals I∆,1 and I−∆,1 ofU∆, and so belongs to some Im,k (|m| ≥ ∆) with I+m,k ⊂ U∆. Furthermore,ξn(ω) ⊂ I+m,k. Indeed, if there was any point of ξn(ω) outside I+m,k, sinceξn(ω) is an interval (because ω is an interval by the induction assumptionand ξn is continuous) and we have a point x1 of ξn(ω) inside Im,k, ξn(ω)had to cover an adjacent interval Im′,k′ of Im,k. But, this is not possiblein this case. We put ω ∈ P ′n and Rn(ω) = Rn−1(ω)

⋃{n}. The return nis called an inessential return and we set I+m,k as its host interval.

ii. ξn(ω) contains at least one interval Im,k, with |m| ≥ ∆. We say that thereturning situation is essential. In this case we consider the sets

ω′m,k = ξ−1n (Im,k)

⋂ω for |m| ≥ ∆

ω′∆−1,1 = ξ−1n ([0, 1]\U∆)

⋂ω

ω′1−∆,1 = ξ−1n ([−1, 0]\U∆)

⋂ω

27

and being A the set of indices (m, k) such that ωm,k 6= ∅ we have

ω\ξ−1n (0) =⋃

(m,k)∈Aω′m,k . (22)

Since n ≥ N ≥ N0 we have that ω satisfies conditions (a) and (b) ofProposition 3.4. So, as refered at Section 3, ξn|ω is a homeomorphismand then, ω′m,k is an interval. Moreover ξn(ω

′m,k) covers the whole Im,k,

except eventually for two extreme intervals. When ξn(ω′m,k) does not

cover entirely Im,k, we join ω′m,k to its neighbourhood in the previous de-

composition (22) and get a new decomposition of ω\ξ−1n (0) into intervalsωm,k with

Im,k ⊂ ξn(ωm,k) ⊂ I+m,k .Now, we put ωm,k ∈ P ′n iff |m| ≤ [αn] − 1, and set I+m,k as its hostinterval. Note that the portion of ω excluded is an interval ωexc containedin U[αn]−1. If |m| ≥ ∆ we set Rn(ωm,k) = Rn−1(ω) ⋃{n} and n is called anessential return for ωm,k. If |m| = ∆−1 then we set Rn(ωm,k) = Rn−1(ω),ωm,k is called an escape component and n an escaping situation for a ∈ωm,k. Escape components will play an important role in the measure ofthe excluded set of parameters that do not satisfy the (FAn) assumption.

At this point it is not difficult to check that every descendant of an ω ∈ Pn−1that belongs to P ′n satisfies (BAn):(a) n is a bound time. In this situation, n = µ + j, j ≥ 1, where µ is a return, and

by (BC), we have that for all a ∈ ω, |ξµ+j(a)− ξj(a)| < e−βj. So,

|ξµ+j(a)| ≥ |ξj(a)| − e−βj (23)

and since a satisfies (BAj), because a satisfies (BAn−1) by the induction as-sumption, the length p of the bound period is strictly less than µ and soj ≤ p ≤ µ− 1 ≤ n− 1. Thus, by (23)

|ξn(a)| = |ξµ+j(a)| ≥ e−αj − e−βj≥ 1− eα−β≥ e−αn since n ≥ N and N is large.

(b) n is a free time. In this case, for every a ∈ ω, |ξn(a)| ≥ e−(∆+1) and then,providing N = N(∆) sufficiently large (N ≥ ∆+1

α), it follows immediately that

|ξn(a)| ≥ e−αn.

28

(c) n is a returning situation for ω.

i. n is an inessential return for ω. Here ξn(ω) does not cover any intervalIm,k. If ω does not satisfy (BAn) then we have x ∈ ξn(ω)\{0} with|x| < e−αn and thus, the host interval I+m,k of ω at time n must be suchthat |m| ≥ [αn] − 1. Therefore |ξn(ω)| ≤ |I+m,k| ≤ 5|Im|m2 ≤ e−αn for Nlarge. But this is not possible since we will prove in Lemma 5.3 that|ξn(ω)| ≥ e−αn2 .

ii. n is an essential returning situation, i.e. n is an essential return or anescaping situation for the descendants ωm,k ∈ P ′n of ω. Once more, it iseasy to check that ωm,k satisfies (BAn), because we have ξn(ωm,k) ⊂ I+m,kwith |m| ≤ [αn] − 1 and then, ξn(ω) ⋂ U[αn] = ∅, which means that ωm,ksatisfies (BAn).

Therefore, definingΩ′n =

⋃{ω : ω ∈ P ′n}we have that Ω′n satisfies (BAn) and (EGn−1). Now we put

Pn = {ω ∈ P ′n : Fn(ω) ≥ (1− αn)} .

Thus, Ωn satisfies (BAn), (EGn−1), (FAn), and so (EGn) (see Section 2).

For each a ∈ Ωn, for every k = 1, . . . , n, a belongs to one and only oneωk ∈ Pk. Such intervals are generated as follows. ω1 = . . . = ωN−1 and at time N ,since ξN(Ω) ⊃ U1, by Proposition 3.6, ν1(a) = N is an essential returning sit-uation for ωN−1. So we subdivide Ω into intervals Ω(m,k) and Ω(m,k) ∈ PN for∆−1 ≤ |m| ≤ [αN ]−1. Since a ∈ Pn there is some (m1, k1), such that ωN = Ω(m1,k1).Now, ωk = Ω(m1,k1) for k = ν1(a) + 1, . . . ν2(a), where ν2(a) is the next essential re-turning situation for Ω(m1,k1). At time ν2(a) we split again Ω(m1,k1) and we get a newcomponent Ω(m1,k1),(m2,k2) of Pν2(a) such that ων2(a) = Ω(m1,k1). Continuing in thisway we will have sequences ν1, . . . , νs and (m1, k1), . . . , (ms, ks) (s = s(a, n)), suchthat

ωνi = Ω(m1,k1),...,(mi,ki)

ωk = ωνi for k = νi + 1, . . . , νi+1 − 1ωνi ⊂ ωνi−1 but ωνi 6= ωνi−1 ,

whereImi,ki ⊂ ξνi(Ω(m1,k1),...,(mi,ki)) ⊂ I+mi,ki .

