Proof of Lemma 3.9

Before proving for $\hat v(\delta )$ , we first prove for fixed $\hat v$ . Let us fix some constant $\hat v> 1$ and small $\delta> 0$ , and assume that W is a connected component of $(T_0^n)^{-1}(\tilde B(\hat v \delta ))$ for some $n \in \mathbb Z_+ $ such that $\text {dist}(W, T_0(0) ) \le \delta $ . Clearly, the distance condition implies that $T_0(\tilde B(\delta )) \cap W \neq \emptyset $ .

To show $|W| < \delta $ , let $x \in \tilde B(\delta ) \cap T_0^{-1}(W)$ and let $W_k = W_k(x)$ be the component of $(T_0^{n + 1- k})^{-1}(\tilde B (\hat v \delta ))$ containing $T_0^k(x)$ . Our W will correspond to $W_1$ , so the proof becomes a question of showing $|W_1| < \delta $ , which we do by following an inductive argument on n.

For the base case $n = 1$ , with $T_0(x) \in W_1 \subset T_0^{-1}(\tilde B (\hat v \delta ))$ and $T_0^2(x) \in \tilde B (\hat v \delta ) $ , notice that it is sufficient to show $T_0^2(0) \in \tilde B (\hat v \delta )$ . Indeed, this would mean $W_0$ satisfies the assumptions of Sublemma 3.11, which implies $W_0 \subset \tilde B (\delta )$ , and since $W_1 = T_0(W_0)$ , we would have $|W_1| < \delta $ by the definition of $\tilde B (\delta )$ . To show this is the case, recall the initial assumption that $\text {dist}(W_1, T_0(0) ) \le \delta $ , which trivially implies $|T_0(x) - T_0(0)| \le \delta $ . Since we have an upper bound on the derivative, we therefore have for sufficiently small $\delta> 0$ that the distance between $T_0^2(x)$ and $T_0^2(0)$ will be small enough to also guarantee $T_0^2(0) \in \tilde B (\hat v \delta )$ .

For the inductive step, assume that the result holds for all $m < n$ . To prove the result for n, define the sets $\{ G_j \}_{j = 0}^{n+1}$ where $G_{n+1} = \tilde B (8^s \hat v \delta )$ and $x \in G_0$ , where we note that $\tilde B (8^s \hat v \delta )$ is a 3-scaled neighbourhood of $\tilde B ( \hat v \delta )$ . There are two possible cases.

In the first case, we have that there is one $0 \le n_1 < n+1$ such that $0 \in G_{n_1}$ . Then, from Sublemma 3.11, this therefore implies $G_{n_1} \subset \tilde B(\delta )$ . If $n_1 = 0$ , then $W_0 \subset G_0 \subset \tilde B(\delta )$ and thus proves the desired result. If $n_1>0$ , then we simply repeat this process as we did in the proof of the previous lemma, giving us the desired result.

In the second case, we have that $T_0^{k}(0) \notin \tilde B(\hat v \delta )$ for $0 \le k < n + 1$ . This immediately implies that $T_0^{n}: G_1 \to G_{n+1}$ is a diffeomorphism. Again, using the Koebe principle, we have that $G_1$ contains a 1-scaled neighbourhood of $W_1$ , and thus $0 \notin G_0$ , $T_0(0) \notin G_1$ and $T_0(x) \in B_{\delta }(T_0(0))$ , thereby completing the proof for fixed $\hat v$ .

Now, to prove there exists a $\hat v (\delta )$ satisfying all the desired properties, observe that for any value of fixed $\hat v$ , we can choose some $\delta _{\hat v}>0$ such that the results hold for any $\delta \le \delta _{\hat v}$ . Then, for integer values of $\hat v> 1$ , we can define the function $\hat v (\delta ) = \hat v$ for when $\delta \in (\delta _{\hat v + 1} , \delta _{\hat v} ]$ . Clearly, $\hat v (\delta ) \to \infty $ as $\delta \to 0$ and, by construction, we know the results we have just proven will hold for $\tilde B (\hat v (\delta ) \delta ) = \tilde B (\hat v \delta ) $ when $\delta \in (\delta _{\hat v + 1} , \delta _{\hat v} ]$ . Furthermore, we can guarantee that $\hat v (\delta ) \delta \to 0$ as $\delta \to 0$ by choosing $\delta _{\hat v}$ sufficiently small for each integer $\hat v$ .

