# Rigidity of Fibonacci Circle Maps with a Flat Piece and Different Critical Exponents

Bertuel Tangue Ndawa

► To cite this version:

Bertuel Tangue Ndawa. Rigidity of Fibonacci Circle Maps with a Flat Piece and Different Critical Exponents. Journal of Dynamical and Control Systems, 2024, 31 (2), pp.13. 10.1007/s10883-024-09687-z. hal-05283582

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# Rigidity of Fibonacci Circle Maps with a Flat Piece and Different Critical Exponents

TANGUE NDAWA Bertuel

University of Ngaoundere

University Institute of Technology

Computer Engineering

Ngaoundere (Cameroon)

bertuelt@yahoo.fr

https://orcid.org/0000-0001-8995-9522

February 25, 2024

To my Dad NDAWA Joseph and my Mom TCHOUMI Epse NDAWA Julienne.

**Abstract** We consider order preserving $C^3$ circle maps with a flat piece, Fibonacci rotation number, and negative Schwarzian derivative where the critical exponents (the degrees of the singularities at the boundary of the flat piece) might be different.

This paper treats the rigidity (geometrical) characteristic of a map of our class. We prove that when the critical exponents belong to $(1, 2)^2$, the geometry of the system is degenerate (double exponentially fast). As a consequence, the renormalization diverges, and the rigidity (geometric) class depends on three ordered pairs.

**Key words:** Circle map, Flat piece, Critical exponent, Geometry, Renormalization, and Rigidity.

## 1 Introduction

We are interested in a certain of weakly order preserving, non-injective (on an interval exactly; called flat piece) $C^3$ circle maps which appear naturally in the study of Cherry flows on the two dimensions torus (see [14],[18]), non-invertible circle continuous maps (see [16]) and of the dependence of the rotation interval on the parameter value for one-parameter families of circle continuous maps (see [21]). We write $S^1 = \mathbb{R}/\mathbb{Z}$ for the circle. There is the natural projection $\pi: \mathbb{R} \to S^1$. This provides a unique (up to integer translation) lift of a map $f$ of our class to a real continuous map $F$, and the rotation number $\rho(f)$ of $f$ is defined as:

$$ \rho(f) := \lim_{n \to \infty} \frac{F^n(x) - x}{n} (\text{mod } 1). $$
The concept of rotation number was first introduced by Poincaré J. H. in 1881, see [20]. Note that when the rotation number $\rho(f)$ is irrational (which is more interesting in the study of renormalization, see [6], pp.19-36, or point 1 of Remark 4.3), the denominators of the nearest rational approximants of $\rho(f)$ are defined as follows:

$$q_0 = 1, q_1 = a_0, \text{ and } q_{n+1} = a_n q_n + q_{n-1} \text{ when } n \ge 2$$

with

$$\rho(f) = [a_0 a_1 \cdots] := \frac{1}{a_0 + \frac{1}{a_1 + \frac{1}{\ddots}}}, a_i \in \mathbb{N}^*, i \in \mathbb{N}.$$

We say that the rotation number $\rho(f)$ is Fibonacci or golden mean if $a_n = 1$ for every $n \in \mathbb{N}$. A circle map with a Fibonacci rotation number will simply be called a Fibonacci circle map. For technical reasons (on the asymptotic behavior of the renormalization operator) the main theorem is proved for Fibonacci circle maps with a flat piece. This is what prompted the title of this paper.

The purpose of this paper is to describe the rigidity (geometrical characteristics) of certain weakly order preserving circle maps with a flat interval and Fibonacci rotation number. In the theory of low-dimensional dynamics, rigidity consists to study (specify) the geometric (rigidity) class under the topological class. Being given a circle map $f$, its topological class is: $\{h \circ f \circ h^{-1}, h \text{ is a circle homeomorphism}\}$. For a fixed irrational rotation number, there is exactly one topological class for $C^2$ circle map with a flat piece. That is, two $C^2$ circle maps $f$ and $g$ with a flat piece and the same irrational rotation number are topologically conjugate in the following sense: there exists a homeomorphism $h$ (called the topological conjugation or conjugacy between $f$ and $g$) such that $h \circ f = g \circ h$), see [14]. Note that if the choice of a topological conjugation could be arbitrary inside the flat piece $U_f$ of $f$ (that is, the image $f(U_f)$ is a point), it still uniquely defined on the attractor $K_f := S^1 \setminus \bigcup_{i=0}^{\infty} f^{-i}(U_f)$ of $f$. Moreover, the equality $h \circ f = g \circ h$ implies that $h(K_f) = K_g$. As a consequence, the geometry of $K_g$ depends on the one of $K_f$, and the regularity of $h$ on $K_f$. For example, if $h$ is differentiable on $K_f$, then $K_f$ and $K_g$ have the same geometry. The question of rigidity is formulated via the inverse of the previous statement: what is the minimal regularity of $h$ so that $K_f$ and $K_g$ have the same geometry? As in [13] (where the authors study the rigidity of some Fibonacci circle maps with a flat piece and the same critical exponent), when $h$ is $C^{1+\beta}$, $\beta > 0$ diffeomorphism, $K_f$ and $K_g$ (so $f$ and $g$) have the same geometry, and the geometry of two systems is rigid, it is not possible to modify it on asymptotical small scales. This result is very surprising according to the class of conjugacy. Indeed, the rigidity has been studied for circle diffeomorphisms [25] [10], critical circle homeomorphisms [2], [23], [7], [1] unimodal maps [15], [4], [3], circle maps with breakpoints [11], [12], for Kleinian groups [17]. And, in all these cases, it turned out that the conjugations are differentiable.

Before continuing to explain our results, we define our class and fix some notations.
**The class of functions:** We consider $S^1$ as an interval $[a, 1]$ where we identify $a$ with $1$ ($a \sim 1$). We consider the class $\mathcal{L}^X$ described as follows:

We fix $\ell_1, \ell_2 > 1$ and denote:

-by $\Sigma^X$ the set

$$ \Sigma^X = \{(x_1, x_2, x_3, x_4, s) \in \mathbb{R}^5 | x_1 < 0 < x_2, x_3 < x_4 < 1, 0 < s < 1\}, $$

- for $r \in \mathbb{N}$, by $Diff^r([0, 1])$ the space of $c^r$ orientation preserving diffeomorphisms of $[0, 1]$.

The space of $C^3$ circle maps with a flat interval is denoted by:

$$ \mathcal{L}^X = \Sigma^X \times (Diff^3([0, 1]))^3. $$

A point

$$ f := (x_1, x_2, x_3, x_4, s, \varphi, \varphi^l, \varphi^r) \in \mathcal{L}^X $$

is defined as follows:

$f : [x_1, 1] \to [x_1, 1]$ (by identifying $x_1$ with 1)

$$ f(x) = \begin{cases} f_{-}(x) = f_{1}(x) = (1 - x_{2})q_{s} \circ \varphi \left( \frac{x_{1} - x}{x_{1}} \right) + x_{2} & \text{if } x \in [x_{1}, 0] \\ f_{+}(x) = \begin{cases} f_{2}(x) = x_{1} \left( \varphi^{l} \left( \frac{x_{3} - x}{x_{3}} \right) \right)^{\ell_{1}} & \text{if } x \in [0, x_{3}] \\ f_{3}(x) = 0 & \text{if } x \in [x_{3}, x_{4}] \\ f_{4}(x) = x_{2} \left( \varphi^{r} \left( \frac{x - x_{4}}{1 - x_{4}} \right) \right)^{\ell_{2}} & \text{if } x \in [x_{4}, 1] \end{cases} & \text{if } x \in [x_{4}, 1] \end{cases} $$

where $q_s : [0, 1] \to [0, 1]$ is the centered ($q_s(0) = 0$) and reduced ($q_s(1) = 1$) diffeomorphic part of $x^{\ell_2}$ parameterized by $s \in (0, 1)$. Namely,

$$ q_s(x) = \frac{[(1 - s)x + s]^{\ell_2} - s^{\ell_2}}{1 - s^{\ell_2}}. $$

The diffeomorphism $q_s$ will play an important role in the study of the asymptotical behavior of the renormalization operator, see Section 4.3. The ordered pair $(\ell_1, \ell_2)$ is called the critical exponents of the map, and the interval $(x_3, x_4)$ is called the flat piece (interval) of the map. Note that the dynamics of $f_-$ and $f_+$ are defined by $f^{q_0}$ and $f^{q_1}$ respectively.

Observe that the $\Sigma^X$-components of a map $f \in \mathcal{L}^X$ are linked as follows:

$$ - f(x_1) = f_-(x_1) = x_2, $$

$$ - f_-(0) = 1 \sim x_1 = f_+(0), $$

$$ - f(x_3) = f_+(x_3) = 0, $$

$$ - f_+(x_4) = f(x_4) = 0, $$

$$ - f(1) = f_+(1) = x_2. $$

Therefore, since we consider circle maps with irrational rotation numbers, it is not possible that $x_2 \in [x_3, x_4]$ where we use the fact that the rotation number of a circle map is rational if and only if the map has a periodic point.

These kinds of maps can be represented as follows:
<figure><img src="image_2.png" /><figcaption>Figure 1: A map $f$ in $\mathcal{L}^X$ with $x_2 < x_3$</figcaption></figure>

or

<figure><img src="image_3.png" /><figcaption>Figure 2: A map $f$ in $\mathcal{L}^X$ with $x_4 < x_2$</figcaption></figure>

The different branches of a map $f$ in $\mathcal{L}^X$ are described as follows:

Let $[a, b]$ be an interval. $A_{[a,b]} : [a, b] \to [0, 1]$ is the map defined by
$x \mapsto (x-a)/(b-a)$ and $B_{[a,b]} : [a, b] \to [0, 1]$ is the map defined by $x \mapsto (x-b)/(a-b)$.
<figure><img src="image_4.png" /><figcaption>Figure 3: Branches of a map $f$ in $\mathcal{L}^X$</figcaption></figure>

## Notations.

**Common notations.** Let $f \in \mathcal{L}^X$ with $U$ as its flat piece.

1. For every $i \in \mathbb{Z}$, instead of $f^i(U)$ we will simply write $\dot{i}$. For example, $\underline{0} = U$. The number $|\dot{i}|$ stands for the length of the interval $\dot{i}$. So, $|\dot{i}| = 0$ when $i > 0$. $|(\dot{i}, \dot{j})|$ denotes the distance between the closest endpoints of these two intervals, while $|[(\dot{i}, \dot{j})]|$ stands for $|\dot{i}| + |(\dot{i}, \dot{j})|$.

2. Let $I$ be an interval. We write $\bar{I}$ to mean the closure of $I$.

3. The scaling ratios $\alpha_n$, $n \in \mathbb{N}$ are defined by

$$ \alpha_n := \frac{|(f^{-q_n}(U), U)|}{|(f^{-q_n}(U), U)| + |f^{-q_n}(U)|} = \frac{|(-q_n, \underline{0})|}{|[-q_n, \underline{0})|}, \quad n \in \mathbb{N}. $$

4. For any sequence $\Gamma_n$ and for any real $d$ we have:

$$ \Gamma_n^{d(\ell_1, \ell_2)} := \begin{cases} \Gamma_n^{d\ell_1} & \text{if } n \equiv 0[2] \\ \Gamma_n^{d\ell_2} & \text{if } n \equiv 1[2] \end{cases} \qquad \Gamma_n^{d(\frac{1}{\ell_1}, \frac{1}{\ell_2})} := \begin{cases} \Gamma_n^{d\frac{1}{\ell_1}} & \text{if } n \equiv 0[2] \\ \Gamma_n^{d\frac{1}{\ell_2}} & \text{if } n \equiv 1[2]. \end{cases} $$

For example, $\alpha_2^{\ell_1, \ell_2} = \alpha_2^{\ell_1}$.
5. Let $x_n$ and $y_n$ be two sequences of positive numbers. We say that $x_n$ is of the order of $y_n$ if there exists a uniform constant $k > 0$ such that, for $n$ big enough $x_n < ky_n$. We will use the notation

$$x_n = O(y_n).$$

**Parameters frequently used.** Let $f \in \mathcal{L}^X$ with $\ell_1, \ell_2 \in (1, 2)$ and Fibonacci rotation number.

1. The geometrical (rigidity) characteristics $(c_u(f), c'_u(f)), (c_+(f), c'_+(f))$ and $(c_s(f), c'_s(f))$ are defined in Proposition 1.2 and Lemma 4.14. Moreover,

- $c_{\iota}^*(f) = c_{\iota}(f)$ or $c_{\iota}'(f), \iota = s, u, +$. For more precision, see Proposition 1.2 or Notation 4.15;

- $\underline{c}_u(f) = \min\{c_u(f), c'_u(f)\}$ and $\bar{c}_u(f) = \max\{c_u(f), c'_u(f)\}$.

2. For every $n \in \mathbb{N}$, $S_{i,n}, y_{i,n}$; $i = 1, 2, 3, 4, 5$ are defined in the subsection 4.3 respectively in (4.4) and (4.5).

3. The vectors $c_{\iota}^{\prime\prime}$; $i = 2, 3, 4, 5$; $\iota = s, u, +$ are defined in Proposition 4.9.

We consider $f \in \mathcal{L}^X$ with $\ell_1, \ell_2 \in (1, 2)$ and Fibonacci rotation number. The study of the renormalization of $f$ generates the geometrical characteristics $(c_u(f), c'_u(f)), (c_+(f), c'_+(f))$ and $(c_s(f), c'_s(f))$ of $K_f$. This study depends on the following result:

**Lemma 1.1.** Let $f$ be a $C^3$ circle map with a flat piece $U$ and Fibonacci rotation number. The scaling ratios $\alpha_n$ go to zero double exponentially fast when $(\ell_1, \ell_2)$ belongs to $(1, 2)^2$.

We will additionally assume in the proof of Lemma 1.1 that, $f$ has a negative Schwarzian derivative. That means,

$$Sf(x) := \frac{D^3 f(x)}{Df(x)} - \frac{3}{2} \left( \frac{D^2 f(x)}{Df(x)} \right)^2 < 0; \quad \forall x, Df(x) \neq 0.$$

As a consequence of Lemma 1.1, the renormalization diverges with three quantitative aspects of the asymptotic divergence: an unstable part, a stable part, and a neutral part. The precise formulation is as follows:

**Proposition 1.2.** Let $(\ell_1, \ell_2) \in (1, 2)^2$. There exist $\lambda_u > 1$, $\lambda_s \in (0, 1)$ such that for $f$ a Fibonacci circle map with a flat piece and the critical exponents $(\ell_1, \ell_2)$ the following holds. There are three geometrical invariant ordered pairs $(c_u(f), c'_u(f)), (c_+(f), c'_+(f))$ and $(c_s(f), c'_s(f))$ such that for all $n \in \mathbb{N}$

$$\ln \alpha_n \asymp c_u^*(f) \lambda_u^{p_n} + c_s^*(f) \lambda_s^{p_n} + c_+^*(f)$$

where $p_n$ is the integer part of $n$, $c_{\iota}^*(f) = c_{\iota}(f)$, $\iota = s, u, +$ if $n$ is even and $c_{\iota}^*(f) = c_{\iota}'(f)$, $\iota = s, u, +$ if $n$ is odd. That is,

$$\ln \alpha_{2p_n} \asymp c_u(f) \lambda_u^{p_n} + c_s(f) \lambda_s^{p_n} + c_+(f)$$

and

$$\ln \alpha_{2p_n+1} \asymp c_u'(f) \lambda_u^{p_n} + c_s'(f) \lambda_s^{p_n} + c_+'(f).$$

Now, we can formulate the result on rigidity as follows:
**Main Theorem:** Let $(\ell_1, \ell_2) \in (1,2)^2$. Let $f$ and $g$ be two circle maps with a flat piece, Fibonacci rotation number, and $(\ell_1, \ell_2)$ as their critical exponents. If $h$ is the topological conjugation between $f$ and $g$, then there exists $\beta \in (0,1)$ such that

$h$ is Holder homeo;

$$h \text{ is a bi-lipschitz homeo} \iff c_u^*(f) = c_u^*(g), c_+(f) + c_+'(f) = c_+(g) + c_+'(g);$$

$$h \text{ is a } C^{1+\beta} \text{ diffeo} \iff c_u^*(f) = c_u^*(g), c_+(f) + c_+'(f) = c_+(g) + c_+'(g),$$

$$c_s^*(f) = c_s^*(g).$$

The rigidity is now completely described. This paper is organized as follows: in §2 and 3 we introduce the basic concepts and well-known results used in this paper. §4 is devoted to renormalization under Lemma 1.1: the scaling ratios $\alpha_n$ go to zero double exponentially fast when $(\ell_1, \ell_2)$ belongs to $(1,2)^2$. The asymptotic study of $\alpha_n$ does not use the main ideas of this paper and thus will be postponed until the last section, §6. The main theorem is proved in §5.

