2.1 Some definitions and Theorem B

We reduce the proof of Theorem A to that of Theorem B stated below. For this, we begin with some suitable definitions.

It will be convenient to explicitly express the constant in the Collet–Eckmann condition of equation (1) and, for this reason, we agree on the following definition.

Definition 2.1. Given $\gamma , C>0$ , we call a parameter $a (\gamma ,C)$ -Collet–Eckmann if

$$ \begin{align*} \vert \partial_x F^n(1;a) \vert \geq C e^{\gamma n} \quad (n \geq 0). \end{align*} $$

The set of all $(\gamma ,C)$ -Collet–Eckmann parameters is denoted $\operatorname {\mathrm {\mathcal {C}\mathcal {E}}}(\gamma ,C)$ .

Our parameter exclusion will be carried out on intervals centred at Collet–Eckmann parameters satisfying the following recurrence assumption.

Definition 2.2. A Collet–Eckmann parameter a is said to have polynomial recurrence (PR) if there exist constants $K=K(a)> 0$ and $\sigma = \sigma (a) \geq 0$ such that

$$ \begin{align*} \vert F^n(0;a) \vert \geq \frac{K}{n^\sigma} \quad (n \geq 1). \end{align*} $$

The set of all PR-parameters is denoted $\operatorname {\mathrm {\mathcal {P}\mathcal {R}}}$ .

Finally, we consider parameters for which the corresponding quadratic maps satisfy the reversed recurrence condition after some fixed time $N \geq 1$ :

$$ \begin{align*} \Lambda_N(\delta_n) = \{a \in \operatorname{\mathrm{\mathcal{N}\mathcal{R}}} : \vert F^n(0;a) \vert \geq \delta_n\ \text{for all}\ n \geq N\}. \end{align*} $$

Clearly, we have that

$$ \begin{align*} \Lambda(\delta_n) = \bigcup_{N \geq 1} \Lambda_N(\delta_n). \end{align*} $$

Theorem A will be deduced from the following theorem.

Theorem B. There exists $\tau \in (0,1)$ such that if $(\delta _n)$ is admissible and

$$ \begin{align*} \sum \frac{\delta_n}{\log n}\tau^{(\log^* n)^3} = \infty, \end{align*} $$

then for all $N \geq 1$ , $\gamma> 0$ , $C> 0$ , and for all $a \in \mathcal {P}\mathcal {R}$ , there exists an interval $\omega _a$ centred at a such that

$$ \begin{align*} \operatorname{\mathrm{\operatorname{Leb}}}(\Lambda_N(\delta_n) \cap \operatorname{\mathrm{\mathcal{C}\mathcal{E}}}(\gamma,C) \cap \omega_a) = 0. \end{align*} $$

2.2 Proof of Theorem A

Using Theorem B, Theorem A is proved by a standard covering argument. Since $\omega _a$ is centred at a, so is the smaller interval $\omega _a' = \omega _a/5$ . By Vitali’s covering lemma, there exists a countable collection $(a_j)$ of PR-parameters such that

$$ \begin{align*} \operatorname{\mathrm{\mathcal{P}\mathcal{R}}} \subset \bigcup_{a \in \operatorname{\mathrm{\mathcal{P}\mathcal{R}}}} \omega'_a \subset \bigcup_{j = 1}^\infty \omega_{a_j}. \end{align*} $$

It now follows directly that

$$ \begin{align*} \operatorname{\mathrm{\operatorname{Leb}}}(\Lambda_N(\delta_n) \cap \operatorname{\mathrm{\mathcal{C}\mathcal{E}}}(\gamma,C) \cap \operatorname{\mathrm{\mathcal{P}\mathcal{R}}}) \leq \sum_{j=1}^\infty \operatorname{\mathrm{\operatorname{Leb}}}(\Lambda_N(\delta_n) \cap \operatorname{\mathrm{\mathcal{C}\mathcal{E}}}(\gamma,C) \cap \omega_{a_j}) = 0, \end{align*} $$

and therefore

$$ \begin{align*} \operatorname{\mathrm{\operatorname{Leb}}} (\Lambda(\delta_n) \cap \operatorname{\mathrm{\mathcal{C}\mathcal{E}}} \cap \operatorname{\mathrm{\mathcal{P}\mathcal{R}}}) &\leq \sum_{N,k,l \geq 1} \operatorname{\mathrm{\operatorname{Leb}}}(\Lambda_N(\delta_n) \cap \operatorname{\mathrm{\mathcal{C}\mathcal{E}}}(k^{-1}\log 2,l^{-1}) \cap \operatorname{\mathrm{\mathcal{P}\mathcal{R}}}) \\ &= 0. \end{align*} $$

Finally, we notice that $\Lambda (\delta _n) \cap \operatorname {\mathrm {\mathcal {C}\mathcal {E}}} \subset \operatorname {\mathrm {\mathcal {P}\mathcal {R}}}$ ; indeed, this is clearly the case since $(\delta _n)$ is assumed to be admissible.

2.3 Outline of proof of Theorem B

The proof of Theorem B will rely on the classical parameter exclusion techniques developed by Benedicks and Carleson [Reference Benedicks and CarlesonBC85, Reference Benedicks and CarlesonBC91], complemented with more recent results. In particular, we allow for perturbation around a parameter in a more general position than $a=2$ . In contrast to the usual application of these techniques, our goal here is the show that what remains after excluding parameters is a set of zero Lebesgue measure. One of the key points in our approach is the introduction of unbounded distortion estimates.

We will carefully study the returns of the critical orbit, simultaneously for maps corresponding to parameters in a suitable interval $\omega \subset [1,2]$ , to a small and fixed interval $(-\delta ,\delta ) = (-e^{-\Delta },e^{-\Delta })$ . These returns to $(-\delta ,\delta )$ will be classified as either inessential, essential, escape, or complete. Per definition of a complete return, we return close enough to $x = 0$ to be able to remove a large portion of $(-\delta _n,\delta _n)$ in phase space. To estimate what is removed in parameter space, we need distortion estimates. This will be achieved by (i) enforcing a $(\gamma ,C)$ -Collet–Eckmann condition and (ii) continuously making suitable partitions in phase space: $(-\delta ,\delta )$ is subdivided into partition elements $I_r = (e^{-r-1},e^{-r})$ for $r>0$ , and $I_r = -I_{-r}$ for $r < 0$ . Furthermore, each $I_r$ is subdivided into $r^2$ smaller intervals $I_{rl} \subset I_r$ , of equal length $\vert I_r\vert /r^2$ . After partitioning, we consider iterations of each partition element individually, and the proof of Theorem B will be one by induction.

We make a few comments on the summability condition appearing in the statement of Theorems A and B. To prove our result, we need to estimate how much is removed at a complete return, but also how long it takes from one complete return to the next. The factor $\tau ^{(\log ^* n)^3}$ is connected to the estimate of what is removed at complete returns and, more specifically, it is connected to distortion; as will be seen, our distortion estimates are unbounded. The factor $(\log n)^{-1}$ is directly connected to the time between two complete returns: if n is the index of a complete return, it will take $\lesssim \log n$ iterations until we reach the next complete return.

In the next section, we prove a couple of preliminary lemmas and confirm the existence of a suitable start-up interval $\omega _a$ centred at $a \in \operatorname {\mathrm {\mathcal {P}\mathcal {R}}}$ , for which the parameter exclusion will be carried out. After that, the induction step will be proved and an estimate for the measure of $\Lambda _N(\delta _n) \cap \operatorname {\mathrm {\mathcal {C}\mathcal {E}}}(\gamma , C) \cap \omega _a$ will be given.

3.1 Outside expansion lemma

The first result we will need is the following version of the classical Mañé hyperbolicity theorem (see [Reference de Melo and van StriendMvS93], for instance).

Lemma 3.1. (Outside expansion)

Given a Collet–Eckmann parameter $a_0$ , there exist constants $\gamma _M,C_M> 0$ such that, for all $\delta> 0$ sufficiently small, there is a constant $\epsilon _M = \epsilon _M(\delta )> 0$ such that, for all $a \in (a_0 - \epsilon _M,a_0 + \epsilon _M)$ , if

$$ \begin{align*} x, F(x;a), F^2(x;a), \ldots, F^{n-1}(x;a) \notin (-\delta,\delta), \end{align*} $$

then

$$ \begin{align*} \vert \partial_x F^n(x;a) \vert \geq \delta C_M e^{\gamma_M n}. \end{align*} $$

Furthermore, if we also have that $F^n(x;a) \in (-2\delta ,2\delta )$ , then

$$ \begin{align*} \vert \partial_x F^n(x;a) \vert \geq C_M e^{\gamma_M n}. \end{align*} $$

A similar lemma for the quadratic family can be found in [Reference Baladi, Benedicks and SchnellmannBBS15, Reference TsujiiTsu93], for instance. The version stated here allows for $\delta $ -independence at a more shallow return to the interval $(-2\delta ,2\delta )$ . To get this kind of annular result constitutes a minor modification of Lemma 4.1 in [Reference TsujiiTsu93]. We refer to Lemma 2.6 in [Reference AspenbergAsp21] and the proof therein, however, for a proof of the above result. This proof is based on Przytycki’s telescope lemma (see [Reference PrzytyckiPrz90] and also [Reference Przytycki, Rivera-Letelier and SmirnovPRLS03]). In contrast to the techniques in [Reference TsujiiTsu93], in the case of the quadratic family, no recurrence assumption is needed.

