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TRANSACTIONS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 363, Number 10, October 2011, Pages 5251–5291S 0002-9947(2011)05208-6Article electronically published on April 21, 2011

ON THE DERIVATIVE OF THE HAUSDORFF DIMENSION

OF THE QUADRATIC JULIA SETS

LUDWIK JAKSZTAS

Abstract. Let d(c) denote the Hausdorff dimension of the Julia set Jc ofthe polynomial fc(z) = z2 + c. The function c �→ d(c) is real-analytic on theinterval (−3/4, 1/4), which is included in the main cardioid of the Mandel-brot set. It was shown by G. Havard and M. Zinsmeister that the derivatived′(c) tends to +∞ as fast as (1/4 − c)d(1/4)−3/2 when c ↗ 1/4. Under nu-merically verified assumption d(−3/4) < 4/3, we prove that d′(c) tends to

−∞ as −(c+ 3/4)3d(−3/4)/2−2 when c ↘ −3/4.

Contents

1. Introduction 52512. Thermodynamical formalism 52533. Position of the Julia set I 52554. Fatou coordinates 52575. Cylinders 52596. Size of cylinders 52637. Invariant measures 52648. Position of the Julia set II 52689. Introduction to the proof 527310. Denominator 527511. Estimation on the set MN 527612. Estimation on the set BN 528513. The end of the proof 5288Appendix A 5289References 5291

1. Introduction

For a polynomial f we define the filled-in Julia set K(f) as the set of points thatdo not escape to infinity under iteration of f , i.e.

K(f) = {z ∈ C : fn(z) � ∞}.The boundary ∂K(f) is called the Julia set of f and is denoted by J(f).

Received by the editors October 19, 2008 and, in revised form, July 11, 2009.2000 Mathematics Subject Classification. Primary 37F45; Secondary 37F35.This work was partially supported by Polish MNiSW grants 2P03A03425, NN201 0222 33, and

EU FP6 Marie Curie RTN CODY at Orleans France.

c©2011 American Mathematical SocietyReverts to public domain 28 years from publication

5251

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

5252 LUDWIK JAKSZTAS

Recall that a rational function f : C → C (in particular, a polynomial) is calledhyperbolic (expanding) if

∃n∈N ∀z∈J(f) |(fn)′(z)| > 1.

Let us consider the family of quadratic polynomials of the form fc(z) = z2 + c.In this case the Julia set J(fc) will be denoted by Jc, while the filled-in Julia setwill be denoted by Kc. We are interested in the function c �→ d(c), where d(c)denotes the Hausdorff dimension of Jc.

We define the Mandelbrot set M as the set of all parameters c for which theJulia set Jc is connected, or equivalently,

M = {c ∈ C : fnc (0) � ∞}.

The function d is real-analytic on each hyperbolic component of Int(M) (consistingof parameters related to hyperbolic maps) as well as on the exterior of M (see[13]). In particular, d is real-analytic on the largest component M0, bounded bythe so-called main cardioid, which consists of the parameters c such that fc has anattracting fixed point z0, i.e. |f ′

c(z0)| < 1.The main cardioid contains the real interval (−3/4, 1/4). The endpoints −3/4,

1/4 are related to the maps with a parabolic fixed point, namely f ′1/4(1/2) = 1 (one

petal) and f ′−3/4(−1/2) = −1 (two petals). Note that d is continuous on [−3/4, 1/4]

(see [11]), and moreover d|R is left-continuous at −3/4 but not right-continuous at1/4 (see [5]).

G. Havard and M. Zinsmeister proved in [7] that for real parameters c < 1/4close enough to 1/4, the derivative of d can be estimated as follows:

1

K

(14− c

)d( 14 )−

32 � d′(c) � K

(14− c

)d( 14 )−

32

,

for some constant K > 1. Since d(1/4) < 3/2 (see [8]), we get d′(c) → +∞ whenc ↗ 1/4.

Figure 1. The filled-in Julia set K− 34.


DERIVATIVE OF THE HAUSDORFF DIMENSION OF THE JULIA SETS 5253

We will prove a similar result for parameters c > −3/4 close to −3/4. The proofwill be carried out under the assumption d(−3/4) < 4/3, which is supported bynumerical experiments: d(−3/4) ≈ 1.23 (see [12]).

Theorem 1.1. There exist c0 > −3/4 and K > 1 such that for every c ∈(−3/4, c0),

−K(34+ c

) 32d(−

34 )−2

� d′(c) � − 1

K

(34+ c

) 32d(−

34 )−2

,

provided d(−3/4) < 4/3.

Thus we get d′(c) → −∞ when c ↘ −3/4. Later on, the quantity 3/4 + c willbe denoted by δc, so c = −3/4 + δc.

The strategy of the proof is the same as in the “1/4” case, but there are veryimportant differences. In our case the dependence of the distance of the points ofthe Julia set to 0 as a function of c in the natural motion, which influences d′, ismuch less clear and must be studied in detail. Moreover, under the assumptiond(−3/4) < 4/3, there is no f−3/4-invariant probability measure equivalent to theHausdorff measure in dimension d(−3/4) (there only exists a σ-finite measure; see[1]). For f1/4 such a measure does exist.

If the estimate d(−3/4) > 4/3 held, the derivative d′(c) would be bounded, whichwould be easy to prove. In fact, in this case we would not need a crucial part ofSection 11. But it would be very hard to verify whether the derivative is positiveor negative. Though there is numerical evidence that this case does not hold, theabove comments may by applicable for other parameters for which there exists aparabolic periodic point.

In Section 2 we obtain a formula for the derivative of the Hausdorff dimensionfor fc. Next, in Sections 3-8 we give some results about the Julia set and invariantmeasures. Results of Sections 6-7 correspond to those obtained in [7], while Section8 is specific to our case. Theorem 1.1 will be proven in Sections 10-13.

2. Thermodynamical formalism

The polynomials fc, c ∈ [−3/4, 1/4], admit a Bottcher coordinate, i.e. thereexists Φc : C \ D → C \Kc holomorphic, bijective, tangent to identity at infinity,and conjugating T (s) = s2 to fc (Φc ◦ T = fc ◦ Φc). If c ∈ (−3/4, 1/4], thenthe sets Jc are Jordan curves, so Φc has homeomorphic extension (Caratheodory’sTheorem); therefore we get conjugation T |∂D to fc|Jc

. For c = −3/4 the mapΦ−3/4 glues together points of period two and their respective preimages, hencethis is only a semiconjugacy.

The map (c, s) �→ Φc(s) gives a holomorphic motion for c ∈ M0 (see [9]). There-fore the functions Φc are quasiconformal, and so also Holder, while c �→ Φc(s) areholomorphic for every s ∈ C \ D (in particular for s ∈ ∂D).

Let us note that f ′c(z) = 2z. Thus if we write Φc(s) = z, then f ′

c(z) = 2Φc(s).Now we use the thermodynamical formalism, which holds for hyperbolic rational

maps, but we will consider only such maps. Write X = ∂D, T (s) = s2, and letϕ : X → R be a Holder continuous function, to be often called a potential function.We will consider potentials of the form ϕ = −t log |2Φc| for c ∈ (−3/4, 1/4), t ∈ R.



The topological pressure can be defined as follows:

P (T, ϕ) := limn→∞

1

nlog

∑x∈T−n(x)

eSn(ϕ(x)),

where Sn(ϕ) =∑n−1

k=0 ϕ ◦ T k, and the limit exists and does not depend on x ∈ ∂D.

If ϕ = −t log |2Φc| and Φc(x) = z, then eSn(ϕ(x)) = |(fnc )

′(z)|−t. Hence

P (T,−t log |2Φc|) = limn→∞

1

nlog

∑z∈f−n

c (z)

|(fnc )

′(z)|−t.

The function t �→ P (T,−t log |2Φc|) is strictly decreasing from +∞ to −∞. Inparticular, there exists a unique t0 such that P (T,−t0 log |2Φc|) = 0. By Bowen’sTheorem (see [10, Corollary 8.1.6] or [15, Theorem 5.12]) we have

(2.1) t0 = d(c).

The Ruelle operator or the transfer operator Lϕ : C0(X) → C0(X) is defined as

Lϕ(u)(x) :=∑

x∈T−1(x)

u(x)eϕ(x).

The Perron-Frobenius-Ruelle theorem [15, Theorem 4.1] asserts that β = eP (T,ϕ)

is a single eigenvalue of Lϕ associated with an eigenfunction hϕ > 0. Moreover,there exists a unique probability measure ωϕ such that L∗

ϕ(ωϕ) = βωϕ, whereL∗ϕ is conjugated to Lϕ. If β = 1 (that is, P (T, ϕ) = 0, and in our case ϕ =

−d(c) log |2Φc|), then μϕ := hϕωϕ is an invariant measure for T .It follows from [15, Proposition 6.11] or [10, Theorem 4.6.5] that for every Holder

ψ and ϕ at every t ∈ R, we have

(2.2)∂

∂tP (T, ψ + tϕ) =

∫∂D

ϕdμψ+tϕ

μψ+tϕ(∂D).

Set ϕc := −d(c) log |2Φc|. We will use the notation ωc := ωϕcand μc := μϕc

(measures supported on the unit circle). The Bottcher coordinate Φc allows usto define related measures supported on Jc, which will be denoted by ωc and μc,respectively.

The measure ωc is called an fc-conformal measure with exponent d(c), i.e. ωc isa Borel probability measure such that for every Borel subset A ⊂ Jc,

ωc(fc(A)) =

∫A

|f ′c|d(c)dωc,

provided fc is injective on A.The measure μc is fc-invariant, equivalent to ωc, and is called the equilibrium

state if it is normalized. But we do not normalize either μc or μc (except in theproof of the proposition below), which will be discussed in Section 7.

Proposition 2.1 ([7, Proposition 2.1]). If c ∈ (−3/4, 1/4), then

(2.3) d′(c) = − d(c)∫∂D

log |2Φc|dμc

∫∂D

∂

∂c(log |2Φc|)dμc.

The proof which we give is different from [7].



Proof. We can assume that the measures μc are normalized. Write P (t, c) :=P (T,−t log |2Φc|). By (2.1) (Bowen’s Theorem) we have that P (t, c) = 0 if andonly if t = d(c), so we will use the implicit function theorem.

From (2.2), because potential function is equal to −t log |2Φc|, we get

( ∂

∂tP)(t(c), c) = −

∫∂D

log |2Φc|dμc,

which is different from zero, because fc is hyperbolic (this is the Lyapunov exponentof fc with respect to μc). So, equation P (t, c) = 0 defines an analytic functiont(c) = d(c), and we have

t′(c) = −( ∂∂cP )(t(c), c)

( ∂∂tP )(t(c), c)

.

Let us compute the c-derivative. Fix c0 ∈ (−3/4, 1/4). For every s ∈ C \ D themap c → Φc(s) is holomorphic, so we can take the Taylor series for −t log |2Φc|centered at c0 (as a function of c):

− t log |2Φc| = −t log |2Φc0 | − t∂

∂c(log |2Φc|)|c0(c− c0) + (higher order terms).

The higher order terms can be omitted because P (t, c) is C2-smooth (see [10, The-orem 4.7.4]), so using (2.2) we get

( ∂

∂cP)(t(c), c) = −t(c)

∫∂D

∂

∂c(log |2Φc|)dμc.

Taking into account the t-derivative and replacing t(c) by d(c), we derive (2.3). �

3. Position of the Julia set I

In this section we give some preliminary results about the position of the Juliaset near the orbit of period two.

When c ↘ −3/4, the points of period two and the attracting fixed point tend to−1/2, which is the parabolic fixed point for f−3/4. So, first we formulate a theoremabout the behavior of a holomorphic map in a neighborhood of such a point (see[1]).

Let us assume that 0 is a parabolic point for f and f ′(0) = 1. Then on aneighborhood V 0, f can be written as follows:

f(z) = z + azp+1 + a2zp+2 + a3z

p+3 + ... ,

where a �= 0 and p � 1 (p is the number of petals).Consider the set {z : azp ∈ R and azp > 0}, which is the union of p rays

beginning at zero and forming angles which are multiples of 2π/p. Denote theserays by L1, L2, ..., Lp.

For 1 � j � p, 0 < r � ∞ and 0 � θ < 2π, let Sj(θ, r) ⊂ V be the set of thosepoints z lying in the open ball B(0, r), for which the angle between the ray Lj andthe interval which joins the points 0 and z does not exceed θ.



Theorem 3.1. For every θ > 0 there exists r > 0 such that for all 1 � j � p and0 < r′ � r,

Sj(θ, r′) ⊂ f(Sj(θ, r′)),

J(f) ∩B(p, r) ⊂p⋃

j=1

Sj(θ, r).

The map f−3/4 has a parabolic fixed point at −1/2, and we have f ′−3/4(−1/2) =

−1. Thus (f2−3/4)

′(−1/2) = 1, and the above theorem can be applied to f2−3/4

(after translation). Now we derive a similar result for f2c when c ∈ [−3/4, c0).

Let us denote the points of period two for fc by p+c and p−c . Hence

p±c = −1

2± i

√δc,

where δc = c+ 3/4. The attracting fixed point will be denoted by αc,

αc =1

2−√1− δc.

For f−3/4 the vertical direction is unstable, while the horizontal is stable, so the

Julia set “tends” to p+c from above in the upper half-plane and to p−c from belowin the lower half-plane. Let us define (we assume that Arg(z) ∈ (−π, π])

S+c (θ, r) := {z ∈ C : |Arg(z − p+c )−

π

2| � θ, |z + 1

2| < r},

S−c (θ, r) := {z ∈ C : |Arg(z − p−c ) +

π

2| � θ, |z + 1

2| < r}.

Lemma 3.2. For every θ > 0 there exist r > 0 and c0 > −3/4 such that ifc ∈ [−3/4, c0), then

Jc ∩B(−1/2, r) ⊂ S+c (θ, r) ∪ S−

c (θ, r).

Proof. Let us take the Taylor series for f2c centered at p+c :

f2c (z) = p+c + (1 + 4δc)(z − p+c )

− i(6√δc − i4δc)(z − p+c )

2

− (2− 4i√δc)(z − p+c )

3 + (z − p+c )4.

(3.1)

Fix θ > 0 (we can assume that θ is small). First we prove that for a suitable chosenc0 > −3/4 and r0 > 0,

(3.2) f2c (S

+c (θ, r)) ⊃ S+

c (θ, r),

where c ∈ [−3/4, c0) and r ∈ (0, r0]. Using (3.1), we will study how to change themodulus |f2

c (z) + 1/2| with respect to |z + 1/2| and the argument Arg(f2c (z)− p+c )

with respect to Arg(z − p+c ), for z ∈ S+c (θ, r).