29

Furthermore, since ξνi|ωνi−1 is a homeomorphism, every ω ∈ Pn is equal to someΩ(m1,k1),...,(ms,ks) for a unique sequence (m1, k1), . . . , (ms, ks) with |mi| ≥ ∆ − 1 andki ≤ m2i .

Next Lemma asserts that escape components return very big compared with U∆.This is important to obtain Lemma 5.3 and |Ωn\Ω′n| estimates in next section.

Lemma 5.1 Suppose that ω ∈ Pν0 is an escape component. Then in the next re-turning situation ν for ω we have that

|ξν(ω)| ≥ e−∆2 .

Proof: Since ω is an escape component at time ν0 it follows that

ξν0(ω) ⊃ Im,k with |m| = ∆− 1 ,

and so, there exists a∗ ∈ ω such that |ξν0(a∗)| = e−∆. Therefore, ξν0+1(a∗) ≥ 1− x∗,where x∗ = 2e−2∆. Thus, if ν = ν0 + 1 then |ξν(ω)| ≥ e−∆2 . Now suppose thatν ≥ ν0 + 2. Writing

ξν0+2(a∗) = f2(ξν0+1(a∗)) + fa∗(ξν0+1(a∗))− f2(ξν0+1(a∗))

and taking into account that f2(ξν0+1(a∗)) ≤ f2(1− x∗), fa∗(y)− f2(y) ≤ 2− a∗ forevery y ∈ [−1, 1] and (21) (2− a∗ ≤ 2x2∗), it follows that

ξν0+2(a∗) ≤ −1 + 4x∗ .

By induction, using the same argument, we can state that for k ≥ 2, providing−1 + 4k−2x∗ ≤ 1 (this is to ensure that we are inside the domain [−1, 1] of ourfamily), we have that

ξν0+k(a∗) ≤ −1 + 4k−1x∗Therefore, if −1 + 4ν−ν0−1x∗ ≤ −12 then ξν(a∗) ≤ −12 and so, |ξν(ω)| ≥ e−

∆2 . So it

remains to complete the proof when −1 + 4ν−ν0−1x∗ > −12 . First, we notice thatwe can assume ξν(ω) ⊂ U1, otherwise we have immediately the conclusion. In theseconditions, we have 2ν−ν0 > e∆, by the value of x∗. Then by Lemma 3.5,

|ξν(ω)| ≥ 1A2

·∣∣∣(f ν−ν0t )

′(x)

∣∣∣ · |ξν0(ω)|, t ∈ ω

≥ 1A2

·∣∣∣(h−1)′(x)

∣∣∣∣∣∣(h−1)′(y)

∣∣∣·∣∣∣(gν−ν0t )

′(h−1(x))

∣∣∣ · e−∆

(∆− 1)2

30

where x = ξν0(t), y = fν−ν0t (x) = ξν(t). By Lemma 3.1 it follows that

|ξν(ω)| ≥ L ·[2− 3π

δ3(2− t)

]ν−ν0· e

−∆

(∆− 1)2

with

L =1

A2

√√√√ 1− (ξν(t))21− (ξν0(t))2

.

Since ξν(ω) ⊂ U1

|ξν(ω)| ≥ 1A2

·√

1− 1e2·[2− 3π

δ3(2− t)

]ν−ν0· e

−∆

(∆− 1)2

≥ 45A2

·[2− 3π

δ3(2− t)

]ν−ν0· e

−∆

(∆− 1)2 . (24)

Now we remark that[2− 3π

δ3(2− t)

]≥ ec0 and then, since |ξν(ω)| ≤ 2, it follows

that ec0(ν−ν0) ≤ 5A22

e∆(∆− 1)2, which implies, for ∆ large, that ν− ν0 ≤ 2∆. Then,by the choice (20) of the parameter a1, we have that

[2− 3π

δ3(2− t)

]ν−ν0 ≥ 12· 2ν−ν0 ,

and thus, from (24) it follows finally that

|ξν(ω)| ≥ 25A2

· 2ν−ν0 · e−∆

(∆− 1)2 .

Now, taking into account that 2ν−ν0 > e∆, we have

ξν(ω)| ≥ 25A2(∆− 1)2

≥ e−∆2 for ∆ large. 2Now we will obtain estimates on the length of |ξk(ω)| at returns k.

Lemma 5.2 Suppose that µ is a return for ω ∈ Pn with host interval Im,k. Set pas the length of its bound period. Then

(a) If µ′ ≤ n is the next return after µ, putting q = µ′ − p− 1, we have:(i) |ξµ′(ω)| ≥ ecq · e(1−6β)|m| · |ξµ(ω)| ≥ 2|ξµ(ω)|.(ii) |ξµ′(ω)| ≥ ecq · e−6β|m| if µ is an essential return.

(b) If µ is the last return up to n and n is a free time then, putting q = n − p,we have

(i) |ξµ′(ω)| ≥ ecq−∆ · e(1−6β)|m| · |ξµ(ω)| ≥ 2|ξµ(ω)|.(ii) |ξµ′(ω)| ≥ ecq−∆ · e−6β|m| if µ is an essential return.