3.2 Phase-parameter distortion

If $t \mapsto F(x;a+t)$ is a family of (analytic) perturbations of $(x;a) \mapsto F(x;a)$ at a, we may expand each such perturbation as

$$ \begin{align*} F(x;a+t) = F(x;a) + t\partial_aF(x;a) + \text{higher order terms}, \end{align*} $$

and it is easy to verify that

$$ \begin{align*} \frac{\partial_a F^n(x;a)}{\partial_x F^{n-1}(F(x;a);a)} = \frac{\partial_a F^{n-1}(x;a)}{\partial_x F^{n-2}(F(x;a);a)} + \frac{\partial_a F(F^{n-1}(x;a);a)}{\partial_x F^{n-1}(F(x;a);a)}. \end{align*} $$

Our concern is with the quadratic family $x \mapsto 1-ax^2 = F(x;a)$ , with a being the parameter value. In particular, we are interested in the critical orbit of each such member and, to this end, we introduce the functions $a \mapsto \xi _j(a) = F^j(0;a)$ for $j \geq 0$ . In view of our notation and the above relationship, we see that

$$ \begin{align*} \frac{\partial_a F^n(0;a)}{\partial_x F^{n-1}(1;a)} = \sum_{k=0}^{n-1} \frac{\partial_aF(\xi_k(a);a)}{\partial_x F^k(1;a)}. \end{align*} $$

Throughout the proof of Theorem B, it will be of importance to be able to compare phase and parameter derivatives. Under the assumption of exponential increase of the phase derivative along the critical orbit, this can be done, as is formulated in the following lemma. The proof is that of Lemma 2.10 in [Reference AspenbergAsp21].

Lemma 3.2. (Phase-parameter distortion)

Let $a_0$ be $(\gamma _0,C_0)$ -Collet–Eckmann, $\gamma _T \in (0,\gamma _0)$ , $C_T \in (0,C_0)$ , and $A \in (0,1)$ . There exist $T, N_T, \epsilon _T> 0$ such that if $a \in (a_0-\epsilon _T,a_0 + \epsilon _T)$ satisfies

$$ \begin{align*} \vert \partial_x F^j(1;a) \vert \geq C_T e^{\gamma_T j} \quad (j = 1,2,\ldots, N_T,\ldots n-1) \end{align*} $$

for some $n-1 \geq N_T$ , then

$$ \begin{align*} (1-A)T \leq \bigg\vert \frac{\partial_a F^n(0;a)}{\partial_x F^{n-1} (1;a)} \bigg\vert \leq (1+A)T. \end{align*} $$

Proof. According to Theorem 3 in [Reference TsujiiTsu00] (see also Theorem 1 in [Reference LevinLev14]),

$$ \begin{align*} \lim_{j \to \infty} \frac{\partial_a F^j(0;a_0)}{\partial_x F^{j-1}(1;a_0)} = \sum_{k=0}^\infty \frac{\partial_a F(\xi_k(a_0);a_0)}{\partial_x F^k(1;a_0)} = T \in \mathbb{R}_{>0}. \end{align*} $$

Let $N_T> 0$ be large enough so that

$$ \begin{align*} \bigg\vert \sum_{k=N_T}^\infty \frac{\partial_a F(\xi_k(a_0);a_0)}{\partial_x F^k(1;a_0)} \bigg\vert \leq \sum_{k=N_T}^\infty \frac{1}{C_0 e^{\gamma_0 k}} \leq \sum_{k=N_T}^\infty \frac{1}{C_T e^{\gamma_T k}} \leq \frac{1}{3}A T. \end{align*} $$

Since $a \mapsto \partial _a F(\xi _k(a);a)/\partial _x F^k(1;a)$ is continuous, there exists $\epsilon _T> 0$ such that given $a \in (a_0-\epsilon _T,a_0+\epsilon _T)$ ,

$$ \begin{align*} \bigg\vert \sum_{k=0}^{N_T-1} \frac{\partial_a F(\xi_k(a);a)}{\partial_x F(1;a)} - T \bigg\vert \leq \frac{1}{2}A T. \end{align*} $$

Assuming $x \mapsto 1-ax^2$ to be $(\gamma _T,C_T)$ -Collet–Eckmann up to time $n> N_T$ , the result now follows since

$$ \begin{align*} \bigg\vert \sum_{k=0}^n \frac{\partial_a F(\xi_k(a);a)}{\partial_x F^k(1;a)} - T \bigg\vert \leq AT.\\[-4pc] \end{align*} $$

3.3 Start-up lemma

With the above two lemmas, we now prove the existence of a suitable interval in parameter space on which the parameter exclusion will be carried out.

Given an admissible sequence $(\delta _n)$ , let $N_A$ be the integer in Definition 1.2. Fix $N_B \geq 1, \gamma _B> 0$ , and $C_B> 0$ , and let $a_0$ be a PR-parameter satisfying a $(\gamma _0,C_0)$ -Collet–Eckmann condition. In Lemma 3.2, we make the choice

$$ \begin{align*} \gamma_T = \min(\gamma_B,\gamma_0,\gamma_M)/20 \quad \text{and} \quad C_T = \min(C_B,C_0)/3. \end{align*} $$

Furthermore, let

$$ \begin{align*} \gamma = \min(\gamma_B,\gamma_0,\gamma_M)/2 \quad \text{and}\quad C = \min(C_B,C_0)/2, \end{align*} $$

and let $m_{-1} = \max (N_A,N_B,N_T)$ .

Proof. Given $x,y \in \xi _{j}(\omega _0)$ , $j \geq 1$ , consider the following distance condition:

(2) $$ \begin{align} \vert x - y \vert \leq \begin{cases} e^{-r}/r^2 &\mbox{if}\ \xi_{j}(\omega_0) \cap I_r \neq \emptyset, \\ S = \epsilon_1 \delta &\mbox{if}\ \xi_{j}(\omega_0) \cap (-\delta,\delta) = \emptyset. \end{cases} \end{align} $$

By making $\epsilon $ smaller, we may assume that equation (2) is satisfied up to time $m_{-1}$ . Moreover, we make sure that $\epsilon $ is small enough to comply with Lemma 3.2. Whenever equation (2) is satisfied, phase derivatives are comparable as follows:

(3) $$ \begin{align} \frac{1}{C_1} \leq \bigg\vert \frac{\partial_x F(x;a)}{\partial_x F(y;b)} \bigg\vert \leq C_1, \end{align} $$

with $C_1> 1$ a constant. This can be seen through the following estimate:

$$ \begin{align*} \bigg\vert \frac{\partial_x F(x;a)}{\partial_x F(y;b)}\bigg\vert = \bigg\vert \frac{-2ax}{-2by} \bigg\vert \leq \frac{a_0 + \epsilon}{a_0-\epsilon}\bigg(\bigg\vert \frac{x-y}{y} \bigg\vert + 1\bigg). \end{align*} $$

If we are outside $(-\delta ,\delta )$ , then

$$ \begin{align*} \bigg\vert \frac{x-y}{y} \bigg\vert \leq \frac{S}{\delta} = \epsilon_1, \end{align*} $$

and if we are hitting $I_r$ with largest possible r,

$$ \begin{align*} \bigg\vert \frac{x-y}{y} \bigg\vert \leq \frac{e^{-r}}{r^2}\frac{1}{e^{-(r+1)}} = \frac{e}{r^2} \leq \frac{e}{\Delta^2}. \end{align*} $$

By making sure that $\epsilon $ , $\epsilon _1$ , and $\delta $ are small enough, $C_1$ can be made as close to $1$ as we want. In particular, we make $C_1$ close enough to $1$ so that

(4) $$ \begin{align} C_1^{-j}C_0 e^{\gamma_0 j} \geq C e^{\gamma j} \quad (j \geq 0). \end{align} $$

As long as the distance condition in equation (2) is satisfied, we will have good expansion along the critical orbits. Indeed, by equations (3) and (4), it follows that, given $a \in \omega _0$ ,

$$ \begin{align*} \vert \partial_x F^j(1;a) \vert &\geq C_1^{-j}\vert \partial_x F^j(1;a_0) \vert \\ &\geq C_1^{-j}C_0 e^{\gamma_0 j} \\ &\geq C e^{\gamma j} \quad (j \geq 0\ \text{such that equation}\ (2)\ \text{is satisfied}). \end{align*} $$

This tells us that, during the time for which equation (2) is satisfied, each $a \in \omega _0$ is $(\gamma ,C)$ -Collet–Eckmann. In particular, since $\gamma> \gamma _T$ and $C> C_T$ , we can apply Lemma 3.2 and, together with the mean value theorem, we have that

$$ \begin{align*} \vert \xi_{j}(\omega_0)\vert &= \vert \partial_a F^j(0;a') \vert \vert \omega_0 \vert \\ &\geq (1-A)T \vert \partial_x F^{j-1}(1;a') \vert \vert \omega_0\vert \\ &\geq (1-A)T C e^{\gamma (j-1)} \vert \omega_0 \vert. \end{align*} $$

Our interval is thus expanding, and we let $m_0 = j$ , with $j \geq m_{-1}$ the smallest integer for which equation (2) is no longer satisfied. This proves statements (i) and (ii).