Proof of Proposition 3.7

For our fixed $L>1$ , it is possible to construct (similarly to how $\hat v(\delta )$ in Lemma 3.9 is constructed) some $L(\delta )$ which, for sufficiently small $\delta>0$ , satisfies $L(\delta )> L $ , $\lim _{\delta \to 0 } L(\delta ) = \infty $ , $\lim _{\delta \to 0 } L(\delta )\delta = 0$ and $\lim _{\delta \to 0 } v(L(\delta ) \delta )/ L(\delta ) = \infty $ . Furthermore, we can choose some $L(\delta )$ such that if $|T_0^n(0) | \le L(\delta ) \delta $ , then

(28) $$ \begin{align} DT_0^{n+1}(T_0(0)) \ge L(\delta)^{1 - \zeta}. \end{align} $$

Indeed, using properties (R1) and (R2) of Rovella maps, we know that if $|T_0^n(0) | \le L(\delta ) \delta $ , then n must be sufficiently large so that $e^{-\alpha n} \le L(\delta ) \delta $ and, thus, $DT_0^{n+1}(T_0(0)) \ge {\unicode{x3bb}} ^{n+1}> L(\delta )^{1 - \zeta }$ . As long as $L(\delta )^{1-\zeta }$ grows sufficiently slowly, then this always holds for all sufficiently small $\delta $ .

For sufficiently small $\delta>0$ , let us denote $N = N(\delta )$ to be the largest time such that we have $A( \underline {0}, x, N) \le \theta / \delta $ . We want to prove the following: for every $j = 1, \ldots , N-1$ , we have $T_0^j(0) \notin \tilde B(L(\delta ) \delta ))$ . Indeed, this can be shown by a simple proof by contradiction. If this is not the case, then there exists some smallest $t \in \{ 1 , \ldots , N-1\} $ satisfying ${T_0^t \in \tilde B(L(\delta )\delta )}$ . Then, by Lemma 3.12, we have

$$ \begin{align*} DT_0^t(0) \ge \frac{v(L(\delta) \delta)}{\mathcal D(L(\delta) \delta)} \end{align*} $$

and, thus, we have

$$ \begin{align*} A( \underline{0}, x, N) \ge \frac{DT_0^t(0)}{|T_0^t(0)|} \ge \frac{v(L(\delta) \delta)}{|\tilde B(L(\delta) \delta)| \mathcal D(L(\delta) \delta)} = \frac{v(L(\delta) \delta)}{v(\delta)\delta}> \frac{\theta}{\delta}, \end{align*} $$

which clearly contradicts our initial assumption.

Now, suppose $T_0^N(0) \in \tilde B(L(\delta ) \delta )$ . In this case, we simply set $M(\delta ) = N$ . Furthermore, since we already know $\delta ' \le L \delta $ , then inequality (23) immediately follows from equation (28). Alternatively, assume the opposite, that is, $T_0^N(0) \notin \tilde B(L(\delta ) \delta )$ . Then, our procedure for defining M becomes more involved.

Let us set $V_k = \tilde B (2^{-k} \delta )$ and let us denote $k_1$ as the first iterate such that $T_0^j(0) \notin V_{k_1}$ for every $0 \le j \le N$ . Clearly, it follows that $\delta /2^{k-1}> L(\delta ) \delta $ . Thus, for sufficiently small $\delta>0$ , we have

$$ \begin{align*} \sum_{k = 0}^{k_1 -1} \bigg( \frac{\delta}{\delta/ 2^k} \bigg)^{\zeta/2} \le \bigg( \frac{\delta}{ \delta/2^{k - 1}} \bigg)^{\zeta/2} \frac{1}{1 - 2^{- \zeta/2}} \le \frac{L(\delta)^{- \zeta/2}}{1 - 2^{- \zeta/2}} \le \frac{\theta}{ 2}. \end{align*} $$

Now, let us set $\nu ^*_{k_1} = -1$ and we set for each $0 \le k < k_1$

$$ \begin{align*} \nu^*_k = \max \{ 0 \le j \le N | T_0^j(0) \in V_k \}. \end{align*} $$

Then, there are two possibilities: either

(29) $$ \begin{align} \sum_{i = \nu^*_0 + 1}^N \frac{DT_0^i(T(0))}{|T_0^i(T(0))|} \ge \frac{\theta}{2 \delta} \end{align} $$

or, alternatively, there exists some $0 \le k < k_1$ such that

(30) $$ \begin{align} \sum_{i = \nu^*_k + 1 }^{s_{k+1}} \frac{DT_0^i(T(0))}{|T_0^i(T(0))|} \ge \frac 1 \delta \bigg( \frac{\delta}{\delta/ 2^k} \bigg)^{\zeta/2}. \end{align} $$

If the former holds, then we again set $M = N$ . As we already have $T_0^n(c) \notin \tilde B(\delta )$ , we therefore know that for every $\nu ^*_0 < j \le N$ , there will exist some constant $C_1>0$ such that

$$ \begin{align*} DT_0^{N+1}(T_0(0)) \ge C_1 \sum_{n = \nu^*_0 + 1}^N \frac{DT_0^n(T(0))}{|T_0^n(0)|} \ge \frac{C_1 \theta}{2 \delta} \ge \bigg( \frac{\delta_*}{\delta} \bigg)^{1 - \zeta}, \end{align*} $$

assuming that $\delta> 0$ is sufficiently small.