31

Proof: Writting|ξµ′(ω)||ξµ(ω)| =

|ξµ′(ω)||ξµ+p+1(ω)| ·

|ξµ+p+1(ω)||ξµ(ω)| ,

it follows from Lemma 3.5 that for some t, t′ ∈ ω,|ξµ′(ω)||ξµ(ω)| ≥

1

A4·∣∣∣(f qt )′(x)

∣∣∣ ·∣∣∣(fp+1t′ )

′(x′)

∣∣∣ ,

where x = ξµ+p+1(t) and x′ = ξµ(t). Thus, part (c) of Proposition 4.5 gives rise to

|ξµ′(ω)||ξµ(ω)| ≥

1

A4·∣∣∣(f qt )′(x)

∣∣∣ · e(1−5β)|m| . (25)

proof of (a): Under the (a) hypothesis, we have f qt (x) = ξµ′(t) ∈ I+m,k with|m| ≥ ∆, and then by part (b) of Proposition 3.3, |(f qt )′(x)| ≥ ec0q. Therefore from(25) we have

|ξµ′(ω)||ξµ(ω)| ≥

1

A4· ec0q · e(1−5β)|m| , (26)

and then, assertion (i) follows from (26) taking ∆ sufficiently large. Statement (ii)follows also from (26) providing ∆ is large enough, taking into account that since µ

is an essential return then ξµ(ω) ⊃ Im,k and then |ξµ(ω)| ≥ e−|m|m2 .proof of (b): Here the proof is analogous to part (a), and the factor e−∆

appears because under these hypothesis, we can only apply part (a) of Proposition3.3 and then we only assure that |f qt )′(x)| ≥ δec0q. 2

Now we state that at returning situations n of ω, the length of ξn(ω) is so largecompared with the length of U[αn], which will allow us to prove in the next sectionthat |Ωn−1\Ω′n| ≤ e−

αn3 .

Lemma 5.3 If n is a returning situation for ω ∈ Pn−1, then

|ξn(ω)| ≥ e−αn2 .

Proof: Since n is a returning situation, then it is not in a bound time of a previousreturn for ω. We take the first ν ≤ n − 1 such that ων = ω. By construction wemust have ξν(ω) ⊃ Im,k.

(i) If ν is an escape time (|m| = ∆ − 1) then by Lemma 5.1 we have that|ξn(ω)| ≥ e−∆2 and thus, the assertion follows providing N = N(∆) is sufficientlylarge so that n ≥ N ⇒ e−∆2 ≥ e−αn2 .

32

(ii) ων is a returning component ([αn] − 1 ≥ |m| ≥ ∆). By Lemma 5.2 takingµ0 = ν, µs+1 = n and (µi)i=1,...s as the returns of ω after ν then, if s = 0

|ξn(ω)| ≥ e−6β|m|−∆

and if s > 0,

|ξn(ω)| = |σ1| ·s−1∏

i=1

|σi+1||σi| ·

|σs+1||σs|

≥ e−6β|m| · 1 · e−∆ .

Thus, in both situations we have

|ξn(ω)| ≥ e− 14 |m|−∆≥ e− 14αν−∆≥ e− 14αn−∆≥ e−αn2 if N = N(∆) is large . 2

Finally we finish this section with the Proposition concerning the distorsion ofξn (Dn−1) over the components of Pn−1.

Proposition 5.4 There exists a constant C = C(α − β) such that: if ω ∈ Pn−1(n ≥ N) and ξn(ω) ⊂ U1 then for every a, b ∈ ω:

(a)|Dn−1(a)||Dn−1(b)| ≤ C (b)

|ξ′n(a)||ξ′n(b)|

≤ C

Proof: Let (µj)j=1,...,s denote the returns of ω up to n − 1 with the correspondinglenghts (pj)j=1,...,s as bound periods, host intervals Im1,k1 , . . . , Ims,ks , µ0 = 1, p0 = −1and µs+1 = n. For j = 1, . . . , s, let σj = ξµj(ω). First we assume that n ≤ µs + ps.By Proposition 3.4 and Lemma 4.2, as

|Dn−1(a)||Dn−1(b)| =

(a

b

)n−1·

n−1∏

i=1

|ξi(a)||ξi(b)|

it suffices to show thatn−1∏

i=1

|ξi(a)||ξi(b)|

33

is bounded by a constant depending only in α− β. So, to prove the Proposition weneed only an upper bound, depending only in α− β, for

S =n−1∑

i=1

|ξi(a)− ξi(b)||ξi(b)| .

Let us split S taking into account the returns, the bound periods and the free orbits.For j = 0, . . . , s let:

S ′j =µj+pj∑

i=µj+1

|ξi(a)− ξi(b)||ξi(b)| and S

′′j =

µj+1−1∑

i=µj+pj+1


and for j = 1, . . . , s take

Rj =|ξµj(a)− ξµj(b)|

|ξµj(b)|.

Let us first estimate S ′′j . For j = 0, . . . , s− 1 and µj + pj + 1 ≤ i ≤ µj+1, by Lemma3.5 for some ti ∈ ω

|ξµj+1(a)− ξµj+1(b)||ξi(a)− ξi(b)| ≥

1

A2·∣∣∣∣|(fµj+1−iti )

′(ξi(ti))

∣∣∣∣ .

Since∣∣∣∣(f

µj+1−it )

′(ξi(t))

∣∣∣∣ ≥ ec0(µj+1−i) by Proposition 3.3 and |ξi(b)| ≥ δ (i is a freetime), we obtain

|ξi(a)− ξi(b)||ξi(b)| ≤

A2

δ· e(i−µj+1) 23 |σj+1| .