To prove statement (iii), let $K_0> 0$ and $\sigma _0 \geq 0$ be the constants associated to $a_0$ for which

$$ \begin{align*} \vert \xi_j(a_0) \vert \geq \frac{K_0}{j^{\sigma_0}} \quad (j \geq 1). \end{align*} $$

In view of equation (2), when we hit $(-\delta ,\delta )$ at some time $j < m_0$ ,

$$ \begin{align*} \vert \xi_j(a) \vert \geq \vert \xi_j(a_0) \vert - \vert \xi_j(\omega_0) \vert \geq \vert \xi_j(a_0) \vert - \frac{e^{-r}}{r^2}. \end{align*} $$

Here, r is such that

$$ \begin{align*} e^{-r-1} \leq \vert \xi_j(a_0) \vert, \end{align*} $$

and therefore, given $\delta $ small enough,

$$ \begin{align*} \vert \xi_j(a) \vert \geq \vert \xi_j(a_0) \vert \bigg(1 - \frac{e}{\Delta^2} \bigg) \geq \frac{K_0/2}{j^{\sigma_0}} \quad (j = 1,2,\ldots,m_0-1).\\[-3.6pc] \end{align*} $$

Since $\Lambda _{N_B}(\delta _n) \subset \Lambda _{m_0}(\delta _n)$ , Theorem B follows if

$$ \begin{align*} \operatorname{\mathrm{\operatorname{Leb}}}(\Lambda_{m_0}(\delta_n) \cap \operatorname{\mathrm{\mathcal{C}\mathcal{E}}}(\gamma_B,C_B) \cap \omega_0) = 0. \end{align*} $$

4.1 Initial iterates

Let $\omega _0 = \Delta _0$ be the start-up interval obtained in Lemma 3.4. Iterating this interval under $\xi $ and successively excluding parameters that do not satisfy the recurrence condition, or the Collet–Eckmann condition, we will inductively define a nested sequence $\Delta _0 \supset \Delta _1 \supset \cdots \supset \Delta _k \supset \cdots $ of sets of parameters satisfying

$$ \begin{align*} \Lambda_{m_0}(\delta_n) \cap \operatorname{\mathrm{\mathcal{C}\mathcal{E}}}(\gamma_B,C_B) \cap \omega_0 \subset \Delta_\infty = \bigcap_{k = 0}^\infty \Delta_k, \end{align*} $$

and our goal is to estimate the Lebesgue measure of $\Delta _\infty $ . This will require a careful analysis of the so-called returns to $(-\delta ,\delta )$ , and we will distinguish between four types of returns: inessential, essential, escape, and complete. At the $k\text {th}$ complete return, we will be in the position of excluding parameters and form the partition that will make up the set $\Delta _k$ . Below, we will describe the iterations from the $k\text {th}$ complete return to the $(k+1)\text {th}$ complete return, and hence the forming of $\Delta _{k+1}$ . Before indicating the partition and giving a definition of the different returns, we begin by considering the first initial iterates of $\xi _{m_0}(\omega _0)$ .

If $\xi _{m_0}(\omega _0) \cap (-\delta ,\delta ) \neq \emptyset $ , then we have reached a return and we proceed accordingly as is described below. If this is not the case, then we are in the situation

$$ \begin{align*} \xi_{m_0}(\omega_0) \cap (-\delta,\delta) = \emptyset \quad \text{and}\quad \vert \xi_{m_0}(\omega_0) \vert \geq S, \end{align*} $$

with S larger than any partition element $I_{rl} \subset (-\delta ,\delta )$ (see Remark 3.5). Since the length of the image is bounded from below, there is an integer $n^* = n^*(S)$ such that, for some smallest $n \leq n^*$ , we have

$$ \begin{align*} \xi_{m_0+n}(\omega_0) \cap (-\delta,\delta) \neq \emptyset. \end{align*} $$

In this case, $m_0 + n$ is the index of the first return. We claim that if $m_0$ is large enough, we can assume a good derivative up to time $m_0 + n$ . To realize this, consider for $j < n$ the distortion quotient

$$ \begin{align*} \bigg\vert\frac{\partial_x F^{m_0+j}(1;a)}{\partial_x F^{m_0+j}(1;b)} \bigg\vert = \bigg\vert\frac{\partial_x F^{m_0-1}(1;a)}{\partial_x F^{m_0-1}(1;b)} \bigg\vert \bigg\vert\frac{\partial_x F^{j+1}(\xi_{m_0}(a);a)}{\partial_x F^{j+1}(\xi_{m_0}(b);b)} \bigg\vert. \end{align*} $$

Since the distance conditions in equation (2) are satisfied up to time $m_0 - 1$ , the first factor in the above right-hand side is bounded from above by the constant $C_1^{m_0-1}$ , with $C_1> 1$ being very close to $1$ (see equation (3)). Furthermore, since $j < n < n^*(S)$ , and since we, by assumption, are iterating outside $(-\delta ,\delta )$ , the second factor in the above right-hand side is bounded from above by some positive constant $C_{S,\delta }$ dependent on S and $\delta $ .

If there is no parameter $a' \in \omega _0$ such that $\vert \partial _x F^{m_0+j}(1;a') \vert \geq C_B e^{\gamma _B(m_0 + j)}$ , then we have already reached our desired result. If, however, there is such a parameter $a'$ , then for all $a \in \omega _0$ , it follows from the above distortion estimate and our choice of $\gamma $ that

$$ \begin{align*} \vert \partial_x F^{m_0+j}(1;a) \vert \geq \frac{C_B e^{\gamma_B (m_0+j)}}{C_1^{m_0-1}C_{S,\delta}} \geq Ce^{\gamma(m_0+j)}, \end{align*} $$

provided $m_0$ is large enough. We conclude that

(5) $$ \begin{align} \vert \partial_x F^{j}(1;a) \vert \geq C e^{\gamma j} \quad (a \in \omega_0,\ j = 0,1,\ldots,m_0+n-1). \end{align} $$

In the case where we have to iterate $\xi _{m_0}(\omega _0)$ further to hit $(-\delta ,\delta )$ , we still let $m_0$ denote the index of the first return.

4.2 The partition

At the $(k+1)\text {th}$ step in our process of excluding parameters, $\Delta _k$ consists of disjoint intervals $\omega _k^{rl}$ , and for each such interval, there is an associated time $m_k^{rl}$ for which either $\xi _{m_k^{rl}}(\omega _k^{rl}) = I_{rl} \subset (-4\delta ,4\delta )$ or $\xi _{m_k^{rl}}(\omega _k^{rl})$ is mapped onto $\pm (\delta ,x)$ , with $\vert x-\delta \vert \geq 3\delta $ . We iterate each such interval individually and let $m_{k+1}^{rl}$ be the time for which $\xi _{m_{k+1}^{rl}}(\omega _k^{rl})$ hits deep enough for us to be able to remove a significant portion of $(-\delta _{m_{k+1}^{rl}},\delta _{m_{k+1}^{rl}})$ in phase space, and let $E_k^{rl}$ denote the corresponding set that is removed in parameter space. We now form the set $\hat {\omega }_k^{rl} \subset \Delta _{k+1}$ and make the partition

$$ \begin{align*} \hat{\omega}_k^{rl} = \omega_k^{rl} \smallsetminus E_k^{rl} = \bigg( \bigcup_{r',l'} \omega_{k+1}^{r'l'} \bigg) \cup T_{k+1} = N_{k+1} \cup T_{k+1}. \end{align*} $$

Here, each $\omega _{k+1}^{r'l'} \subset N_{k+1}$ is such that $\xi _{m_{k+1}^{rl}}(\omega _{k+1}^{r'l'}) = I_{r'l'} \subset (-4\delta ,4\delta )$ , and $T_{k+1}$ consists of (at most) two intervals whose image under $\xi _{m_{k+1}^{rl}}$ is $\pm (\delta ,x)$ , with $\vert x-\delta \vert \geq 3 \delta $ .

Remark 4.1. At most four intervals $\omega _{k+1}^{r'l'} \subset N_{k+1}$ will be mapped onto an interval slightly larger than $I_{r'l'}$ , that is,

$$ \begin{align*} I_{r'l'} \subset \xi_{m_{k+1}^{rl}}(\omega_{k+1}^{r'l'}) \subset I_{r'l'}\cup I_{r"l"}, \end{align*} $$

with $I_{r'l'}$ and $I_{r"l"}$ adjacent partition elements.

4.3 The different returns to $(-\delta ,\delta )$

At time $m_{k+1}^{rl}$ , we say that $\omega _k^{rl}$ has reached the $(k+1)\text {th}$ complete return to $(-\delta ,\delta )$ . In between the two complete returns of index $m_{k}^{rl}$ and $m_{k+1}^{rl}$ , we might have returns which are not complete. Given a return at time $n> m_k^{rl}$ , we classify it as follows.

1. (i) If $\xi _n(\omega _k^{rl}) \subset I_{r'l'} \cup I_{r"l"}$ , with $I_{r'l'}$ and $I_{r"l"}$ adjacent partition elements ( $r' \geq r"$ ), and if $\vert \xi _n(\omega _k^{rl}) \vert < \vert I_{r'l'}\vert $ , we call this an inessential return. The interval $I_{r'l'}\cup I_{r"l"}$ is called the host interval.

2. (ii) If the return is not inessential, it is called an essential return. The outer most partition element $I_r$ contained in the image is called the essential interval.

3. (iii) If $\xi _n(\omega _k^{rl}) \cap (-\delta ,\delta ) \neq \emptyset $ and $\vert \xi _n(\omega _k^{rl}) \smallsetminus (-\delta ,\delta ) \vert \geq 3\delta $ , we call this an escape return. The interval $\xi _n(\omega _k^{rl}) \smallsetminus (-\delta ,\delta )$ is called the escape interval.

4. (iv) Finally, if a return satisfies $\xi _n(\omega _k^{rl}) \cap (-\delta _n/3,\delta _n/3) \neq \emptyset $ , it is called a complete return.

We use these terms exclusively, that is, an inessential return is not essential, an essential return is not an escape, and an escape return is not complete.

Given $\omega _k^{rl} \subset \Delta _k$ , we want to find an upper bound for the index of the next complete return. In the worst case scenario, we encounter all of the above kinds of returns, in the order

$$ \begin{align*} \text{complete} \to \text{inessential} \to \text{essential} \to \text{escape} \to \text{complete}. \end{align*} $$

Given such behaviour, we show below that there is an absolute constant $\kappa>0$ such that the index of the $(k+1){\text {th}}$ complete return satisfies $m_{k+1}^{rl} \leq m_k^{rl} + \kappa \log m_k^{rl}$ .