The sum of p+c and the first term, namely z + 4δc(z − p+c ), does not change theargument, but it increases the modulus. The next two terms increase the modulus(Arg(z − p+c ) is close to π/2), whereas the fourth has a smaller influence than thethird.

In order to get (3.2) it is enough to see that if z ∈ S+c (θ, r) and |Arg(z −

p+c )| = θ, then the sum starting with the second term increases the modulus ofthe argument. Indeed, for small δc we can assume that the second and third terms



increase |Arg(z − p+c )| = θ, while the fourth has a smaller influence than the third(we can suitably choose r0), so we get (3.2). For S−

c (θ, r) we proceed analogously.Now, since by Theorem 3.1 the statement holds for c = −3/4, by the continuity

of Jc in the Hausdorff metric (see [4]) we can find r > 0 such that

Jc ∩ {z ∈ C : r/2 � |z + 1/2| � r} ⊂ S+c (θ, r) ∪ S−

c (θ, r),

for parameters from a suitable interval [−3/4, c0). It follows from (3.2) that thepoints z ∈ Jc can “escape” from B(−1/2, r/2) only through S+

c (θ, r) ∪ S−c (θ, r).

So, because a set of preimages of any piont is dense in Jc, the statement holds. �

Since |1/2+αc| ≈ δc/2 is small with respect to | Im(z)| �√δc, the above lemma

implies the following (we assume Arg(z) ∈ (−π, π]):

Corollary 3.3. For every ε > 0 there exists a neighborhood V of −1/2 and c0 >−3/4 such that if z ∈ V ∩ Jc, c ∈ [−3/4, c0) and z �= −1/2, then the ratio of eachtwo of the quantities

| Im(z)|, |z − αc|, |z + 1/2| = 1

2|f ′

c(z) + 1|, 1

2(π − |Arg(z)|),

belongs to (1− ε, 1 + ε).

Lemma 3.4. For every c ∈ [−3/4, 1/4] we have

B(0, 1/2) ⊂ Kc.

Moreover if c ∈ (−3/4, 1/4), then B(0, 1/2) ∩ Jc = ∅, whereas for c ∈ {−3/4, 1/4}we have B(0, 1/2) ∩ Jc = {−1/2, 1/2}.

Proof. See Appendix A. �

Note that if c ∈ (−3/4, 1/4), then it follows from Lemma 3.4 that fc is expandingfor n = 1, since |f ′

c(z)| = |2z| > 1.

4. Fatou coordinates

In this section we introduce so-called Fatou coordinates, in which f2c (after

conjugation by an affine map) is close to translation by 2 on the set Jc nearp+c = − 1

2 + i√δc. Since we consider the second iteration of fc, p

+c is the repelling

fixed point, (f2c )

′(p+c ) = 1 + 4δc > 1. We have

f2c (z) = z4 + 2z2(−3

4+ δc)−

3

16− 1

2δc + δ2c .

Now we make some modifications. Let us first conjugate f2c by translation z �→

z−( 12−√1− δc) (the fixed point αc =

12−

√1− δc moves to 0) and next by rotation

z �→ −iz (90◦ to the right). We get

Fc(z) := (1− 2√1− δc)

2z + 6i(1−

√1− δc −

2

3δc)z2 + 2(2

√1− δc − 1)z3 − iz4.

Now the fixed point p+c corresponds to√δc + i(1−

√1− δc), which is close to

√δc.

So we move it precisely to√δc, conjugating Fc by a map z �→ acz. The coefficient

ac must be equal to√δc/(

√δc + i(1 −

√1− δc)), and we have |ac − 1| = O(

√δc).

The family of functions obtained in this way will be denoted by Fc.Thus the functions f2

c are conjugated to Fc by affine maps hc (i.e. hc ◦ f2c =

Fc ◦ hc), and hc converges to z �→ −i(z + 12 ) when c ↘ − 3

4 . In order to avoid any



misunderstanding, arguments of Fc (and the Fatou coordinates) will be denoted byz, while arguments of fc will be denoted by z. Thus, we have

Fc(z) = (1− 2√1− δc)

2z + 6aci(1−

√1− δc −

2

3δc)z2

+ 2a2c(2√1− δc − 1)z3 − ia3c z

4.

(4.1)

Let us define the Fatou coordinates as follows:

Zc(z) :=1

2δclog

( z2 − δcz2

).

The functions of the Zc are multi-valued, so we consider principal value of thelogarithm, log 1 = 0. Note that if δc → 0, then Zc converges to the Fatou coordinatefor c = −3/4 (parabolic point with two petals):

Zc(z) =1

2δclog

( z2 − δcz2

)=

1

2δclog

(1− δc

z2

)→ − 1

2z2=: Z−3/4(z).

The inverse functions are given by

(4.2) z = Z−1c (Z) =

( δc1− e2δcZ

) 12

.

These functions are injective in the horizontal strips of width π/δc in the left half-plane. Moreover, the image of the strip {Z : | Im(Z)| < π/(2δc),Re(Z) < 0}contains the angle {z : |Arg(z −

√δc)| < π/4}.

Lemma 4.1. For every ε > 0 there exist r > 0 and c0 > −3/4 such that if

z ∈ J(Fc) ∩B(0, r) ∩ {z ∈ C : Re(z) � 0} and c ∈ [−3/4, c0), then∣∣Zc(z) + 2− Zc(Fc(z))∣∣ < ε.

So, we can assume that Fc is in Zc as close to translation by 2 as we want.

Proof. Using functions Z−1c we can conjugate the translation Z �→ Z + 2 to a family

of odd functions Gc, which can be defined on the whole plane C. In order to provethat Fc is close to the translation in Zc, it is enough to show that Fc and Gc areclose to each other.

Let us take the Taylor series for Gc centered at√δc and for Fc at

√δc. Since

(Gc)′(z) = e4δc

( δcz − e4δc(z − δc)

) 32

,

we can get (in a fixed neighborhood)

Gc(z) =√δc + e4δc(z −

√δc)

+(6√δc +O(δ3/2c )

)(z −

√δc)

2

+(2 +O(δc)

)(z −

√δc)

3 +O((z −

√δc)

4).

(4.3)

The constant implicit in O((z −√δc)

4) does not depend on c ∈ [−3/4, c0) because

of convergence of Gc when c ↘ −3/4.



Using the fact that F ′c(√δc) = (f2

c )′(p+c ) = (1 + 4δc) and taking derivatives of

(4.1), we get

Fc(z) =√δc +

(1 + 4δc

)(z −

√δc)

+(6√δc +O(δc)

)(z −

√δc)

2

+(2 +O(

√δc)

)(z −

√δc)

3 +O((z −

√δc)

4),

(4.4)

where the constants implicit in O are independent of the parameter.Let us compare Gc(z)− z with Gc(z)− Fc(z). Using (4.3) and (4.4), we have

Gc(z)− z =(4δc +O(δ2c )

)(z −

√δc)

+(6√δc +O(δ3/2c )

)(z −

√δc)

2

+(2 +O(δc)

)(z −

√δc)

3 +O((z −

√δc)

4),

Fc(z)− Gc(z) = O(δ2c )(z −√

δc) +O(δc)(z −√δc)

2

+O(√δc)(z −

√δc)

3 +O((z −

√δc)

4).

We can find a neighborhood of 0 and c0 > − 34 such that the terms of Fc(z) −

Gc(z) are small with respect to the related terms of Gc(z)− z (the “tail” is smallwith respect to the third term). Because the argument of rotation implicit in the

affine maps conjugating f2c to Fc is close to −π/2, it follows from Lemma 3.2 that

Arg(z −√δc) is close to 0. Thus Fc(z)− Gc(z) is small with respect to Gc(z)− z,

so because Gc is translation in the coordinates Zc, the lemma follows. �

5. Cylinders

In this section we will define a partition of a neighborhood of the points of periodtwo. Pieces of this partition will be called cylinders. We will also prove some oftheir properties.

First we define a partition of a subset of the the unit circle ∂D. Let us considertwo closed arcs: I+ consisting of points with arguments belonging to [2π 3

12 , 2π512 ],

and I−, with arguments belonging to [2π 712 , 2π

912 ]. Note that periodic points p+0 ,

p−0 (with arguments 2π 13 , 2π

23 ) are included in I+, I−, respectively.

Let C1 be the set which consists of four closed arcs: points with argumentsbelonging to [2π 6

24 , 2π724 ], [2π

924 , 2π

1024 ] (which are included in I+), and [2π 14

24 , 2π1524 ],

[2π 1724 , 2π

1824 ] (included in I−). Next, for n > 1, let Cn be the component of the

preimage of Cn−1 under T (s) = s2 which is included in I+ ∪ I− (four connectedcomponents). So we have defined cylinders Cn for n � 1. Note that T mapsCn onto Cn−1 bijectively, interiors of Ci and Cj are disjoint provided i �= j, andI+ ∪ I− \ {p+0 , p−0 } =

⋃∞n=1 Cn.

The fact that the function Φc|∂D conjugates T |∂D to fc|Jcallows us to define the

corresponding partition of a neighborhood of points p+c , p−c onto the sets Cn(c) ⊂ Jc,

n � 1, with properties as before: fc maps Cn(c) onto Cn−1(c) bijectively, etc.Let us also denote by C+

n (c) and C−n (c) the components of Cn(c) (each consists

of two connected components) which are included in upper and lower half-plane,respectively.



For N ∈ N let us define the set of points which are “near” the points of periodtwo:

MN :=⋃

n>N

Cn.

The rest, i.e. points which are “far”, is denoted by

BN := ∂D \MN .

Related subsets of Jc are denoted by MN (c) and BN (c).Instead of the diameters of the cylinders, we use a quantity which will be called

the size of the cylinder and denoted by |Cn(c)|. Let I+1 ⊂ ∂D be the set of points ofarguments belonging to [2π 1

4 , 2π13 ] (I

+1 ⊂ I+, and each cylinder Cn(c) has exactly

one connected component included in Φc(I+1 )). Fix n � 1, and let z1 be the point

of the largest argument of external ray among all belonging to Cn(c) ∩ Φc(I+1 ) (z1

is the image of the point from I+1 of the largest argument). We define

|Cn(c)| :=1

2|f2

c (z1)− z1|.

The point z1 is a common point of Cn(c) and Cn+1(c), while f2c (z1) is a common

point of Cn−2(c) and Cn−1(c). So there are components of two cylinders betweenz1 and f2

c (z1), and this is the reason why we put “1/2” in the definition.Since the set

⋃n∈N

fnc (0) is included in the real line, Lemma 3.2 and the Koebe

Distortion Theorem (see [6]) imply the following corollary:

Corollary 5.1. There exists a constant K > 1 such that for every c ∈ [−3/4, 1/4],n � 1 and z ∈ Cn(c),

(1) 1K |f2

c (z)− z| � 2|Cn(c)| � K|f2c (z)− z|,

(2) 1K diam(C±

n (c)) � |Cn(c)| � K diam(C±n (c)),

(3) 1K |Cn−k(c)| � |Cn(c)| · |(fk

c )′(z)| � K|Cn−k(c)|, provided 0 < k < n.

Later on we will show that the constant in the first statement can be arbitrarilyclose to 1 (for suitably large n, and c close to −3/4).

Now we use the Fatou coordinates which were introduced in the previous section.Let Cn(c) ⊂ J(Fc) denote the image of Cn(c) under the map conjugating f2

c to Fc

(C±n (c) we define analogously).

Lemma 5.2. For every K1 > 1, K2 > 1 there exist N ∈ N, c0 > −3/4 such that

for every n � N and c ∈ [−3/4, c0), if z ∈ Cn(c), then

(1) −K1n < Re(Zc(z)) < −(1/K1)n,(2) |Zc(z)| < −K2 Re(Zc(z)).

Proof. Fix c0 > −3/4 and a neighborhood U 0, such that for c ∈ [− 34 , c0) the

function Fc, in the coordinates Zc, is close to translation on U .Let us take a repelling periodic point z−3/4 ∈ U for the map F−3/4. Using the

Implicit Function Theorem we get a family of repelling periodic points zc ∈ J(Fc)depending analytically on the parameter. Let N be the number for which zc ∈CN (c). We can also assume that zc ∈ U for c ∈ [− 3

4 , c0).



So, in Zc we have a family of points such that Zc(zc) → Z−3/4(z−3/4) when

c ↘ − 34 . Thus there exist K1 > 1, K2 > 1 for which

−K1N < Re(Zc(zc)) < −(1/K1)N and |Zc(zc)| < K2 Re(Zc(zc)),

provided c ∈ [−3/4, c0) (zc ∈ CN (c)).Since the diameter of Cn(c) is comparable with the size (Corollary 5.1 (2)) and

Fc is close to translation in Zc, the diameters of Zc(Cn(c)) are uniformly bounded(Z−1

c is univalent on strips of width π/δc in the left half-plane, so we can control

distortion). Thus the above inequalities hold for each point from CN (c), and since

Fc is close to the translation, they also hold for cylinders Cn(c), n > N (possiblychanging K1 and K2). So, we have already proven that there exist K1 and K2 forwhich the statement holds.

In order to improve K1 and K2 it is enough to consider the sufficiently smallneighborhood U 0 and cylinders Cn(c), n > N + n, where n is sufficientlylarge. �Lemma 5.3. For every α > 0 there exist K > 1, c0 > −3/4 such that for everyc ∈ [−3/4, c0) and z ∈ Cn(c),

(1) if nδc > α, then (1/K)δc1/2 < |z − αc| < Kδc

1/2,(2) if nδc � α, then (1/K)n−1/2 < |z − αc| < Kn−1/2.

In particular, for every r > 0 there exist N ∈ N, c0 > −3/4 such that for c ∈[−3/4, c0) we have MN (c) ⊂ B(αc, r) (and MN (c) ⊂ B(−1/2, r)).

Proof. Fix α > 0. Since f2c and Fc are affinely conjugated, it is enough to prove

similar estimates with |z| instead of |z − αc| (conjugation maps αc onto 0). Hencewe need to estimate |z|. It follows from (4.2) that

(5.1) |z| =∣∣∣ δc1− e2δcZ

∣∣∣12

.

Let us consider the first statement. It follows from Lemma 5.2 (1) and assump-tion nδc > α that

δc Re(Z) < −δc(1/K1)n < −(1/K1)α,

and then ∣∣e2δcZ∣∣ = e2δc Re(Z) < e−2 1K1

α � K2 < 1.