So,

S ′′j ≤A2

δ|σj+1|

µj+1−1∑

i=µj+1

e(i−µj+1)23

≤ L1 |σj+1|δ

= L1|Imj+1|

δ

|σj+1||Imj+1|

≤ L2 |σj+1||Imj+1|.

For j = 0, . . . , s we have

Rj =|ξµj(a)− ξµj(b)|

|ξµj(b)|≤ |σj|

e−(mj+1)

≤ 2|σj||Imj |. (27)

34

Now, let us bound S ′j. For µj + 1 ≤ i ≤ µj + pj,

|ξi(a)− ξi(b)||ξi(b)| =

|ξi(a)− ξi(b)||ξi(b)− ξi−µj(b)|

|ξi(b)− ξi−µj(b)||ξi(b)| . (28)

Since by (BC)|ξi(b)− ξi−µj(b)| < e−β(i−µj)

and also

|ξi(b)| ≥ |ξi−µj(b)| − e−β(i−µj)≥ e−α(i−µj)(1− e(α−β)(i−µj))≥ (1− eα−β)e−α(i−µj) ,

so, the second factor in (28) is less or equal than

L3e(α−β)(i−µj) , where L3 = (1− eα−β)−1 . (29)

For the first factor, by Lemma 3.5, we have for some ti ∈ ω|ξi(a)− ξi(b)||ξµj(a)− ξµj(b)|

≤ A2∣∣∣∣(f

i−µj−1b )

′(ft(ξµj(t)))

∣∣∣∣∣∣∣f ′t(ξµj(t))

∣∣∣

and for some x ∈ U+m,

|ξi(b)− ξi−µj(b)| =∣∣∣∣(f

i−µj−1b )

′(fb(x))

∣∣∣∣ ·∣∣∣fb(ξµj(b))− fb(0)

∣∣∣

= b∣∣∣ξµj(b)

∣∣∣2 ·

∣∣∣∣(fi−µj−1b )

′(fb(x))

∣∣∣∣

and so from Lemma 4.4

|ξi(a)− ξi(b)||ξi(b)− ξi−µj(b)|

≤ A2B3 2|t|b

|ξµj(t)||ξµj(b)|

|ξµj(a)− ξµj(b)||ξµj(b)|

≤ L4|ξµj(a)− ξµj(b)|

|ξµj(b)|, L4 = L4(α− β) . (30)

Therefore, from (27), (29) and (30) it follows that

S ′j ≤ L5|σj||Imj |

35

where we have used that α < β. So, for j = 1, . . . , s

S ′′j−1 + S′j + Rj ≤ L6

|σj||Imj |

where L6 = L6(α− β) .

Now let Nm beNm =

{i ∈ {1, . . . , s} | mi = m

}

If Nm 6= ∅ and rm = max Nm, since |σj+1| ≥ 2|σj| by Lemma 5.2 (a),∑

j∈Nm

|σj||Im| ≤ 2

|σrm||Im| ≤ 2

|I+m,k||Im|

≤ 10m2

.

Hence

S =s∑

j=1

S ′′j−1 + S′j + Rj

≤ L6∑

m:Nm 6=∅

∑

j∈Nm

|σj||Im|

≤ 10L6∑

m:Nm 6=∅

1

m2.

≤ 10L6∞∑

m=∆

1

m2< ∞ .

Now, it remains to deal with the case n ≥ µs + ps + 1. If |ξn−1(ω)| ≤ e−2∆ then,once more by Lemma 3.5

|ξn−1(a)− ξn−1(b)||ξi(a)− ξi(b)| ≥

1

A2·∣∣∣(fn−1−iti )

′(ξi(ti))

∣∣∣ ti ∈ ω .

Here we do not have the guarantee that ξn−1(ti) ∈ U∆−1 and so we can only applypart (a) of Proposition 3.3 and obtain that

∣∣∣(fn−1−iti )′(ξi(ti))

∣∣∣ ≥ δec0(n−1−i). Thus

|ξi(a)− ξi(b)||ξi(b)| ≤ A

2 · e−2∆

δ2· e− 23 (n−1−i) ,

and then

S ′′s,1 =n−1∑

i=µs+ps+1

|ξi(a)− ξi(b)||ξi(b)| ≤ L7 .

36

So, we have the Proposition proved. When we have |ξn−1(ω)| ≥ e−2∆, there is a firstinteger k1 ≥ µs + ps + 1 with |ξk2(ω)| ≥ e−2∆ and for i = µs + ps + 1, . . . , k1 − 1|ξi(ω)| ≤ e−2∆. With the same argument as before we obtain bound distorsion

|Dk1−1(a)||Dk1−1(b)|

≤ L8 .

At this point we consider the interval [k1, k2 − 1] (eventually empty) whose times iverify ξi(b) 6∈ U1. Then

S ′′s,2 =k2−1∑

i=k1


≤ 3k2−1∑

i=k1

|ξi(a)− ξi(b)| (ξi(b) 6∈ U1)

≤ 3k2−1∑

i=k1

5A2

4e−(n−i)

23 |ξn(ω)| by Lemma 3.5 and Proposition 3.3 (c)

< ∞ .If k1 = n the Proposition is proved; otherwise (if there is no times i ≥ k1 withξi(ω) 6∈ U1 we take k2 = k1) writing

|Dn−1(a)||Dn−1(b)| =

|(fn−k2a )′(x)||(fn−k2b )

′(y)|

· |Dk2−1(a)||Dk2−1(b)|where x = ξk2(a) and y = ξk2(b), we need only to bound the first factor, which wedivide in two cases.