Suppose that equation (29) is not the case. Then, we instead set $M = \nu ^*_k$ . By definition, $T_0^{\nu ^*_k} \in \tilde B(\delta _k)$ . By Lemma 3.13, there exists some constant $C_2>0$ such that

$$ \begin{align*} \frac{DT_0^{s_k}(T_0(0))}{|\tilde B(\delta_k)|} \ge C_2 \sum_{n = \nu^*_k + 1}^{\nu^*_{k+1}} \frac{DT_0^n(T(0))}{|T_0^n(0)|} \ge \frac{C_2}{\delta}\bigg( \frac{\delta}{\delta_k} \bigg)^{ \zeta / 2}. \end{align*} $$

Now, using the fact that $\nu ^*_k> \nu ^*_{k+1}$ and the fact that $T_0^{\nu ^*_k}(0) \in \tilde B(\delta _k/2)$ , we therefore have $DT_0(T_0^{\nu ^*_k}(0)) \cdot |\tilde B(\delta _k)| \ge C_3 \delta _k$ , with $C_3>0$ being some constant. Combining this altogether, we have

$$ \begin{align*} DT_0^{\nu^*_k + 1}(T_0(0)) = DT_0^{\nu^*_k}(T(0)) \cdot DT_0(T_0^{\nu^*_k}(T(0))) \ge C_2 C_3 \frac{\delta_k}{\delta}\bigg( \frac{\delta}{\delta_k} \bigg)^{ \zeta / 2}. \end{align*} $$

Using the fact that $\delta _k/\delta \ge 2 \delta _{k_1}/\delta \ge L(\delta )$ , then if $\delta $ is sufficiently small, then we obtain our desired result.

Proof of Proposition 3.8

Let us take some constant $\delta _0>0$ sufficiently small such that for all $\delta \in (0, \delta _0)$ , Lemma 3.12 can be applied and that $v(\delta )> 2$ . Because we already know that $T_0^n(x) \in \tilde B(2 \delta )$ , we therefore know that there exists a sequence of iterates $n_1 < n_2 < \cdots < n_m = n$ such that $n_1$ is the first iterate when $T_0^{n_1}(x) \in \tilde B(\delta _0)$ and such that, for every $i = 1, 2, \ldots , m - 1$ , the iterate $n_{i + 1}>n_i$ is the minimal integer such that

$$ \begin{align*} |T_0^{n_{i+1}+1}(x) - T_0(0)| \le \min \{ \delta_0, 2 |T_0^{n_{i}+1}(x) - T_0(0)| \}. \end{align*} $$

Additionally, for every $i = 1, 2, \ldots , m - 1$ , let $\rho _i = |T_0^{n_{i}+1}(x) - T_0(0)|$ , which means $\rho _i \in [\delta , \delta _0)$ for every $i = 1, 2, \ldots , m - 1$ .

Now, for every $i = 1, 2, \ldots , m - 1$ , consider the segment of the orbit $T_0^{n_i+1}(x), \ldots , T_0^{n_{i+1}}(x)$ . If we apply Lemma 3.12 with respect to $\rho _i$ instead of $\delta $ , then we have

$$ \begin{align*} DT_0^{n_{i+1} - n_i}(T_0^{n_i}(x)) &= DT_0(T_0^{n_i}(x)) \cdot DT_0^{n_{i+1} - n_i - 1}(T_0^{n_i + 1}(x))\\ & \ge \mathcal{D}(\rho_i) \frac{v(\rho_i)}{\mathcal{D}(\rho_i)} \ge 2 \frac{\mathcal{D}(\rho_i)}{\mathcal{D}(\rho_i)}. \end{align*} $$