So we have0 < K3 � |1− e2δcZ | < 1,

and using (4.2) we obtain

δc1/2 < |z| � K

−1/23 δc

1/2,

which gives the first statement.Now we prove the second statement. We can assume that n > N for some fixed

N ∈ N. By Lemma 5.2 (parts (2) and (1)) and next, by using assumption nδc � α,we get

K4δc|Z| < δc|Re(Z)| < K5δcn � K5α.

Hence we can assume that Re(δcZ) ∈ (−K.0), where K > 0 is a constant indepen-dent of the parameter. Since we also have |Re(Z)| > K4|Z| (then 1− e2δcZ = 0 ⇔Z = 0),

1

K6|2δcZ| < |1− e2δcZ | < K6|2δcZ|,



for suitable constant K6 > 1. Therefore, if Z = Zc(z), then (5.1) gives

K−1/26

∣∣∣ δc2δcZ

∣∣∣1/2 < |z| < K1/26

∣∣∣ δc2δcZ

∣∣∣1/2.Lemma 5.2 implies −K7n < |Z| < −(1/K7)n. Therefore we obtain

K−18 n−1/2 < |z| < K8n

−1/2,

which completes the proof. �

Corollary 5.4. For every N ∈ N there exists ε > 0 such that if c ∈ [−3/4, 0], then(Jc ∩B

(0,

1

2+ ε

))⊂

(MN (c) ∪ −MN (c)

).

Proof. Because the maps conjugating f2c and Fc converge to z �→ −i(z+ 1

2 ), Lemma3.2 gives us

J(Fc) ∩ {z : Re(z) > 0, |z| < r} ⊂ {z : |Arg(z −√δc)| < θ, |z| < r}.

Note that for every K we can find r and θ, so that the set {z : |Arg(z−√δc)| <

θ, |z| < r} is included in the half-plane {Z : Re(Z) < −K} in the Fatou coordinates(this is not true for a disk centered at

√δc).

Hence by Lemma 5.2, in some neighborhood of the fixed point√δc there are

only cylinders Cn(c) with large indexes. Thus, also in some neighborhood of p+cand −1/2, cylinders Cn(c) have indexes greater than some large N .

It follows from Lemma 3.4 and the continuity of Jc in the Hausdorff metric, thatfor small ε the ball B(0, 1

2 + ε) can intersect Jc only close to ±1/2. But close to−1/2 there is MN (c), while close to 1/2 we have −MN (c). �

Lemma 5.5. (1) For every K > 0 and λ > 1 there exist r > 0, c0 > −3/4 suchthat if z, z′ ∈ B(αc, r), c ∈ [−3/4, c0) and |z − z′| < K|f2

c (z′)− z′|, then

1

λ|f2

c (z′)− z′| < |f2

c (z)− z| < λ|f2c (z

′)− z′|.

(2) For every λ > 1 there exist N ∈ N, c0 > −3/4 such that if z ∈ Cn(c), n > N ,and c ∈ [−3/4, c0), then we have

2

λ|Cn(c)| < |f2

c (z)− z| < 2λ|Cn(c)|

for every z′ ∈ Cn(c).

Proof. Put ζ := z−z′. Taking the Taylor series for f2c centered at z and subtracting

z = z′ + ζ from both sides, we get

f2c (z)− z = f2

c (z′)− z′ + ((f2

c )′(z′)− 1)ζ +O(ζ2).

Suitably choosing r and c0 we can assume that (f2c )

′(z′) is close to 1 and |ζ2| issmall relative to |ζ|. Hence, by assumption |ζ| < K|f2

c (z′)− z′| we obtain the first

statement.By Lemma 5.3 we can choose N so thatMN (c) ⊂ B(αc, r). Next using Corollary

5.1 we have 1K diam(C±

n (c)) � K|f2c (z) − z| for every z ∈ Cn(c). Taking as z′ the

point z1 ∈ Cn(c) used in the definition of the size of the cylinder, we can obtainthe second statement from the first. �



6. Size of cylinders

Lemma 6.1. There exist c0 > −3/4, K > 1 such that if c ∈ [−3/4, c0) andz ∈ Cn(c), n � 1, then

1

K|z − αc|3e−2Knδc � |Cn(c)| � K|z − αc|3e−

2K nδc .

Moreover, we can assume that K is arbitrarily close to 1 if z ∈⋃

n>N Cn(c) forsuitably chosen c0 > −3/4, N ∈ N.

Proof. It is enough to consider cylinders Cn(c) with indexes greater than some fixedN ∈ N, because then only finitely many of them remain and we may just changethe constant.

Let us consider a neighborhood U 0 on which Fc is sufficiently close to trans-lation in the coordinates Zc, for c ∈ [− 3

4 , c0). Changing c0 if necessary, we can

assume that there exists N ∈ N such that for every n > N we have Cn(c) ⊂ U (see

Lemma 5.3), and if z ∈ Cn(c), then Re(z) < −K1n for some K1 > 0 (see Lemma5.2 (1)).

Note that because of Corollary 5.1 (2), the diameters of Zc(Cn(c)) are uniformly

bounded (the diameter of Cn(c) is comparable with its size, and Fc in Zc is closeto a translation). The function Z−1

c is univalent on strips of width π/δc in the left

half-plane, so we can assume that distortion of Z−1c is close to 1 on Zc(Cn(c)) for

n > N (where N is large) and parameters close to − 34 .

Thus we obtain that for every K > 1 there exist N and c0 for which

1

K|(Z−1

c )′(Z)| < 1

2|Fc(z)− z| < K|(Z−1

c )′(Z)|,

provided Z−1c (Z), z ∈ Cn(c), n > N , and c ∈ [− 3

4 , c0).

Because the maps conjugating f2c and Fc converge to z �→ −i(z+ 1

2 ), the quantity

|Fc(z)− z| can be replaced by |f2c (z)− z|. Therefore, taking as z the point z1 used

in the definition of the size of a cylinder, we get

(6.1)1

K|(Z−1

c )′(Z)| < |Cn(c)| < K|(Z−1c )′(Z)|.

By equation (4.2), we conclude that

|(Z−1c )′(Z)| =

∣∣∣( δc1− e2δcZ

) 32

e2δcZ∣∣∣ = |z|3e2δc ReZ .

Combining this with (6.1), we get

K−1|z|3e2δc ReZ < |Cn(c)| < K|z|3e2δc ReZ .

So it remains to replace |z| by |z−αc| and estimate Re(Z) by −n up to a constant(Lemma 5.2). �

Proposition 6.2. For every α > 0 there exist K > 1, c0 > −3/4 such that forevery c ∈ [−3/4, c0) and z ∈ Cn(c),

(1) hyperbolic estimate: if nδc > α, then

1

Kδ3/2c e−2Knδc � |Cn(c)| � Kδ3/2c e−

2K nδc ,



(2) parabolic estimate: if nδc � α, then

1

Kn−3/2 � |Cn(c)| � Kn−3/2.

Proof. Combining Lemma 6.1 with the first point of Lemma 5.3 we obtain the firststatement.

The estimate from the second point of Lemma 5.3 gives

1

K1n−3/2e−2Knδc < |Cn(c)| < K1n

−3/2e−2 1K nδc .

Therefore, by assumption nδc � α we get

e−2KαK−11 n−3/2 < |Cn(c)| < K1n

−3/2e0,

and the proof is finished. �Remark 6.3. In the rest of the paper we will assume that α is fixed. The precisevalue of α (and the constant K in the above proposition) does not matter to us,but the optimal value of the constant in Lemma 6.1 is important.

Corollary 6.4. There exist K,K1,K2 > 0 and c0 > −3/4 such that for everyn � 1 and c ∈ [−3/4, c0), we have

K1δ3/2c e−Knδc � |Cn(c)| � K2n

−3/2.

Proof. For every λ > 0 there exists C > 0 such that x3/2 � Ceλx for x � 0. If wesubstitute nδc in place of x, then we get

δ3/2c e−λnδc � Cn−3/2.

Combining this inequality with Proposition 6.2 completes the proof. �

7. Invariant measures

In this section, in our case (the maps fc : Jc → Jc and the measures ωc) werecall the construction of σ-finite (finite for fc-hyperbolic) invariant measures (see[14]). These measures have already been denoted by μc. This idea was also used in[3] and [1], in the case where maps admit a parabolic fixed point.

The unit circle admits a natural partition with “Markov property”, correspond-ing to the dyadic development of the argument (divided by 2π). We denote cylin-ders of order 1 by B(0) (the closed arc between arguments 0 and π) and B(1) (theclosed arc between π and 2π). If B(j1, ..., jn) is a cylinder of order n � 1, thenits preimages under T (s) = s2 (cylinders of order n + 1) included in the upperand lower half-plane are denoted by B(0, j1, ..., jn) and B(1, j1, ..., jn), respectively.Using the Bottcher coordinate we can define the corresponding partition on Jc, forc ∈ [−3/4, 0] (we will use the same notation).

Now we define the partition B of Jc and the jump transformation, which helpsus construct invariant measures. We claim that for every parameter c ∈ [− 3

4 , 0]

and every piece B of this new partition, the iteration fkc mapping B onto Jc, as

well as each branch of f−nc on B, have uniformly bounded distortion. There is

no uniformity in a neighborhood of points of period two p±c (parabolic point forc = − 3

4 ), so let us denote by Dn the family of cylinders of order n which containthese points (each Dn consists of two cylinders), while the family of cylinders oforder n which belong to B will be denoted by Bn (Dn and Bn are disjoint). Thesets B(0) and B(1) contain points of period two, therefore belong to D1, and B1



is empty. D2 consists of B(0, 1) and B(1, 0), while B(0, 0) and B(1, 1) belong toB2. In general, for n � 1, Bn+1 consists of these two cylinders obtained by cuttingcylinders belonging to Dn which do not contain p±c . Thus, the families

B = {Bi}∞i=2 and {Dn} ∪ {Bi}ni=2

form partitions of Jc (in the first case without the points p±c ).Define the jump transformation f∗

c (z) : Jc → Jc by

f∗c (z) := fk

c (z),

provided z ∈ B and B ∈ Bk. It follows from [14, Lemma 4, Lemma 5] or [3, Lemma3.5, Proposition 3.6] that there exists the f∗

c -invariant probability measure μ∗c such

that the Radon-Nikodym derivative satisfies

(7.1) D−1 � dμ∗c

dωc� D,

and the constant D > 1 depends only on the distortion of fnc on the pieces of B.

Write h∗c := dμ∗

c/dωc.For every cylinder B = B(j1, ..., jn), let f

−nc,B be the inverse branch of fn

c whichmaps Jc onto B. Define

ΔB(z) :=dωcf

−nc,B

dωc(z) = |(f−n

c,B)′(z)|d(c).

It follows from [14, Theorem 2] or [3, Theorem 3.7] that there exists a unique (upto multiplicative constant) σ-finite fc-invariant measure μc equivalent to ωc whoseRadon-Nikodym derivative is given by the formula

(7.2)dμc

dωc(z) = h∗

c(z) +

∞∑n=1

∑B∈Dn

h∗c(f

−nc,B(z))ΔB(z).

Hence if A ⊂ Jc, then

(7.3) μc(A) = μ∗c(A) +

∞∑k=1

∑B∈Dk

μ∗c(f

−kc,B(A)).

Because the maps fc are hyperbolic for c ∈ (− 34 , 0], the measures μc are finite,

although we will see that μc(Jc) → ∞ when c ↘ − 34 (the week limit of μc is equal

to μ−3/4, which is only σ-finite under the assumption d(− 34 ) <

43 ). Hence we can

normalize μc, but it will be more comfortable to consider densities given by theformula (7.2).

Recall that the measures on the unit circle ∂D which correspond to μc and ωc

have already been denoted by μc, ωc (i.e. if A ⊂ ∂D, then μc(A) = μc(Φ−1c (A))

and ωc(A) = ωc(Φ−1c (A))). The theorems will be formulated in the case of μc and

ωc, while some of the proofs will be carried out using μc and ωc.

Lemma 7.1. There exists K > 1 such that for every n � 1 and c ∈ [−3/4, 0],

1

K

∞∑k=n

ωc(Ck) � μc(Cn) � K

∞∑k=n

ωc(Ck).

Proof. Fix n � 1; (7.3) leads to

(7.4) μc(Cn(c)) = μ∗c(Cn(c)) +

∞∑k=1

∑B∈Dk

μ∗c(f

−kc,B(Cn(c))).



Since the distortion of fnc on the pieces B is uniformly bounded, it follows from

(7.1) that we can consider ωc instead of μ∗c .

For every k � 1 there are two maps f−kc,B (B ∈ Dk). The image of C+

n (c)

under the first of them is equal to C+n+k(c) or C−

n+k(c), so the image of C−n (c) we

denote by E−n+k(c). The image of C−

n (c) under the second map is equal to C−n+k(c)

or C+n+k(c), and the image of C+

n (c) will be denoted by E+n+k(c). Thus, as the

image of Cn(c) under the maps f−kc,B , we get the whole cylinder Cn+k(c) and the set

E−n+k(c) ∪ E+

n+k(c).

Note that the image under f−kc,B of the Julia set included in the half-plane (upper

or lower) that contains the set C±n (c) that is mapped onto E±

n+k(c) is a cylinder

B′ ∈ Bk+1. Thus the distortion of f−kc,B restricted to Jc included in this half-plane

is uniformly bounded (independent of k � 1). So, because the interiors of cylindersB ∈ B are disjoint, for some K > 1 we get

ωc

( ∞⋃k=1

E−n+k(c) ∪ E+

n+k(c))< Kωc(Cn(c)).

The union of all images f−kc,B(Cn(c)) has the following form:

∞⋃k=1

⋃B∈Dk

f−kc,B(Cn(c)) =

∞⋃k=1

Cn+k(c) ∪( ∞⋃k=1

E−n+k(c) ∪E+

n+k(c)).

Hence, using (7.4), we obtain the assertion. �Now we are in a position to estimate the invariant measures of the cylinders

(compare [7, Lemma 3.7]).

Proposition 7.2. There exist K > 1 and c0 > −3/4 such that for every c ∈[−3/4, c0),

(1) if nδc > α, then

1

Kδ

32d(c)−1c e−Knδc � μc(Cn) � Kδ

32d(c)−1c e−

1K nδc ,

(2) if nδc � α, then

1

Kn− 3

2d(c)+1 � μc(Cn) � Kn− 32d(c)+1,

(3) if n � 1, then

1

Kδ

32d(c)−1c e−Knδc � μc(Cn) � Kn− 3

2d(c)+1.