1. ξk2(a) ≥ 12 . Then, since ξk2(b) < 1e , we have |ξk2(a) − ξk2(b)| ≥ 110 . Thereforeby Lemma 3.5 and Proposition 3.3

|ξn(ω)| ≥ 4A2

5· e(n−k2) 23 · 1

10.

So, since |ξn(ω)| ≤ 1, it follows that

n− k2 ≤ 32

log 8A2

and then attending to |(fn−k2a )′(x)| ≤ 4n−k2 and |(fn−k2b )′(y)| ≥ 4

5e

23(n−k2) we

have for some constant L9,

|(fn−k2a )′(x)||(fn−k2b )

′(y)|

≤ L9 .

37

2. |ξk2(a)| ≤ 12 . Since

|(fn−k2a )′(x)||(fn−k2b )

′(y)|

= L ·∣∣∣(gn−k2a )

′((h−1(x))

∣∣∣∣∣∣(gn−k2b )

′(h−1(x))

∣∣∣. (31)

where

L =

√√√√1− (fka (x))21− x2 ·

√√√√ 1− y21− (fkb (y))2

≤ 43

. (32)

Lemma 3.1 says us that∣∣∣(gn−k2a )

′((h−1(x))

∣∣∣∣∣∣(gn−k2b )

′(h−1(x))

∣∣∣≤

[2 + 3π

δ3(2− a)

[2− 3πδ3

(2− a)

]n−k2(33)

Since |ξk1(ω)| ≥ e−2∆ we can conclude like in case 1, that n − k1 ≤ 4∆(providing ∆ is sufficiently large) and then n− k2 ≤ n− k1 ≤ 4∆. So by thechoice (20) of the parameter a1 we have immediately from (31), (32) and (33)that

|(fn−k2a )′(x)||(fn−k2b )

′(y)|

≤ 83

.2

6 Estimates of the excluded set

In this section we are going to see that the construction of the sets (Ωn)n leadsto have for n ≥ N ,

|Ωn−1\Ω′n| ≤ e−²n|Ωn−1| and |Ω′n\Ωn| ≤ e−²n|Ω| .Therefore, since |Ωn−1\Ωn| = |Ωn−1\Ω′n| + |Ω′n\Ωn|, we have the recurrencerelation for n ≥ N :

|Ωn| ≥ (1− e−²n)|Ωn−1| − e−²n|Ω|and then, as explained in the Section 2,

|Ω∞| ≥ (1− ρ)|Ω| .We begin estimating |Ωn−1\Ω′n|.

38

Proposition 6.1 For n ≥ N , we have:|Ωn−1\Ω′n| ≤ e−²n |Ωn−1| .

Proof: Take ω ∈ Pn−1 and denote by ωexc the set {a ∈ ω : a 6∈ Ω′n}. If n is nota returning situation for ω, or n is an inessential return then by constructionωexc = ∅. When n is a returning situation but not an inessential return for ω,then ωexc is an interval such that ξn(ωexc) ⊂ U[αn−1] and then

|ξn(ωexc)| ≤ 2e−αn+1 . (34)Now denoting by ω̂ the interval ξ−1n (U1)

⋂ω (remember that ξn|ω is a homem-

orphism) we have that

|ωexc||ω| ≤

|ωexc||ω̂|

=|ξ′n(t1)||ξ′n(t2)|

· |ξn(ωexc)||ξn(ω̂)| .

where t1 ∈ ω̂ and t2 ∈ ωexc ⊂ ω̂ (U[αn−1] ⊂ U1 for N large). By (34) andProposition 5.4 it follows that

|ωexc||ω| ≤ 2C ·

e−αn+1

|ξn(ω̂)| . (35)

Now by Lemma 5.3, |ξn(ω̂)| ≥ e−αn2 (notice that ω̂ 6= ω implies that|ξn(ω̂)| ≥ 1e − e−∆, and so |ξn(ω̂)| ≥ e−

αn2 for N large). Thus, from (35)

we obtain

|ωexc||ω| ≤ 2Ce

−αn2

+1

≤ e−αn3 . (36)Finally, attending to the fact that Pn−1 is a partition of Ωn−1 it follows that

|Ωn−1\Ω′n| =∑

ω∈Pn−1|ωexc|

≤ ∑ω∈Pn−1

e−αn3 |ω| by (36)

≤ e−αn3 |Ωn−1| .2Actually, to complete this work we are left to prove that |Ω′n\Ωn| ≤ e−²n|Ω|,

i.e. |{a ∈ Ω′n : Fn(a) ≤ (1− αn)}| ≤ e−²n|Ω|. First, we prove a Lemma whichstates that essential returns are quite frequent.

39

Lemma 6.2 Suppose ν0 is an essential return for ω ∈ Pν0 with|ξν0(ω)| ⊃ Im0,k0 (|m0| ≥ ∆). Then the next essential return situation ν1satisfies

ν1 − ν0 < 5|m0| .

Proof: Let s denote the number of inessential returns µ1 < . . . < µs be-tween ν0 and ν1, with I

+m,1, . . . , I

+ms,ks

as host intervals. We also put µ0 = ν0,µs+1 = ν1, and, for i = 0, . . . , s + 1, σi = ξµi(ω), qi = µi+1 − (µi + pi + 1),where as usually, for i = 0, . . . , s, is the length of the bound period associatedto the return µi.