4.4 Induction assumptions

Up until the start time $m_0$ , we do not want to assume anything regarding recurrence with respect to our recurrence rate $(\delta _n)$ . Since the perturbation is made around a PR-parameter $a_0$ , we do however have the following polynomial recurrence to rely on (Lemma 3.4).

1. (PR) $\vert F^j(0;a) \vert \geq K/j^\sigma $ for all $a \in \omega ^{rl}_k$ and $j = 1,2,\ldots ,m_0-1$ .

After $m_0$ we start excluding parameters according to the following basic assumption.

1. (BA) $\vert F^j(0;a) \vert \geq \delta _j/3$ for all $a \in \omega _k^{rl}$ and $j = m_0,m_0+1,\ldots ,m_k^{rl}$ .

Since our sequence $\delta _j$ is assumed to be admissible, we will frequently use the fact that $\delta _j/3 \geq 1/(3j^{\overline {e}})$ .

From equation (5), we know that every $a \in \omega ^{rl}_k$ is $(\gamma ,C)$ -Collet–Eckmann up to time $m_0$ , and this condition is strong enough to ensure phase-parameter distortion (Lemma 3.2). We will continue to assume this condition at complete returns, but in between two complete returns, we will allow the exponent to drop slightly due to the loss of derivative when returning close to the critical point $x = 0$ . We define the basic exponent conditions as follows.

1. (BE)(1) $\vert \partial _x F^{m_k^{rl}-1}(1;a) \vert \geq C e^{\gamma (m_k^{rl}-1)}$ for all $a \in \omega ^{rl}_k $ .

2. (BE)(2) $\vert \partial _x F^j(1;a) \vert \geq C e^{(\gamma /3) j}$ for all $a \in \omega ^{rl}_k$ and $j = 0,1,\ldots ,m_k^{rl}-1$ .

Assuming (BA) and (BE)(1,2) for $a \in \omega _k^{rl} \subset \Delta _k$ , we will prove it for $a' \in \omega _{k+1}^{r'l'} \subset \Delta _{k+1} \subset \Delta _k$ . Before considering the iteration of $\omega _k^{rl}$ , we define the bound period and the free period, and prove some useful lemmas connected to them. For technical reasons, these lemmas will be proved using the following weaker assumption on the derivative. Given a time $n \geq m^{rl}_k$ , we consider the following condition.

1. (BE)(3) $\vert \partial _x F^j(1;a) \vert \geq C e^{(\gamma /9)j}$ for all $a \in \omega _k^{rl}$ and $j = 0,1,\ldots ,n-1$ .

Notice that $\gamma /9> \gamma _T$ and hence we will be able to apply Lemma 3.2 at all times.

To rid ourselves of cumbersome notation, we drop the indices from this point on and write $\omega = \omega _k^{rl}$ and $m = m_k^{rl}$ .

4.5 The bound and free periods

Assuming we are in the situation of a return for which $\xi _n(\omega ) \subset I_{r+1} \cup I_r \cup I_{r-1} \subset (-4\delta ,4\delta )$ , we are relatively close to the critical point, and therefore the next iterates $\xi _{n+j}(\omega )$ will closely resemble those of $\xi _j(\omega )$ . We quantify this and define the bound period associated to this return as the maximal p such that

1. (BC) $\vert \xi _\nu (a) - F^\nu (\eta ;a) \vert \leq \vert \xi _\nu (a) \vert /(10\nu ^2)$ for $\nu = 1,2,\ldots ,p$

holds for all $a \in \omega $ and all $\eta \in (0,e^{-\vert r-1\vert })$ . We refer to (BC) as the binding condition.

The bound period is of central importance and we establish some results connected to it (compare with [Reference Benedicks and CarlesonBC85]). An important fact is that during this period, the derivatives are comparable in the following sense.

Lemma 4.5. (Bound distortion)

Let n be the index of a return for which $\xi _n(\omega ) \subset I_{r+1} \cup I_r \cup I_{r-1}$ , and let p be the bound period. Then, for all $a \in \omega $ and $\eta \in (0,e^{-\vert r-1\vert })$ ,

$$ \begin{align*} \frac{1}{2} \leq \bigg\vert \frac{\partial_x F^j(1-a\eta^2;a)}{\partial_xF^j(1;a)} \bigg\vert \leq 2 \quad (j = 1,2,\ldots,p). \end{align*} $$

Proof. It is enough to prove that

(6) $$ \begin{align} \bigg \vert \frac{\partial_x F^j(1-a\eta^2;a)}{\partial_x F^j(1;a)} - 1 \bigg\vert \leq \frac{1}{2}. \end{align} $$

The quotient can be expressed as

$$ \begin{align*} \frac{\partial_x F^j(1-a\eta^2;a)}{\partial_x F^j(1;a)} = \prod_{\nu=1}^j \bigg( \frac{F^\nu(\eta;a) - \xi_\nu(a)}{\xi_\nu(a)} + 1\bigg), \end{align*} $$

and applying the elementary inequality

$$ \begin{align*} \bigg\vert \prod_{\nu = 1}^j (u_n + 1) - 1\bigg\vert \leq \exp \bigg(\sum_{\nu=1}^j \vert u_n \vert \bigg) -1, \end{align*} $$

valid for complex $u_n$ , equation (6) now follows since

$$ \begin{align*} \sum_{\nu = 1}^j \frac{\vert F^\nu(\eta;a) - \xi_\nu(a)\vert}{\vert \xi_\nu(a) \vert} \leq \frac{1}{10}\sum_{\nu=1}^j \frac{1}{\nu^2} \leq \log \frac{3}{2}.\\[-3.8pc] \end{align*} $$

The next result gives us an estimate of the length of the bound period. As will be seen, if (BA) and (BE)(3) are assumed up to time $n \geq m = m_k^{rl}$ , the bound period is never longer than n, and we are therefore allowed to use the induction assumptions during this period. In particular, in view of the above distortion result and (BE)(3), we inherit expansion along the critical orbit during the bound period; making sure $m_0$ is large enough, and using (BA) together with the assumption that $(\delta _n)$ is admissible, we have

(7) $$ \begin{align} \vert \partial_x F^{n + j}(1;a) \vert &= 2a \vert \xi_n(a) \vert \vert \partial_x F^{n-1}(1;a) \vert \vert \partial_x F^j(1-a\xi_n(a)^2;a) \vert \nonumber \\ &\geq \frac{2}{3n^{\overline{e}}} C^2e^{(\gamma/9)(n+j-1)} \nonumber \\ &= \frac{2}{3}C^2 e^{-\gamma/9} \exp\bigg\{ \bigg(\frac{\gamma}{9} - \frac{\overline{e}\log n}{n+j}\bigg)(n+j)\bigg\} \nonumber \\ &\geq C_T e^{\gamma_T (n+j)} \quad (j = 0,1,\ldots,p). \end{align} $$

This above estimate is an a priori one, and will allow us to use Lemma 3.2 in the proof of Lemma 4.10.

Lemma 4.6. (Bound length)

Let n be the index of a return such that $\xi _n(\omega ) \subset I_{r+1} \cup I_r \cup I_{r-1}$ , and suppose that (BA) and (BE)(3) are satisfied up to time n. Then there exists a constant $\kappa _1> 0$ such that the corresponding bound period satisfies

(8) $$ \begin{align} \kappa_1^{-1}r \leq p \leq \kappa_1 r. \end{align} $$

Proof. By the mean value theorem and Lemma 4.5, we have that

(9) $$ \begin{align} \vert \xi_j(a) - F^j(\eta;a) \vert &= \vert F^{j - 1}(1;a) - F^{j-1}(1-a\eta^2;a) \vert \nonumber \\ &= a\eta^2 \vert \partial_x F^{j-1}(1-a\eta^{\prime2};a)\vert \\ &\geq \frac{a\eta^2}{2} \vert \partial_x F^{j-1}(1;a) \vert, \nonumber \end{align} $$

as long as $j \leq p$ . (Here, $0 < \eta ' < \eta $ .) Furthermore, as long as we also have $j \leq (\log n)^2$ , say, we can use the induction assumptions: using (BE)(3), we find that

$$ \begin{align*} \frac{1}{2} e^{-2(r+1)} C e^{(\gamma/9)(j-1)} \leq \frac{a\eta^2}{2} \vert \partial_x F^{j-1}(1;a)\vert \leq \frac{\vert \xi_j(a)\vert}{10j^2} \leq 1. \end{align*} $$

Taking the logarithm, using (BA), and making sure that $m_0$ is large enough, we therefore have

$$ \begin{align*} j \leq 1 + \frac{9}{\gamma}(2r + 2 + \log 2 - \log C) \lesssim r \lesssim \log n \leq (\log n)^2, \end{align*} $$

as long as $j \leq p$ and $j \leq (\log n)^2$ . This tells us that $j \leq p$ must break before $j \leq (\log n)^2$ ; in particular, there is a constant $\kappa _1> 0$ such that $p \leq \kappa _1 r$ .

For the lower bound, consider $j = p+1$ and the equality in equation (9). With $a \in \omega $ being the parameter for which the inequality in the binding condition is reversed, using Lemma 4.5, we find that

$$ \begin{align*} \frac{\vert \xi_{p+1}(a)\vert}{10(p+1)^2} \leq \vert \xi_{p+1}(a) - F^{p+1}(\eta;a)\vert \leq 4e^{-2r}\vert \partial_x F^p(1;a) \vert \leq 4e^{-2r} 4^p. \end{align*} $$

Using the upper bound for p, we know that (BA) (or (PR)) is valid at time $p+1$ , and hence

$$ \begin{align*} \frac{\vert \xi_{p+1}(a) \vert}{10(p+1)^2} \geq \frac{1}{30(p+1)^{2+\hat{e}}}, \end{align*} $$

where $\hat {e} = \max (\overline {e},\sigma )$ . Therefore,

$$ \begin{align*} \frac{1}{30(p+1)^{2 + \hat{e}}} \leq 4 e^{-2r}4^p, \end{align*} $$

and taking the logarithm proves the lower bound.