## 2 Technical tools

### Remark 2.1.

1. The sequences used in this paper are $((\ell_1, \ell_2), (\ell_1, \ell_2))$ equivalent on the parity in the following sense: if $\eta_n$ is a sequence, then $\eta_{2n}$ depends on $(\ell_1, \ell_2)$ by a function $\Psi_{2n}$ if and only if $\eta_{2n+1}$ depends on $(\ell_2, \ell_1)$ by the same function. That is,

$$\eta_{2n} = \Psi_{2n}(\ell_1, \ell_2) \iff \eta_{2n+1} = \Psi_{2n+1}(\ell_2, \ell_1).$$

As a consequence, a statement or proof presented for $n$ even deduces by himself the case $n$ odd and vice-versa.

2. The renormalization operator $\mathcal{R}$ maps $(\ell_1, \ell_2)$ to $(\ell_2, \ell_1)$. Namely, for a fixed $f \in \mathcal{L}^X$ with critical exponents $(\ell_1, \ell_2)$, if $\mathcal{R}f$ exists, then it belongs to $\mathcal{L}^X$ with critical exponents $(\ell_2, \ell_1)$. Thus, because the asymptotic study of the operator $\mathcal{R}$ on $f$ is equivalent to that of $\mathcal{R}^2$ on $(f, \mathcal{R}(f))$, then we are interested at the operator $\mathcal{R}^2$ which preserves the critical exponents.

**Fact 2.2.** Let $f \in \mathcal{L}^X$ with $U$ as its flat piece. Let $l(U)$ and $r(U)$ be the left and right endpoints of $U$ respectively. There are a left-sided neighborhood $I^{l_1}$ of $l_1(U)$, a right-sided neighborhood $I^{l_2}$ of $l_2(U)$ and three positive constants $K_1, K_2, K_3$ such that the following holds:

1. Let $y \in I^{l_i}$; $i = 1,2$. Then

$$K_1|l_i(U) - y|^{l_i} \le |f(l_i(U)) - f(y)| \le K_2|l_i(U) - y|^{l_i},$$

$$K_1|l_i(U) - y|^{l_i-1} \le \frac{df}{dx}(y) \le K_2|l_i(U) - y|^{l_i-1}.$$

2. Let three points with $y$ between $x$ and $z$ be arranged so that, of the three, the point $z$ is the closest to the flat piece. If the interval $(x, z)$ does intersect the flat piece $U$ (that is, $f$ is a diffeomorphism on $(x, z)$), then

$$\frac{|f(x) - f(y)|}{|f(x) - f(z)|} \le K_3 \frac{|x - y|}{|x - z|}.$$
**Remark 2.3.**

1. The first part of Fact 2.2 implies that for $i = 1, 2$; $f_{|I_i}(x) \approx k_i x^{\ell_i}$.

2. Let us note that, Fact 2.2 has also been applied in control theory, see [24].

# 3 Basic results

Let $f \in \mathcal{L}^X$ with $U$ as its flat piece.

**Proposition 3.1.** Let $n \ge 1$.

* The set of “long” intervals consists of the intervals

$$A_n := \{(q_{n-1} + i + 1, i + 1); 0 \le i < q_n\}.$$

* The set of “short” intervals consists of the intervals

$$B_n := \{(i + 1, i + 1 + q_n); 0 \le i < q_{n-1}\}.$$

The set $P_n := A_n \cup B_n$ covers the circle modulo the end points and is called $n^{th}$ dynamical partition. The dynamic partition produced by the first $q_{n+1} + q_n$ pre-images of $U$ is denoted $P^n$. It consists of

$$\wp_n := \{-i; 0 \le i \le q_{n-1} + q_n - 1\}$$

together with the gaps between these sets. As in the case of $P_n$ there are two kinds of gaps, “long” and “short”:

* The set of “long” intervals consists of the intervals

$$A^n := \{I_i^{n-1} := (-i, -q_{n-1} - i); 0 \le i < q_{n-1}\}.$$

* The set of “short” intervals consists of the intervals

$$B^n := \{I_i^n := (-q_n - i, -i); 0 \le i < q_{n-1}\}.$$

*Proof.* This comes from §1.4 in [9]. □

<figure><img src="image_5.png" /></figure>

**Proposition 3.2.** The sequence $|\langle 0, q_n \rangle|$ tends to zero uniformly and at least exponentially fast.

**Proposition 3.3.** If $A$ is a pre-image of $U$ belonging to $P^n$ and if $B$ is one of the gaps adjacent to $A$, then $|A|/|B|$ is bounded away from zero by a constant that does not depend on $n$, $A$ or $B$.

The proofs of these two results can be found in [9] (proof of Proposition 1 and Proposition 2 respectively).
**Cross-Ratio Inequality (CRI)** The cross ratio inequality was introduced and proved by several authors, see [21, 5, 26]. We shall adapt the one introduced in [21] and redefined in [8] to our needs.

Let $f \in \mathcal{L}^X$ with $U$ as its flat piece and negative Schwarzian derivative. Let $I, J \subset [x_1, 1]$ be two intervals of finite and nonzero lengths such that $\bar{I} \cap \bar{J} = \emptyset = (\bar{J} \cup \bar{I}) \cap \bar{U}$, and $J$ is on the right of $I$. We define their cross-ratio as

$$ \mathbf{Cr}(I, J) := \frac{|I||J|}{|[I, J)]||(I, J]|} $$

The distortion $\mathcal{DCr}$ of the cross-ratio by $f$ is given by

$$ \mathcal{DCr}(I, J; f) := \frac{\mathbf{Cr}(f(I), f(J))}{\mathbf{Cr}(I, J)}. $$

When there is no ambiguity we write $\mathcal{DCr}(I, J)$ instead of $\mathcal{DCr}(I, J; f)$.

Let $n \in \mathbb{N}$ such that

1. $f^n$ is a diffeomorphism on $(I, J)$;

2. Each point of the circle belongs to at most $k$ intervals $f^i([I, J])$.

Then,

$$ \mathcal{DCr}(I, J; f^n) = \prod_{i=0}^{n-1} \mathcal{DCr}(f^i(I), f^i(J); f) < 1. $$

**Remark 3.4.** Let $f \in \mathcal{L}^X$ with $U$ as its flat piece. Given $n \ge 1$, and let $T \subset [x_1, 1]$ be an interval. Then $f^n: T \to f^n(T)$ is diffeomorphism if and only if for every $0 \le i \le n-1$, $f^i(T) \cap \bar{U} = \emptyset$.

**Lemma 3.5.** For all $i = 1, \dots, q_{n-1} - 1$, the parameter sequence

$$ \varrho_n(i) := \frac{|-q_n + i|}{|(|q_{n-1} + i, -q_n + i]|} $$

is bounded away from zero.

*Proof.* Observe that if $0 \le i < q_{n-2}$, then $\varrho_n(i)$ is larger than

$$ \mathbf{Cr}(|-q_{n-2} + i, q_{n-1} + i|, -q_n + i) $$

which by **CRI** with $f^{q_{n-2}-i}$ is greater than

$$ \frac{|-q_{n-1}|}{|(|q_n, -q_{n-1}||} $$

times a uniform constant. Moreover, the above ratio is bigger than

$$ \frac{|-q_{n-1}|}{|(-q_{n+1}, -q_{n-1})|} $$

which by Proposition 3.3 go away from zero.

Now, suppose that $q_{n-2} \le i := q_{n-2} + j \le q_{n-1} - 1$. Then,

$$ \varrho_n(i) \ge \frac{|-q_{n-1} + j|}{|(-q_{n+1} + j, -q_{n-1} + j)|} $$

which by Proposition 3.3 go away from zero.

This completes the proof. $\square$

<figure><img src="image_6.png" /></figure>
A proof of the following Proposition can be found in [6] theorem: 3.1 pp .285.

**Proposition 3.6 (Koebe principle).** Let $f \in \mathcal{L}^X$. For every $\zeta, \alpha > 0$ there exists a constant $\zeta(\zeta, \alpha) > 0$ such that the following holds. Let $T$ and $M \subset T$ be two intervals and let $S, D$ be the left and the right component of $T \setminus M$ and $n \in \mathbb{N}$. Suppose that:

$$
1. \sum_{i=0}^{n-1} f^i(T) < \zeta,
$$

2. $f^n : T \to f^n(T)$ is a diffeomorphism,

$$
3. \frac{|f^n(M)|}{|f^n(S)|}, \frac{|f^n(M)|}{|f^n(D)|} < \alpha.
$$

Then,

$$
\frac{1}{\zeta(\zeta, \alpha)} \le \frac{D f^n(x)}{D f^n(y)} \le \zeta(\zeta, \alpha), \quad \forall x, y \in M.
$$

That is,

$$
\frac{1}{\zeta(\zeta, \alpha)} \cdot \frac{|A|}{|B|} \le \frac{f^n(A)}{f^n(B)} \le \zeta(\zeta, \alpha) \cdot \frac{|A|}{|B|}, \quad \forall A, B \text{ (intervals)} \subseteq M
$$

where

$$
\zeta(\zeta, \alpha) = \frac{1 + \alpha}{\alpha} e^{C\zeta}
$$

and $C \ge 0$ only depends on $f$.

4 Renormalization

The key of this part is to set up for large enough *n* a right algebraic relation
between the (*n* + 1)th renormalization and *n*th renormalization of the form:
*R*<sup>*n*+1</sup>*f* = *LR*<sup>*n*</sup>*f* + *C*<sub>0</sub>(*f*) with *f* a map in our class with Fibonacci rotation
number. The technique used requires a double change of variables on *Σ*<sup>*X*</sup>.
Let us mention once again that Lemma 1.1 (the scaling ratios *α*<sub>*n*</sub> go to zero
double exponentially fast when (*ℓ*<sub>1</sub>, *ℓ*<sub>2</sub>) belongs to (1, 2)<sup>2</sup>) will be used frequently
throughout this part.

4.1 Renormalization operator

Let $f \in \mathcal{L}^X$ with $[a_0, a_1, \dots]$ as its rotation number. If $1 \le a_0, a_1 < \infty$, then $f$ can be renormalized in the following sense: let $x_{1,1} = f^{a_1-1}(x_2)/x_1$, let $h: [x_1, f_4^{a_1-1}(x_2)) \to [x_{1,1}, 1)$ be the map $x \mapsto x/x_1$, and let $\mathcal{R}f: [x_{1,1}, 1) \to [x_{1,1}, 1)$ be the map defined by

$$
\mathcal{R}f(x) = \begin{cases} \mathcal{R}f_{-}(x) = h \circ f_{+} \circ h^{-1}(x) & \text{if } x \in [x_{1,1}, 0) \\ \mathcal{R}f_{+}(x) = h \circ f_{+}^{a_1} \circ f_{-} \circ h^{-1}(x) & \text{if } x \in (0, 1) \end{cases} \tag{4.1}
$$

More precisely,

$$
\mathcal{R}f(x) = \begin{cases} \mathcal{R}f_{-}(x) = h \circ f_{2} \circ h^{-1}(x) & \text{if } x \in [x_{1,1}, 0) \\ \mathcal{R}f_{+}(x) = h \circ f_{2} \circ f_{3}^{a_{1}-1} \circ f_{1} \circ h^{-1}(x) & \text{if } x \in (0, 1] \end{cases}
$$
Observe that

$$ \mathcal{R}f = h \circ p\mathcal{R}f \circ h^{-1} $$

where

$$ p\mathcal{R}f(x) = \begin{cases} \mathcal{R}f_{-}(x) = f_{+}(x) & \text{if } x \in (x_{1}, 0) \\ \mathcal{R}f_{+}(x) = f_{+}^{a_{1}} \circ f_{-} & \text{if } x \in (0, f_{4}^{a_{1}-1}(x_{2})] \end{cases} $$

Note that $\mathcal{R}f$ is nothing else than $p\mathcal{R}$ rescaled and flipped. The function $h$ is a normalization operator, $p\mathcal{R}$ is the first return (Poincaré) map of $f$ to the interval $(x_1, f_4^{a_1-1}(x_2)]$ and $\mathcal{R}f$ is called the first renormalization of $f$.

By $\mathcal{L}_0^X$ we denote the subset of $\mathcal{L}^X$ consisting of maps renormalizable.

Before explaining more precisely $\mathcal{R}f$ for a fixed $f$ in $\mathcal{L}_0^X$, we introduce the zoom map on an interval $[a, b]$. The zoom map $Z_{[a,b]} : \text{Diff}([0,1]) \to \text{Diff}([0,1])$ is defined by $\psi \mapsto A_{[\psi(a),\psi(b)]} \circ \psi \circ A_{[a,b]}^{-1}$ where $A_{[a,b]} : [a,b] \to [0,1]$ is the map $x \mapsto (x-a)/(b-a)$. We are now ready to describe more precisely the function $\mathcal{R}f$ for a fixed $f \in \mathcal{L}_0^X$.

**Proposition 4.1.** Let $f \in \mathcal{L}_0^X$ with $[a_0, a_1, \dots]$ as its rotation number, and $(\ell_1, \ell_2)$ as its critical exponents. If $1 \le a_0, a_1 < \infty$, then the renormalization map $\mathcal{R}f$ of $f$ belongs to $\mathcal{L}$. More precisely,

1. $\rho(\mathcal{R}f) = [a_1, a_2, \dots]$,

2. $\mathcal{R}f := (x_{1,1}, x_{2,1}, x_{3,1}, x_{4,1}, s_1, \varphi_1, \varphi_1^l, \varphi_1^r)$ with

$$ x_{1,1} = \frac{f_4^{a_1-1}(x_2)}{x_1}, $$

$$ x_{2,1} = \left[ \varphi^l \left( \frac{x_3 - f_4^{a_1-1}(x_2)}{x_3} \right) \right]^{\ell_1}, $$

$$ x_{3,1} = 1 - \varphi^{-1} \circ q_s^{-1} \left( \frac{f_4^{-a_1+1}(x_4) - x_2}{1 - x_2} \right), $$

$$ x_{4,1} = 1 - \varphi^{-1} \circ q_s^{-1} \left( \frac{f_4^{-a_1+1}(x_3) - x_2}{1 - x_2} \right), $$

$$ s_1 = \varphi^l \left( \frac{x_3 - f_4^{a_1-1}(x_2)}{x_3} \right), $$

$$ \varphi_1 = Z_{[1-f_4^{a_1-1}(x_2)/x_{3,1}]}(\varphi^l), $$

$$ \varphi_1^l = Z_{[(x_2-x_4)/(1-x_4),1]} (\varphi^r)^{a_1-1} \circ Z_{[1-x_{3,1},1]} (q_s \circ \varphi), $$

$$ \varphi_1^r = Z \left[ 0, \frac{x_3 - f_4^{a_1-1}(x_2)}{x_3} \right] (\varphi^l) \circ Z \left[ \frac{x_2 - x_4}{1 - x_4}, 1 \right] (\varphi^r)^{a_1-1} \circ Z_{[0,1-x_{4,1}]} (q_s \circ \varphi). $$

*Proof.*

1. Point 1 is a general result of the renormalization of circle maps.

2. The graph of $p\mathcal{R}$ is described in the quadrant in blue dashed lines of the following figure.
<figure><img src="image_7.png" /><figcaption>Figure 4: First return map of a map $f$ in $\mathcal{L}^X$</figcaption></figure>

By combining (4.1), the descriptions of the branches of $f$ (see, Figure 3) and the graph of $p\mathcal{R}$ (see, Figure 4) point 2 follows. $\square$

**Definition 4.2.** A map $f$ in $\mathcal{L}^X$ is infinitely renormalizable if for every $n \in \mathbb{N}$, $R^n f \in \mathcal{L}_0^X$. The class of infinitely renormalizable functions will be denoted by $W^X$.

The following observation explains more the $n$th renormalization, for every $n$ in $\mathbb{N}$.

**Remark 4.3.** Let $f$ be a map in $W^X$ with $\rho(f) = [a_0, a_1, \dots]$ as its rotation number, and $(\ell_1, \ell_2)$ as its critical exponents. For every $n \in \mathbb{N}$, we have:

1. $\mathcal{R}^n f$ belongs to $W^X$ with $\rho(\mathcal{R}^n f) = [a_n, a_{n+1}, \dots]$. As a consequence, $W^X$ corresponds to the subset of $\mathcal{L}^X$ with an irrational rotation number.