First assume s > 0. Then by part (a) of Lemma 5.2

|σ1| ≥ e−6β|m0| (37)and

|σi+1||σi| ≥ e

c0qie(1−6β)|mi| for 0 ≤ i ≤ s− 1 . (38)

Writing

|σs| = |σ1| ·s−1∏

i=1

|σi+1||σi|

and taking into account that σs ⊂ Ims,ks , with |ms| ≥ ∆ (thus |σs| ≤ e−(∆+1)),(37) and (38), it follows that

exp

{−6β|m0|+

s−1∑

i=0

coqi +s−1∑

i=1

(1− 6β)|mi|}

≤ exp {−(∆ + 1)}

and thus,

s−1∑

i=0

coqi +s−1∑

i=1

(1− 6β)|mi| ≤ 6β|m0| − (∆ + 1) . (39)

Now, attending to part (b) of Lemma 5.2, we have

|σs+1||σs| ≥ e

c0qs−∆ · e(1−6β)|ms| . (40)

So, since by part (a)− (i) of the same Lemma |σs| ≥ 2s−1|σ1| ≥ |σ1| we havethat

2 ≥ |σs+1| ≥ |σ1| · |σs+1||σs|

40

and then, attending to (37) and (40) we obtain

exp {−6β|m0|+ c0qs −∆ + (1− 6β)|ms|} ≤ 2 .and thus,

c0qs + (1− 6β)|ms| ≤ ∆ + 1 + 6β|m0| . (41)Hence, by (39) and (41)

s∑

i=0

coqi +s∑

i=1

(1− 6β)|mi| ≤ 12β|m0| . (42)

Therefore, since pi ≤ 3|mi| ≤ 4(1− 6β)|mi|, it follows that

ν1 − ν0 ≤s∑

i=0

pi + qi

≤ 3|m0|+ 4[

s∑

i=0

coqi +s∑

i=1

(1− 6β)|mi|]

.

Now, by (42) we obtain

ν1 − ν0 ≤ 3|m0|+ 48β|m0|< 4|m0| since β < 1

48.

It remains to prove the case s = 0. In this situation, Lemma 5.2, part (b)−(ii)says us that

|σ1||σ0| ≥ e

−6β|m0| · ec0q0−∆

and then, attending to the fact that |σ1| ≤ 2 we havec0q0 ≤ 6β|m0|+ ∆ + 1

which implies, since c0 ≥ 23 , that q0 ≤ 10β|m0| + 32∆, providing ∆ is largeenough. Therefore

ν1 − ν0 ≤ p0 + q0≤ 3|m0|+ 10β|m0|+ 3

2∆

< 4|m0|+ ∆< 5|m0| .2

41

For a ∈ Ω′n with a ∈ ωk ∈ Pk (k = 1, . . . , n − 1), denote by ν1 < . . . < νsthe returning situations of a, with host intervals Im1,k1 , . . . , Ims,ks , ν0 = 1 andνs+1 = n. Let us split the orbit {ξk(a) : k = 1, . . . , n−1} into times Oi = Oi(a)as follows:

O0 = {ν`0 , . . . , ν`1 − 1} (`0 = 0)O1 = {ν`1 , . . . , ν`2 − 1}...

......

Or = {ν`r , . . . , ν`r+1 − 1} (`r+1 = s + 1)where

|m`| = ∆− 1 for `0 ≤ ` ≤ `1|m`| ≥ ∆ for `1 ≤ ` ≤ `2

......

...|m`| = ∆− 1 for `2i ≤ ` ≤ `2i+1 − 1|m`| ≥ ∆ for `2i+1 ≤ ` ≤ `2i+2 − 1

......

...|m`| = ∆− 1 for `r ≤ ` ≤ `r+1 if r is even|m`| ≥ ∆ for `r ≤ ` ≤ `r+1 if r is odd .

For i = 0, . . . ,[

r2

], O2i begins with an escaping situation ν`2i (here, and in the

rest of this section we consider, by convenience of exposition, 1 and n returningand escaping situations for any element of Ω′n), after this escaping situationall the return situations ν ∈ O2i are escaping situations, and so, O2i is a pieceof free orbit. This periods O2i are called escape periods. Therefore, denotingby |Oi| the length of Oi, we have, putting

Tn(a) =r1∑

i=0

|O2i+1| r1 =[r − 1

2

]

that n− T (a) ≥ Fn(a). So, we can get|{a ∈ Ω′n : Fn(a) ≤ (1− α)n}| ≤ e−²n|Ω|

providing that

|{a ∈ Ω′n : Tn(a) ≥ αn}| ≤ e−²n|Ω| .This last inequality will be obtained by an estimation on the deviation of theexpected value of Tn:

∫

Ω′neγTn(a)da (43)

42

for a suitable γ. From now on our purpose is to obtain (43) (Lemma 6.4) andconsequently, the required recurrence relation on the sets (Ωn)n.

We begin introducing a new point of view over the value Tn(a) just defined.Being 1 = e0 < e1 < . . . < es1 < es1+1 = n, the escaping situations, we definefor i = 1, . . . s1 + 1

Ei(a) = ei − µ̂i ,where µ̂i is the next returning situation after ei−1. When µ̂i is an escapingsituation, Ei = 0 and then, {ei−1, . . . , µ̂i} is a portion of escape period. On theother hand, if µ̂i is a return then {eii , . . . , µ̂i − 1} is a piece of escape periodand {µ̂i, . . . , ei − 1} is equal to some O2j+1 (j = j(i)) and thus, Ei = |O2j+1|.Hence,

Tn(a) = E1(a) + . . . + Es1+1(a) .

Notice that Ei (i = 1, . . . s1 + 1) is constant on each component ωei and thuswe can write

Tn(ω) = E1(ωe1) + . . . + Es1+1(ωes1+1) .

At this point, we have for every ω ∈ P ′n, s1 = s1(ω) < n, escaping situ-ations 1 = e0 < e1 < . . . < es1 = n, (ei = ei(ω)), returning situations afterescapes N = µ̂1 < . . . < µ̂s1 < µ̂s1+1 = n (µ̂i = µ̂i(ω)), a sequence of escapecomponents Ω = ω0 ⊃ ω1 ⊃ . . . ⊃ ωs1 ⊃ ωs1+1 = ω (ωi = ωei) and

Tn(ω) =s1(ω)∑

i=1

Ei(ωi) .