The next result will concern the growth of $\xi _n(\omega )$ during the bound period.

Lemma 4.8. (Bound growth)

Let n be the index of a return such that $\xi _n(\omega ) \subset I$ with $I_{rl} \subset I \subset I_{r+1} \cup I_r \cup I_{r-1}$ , and suppose that (BA) and (BE)(3) are satisfied up to time n. Then there exists a constant $\kappa _2> 0$ such that

$$ \begin{align*} \vert \xi_{n+p+1}(\omega) \vert \geq \frac{1}{r^{\kappa_2}}\frac{\vert\xi_n(\omega) \vert}{\vert I \vert}. \end{align*} $$

Proof. Denote $\Omega = \xi _{n+p+1}(\omega ) $ and notice that for any two given parameters $a,b \in \omega $ , we have

(10) $$ \begin{align} \vert \Omega \vert &\geq \vert F^{n + p + 1}(0;a) - F^{n + p + 1}(0;b) \vert \nonumber \\ &= \vert F^{p + 1}(\xi_{n}(a);a) - F^{p + 1}(\xi_{n}(b);b) \vert \nonumber \\ &\geq \vert F^{p+1}(\xi_{n}(a);a) - F^{p+1}(\xi_{n}(b);a) \vert \nonumber \\ &\quad - \vert F^{p+1}(\xi_{n}(b);a) - F^{p+1}(\xi_{n}(b);b)\vert. \end{align} $$

Due to the exponential increase of the phase derivative along the critical orbit, the dependence on the parameters is inessential in the following sense:

(11) $$ \begin{align} \vert F^{p+1}(\xi_{n}(b);a) - F^{p+1}(\xi_{n}(b);b)\vert \leq e^{-(\gamma /18)n} \vert \xi_n(\omega)\vert. \end{align} $$

To realize this, first notice that we have the following (somewhat crude) estimate for the parameter derivative:

$$ \begin{align*} \vert \partial_a F^j(x;a) \vert \leq 5^j \quad (j=1,2,\ldots). \end{align*} $$

Indeed, $\vert \partial _a F(x;a) \vert \leq 1 < 5$ , and by induction,

$$ \begin{align*} \vert \partial_a F^{j+1}(x;a)\vert &= \vert \partial_a (1-aF^j(x;a)^2) \vert \\ &= \vert -F^j(x;a)^2 - 2aF^j(x;a)\partial_a F^j(x;a) \vert \\ &\leq 1 + 4\cdot5^j \\ &\leq 5^{j+1}. \end{align*} $$

Using the mean value theorem twice, Lemma 3.2, and (BE)(3), we find that

$$ \begin{align*} \vert F^{p+1}(\xi_{n}(b);a) - F^{p+1}(\xi_{n}(b);b)\vert \leq [(1-A)T]^{-1} 5^{p+1}C^{-1}e^{-(\gamma/9) (n-1)}\vert \xi_n(\omega)\vert. \end{align*} $$

In view of equation (8) and (BA), making $m_0$ larger if needed, the inequality in equation (11) can be achieved.

Assume now that at time $p+1$ , (BC) is broken for parameter a, and let b be an endpoint of $\omega $ such that

$$ \begin{align*} \vert \xi_n(a) - \xi_n(b) \vert \geq \frac{\vert \xi_n(\omega) \vert}{2}. \end{align*} $$

Continuing the estimate of $\vert \Omega \vert $ , using equation (11), we find that

(12) $$ \begin{align} \vert \Omega \vert &\geq \vert F^{p}(1-a\xi_{n}(a)^2;a) - F^{p}(1-a\xi_{n}(b)^2;a)\vert \nonumber \\ &\quad - \vert F^{p+1}(\xi_{n}(b);a) - F^{p+1}(\xi_{n}(b);b)\vert \nonumber \\ &\geq (a \vert \xi_{n}(a) + \xi_{n}(b)\vert \vert \partial_x F^{p}(1-a\xi_{n}(a')^2;a)\vert - 2e^{-(\gamma /18)n}) \frac{\vert \xi_n(\omega) \vert}{2} \nonumber \\ &\geq (2ae^{-r}\vert \partial_xF^{p}(1-a\xi_{n}(a')^2;a) \vert - 2e^{-(\gamma /18)n} ) \frac{\vert \xi_n(\omega)\vert}{2}. \end{align} $$

Using Lemma 4.5 twice and the equality in equation (9) (with $p+1$ instead of p) together with (BC) (now reversed inequality), we continue the estimate in equation (12) to find that

(13) $$ \begin{align} \vert \Omega \vert &\geq \bigg( 2ae^{-r}\frac{1}{4 a\eta^2}\frac{\vert \xi_{p+1} (a) \vert}{10(p+1)^2} - 2e^{-(\gamma/18)n} \bigg) \frac{\vert \xi_n(\omega)\vert}{2} \nonumber \\ &\geq \bigg( e^r \frac{\vert \xi_{p+1}(a) \vert}{20 (p+1)^2} - 2e^{-(\gamma/18)n} \bigg) \frac{\vert \xi_n(\omega)\vert}{2}. \end{align} $$

In either case of $p \leq m_0$ or $p> m_0$ , we have that (using (BA), (PR), and the assumption that our recurrence rate is admissible)

$$ \begin{align*} \frac{\vert \xi_{p+1}(a) \vert}{(p+1)^2} \geq \frac{K}{3(p+1)^{2+\hat{e}}}, \end{align*} $$

where $\hat {e} = \max (\overline {e},\sigma )$ . We can make sure that the second term in the parenthesis in equation (13) is always less than a fraction, say $1/2$ , of the first term and therefore, using (BC), equation (8), and that $e^r \geq 1/(2r^2 \vert I \vert )$ , we finish the estimate as follows:

(14) $$ \begin{align} \vert \Omega \vert &\geq \frac{K}{240}\frac{1}{(p+1)^{2+\hat{e}}} \vert \xi_n(\omega)\vert e^r \nonumber \\ &\geq \frac{K}{480}\frac{1}{r^2 (p+1)^{2 + \hat{e}}} \frac{\vert \xi_n(\omega) \vert}{\vert I \vert}\nonumber \\ &\geq \frac{K}{480 (2\kappa_1)^{2+\hat{e}}}\frac{1}{r^{4+\hat{e}}} \frac{\vert \xi_n(\omega) \vert}{\vert I \vert} \nonumber \\ &\geq \frac{1}{r^{\kappa_2}} \frac{\vert \xi_n(\omega) \vert}{\vert I \vert}, \end{align} $$

where we can choose $\kappa _2 = 5 + \hat {e}$ as long as $\delta $ is sufficiently small.

Remark 4.9. Using the lower bound for p, the upper bound

$$ \begin{align*} \vert \xi_{n+p+1}(\omega) \vert \leq \frac{1}{r} \frac{\vert \xi_n(\omega)\vert}{\vert I \vert} \end{align*} $$

can be proved similarly.

This finishes the analysis of the bound period, and we continue with describing the free period. A free period will always follow a bound period, and during this period, we will be iterating outside $(-\delta ,\delta )$ . We let L denote the length of this period, that is, L is the smallest integer for which

$$ \begin{align*} \xi_{n+p+L}(\omega) \cap (-\delta,\delta) \neq \emptyset. \end{align*} $$

The following lemma gives an upper bound for the length of the free period, following the bound period of a complete return or an essential return.

Lemma 4.10. (Free length)

Let $\xi _n(\omega ) \subset I_{r+1} \cup I_r \cup I_{r-1}$ with n being the index of a complete return or an essential return, and suppose that (BA) and (BE)(3) are satisfied up to time n. Let p be the associated bound period and let L be the free period. Then there exists a constant $\kappa _3> 0$ such that

$$ \begin{align*} L \leq \kappa_3 r. \end{align*} $$

Proof. Assuming $j \leq L$ and $j \leq (\log n)^2$ , similar calculations as in the proof of Lemma 4.8 gives us parameter independence (see equation (11) and notice that from equation (7), we are allowed to apply Lemma 3.2); using Lemmas 4.8 and 3.1, we find that

$$ \begin{align*} 2 \geq \vert \xi_{n+p+j}(\omega) \vert \geq \frac{\delta C_M}{2}e^{\gamma_M(j-1)}\frac{1}{r^{\kappa_2}}. \end{align*} $$

Taking the logarithm, using (BA), and making sure that $m_0$ is large enough, we therefore have

$$ \begin{align*} j \leq 1 + \frac{1}{\gamma_M}(\kappa_2 \log r + \Delta + \log 4 - \log C_M) \lesssim r \lesssim \log n < \frac{1}{2}(\log n)^2, \end{align*} $$

as long as $j \leq L$ and $j \leq (\log n)^2$ . This tells us that $j \leq L$ must break before $j \leq (\log n)^2$ ; in particular, there is a constant $\kappa _3> 0$ such that $L \leq \kappa _3 r$ .

Remark 4.11. If the return $\xi _{n+p+L}(\omega )$ is inessential or essential, then there is no $\delta $ -dependence in the growth factor; more generally, if the prerequisites of Lemma 4.8 are satisfied, then

$$ \begin{align*} \vert \xi_{n+p+L}(\omega) \vert \geq \frac{C_M}{2}e^{\gamma_M(L-1)} \frac{1}{r^{\kappa_2}}\frac{\vert \xi_{n}(\omega)\vert}{\vert I \vert}. \end{align*} $$

Before considering iterations of $\omega = \omega _k^{rl} \subset \Delta _k$ from $m = m_k^{rl}$ to $m_{k+1}^{rl}$ , we make the following observation that as long as (BA) is assumed in a time window $[n,2n]$ , the derivative will not drop too much.