Proof. The bounded distortion implies that ωc(Cn(c)) can be estimated by|Cn(c)|d(c), so using Lemma 7.1 we get that there exist K > 1 such that for everyn � 1,

1

K

∞∑k=n

|Ck(c)|d(c) < μc(Cn(c)) < K∞∑

k=n

|Ck(c)|d(c).

Statements (1) and (3) can be obtained by using Proposition 6.2 (1) and Corol-lary 6.4. Namely, we get a series of the form

(7.5)∞∑

k=m

δ32d(c)c e−Ckδc = δ

32d(c)c

e−Cnδc

1− e−Cδc� δ

32d(c)−1c e−Cnδc



(C > 0) and

(7.6)∞∑

k=m

k−32d(c) � n− 3

2d(c)+1,

where A � B denotes A/K < B < KA for a certain constant K which is indepen-dent of the parameter c ∈ [− 3

4 , c0).The right-hand side inequality in statement (2) follows from statement (3). By

(7.5) and (7.6) for m = [α/δc] + 1 we can conclude that

1

K1

∞∑k=[α/δc]+1

δ32d(c)c e−Kkδc >

e−Kα

K2

( 1

δc

)− 32d(c)+1

>∞∑

k=[α/δc]+1

1

K3k−

32d(c).

Hence the sum of “hyperbolic” estimates can be bounded below by the sum of“parabolic” ones. Therefore the left-hand side inequality follows from (7.6) andProposition 6.2. �

Lemma 7.3. There exists D > 1, and for every N ∈ N there exists λ(N) > 1 suchthat if c ∈ [−3/4, 0], then

D−1 � dμc

dωc|BN

� λ(N).

Moreover, dμc/dωc can also be uniformly estimated on the whole set ∂D if c ∈ [c0, 0],where −3/4 < c0 � 0. In this case the upper bound depends on c0.

Proof. It follows from (7.1) that h∗c can be bounded below by D−1. Thus using

(7.2), we get the left-hand side inequality.Fix N ∈ N. Note that the distance between BN (c) and the set of images of

the critical point can be estimated from below by a constant, independent of c ∈[−3/4, 0] (but dependent on N). Therefore, distortion of the inverse branches f−n

c,B

is uniformly bounded on BN (c). Hence there exists a constant K(N) such that forevery x, y ∈ BN (c), we have

∞∑n=1

∑B∈Dn

ΔB(x) < K(N)∞∑n=1

∑B∈Dn

ΔB(y).

The measures ωc are a probability, so it follows from (7.1) and (7.2) that theabove series converge, and next that the Radon-Nikodym derivatives dμc/dωc areuniformly bounded above.

In the case of the whole set Jc and c ∈ [c0, 0], we proceed analogously, since Jcis separated from the set of images of the critical point. �

Corollary 7.4. For every N � 1 there exist K > 1 and c0 > − 34 such that if

c ∈ (− 34 , c0), then

1

Kδ

32d(c)−2c � μc(MN ) � Kδ

32d(c)−2c .

Moreover, the same estimate holds for ∂D, provided d(− 34 ) <

43 .



Proof. Fix N � 1. By Proposition 7.2 and using (7.5), (7.6) we get

K1δ32d(c)−2c �

∞∑n=N+1

1

Kδ

32d(c)−1c e−Knδc � μc(MN )

�[α/δc]∑

n=N+1

Kn− 32d(c)+1 +

∞∑n=[α/δc]+1

Kδ32d(c)−1c e−

1K nδc

� K2(α

δc)−

32d(c)+2 +K2δ

32d(c)−2c e−

1K

αδc

δc � K3δ32d(c)−2c .

It follows from the assumption d(− 34 ) <

43 that the exponent 3

2d(c)−2 is negative;hence μc(MN ) → ∞ when c ↘ −3/4. Lemma 7.3 implies that μc(BN ) is uniformlybounded, so the above estimate (possibly with a different constant) holds for ∂D.

�

8. Position of the Julia set II

In Section 3 we have proven that in some neighborhood of − 12 the Julia set

is included in the angles S+c (θ, r), S−

c (θ, r) (see Lemma 3.2). Now we give moreprecise results. The main result is that the Julia set is close to the circle S(−1/2−√1− δc, 1) = S(αc − 1, 1). Notice that p±c = − 1

2 ± i√δc and αc = 1

2 −√1− δc

belong to S(αc − 1, 1).

−1/2

r=1p

p

+c

c−

cc−1 0

Figure 2. The arc of the circle S(αc − 1, 1) and the periodic points.

Let us fix θ and choose r and c0 as in Lemma 3.2. If z ∈ S(αc−1, 1), then we canwrite z = αc − 1 + eit for some t ∈ (−π, π]. Since we consider points from the setS+c (θ, r)∪S−

c (θ, r), we can assume that |t| < r and |t| � tc, where αc−1+eitc = p+c(hence tc � 0). If z /∈ S(αc − 1, 1), then z can be written in the form z = z0 + ζ,where z0 ∈ S(αc − 1, 1). So in this case we have z = αc − 1 + eit + ζ.

Let us note that if z = z0 + ζ, then

(8.1) fc(z) = fc(z0) + (2z0)ζ + ζ2.



We begin with a lemma which gives us information about the iteration of pointsbelonging to S(αc − 1, 1). This will help us draw a conclusion about the Julia set.

Lemma 8.1. For every ε > 0 there exist r > 0, c0 > −3/4 such that if z0 =αc − 1 + eit, where |t| ∈ [tc, r), and c ∈ [−3/4, c0), then

(1) if Im(z0) > 0, then Arg(fc(z0)− z0) = −π/2 + t/2,if Im(z0) < 0, then Arg(fc(z0)− z0) = π/2 + t/2.

(2)∣∣(f2

c (z0)− z0) + 2(fc(z0)− z0)∣∣ < ε|fc(z0)− z0|.

(3) dist(fc(z0), S(αc − 1, 1)

)< ε|f2

c (z0)− z0|.

Proof. Ad. (1) Let z0 = αc − 1 + eit = −1/2−√1− δc + eit. Then

fc(z0) − z0 =(34− δc + c

)+ 1 + 2

√1− δc − 2

√1− δce

it − eit + e2it − e−it.

Since 3/4− δc + c = 0, we get

(8.2) fc(z0)− z0 = (1 + 2√1− δc)(1− eit) + (e2it − e−it).

Note that the complex numbers (e2it − e−it), (1 − eit) are related to some chordsof the unit circle, where the first is subtended by three times greater angle. Thearguments of those numbers differ by π, and if we assume that Im(z) > 0 (t > 0),then the arguments will be equal to (π/2 + t/2) and (−π/2 + t/2), respectively.

Since p+c = −1/2−√1− δc + eitc and fc(p

+c ) = p−c = p+c , it follows from (8.2) that

(1 + 2√1− δc)(1− eitc) + (e2itc − e−itc) = 0.

Next, increasing t (t > tc � 0) we get

|(1 + 2√1− δc)(1− eit)| > |e2it − e−it|,

because the length of the shorter chord grows relatively faster. Therefore we have

Arg((1 + 2

√1− δc)(1− eit) + (e2it − e−it)

)= −π

2+

t

2

(not π/2 + t/2), which combined with (8.2) proves the first statement (in the caseIm(z) < 0 we proceed analogously).

Ad. (2) Set ζ = fc(z0)−z0. Because z0 ∈ S(αc−1, 1), we may put z = fc(z0) =z0 + ζ into (8.1), and then

fc(z) = f2c (z0) = fc(z0) + 2ζz0 + ζ2.

Since fc(z0) = z0 + ζ, we have

(8.3) f2c (z0) = z0 + ζ + 2ζz0 + ζ2.

For suitable r and c0, z0 = −1/2−√1− δc + e−it can be made sufficiently close to

−1/2, while ζ2 is small with respect to ζ. Hence, by (8.3) for every ε we can get|f2

c (z0)− (z0 + ζ − ζ)| < (ε/2)|ζ|, so∣∣(f2c (z0)− z0

)+ 2i Im(ζ)

∣∣ < (ε/2)|ζ|.

Since ζ − i Im(ζ) = Re(ζ) = |ζ| sin(t/2), then decreasing r if necessary, we canassume that 2(ζ − i Im(ζ)) < (ε/2)|ζ|, and then∣∣(f2

c (z0)− z0) + 2ζ∣∣ < ε|ζ|,

which gives the second statement because ζ = fc(z0)− z0.



Ad. (3) Since Arg(ζ) = −π/2+ t/2, the vector ζ at the point −1/2−√1− δc+

e−it makes the angle 3t/2 with the tangent to the circle S(αc − 1, 1). So, becausefc(z0) = z0 + ζ changing r if necessary, we can get

dist(fc(z0), S(αc − 1, 1)

)< ε|ζ|.

By the second statement |2ζ| < |f2c (z0)− z0|+ ε|ζ|. Therefore

dist(fc(z0), S(αc − 1, 1)

)< ε

1

2

(∣∣f2c (z0)− z0

∣∣+ ε|ζ|)< ε

∣∣f2c (z0)− z0

∣∣,which gives the third statement. �

If z ∈ S+c (θ, r) ∪ S−

c (θ, r), then in the remainder of this section we will denoteby z0 the closest point from S(αc − 1, 1). Moreover z − z0 will be denoted by ζ.Thus we have z = z0 + ζ and |ζ| = dist(z, S(αc − 1, 1)).

Now we prove the main results of this section.

Proposition 8.2. There exist K > 0 and c0 > −3/4 such that for every n ∈ N,

dist(z, S(αc − 1, 1)) � K|Cn(c)|,

provided z ∈ Cn(c), c ∈ [−3/4, c0).

Proof. It is enough to consider cylinders Cn(c) with indexes greater than somefixed N ∈ N, because there remains finitely many. Thus, we can assume thatz ∈ Jc ∩ (S+

c (θ, r) ∪ S−c (θ, r)) for some θ and r, whereas c belongs to the short

interval [−3/4, c0).If |ζ| � |f2

c (z0) − z0|, then the assertion follows from Lemma 5.5. It suffices totake z0 = z′ in the first point of the lemma, and next apply the second.

So we can assume that |f2c (z0) − z0| � |ζ| and that (2z0) is close to −1, while

|ζ2| is small with respect to |ζ|. Hence, (8.1) leads to

(8.4) |fc(z)− (fc(z0)− ζ)| < ε|ζ|,

for arbitrarily chosen ε.By Lemma 8.1 (3), distance between fc(z0) and S(αc − 1, 1) is small relative to

|f2c (z0)− z0|, so by assumption |f2

c (z0)− z0| � |ζ| is also relative to |ζ|. Thereforethe points z0 + ζ and fc(z0) − ζ are in different components of C \ S(αc − 1, 1).Then by (8.4) (fc(z) ≈ fc(z0)− ζ); in different components are z and fc(z), and so

also z and fc(z). Since |ζ| = dist(z, S(αc − 1, 1)), we get

|ζ| < |fc(z)− z|.

If z ∈ Cn(c), then fc(z) ∈ Cn−1(c), and using Corollary 5.1 we obtain

|ζ| < diam(Cn(c)) + diam(Cn−1(c)) < K|Cn(c)|,

and the proof is finished. �

Proposition 8.3. For every ε > 0 there exist N ∈ N and c0 > −3/4 such that ifc ∈ [−3/4, c0), then

(1) MN (c) ∩B(−3/2 + ε, 1− ε) = ∅,(2) MN (c) ∩ {z ∈ C : Re(z) > −1/2} = ∅.



Proof. Let us fix ε > 0 and consider the first statement.If z ∈ Cn(c), then applying Lemma 6.1 to Proposition 8.2, for some constant K

and set of parameters [−3/4, c0), we have

(8.5) dist(z, S(αc − 1, 1)) < K|z − αc|3.

Moreover, using Corollary 3.3, |z − αc| can be replaced by | Im(z)|. Next, if z′ ∈S(−3/2 + ε, 1− ε), we get

C(ε)| Im(z′)|2 � dist(z′, S(−3/2, 1)) � dist(z′, S(αc − 1, 1)).

So, using (8.5) we see that the distance from z to S(αc−1, 1) is of higher order thanthe distance between the circle S(−3/2 + ε, 1 − ε) and S(−3/2, 1) on the suitablelevel, and so even more between S(−3/2+ ε, 1− ε) and S(αc− 1, 1). Consequently,for suitable N and c0, we have MN (c) ∩B(−3/2 + ε, 1− ε) = ∅.

It remains to prove the second statement. If z′ ∈ S(αc − 1, 1), then

Re(z′) < αc −1

2Im2(z′) < −1

2+ 2

δc2

− 1

2Im2(z′).

Let us assume that Im2(z′) > 4δc, hence δc − 12 Im

2(z′) < − 14 Im

2(z′). Thus we get

1

4Im(z′)2 < −1

2− Re(z′) = dist

(z′,

{z : Re(z) = −1

2

}).

Using (8.5) with |z − αc| replaced by Im(z), analogously as before, we can obtainthe assertion for z ∈ MN (c) provided Im2(z) > 4δc.

For parameters close to −3/4, the set of points for which those conditions aresatisfied contains a cylinder “fundamental domain” (for c = −3/4 it finishes theproof, because δ−3/4 = 0). So, the trajectory of every point z ∈ MN (c) meets theset for which the proposition is already proven. We can also assume that it happensfor an even iteration. Hence we have Re(f2k

c (z)) � −1/2 for some k, and the wholetrajectory from z to f2k

c (z) is included in S+c (θ, r) ∪ S−

c (θ, r). Therefore, becausefc is univalent on this set, in order to finish the proof, it is enough to show that if| Im(z)| >

√δc, then

Re(z) = −1

2⇒ Re(f2

c (z)) > −1

2.

Indeed, it means that we cannot leave the set {z : Re(z) > −1/2} because forv >

√δc we have

Re(f2c

(− 1

2+ iv

))= −1

2+ (v2 − δc)

2 > −1

2,


The remainder of this section is devoted to formulate some corollaries. The mostimportant is the first statement of Corollary 8.4, because the value of the constantfrom the right-hand side is strongly related to the constant which will be obtainedin Lemma 11.7, which is the key to the proof of Theorem 1.1.