2. The critical exponents of $\mathcal{R}^n f$ is $(\ell_1, \ell_2)$ if $n$ is even and $(\ell_2, \ell_1)$ if $n$ is odd. More precisely, if $n$ is even,

$$ \mathcal{R}^n f := (x_{1,n}, x_{2,n}, x_{3,n}, x_{4,n}, s_n, \varphi_n, \varphi_n^l, \varphi_1^r) \in W^X $$

and

$$ \mathcal{R}^n f(x) = \begin{cases} \mathcal{R}^n f_-(x) & \text{if } x \in [x_{1,n}, 0) \\ \mathcal{R}^n f_+(x) & \text{if } x \in (0, 1] \end{cases} $$
with

$$
\begin{align}
\mathcal{R}^n f_-(x) &= f_{1,n}(x) \nonumber \\
&= (1 - x_{2,n}) q_{s_n} \circ \varphi_n \left( \frac{x_{1,n} - x}{x_{1,n}} \right) + x_{2,n} \quad \text{if } x \in [x_{1,n}, 0) \tag{4.2}
\end{align}
$$

and

$$
\mathcal{R}^n f_+(x) = \begin{cases}
f_{2,n} = x_{1,n} \left( \varphi_n^l \left( \frac{x_{3,n} - x}{x_{3,n}} \right) \right)^{\ell_1} & \text{if } x \in [0, x_{3,n}] \\
f_{3,n} = 0 & \text{if } x \in [x_{3,n}, x_{4,n}] \\
f_{4,n} = x_{2,n} \left( \varphi_n^r \left( \frac{x - x_{4,n}}{1 - x_{4,n}} \right) \right)^{\ell_2} & \text{if } x \in [x_{4,n}, 1]
\end{cases}
$$

3. The dynamic of $\mathcal{R}^n f$ is controlled by the couple $(q_{n-1}, q_n)$ in the following sense: $\mathcal{R}^n f|_{[x_{1,n},0)}$ and $\mathcal{R}^n f|_{(0,1]}$ are respectively a rescaled version of $f^{q_{n-1}}$ and $f^{q_n}$. Namely, the quadruplet $(x_{1,n}, x_{2,n}, x_{3,n}, x_{4,n})$ corresponds to $(q_n + 1, q_{n+1} + 1, r(-q_n + 1), l(-q_n + 1))$ if $n$ is even and corresponds to $(\underline{q}_n + 1, \underline{q}_{n+1} + 1, \underline{l}(-\underline{q}_n + 1), \underline{r}(-\underline{q}_n + 1))$ if $n$ is odd.

4.2 Asymptotic Distortions

This part is devoted to proving the following result.

**Proposition 4.4.** Let $f \in W^X$ with $(\ell_1, \ell_2)$ as its critical exponents. We have

$$
dist(\varphi_n^l) = O\left(\alpha_{n-1}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}\right), \quad dist(\varphi_n) = O\left(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}\right), \quad dist(\varphi_n^r) = O\left(\alpha_n^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}\right)
$$

Proof. It is based on the Koebe principle (Proposition 3.6). Let $[a_0, a_1, \dots]$ be the rotation number of $f$.

Distortion of $\varphi_n^l$:

$$
-T = \left[-q_{n+1} + 1 + (a_n - 1)q_n, -q_n + 1\right],
$$

$$
-M = (\underline{1}, \underline{-q_n + 1}),
$$

$$
-S = \underline{-q_{n+1} + 1 + (a_n - 1)q_n, 1},
$$

$$
-D = \underline{-q_n + 1}.
$$

Observe that

$$
\varphi_n^l = Z_M f^{q_n - 1}.
$$

We claim that

1. The members of the family $\{f^i(T), 0 \le i \le q_n - 2\}$ are pairwise disjoint,

2. $f^{q_n-1} : T \to f^{q_n-1}(T)$ is diffeomorphism,

3. $\frac{|f^{q_n-1}(M)|}{|f^{q_n-1}(S)|}, \frac{|f^{q_n-1}(M)|}{|f^{q_n-1}(D)|} = O\left(\alpha_{n-1}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}\right)$.

Points 1 and 2 come respectively from Proposition 3.1 and Remark 3.4. For
point 3, observe that by Proposition 3.3 and point 2 of Fact 2.2,

$$
\frac{|f^{q_n}(M)|}{|f^{q_n}(S)|} = \frac{|(q_n + 1, \underline{1})|}{|[-q_{n-1} + 1, q_n + 1]|} < \frac{1}{K} \cdot \frac{|(-q_{n-1} + 1, \underline{1})|}{|[-q_{n-1} + 1, 1]|} < \frac{1}{K} \cdot \alpha_{n-1}.
$$
Therefore,

$$ \frac{|f^{q_n-1}(M)|}{|f^{q_n-1}(S)|} = O(\alpha_{n-1}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}). $$

Thus, since

$$ \frac{|f^{q_n-1}(M)|}{|f^{q_n-1}(D)|} < \frac{|f^{q_n-1}(M)|}{|f^{q_n-1}(S)|} $$

then by Proposition 3.6 we get the desired distortion estimate for $\varphi_n^l$.

**Distortion of $\varphi_n$:**

$$ - T = [-q_{n-1} + 1, -q_n + (a_{n-1} - 1)q_{n-1} + 1], $$

$$ - M = (q_n + (a_{n-1} - 1)q_{n-1} + 1, 1), $$

$$ - S = [-q_{n-1} + 1, q_n + (a_{n-1} - 1)q_{n-1} + 1], $$

$$ - D = (1, -q_n + (a_{n-1} - 1)q_{n-1} + 1]. $$

Observe that

$$ \varphi_n = Z_M f^{q_{n-1}-1}. $$

We claim that

1. The members of the family $\{f^i(T), 0 \le i \le q_n - 2\}$ are pairwise disjoint,

2. $f^{q_{n-1}} : T \to f^{q_{n-1}}(T)$ is diffeomorphism,

3. $\frac{|f^{q_{n-1}-1}(M)|}{|f^{q_{n-1}-1}(S)|}, \frac{|f^{q_{n-1}-1}(M)|}{|f^{q_{n-1}-1}(D)|} = O\left(\alpha_{n-1}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}\right)$.

Points 1 and 2 come respectively from Proposition 3.1 and the Remark 3.4.

For point 3, observe that by Proposition 3.3 and point 3 of Fact 2.2,

$$ \frac{|f^{q_{n-1}}(M)|}{|f^{q_{n-1}}(D)|} = \frac{|(q_{n+1} + 1, q_{n-1} + 1)|}{|(q_{n-1} + 1, -q_{n-2} + 1)|} < \frac{1}{K} \cdot \frac{|(1, -q_{n-2} + 1)|}{|(1, -q_{n-2} + 1)|} < \frac{\alpha_{n-2}}{K}. $$

Therefore,

$$ \frac{|f^{q_{n-1}-1}(M)|}{|f^{q_{n-1}-1}(D)|} = O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}). $$

Thus, since

$$ \frac{|f^{q_{n-1}-1}(M)|}{|f^{q_{n-1}-1}(S)|} < \frac{|f^{q_{n-1}-1}(M)|}{|f^{q_{n-1}-1}(D)|}, $$

then by the Proposition 3.6 we get the desired distortion estimate for $\varphi_n$. The Distortion of $\varphi_n^r$ are obtained by doing similar calculations with:

**Distortion of $\varphi_n^r$:**

$$ - T = [-q_n + 1, -q_{n-2} + 1), $$

$$ - M = (-q_n + 1, (a_n - 1)q_n + q_{n-1} + 1), $$

$$ - S = -q_n + 1, $$

$$ - D = ((a_n - 1)q_n + q_{n-1} + 1, -q_{n-2} + 1], $$

Observe that

$$ \varphi_n = Z_M f^{q_{n-1}}. $$

□

**Remark 4.5.** Proposition 4.4 shows that $\varphi_n^l$, $\varphi_n$ and $\varphi_n^r$ go to $\text{Id}_{[0,1]}$ when $n$ goes to infinity. As a consequence, the asymptotic behavior of $\mathbb{R}^n f$ depends only on $\Sigma^X$.
**Fibonacci circle maps with a flat piece.** Let $f \in W^X$ with $\rho(f) = [a_0, a_1, \dots]$. Recall that for every $n \in \mathbb{N}$,

$$ \mathcal{R}^n f := (x_{1,n}, x_{2,n}, x_{3,n}, x_{4,n}, s_n, \varphi_n, \varphi_n^l, \varphi_n^r) \in W^X. $$

By the equalities in (4.1), Figure 1 and Figure 2, it is easy to prove that

$$ x_{2,n} < x_{3,n} \iff a_n = 1. $$

In the following of this paper, we denote by $W_{[1]}^X$ the subclass of $W^X$ consisting of maps with Fibonacci rotation number, and their critical exponents belonging to $(1, 2)^2$.

## 4.3 The Asymptotic of Renormalization

The set $\Sigma^X$ can be redefined as follows:

**changes of variables**

($X \to S$). Let

$$ \begin{align} S_{1,n} &= \frac{x_{3,n} - x_{2,n}}{x_{3,n}}, \quad S_{2,n} = \frac{1 - x_{4,n}}{1 - x_{2,n}}, \tag{4.3} \\ S_{3,n} &= \frac{x_{3,n}}{1 - x_{4,n}}, \quad S_{4,n} = -\frac{x_{2,n}}{x_{1,n}}, \quad S_{5,n} = s_n. \nonumber \end{align} $$

That is,

$$ \left\{ \begin{aligned} x_{1,n} &= \frac{S_{3,n}(1 - S_{1,n})S_{2,n}}{(1 + S_{3,n}(1 - S_{1,n})S_{2,n})S_{4,n}} \\ x_{2,n} &= \frac{S_{3,n}(1 - S_{1,n})S_{2,n}}{1 + S_{3,n}(1 - S_{1,n})S_{2,n}} \\ x_{3,n} &= \frac{S_{3,n}S_{2,n}}{1 + S_{3,n}(1 - S_{1,n})S_{2,n}} \\ x_{4,n} &= 1 - \frac{S_{2,n}}{1 + S_{3,n}(1 - S_{1,n})S_{2,n}} \end{aligned} \right. \tag{4.4} $$

($S \to Y$). Let

$$ \begin{align} y_{1,n} &= S_{1,n}, \quad y_{2,n} = \ln S_{2,n}, \quad y_{3,n} = \ln S_{3,n}, \tag{4.5} \\ y_{4,n} &= \ln S_{4,n}, \quad y_{5,n} = \ln S_{5,n}. \nonumber \end{align} $$

**Notation 4.6.** Considering the respective changes of variables ($X \to S$) and ($S \to Y$), the space $\mathcal{L}^X$ becomes successively $\mathcal{L}^S$ and $\mathcal{L}^Y$. It will be clear which parametrization of our space we are using. The space will then be simply denoted by $\mathcal{L}$ instead of $\mathcal{L}^X$, $\mathcal{L}^S$ or $\mathcal{L}^Y$. Similarly, we will denote by $W_{[1]}$ the class of Fibonacci circle maps with a flat piece, and their critical exponents belonging to $(1, 2)^2$.

**Proposition 4.7.** Let $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents. For every $n \in 2\mathbb{N}$,

$$ R(S_{1,n}, S_{2,n}, S_{3,n}, S_{4,n}, S_{5,n}, \varphi_n, \varphi_n^l, \varphi_n^r) $$

corresponds to

$$ (S_{1,n+1}, S_{2,n+1}, S_{3,n+1}, S_{4,n+1}, S_{5,n+1}, \varphi_{n+1}, \varphi_{n+1}^l, \varphi_{n+1}^r) $$

where
$$1. S_{1,n+1} = 1 - \frac{(\varphi_n^l(S_{1,n}))^{\ell_1}}{\ell_2(1 - \varphi_n^{-1} \circ q_{s_n}^{-1}(1 - S_{2,n}))},$$

$$2. S_{2,n+1} = \frac{\varphi_n^{-1} \circ q_{s_n}^{-1}(S_{1,n}S_{2,n}S_{3,n})}{1 - (\varphi_n^l(S_{1,n}))^{\ell_1}},$$

$$3. S_{3,n+1} = \frac{1 - \varphi_n^{-1} \circ q_{s_n}^{-1}(1 - S_{2,n})}{\varphi_n^{-1} \circ q_{s_n}^{-1}(S_{1,n}S_{2,n}S_{3,n})},$$

$$4. S_{4,n+1} = \frac{[\varphi_n^l(S_{1,n})]^{\ell_1}}{S_{4,n}},$$

$$5. S_{5,n+1} = \varphi_n^l(S_{1,n}),$$

$$6. \varphi_{n+1} = Z_{[S_{1,n},1]}(\varphi_n^l),$$

$$7. \varphi_{n+1}^l = \varphi_n^r \circ Z_{[\varphi_n^{-1} \circ q_{s_n}^{-1}(1-S_{2,n}),1]}(q_{s_n} \circ \varphi_n),$$

$$8. \varphi_{n+1}^r = Z_{[0,S_{1,n}]}(\varphi_n^l) \circ Z_{[0,\varphi_n^{-1} \circ q_{s_n}^{-1}(S_{1,n}S_{2,n}S_{3,n})]}(q_{s_n} \circ \varphi_n).$$

*Proof.* It follows from equalities in (4.3) and Proposition 4.1. □

**Lemma 4.8.** $S_{1,n} = \frac{x_{3,n} - x_{2,n}}{x_{3,n}} = O(\alpha_{n+1})$.

*Proof.* Observe that,

$$S_{1,n} = \frac{|(-q_{n+1} + 1, q_{n+1} + 1)|}{|(-q_n + 1, 1)|} = O\left(\frac{|(1, q_{n+2} + 1)|}{|(1, q_n + 1)|}\right) = O(\alpha_{n+1})$$

where we use Proposition 3.6 and Proposition 3.3. □

Since $\alpha_n$ goes to zero, then by Lemma 4.8 it follows that $y_{1,n} = S_{1,n}$ goes to zero. As a consequence, the asymptotic behavior of renormalization ultimately depends on $y_{2,n}$, $y_{3,n}$, $y_{4,n}$ and $y_{5,n}$. Let

$$w_n(f) := \begin{pmatrix} y_{2,n} \\ y_{3,n} \\ y_{4,n} \\ y_{5,n} \end{pmatrix}.$$

We have the following result.

**Proposition 4.9.** Let $(\ell_1, \ell_2) \in (1,2)^2$. Then there exists $\lambda_u > 1$, $\lambda_s \in (0,1)$, $E^u, E^s, E^+$, $w_{fix} \in \mathbb{R}^4$ such that the following holds. Given $f \in W_{[1]}$ with critical exponents $(\ell_1, \ell_2)$, there exists $c_u(f), c'_u(f) < 0$, $c_s(f), c'_s(f), c_+(f)$ and $c'_+(f)$ such that for all $n := 2p_n \in 2\mathbb{N}^*$,

$$w_n(f) = c_u(f)\lambda_u^{p_n} E^u + c_s(f)\lambda_s^{p_n} E^s + c_+(f)E^+ + w_{fix} + O(e, c_u(f), \lambda_u, n),$$

and

$$w_{n+1}(f) = c'_u(f)\lambda_u^{p_n} E^u + c'_s(f)\lambda_s^{p_n} E^s + c'_+(f)E^+ + w_{fix} + O(e, c'_u(f), \lambda_u, n)$$

where $O(e, c_u(f), \lambda_u, n)$ is the vector consisting of all

$$O\left(\left(e^{c_u(f)\lambda_u^{p_n-4}}\right)^{1/\ell}\right)$$
with

$\bar{l} := \max\{\ell_1, \ell_2\}.$

Also,

$dist(\varphi_n) = O\left(e^{\frac{c_u(f)\lambda_u^{p_n-2}}{\ell_1}}\right),$

$dist(\varphi_n^l) = O\left(e^{\frac{c_u(f)\lambda_u^{p_n-1}}{\ell_2}}\right),$

$dist(\varphi_n^r) = O\left(e^{\frac{c_u(f)\lambda_u^{p_n}}{\ell_1}}\right).$

The rest of the section will be devoted largely to proving this proposition. In particular, we are going to show that:

$$1. \quad \lambda_u = \frac{1}{2\ell_1\ell_2} \left( \ell_1 + \ell_2 + 1 + \sqrt{\ell_1^2 + (2-2\ell_2)\ell_1 + \ell_2^2 + 2\ell_2 + 1} \right),$$

$$2. \quad \lambda_s = \frac{1}{2\ell_1\ell_2} \left( \ell_1 + \ell_2 + 1 - \sqrt{\ell_1^2 + (2-2\ell_2)\ell_1 + \ell_2^2 + 2\ell_2 + 1} \right),$$

$$3. \quad E^+ = \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix}^t,$$

$$4. \quad E^s(\ell_1, \ell_2, \lambda) = \begin{pmatrix} e_2^s \\ e_3^s(\ell_1, \ell_2, \lambda) \\ e_4^s(\ell_1, \ell_2, \lambda) \\ e_5^s(\ell_1, \ell_2, \lambda) \end{pmatrix},$$

$$5. \quad E^u(\ell_1, \ell_2, \lambda) = \begin{pmatrix} e_2^u \\ e_3^u(\ell_1, \ell_2, \lambda) \\ e_4^u(\ell_1, \ell_2, \lambda) \\ e_5^u(\ell_1, \ell_2, \lambda) \end{pmatrix}$$

where

$e_2^s = e_2^u = 1,$

$$\lambda = \sqrt{\ell_1^2 + (2 - 2\ell_2)\ell_1 + \ell_2^2 + 2\ell_2 + 1}$$

$$e_3^s(\ell_1, \ell_2, -\lambda) = \frac{(\ell_1\ell_2 - \ell_2 - 1)\lambda - ((\ell_1 - 1)(\ell_1 - \ell_2 - 1) + 1)\ell_2 - \ell_1 - 1}{(\ell_1 + \ell_2 + 1)\lambda + (\ell_1 + 1)^2 + (\ell_2 + 1)^2 - 1} \\ = e_3^u(\ell_1, \ell_2, \lambda),$$

$$e_4^s(\ell_1, \ell_2, \lambda) = \frac{(\ell_1^2 + \ell_1)\lambda + (2\ell_1\ell_2 + 1)(\ell_1 - 1)\ell_1\ell_2 - \ell_1(\ell_1 + 1)^2}{((\ell_1 + 1)^2 + (1 - \ell_1)\ell_1\ell_2^2 + \ell_2)\lambda + D} \\ = e_3^u(\ell_1, \ell_2, -\lambda)$$

and

$e_5^s(\ell_1, \ell_2, \lambda) = \ell_2\lambda_s = e_5^u(\ell_1, \ell_2, -\lambda)$

with

$$D = (\ell_1^3 + (\ell_1^2 - \ell_1)\ell_2 - \ell_1 - 1)\ell_2^2 - (\ell_1 + 1)^3 + (\ell_1^2 - \ell_1 - 2)\ell_2.$$
### 4.3.1 Asymptotic affine behavior of renormalization

Let $f \in W_{[1]}$. We are going to show that $R^{n+1}f = LR^n f + C_0(f)$ for large enough $n$. We start with the following result.