Lemma 4.12. Suppose that a is a parameter such that

(15) $$ \begin{align} \vert \partial_x F^j(1;a) \vert \geq C e^{\gamma' j} \quad (j = 0,1,\ldots,n-1), \end{align} $$

with $\gamma ' \geq \gamma /3$ . Then, if (BA) is satisfied up to time $2n$ , we have

$$ \begin{align*} \vert \partial_x F^{n+j}(1;a) \vert \geq Ce^{(\gamma'/3)(n+j)} \quad (j = 0,1, \ldots,n-1). \end{align*} $$

In other words, if (BA) and (BE)(1) [(BE)(2)] are satisfied up to time n, then (BE)(2) [(BE)(3)] is satisfied up to time $2n$ , as long as (BA) is.

Proof. The proof is based on the fact that we trivially have no loss of derivative during the bound and free periods. Indeed, suppose $\xi _{n'}(a) \sim e^{-r}$ , with $n' \geq n$ , and let p be the bound period (here we use pointwise binding, see Remark 4.4) and L the free period. Moreover, we assume that $n' + p + L < 2n$ ; in particular, this implies $p < n$ and we can use equation (15) during this period. Introducing $D_p = \vert \partial _x F^p(1;a) \vert $ and using similar calculations as in Lemma 4.8 (e.g. the equality in equation (9) and reversed inequality in (BC)), we find that

$$ \begin{align*} e^{-2r} D_p \gtrsim a \eta^2 \vert \partial_x F^p(1-a\eta^2;a) \vert \geq \frac{\vert \xi_{p+1}(a) \vert}{10(p+1)^2} \gtrsim \frac{1}{(p+1)^{2 + \hat{e}}}, \end{align*} $$

where we used (BA) (or (PR)). Since $p < n$ , we are free to use equation (15) and therefore, the above inequalities yield

$$ \begin{align*} e^{-r} D_p \gtrsim D_p^{1/2} \frac{1}{\sqrt{(p+1)^{2 + \hat{e}}}} \gtrsim \frac{e^{(\gamma'/2)p}}{\sqrt{(p+1)^{2 + \hat{e}}}} \geq C_M^{-1}, \end{align*} $$

provided $\delta $ is small enough. Here, in the last inequality, we used the lower bound in equation (8) (see Remark 4.7). Assuming $\xi _{n'+p+L}(a)$ is a return (and that $n'+p+ L < 2n$ ), we therefore have

$$ \begin{align*} \vert \partial_x F^{p+L}(\xi_{n'}(a);a) \vert &\geq 2a\vert \xi_{n'}(a) \vert \vert \partial_x F^p(1-a\xi_{n'}(a)^2;a) \vert \vert \partial_x F^{L-1}(\xi_{n'+p+1}(a);a)\vert \\ &\gtrsim e^{-r} D_p C_M e^{\gamma_M(L-1)} \\ &\geq 1. \end{align*} $$

We conclude that the combination of a return, a bound period, and a free period does not decrease the derivative.

Let us now follow a parameter a satisfying equation (15) and (BA) up to time $2n$ . If the iterates $\xi _{n+j}(a)$ are always outside $(-\delta ,\delta )$ , then

$$ \begin{align*} \vert \partial_x F^{n+j}(1;a) \vert &= \vert \partial_x F^{n-1}(1;a) \vert \vert \partial_x F^{j+1}(\xi_n(a);a) \vert \\ &\geq Ce^{\gamma'(n-1)} \delta C_M e^{\gamma_M(j+1)} \\ &\geq Ce^{(\gamma'/3)(n+j)} \delta C_M e^{(2\gamma'/3)(n+j)} \\ &\geq Ce^{(\gamma'/3)(n+j)} \quad (j = 0,1,\ldots,n-1), \end{align*} $$

provided $m_0$ is big enough.

Otherwise, the worst case is if we have a short free period followed by a return, a bound period, a free period, and so on, and which ends with a return together with a short bound period. In this case, using the above argument, the estimate is as follows:

$$ \begin{align*} \vert \partial_x F^{n+j}(1;a) \vert &\geq \vert \partial_x F^{n-1}(1;a) \vert \cdot C_M \cdot 1 \cdot 1 \cdots 1 \cdot 2a\vert \xi_{n+j}(a) \vert \cdot C \\ &\geq Ce^{\gamma'(n-1)} C_M C 2a \frac{\delta_{n+j}}{3} \\ &\geq Ce^{(\gamma'/3)(n+j)} C_M C \frac{2}{3a} e^{(\gamma'/3)n - \overline{e}\log(2n)} \\ &\geq Ce^{(\gamma'/3)(n+j)} \quad (j = 0,1,\ldots,n-1), \end{align*} $$

provided $m_0$ is big enough. This proves the lemma.

4.6 From the $k\text {th}$ complete return to the first inessential return

If $\omega \subset T_k$ , then we have already reached an escape situation and proceed accordingly as is described below in the section about escape. We therefore assume $\omega \subset N_k$ and $\xi _m(\omega ) = I_{r_0l} \subset (-4\delta ,4\delta )$ .

If it happens that for some $j \leq p$ ,

$$ \begin{align*} \xi_{m+j}(\omega) \cap (-\delta_{m+j}/3,\delta_{m+j}/3) \neq \emptyset, \end{align*} $$

then we stop and consider this return complete. If not, we notice that $\xi _{m+p}(\omega )$ can not be a return, unless it is escape or complete; indeed, we would otherwise have $\vert \xi _{m+p+1}(\omega ) \vert < \vert \xi _{m+p}(\omega )\vert $ , due to the fact that we return close to the critical point, and thus contradict the definition of the bound period. We therefore assume that $\xi _{m+p}(\omega )$ does not intersect $(-\delta ,\delta )$ .

Up until the next return, we will therefore experience an orbit outside of $(-\delta ,\delta )$ , that is, we will be in a free period. After the free period, our return is either inessential, essential, escape, or complete. In the next section, we consider the situation of an inessential return.

4.7 From the first inessential return to the first essential return

Let $i_1 = m + p_0 + L_0$ denote the index of the first inessential return to $(-\delta ,\delta )$ . We will keep iterating $\xi _{i_1}(\omega )$ until we once again return. If this next return is again inessential, we denote its index by $i_2 = i_1 + p_1 + L_1$ , where $p_1$ and $L_1$ are the associated bound period and free period, respectively. Continuing like this, let $i_j$ be the index of the $j\text {th}$ inessential return.

The following lemma gives an upper bound for the total time spent doing inessential returns (compare with Lemma 2.3 in [Reference Benedicks and CarlesonBC91]).

Lemma 4.13. (Inessential length)

Let $\xi _n(\omega ) \subset I_{r+1} \cup I_r \cup I_{r-1}$ with n being the index of a complete return or an essential return, and suppose that (BA) and (BE)(2) are satisfied up to time n. Then there exists a constant $\kappa _4> 0$ such that the total time o spent doing inessential returns satisfies

$$ \begin{align*} o \leq \kappa_4 r. \end{align*} $$

Proof. Let $i_1 = n + p + L$ be the index of the first inessential return, that is, $\xi _{i_1}(\omega ) \subset I_{r_1}$ , with $I_{r_1}$ being the host interval. From Lemmas 4.6 and 4.10, together with (BA), we have that

$$ \begin{align*} i_1 = n + p + L \leq n + (\kappa_1 + \kappa_3)r \leq 2n, \end{align*} $$

provided $m_0$ is large enough. We can therefore apply Lemma 4.12 and conclude that (BE)(3) is satisfied at time $i_1$ . To this first inessential return, we associate a bound period of length $p_1$ (satisfying $p_1 \leq \kappa _1 r_1$ due to the fact that (BE)(3) is satisfied at time $i_1$ ) and a free period of length $L_1$ . We let $i_2 = i_1 + p_1 + L_1$ denote the index of the second inessential return. Continuing like this, we denote by $i_j = i_{j-1} + p_{j-1} + L_{j-1}$ the index of the $j\text {th}$ inessential return. With $o_j$ denoting the total time spent doing inessential returns up to time $i_j$ , we have that $o_j = i_{j} - i_1 = \sum _{k = 1}^{j-1} (p_k + L_k)$ . Suppose that the return with index $i_s$ is the first that is not inessential. We estimate $o = o_s$ as follows. Suppose that $o_j$ is as above and that $p_k \leq \kappa _1 r_k$ for $k = 1,2,\ldots ,j-1$ . Using Remark 4.11, we find that

(16) $$ \begin{align} \frac{\vert \xi_{i_{k+1}}(\omega)\vert}{\vert \xi_{i_k}(\omega)\vert} \geq \frac{C_M e^{\gamma_M (L_k-1)}}{2 r_k^{\kappa_2}\vert I_{r_k}\vert} \geq \frac{C_M}{2} \frac{e^{\gamma_M (L_k-1) + r_k}}{r_k^{\kappa_2}}, \end{align} $$

and therefore,

(17) $$ \begin{align} 2 \geq \vert \xi_{i_j}(\omega) \vert = \vert \xi_{i_1}(\omega)\vert \prod_{k=1}^{j-1} \frac{\vert \xi_{i_{k+1}}(\omega)\vert}{\vert \xi_{i_k}(\omega) \vert} \geq \frac{\delta C_Me^{\gamma_M}}{2 r^{\kappa_2}} \prod_{k=1}^{j-1} \frac{C_M}{2} \frac{e^{\gamma_M (L_k-1) + r_k}}{r_k^{\kappa_2}}. \end{align} $$

Here, the $\delta $ is added to make sure that the estimate also holds for the last free orbit, when the return can be escape or complete. This gives us a rather poor estimate, but since $p \lesssim r$ , it is good enough.