Corollary 8.4. For every ε > 0 there exist N ∈ N, c0 > −3/4 such that ifz ∈ MN (c) and c ∈ [−3/4, c0), then

(1)∣∣f ′

c(z)∣∣2 − 1 �

(6 + ε

)∣∣z + 12

∣∣2,(2)

∣∣f ′c(z)

∣∣− 1 �(3 + ε

)∣∣z + 12

∣∣2.



Proof. Now we prove the first statement. In order to estimate |f ′c(z)|2 − 1, let us

consider the triangle with vertexes 0,−1, 2z. By the law of cosines, we get

(8.6) |f ′c(z)|2 = |2z|2 = 1 + |2z + 1|2 − 2 · |2z + 1| cos(Arg(2z + 1)).

If z ∈ MN (c), then Proposition 8.3 leads to

−|z + 1

2 |2(1− ε)

� cos(Arg(2z + 1)) � 0,

where the right-hand side inequality follows from the second point, while the left-hand side follows from the first. Indeed, for z ∈ S(−3/2 + ε, 1 − ε), we have|z+1/2| = −2(1−ε) cos(Arg(z+1/2)) (in polar coordinates the circle S(−1+ε, 1−ε)satisfies r = −2(1− ε) cosϕ). Hence, (8.6) leads to

|f ′c(z)|2 − 1 � 4|z + 1/2|2 + 2|z + 1/2|2 1

1− ε� (6 + 3ε)|z + 1/2|2,

and the first statement is proved.In order to prove the second statement, it is enough to divide both sides by

|f ′c(z)|+ 1 and use the fact that |f ′

c(z)|+ 1 � 2. �

Corollary 8.5. For every ε > 0 there exist N ∈ N, c0 > −3/4 such that ifz ∈

⋃n>N Cn(c) and c ∈ [−3/4, c0), then

(1) (1− ε)|Cn(c)| � | Im(fc(z))| − | Im(z)| � (1 + ε)|Cn(c)|,(2) 2(1− ε)|Cn(c)| � |Arg(f2

c )′(z)| � 2(1 + ε)|Cn(c)|.

Moreover, Arg(f2c )

′(z) > 0 ⇔ Im(z) > 0.

Proof. Let us fix ε > 0. We will assume that z ∈ MN (c) for suitable N ∈ N,c ∈ [−3/4, c0) and Im(z) > 0 (recall that z = z0 + ζ, where z0 ∈ S(αc − 1, 1)).

Now we consider the first statement. Subtracting z0 from both sides of (8.1) andtaking imaginary parts, we get

(8.7) Im(fc(z)− z0) = Im(fc(z0)− z0) + Im(2z0ζ + ζ2).

Let us estimate the expression from the right-hand side. Lemma 8.1 (2) implies

(1− ε)|f2c (z0)− z0| < 2|fc(z0)− z0| < (1 + ε)|f2

c (z0)− z0|.

By Lemma 5.5 and Proposition 8.2, we have

(1− ε)2|Cn(c)| < |fc(z0)− z0| < (1 + ε)2|Cn(c)|.

Using Lemma 8.1 (1) we can assume that Arg(fc(z0)−z0) is close to −π/2 (becauseIm(z0) > 0). Hence the above inequalities give us

(8.8) (1− ε)3|Cn(c)| < − Im(fc(z0)− z0) < (1 + ε)2|Cn(c)|.

Next, because 2z0 is close to −1, |ζ| small relative to |ζ2|, and Arg(ζ) close to 0or ±π, we can assume that | Im(2z0ζ + ζ2)| < ε|ζ|. Thus, Proposition 8.2 leads to

(8.9) | Im(2z0ζ + ζ2)| < εK|Cn(c)|.

For suitable N and c0, it follows from (8.7), (8.8), and (8.9) that

−(1 + ε)2|Cn(c)| − εK|Cn(c)| < Im(fc(z)− z0) < −(1− ε)3|Cn(c)|+ εK|Cn(c)|.



Since Arg(ζ) is close to 0 or ±π, and |ζ| < K|Cn(c)|, we can assume that

| Im(z + z0)| = | Im(ζ)| < εK|Cn(c)|.

So, multiplying the above inequalities by −1, we get

|Cn(c)| − 3ε(K + 1)|Cn(c)| < − Im(fc(z))− Im(z) < |Cn(c)|+ 3ε(K + 1)|Cn(c)|

which, because of assuption Im(z) > 0, gives the first statement.For suitable N and c0, both fc(z) and z are close to −1, so we can estimate

Arg(fc(z)) and Arg(z) by −π−Im(fc(z)) and π−Im(z), respectively, with arbitraryprecision. So, because we have Arg(f2

c )′(z) = Arg(fc(z)) + Arg(z), the second

statement follows from the first.If Im(z) > 0, then Im(fc(z)) < 0. Thus Arg(f2

c (z)) is greater than zero. ForIm(z) < 0 we proceed analogously. �

9. Introduction to the proof

Beginning with the next section we will estimate integrals from the formula (2.3).Also in Section 10 we will deal with the integral being in the denominator:∫

∂D

log |2Φc|dμc.

It is rather easy to prove that it can be bounded above and below by constants,independently of c ∈ [−3/4, 0]. Therefore, we will need the following estimate ofthe second integral (K > 0):

1

Kδ

32d(−

34 )−2

c �∫∂D

∂

∂c(log |2Φc|)dμc � Kδ

32d(−

34 )−2

c .

But, first we give a slightly different result, namely

(9.1)1

Kδ

32d(c)−2c �

∫∂D

∂

∂c(log |2Φc|)dμc � Kδ

32d(c)−2c ,

and the last step of the proof is devoted to replacing d(c) by d(− 34 ).

Let us define Φc :=∂∂cΦc and observe that

(9.2)∂

∂c(log |2Φc|) = Re

( Φc

Φc

).

Now we give some important properties of the function Φc. We know that Φc

conjugates T (s) = s2 to fc(z) = z2 + c; hence

Φc(s2) = Φ2

c(s) + c.

Differentiating both sides with respect to c, we obtain

Φc(s2) = 2Φc(s)Φc(s) + 1;

thus

Φc(s) = − 1

2Φc(s)+

1

2Φc(s)Φc(s

2).



Replacing s by s2, s4,..., s2m−1

, we can get

Φc(s) = −m−1∑k=0

1

2Φc(s) · 2Φc(s2) · ... · 2Φc(s2k)

+1

2Φc(s) · 2Φc(s2) · ... · 2Φc(s2m−1)

Φc(s2m).

If Φc(s) = z, then 2Φc(s) = f ′c(z). Therefore the above formula can be written as

follows:

(9.3) Φc(s) = −m∑

k=1

1

(fkc )

′(z)+

1

(fmc )′(z)

Φc(Tm(s)).

Note that for every c ∈ (− 34 ,

14 ) we have |(fn

c )′(z)| � K(c) > 1 (see Lemma 3.4), so

Φc(s) = −∞∑k=1

1

(fkc )

′(z).

If z ∈ Cn(c), then taking (9.3) for m = n, we divide Φc(s) onto two parts,the finite sum “until living the set M0(c)” and the “tail”. The finite sum will be

denoted by Ψc(z). Hence, if z ∈ Cn(c) we have

(9.4) Ψc(z) = −n∑

k=1

1

(fkc )

′(z).

If for the points p±c = Φc(e± 2

3πi) we set Ψc(p±c ) = Φc(e

± 23πi), then Ψc(z) will be

defined on the whole set M0(c), for c ∈ (− 34 ,

14 ).

On each cylinder Cn(c), we can also define Ψc(z) as the derivative of some implicitfunction. Let z ∈ Cn(c) and write F (c, z) = fn

c (z). Then we have Φc(Tn(s)) =

F (c,Φc(s)), and taking the c-derivative we get

Φc(Tn(s)) =

∂F

∂c(c,Φc(s)) +

∂F

∂z(c,Φc(s))Φc(s).

Since ∂F∂z (c,Φc(s)) =

∂F∂z (c, z) = (fn

c )′(z) �= 0, the following holds:

Φc(s) = −∂F∂c (c, z)∂F∂z (c, z)

+1

(fnc )

′(z)Φc(T

n(s)).

So, by (9.3) and (9.4) Ψc(z) is equal to the quotient of the derivatives, and henceis equal to the derivative of the implicit function Ψc defined by F (c,Ψc(z)) =

fnc (Ψc(z)) = p for suitable p (that is, p = fn

c0(z), if we want to get Ψc0(z)).

In Section 11, we will study the function Re(Ψc/Φc) (i.e. Re(Φc/Φc) withoutthe “tail”). It will be proven that on the set MN it can be estimated from aboveand below by positive constants. Notice that the main problem of the proof is toget the lower bound.

Since μc(MN ) is comparable to δ32d(c)−2c (see Corollary 7.4), it follows that the

estimates (9.1) hold for the integral∫MN

Re(Ψc/Φc)dμc. Next, we will prove that

the integral over BN is bounded (Section 12) and that the “tail” is not important(Section 13), which will give us (9.1).



10. Denominator

Now we estimate the integral from the denominator of the formula (2.3).

Proposition 10.1. There exists K > 1 such that for every c ∈ [−3/4, 0],

1

K�

∫∂D

log |2Φc|dμc � K.

Proof. Let us fix N , divide ∂D onto BN and MN , and assume that c belongs to ashort interval [−3/4, c0).

Now we estimate the integral restricted to BN . Note that if c ∈ [−3/4, 1/4], thenJc ⊂ B(0, 2). Hence for every z ∈ Jc we have log |f ′

c(z)| < log 4. Next, by Corollary5.4, for fixed N ∈ N, there exists ε > 0 such that(

|f ′c(z)| < 1 + 2ε

)⇒

(z ∈ MN (c) ∪ −MN (c)

).

Hence, if z ∈ BN (c) \ −MN (c), then log(1 + 2ε) � log |f ′c(z)| < log 4. It follows

form Lemma 7.3 that dμc/dωc is bounded above and below by λ(N) and D−1,respectively. Thus we have

D−1 log(1 + 2ε) · ωc(BN \ (−MN )) �∫BN

log |2Φc|dμc � λ(N) log 4 · ωc(BN ).

So, there exists C > 1 such that

(10.1)1

C�

∫BN

log |2Φc|dμc � C.

There remains to estimate the integral restricted to MN . By Lemma 3.4 thisintegral is bounded below by 0, so we deal with the upper bound. Let zn be asequence of points such that zn ∈ Cn(c). Then, for some constant K1 > 0 we have

∫MN

log |2Φc|dμc � K1

∞∑n=N+1

(|f ′c(zn)| − 1)μc(Cn(c)).

By Corollaries 8.4 and 3.3, |f ′c(zn)|−1 can be estimated from above by K2|zn−αc|2.

Next by using Lemma 5.3 it can also be estimated from above by K3δc or K3n−1.

So, it follows from Proposition 7.2 that the above integral can be bounded by

K4

[α/δc]∑n=N+1

n−1n− 32d(c)+1 +K4

∞∑n=[α/δc]+1

δc · δ32d(c)−1c e−

1K nδc

� K4

∞∑n=N+1

n− 32d(c) +K4δ

32d(c)c

∞∑n=[α/δc]+1

e−1K nδc � K5 +K6δ

32d(c)−1c � K7.

Thus we get

0 �∫MN

log |2Φc|dμc � K7.

Combining this and (10.1) we get the assertion for c ∈ [−3/4, c0).If c ∈ [c0, 0], then it follows from Lemma 7.3 that the measures μc are uniformly

bounded on ∂D. So, because |f ′c| is uniformly separate from zero on the set Jc (see

Lemma 3.4), the assertion follows. �



11. Estimation on the set MN

In this section we estimate the value of the function Re(Ψc/Φc) on the set MN .We prove that it can be bounded above and below by positive constants.

Since we investigate the behavior near the periodic point of period two, it isuseful to write Ψc in the following form:

(11.1) Ψc(z) = −m∑

k=1

( 1

(f2k−1c )′(z)

+1

(f2kc )′(z)

)= −

m∑k=1

f ′c(f

2k−1c (z)) + 1

(f2kc )′(z)

,

where z ∈ Cn(c) and n = 2m. If n = 2m+ 1, we must add 1(fn

c )′(z) .

If z ∈ Cn(c) and n1 < n, then we have

Ψc(z) = −n1∑k=1

1

(fkc )

′(z)−

n∑k=n1+1

1

(fn1c )′(z)

· 1

(fk−n1c )′(fn1

c (z)).

Thus, we get the formula which is similar to (9.3), namely

(11.2) Ψc(z) = −n1∑k=1

1

(fkc )

′(z)+

1

(fn1c )′(z)

Ψc(fn1c (z)),

where z ∈ Cn(c) and n1 < n.

Now we prove a lemma which says that the function Ψc is almost constant onthe cylinders C±

n (c).

Lemma 11.1. For every ε > 0 there exist N ∈ N, c0 > −3/4 such that if x, y ∈C+

n (c) (or x, y ∈ C−n (c)), n > N , and c ∈ (−3/4, c0), then

|Ψc(x)− Ψc(y)| < ε.

Proof. Since Ψc(x) = Ψc(x) and the sets C+n (c), C−

n (c) are placed symmetricallywith respect to the real axis, it is enough to prove the lemma for x, y ∈ C+

n (c).First note that if z ∈ Cn(c), 2k < n, then it follows immediately from Corollary

5.1 (3) that for some constant K1 > 1,

(11.3) K−11

|Cn−2k(c)||Cn(c)|

� |(f2kc )′(z)| � K1

|Cn−2k(c)||Cn(c)|

.

Let x, y ∈ C+n (c). We will estimate the difference between the terms of the sum

(11.1) for x and y (under the assumption n− 2k > n), namely

(11.4)

∣∣∣∣f′c(f

2k−1c (x)) + 1

(f2kc )′(x)

− f ′c(f

2k−1c (y)) + 1

(f2kc )′(y)

∣∣∣∣.After reducing to the common denominator, the numerator has the following form:

|(f ′c(f

2k−1c (x)) + 1)(f2k

c )′(y)− (f ′c(f

2k−1c (y)) + 1)(f2k

c )′(x)|.Adding and subtracting (f ′

c(f2k−1c (x)) + 1)(f2k

c )′(x), we can get

(11.5)∣∣(f ′

c(f2k−1c (x)) + 1)((f2k

c )′(y)− (f2kc )′(x)

)−(f ′c(f

2k−1c (y))− f ′

c(f2k−1c (x))

)(f2k

c )′(x)∣∣.

First, note that

(11.6) |f ′c(f

2k−1c (x))− f ′

c(f2k−1c (y))| = 2|f2k−1

c (x)− f2k−1c (y)| � K2|Cn−2k+1(c)|.