**Lemma 4.10.** Let $f \in W_{[1]}$. For large enough even $n$, we have:

$$1. \quad \frac{x_{2,n+1}}{x_{1,n}} = O(\alpha_{n+1}),$$

$$2. \quad \frac{x_{3,n+1}}{x_{1,n}} = O(\alpha_{n+1}),$$

$$3. \quad \frac{x_{1,n} - x_{4,n+1}}{x_{1,n}} = O(\alpha_n),$$

$$4. \quad s_n = S_{5,n} = O(\alpha_n).$$

*Proof.* The two first points of Lemma 4.10 come directly from point 2 of Remark 4.3. Indeed,

$$\frac{x_{2,n+1}}{x_{1,n}} = \frac{|(1, q_{n+2} + 1)|}{|(1, \underline{q_n + 1})|} < \frac{|(1, -q_{n+1} + 1)|}{|(1, \underline{-q_{n+1} + 1})|} < \alpha_{n+1}.$$

$$\frac{x_{3,n+1}}{x_{1,n}} = \frac{|(1, -q_{n+1} + 1)|}{|(1, \underline{q_n + 1})|} < \frac{|(1, -q_{n+1} + 1)|}{|(1, \underline{-q_{n+1} + 1})|} < \alpha_{n+1}.$$

For point 3, observe that

$$(-q_{n+1} + 1, q_n + 1) \subset (1, q_n + 1).$$

Therefore,

$$\frac{x_{1,n} - x_{4,n+1}}{x_{1,n}} = \frac{|(-q_{n+1} + 1, q_n + 1)|}{|(1, \underline{q_n + 1})|} = O\left(\frac{|(-q_n + 1, q_{n+1} + 1)|}{|(q_{n-1} + 1, \underline{q_{n+1} + 1})|}\right) = O(\alpha_n)$$

where we use Proposition 3.6 and Proposition 3.3.

Let $(\ell_1, \ell_2)$ be the critical exponents of $f$. By Lemma 4.8 and Proposition 4.4, we get

$$s_n = S_{5,n} = S_{1,n-1} \frac{\varphi_n^l(S_{1,n-1})}{S_{1,n-1}} = O(\alpha_n) (1 + O(\alpha_{n-1}^{\ell_1})) = O(\alpha_n),$$

and the lemma is proved. $\square$

**Proposition 4.11.** Let $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents. For large enough even $n$, the followings hold.

$$1. \quad S_{1,n+1} = 1 - \frac{\ell_2 S_{1,n}^{\ell_1}}{S_{2,n}} \left(1 + O\left(\alpha_{n-1}^{1/\ell_2}\right)\right),$$

$$2. \quad S_{2,n+1} = \frac{S_{1,n} S_{2,n} S_{3,n}}{\ell_2 S_{5,n}^{\ell_2-1}} \left(1 + O\left(\alpha_{n-2}^{1/\ell_1}\right)\right),$$

$$3. \quad S_{3,n+1} = \frac{S_{5,n}^{\ell_2-1}}{S_{1,n} S_{3,n}} \left(1 + O\left(\alpha_{n-2}^{1/\ell_1}\right)\right),$$

$$4. \quad S_{4,n+1} = \frac{S_{1,n}^{\ell_1}}{S_{4,n}} \left(1 + O\left(\alpha_{n-1}^{1/\ell_2}\right)\right),$$
$$5. S_{5,n+1} = S_{1,n} \left(1 + O\left(\alpha_{n-1}^{1/\ell_2}\right)\right).$$

The case n odd is just a permutation between $\ell_1$ and $\ell_2$ in the previous equalities, as mentioned in point 1 of Remark 2.1.

*Proof.* Observe that

$$f^{q_{n-1}}\left(\left(\frac{1}{\ell_1}, \frac{-q_{n+1}}{\ell_1}\right)\right) = \left(\frac{q_{n-1}}{\ell_1}, \frac{1}{\ell_1}, \frac{-q_n}{\ell_1}\right).$$

On the other hand, by (4.2) we have

$$\mathcal{R}^n f_{|(x_1,n,0)} = (1 - x_{2,n}) q_{s_n} \circ \varphi_n \left( \frac{x_{1,n} - x}{x_{1,n}} \right) + x_{2,n}.$$

Then by the value theorem there is $\theta_1, \theta_2 \in (0,1)$ such that,

$$\frac{|(q_{n-1} + 1, -q_n + 1)|}{|(1, -q_{n+1} + 1)|} = (1 - x_{2,n})\varphi'_n(\theta_1)q'_s(\theta_2). \quad (4.6)$$

Remark that $q'_{s_n}(\theta_2) := Dq_{s_n}(\theta_2) = \ell_2 O(1)$ and note that by Proposition 4.4 $\varphi'_n(\theta_1) = 1 + O(\alpha_{n-2}^{1/\ell_1})$. Thus, by point 1 of Remark 4.3 and equality (4.6), we obtain

$$S_{2,n} = \frac{1 - x_{4,n}}{1 - x_{2,n}} = \ell_2 x_{3,n+1} (1 + O(\alpha_{n-2}^{1/\ell_1})).$$

Therefore,

$$x_{3,n+1} = \frac{S_{2,n}}{\ell_2} (1 + O(\alpha_{n-2}^{1/\ell_1})). \quad (4.7)$$

By Proposition 4.1 and Proposition 4.11 we obtain

$$x_{2,n+1} = (\varphi_n^l(S_{1,n}))^{\ell_1} = S_{1,n}^{\ell_1} \left( \frac{\varphi_n^l(S_{1,n})}{S_{1,n}} \right)^{\ell_1} = S_{1,n}^{\ell_1} (1 + O(\alpha_{n-1}^{1/\ell_2})). \quad (4.8)$$

It follows from (4.7) and (4.8) that

$$S_{1,n+1} = 1 - \frac{x_{2,n+1}}{x_{3,n+1}} = 1 - \ell_2 \frac{S_{1,n}^{\ell_1}}{S_{2,n}} (1 + O(\alpha_{n-1}^{1/\ell_2})).$$

By Proposition 4.1 we have

$$(1 - x_{2,n}) q_{s_n} \circ \varphi_n (1 - x_{4,n+1}) = x_{3,n} - x_{2,n}.$$

That is,

$$(1 - x_{4,n+1}) \frac{q_{s_n} \circ \varphi_n (1 - x_{4,n+1})}{1 - x_{4,n+1}} = \frac{x_{3,n} - x_{2,n}}{1 - x_{2,n}}.$$

Thus, by point 3 of Lemma 4.10 and Proposition 4.4 we get

$$\ell_2 S_{5,n}^{\ell_2-1} (1 + O(\alpha_{n-2}^{1/\ell_1})) (1 - x_{4,n+1}) = \frac{x_{3,n} - x_{2,n}}{1 - x_{2,n}} \\ = S_{1,n} S_{2,n} S_{3,n}.$$

Therefore,

$$1 - x_{4,n+1} = \frac{S_{1,n} S_{2,n} S_{3,n}}{\ell_2 S_{5,n}^{\ell_2-1}} (1 + O(\alpha_{n-2}^{1/\ell_1})). \quad (4.9)$$
By point 1 of Lemma 4.10 and equality (4.9), we have

$$S_{2,n+1} = \frac{1 - x_{4,n+1}}{1 - x_{2,n+1}} = \frac{S_{1,n}S_{2,n}S_{3,n}}{\ell_2 S_{5,n}^{\ell_2-1}}(1 + O(\alpha_{n-2}^{1/\ell_1})).$$

From (4.7) and (4.9) we obtain

$$S_{3,n+1} = \frac{x_{3,n+1}}{1 - x_{4,n+1}} = \frac{S_{5,n}^{\ell_2-1}}{S_{1,n}S_{3,n}}(1 + O(\alpha_{n-2}^{1/\ell_1})).$$

By Proposition 4.1 and equality (4.8), it follows that

$$S_{4,n+1} = \frac{x_{2,n+1}}{-x_{1,n+1}} = \frac{x_{2,n+1}}{S_{4,n}} = \frac{S_{1,n}^{\ell_1}}{S_{4,n}}(1 + O(\alpha_{n-1}^{1/\ell_2}))$$

and finally, by Proposition 4.7 and Proposition 4.4, we have

$$S_{5,n+1} = S_{1,n} \left( \frac{\varphi_n^l(S_{1,n})}{S_{1,n}} \right) = S_{1,n} (1 + O(\alpha_{n-1}^{1/\ell_2})).$$

□

**Corollary 4.12.** Let $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents. For large enough $n \in \mathbb{N}$, we have

$$\frac{\ell_2 S_{1,2n}^{\ell_1}}{S_{2,2n}} = 1 + O\left(\alpha_{2n-1}^{\frac{1}{\ell_2}}\right) \quad \text{and} \quad \frac{\ell_1 S_{1,2n+1}^{\ell_2}}{S_{2,2n+1}} = 1 + O\left(\alpha_{2n}^{\frac{1}{\ell_1}}\right).$$

*Proof.* By point 1 of Proposition 4.11 and Lemma 4.8

$$\frac{\ell_2 S_{1,n}^{\ell_1}}{S_{2,n}} = (1 - S_{1,n+1})(1 + O(\alpha_{n-1}^{1/\ell_2})) = 1 + O(\alpha_{n-1}^{1/\ell_2}).$$

□

By Proposition 4.11 and Corollary 4.12, we have the following result.

**Proposition 4.13.** Let $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents. For large enough even $n$, the following equality holds.

$$w_{n+2} = L_{(\ell_1, \ell_2)} w_{n+1} + w_{(\ell_1, \ell_2)}^* + \underline{Q}(n, \ell, \alpha) \quad (4.10)$$

where

$$L_{(\ell_1, \ell_2)} = \begin{pmatrix} 1 + \frac{1}{\ell_1} & 1 & 0 & 1 - \ell_2 \\ -\frac{1}{\ell_1} & -1 & 0 & \ell_2 - 1 \\ 1 & 0 & -1 & 0 \\ \frac{1}{\ell_1} & 0 & 0 & 0 \end{pmatrix}, \\ w_{(\ell_1, \ell_2)}^* = \begin{pmatrix} -\left(1 + \frac{1}{\ell_1}\right) \log \ell_2 \\ \frac{1}{\ell_1} \log \ell_2 \\ -\log \ell_2 \\ -\frac{1}{\ell_1} \log \ell_2 \end{pmatrix}$$

and

$$\underline{Q}(n, \ell, \alpha) = \left( O\left(\alpha_{n-2}^{1/\ell}\right), O\left(\alpha_{n-2}^{1/\ell}\right), O\left(\alpha_{n-2}^{1/\ell}\right), O\left(\alpha_{n-2}^{1/\ell}\right) \right)^t.$$
As noted in the introduction, we study the operator $R^2$ on $(f, \mathcal{R}f)$ rather than $R$ on $f$. By (4.10), we have

$$w_{n+2} = \bar{L}_{(\ell_1, \ell_2)} w_n + \bar{w}^*_{(\ell_1, \ell_2)} + \underline{O}(n, \bar{l}, \alpha) \quad (4.11)$$

where

$$\bar{L} = \bar{L}_{(\ell_1, \ell_2)} = L_{(\ell_2, \ell_1)} L_{(\ell_1, \ell_2)}$$

and

$$\bar{w}^*_{(\ell_1, \ell_2)} = L_{(\ell_2, \ell_1)} w^*_{(\ell_1, \ell_2)} + w^*_{(\ell_2, \ell_1)}.$$

The next step is to explore equality (4.11) in order to conclude the proof of Proposition 4.9.

### 4.3.2 Asymptotic in Y-Coordinates

Let $w_{fix}$ be the fixed point of the equation $\bar{L}w + \bar{w}^*_{(\ell_1, \ell_2)} = w$. We have the following result.

**Lemma 4.14.** Let $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents. Then,

$$w_n(f) = c_u(f)\lambda_u^{p_n} E^u + c_s(f)\lambda_s^{p_n} E^s + c_+(f)E^+ + w_{fix} + O(n, \ell_1, \bar{l}, \alpha_0);$$

$$w_{n+1}(f) = c'_u(f)\lambda_u^{p_n} E^u + c'_s(f)\lambda_s^{p_n} E^s + c'_+(f)E^+ + w_{fix} + O(n, \ell_2, \bar{l}, \alpha_0)$$

where $O(n, \ell_1, \bar{l}, \alpha_0)$ and $O(n, \ell_2, \bar{l}, \alpha_0)$ are the vectors whose components are equal to

$$O\left(\alpha_0^{\left(\frac{2}{\bar{l}_1}\right)^{p_n-4}}\right)^{\frac{1}{\bar{l}}} \quad \text{and} \quad O\left(\alpha_0^{\left(\frac{2}{\bar{l}_2}\right)^{p_n-4}}\right)^{\frac{1}{\bar{l}}}$$

respectively.