Taking the logarithm of equation (17), we find that

$$ \begin{align*} \sum_{k=1}^{j-1} (\log C_M - \log 2 + \gamma_M (L_k-1) + r_k - \kappa_2 \log r_k ) \leq \kappa_2 \log r + \Delta + \operatorname{const.} \end{align*} $$

Provided $\delta $ is small enough, we have $r_k \geq 4\kappa _2 \log r_k$ and $r_k \geq -\log \delta> -2(\log C_M + \gamma _M + \log 2)$ . Therefore, using $p_k \leq \kappa _1 r_k$ , we find that

$$ \begin{align*} o_j = i_j - i_1 = \sum_{k=1}^{j-1}( p_k + L_k ) \leq \kappa_4 r, \end{align*} $$

with $\kappa _4$ being an absolute constant. In particular,

$$ \begin{align*} i_j = i_1 + o_j \leq 2n, \end{align*} $$

and therefore (BE)(3) is still valid at time $i_j$ . Consequently, the associated bound period satisfies $p_j \leq \kappa _1 r_j$ , and the above argument can therefore be repeated. With this, we conclude that $o_s \leq \kappa _4 r$ .

We proceed in the next section with describing the situation if our return is assumed to be essential.

4.8 From the first essential return to the first escape return

With $n_1$ denoting the index of the first essential return, we are in the following situation:

$$ \begin{align*} &\xi_{n_1}(\omega) \cap I_{rl} \neq \emptyset, \quad \vert \xi_{n_1}(\omega)\vert \geq \vert I_{rl}\vert,\\ &\text{and}\quad \xi_{n_1}(\omega) \subset (-4\delta,4\delta) \smallsetminus (-\delta_{n_1}/3,\delta_{n_1}/3) \end{align*} $$

for some $r,l$ . At this point, to not lose too much distortion, we will make a partition of as much as possible, and keep iterating what is left. That is, we will consider iterations of larger partition elements $I_r = (e^{-r-1},e^{-r}) \subset (-4\delta ,4\delta )$ , and we establish an upper bound for the number of essential returns needed to reach an escape return or a complete return.

Let $\Omega _1 = \xi _{n_1}(\omega )$ and let $I_1 = I_{r_1} \subset \Omega _1$ for smallest such $r_1$ . (In fact, we extend $I_1$ to the closest endpoint of $\Omega _1$ , and therefore have $I_1 \subset I_{r_1} \cup I_{r_1-1}$ .) If there is no such r, we instead let $I_1 = \Omega _1$ . Moreover, let $\omega ^1$ be the interval in parameter space for which $\xi _{n_1}(\omega ^1) = I_{1}$ . The interval $I_1$ is referred to as the essential interval, and this is the interval we will iterate. If $\hat {\omega } = \omega \smallsetminus \omega ^1$ is non-empty, we make a partition

$$ \begin{align*} \hat{\omega} = \bigcup_{r,l} \omega^{rl} \subset \Delta_{k+1}, \end{align*} $$

where each $\omega ^{rl}$ is such that $I_{rl} \subset \xi _{n_1}(\omega ^{rl}) = I_{rl}\cup I_{r'l'} \subset (-4\delta ,4\delta )$ . (If there is not enough left for a partition, we extend $I_1$ further so that $I_1 \subset I_{r_1+1}\cup I_{r_1} \cup I_{r_1-1}$ .) Notice that, since the intervals $I_r$ are dyadic, the proportion of what remains after partitioning satisfies

(18) $$ \begin{align} \frac{\vert I_1 \vert}{\vert \Omega_1 \vert} \geq 1 - \frac{1}{e} \geq \frac{1}{2}. \end{align} $$

We associate for each partitioned parameter interval $\omega ^{rl}$ the complete return time $n_1$ (even though nothing is removed from these intervals). From the conclusions made in the previous sections, we know that

$$ \begin{align*} n_1 = m + p_0 + L_0 + o_0 \leq m + (\kappa_1 + \kappa_3 + \kappa_4)r_0 \leq 2m, \end{align*} $$

provided $m_0$ is large enough. In particular, Lemma 4.12 tells us that (BE)(2) is satisfied up to time $n_1$ for all $a \in \omega $ . At this step, to make sure that (BE)(1) is satisfied for our partitioned parameter intervals $\omega ^{rl} \subset \Delta _{k+1}$ , we make the following rule (compare with the initial iterates at the beginning of the induction step). If there is no $a' \in \omega $ such that

$$ \begin{align*} \vert \partial_x F^{n_1-1}(1;a')\vert \geq C_B e^{\gamma_B(n_1-1)}, \end{align*} $$

then we remove the entire interval. If there is such a parameter, however, using Lemma 5.1, we have that

$$ \begin{align*} \vert \partial_x F^{n_1 - 1}(1;a) \vert &\geq D_1^{-(\log^* m)^2} \vert \partial_x F^{n_1-1}(1;a')\vert \\ &\geq C_B \exp\bigg\{ \bigg(\gamma_B - \frac{(\log^* m)^2}{n_1-1}\log D_1 \bigg)(n_1-1) \bigg\} \\ &\geq Ce^{\gamma(n_1-1)}, \end{align*} $$

provided $m_0$ is large enough.

With the above rules applied at each essential return to come, we now describe the iterations. Since $\xi _{m}(\omega ) = I_{r_0l}$ , using Lemma 4.8, we know that the length of $\Omega _1$ satisfies

$$ \begin{align*} \vert \Omega_1 \vert \geq \frac{C_M e^{\gamma_M}}{2}\frac{1}{r_0^{\kappa_2}} \geq \frac{1}{r_0^{\kappa_2 + 1}}. \end{align*} $$

Notice that since $e^{-r_1+1} \geq \vert \Omega _1 \vert $ , we have that $r_1 \leq 2\kappa _2 \log r_0$ . Iterating $I_{1}$ with the same rules as before, we will eventually reach a second non-inessential return, and if this return is essential, we denote its index by $n_2$ . This index constitutes the addition of a bound period, a free period, and an inessential period: $n_2 = n_1 + p_1 + L_1 + o_1$ . Similarly as before, we let $\Omega _2 = \xi _{n_2}(\omega ^1)$ , and let $I_{2} \subset \Omega _2$ denote the essential interval of $\Omega _2$ . Let $\omega ^2 \subset \omega ^1$ be such that $\xi _{n_2}(\omega ^2) = I_{2}$ , and make a complete partition of $\omega ^1 \smallsetminus \omega ^2$ . By applying Lemma 4.8 again, we find that

$$ \begin{align*} \vert \Omega_2 \vert \geq \frac{1}{r_1^{\kappa_2 + 1}} \geq \frac{1}{(2\kappa_2 \log r_0)^{\kappa_2 + 1}}. \end{align*} $$

If we have yet to reach an escape return or a complete return, let $n_j$ be the index of the $j\text {th}$ essential return, and realize that we are in the following situation:

(19) $$ \begin{align} \xi_{n_j}(\omega^j) = I_j \subset \Omega_j = \xi_{n_j}(\omega^{j-1})\quad \text{and} \quad \vert \Omega_j \vert \geq \frac{1}{r_{j-1}^{\kappa_2 + 1}}. \end{align} $$

Introducing the function $r \mapsto 2\kappa _2 \log r = \varphi (r)$ , we see from the above that $r_j \leq \varphi ^j(r_0)$ . The orbit $\varphi ^j(r_0)$ will tend to the attracting fixed point $\hat {r} = -2\kappa _2 W(-1/(2\kappa _2))$ , where W is the Lambert W function. The following simple lemma gives an upper bound for the number of essential returns needed to reach an escape return or a complete return.

Lemma 4.14. Let $\varphi (r) = 2\kappa _2 \log r$ , and let $s = s(r)$ be the integer defined by

$$ \begin{align*} \log_{s} r \leq 2\kappa_2 \leq \log_{s-1} r. \end{align*} $$

Then,

$$ \begin{align*} \varphi^s(r) \leq 12\kappa_2^2. \end{align*} $$

Proof. Using the fact that $3 \leq 2\kappa _2 \leq \log _j r$ , for $j = 0,1,\ldots , s-1$ , it is straightforward to check that

(20) $$ \begin{align} \varphi^j(r) \leq 6 \kappa_2 \log_j r. \end{align} $$

Therefore,

$$ \begin{align*} \varphi^s(r) \leq 2\kappa_2 \log( 6 \kappa_2 \log_{s-1}r ) = 2\kappa_2 ( \log 3 + \log 2\kappa_2 + \log_s r) \leq 12 \kappa_2^2.\\[-3pc] \end{align*} $$

Given $s = s(r_0)$ as in the above lemma, we have that $r_s \leq \varphi ^s(r_0) \leq 12\kappa _2^2$ . By making sure $\delta $ is small enough, we therefore conclude that

$$ \begin{align*} \vert \Omega_{s+1} \vert \geq \frac{1}{(12\kappa_2^2)^{\kappa_2 + 1}} \geq 4\delta. \end{align*} $$

To express s in terms of $r_0$ , we introduce the so-called iterated logarithm, which is defined recursively as

$$ \begin{align*} \log^* x = \begin{cases} 0 &\mbox{if}\ x \leq 1, \\ 1 + \log^* \log x &\mbox{if}\ x> 1. \end{cases} \end{align*} $$

That is, $\log ^* x$ is the number of times one has to apply to logarithm to x for the result to be less than or equal to one.