Next, let us estimate |(f2kc )′(y)−(f2k

c )′(x)|. If x, y ∈ Cn(c), then f2kc (x), f2k

c (y) ∈Cn−2k(c). Because a suitable branch of the inverse function of f2k

c is defined onthe disk of radius Im(f2k

c (x)) centered at x, by the Koebe Distortion Theorem (see[6]), we get∣∣∣∣ (f

2kc )′(x)

(f2kc )′(y)

∣∣∣∣ < 1 +K3|Cn−2k(c)|Im(f2k

c (x)), Arg

( (f2kc )′(x)

(f2kc )′(y)

)< K4

|Cn−2k(c)|Im(f2k

c (x)),

and then ∣∣∣∣ (f2kc )′(x)

(f2kc )′(y)

− 1

∣∣∣∣ < K5|Cn−2k(c)|Im(f2k

c (x)).

Combining this with (11.3), we obtain

(11.7) |(f2kc )′(x)− (f2k

c )′(y)| < K6|Cn−2k(c)|Im(f2k

c (x))· |Cn−2k(c)|

|Cn(c)|.

If we assume n− 2k > n for suitable n, then it follows from Corollaries 8.5 and 3.3that

Im(f2kc (x)) > | Im(f2k−1

c (x))| > K7|f ′c(f

2k−1c (x)) + 1|.

Thus, changing the constant if necessary, Im(f2kc (x)) can be replaced by |f ′

c(f2kc (x))

+ 1| in (11.7).Using (11.6), (11.7) (after replacing) and (11.3), the expression (11.5) can be

estimated from above by

K8|f ′c(f

2k−1c (x)) + 1| · |Cn−2k(c)|

|f ′c(f

2k−1c (x)) + 1|

|Cn−2k(c)||Cn(c)|

+K9|Cn−2k+1(c)| ·|Cn−2k(c)||Cn(c)|

.

The quotients of the sizes of the consecutive cylinders are bounded by a constant,so using (11.3) we can estimate (11.4) from above by

(K8

|Cn−2k(c)|2|Cn(c)|

+K10|Cn−2k(c)|2|Cn(c)|

)·(K1

|Cn(c)||Cn−2k(c)|

)2

.

Thus we get ∣∣∣∣f′c(f

2k−1c (x)) + 1

(f2kc )′(x)

− f ′c(f

2k−1c (y)) + 1

(f2kc )′(y)

∣∣∣∣ � K11|Cn(c)|,

provided n− 2k > n.There remains to estimate n terms of (9.4), but each derivative can be bounded,

as in (11.3), by the quotient of the sizes of the cylinders. So finally, Corollary 6.4leads to

|Ψc(x)− Ψc(y)| �n− n

2K11|Cn(c)|+ nK12

n− 32

n− 32

� K13n− 1

2 +K12n52n− 3

2 ,

which tends to zero independently of c ∈ (−3/4, c0). �

Note that Lemma 11.1 still holds if we admit points from two consecutive cylin-

ders C+n (c), C+

n−1(c), for instance z and fc(z). Therefore, since Ψc(z) = Ψc(z), weget the following corollary.



Corollary 11.2. For every ε > 0 there exist N ∈ N and c0 > −3/4 such that

|Ψc(z)− Ψc(fc(z))| < ε,

provided z ∈ MN (c) and c ∈ (−3/4, c0).

Before passing to estimations of Ψc(z) and Re(Ψc(z)/Φc(s)), let us write another

formula which gives dependence between Ψc(z) and Ψc(fc(z)). Equation (11.2) forn1 = 1 leads to

(11.8) Ψc(z) =1

f ′c(z)

(− 1 + Ψc

(fc(z)

)).

Lemma 11.3. For every ε > 0 there exist N ∈ N, c0 > −3/4 such that if z ∈MN (c) and c ∈ (−3/4, c0), then

(1) Im(z) and Im(Ψc(z)) are of the same sign,

(2) −ε < Re(Ψc(z)) <12 + ε.

Proof. Let us note that Ψc(p±c ) = ∂

∂c (−12 ± i

√δc) = i ±1

2√δc, so we may consider

points which belong to some cylinder Cn(c), and since Ψc(z) = Ψc(z), without lossof generality we can assume that Im(z) > 0.

Now we prove the first statement and the left-hand side inequality from the sec-ond. There exists n such that for z ∈

⋃n>n Cn(c) the arguments of the derivatives

of even iterates can be estimated by the sizes of the cylinders up to some multi-plicative constant (see Corollary 8.5). So, if z ∈ Cn(c) and n− 2k > n, then usingCorollary 6.4 for some constant K1, we get

0 < Arg(f2kc )′(z) � K1

∑n>n

n−3/2,

and this sum can be arbitrarily small for sufficiently chosen n. Using Proposition8.3 we can assume that

0 � Re(−f ′c(f

2k−1c (z))− 1)

and that Arg(−f ′c(f

2k−1c (z))− 1) is close to π/2. Hence,

(11.9) Re(− f ′

c(f2k−1c (z)) + 1

(f2kc )′(z)

)> 0,

and for some K2 > 0,

(11.10) Im(− f ′

c(f2k−1c (z)) + 1

(f2kc )′(z)

)> K2

Im(f2k−1c (z))

|(f2kc )′(z)| > 0,

provided z ∈ C+n (c) and n− 2k > n.

Let us assume that both n and n are even or odd (if not, then it is enough toincrease n by one). By (11.2), grouping terms in pairs, we get

(11.11) Ψc(z) = −n−n

2∑k=1

f ′c(f

2k−1c (z)) + 1

(f2kc )′(z)

+1

(fn−nc )′(z)

Ψc(fn−nc (z)).

The modulus |Ψc(fn−nc (z))| can be estimated independently of the parameter by a

constant (depending on n). So, by bounded distortion (see Corollary 5.1 (3)), we



get the following estimate of the “tail” of Ψc(z):∣∣∣∣ 1

(fn−nc )′(z)

Ψc

(fn−nc (z)

)∣∣∣∣ < |Cn(c)||Cn(c)|

K2(n) < K3(n)|Cn(c)|,

where K3(n) depends on n. By Lemma 6.1, |Cn(c)| can be bounded above byK4 Im

3(z). Thus for every ε for suitable N and c0 (n is fixed) we can get

(11.12)

∣∣∣∣ 1

(fn−nc )′(z)

Ψc

(fn−nc (z)

)∣∣∣∣ < K5(n) Im3(z) < ε Im(z),

where z ∈ Cn(c), n > N and c ∈ (−3/4, c0). Combining this, (11.9) and (11.11),

we obtain −ε < Re(Ψc(z)).

The estimate 0 < Im(Ψc(z)) follows from (11.10), (11.11), and the fact that the“tail” (11.12) can be bounded by expression (11.10) for k = 1, which is comparableto Im(z).

It only remains to prove that Re(Ψc(z)

)< 1

2 + ε. It follows from Corollary 11.2that we can assume

(11.13)∣∣Re (Ψc(fc(z))

)− Re

(Ψc(z)

)∣∣ < ε.

We are going to obtain a contradiction. Suppose that Re(Ψc(fc(z))) > 1/2 + ε forsome z ∈ C+

n (c). Hence

Re(− 1 + Ψc(fc(z))

)> −1/2 + ε.

Since Im(fc(z)) < 0, using the already proved part of the lemma we get

Im(− 1 + Ψc(fc(z))

)< 0.

Since |f ′c(z)| > 1 and we may assume that π/2 < Arg(f ′

c(z)) < π, formula (11.8)leads to

Re(Ψc(z)

)= Re

( 1

f ′c(z)

(− 1 + Ψc(fc(z))

))< −(−1/2 + ε) = 1/2− ε,

which gives us a contradiction with (11.13), because we have assumed that

Re(Ψc(fc(z))) > 1/2 + ε, and therefore Re(Ψc(fc(z)))− Re(Ψc(z)) > 2ε. �

Let us denote by α(z) the following quantity (Arg(z) ∈ (−π, π]):

α(z) =π − |Arg(z)|

2.

Note that α(z) is close to Im(z) (see Corollary 3.3).

Now we will study a relation between the value of the product |Ψc(z)| ·α(z) andthe argument Arg(Ψc(z)). Next, we will obtain a fundamental relation between

|Ψc(z)|·α(z) and Re(Ψc(z)/Φc(s)). We will proceed under the technical assumption

that |Ψc(z)| · α(z) > 1/100. We will see (Proposition 11.6) that in order to get the

crucial lower bound for Re(Ψc(z)/Φc(s)) by a positive constant, we need much

more, namely |Ψc(z)| · α(z) � C > 1/6. Hopefully, it will be proven (Lemma 11.7)

that |Ψc(z)| · α(z) > 1/4− ε.

Lemma 11.4. For every ε > 0 there exist N ∈ N, c0 > −3/4 such that if z ∈MN (c), c ∈ (−3/4, c0) and we have

|Ψc(z)| =β

α(z),



where β > 1/100, then

π

2+

α(z)

β

(β − 1

2− ε

)�

∣∣Arg(Ψc(z)

)∣∣ � π

2+

α(z)

β

(β − 1

2+ ε

).

Moreover, if Im(z) > 0, then the modulus can be omitted.

Proof. Let us fix ε > 0. We will assume that Im(z) > 0. Note that for suitable

chosen N and c0, α(z) is small, so because β > 1/100, we can assume that |Ψc(z)|is as large as we want.

By (11.8) we have

Arg(Ψc(z)

)=

(Arg(Ψc(fc(z))− 1)−Arg(f ′

c(z)))(mod 2π).

In this case we must add 2π; hence

(11.14) Arg(Ψc(z)

)−Arg

(Ψc(fc(z))− 1

)= −

(π − 2α(z)

)+ 2π.

By Lemma 11.3 the real part of Ψc is bounded. Because the modulus is large,it means that Arg(Ψc) is close to ±π/2. Again using Lemma 11.3 (under the

assumption Im(z) > 0) we get Arg(Ψc(z)) ≈ π/2 and Arg(Ψc(fc(z))) ≈ −π/2.

Since |Ψc| is large, we deduce that addition −1 changes the argument of Ψc(fc(z))about the inverse of the modulus. More precisely, for every ε > 0 we can choose Nand c0 so that

− 1 + ε

|Ψc(fc(z))|� Arg

(Ψc(fc(z))− 1

)−Arg

(Ψc(fc(z))

)� − 1− ε

|Ψc(fc(z))|.

Taking into account (11.14), we get

(11.15)(π + 2α(z)

)− 1 + ε

|Ψc(fc(z))|� Arg

(Ψc(z)

)−Arg

(Ψc(fc(z))

)

�(π + 2α(z)

)− 1− ε

|Ψc(fc(z))|.

Using Corollary 11.2 we can assume that the difference between Ψc(z) and

Ψc(fc(z)) is bounded by an arbitrary ε > 0. Therefore

∣∣Arg(Ψc(z))− Arg(Ψc(fc(z)))∣∣ < ε · 1

|Ψc(fc(z))|,

because the difference of the arguments can be estimated by the inverse of themodulus multiplied by the modulus of the difference (bounded by ε). So, (11.15)leads to

(π + 2α(z)

)− 1 + 2ε

|Ψc(fc(z))|� 2Arg

(Ψc(z)

)�

(π + 2α(z)

)− 1− 2ε

|Ψc(fc(z))|.

Replacing |Ψc(z)| by β/α(z) we get

π + 2α(z)

ββ − α(z)

β(1 + 2ε) � 2Arg

(Ψc(z)

)� π + 2

α(z)

ββ − α(z)

β(1 − 2ε),





|Ψc(z)| �1/2 + ε

α(z),

provided z ∈ MN (c), c ∈ (−3/4, c0).

Proof. Let z ∈ MN (c). We can assume that |Ψc(z)| > 1/100α(z) , otherwise there is

nothing to prove. For suitable c0 and N the modulus |Ψc(z)| is large, so by Lemma11.3 (the left-hand side inequality for ε/3) we can deduce that

|Arg(Ψc(z))| �π

2+

ε/2

|Ψc(z)|.

It follows from Lemma 11.4 (the left-hand side inequality for ε/2) that

π

2+

1

|Ψc(z)|

(α(z)|Ψc(z)| −

1

2− ε

2

)� |Arg(Ψc(z))|.

Hence, combining the above inequalities we get

α(z)|Ψc(z)| − 1/2− ε/2 � ε/2,

and the proof is complete. �

Proposition 11.6. For every ε > 0 there exist c0 > −3/4, N ∈ N such that ifz ∈ MN (c), c ∈ (−3/4, c0), and we have

|Ψc(z)| =β

α(z),

where β > 1/100, then

(6β − 1− ε

)� Re

( Ψc(z)

Φc(s)

)�

(6β − 1 + ε

).

Proof. Let us fix ε > 0 and assume that Im(z) > 0. By definition we have π −2α(z) = Arg(z) = Arg(Φc(s)), so Lemma 11.4 leads to

−π

2+

α(z)

β

(2β + β − 1

2− ε

)� Arg

( Ψc(z)

Φc(s)

)� −π

2+

α(z)

β

(2β + β − 1

2+ ε

).

Since β is bounded (see Corollary 11.5) we may assume that α(z) is small (forsuitable N and c0). Thus we can get the following:

α(z)

β

(3β − 1

2− ε

)− α(z)

βε � cos

(Arg

( Ψc(z)

Φc(s)

))� α(z)

β

(3β − 1

2+ ε

).

The real part is equal to the modulus multiplied by the cosine of the argument.Therefore, because for suitable N and c0 we have

1

2� |Φc(s)| �

1

2− ε,

it follows that

(2− ε)(3β − 1

2− 2ε

)� Re

( Ψc(z)

Φc(s)

)� 2

(3β − 1

2+ ε

),




Lemma 11.7. For every ε > 0 there exist N ∈ N and c0 > −3/4 such that

|Ψc(z)| >1/4− ε

α(z),

provided z ∈ MN (c), c ∈ (−3/4, c0).

Proof. It follows from Corollary 3.3 that α(z) can be replaced by Im(z). We will

also consider Im(Ψc(z)) instead of |Ψc(z)| (the real part is bounded). Thus, we aregoing to show the following:

(11.16) Im(Ψc(z)) >1/4− ε

Im(z).

Fix ε > 0 (small), and let z ∈ Mn(c) (n will be increased if necessary) andc ∈ (−3/4, c0). We also assume that Im(z) > 0, so we will be able to omit some ofthe moduli.