*Proof.* Consider the sequence $(v_n)_{n \in \mathbb{N}}$ defined by $v_n = w_n - w_{fix}$. From equality (4.11) and Proposition 6.11, it follows that

$$v_{n+2} = \bar{L}v_n + \epsilon_n$$

where $\epsilon_n = O(n, \ell_1, \bar{l}, \alpha_0)$. So,

$$v_n = \bar{L}^{p_n} v_0 + \sum_{k=0}^{p_n-2} \bar{L}^{p_n-k-1} \epsilon_{2k}. \quad (4.12)$$

Observe that

$$\begin{aligned} v_0 &= c_{u,0}E^u + c_{s,0}E^s + c_{+,0}E^+ + c_{0,0}E^0, \\ \epsilon_n &= \epsilon_{u,n}E^u + \epsilon_{s,n}E^s + \epsilon_{+,n}E^+ + \epsilon_{0,n}E^0. \end{aligned} \quad (4.13)$$

We consider the following quantities

$$\begin{aligned} C_u(f) &= c_{u,0} + \sum_{k=0}^{\infty} \frac{\epsilon_{u,2k}}{\lambda_u^{k+1}}, \\ C_s(f) &= c_{s,0} + \sum_{k=0}^{\infty} \frac{\epsilon_{s,2k}}{\lambda_s^{k+1}}, \\ C_+(f) &= c_{+,0} + \sum_{k=0}^{\infty} \epsilon_{+,2k}. \end{aligned} \quad (4.14)$$

By introducing 4.13 and 4.14 in (4.12), we get:

$$v_n = \left( c_{u,0} + \sum_{k=0}^{p_n-2} \frac{\epsilon_{u,2k}}{\lambda_u^{k+1}} \right) \lambda_u^{p_n} E^u + \left( c_{s,0} + \sum_{k=0}^{p_n-2} \frac{\epsilon_{s,2k}}{\lambda_s^{k+1}} \right) \lambda_s^{p_n} E^s +$$
$$
\begin{align*}
&+ \left( c_{+,0} + \sum_{k=0}^{p_{n-2}} \epsilon_{+,2k} \right) E^+ + \epsilon_{0,n-2} E^0 \\
&= c_u(f) \lambda_u^{p_n} E^u + c_s(f) \lambda_s^{p_n} E^s + c_+(f) E^+ + \\
&\qquad + \left( \sum_{k=p_n}^{\infty} \frac{\epsilon_{u,n}}{\lambda_u^{k+1}} \right) \lambda_u^{p_n} E^u + \left( \sum_{k=p_n}^{\infty} \frac{\epsilon_{s,k}}{\lambda_s^{k+1}} \right) \lambda_s^{p_n} E^s + \\
&\qquad + \left( \sum_{k=p_n}^{\infty} \epsilon_{+,k} \right) E^+ + \epsilon_{0,n-2} E^0 \\
&= c_u(f) \lambda_u^{p_n} E^u + c_s(f) \lambda_s^{p_n} E^s + c_+(f) E^+ + O(n-2, \ell_1, \bar{\ell}, \alpha_0).
\end{align*}
$$

In fact, the three sums are estimated by

$$
O \left( \alpha_0^{\left( \frac{2}{\ell_1} \right)^{p_n-2}} \right)^{\frac{1}{\ell}} .
$$

To convince oneself of that, it suffices to remark that for large enough $k$, the
following inequality holds:

$$
\left( \frac{\alpha_0 \left( \frac{2}{\ell_1} \right)^{k-1}}{\alpha_0 \left( \frac{2}{\ell_1} \right)^{k-2}} \right)^{\frac{1}{\ell}} < \frac{1}{2} \lambda_u.
$$

Therefore,

$$
\begin{align*}
\left| \sum_{k=p_n}^{\infty} \frac{\epsilon_{u,2k}}{\lambda_u^{k+1}} \right| \lambda_u^{p_n} &= O \left( \sum_{k=p_n}^{\infty} \frac{\left( \alpha_0 \left( \frac{2}{\ell_1} \right)^{k-1} \right)^{\frac{1}{\ell}}}{\lambda_u^k} \lambda_u^{p_n-1} \right) \\
&= O \left( \frac{\left( \alpha_0 \left( \frac{2}{\ell_1} \right)^{p_n-2} \right)^{\frac{1}{\ell}}}{\lambda_u^{p_n-1}} \lambda_u^{p_n} \sum_{k=p_n}^{\infty} \left( \frac{1}{2} \right)^{k-p_n} \right) \\
&= O \left( \left( \alpha_0 \left( \frac{2}{\ell_1} \right)^{p_n-2} \right)^{\frac{1}{\ell}} \right).
\end{align*}
$$

By analogy, we have the other estimates.

<figure><img src="image_8.png" /></figure>

**Notation 4.15.** We can write $w_n(f)$ (see Lemma 4.14) as follows:

$$
w_n(f) = c_u^*(f)\lambda_u^{p_n} E^u + c_s^*(f)\lambda_s^{p_n} E^s + c_+^*(f)E^+ + w_{fix} + O^*(n, \ell_1, \bar{\ell}, \alpha_0)
$$

with $c_u^*(f) = c_u(f)$ if $n$ is even and $c_u^*(f) = c'_u(f)$ if $n$ is odd. Also, $c_u^*(f) = c'_u(f)$ when $c_u^*(f) = c_u(f)$ and vice-versa. We consider the same convention for $c_s^*(f)$ and $c_+^*(f)$. The vector $O^*(n, \ell_1, \bar{\ell}, \alpha_0)$ is defined from Lemma 4.14.
**Lemma 4.16.** Let $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents. For every $n := 2p_n \in \mathbb{N}$, we have

$$ \alpha_n = O\left(e^{c_u(f)\lambda_u^{p_n}(e_2^u + e_3^u)}\right), \quad \alpha_{n+1} = O\left(e^{c'_u(f)\lambda_u^{p_n}(e_2^u + e_3^u)}\right) \text{ and } c'_u(f), c_u(f) < 0. $$

*Proof.* We claim that $1 - x_{2,n}$ and $x_{4,n}$ are comparable. Observe that:

$$ \frac{1 - x_{2,n}}{x_{4,n}} = \frac{\frac{x_{3,n} - x_{2,n}}{x_{4,n} - x_{3,n}} + 1 + \frac{1 - x_{4,n}}{x_{4,n} - x_{3,n}}}{\frac{x_{3,n}}{x_{4,n} - x_{3,n}} + 1}, $$

and for large enough $n$,

$$ \alpha_n = \frac{x_{3,n}}{x_{4,n}}, \quad \frac{x_{3,n} - x_{2,n}}{x_{4,n} - x_{3,n}} < \frac{x_{3,n}}{x_{4,n} - x_{3,n}} < 2\alpha_n. $$

Therefore,

$$ \frac{1 - x_{2,n}}{x_{4,n}} = O(1 + \alpha_n). $$

On the other hand,

$$ \frac{1 - x_{2,n}}{x_{4,n}} > \frac{1}{\frac{x_{3,n}}{x_{4,n} - x_{3,n}} + 1} > \frac{1}{2\alpha_n + 1}, $$

and the claim follows. As a consequence,

$$ \alpha_n := \frac{x_{3,n}}{x_{4,n}} = S_{2,n}S_{3,n} \frac{1 - x_{2,n}}{x_{4,n}} = O(1)S_{2,n}S_{3,n}. $$

That is,

$$ \alpha_n = O(S_{2,n}S_{3,n}). \qquad (4.15) $$

By combining equality (4.15) and Lemma 4.14) the lemma is proved. $\square$

*Proof of Proposition 4.9.* The first part of Proposition 4.9 comes from Proposition 6.11, Lemma 4.14 and Lemma 4.16, and the second part comes from Proposition 4.4 and Lemma 4.16. $\square$

**Corollary 4.17.** Let $f$ be a Fibonacci circle map with a flat piece, and $(\ell_1, \ell_2)$ as its critical exponents. Let $(\ell_n)$ be the sequence defined by $\ell_{2k+1} = \ell_1$ and $\ell_{2k} = \ell_2$ for every $k \in \mathbb{N}$. If the sequence

$$ \Psi_{0,n} = \frac{S_{1,n}^{\ell_n}}{S_{2,n}} $$

and the logarithm of sequences

$$ \Psi_{1,n} = \frac{S_{2,n}}{\ell_{n+1}(\varphi_n^{-1} \circ q_{s_n}^{-1}(1-S_{2,n}))} \cdot \frac{(\varphi_n^l(S_{1,n}))^{\ell_n}}{S_{1,n}^{\ell_n}}, $$

$$ \Psi_{2,n} = \frac{\ell_{n+1} s_{5,n}^{\ell_n - 1} \varphi_n^{-1} \circ q_{s_n}^{-1}(S_{1,n} S_{2,n} S_{3,n})}{S_{1,n} S_{2,n} S_{3,n}} \cdot \frac{1}{1 - \varphi_n^l(S_{1,n})}, $$
$$ \Psi_{3,n} = \frac{S_{1,n}S_{3,n}(1 - \varphi_n^{-1} \circ q_{s_n}^{-1}(1 - S_{2,n}))}{S_{5,n}^{\ell_{n+1}-1} \varphi_n^{-1} \circ q_{s_n}^{-1}(S_{1,n}S_{2,n}S_{3,n})}, \\ \Psi_{4,n} = \frac{\varphi_n^l(S_{1,n})}{S_{1,n}}, $$

are uniformly bounded, then

1. the sequence $w_n(f)$ is bounded if $\lambda_u \le 1$;

2. one component of the vectors $w_n(f)$ tend to infinity exponentially when $\lambda_u > 1$.

The different $\lambda_u$-zones are represented as follows.

<figure><img src="image_9.png" /><figcaption>Figure 5: Bifurcation of $\lambda_u$</figcaption></figure>

*Proof.* From Proposition 4.7, we have

$$ \begin{align*} S_{1,1} &= 1 - \frac{\ell_2 S_1^{\ell_1}}{S_2} \cdot \left[ \frac{S_2}{\ell_2(1 - \varphi^{-1} \circ q_s^{-1}(1 - S_2))} \cdot \frac{(\varphi^l(S_1))^{\ell_1}}{S_1^{\ell_1}} \right], \\ S_{2,1} &= \frac{S_1 S_2 S_3}{\ell_2 S_5^{\ell_2-1}} \cdot \left[ \frac{\ell_2 S_5^{\ell_2-1} \varphi^{-1} \circ q_s^{-1}(S_1 S_2 S_3)}{S_1 S_2 S_3} \cdot \frac{1}{1 - (\varphi^l(S_1))^{\ell_1}} \right], \\ S_{3,1} &= \frac{S_5^{\ell_2-1}}{S_1 S_3} \cdot \left[ \frac{S_1 S_3(1 - \varphi^{-1} \circ q_s^{-1}(1 - S_2))}{S_5^{\ell_2-1} \varphi^{-1} \circ q_s^{-1}(S_1 S_2 S_3)} \right], \\ S_{4,1} &= \frac{S_1^{\ell_1}}{S_4} \cdot \left[ \left( \frac{\varphi^l(S_1)}{S_1} \right)^{\ell_1} \right], \\ S_{5,1} &= S_1 \cdot \left[ \frac{\varphi^l(S_1)}{S_1} \right], \\ \varphi &= Z_{[S_1,1]}(\varphi^l), \\ \varphi_1^l &= \varphi^r \circ Z_{[\varphi^{-1} \circ q_s^{-1}(1-S_2),1]}(q_s \circ \varphi), \\ \varphi_1^r &= Z_{[0,S_1]}(\varphi^l) \circ Z_{[0,\varphi^{-1} \circ q_s^{-1}(S_1 S_2 S_3)]}(q_s \circ \varphi). \end{align*} $$

And the result follows by Proposition 4.9. □
## 4.4 Asymptotic in X-coordinates

**Lemma 4.18.** Let $f \in W_{[1]}$ with critical exponents $(\ell_1, \ell_2)$. Then for large enough $n := 2p_n \in \mathbb{N}$, we have

$$-x_{1,n} = e^{c_u(f)\lambda_u^{p_n}(e_2^u + e_3^u - e_4^u)} + c_s(f)\lambda_s^{p_n}(e_2^s + e_3^s - e_4^s) - c_+ + O((e^{c_u(f)\lambda_u^{p_n-4}(e_2^u + e_3^u)}})^{1/\ell}$$

$$x_{2,n} = e^{c_u(f)\lambda_u^{p_n}(e_2^u + e_3^u)} + c_s(f)\lambda_s^{p_n}(e_2^s + e_3^s) + O((e^{c_u(f)\lambda_u^{p_n}(e_2^u + e_3^u)}})^{1/\ell}$$

$$x_{3,n} = e^{c_u(f)\lambda_u^{p_n}(e_2^u + e_3^u)} + c_s(f)\lambda_s^{p_n}(e_2^s + e_3^s) + O((e^{c_u(f)\lambda_u^{p_n-4}(e_2^u + e_3^u)}})^{1/\ell}$$

$$1 - x_{4,n} = e^{c_u(f)\lambda_u^{p_n}e_2^u} + c_s(f)\lambda_s^{p_n}e_2^s + O((e^{c_u(f)\lambda_u^{p_n-4}(e_2^u + e_3^u)}})^{1/\ell}$$

*Proof.* Recall that (see 4.4)

$$\left\{
\begin{aligned}
x_{1,n} &= \frac{S_{3,n}(1 - S_{1,n})S_{2,n}}{(1 + S_{3,n}(1 - S_{1,n})S_{2,n})S_{4,n}} \\
x_{2,n} &= \frac{S_{3,n}(1 - S_{1,n})S_{2,n}}{1 + S_{3,n}(1 - S_{1,n})S_{2,n}} \\
x_{3,n} &= \frac{S_{3,n}S_{2,n}}{1 + S_{3,n}(1 - S_{1,n})S_{2,n}} \\
x_{4,n} &= 1 - \frac{S_{2,n}}{1 + S_{3,n}(1 - S_{1,n})S_{2,n}}
\end{aligned}
\right.$$

Proposition 3.3 implies that $S_{3,n}S_{2,n} = O(\alpha_n)$ and point 4 of Lemma 4.10 shows that $S_{1,n} = O(\alpha_{n+1})$. Thus, by Proposition 4.9 the lemma follows. □

## 5 Rigidity

Let $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents. We have the following parameters which will be used frequently throughout this section:

- $(c_u(f), c'_u(f)), (c_+(f), c'_+(f))$ and $(c_s(f), c'_s(f))$, see Lemma 4.14;

- $c'_i(f) = c_i(f)$ or $c'_i(f), i = s, u, +, $ see Notation 4.15;

- $\underline{c}_u(f) = \min\{c_u(f), c'_u(f)\}$ and $\bar{c}_u(f) = \max\{c_u(f), c'_u(f)\}$;

- $\lambda_s(f) = \lambda_s(\ell_1, \ell_2)$, $\lambda_u(f) = \lambda_u(\ell_1, \ell_2)$, and $e'_i(f), i = 2, 3, 4, 5; i = s, u, +, $ see Proposition 4.9.

Note that if there is no ambiguity, a parameter will be written without the expression “(f)”. For example, $\lambda_u$ stands for $\lambda_u(f)$.

Let $f, g \in W_{[1]}$, and let $h$ be a conjugacy between $f$ and $g$. Observe that $h(U_f) = U_g$, moreover, the choice of $h$ inside $U_f$ could be arbitrary. However, $h|_{K_f}$ is uniquely defined. Being interested in the geometry of $K_f$, we will only study $h|_{K_f}$ that we will denote $h$ yet. The main question treated in this section is: when do $K_f$ and $K_g$ have the same geometry? In other words, for fixed $f$ in $W_{[1]}$, what is the geometry class of $f$ ($K_f$)? Before answering this question, we give bi-lipschitz class of $f$. This section is devoted to proving the following result:

**Theorem 5.1.** Let $f, g \in W_{[1]}$ with $(\ell_1, \ell_2)$ as their critical exponents. If $h$ is the topological conjugation between $f$ and $g$, then for $\beta = \frac{\bar{c}_u(f)(e_2^u + e_3^u)(\lambda_u - 1)}{2\bar{\ell}\lambda_u^2} \in$
(0, 1), the following holds:

h is Holder homeo;

h is a bi-lipschitz homeo $\iff c_u^*(f) = c_u^*(g), c_+(f) + c'_+(f) = c_+(g) + c'_+(g);$

h is a $C^{1+\beta}$ diffeo $\iff c_u^*(f) = c_u^*(g), c_+(f) + c'_+(f) = c_+(g) + c'_+(g),$ \\
$c_s^*(f) = c_s^*(g).$

## 5.1 Preliminaries

**Proposition 5.2.** If $f, g \in W_{[1]}$ with different unstable eigenvalues, then the topological conjugation $h$ between $f$ and $g$ is not Holder.

*Proof.* Observe that for every $n$ even,

$$ \frac{f(U_f) - f^{q_{n+1}+1}(U_f)}{f^{q_n+1}(U_f) - f(U_f)} = \frac{x_{2,n}(f)}{-x_{1,n}(f)} = S_{4,n}(f). $$

Thus, by Proposition 4.9 we obtain,

$$ f(U_f) - f^{q_{n+1}+1}(U_f) = \prod_{k \le n} S_{4,k}(f) \sim e^{c_u(f)e_u^u(f)\lambda_u^{p_n}(f) + \frac{n}{2}(c_+(f)+c'_+(f))}. \quad (5.1) $$

In the same way

$$ g(U_g) - g^{q_{n+1}+1}(U_g) = \prod_{k \le n} S_{4,k}(g) \sim e^{c_u(g)e_u^u(g)\lambda_u^{p_n}(g) + \frac{n}{2}(c_+(g)+c'_+(g))}. \quad (5.2) $$

Recall that $h(U_f) = U_g$. Therefore, $h(f(U_f)) = g(U_g)$. That is, $h(f^n(U_f)) = g^n(U_g)$, for every $n \in \mathbb{N}$. As a consequence, for every $\beta \in (0, 1)$

$$ \lim_{n \to \infty} \frac{g(U_g) - g^{q_{n+1}+1}(U_g)}{(f(U_f) - f^{q_{n+1}+1}(U_f))^\beta} = \begin{cases} 0 & \text{si } \lambda_u(f) < \lambda_u(g) \\ \infty & \text{si } \lambda_u(f) > \lambda_u(g) \end{cases} $$

and the proposition is proved. $\square$

<figure><img src="image_10.png" /></figure>

## 5.2 Bi-lipschitz homeomorphism Conjugacy

**Proposition 5.3.** Let $f, g \in W_{[1]}$ with critical exponents $(\ell_1, \ell_2)$. If $h$ is a topological conjugation between $f$ and $g$, then

h is a bi-lipschitz homeo $\implies c_u^*(f) = c_u^*(g), c_+(f) + c'_+(f) = c_+(g) + c'_+(g).$

*Proof.* The proposition comes directly from (5.1) and (5.2). $\square$

<figure><img src="image_11.png" /></figure>

In the following, we use the below notations. For all $n \in \mathbb{N}$,

* $A_n(f) = (f(U_f), f^{q_n+1}(U_f));$

* $B_n(f) = (f^{-q_n+1}(U_f), f(U_f));$

* $C_n(f) = f^{-q_n+1}(U_f);$

* $D_n(f) = (f^{q_{n-1}+1}(U_f), f^{-q_n+1}(U_f)).$

And their iterates
• $A_n^i(f) = f^i(A_n(f))$ for $0 \le i < q_{n-1}$;

• $B_n^i(f) = f^i(B_n(f))$ for $0 \le i < q_n$;

• $C_n^i(f) = f^i(C_n(f))$ for $0 \le i < q_n$;

• $D_n^i(f) = f^i(D_n(f))$ for $0 \le i < q_n$.