Since s satisfies $\log _s r_0 \leq 2\kappa _2 \leq \log _{s-1} r_0$ and since $2\kappa _2> 1$ , we have

(21) $$ \begin{align} s \leq \log^* r_0 \leq \log^* m. \end{align} $$

We finish by giving an upper bound for the index of the first escape return (or $(k+1){\text {th}}$ complete return), that is, we wish to estimate

$$ \begin{align*} n_{s+1} = m + \sum_{j=0}^{s} (p_j + L_j + o_j).\end{align*} $$

From Lemmas 4.6, 4.10, and 4.13, we have that

$$ \begin{align*} p_j \leq \kappa_1 r_j,\quad L_j \leq \kappa_3 r_j,\quad \text{and}\quad o_j \leq \kappa_4 r_j. \end{align*} $$

Together with the inequalities $r_j \leq \varphi ^j(r_0)$ and equation (20), we find that

$$ \begin{align*} \sum_{j=0}^{s}( p_j + L_j + o_j) &\lesssim r_0 + \sum_{j=1}^{s} \varphi^j(r_0) \\ &\lesssim r_0 + \sum_{j=1}^{s} \log_j r_0 \\ &\lesssim r_0. \end{align*} $$

Using (BA), we conclude that $n_{s+1} - m \lesssim \log m$ , provided $m_0$ is large enough.

4.9 From the first escape return to the $(k+1)\text {th}$ complete return

Keeping the notation from the previous section, $\Omega _{s+1} = \xi _{n_{s+1}}(\omega ^s)$ is the first escape return, satisfying

$$ \begin{align*} \Omega_{s+1} \cap (-\delta,&\delta) \neq \emptyset, \quad \Omega_{s+1} \cap (-\delta_{n_s}/3,\delta_{n_s}/3) = \emptyset \\ &\kern-2pt\text{and} \quad \vert \Omega_{s+1} \smallsetminus (-\delta,\delta) \vert \geq 3\delta. \end{align*} $$

We will keep iterating $\omega ^s$ until we get a complete return, and we show below that this must happen within finite (uniform) time. To not run into problems with distortion, we will, as in the case of essential returns, whenever possible make a partition of everything that is mapped inside of $(-\delta ,\delta )$ , and the corresponding parameter intervals will be a part of $\Delta _{k+1}$ ; that is, at time $n_{s+1+j} = n_{s+1}+j$ let $I_{s+1+j} = \Omega _{s+1+j} \smallsetminus (\Omega _{s+1+j} \cap (-\delta ,\delta ))$ , let $\omega ^{s+1+j}$ be such that $\xi _{n_{s+1+j}}(\omega ^{s+1+j}) = I_{s+1+j}$ , and make a partition of $\omega ^{s+j}\smallsetminus \omega ^{s+1+j}$ . As in the case of essential returns, we associate to each partitioned parameter interval the complete time $n_{s+1+j}$ and, as before, we make sure that at these times, (BE)(1) is satisfied.

Let $\omega _e = \omega _L \cup \omega _M \cup \omega _R$ be the disjoint union of parameter intervals for which

$$ \begin{align*} \xi_{n_{s+1}}(\omega_L) = (\delta,2\delta), \quad \xi_{n_{s+1}}(\omega_M) = (2\delta,3\delta)\quad \text{and} \quad \xi_{n_{s+1}}(\omega_R) = (3\delta,4\delta). \end{align*} $$

Clearly, it is enough to show that $\omega _e$ reaches a complete return within finite time. Let $t_*$ be the smallest integer for which

$$ \begin{align*} C_M e^{\gamma_M t_*} \geq 4. \end{align*} $$

If $\delta $ is small enough, and if $\vert \omega _0 \vert = 2\epsilon $ is small enough, we can make sure that

$$ \begin{align*} \xi_{n_{s+1} + j}(\omega_e) \cap (-2\delta,2\delta) = \emptyset \quad (1 \leq j \leq t_*). \end{align*} $$

Suppose that, for some $j \geq t_*$ , $\xi _{n_{s+1+j}}(\omega _e) \cap (-\delta ,\delta ) \neq \emptyset $ , and that this return is not complete. Assuming that $\omega _L$ returns, we can not have $\xi _{n_{s+1+j}}(\omega _L) \subset (-2\delta ,2\delta )$ . Indeed, if this was the case, then (using Lemma 3.1 and parameter independence)

$$ \begin{align*} \vert \xi_{n_{s+1+j}}(\omega_L)\vert> 2 \vert \xi_{n_{s+1}}(\omega_L)\vert > 2\delta, \end{align*} $$

which contradicts the return not being complete. We conclude that after partitioning what is mapped inside of $(-\delta ,\delta )$ , what is left is of size at least $\delta $ , and we are back to the original setting. In particular, $\omega _M$ did not return to $(-\delta ,\delta )$ . Repeating this argument, $\omega _L$ and $\omega _R$ will return, but $\omega _M$ will stay outside of $(-\delta ,\delta )$ . (Here, we abuse the notation: if $\omega _L$ returns, we update it so that it maps onto $(\delta ,2\delta )$ , and similarly if $\omega _R$ returns.) Due to Lemma 3.1, we therefore have

$$ \begin{align*} 2 \geq \vert \xi_{n_{s+1+j}}(\omega_M) \vert \gtrsim \vert \xi_{n_{s+1}}(\omega_M) \vert \delta C_M e^{\gamma_M j} \geq \delta^2 C_M e^{\gamma_M j} \quad (j \geq 0), \end{align*} $$

and clearly we must reach a complete return after $j = t$ iterations, with

$$ \begin{align*} t \lesssim \frac{2\Delta - \log C_M}{\gamma_M}. \end{align*} $$

With this, we conclude that if $m_0$ is large enough, then there exists a constant $\kappa> 0$ such that

(22) $$ \begin{align} m_{k+1}^{rl} \leq m_k^{rl} + \kappa \log m_k^{rl}. \end{align} $$

We finish by estimating how much of $\Omega _{s+1+j}$ is being partitioned at each iteration. By definition of an escape return, we have that $\vert \Omega _{s+1}\vert \geq 3\delta $ , and since it takes a long time for $\omega _e$ to return, the following estimate is valid:

(23) $$ \begin{align} \frac{\vert I_{s+1+j} \vert}{\vert \Omega_{s+1+j} \vert} \geq \frac{\vert \Omega_{s+1+j} \vert - \delta}{\vert \Omega_{s+1+j} \vert } \geq 1 -\frac{1}{3} = \frac{2}{3}. \end{align} $$

4.10 Parameter exclusion

We are finally in the position to estimate how much of $\omega $ is being removed at the next complete return. Up until the first free return, nothing is removed (unless we have a bound return, for which we either remove nothing, or remove enough to consider the return complete). Let E be what is removed in parameter space, and write $\omega = \omega ^0$ . Taking into account what we partition in between $m_k$ and $m_{k+1}$ , we have that

$$ \begin{align*} \frac{\vert E \vert}{\vert \omega^0 \vert} = \frac{\vert E \vert}{\vert \omega^{s+t} \vert} \prod_{\nu = 0}^{t-1} \frac{\vert \omega^{s+1+\nu}\vert}{\vert \omega^{s+\nu}\vert} \prod_{\nu=0}^{s-1} \frac{\vert \omega^{1 + \nu}\vert}{\vert \omega^\nu \vert}. \end{align*} $$

Using the the mean value theorem, we find that for each factor in the above expression,

$$ \begin{align*} \frac{\vert \omega^j\vert}{\vert \omega^{j-1}\vert} &= \frac{\vert a_j - b_j \vert}{\vert a_{j-1} - b_{j-1} \vert} \\ &= \frac{\vert a_j - b_j \vert}{\vert \xi_{n_j}(a_j) - \xi_{n_j}(b_j)\vert} \frac{\vert \xi_{n_j}(a_{j-1}) - \xi_{n_j}(b_{j-1})\vert}{\vert a_{j-1} - b_{j-1}\vert} \frac{\vert \xi_{n_j}(a_j) - \xi_{n_j}(b_j)\vert}{\vert \xi_{n_j}(a_{j-1}) - \xi_{n_j}(b_{j-1})\vert} \\ &= \frac{1}{\vert \partial_a \xi_{n_j}(c_j)\vert} \vert \partial_a \xi_{n_j}(c_{j-1})\vert \frac{\vert I_{j}\vert}{\vert \Omega_j \vert} \\ &= \frac{\vert \partial_x F^{n_j-1}(1;c_j)\vert}{\vert \partial_a \xi_{n_j}(c_j)\vert} \frac{\vert \partial_a \xi_{n_j}(c_{j-1})\vert}{\vert \partial_x F^{n_j-1}(1;c_{j-1})\vert} \frac{\vert \partial_x F^{n_j-1}(1;c_{j-1})\vert}{\vert \partial_x F^{n_j-1}(1;c_j)\vert} \frac{\vert I_{j}\vert}{\vert \Omega_j \vert}. \end{align*} $$

Making use of Lemmas 3.2 and 5.1, we find that

$$ \begin{align*} \frac{\vert \omega^j \vert}{\vert \omega^{j-1}\vert} : \frac{\vert I_{j} \vert}{\vert \Omega_j \vert} \sim D_A D_1^{(\log^* m_k)^2}, \end{align*} $$

and therefore, using equations (21), (18), and (23), there is, provided $m_0$ is large enough, an absolute constant $0 < \tau < 1$ such that

$$ \begin{align*} \frac{\vert E \vert}{\vert \omega^0 \vert} \geq \frac{(\delta_{m_{k+1}}/3)}{1}\bigg(\frac{1}{3}D_A^{-1} D_1^{-(\log^* m_k)^2}\bigg)^{t + \log^* m_k} \geq \delta_{m_{k+1}}\tau^{(\log^* m_{k+1})^3}. \end{align*} $$

In particular, for the remaining interval $\hat {\omega } = \omega \smallsetminus E$ , we have that

(24) $$ \begin{align} \vert \hat{\omega} \vert \leq \vert \omega \vert (1- \delta_{m_{k+1}} \tau^{(\log^* m_{k+1})^3}). \end{align} $$