Completing the sequences for i = s1 + 2, . . . , n by

ei = n = es1+1

µ̂i = n = µs1+1

ωi = ω = ωs1+1

Ei(ωi) = 0 = ei − µ̂i

and setting for 1 ≤ j ≤ n

Tj(ω) =j∑

i=1

Ei(ωi)

the value for j = n is consistent with the previous definition. Now, we considerfor i = 0, . . . , n the sets of parameter values

Qi =⋃{ωi : ω ∈ P ′n}

43

and its partition

Qi = {ωi : ω ∈ P ′n} .

We have Q0 = Ω, Q0 = {Ω}, Qn = Ω′n and Qn = P ′n. Moreover Ti (i =1, . . . , n) depends only on the set Qi. For a component ω

′ ⊂ Qi−1 we define

< ω′ > = ω′⋂

Qi

andQ(ω′) = {ω′′ ∈ Qi : ω′′ ⊂ ω′} .

Lemma 6.3 For i = 1, . . . , n and ωi−1 ∈ Qi−1 we have for γ = 140 :∫

eγEi(a)da ≤ e²|ωi−1| .

Proof: Let ei−1 be the escape time of ωi−1 and µ̂i the next returning situation.If µ̂i = n then for every a ∈ ωi−1, Ei(a) = ei(a)− µ̂i = 0, since n ≥ ei(a) ≥ µ̂i,and thus, the statement is obvious. Now suppose that µ̂i < n. Then we havean essential returning situation for ωi−1 at time µ̂i (µ̂i is not an inessentialreturn for ωi−1 since by Lemma 5.1 |ξµ̂i(ωi−1)| ≥ e−

∆2 and thus, ξµ̂i(ω

i−1)is not contained in U∆). Hence in the construction of the set Ωµ̂i , ω

i−1 issubdivided into disjoint intervals ωm,k and for a surviver of the exclusion ofparameters by (BAµ̂i), ωm,k ∈ P ′µ̂i , we have the following Claim.

Claim 2 If t > 10|m|,

|{a ∈< ωm,k >: Ei(a) = t}| ≤ e−2γt|ωm,k|.

Proof: We begin remarking that if ωm,k is an escape component, then{a ∈< ωm,k >: Ei(a) = t} = ∅, and so the inequality is obvious. Then for acomponent ω ∈ Q(ωm,k) with escape situation ei = µ̂i + t, denoting byµ̂i = ν0(ω) < ν1(ω) < . . . < νs(ω) the essential returning situations thatare not escaping situations of ω, by ωi = ω(m0,k0),(m1,k1),...,(mi,ki) the componentin Pνi(ω) that contains ω and

Imi,ki ⊂ ξνi(ωi) ⊂ I+mi,ki (|mi| ≥ ∆) , (44)

44

we have that, by construction, ωs is well determined by the knowledge of thesequence (m1, k1), . . . , (ms, ks). For a fixed component ωs, it follows that

|ωs||ω0| =

s∏

i=1

|ωi||ωi−1|

≤s∏

i=1

|ωi||ω̂i−1| , ω̂i−1 = ξ

−1νi

(ωi−1)⋂

U1 .

Now, by The Mean Value Theorem and Proposition 5.4, it follows that

|ωs||ω0| ≤

s∏

i=1

C|ξνi(ωi)||ξνi(ωi−1)|

.

Using (44) and Lemma 5.2 we obtain

|ωs||ω0| ≤

s∏

i=1

5C

|mi|2 ·e−|mi|

e−6β|mi−1|

≤s∏

i=1

exp{6β|mi−1| − |mi|} for large ∆

So,

|ωs| ≤ exp{−(1− 6β)

s∑

i=1

|mi|+ 6β|m|}|ωm,k| (45)

For a fixed M and s let us denote by ηs(M) the number of componentsω(m0,k0),(m1,k1),...,(ms,ks) ⊂ ωm,k without any escaping situation after s essen-tial returning situations and |m1|+ · · ·+ |ms| = M , and by η(M) the sum ofall ηs(M), s ≥ 0. Then, denoting by XM the set of parameters a ∈ ωm,k suchthat after s essential returning situations a has no any escaping situation andbelongs to some ω(m0,k0),(m1,k1),···,(ms,ks) with |m1|+ · · ·+ |ms| = M , we have

|XM | ≤ η(M) exp {−7M/8 + 6β|m|} |ωm,k| . (46)Now, for fixed u1 ≥ ∆, · · · , us ≥ ∆ we have at most 2su21 · · · u2s ≤ 2se(u1+···+us)/20components of the form ω(m0,k0),(m1,k1),···,(ms,ks) with |m1| = u1, · · · , |ms| = us,when ∆ is a sufficiently large number such that |ui| ≥ ∆ ⇒ u2i ≤ eui/20. Sinceui ≥ ∆, we have that s∆ ≤ u1 + · · ·+ us, and so

ηs(M) ≤ ρs(M)2M/∆eM/20≤ ρs(M)eM/16 for large ∆ (47)

45

where ρs(M) is the number of solutions of the equation

M = u1 + · · ·+ us, ui ∈ Z+

We have

ρs(M) =

(M + s− 1

s− 1

),

and then, since s ≤ M/∆,

ρs(M) ≤(M + M/∆− 1

M/∆− 1

). (48)

We have by the Stirling formula for every integer v

√2πv · vv · e−v ≤ v! ≤

√2πv · vv · e−v

(1 +

1

4v

),

and so, from (48)

ρs(M) ≤ (M + M/∆− 1)(M+M/∆−1)

MM(M/∆− 1)(M/∆−1)≤ (1 + L1(∆))M ,

where L1(∆) −→ 0 as ∆ −→ ∞. So, taking ∆ sufficiently large, we can putby (47)

η(M) ≤M/∆∑

s=1

ρs(M)eM/16

≤ M∆

(1 + L1(∆))MeM/16

≤ eM/8 .