Observe that for all $n \in \mathbb{N}$,

$$ \mathcal{P}_n = \{A_n^i(f), B_n^j(f), C_n^j(f), D_n^j(f) | 0 \le i < q_{n-1}, 0 \le j < q_n\}. $$

Since $h(U_f) = U_g$, it follows that

• $h(A_n^i(f)) = A_n^i(g)$ for $0 \le i < q_{n-1}$;

• $h(B_n^i(f)) = B_n^i(g)$ for $0 \le i < q_n$;

• $h(C_n^i(f)) = C_n^i(g)$ for $0 \le i < q_n$;

• $h(D_n^i(f)) = D_n^i(g)$ for $0 \le i < q_n$.

Let $n \in \mathbb{N}$, $Dh_n : [0, 1] \to \mathbb{R}^+$ is the function defined by

$$ Dh_n(x) = \frac{|h(I)|}{|I|} $$

with $I \in \mathcal{P}_n$ and $x \in I$. Observe that if $T = \cup I_i$ with $I_i \in \mathcal{P}_n$, then for all $m \ge n$,

$$ h(T) = \int_T Dh_m. \qquad (5.3) $$

**Lemma 5.4.** Let $n \in \mathbb{N}$ large enough. If $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents, then there is $C > 0$ such that for every interval $I \in \mathcal{P}_n$, the following inequality holds:

$$ |I| \ge \frac{1}{C} e^{c_u \lambda_u^{p_n} e_2^u \frac{2\lambda_u}{\lambda_u - 1}} $$

where $c_u = \min\{c_u, c'_u\}$, and $p_n$ is the integer part of $n/2$.

*Proof.* Suppose that $I \ne C_n^i$. Let $J \in \mathcal{P}_{n-1}$ and $I \subset J$. By the construction of $\mathcal{P}_n$ from $\mathcal{P}_{n-1}$, it follows that

$$ \frac{|I|}{|J|} \ge (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \min\{1 - x_{4,n+1}, x_{3,n+1}, S_{1,n}\} $$

where we used the mean value theorem and Proposition 4.4.

Also, from Corollary 4.12, Lemma 4.18 and Proposition 4.9 we get:

$$ S_{1,n} \ge e^{\frac{c_u^*(f)}{\ell} \lambda_u^{p_n} e_2^u(f) - K \lambda_s^{p_n}}; $$

$$ x_{3,n+1} \ge e^{\bar{c}_u^*(f) \lambda_u^{p_{n+1}} (e_2^u(f) + e_3^u(f)) - K \lambda_s^{p_n}}; $$

$$ 1 - x_{4,n+1} \ge e^{\bar{c}_u^*(f) \lambda_u^{p_{n+1}} e_2^u(f) - K \lambda_s^{p_n}}. $$

Thus, from points 1, 2, 4 and 5 of Proposition 4.9, we obtain

$$ \frac{|I|}{|J|} \ge e^{c_u(f) e_2^u(f) \lambda_u^{p_{n+1}} - K \lambda_s^{p_n}}. $$

Therefore, if $J \ne C_{n-1}^j$, we can repeat these estimates and we get

$$ |I| \ge e^{c_u(f) e_2^u(f) \sum_{i=3}^{n+1} \lambda_u^{p_i} - K \sum_{i=1}^{n} \lambda_s^{p_i}} \ge e^{2c_u(f) e_2^u(f) \sum_{i=1}^{p_{n+1}} \lambda_u^{p_{2i}} - K \sum_{i=1}^{n} \lambda_s^{p_i}}. $$

Moreover, by Proposition 3.3 we have $C_n^j > K' B_n^j$ for some $K' > 0$. And the lemma follows. □
**Lemma 5.5.** Let $n \in \mathbb{N}$ large enough. If $f \in W_{[1]}$ with $(\ell_1, \ell_2)$ as its critical exponents, then there is $C > 0$ such that for every interval $I \in \mathcal{P}_n$ the following inequality holds.

$$|I| \le \frac{1}{C} e^{\bar{c}_u(f)} (e_2^u(f) + e_3^u(f)) \lambda_u^{p_n}$$

where $\bar{c}_u = \max\{c_u, c'_u\}$, and $p_n$ is the integer part of $n/2$.

*Proof.* By mean value theorem and Lemma 4.4 we get

1. $|A_n^i| = x_{2,n-1}(1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}))$;

2. $|B_n^i| = x_{3,n}(1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}))$;

2. $|D_n^i| = (1 - x_{4,n})(1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}))$.

And by Lemma 4.18 there exists $K > 0$ such that

1. $x_{2,n-1} \le Ke^{c_u^*(f)}e_2^u(f)\lambda_u^{p_n-1}$;

2. $x_{3,n} \le Ke^{\bar{c}_u^*(f)}(e_2^u(f)+e_3^u(f))\lambda_u^{p_n}$;

3. $(1 - x_{4,n}) \le Ke^{\bar{c}_u^*(f)}e_2^u(f)\lambda_u^{p_n}$.

Since $\lambda_u e_2^u(f) > e_2^u(f) > e_2^u(f) + e_3^u(f)$, the lemma follows. $\square$

**Lemma 5.6.** Let $n \in \mathbb{N}$ large enough. If $f, g \in W_{[1]}$ with critical exponent $(\ell_1, \ell_2)$ such $c_u^*(f) = c_u^*(g)$ and $c_+(f) + c'_+(f) = c_+(g) + c'_+(g)$ then

$$\log \frac{Dh_{n+1}}{Dh_n} = O\left(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}} + |c_s^*(g) - c_s^*(f)|\lambda_s^{p_n}\right).$$

*Proof.* From Lemma 4.16 it follows that, if $c_u^*(f) = c_u^*(g)$, then $\alpha_n(f) = \alpha_n(g)$. Thus, by Lemma 4.4 and Lemma 4.10 we get

$$\begin{align*}
\frac{Dh_{n+1}|_{A_{n+1}^i(f)}}{Dh_n|_{A_{n+1}^i(f)}} &= \frac{\frac{|A_{n+1}^i(g)|}{|A_{n+1}^i(f)|}}{\frac{|B_n^i(g)|}{|B_n^i(f)|}} = \frac{\frac{|A_{n+1}^i(g)|}{|B_n^i(g)|}}{\frac{|A_{n+1}^i(f)|}{|B_n^i(f)|}} \\
&= \frac{\frac{|A_{n+1}(g)|}{|B_n(g)|}}{\frac{|A_{n+1}(f)|}{|B_n(f)|}} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= \frac{\frac{x_{2,n}(g)}{x_{3,n}(g)}}{\frac{x_{2,n}(f)}{x_{3,n}(f)}} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= \frac{1 - S_{1,n}(g)}{1 - S_{1,n}(f)} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}))
\end{align*}$$

Observe that if $i \ge q_{n-1}$, then $B_{n+1}^i(f) = D_{n+1}^{i-q_{n-1}}(f)$. Therefore,

$$\frac{Dh_{n+1}|_{B_{n+1}^i(f)}}{Dh_n|_{B_{n+1}^i(f)}} = 1.$$
For $i < q_{n-1}$, with a similar calculation as before, we have

$$ 
\begin{aligned}
\frac{Dh_{n+1|B_{n+1}^i(f)}}{Dh_{n|B_{n+1}^i(f)}} &= \frac{|B_{n+1}(g)|}{|A_n(g)|} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= \frac{x_{3,n+1}(g)}{x_{3,n+1}(f)} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) e^{(c_s^*(g)-c_s^*(g))(e_2^s+e_3^s)\lambda_s^{p_n}} \\
&= 1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}} + |c_s^*(g) - c_s^*(f)|\lambda_s^{p_n})
\end{aligned}
$$

where we used Lemma 4.18.

Let us remark also that for $i \ge q_{n-1}$, $C_{n+1}^i(f) = C_{n+1}^{i-q_{n-1}}(f)$. Thus

$$ \frac{Dh_{n+1|C_{n+1}^i(f)}}{Dh_{n|C_{n+1}^i(f)}} = 1 $$

Now, when $i < q_{n-1}$, with a similar calculation as before we obtain

$$ 
\begin{aligned}
\frac{Dh_{n+1|C_{n+1}^i(f)}}{Dh_{n|C_{n+1}^i(f)}} &= \frac{|C_{n+1}(g)|}{|A_n(g)|} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= \frac{x_{4,n+1}(g) - x_{3,n+1}(g)}{x_{4,n+1}(f) - x_{3,n+1}(f)} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= \frac{1 - [x_{3,n+1}(g) + (1 - x_{4,n+1}(g))]}{1 - [x_{3,n+1}(f) + (1 - x_{4,n+1}(f))]} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= \frac{1 - \alpha_{n+1}(g)}{1 - \alpha_{n+1}(f)} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= 1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})
\end{aligned}
$$

where we used Lemma 4.10 and Lemma 4.16.

Let $\Delta_n(f)$ be defined by $\Delta_n(f) = (f^{-q_{n+1}}(U), f^{q_{n+1}+1}(U))$ and its iterates given by $\Delta_n^i(f) = f^i(\Delta_n(f))$, $i < q_n$. Observe that if $i \ge q_{n-1}$, then $D_{n+1}^i(f) = \Delta_n^{i-q_{n-1}}(f)$. Thus, since $S_{1,n} = O(\alpha_{n+1})$ (see point 4 of Lemma 4.10), it follows that

$$ 
\begin{aligned}
\frac{Dh_{n+1|D_{n+1}^i(f)}}{Dh_{n|D_{n+1}^i(f)}} &= \frac{|\Delta_n(g)|}{|B_n(g)|} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= \frac{S_{1,n+1}(g)}{S_{1,n+1}(f)} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= 1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}).
\end{aligned}
$$
For $i < q_{n-1}$, we obtain

$$
\begin{aligned}
\frac{Dh_{n+1}|_{D_{n+1}^i(f)}}{Dh_{n|D_{n+1}^i(f)}} &= \frac{|D_{n+1}(g)|}{|A_n(g)|} \frac{|D_{n+1}(g)(f)|}{|A_n(f)|} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= \frac{1 - x_{4,n+1}(g)}{1 - x_{4,n+1}(f)} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\
&= (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}))e^{(c_s^*(g)-c_s^*(g))e_2^{\ell_2} \lambda_s^{p_n}} \\
&= 1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}} + |c_s^*(g) - c_s^*(f)| \lambda_s^{p_n}).
\end{aligned}
$$

This ends the proof of the lemma.

<figure><img src="image_12.png" /></figure>

Note that, every boundary point of an interval in $P_n$ is in the orbit $o(f(U))$ of the critical points $f(u) = 0$. So, if $x \notin o(f(U))$ then $Dh_n$ is well defined. Lemma 5.6 implies that

$$D(x) = \lim_{n \to \infty} Dh_n(x)$$

exists. Moreover,

$$0 < \inf_x D(x) < \sup_x D(x) < \infty. \qquad (5.4)$$

**Lemma 5.7.** If $f, g \in W_{[1]}$ with critical exponents $(\ell_1, \ell_2)$ and $c_u^*(f) = c_u^*(g)$, then

$$D : [0, 1] \setminus o(f(U)) \to (0, \infty)$$

is continuous. In particular, for all $n \in \mathbb{N}$,

$$\log \frac{D(x)}{Dh_n(x)} = O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}} + |c_s^*(g) - c_s^*(f)|\lambda_s^{p_n}).$$

*Proof.* Fix $n_0 \in \mathbb{N}$, then by Lemma 5.6 we have

$$
\begin{aligned}
\left| \log \frac{D(x)}{Dh_{n_0}(x)} \right| &\le \sum_{k \ge 0} \log \frac{Dh_{n_0+k+1}(x)}{Dh_{n_0+k}(x)} \\
&\le \sum_{k \ge 0} O(\alpha_{k-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}} + |c_s^*(g) - c_s^*(f)| \lambda_s^{p_k}) \\
&\le O(\alpha_{n_0-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}} + |c_s^*(g) - c_s^*(f)| \lambda_s^{p_{n_0}}).
\end{aligned}
$$

Therefore, if $x, y \in I \in P_{n_0}$, we have

$$
\begin{aligned}
\left| \log \frac{D(x)}{D(y)} \right| &\le \left| \log \frac{D(x)}{Dh_{n_0}(x)} \right| + \left| \log \frac{D(y)}{Dh_{n_0}(x)} \right| \\
&\le O(\alpha_{n_0-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}} + |c_s^*(g) - c_s^*(f)| \lambda_s^{p_{n_0}})
\end{aligned}
$$

which means that $D$ is continuous.

<figure><img src="image_13.png" /></figure>

Let $T$ be an interval in $[0, 1]$, then by formula (5.3),

$$|h(T)| = \lim_{n \to \infty} |h_n(T)| = \lim_{n \to \infty} \int_T Dh_n \le \int_T \sup_x D(x) \le \sup_x D(x)|T|. \quad (5.5)$$
Also,

$$|h(T)| = \lim_{n \to \infty} |h_n(T)| = \lim_{n \to \infty} \int_T Dh_n \ge \inf_x D(x) \ge \inf_x D(x)|T|. \quad (5.6)$$

**Proposition 5.8.** Let $f, g \in W_{[1]}$. Then,

$$h \text{ is a bi-lipschitz homeo} \iff c_u^*(f) = c_u^*(g), c_+(f) + c_+'(f) = c_+(g) + c_+'(g).$$

*Proof.* If $c_u^*(f) = c_u^*(g)$ and $c_+(f) + c_+'(f) = c_+(g) + c_+'(g)$ then by inequalities (5.5), (5.6), Lemma 5.7 and inequalities in (5.4), $h$ is a bi-Lipschitz. $\square$

As consequence of Proposition 5.8, we have:

**Corollary 5.9.** Let $f, g \in W_{[1]}$ with critical exponents $(\ell_1, \ell_2)$. If $c_u^*(f) = c_u^*(g)$ and $c_+(f) + c_+'(f) = c_+(g) + c_+'(g)$, then $K_f$ and $K_g$ have the same Hausdorff dimension $1$.

## 5.3 $C^{1+\beta}$ diffeomorphism Conjugacy

The rigidity class is described as:

**Proposition 5.10.** Let $f, g \in W_{[1]}$ with $(\ell_1, \ell_2)$ as their critical exponents. If $h$ is the conjugacy between $f$ and $g$, then for $\beta = \frac{\bar{c}_u(f)}{c_u(f)} \frac{(e_2^u + e_3^u)(\lambda_u - 1)}{2\bar{l}\lambda_u^2} \in (0, 1)$, the following biconditional holds

$$h \text{ is a } C^{1+\beta} \text{ diffeo} \iff \begin{aligned}[t] & c_u^*(f) = c_u^*(g), c_+(f) + c_+'(f) = c_+(g) + c_+'(g) \\ & \text{and } c_s^*(f) = c_s^*(g). \end{aligned}$$

*Proof.* Suppose that $h$ is a $C^{1+\beta}$ diffeo, then by Proposition 5.3, $c_u^*(f) = c_u^*(g)$ and $c_+(f) + c_+'(f) = c_+(g) + c_+'(g)$. Thus, by Proposition 4.9 and Lemma 4.16 we get

$$\begin{align*} \frac{Dh(f^{q_{n+1}+1}(U_f))}{Dh(f^{q_n+1}(U_f))} &= \frac{Dg^{q_{n-1}}(g^{q_n}(0))}{Df^{q_{n-1}}(f^{q_n}(0))} \\ &= \frac{Dq_{s_n(g)}(0) \frac{1-x_{2,n}(g)}{x_{1,n}(g)}}{Dq_{s_n(g)}(0) \frac{1-x_{2,n}(g)}{x_{1,n}(g)}} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\ &= \frac{s_n^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}(g)(1+O(s_n(g))) \frac{1-x_{2,n}(g)}{x_{1,n}(g)}}{s_n^{\frac{1}{\ell_1}, \frac{1}{\ell_2}}(f)(1+O(s_n(f))) \frac{1-x_{2,n}(g)}{x_{1,n}(g)}} (1 + O(\alpha_{n-2}^{\frac{1}{\ell_1}, \frac{1}{\ell_2}})) \\ &= e^{(c_s^*(f)-c_s^*(g))\lambda_s^{p_n}(e_2^s+e_3^s-e_4^s)+O((e^{c_u^*(f)}\lambda_u^{p_n-4}(e_2^u+e_3^u))^{1/\bar{l}})} \end{align*}$$

1.

$$\mathrm{Dim}_H K_f := \inf_{s \ge 0} \left\{ H_s(K_f) := \inf_{\delta > 0} \left\{ \sum_{|U_i|^s} K_f \subset \cup U_i, \ |U_i| < \delta \right\} = 0 \right\}.$$
Let us note that, by point 5 of Proposition 4.9 $e_2^s + e_3^s - e_4^s \neq 0$. Moreover, since $h$ is a $C^{1+\beta}$ diffeo, then by Lemma 5.5, there exists $K, C' > 0$ such that

$$ \frac{Dh(f^{q_{n+1}+1}(U_f))}{Dh(f^{q_n+1}(U_f))} \leq K |(f^{q_{n+1}+1}(U_f), f^{q_n+1}(U_f))|^\beta \leq \frac{K}{C'} e^{\beta \bar{c}_u(f)(e_2^u+e_3^u)\lambda_u^{p_n}} $$

where we used Lemma 5.5. The two above estimates imply that $c_s^*(f) = c_s^*(g)$.