Step 1. First we estimate how large the growth of the imaginary part must be, thatis Im(Ψc(z)− Ψc(f

2c (z))), in order that

(11.17) Im(Ψc(f2c (z))) =

b

Im(f2c (z))

⇒ Im(Ψc(z)) �b

Im(z),

where z ∈ Mn(c) and c ∈ (−3/4, c0).

If Im(Ψc(f2c (z))) = b/ Im(f2

c (z)), then it is enough to verify when

Im(Ψc(z)− Ψc(f

2c (z))

)� b

Im(z)− b

Im(f2c (z))

.

We want to estimate the expression from the right-hand side of the above inequality,which is equal to

b · Im(f2c (z))− Im(z)

Im(z) Im(f2c (z))

.

For suitably n and c0, by Lemma 5.5 we get

Im(f2c (z))− Im(z) � |f2

c (z)− z| < 2(1 + ε)|Cn(c)|.Next, replacing |z−αc| by Im(z) in Lemma 6.1 and using the fact that the constantscan be arbitrarily close to one, we get

(11.18) Im(f2c (z))− Im(z) < 2(1 + ε)|Cn(c)| < 2(1 + 2ε) Im3(z).

So, implication (11.17) is satisfied under the following assumption:

(11.19) Im(Ψc(z)− Ψc(f

2c (z))

)� 2b(1 + 2ε)

Im2(z)

Im(f2c (z))

.

Step 2. Now we estimate the growth of the imaginary part of Ψc(z). Let us subtract

Ψc(f2c (z)) from both sides of (11.2) for n1 = 2. We obtain

(11.20) Ψc(z)− Ψc(f2c (z)) = −f ′

c(fc(z)) + 1

(f2c )

′(z)− (f2

c )′(z)− 1

(f2c )

′(z)Ψc(f

2c (z)).

Let us consider the first fraction from the right-hand side. Arg((f2c )

′(z)) is closeto zero (see Corollary 8.5), while it follows from Lemma 3.2 that Arg(−f ′

c(fc(z))−1) = π +Arg(fc(z) +

12 ) is close to π/2, so choosing n and c0 to ε, we can get∣∣∣Arg

(− f ′

c(fc(z)) + 1

(f2c )

′(z)

)− π

2

∣∣∣ < ε.



Thus, the sine of the considered expression can be estimated from below by (1 −ε2/2). Since we can also assume that |(f2

c )′(z)| < (1 + ε

2 ), we have

(11.21) Im(− f ′

c(fc(z)) + 1

(f2c )

′(z)

)� |f ′

c(fc(z)) + 1|1 + ε

2

(1− ε2

2

)

>2 Im(z)

1 + ε2

(1− ε2

2

)> (2− 2ε) Im(z).

In order to estimate the second fraction from (11.20), we replace (f2c )

′(z)− 1 by|f ′

c(z)|2 − 1. If z ∈ C+n (c), then z ∈ C−

n (c) and fc(z) ∈ C−n−1(c). Hence∣∣f ′

c(z)− f ′c(fc(z))

∣∣ = 2|z − fc(z)| < K1|Cn(c)|.

Next multiplying both sides by |f ′c(z)| and using Lemma 6.1 (and Corollary 3.3),

we get ∣∣|f ′c(z)|2 − (f2

c )′(z)

∣∣ < K2|Cn(c)| � K3|z − αc|3 � K4 Im3(z).

Therefore, the error which we make by replacing (f2c )

′(z) − 1 by |f ′c(z)|2 − 1 is

bounded by K4 Im(z)3 < ε Im2(z). Note that Corollary 8.4 (1) implies

|f ′c(z)|2 − 1 � (6 + ε) Im2(z).

Since |(f2c )

′(z)| � 1, we obtain

(11.22)∣∣∣ (f2

c )′(z)− 1

(f2c )

′(z)

∣∣∣ � (6 + ε) Im2(z) + ε Im2(z) = (6 + 2ε) Im2(z).

Combining (11.21), (11.22) and formula (11.20), we derive the following estimate:

Im(Ψc(z)− Ψc(f

2c (z))

)� Im

(− f ′

c(fc(z)) + 1

(f2c )

′(z)

)−∣∣∣ (f2

c )′(z)− 1

(f2c )

′(z)Ψc(f

2c (z))

∣∣∣� (2− 2ε) Im(z)− (6 + 2ε) Im2(z)|Ψc(f

2c (z))|.

Step 3. Let us assume that |Ψc(f2c (z))| �

1/4−ε/2Im(f2

c (z)). The above estimate leads to

Im(Ψc(z)− Ψc(f

2c (z))

)� (2− 2ε) Im(z)− (6 + 2ε)

(14− ε

2

) Im2(z)

Im(f2c (z))

� (2− 2ε)Im2(z)

Im(f2c (z))

−(32− 5

2ε) Im2(z)

Im(f2c (z))

>1

2(1 + ε)

Im2(z)

Im(f2c (z))

.

Since for b � 1/4− ε/2 we have

1

2(1 + ε) >

1

2(1− 2ε)(1 + 2ε) � 2b(1 + 2ε),

the assumption (11.19) is satisfied. It means that for every b � 1/4 − ε/2 thegrowth is sufficiently large in order to satisfy (11.17), provided z ∈ Mn(c) andc ∈ (−3/4, c0).

It follows from (11.18) that Im(z) − Im(f2c (z)) tends to 0 faster than Im(z).

Hence

1/4− ε/2

Im(f2c (z))

− 1/4− ε

Im(z)=

( 14 − ε)(Im(z)− Im(f2c (z))) +

ε2 Im(z)

Im(f2c (z)) Im(z)

>ε/4

Im(f2c (z))

.



Using Lemma 11.1 we can assume that Im(Ψc(f2c (z)))− Im(Ψc(z)) is smaller than

ε/4Im(f2

c (z)). Thus we get

Im(Ψc(f2c (z))) >

1/4− ε/2

Im(f2c (z))

⇒ Im(Ψc(z)) >1/4− ε

Im(z).

Because (11.17) is satisfied for b � 1/4− ε/2, we obtain

(11.23) Im(Ψc(f2c (z))) >

1/4− ε

Im(f2c (z))

⇒ Im(Ψc(z)) >1/4− ε

Im(z)

for z ∈ Mn(c) and c ∈ (−3/4, c0).

Step 4. By Lemma 5.3 there exists η > 0 such that for every z′ ∈ C+n−1(c)∪C+

n (c),c ∈ (−3/4, c0), we have Im(z′) > η. Again by Lemma 5.3, N and c0 can be chosenso that, for every z ∈ C+

n (c), n > N , and c ∈ (−3/4, c0), we get Im(z) < εη. So,under these assumptions,

(11.24) Im(z′) >Im(z)

ε.

We can also assume that both n and N are even or odd.To finish the proof it remains to show that (11.16) holds for every z ∈ C+

N+1(c)∪C+

N+2(c), because by (11.23) it will also be satisfied for f−2c (z) ∈ C+

N+3(c)∪C+N+4(c),

and thus for every z ∈ MN (c).Let z′ = fN−n+2

c (z). We can assume that (11.16) does not hold for any pointof the form f2k+2

c (z), 0 � k � N−n2 . Therefore, growth of the imaginary part is

sufficiently large to satisfy (11.17) for every b � 1/4− ε/2 (see Step 3). Hence

N−n2∑

k=0

Im(Ψc(f2kc (z))− Ψc(f

2k+2c (z))) +

1/4− ε/2

Im(z′)� 1/4− ε/2

Im(z).

Thus we get

Im(Ψc(z)) �1/4− ε/2

Im(z)− 1/4− ε/2

Im(z′)+ Im(Ψc(z

′)).

Im(Ψc(z′)) can be omitted, because by Lemma 11.3 it is greater than zero. Using

(11.24), we obtain

Im(Ψc(z)) �1/4− ε/2

Im(z)− ε

1/4− ε/2

Im(z)>

1/4− ε

Im(z),


Combining Corollary 11.5, Lemma 11.7 and Proposition 11.6 we conclude:


1

2− ε � Re

( Ψc(z)

Φc(s)

)� 2 + ε,

provided z ∈ MN (c), c ∈ (−3/4, c0).



12. Estimation on the set BN

In this section, in much the same way as G. Havard and M. Zinsmeister in [7,Proposition 4.1], we will estimate from above the following integral:

∫BN

∣∣∣ Φc

Φc

∣∣∣dμc.

If we simply repeat the estimations from [7], we get the upper bound of the form

λ(N)δ32d(c)−2c . But using results from Section 11, we show that this integral can be

estimated by a constant (depending on N).Let us begin with the following lemma.

Lemma 12.1. There exists c0 > −3/4, and for every N ∈ N there is a constantλ(N) > 0 such that if z ∈ Jc and c ∈ [−3/4, c0), then

fnc (z) ∈ BN (c) ⇒ 1

|(fnc )

′(z)| �λ(N)

n3/2.

Proof. Let us fix N ∈ N. By Corollary 5.4 there exists ε > 0 such that (Jc ∩B(0, 1/2 + ε)) ⊂ (MN ∪ −MN ).

If a point lies outside the set B(0, 1/2 + ε), then the modulus of the derivativeis greater than 1 + 2ε. Hence |(fn

c )′(z)| > (1 + 2ε)k, where k denotes the num-

ber of points from the trajectory which lie outside B(0, 1/2 + ε) (by Lemma 3.4,|(fn

c )′(z)| � 1 for every z ∈ Jc).

If for some λ > 0 we have (1 + 2ε)k > λ−1n3/2, then the assertion holds. Hencewe assume that k < K log(n) for some K > 0 (which depends on ε and λ) andthat n is larger than some n0. Thus, the number of points in B(0, 1/2 + ε) canbe bounded below by n/2 (for suitably large n0), whereas the maximal numberof consecutive points in B(0, 1/2 + ε) can be bounded below by n/2k. Therefore,bounded distortion implies

|(fnc )

′(z)| > K1(1 + 2ε)k|CN (c)|

|CN+[n/2k](c)|,

and then by Corollary 6.4 we get

|(fnc )

′(z)| > K2(1 + 2ε)k(N + [n/2k])3/2 > K3(1 + 2ε)k

2k3/2n3/2 > K4n

3/2,

which finishes the proof. �

Let us fix N . Then for every N0 ∈ N we define a family of sets {AN0,n}n�0

which form a partition of BN . Let

AN0,n = T−N0(CN+n) ∩ BN for n � 1, andAN0,0 = T−N0(BN ) ∩ BN .

The sets AN0,n(c) we define as the images of AN0,n under Φc.

Lemma 12.2. For every N ∈ N, there exists λ(N) > 0 such that for all c ∈[−3/4, 0],

μc(AN0,n) � λ(N) ·N0 · ωc(CN+n),

where {AN0,n}n�0 is the partition of BN .



Proof. The proof will be carried out for the sets AN0,n(c). Let us write the preimagef−N0c (CN+n(c)) in the following form (the first sum is related to the number of stepsback, which remain in the set MN (c)):

f−N0c (CN+n(c)) = CN+n+N0

(c) ∪N0−1⋃i=0

⋃ν

f−N0+ic,ν (CN+n+i(c)),

where the ν range over the set of all inverse branches of fN0−ic , which leaves the

set MN (c) in one step. Since we are interested in the intersection BN (c) andf−N0c (CN+n(c)) we can omit CN+n+N0

(c), so using Lemma 7.3 we are able toestimate μc(AN0,n(c)) from above by λ(N)ωc(AN0,n(c)).

Because of bounded distortion of f−N0+ic,ν on the set M0(c) (it follows from the

choice of the inverse branches), we have

ωc(f−N0+ic,ν (CN+n+i(c))) < KCνωc(CN+n+i(c)),

where Cν = ωc(f−N0+ic,ν (Jc)). Hence,

ωc(⋃ν

f−N0+ic,ν (CN+n+i(c))) < K

⋃ν

Cν · ωc(CN+n+i(c)) < Kωc(CN+n+i(c)).

So we get

μc(AN0,n(c)) < λ(N) · ωc(AN0,n(c))

< λ(N) · ωc

(N0−1⋃i=0

⋃ν

f−N0+ic,ν

(CN+n+i(c)

))

< λ(N)K

N0−1∑i=0

ωc(CN+n+i(c)) < λ(N)KN0 · ωc(CN+n(c)),

and the lemma follows. �

Now we formulate the main result of this section.

Proposition 12.3. There exists c0 > −3/4 such that for every N ∈ N there is aconstant λ(N) > 0 such that ∫

BN

∣∣∣ Φc

Φc

∣∣∣dμc � λ(N),

provided c ∈ (−3/4, c0).

Proof. It follows from Lemma 3.4 that |Φc| � 1/2, so it is enough to estimate the

integral of |Φc|.Let us fix N ∈ N and N0 ∈ N. The sets {AN0,n}n�0 form the partition of BN ;

hence

(12.1)

∫BN

|Φc|dμc =

∞∑n=0

∫AN0,n

|Φc|dμc.

Now we will study∫AN0,n

|Φc|dμc. Using (9.3) for m = N0 + n, we get

∫AN0,n

∣∣Φc

∣∣dμc �∫AN0,n

∣∣∣−N0+n∑k=1

1

(fkc )

′(z)+

1

(fN0+nc )′(z)

Φc

(TN0+n(s)

)∣∣∣dμc(s),



and then∫AN0,n

∣∣Φc

∣∣dμc �∫AN0,n

1

|(fN0+nc )′(z)|

∣∣Φc(TN0+n(s))

∣∣dμc(s)

+

∫AN0,n(c)

∣∣∣∣ 1

(fN0c )′(z)

n∑k=1

1

(fkc )

′(fN0c (z))

+

N0∑k=1

1

(fkc )

′(z)

∣∣∣∣dμc(z).

If z ∈ AN0,n, then fN0+nc (z) ∈ BN . Hence Lemma 12.1 leads to

∫AN0,n

∣∣Φc

∣∣dμc �λ1(N)

(N0 + n)3/2

∫AN0,n

∣∣Φc(TN0+n(s))

∣∣dμc(s)

+

∫AN0,n(c)

1

|(fN0c )′(z)|

∣∣Ψc(fN0c (z))

∣∣dμc(z) +N0 · μc(AN0,n).

Next, since μc is T -invariant and TN0+n(AN0,n) ⊆ BN , we get∫AN0,n

∣∣Φc

∣∣dμc �λ1(N)

(N0 + n)3/2

∫BN

|Φc|dμc

+ supz∈CN+n(c)

∣∣Ψc(z)∣∣ · μc(AN0,n(c)) +N0 · μc(AN0,n).