Therefore, by (46) it follows that

|XM | ≤ exp {−3M/4 + 6β|m|} |ωm,k|.

TakingYM,t = XM ∩ {a ∈< ωm,k >: Ei(a) = t}

46

we have that YM,t = ∅ if t ≥ 5|M | + 5|m|. Indeed, for a ∈ YM,t, witha ∈ ω(m1,k1),···,(ms,ks) and escaping situation νs+1 = ei = µ̂i + t

t = ei − µ̂i =s∑

i=0

νi+1 − νi

≤ 5s∑

i=0

|mi| by Lemma 6.2

≤ 5M + 5|m| .

Thus

m {a ∈< ωm,k >: Ei(a) = t} =∑

5M+5|m|≥t|YM,t|

≤ e6β|m||ωm,k|∑

M≥t/5−|m|e−3M/4

≤ L1e6β|m||ωm,k| exp {−3t/20 + 3|m|/4} .

Finally, since t > 10|m|,

m {a ∈< ωm,k >: Ei(a) = t} ≤ e−3t/20+|m| for large ∆≤ e−2γt|ωm,k| . 2

Now denoting by I the set of index (m, k) such that ωm,k ∈ Pµ̂i we have∫

eγEi(a)da =

∑

(m,k)∈I

∫

eγEi(a)da

=∑

(m,k)∈I

∑

t≥0eγt |{a ∈< ωm,k >: Ei(a) = t}|

= S1 + S2 + S3

where

S1 =∑

(m,k)∈I|{a ∈< ωm,k >: Ei(a) = 0}|

S2 =∑

(m,k)∈I

10|m|∑

t=1

eγt |{a ∈< ωm,k >: Ei(a) = t}|

S3 =∑

(m,k)∈I

∑

t>10|m|eγt |{a ∈< ωm,k >: Ei(a) = t}| .

47

We have that

S1 ≤∑

(m,k)∈I|< ωm,k > |

≤ ∑(m,k)∈I

|ωm,k|

≤ |ωi−1| (49)and, for S2, since for a ∈< ωm,k >, Ei(a) > 0 only if ωm,k is not an escapingcomponent, i.e. |m| ≥ ∆, it follows that

S2 ≤∑

(m,k)∈I|m|≥∆

10|m|∑

t=1

e10γ|m||< ωm,k > |

≤ ∑(m,k)∈I|m|≥∆

10|m|e10γ|m||ωm,k| . (50)

Attending to the fact thatωm,kωi−1 ≤

ωm,kω̂i−1 , ω̂

i−1 = ξ−1µ̂i (U1)⋂

ωi−1, the Mean ValueTheorem and Proposition 5.4,

|ωm,k||ωi−1| ≤ C ·

|ξµ̂i(ωm,k)||ξµ̂i(ωi−1)|

.

Therefore, taking into account that for |m| ≥ ∆, |ξµ̂i(ωm,k)| ≤ 5e−|m|m2

, since

ξµ̂i(ωm,k) ⊂ I+m,k, and |ξµ̂i(ωi−1)| ≥ e−∆2 by Lemma 5.1, we obtain from (50)

that

S2 ≤ |ωi−1| ·∞∑

|m|=∆

m2∑

k=1

50C

|m| e−|m|+∆

2

≤ |ωi−1| ·∞∑

|m|=∆50C|m|e−|m|+∆2

≤ L1(∆) · |ωi−1| , (51)where L1(∆) −→ 0 as ∆ −→ +∞. Finally, for S3, by Claim 2

S3 ≤∑

(m,k)∈I

∑

t>10|m||e−γt|< ωm,k > |

≤ ∑(m,k)∈I

L2e−10γ|m||ωm,k| L2 = (1− e−γ)−1

≤ L2e−10γ∆|ωi−1|, (52)

48

Thus from (49), (51), (52) we have∫

eγEi(a)da ≤ (1 + L3(∆)) · |ωi−1|

where L3(∆) −→ 0 as ∆ −→ +∞ and so the result is proved choosing ∆ largeenough. 2

Lemma 6.4 ∫

Ω′neγTn(a)da ≤ e²n|Ω| .

Proof: The proof follows proving by induction that:∫

QjeγTj(a)da ≤ e²j|Ω| ∀j ≥ 1 ,

since Qn = Ω′n. For j = 1 the assertion is Lemma 6.3 for i = 1. Now assume

that the assertion is valid for j − 1. We have∫

QjeγTj(a)da =

∑

ω′∈Qj−1

∫

eγTj(a)da

=∑

ω′∈Qj−1eγTj−1(ω

′)∫

ω′eγEj(a)da

≤ ∑ω′∈Qj−1

eγTj−1(ω′)e²|ω′| by Lemma 6.3

= e²∫

Qj−1eγTj−1(a)da

≤ e²j|Ω| by induction assumption. 2Finally we can obtain the final Proposition of this work.

Proposition 6.5

|{a ∈ Ω′n : Tn(a) ≥ αn}| ≤ e−²n|Ω| .

Proof: Since

eγαn · |{a ∈ Ω′n : Tn(a) ≥ αn}| ≤∫

Ω′neγTn(a)da

and 2² ≤ γα the result follows immediately from the previous Lemma. 2

49

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