Let us suppose that $c_s^*(f) = c_s^*(g)$, $c_u^*(f) = c_u^*(g)$ and $c_+(f) + c'_+(f) = c_+(g) + c'_+(g)$.

We are going to show first that, under conditions $c_u^*(f) = c_u^*(g)$ and $c_+(f) + c'_+(f) = c_+(g) + c'_+(g)$, $h$ is $C^1$ diffeo.

Since by (5.5) $h$ is differentiable with $Dh(x) = D(x)$, for all $x$, then it remains to prove that $D$ can be extended to a continuous function. This is possible if and only if for every $k \geq 0$

$$ \lim_{x \to c_k^-} D(x) = \lim_{x \to c_k^+} D(x) $$

where $c_k = f^k(f(U))$.

Let us fix $k \geq 0$, and let $n \in \mathbb{N}$ big enough such that $q_{n-1} > k$. Observe that

$$ D_-(c_k) := \lim_{x \to c_k^-} D(x) = \lim_{n \to \infty} \frac{|B_{2n}^k(g)|}{|B_{2n}^k(f)|}, $$

and

$$ D_+(c_k) := \lim_{x \to c_k^+} D(x) = \lim_{n \to \infty} \frac{|A_{2n}^k(g)|}{|A_{2n}^k(f)|}. $$

By Proposition 4.4 and Lemma 4.18, we get

$$ \frac{D_-(c_k)}{D_+(c_k)} = \lim_{n \to \infty} \frac{\frac{|B_{2n}^k(g)|}{|A_{2n}^k(g)|}}{\frac{|B_{2n}^k(f)|}{|A_{2n}^k(f)|}} = \lim_{n \to \infty} \frac{\frac{|B_{2n}(g)|}{|A_{2n}(g)|}}{\frac{|B_{2n}(f)|}{|A_{2n}(f)|}} = \lim_{n \to \infty} \frac{\frac{x_{3,2n}(g)}{-x_{1,2n}(g)}}{\frac{x_{3,2n}(f)}{-x_{1,2n}(f)}} = \lim_{n \to \infty} e^{\lambda_n^s} = 1. $$

Now, let $x, y \in K_f$ and choose $n$ maximal such that there exists $I \in \mathcal{P}_n$ and $I$ contains $[x, y]$. Since $B_{n+1}^{i+q_{n-1}} = D_n^i$, then by maximality of $n$, $I \neq D_n^i$. Also, as $x, y \in K_f$, then $I \neq C_n^i$. So, either $I = A_n^i$ or $I = B_n^i$. In the case where $I = A_n^i$, then we can assume that $x \in D_{n+1}^i$ and $y \in B_{n+1}^i$. Thus, by Lemma 3.5 and by Lemma 5.4 we obtain

$$ |x - y| \geq K |A_n^i| \geq \frac{K}{C} e^{c_u \lambda_u^{p_n} e_2^u \frac{2\lambda_u}{\lambda_u - 1}}. \quad (5.7) $$

Observe that by Lemma 5.7 and Lemma 4.16 we have

$$ \left| \log \frac{Dh(x)}{Dh(y)} \right| = O\left(e^{\frac{1}{\ell} c_u^* \lambda_u^{p_n-2} (e_2^u + e_3^u)}\right) = O\left(e^{\frac{1}{\ell} \bar{c}_u \lambda_u^{p_n-2} (e_2^u + e_3^u)}\right). \quad (5.8) $$

By (5.7) and (5.8), we have

$$ \frac{|Dh(x) - Dh(y)|}{|x - y|^\beta} = O\left(e^{\lambda_u^{p_n-2} \left[ \frac{1}{\ell} \bar{c}_u(e_2^u + e_3^u) - \beta c_u e_2^u \frac{2\lambda_u^2}{\lambda_u - 1} \right]}\right) = O(1) $$
for $0 < \beta < \frac{\bar{c}_u(e_2^u + e_3^u)(\lambda_u - 1)}{2\underline{c}_u \bar{l} e_2^u \lambda_u^2} = \frac{\bar{c}_u(1 + e_3^u)(\lambda_u - 1)}{2\underline{c}_u \bar{l} \lambda_u^2}$, see point 5 of Proposition 4.9.

For $I = B_n^i$, then we can choose $(x, y)$ such that $x \in A_{n+1}^i$ and $y \in B_{n+1}^{i+q_{n+1}}$.

If $x \in (f^{q_{n+3}+i+1}(U_f), f^{i+1}(U_f))$, then for $m = n + 1$, Lemma 3.5 and by Lemma 5.4 imply that

$$|x - y| \geq K |A_m^i| \geq \frac{K}{C} e^{\frac{c_u}{\bar{l}} \lambda_u^{p_m} e_2^u \frac{2\lambda_u}{\lambda_u - 1}}. \quad (5.9)$$

Otherwise, let $k > 0$, $m = n+2k$ maximal such that $x \in A_m^{i+q_{n+1}}$ and $y \in B_m^{i+q_{n+1}}$. By Lemma 5.4, we get

$$|x - y| \geq \frac{1}{2} \min\{|A_m^{i+q_{n+1}}, |B_m^{i+q_{n+1}}|\} \geq \frac{1}{C} e^{\frac{c_u}{\bar{l}} \lambda_u^{p_m} e_2^u \frac{2\lambda_u}{\lambda_u - 1}}. \quad (5.10)$$

On the other hand, for all $x, y \in B_n^i$,

$$\left|\log \frac{Dh(x)}{Dh(y)}\right| = O\left(e^{\frac{1}{\bar{l}} c_u^* \lambda_u^{p_m-2} (e_2^u + e_3^u)}\right) = O\left(e^{\frac{1}{\bar{l}} \bar{c}_u \lambda_u^{p_m-2} (e_2^u + e_3^u)}\right) \quad (5.11)$$

where we use Lemma 5.6 and Lemma 4.16.

Thus, by (5.9) (or (5.10)) and (5.11) we get

$$\frac{|Dh(x) - Dh(y)|}{|x - y|^{\beta}} = O\left(e^{\lambda_u^{p_m-2} \left[ \frac{1}{\bar{l}} \bar{c}_u (e_2^u + e_3^u) - \beta \underline{c}_u e_2^u \frac{2\lambda_u^2}{\lambda_u - 1} \right]}\right) = O(1);$$

for $0 < \beta < \frac{\bar{c}_u(e_2^u + e_3^u)(\lambda_u - 1)}{2\underline{c}_u \bar{l} e_2^u \lambda_u^2} = \frac{\bar{c}_u(1 + e_3^u)(\lambda_u - 1)}{2\underline{c}_u \bar{l} \lambda_u^2}$, see point 5 of Proposition 4.9. $\square$

## 5.4 Holder Conjugacy

**Proposition 5.11.** Let $f, g \in W_{[1]}$ with $(\ell_1, \ell_2)$ as their critical exponents, then the topological conjugation $h$ between $f$ and $g$ is a $C^\beta$ circle map where
$$\beta = \frac{\bar{c}_u(g)(e_2^u + e_3^u)(\lambda_u - 1)}{\underline{c}_u(f) \frac{2\bar{l}\lambda_u}{2\bar{l}\lambda_u}}.$$

*Proof.* Let $x, y \in K_f$ and choose $n$ maximal such that there exists $I(f) \in \mathcal{P}_n$ and $I(f)$ contains $[x, y]$. Since $B_{n+1}^{i+q_{n-1}}(f) = D_n^i(f)$, then by maximality of $n$, $I(f) \neq D_n^i(f)$. Also, as $x, y \in K_f$, then $I(f) \neq C_n^i(f)$. So, either $I(f) = A_n^i(f)$ or $I(f) = B_n^i(f)$. The assumptions on $Dh$ in the previous section are the same on $h$ in this section. Thus, the proof is done as in the previous section where we show that $Dh$ the derivative of $h$ is Holder. For more details, we have:

In the case where $I(f) = A_n^i(f)$, then we can assume that $x \in D_{n+1}^i(f)$ and $y \in B_{n+1}^i(f)$. Thus, by Lemma 3.5 and Lemma 5.4, we obtain

$$|x - y| \geq K |A_n^i(f)| \geq \frac{K}{C} e^{\frac{c_u}{\bar{l}} \lambda_u^{p_n} e_2^u \frac{2\lambda_u}{\lambda_u - 1}}. \quad (5.12)$$

Also, by Lemma 5.5 we get

$$|h(x) - h(y)| \leq |h(I(f))| = |h(I(g))| = O\left(e^{\frac{1}{\bar{l}} \bar{c}_u(g) \lambda_u^{p_n} (e_2^u + e_3^u)}\right). \quad (5.13)$$
By (5.12) and (5.13) we have

$$ \frac{|h(x) - h(y)|}{|x - y|^{\beta}} = O\left(e^{\lambda_{u}^{p_n} \left[ \frac{1}{\ell} \bar{c}_u(g)(e_2^u + e_3^u) - \beta c_u(f)e_2^u \frac{2\lambda_u}{\lambda_u - 1} \right]}\right) = O(1) $$

for

$$ 0 < \beta < \frac{\bar{c}_u(g)}{c_u(f)} \frac{(e_2^u + e_3^u)(\lambda_u - 1)}{2\bar{\ell}e_2^u\lambda_u} = \frac{\bar{c}_u(g)}{c_u(f)} \frac{(1 + e_3^u)(\lambda_u - 1)}{2\bar{\ell}\lambda_u}, $$

see point 5 of Proposition 4.9.

Now for $I = B_n^i$, then we can choose $(x, y)$ such that $x \in A_{n+1}^i$ and $y \in B_{n+1}^{i+q_{n+1}}$.

If $x \in (f^{q_{n+3}+i+1}(U_f), f^{i+1}(U_f))$ (Let us put $m := n+1$) then by Lemma 3.5 and Lemma 5.4, we obtain

$$ |x - y| \geq K |A_m^i| \geq \frac{K}{C} e^{\frac{1}{\lambda_u} \lambda_u^{p_m} e_2^u \frac{2\lambda_u}{\lambda_u - 1}}. \quad (5.14) $$

Otherwise, let $k > 0$ ($m := n + 2k$) maximal such $x \in A_m^{i+q_{n+1}}(f)$ and $y \in B_m^{i+q_{n+1}}(f)$. By Lemma 5.4 we get

$$ |x - y| \geq \frac{1}{2} \min\{|A_m^{i+q_{n+1}}(f)|, |B_m^{i+q_{n+1}}(f)|\} \geq \frac{1}{C} e^{\frac{1}{\lambda_u} \lambda_u^{p_m} e_2^u \frac{2\lambda_u}{\lambda_u - 1}}. \quad (5.15) $$

Thus, in the both cases, by Lemma 5.6 and Lemma 4.16 we get

$$ |h(x) - h(y)| \leq |h(I(f))| = |h(I(g))| = O\left(e^{\frac{1}{\ell} \bar{c}_u(g) \lambda_u^{p_n} (e_2^u + e_3^u)}\right) \quad (5.16) $$

and by (5.14) (or (5.15)) and (5.16) we obtain

$$ \frac{|h(x) - h(y)|}{|x - y|^{\beta}} = O\left(e^{\lambda_u^{p_m} \left[ \frac{1}{\ell} \bar{c}_u(g)(e_2^u + e_3^u) - \beta c_u(f)e_2^u \frac{2\lambda_u}{\lambda_u - 1} \right]}\right) = O(1) $$

for

$$ 0 < \beta < \frac{\bar{c}_u(g)}{c_u(f)} \frac{(e_2^u + e_3^u)(\lambda_u - 1)}{2e_2^u \bar{\ell} \lambda_u} = \frac{\bar{c}_u(g)}{c_u(f)} \frac{(1 + e_3^u)(\lambda_u - 1)}{2\bar{\ell}\lambda_u}, $$

This concludes the proof.

□

# 6 Proof of Lemma 1.1

We use the formalism presented in [9] in which the authors found a transition between degenerate geometry (i.e. the scalings $\alpha_n$ go to zero) and the bounded geometry (i.e, the sequence $\alpha_n$ is bounded away from zero) for the critical exponents $(\ell, \ell)$ with $\ell > 1$. These results were generalized in [22] for some $(\ell_1, \ell_2)$ belonging to $[1, \infty[$ and under the following assumption: the coefficients before the powers of $x$ in the asymptotics of a map $f$ in our class near the ends of the flat interval (see point 1 of Remark 2.3) are the same. More precisely, in [22] the author uses the assumption in the proof of lemma 2 pp. 658, whose the Lemma 6.7 is a reformulation for Fibonacci circle map with a flat piece. In the generic case, the assumption will fail for the first return maps, so it does not hold for infinitely renormalizable maps. That is why we study the asymptotic
behavior of $\alpha_n$ in our case. This proof can be adapted in [22]. Other results on the geometry of circle maps with a flat interval can be found in [8] and [19].

Let us put together sequences which are frequently used in this section.

$$ \alpha_n = \frac{|(-q_n, \underline{0})|}{|[-q_n, \underline{0})|}, \quad \sigma_n = \frac{|(0, q_n)|}{|(q_{n-1}, \underline{0})|} \quad \text{and} \quad s_n := \frac{|[-q_{n-2}, \underline{0})|}{|\underline{0}|} $$

# 6.1 Preliminaries

## 6.1.1 Cross-Ratio Inequalities

**Notation 6.1.** We denote by $\mathbb{R}_<^4$, the subset of $\mathbb{R}^4$ defined by

$$ \mathbb{R}_<^4 := \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4, \text{ such that } x_1 < x_2 < x_3 < x_4\}. $$

The following result comes from 1.5 Lemma in [5].

**Proposition 6.2.** *The Cross-Ratio Inequality.*

Let $f \in \mathcal{L}^X$ with $U$ as its flat piece. Let $(a, b, c, d) \in \mathbb{R}_<^4$. The cross-ratio $\mathbf{Poin}$ is defined by

$$ \mathbf{Poin} (a, b, c, d) := \frac{|d - a||b - c|}{|c - a||d - b|}. $$

The distortion of the cross-ratio $\mathbf{Poin}$ are given by

$$ \mathcal{DPoin} (a, b, c, d) := \frac{\mathbf{Poin} (f(a), f(b), f(c), f(d))}{\mathbf{Poin} (a, b, c, d)}. $$

Let us consider a set of $n + 1$ quadruples $\{a_i, b_i, c_i, d_i\}$ with the following properties:

1. Each point of the circle belongs to at most $k$ intervals $(a_i, d_i)$.

2. The intervals $(b_i, c_i)$ do not intersect $U$.

Then

$$ \prod_{i=0}^{n} \mathcal{DPoin} (a_i, b_i, c_i, d_i) \geq 1. $$

**Remark 6.3.** Let $I$ and $J$ be two intervals of finite and nonzero lengths such that $\bar{I} \cap \bar{J} = \emptyset$. We assume that, $J$ is on the right of $I$ and we put $I := [a, b]$ and $J := [c, d]$. Then

$$ \mathbf{Poin} (I, J) := \frac{|(I, J)||[I, J]|}{|[I, J)||(I, J]|} = \mathbf{Poin} (a, b, c, d). $$

## 6.1.2 Basic Lemmas

**Lemma 6.4.** *The ratio*

$$ \frac{|(2q_{n-1}, q_{n-1})|}{|(2q_{n-1}, \underline{0})|} $$

is uniformly bounded away from zero.

The proof can be found in [9] (proof of Lemma 1.2).
**Lemma 6.5.** The sequence

$$ \frac{|(\underline{1}, q_n + 1)|}{|(q_{n-1} + 1, \underline{1})|} $$

is bounded.