Using (12.1), and Lemma 12.2 we obtain

(12.2)

∫BN

|Φc|dμc �∞∑

n=0

(λ1(N)

(N0 + n)3/2

)∫BN

|Φc|dμc

+ λ2(N)N0

∞∑n=0

(sup

z∈CN+n(c)

∣∣Ψc(z)∣∣ωc(CN+n(c))

)+ λ2(N)N2

0

∞∑n=0

ωc(CN+n).

Let us consider the second expression of the above estimate. Let zN+n be anarbitrary point from CN+n(c). Then by Corollary 11.5 and Lemma 12.2 we have

∞∑n=0

supz∈CN+n(c)

∣∣Ψc(z)∣∣ωc(CN+n(c)) �

∞∑n=0

K1

|zN+n − αc|ωc(CN+n(c))

� K2

∞∑n=0

|CN+n(c)|d(c)|zN+n − αc|

� K3

∞∑n=0

|zN+n − αc|3d(c)−1e−2K (N+n)δc .

Next, applying Lemma 5.3, it can be bounded above by

K4

[α/δc]∑n=N

n− 32d(c)+

12 +K5

∞∑n=[α/δc]+1

δ32d(c)−

12

c e−2K nδc � K6 +K7δ

32d(c)−

32

c � K8.

Thus, estimate (12.2) leads to∫BN

|Φc|dμc � K9λ1(N)

N1/20

∫BN

|Φc|dμc + λ2(N)N0K8 + λ2(N)N20 ωc(MN−1),

hence (1−K9

λ3(N)

N1/20

)∫BN

|Φc|dμc < λ2(N)N0K8 + λ2(N)N20 .

So, by chosen suitably large N0, we get the assertion. �



13. The end of the proof

Proposition 13.1. There exist K > 1, N ∈ N and c0 > −3/4, such that for everyc ∈ (−3/4, c0),

1

Kδ

32d(c)−2c �

∫MN

Re( Φc

Φc

)dμc � Kδ

32d(c)−2c ,

provided d(−3/4) < 4/3.

Proof. Let s ∈ Cn, so Φc(s) = z ∈ Cn(c). Following (9.3) we have

Φc(s) = Ψc(z) +1

(fnc )

′(z)Φc

(Tn(s)

).

Hence the integral can be written as follows (MN =⋃

n>N Cn):

∫MN

Re

(Ψc(z)

Φc(s)

)dμc(s) +

∑n>N

∫Cn

Re

(1

(fnc )

′(z)·Φc

(Tn(s)

)Φc(s)

)dμc(s).

Let us fix ε > 0 (small). For suitably chosen N and c0, by Corollary 11.8 we get

(13.1)(12− ε

)μc(MN )−

∑n>N

∫Cn

∣∣∣∣ 1

(fnc )

′(z)

Φc

(Tn(s)

)Φc(s)

∣∣∣∣dμc(s)

�∫MN

Re( Φc

Φc

)dμc �

(2 + ε

)μc(MN ) +

∑n>N

∫Cn

∣∣∣∣ 1

(fnc )

′(z)

Φc

(Tn(s)

)Φc(s)

∣∣∣∣dμc(s).

Now we estimate the “tail”, i.e. the sum of the integrals. Let zn be an arbitrarypoint from the cylinder Cn(c). Bounded distortion implies

∑n>N

∫Cn

∣∣∣∣ 1

(fnc )

′(z)

Φc

(Tn(s)

)Φc(s)

∣∣∣∣dμc(s) <∑n>N

K1

|(fnc )

′(zn)|

∫Cn

∣∣Φc

(Tn(s)

)∣∣dμc(s).

Using the fact that μc is fc-invariant and Tn(Cn) ⊂ B0, and next Proposition 12.3and Lemma 12.1, we can continue the estimations as follows:

<

∫B0

∣∣Φc(s)∣∣dμc(s)

∑n>N

(K1

|(fnc )

′(zn)|

)< λ1(0)

∑n>N

(K1

|(fnc )

′(zn)|

)

< λ1(0)∑n>N

K1λ2(N)

n3/2< K2λ1(0)λ2(N) < K3(N).

Combining this with (13.1) we conclude that

(13.2)(12−ε

)μc(MN )−K3(N) �

∫MN

Re( Φc

Φc

)dμc �

(2+ε

)μc(MN )+K3(N).

Thus the assertion follows from Corollary 7.4, because the assumption d(−3/4) <

4/3 implies δ32d(c)−2c → ∞. �

Proof of Theorem 1.1. Propositions 13.1 and 12.3 imply that

1

K1δ

32d(c)−2c �

∫∂D

Re( Φc

Φc

)dμc � K1δ

32d(c)−2c



for some constant K1 > 1 and c ∈ (−3/4, c0). Thus, equation (9.2), Proposition10.1 and the formula (2.3) give

(13.3) −K2δ32d(c)−2c � d′(c) � − 1

K2δ

32d(c)−2c ,

for some constant K2 � 1.In order to finish the proof, we have to replace d(c) by d(−3/4) in the exponents

(possibly changingK2). From (13.3) we already know that d(c) is a decreasing func-tion on (−3/4, c0), and it follows from [11] or [2] that d(c) → d(−3/4). Therefore

δd(−3/4)3/2−2c < δ

d(c)3/2−2c , and d(c) can be replaced by d(−3/4) in the left-hand

side inequality. Thus we have to improve the estimate from above.

We need to show that there exists a constant C > 0 such that Cδd(c)3/2−2c <

δd(−3/4)3/2−2c or, equivalently, C < δ

d(−3/4)−d(c)c . Let us estimate d(−3/4) − d(c)

from above. It follows from (13.3) (the left-hand side inequality) that

d(−3/4)− d(c) = −∫ c

−3/4

d′(ζ)dζ � K2

∫ c

−3/4

(34+ ζ

) 32d(ζ)−2

dζ.

Since −3/4 < ζ � c < c0, we have d(ζ) � d(c), so d(ζ) can be replaced by d(c) inthe exponent under the integral. Therefore

d(−3/4)− d(c) � K2

∫ c

−3/4

(34+ s

) 32d(c)−2

ds � K3

(34+ c

) 32d(c)−1

= K3δ32d(c)−1c ,

and next

δd(− 3

4 )−d(c)c > δK3δ

32d(c)−1

cc = eK3 log(δc)δ

d(c) 32−1

c .

Since log(δc)δ32d(c)−1c → 0 when δc → 0, changing c0 if necessary, we can get

δd(− 3

4 )−d(c)c > C > 0 for c ∈ (−3/4, c0). The proof is finished. �

Appendix A

Lemma A.1. For every c ∈ [−3/4, 1/4] we have

B(0, 1/2) ⊂ Kc.

Moreover, if c ∈ (−3/4, 1/4), then B(0, 1/2) ∩ Jc = ∅, whereas for c ∈ {−3/4, 1/4}we have B(0, 1/2) ∩ Jc = {−1/2, 1/2}.

Proof. Note that for c = 0 the lemma is obvious because J0 = D.In order to prove that B(0, 1/2) ⊂ Kc, it is enough to show that

(A.1) f2c (B(0, 1/2)) ⊂ B(0, 1/2).

Let us parameterize ∂B(0, 1/2) by eit/2, and compute where |f2c | attains its

maximum. Hence, we solve the following:

∂

∂t

∣∣∣f2c

(12eit

)∣∣∣2 =∂

∂t

∣∣∣ 116

e4it +1

2c · e2it + c2 + c

∣∣∣2 = 0.



Decomposing f2c into real and imaginary parts, we get

(A.2)∣∣∣f2

c

(12eit

)∣∣∣2 =( 1

16cos 4t+

1

2c · cos 2t+ c2 + c

)2

+( 1

16sin 4t+

1

2c · sin 2t

)2

=1

256+

c2

4+ c2

(c+ 1

)2

+(c2 + c

)(18cos 4t+ c cos 2t

)+

c

16

(cos 4t cos 2t+ sin 4t sin 2t

)

=1

256+

c2

4+ c2

(c+ 1

)2+(c2 + c

)(18cos 4t+ c cos 2t

)+

c

16cos 2t.

Taking the t-derivative, we have

−(c2+ c

)(12sin 4t+2c sin 2t

)− c

8sin 2t = −

(c2+ c

)sin 2t

(cos 2t+2c+

1/8

c+ 1

).

Thus for c ∈ [−3/4, 0) ∪ (0, 1/4] the derivative is equal to 0 if and only if

(A.3) sin 2t = 0 or cos 2t = −2c− 1/8

c+ 1.

For c ∈ [−3/4, 1/4] we have −2c − 1/8c+1 ∈ (−1, 1], where the value 1 is attained

only for c = −3/4. Hence, if c ∈ (−3/4, 0) ∪ (0, 1/4], then the derivative has eightdifferent zeros (each equality in (A.3) gives four), whereas in the case c = −3/4,four zeros (0 and π have multiplicity 3).

Let us first consider the case c ∈ (0, 1/4]. The modulus |f2c | attains its maximum

if t = kπ/2. For k-odd we get

∣∣f2c (e

ik π2 /2)

∣∣ = ∣∣∣ 116

+1

2c+ c2

∣∣∣ � 1

4<

1

2,

while for k-even, ∣∣f2c (e

ik π2 /2)

∣∣ = ∣∣∣ 116

+3

2c+ c2

∣∣∣ � 1

2,

where equality is attained only for the parameter c = 1/4 (k-even). Thus weget (A.1) and the assertion for c ∈ (0, 1/4], because eik

π2 /2 = ±1/2 ∈ J1/4 (the

parabolic point and its preimage).If c ∈ [−3/4, 0), then the maximum is attained at the points eit/2 where t satisfies

the second equality of (A.3). Since cos 4t = 2 cos2 2t− 1, it follows from (A.2) that|f2

c |2 can be written as a function of c and cos 2t. Thus the value of |f2c |2 does not

depend on the choice of the solution of (A.3). Set w(c) := |f2c (e

itc/2)|2, where tc is

a solution of (A.3), and substitute in (A.2) −2c− 1/8c+1 in place of cos 2tc. We have

w(c) =1

256+

c2

4+ c2

(c+ 1

)2+(c2 + c

)(− c2 − 1

8+

1

256

1

(c+ 1)2

)

+c

16

(− 2c− 1/8

c+ 1

)= c3 + c2 − 1

8c+

1

256· 1

c+ 1.

We need to show that w(c) � 1/4, where c ∈ [−3/4, 0). Taking the c-derivative,we get

w′(c) =(3c2 + 2c− 1

8

)− 1

256

1

(c+ 1)2.



Let us multiply both sides by (c+1)2

c+3/4 and write the result as a function of δc = c+3/4.

(c+ 1)2

c+ 3/4w′(c) =

(δc + 1/4)2

δcw′(δc − 3/4) = 3δ3c − δ2c − δc −

1

8=: v(δc).

Since v is a polynomial of degree three, the leading coefficient is positive, and sincewe have v(−1/4) = 1/64 > 0, v(0) = −1/8 < 0, and v(3/4) = −11/64 < 0, v has nozeros between 0 and 3/4. Hence we conclude that w′(c) is negative in (−3/4, 0) andw attains its maximal value at c = −3/4, w(−3/4) = 1/4. Thus we get |fc|2 � 1/4,hence B(0, 1/2) ⊂ Kc. Since the equality is attained only for ±1/2 and these pointsbelong to J−3/4 (the parabolic point and its preimage), the lemma follows. �

References

[1] J. Aaronson, M. Denker, M. Urbanski, Ergodic theory for Markov fibered systems and para-bolic rational maps, Transactions of A.M.S. 337 (1993), 495–548. MR1107025 (94g:58116)

[2] X. Buff, L. Tan, Dynamical convergence and polynomial vector fields, Journal of DifferentialGeometry 77 (2007), 1–41. MR2344353 (2008k:37096)

[3] M. Denker, M. Urbanski, Absolutely continuous invariant measures for expansive rationalmaps with rationaly indifferent periodic points, Forum Math. 3 (1991), 561–579. MR1129999(92k:58154)

[4] A. Douady, Does a Julia set depend continuously on the polynomial?, Proceedings of Sym-posia in Applied Mathematics 49 (1994), 91–135. MR1315535

[5] A. Douady, P. Sentenac, M. Zinsmeister, Implosion parabolique et dimension de Hausdorff,C. R. Acad. Sci. Paris, Ser. I 325 (1997), 765–772. MR1483715 (98i:58195)

[6] G. M. Goluzin, Geometric theory of functions of a complex variable, Translations of Mathe-matical Monographs, vol. 26, AMS, 1969. MR0247039 (40:308)

[7] G. Havard, M. Zinsmeister, Le chou-fleur a une dimension de Hausdorff inferieure a 1,295,Preprint, 2000.

[8] G. Havard, M. Zinsmeister, Thermodynamic formalism and variations of the Hausdorff di-mension of quadratic Julia sets, Commun. Math. Phys. 210 (2000), 225–247. MR1748176(2002f:37082)

[9] J. H. Hubbard, Teichmuller Theory, vol. 1, Matrix Editions, 2006. MR2245223 (2008k:30055)[10] F. Przytycki, M. Urbanski Fractals in the Plane, Ergodic Theory Methods, London Math.

Soc. Lecture Note Series, vol. 371, Cambridge University Press, 2010. MR2656475[11] C. McMullen, Hausdorff dimension and conformal dynamics II: Geometrically finite rational

maps, Comm. Math. Helv. 75 (2000), 535–593. MR1789177 (2001m:37089)[12] C. McMullen, Hausdorff dimension and conformal dynamics III: Computation of dimension,

Amer. J. Math. 120 (1998), 471–515. MR1637951 (2000d:37055)[13] D. Ruelle, Repellers for real analytic maps, Ergodic Theory and Dynam. Syst. 2 (1982),

99–107. MR684247 (84f:58095)[14] F. Schweiger, Number theretical endomorphisms with σ-finite invariant measures, Isr. J.

Math. 21 (1975), 308–318. MR0384735 (52:5608)[15] M. Zinsmeister, Thermodynamic Formalism and Holomorphic Dynamical Systems,

SMF/AMS Texts and Monographs, vol. 2, 1996. MR1462079 (98k:58150)

Faculty of Mathematics and Information Sciences, Warsaw University of Technol-

ogy, Pl. Politechniki 1, 00-661 Warsaw, Poland

E-mail address: [email protected]


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