*Proof.* We will show that the sequence

$$ \frac{|(q_{n-1} + 1, \underline{1})|}{|(\underline{1}, q_n + 1)|} $$

is uniformly bounded away from zero. Let us observe that, the previous ratio is larger than

$$ \mathrm{Cr}((\underline{q_{n-1} + 1}, \underline{1}), -\underline{q_{n-1} + 1}) $$

which by CRI with $f^{q_{n-1}-1}$ is greater than

$$ \frac{|(2q_{n-1}, q_{n-1})|}{|(2q_{n-1}, 0)|} $$

times a constant. This last ratio is uniformly bounded away from zero, see Lemma 6.4. □

## 6.2 A priori Bounds of $\alpha_n$

**Proposition 6.6.** Let $n \in \mathbb{N}$ and $(\ell_1, \ell_2) \in (1, 2)^2$.

For all $\alpha_n$,

$$ \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} < 0.55; $$

for at least every other $\alpha_n$

$$ \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \le 0.3. $$

If

$$ \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} > 0.3, $$

then either,

$$ \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} < 0.44 \quad \text{or} \quad \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} < 0.16. $$

*Proof.* Let

$$ \gamma_{1,n} = |(-q_n, 0)|, \quad \gamma_n = \frac{\gamma_{1,n}}{\gamma_{1,n-1}}, $$

$$ \gamma_n^{(\ell_1|\ell_2)} := \frac{k_2}{k_1} \cdot \frac{\gamma_{1,n}^{\ell_1}}{\gamma_{1,n-1}^{\ell_2}} \quad \text{if } n \in 2\mathbb{Z} \text{ and } \gamma_n^{(\ell_1|\ell_2)} := \frac{k_1}{k_2} \cdot \frac{\gamma_{1,n}^{\ell_2}}{\gamma_{1,n-1}^{\ell_1}} \quad \text{if } n \in 2\mathbb{Z} + 1 $$

where $k_1$ and $k_2$ come from Fact 2.2.

These notations simplify the formalization of the following lemma which will play an important (essential) role in the proof of Proposition 6.6.

**Lemma 6.7.** For large enough $n \in \mathbb{N}$, the following inequality holds

$$ \frac{(\alpha_n^{\ell_1, \ell_2} + \alpha_{n-1}^{\ell_1, \ell_2} \gamma_n^{(\ell_1|\ell_2)})(1 + \gamma_n^{(\ell_1|\ell_2)})}{(1 + \alpha_{n-1}^{\ell_1, \ell_2} \gamma_n^{(\ell_1|\ell_2)})(\alpha_n^{\ell_1, \ell_2} + \gamma_n^{(\ell_1|\ell_2)})} \le s_n \alpha_{n-2}. \quad (6.1) $$
Proof. Let $n \in 2\mathbb{N} + 1$ large enough. By point 1 of Remark 2.3, the left hand side of (6.1) is equal to the cross-ratio

$$ \mathbf{Poin}(-q_n + 1, -q_{n-1} + 1). $$

Applying $f^{q_{n-1}-1}$, by the expanding cross-ratio property, we get the inequality. $\square$

The left hand side is a function of the three variables $\alpha_n^{\ell_1, \ell_2}$, $\alpha_{n-1}^{\ell_1, \ell_2}$ and $\gamma_n^{(\ell_2|\ell_1)}$. Observe that the function increases monotonically with each of the first two variables. However, relatively to the third variable, the function reaches a minimum. To see this, take the logarithm of the function and check that the first derivative is equal to zero only when

$$ (\gamma_n^{(\ell_2|\ell_1)})^2 = \frac{\alpha_n^{\ell_1, \ell_2}}{\alpha_{n-1}^{\ell_1, \ell_2}}. $$

By substituting this for $\gamma_n^{(\ell_2|\ell_1)}$ we get that

$$ \left( \frac{\alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} + \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}}{1 + \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}} \right)^2 \le s_n \alpha_{n-2}. \quad (6.2) $$

Let be the sequence $y'_n$ defined by

$$ y'_{n} = \min\{\alpha_{n}^{\frac{\ell_{1}}{2}, \frac{\ell_{2}}{2}}, \alpha_{n-1}^{\frac{\ell_{1}}{2}, \frac{\ell_{2}}{2}}\}. $$

Since $\alpha_{n-2} \le \alpha_{n-2}^{\frac{\ell_1}{2}, \frac{\ell_2}}$. Substituting the above variable into inequality (6.2) gives rise to a quadratic inequality $y'_n$ whose only root in the interval $(0, 1)$ is given by

$$ \frac{\sqrt{s_n \alpha_{n-2}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}}}{1 + \sqrt{1 - s_n \alpha_{n-2}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}}}. $$

That is,

$$ y'_{n} = \min\{\alpha_{n}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}, \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}\} \le \frac{\sqrt{s_n \alpha_{n-2}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}}}{1 + \sqrt{1 - s_n \alpha_{n-2}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}}}. \quad (6.3) $$

We use the elementary lemma (Lemma 3.4 in [9]).

**Lemma 6.8.** *The function*

$$ h_n(z) = \frac{\sqrt{s_n z}}{1 + \sqrt{1 - s_n z}} $$

moves points to the left, $h(z) < z$, if $z \ge 0.3$ and $n$ is large enough.

**Lemma 6.9.** *There is a subsequence of $\alpha_n$ including at least every other $\alpha_n$, such that*

$$ \limsup \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \le 0.3. $$
Proof. We select the subsequence.

1. The initial term: There exists $n-2 \in \mathbb{N}$, such that $\alpha_{n-2}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \le 0.3$. This comes directly from the properties of the function $h_n$ (Lemma 6.8) and inequality (6.3).

2. The next element. Suppose that $\alpha_{n-2}$ has been selected. If

$$y'_{n} = \min\{\alpha_{n}^{\frac{\ell_{1}}{2}, \frac{\ell_{2}}{2}}, \alpha_{n-1}^{\frac{\ell_{1}}{2}, \frac{\ell_{2}}{2}}\} = \alpha_{n-1}^{\frac{\ell_{1}}{2}, \frac{\ell_{2}}{2}} \quad \text{or} \quad \alpha_{n-1}^{\frac{\ell_{1}}{2}, \frac{\ell_{2}}{2}} \le 0.3,$$

then, we select $\alpha_{n-1}$ as the next term. Otherwise, $\alpha_n$ is the next term. Thus, the sequence is constructed. $\square$

**Corollary 6.10.** For the whole sequence $\alpha_n$ we have

$$\limsup \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \le (0.3)^{\frac{1}{2}}.$$

Moreover, if $\alpha_{n-1}$ does belong to the subsequence $\alpha_n$ defined by Lemma 6.9 then either

$$\alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} < 0.44 \quad \text{or} \quad \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} < 0.16.$$

*Proof.* Observe that the function

$$H : (s,t) \in \mathbb{R}_+^2 \mapsto F(s,t) = \frac{s+t}{1+st}$$

is and for fixed $s$, the function $F(s, \cdot)$ reaches its minimum at zero by taking the value $s$. Therefore, for every $s, t \ge 0$,

$$s, t \le \frac{s+t}{1+st}.$$

So,

$$\alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}, \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \le \frac{\alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} + \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}}{1 + \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}}. \quad (6.4)$$

Thus, according to that $\alpha_{n-2}$ is an element of the sequence of Lemma 6.9, it follows from (6.2) that the right member of (6.4) is estimated as follows

$$\frac{\alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} + \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}}{1 + \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}} \le \sqrt{s_n \alpha_{n-2}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}} \approx \sqrt{\alpha_{n-2}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}} \le \sqrt{0.3}. \quad (6.5)$$

Now, suppose that $\alpha_{n-1}$ does not belong to the subsequence chosen in the proof of Lemma 6.9, then

$$\min\{\alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}, \alpha_{n-1}^{\frac{\ell_1}{2}, \frac{\ell_2}{2}}\} = \alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \le 0.3. \quad (6.6)$$

Thus, if $\alpha_n^{\frac{\ell_1}{2}, \frac{\ell_2}{2}} \ge 0.16$, then by combining this with (6.5) and (6.6), we obtain the desired estimate. $\square$

This ends the prove of Proposition 6.6. $\square$
## 6.3 Recursive formula of $\alpha_n$

**Proposition 6.11.** Fix $\ell_1, \ell_2 > 1$. Let $n \in \mathbb{N}$ large enough, we have

$$ \alpha_{2n}^{\ell_1} \le M_{2n}(\ell_1)\alpha_{2n-2}^2 \quad \text{and} \quad \alpha_{2n+1}^{\ell_2} \le M_{2n+1}(\ell_2)\alpha_{2n-1}^2 \quad (6.7) $$

where

$$ M_n(\ell) = s_{n-1}^2 \cdot \frac{2}{\ell} \cdot \frac{1}{1 + \sqrt{1 - \frac{2(\ell-1)}{\ell} s_{n-1} \alpha_{n-1}}} \cdot \frac{1}{1 - \alpha_{n-2}} \cdot \frac{\sigma_n}{\sigma_{n-2}} $$

*Proof.* We treat the case even $n$ and the case odd $n$ is treated in a similar way. Recall that,

$$ \alpha_n = \frac{|(-q_n, \underline{0})|}{|(-q_n, \underline{0})|} $$

For large enough even $n$, applying $f$ to the equality we have

$$ \alpha_n^{\ell_1} = \frac{|(-q_n + 1, \underline{1})|}{|(-q_n + 1, \underline{1})|} $$

which is certainly less than the cross-ratio

$$ \mathbf{Poin}(-q_n + 1, (\underline{1}, -q_{n-1} + 1]). $$

Since the cross-ratio **Poin** is expanded by $f^{q_{n-1}-1}$, then

$$ \alpha_n^{\ell_1} < \delta_n(1)s_n(1) \quad (6.8) $$

with

$$ \delta_n(k) := \frac{|(-q_n + kq_{n-1}, kq_{n-1})|}{|(-q_n + kq_{n-1}, \underline{kq_{n-1}})|} $$

and

$$ s_n(k) := \frac{|(-q_n + kq_{n-1}, \underline{0})|}{|(-q_n + kq_{n-1}, \underline{0})|} $$

By multiplying and dividing the right member of (6.8) by $\alpha_{n-2}^2$, we get

$$ \alpha_n^{\ell_1} \le s_n \nu_{n-2} \mu_{n-2} \alpha_{n-2}^2 \quad (6.9) $$

with

$$ \nu_{n-2} := \frac{|(-q_{n-2}, \underline{0})|}{|(-q_{n-1}, \underline{0})|} \cdot \frac{|(-q_{n-2}, \underline{0})|}{|(-q_{n-2}, q_{n-1})|} $$

and

$$ \mu_{n-2} := \frac{|(-q_{n-2}, q_{n-1})|}{|(-q_{n-2}, \underline{0})|} $$

It remains to estimate $\nu_{n-2}$ and $\mu_{n-2}$ to end this part. For $\nu_{n-2}$, observe that

$$ |(-q_{n-2}, \underline{0})| \le |(q_{n-3}, \underline{0})| $$

so that,

$$ \nu_{n-2} = \frac{1}{\sigma_{n-1}\sigma_{n-2}} \cdot \frac{1}{1 - \alpha_{n-2}} \quad (6.10) $$

For the estimate of $\mu_{n-2}$, we use the following lemma (Lemma 3.1 in [9]).
**Lemma 6.12.** Let $\ell \in (1,2)$. For all numbers $x > y$, we have the following inequality:

$$ \frac{x^{\ell} - y^{\ell}}{x^{\ell}} \geq \left( \frac{x-y}{x} \right) \left[ \ell - \frac{\ell(\ell-1)}{2} \left( \frac{x-y}{x} \right) \right]. $$

Now, apply $f$ into the intervals defining the ratio $\mu_{n-2}$. By Lemma 6.12, the resulting ratio is larger than

$$ \mu_{n-2}(\ell_1 - \frac{\ell_1(\ell_1 - 1)}{2}\mu_{n-2}). $$

The cross-ratio Poin

$$ \text{Poin}(-q_{n-2} + 1, (q_{n-1} + 1, 1)); $$

that is,

$$ \frac{|(-q_{n-2} + 1, q_{n-1} + 1)||[-q_{n-2} + 1, 1]|}{|[-q_{n-2} + 1, q_{n-1} + 1]||(-q_{n-2} + 1, 1)|} $$

is larger again. Thus, by expanding cross-ratio property on $f^{q_{n-2}}$, we obtain:

$$ \mu_{n-2}(\ell_1 - \frac{\ell_1(\ell_1 - 1)}{2}\mu_{n-2}) \leq s_{n-1}\sigma_n\sigma_{n-1}. $$

By solving this quadratic inequality, we obtain

$$ \mu_{n-2} < \frac{2}{\ell_1} \cdot \frac{1}{1 + \sqrt{1 - \frac{2(\ell_1 - 1)}{\ell_1} s_{n-1}\sigma_n\sigma_{n-1}}} s_{n-1}\sigma_n\sigma_{n-1}. \quad (6.11) $$

Since, $\sigma_n\sigma_{n-1} < \alpha_{n-1}$, the first inequality in (6.7) follows by combining inequalities (6.9), (6.10) and (6.11). $\square$

$\alpha_n$ goes to zero

Technical reformulation of Proposition 6.11. Let $W_n$ be a sequence defined by

$$ M_n(\ell) = W_n(\ell) \frac{\sigma_n}{\sigma_{n-2}}. $$

Let

$$ M'_{n}(\ell) := M_{n}(\ell)\alpha_{n-2}^{2-\ell} \quad \text{and} \quad W'_{n}(\ell) := W_{n}(\ell)\alpha_{n-2}^{2-\ell}. $$

For large enough even $n$, the recursive formula (6.7) can be written in the form:

$$ \alpha_n^{\ell_1} \leq W_n'(\ell_1) \frac{\sigma_n}{\sigma_{n-2}} \alpha_{n-2}^{\ell_1}. $$

so,

$$ \alpha_n^{\ell_1} \leq \prod_{k=2}^{k=n} W_k'(\ell_1) \frac{\sigma_n}{\sigma_0} \alpha_0^{\ell_1}. $$

$\prod_{k=2}^{k=n} W_k'(\ell_1)$ goes to zero: Observe that the size of $W_n'(\ell_1)$ is given by the study of the function

$$ W'_{n}(x, y, \ell_1) = \frac{1}{\frac{\ell_1}{2} + \frac{\ell_1}{2} \sqrt{1 - \frac{2(\ell_1 - 1)}{\ell_1} x^{\frac{2}{\ell_2}}}} \cdot \frac{y^{\frac{4}{\ell_1} - 2}}{1 - y^{\frac{2}{\ell_1}}}. $$

The meaning of variation of $W'_k(x, y, \ell_1)$ relative to the third variable is given by the following lemma (Lemma 3.2 in [9]).
**Lemma 6.13.** For any $0 < y < \frac{1}{\sqrt{e}}$, $x \in (0,1)$ and $\ell_1 \in (1,2]$ the function $W'_n(x, y, \ell_1)$ is increasing with respect to $\ell_1$.

**Analyse the asymptotic size of $W'_i(2)$.**

Since the hypotheses of Lemma 6.13 are satisfied (Proposition 6.6), it enough
to verify that the convergence of $\prod_{i=1}^n W'_i(2)$.

- If $\alpha_{n-2} < (0.3)^{\ell_1}$, then $W'(2) < W'(0.55, 0.16, 2) < 0, 9$.

- If not, then by Proposition 6.6, $W'(2) < W'(0.3, 0.44, 2) < 0.98$ or else,
$W'_{n+1}(2)W'_n(2) < W'(0.55, 0.16, 2)W'(0.16, 0.55, 2) < 0.85$.

As a consequence, we have:

**Corollary 6.14.** Let $\ell_1, \ell_2 \in (1,2)$. Then $\alpha_n$ go to zero least double exponentially fast.

**Acknowledgment:** The author would like to thank the referee whose valuable comments helped to improve the exposition of the paper.

I sincerely thank Prof. M. Martens and Prof. Dr. L. Palmisano for introducing me (at Institute of Mathematics and Physical Sciences (IMSP)) to the subject of this paper, for valuable advice, continuous encouragement and helpful discussions. Part of the research for this paper took place at National Advanced School of Engineering, Yaounde (ENSPY) and University of Dauala (UD). The author would like to thank the IMSP, ENSPY, UD and in particular Prof. Joel TOSSA, Prof. Carlos OGOUYANDJOU, Prof. Léonard TODJI-HOUNDE, Prof. Thomas BOUETOU BOUETOU, Prof. Bridinette THIOD-JIO SENDJA, épse FAN DIO, Prof. Ferdinand NGAKEU, Prof. Louis Aimé FONO and Prof. Anatole TEMGOUA for continuous encouragement or their hospitality.

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