Additionally, if we set $\rho _0 = \delta " = \max \{ |T_0(x)|, \delta \}$ , then this gives us $DT_0^{n_1}(x) \ge 2 \kappa _0 / \mathcal {D}(\rho _0)$ . Indeed, if we have that $\delta " < \delta _0$ , then we can just apply Lemma 3.12. However, if $\delta " \ge \delta _0$ , then we can just use the non-uniform expansion. Therefore, combining this all together, we obtain

$$ \begin{align*} DT_0^n(x) &\ge \frac{2\kappa_0}{\mathcal{D}(\rho_0)} \prod_{i = 1}^{m - 1} \frac{\mathcal{D}(\rho_i)}{\mathcal{D}(\rho_i)} = \frac{2\kappa_0}{\mathcal{D}(\rho_{m-1})} \prod_{i = 1}^{m - 1} \frac{\mathcal{D}(\rho_i)}{\mathcal{D}(\rho_{i-1})} \\ &\ge \frac{2\kappa_0}{\mathcal{D}(\rho_{m-1})} \frac{\mathcal{D}(\rho_i)}{\mathcal{D}(\rho_{i-1})} \ge \frac{2\kappa_0}{\mathcal{D}(\rho_{m-1})} \bigg( \frac{\rho_i}{\rho_{i-1}} \bigg)^{1 - 1/s} \ge \frac{\kappa_0}{\mathcal{D}(\delta)} \bigg( \frac{\delta}{\delta " } \bigg)^{1 - 1/s}.\\[-3.4pc] \end{align*} $$

Proof of Proposition 3.14

Let us assume our $\delta _0 $ is sufficiently small so that for all ${0 < \delta \le \delta _0}$ , we have $\Lambda _0(\delta )> 1$ . Let $s_1$ be the first time that $\nu _1 \ge M(\delta )$ and ${T_\omega ^{\nu _1}(x) \in \tilde B(\delta _0)}$ . If $n = \nu _1$ , then the desired result follows from Lemma 3.18.

Alternatively, if $\nu _1 < n$ , then we set $\rho _1 = |T_\omega ^{\nu _1}(x)|$ . By inequality (42), we have

(44) $$ \begin{align} DT_\omega^{\nu_1 + 1}(x) \ge \Lambda_0(\delta)^{\zeta_3} \bigg( \frac{\rho_1}{\delta} \bigg)^{1 - 1/s}. \end{align} $$

Now, let $\nu _2> \nu _1$ be the smallest integer such that $\nu _2 \ge M(\delta _1)$ and $T_\omega ^{\nu _2}(x) \in \tilde B(\delta _0)$ . If $\nu _2 = n$ , then we stop, otherwise, we define $\rho _2 = |T_\omega ^{\nu _2}(x)|$ then find $\nu _3$ . We continue this process until we reach some $\nu _k = n$ and then apply inequality (42) to obtain for all ${i = 1, \ldots , k - 2}$ ,

$$ \begin{align*} \prod_{j =\nu_i+1}^{\nu_{i + 1}} DT_{\sigma^j \omega}(T_\omega^j(x)) \ge \Lambda_0(\rho_i)^{\zeta_3} \bigg( \frac{\rho_{i + 1}}{\rho_i} \bigg)^{1 - 1/s}, \end{align*} $$

and using Lemma 3.18, we have

$$ \begin{align*} \prod_{j = \nu_{k -1}+ 1}^{n -1}DT_{\sigma^j \omega}(T_\omega^j(x)) \ge \frac{\Lambda_0(\rho_{k-1})^{\zeta_3}}{\mathcal{D}(\delta)}. \end{align*} $$

Thus, combining the above inequalities together, we have

$$ \begin{align*} DT_\omega^n(x) &= DT_\omega^{n+1}(x) \bigg( \prod_{i = 1}^{k - 2} \prod_{j = \nu_i + 1}^{\nu_{i + 1}} DT_{\sigma^j \omega}(T_\omega^j(x)) \bigg) \prod_{j = \nu_{k-1}+1}^{n -1} DT_{\sigma^j \omega}(T_\omega^j(x)) \\ & \ge \frac{\Lambda_0(\delta)^{\zeta_3}}{\mathcal{D}(\delta)} \prod_{i = 1}^{k - 2} \Lambda_0(\rho_j)^{\zeta_3} \bigg( \frac{\rho_{k-1}}{\delta} \bigg)^{1 - 1/s} \ge \frac{\Lambda_0(\delta)^{\zeta_3}}{\mathcal{D}(\delta)}. \end{align*} $$

Thus, the result holds.

Proof of Proposition 3.1

Statement (1): this follows directly from Lemma 3.17. Indeed, from Lemma 3.17, we know that because $x, T_\omega (x), \ldots , T_\omega ^{n-1}(x) \notin \tilde B(\delta )$ and $T_\omega ^n(x) \in \tilde B(2\delta )$ , then

$$ \begin{align*} DT_\omega^n(x) \ge \frac{\kappa}{\mathcal{D} (\delta)} \bigg( \frac{\delta}{\delta_3} \bigg)^{1 - 1/s} e^{\hat \eta n}, \end{align*} $$

where $\delta _3 = \max \{ |x - T_0(0)|, \delta \}$ . The smallest value $( {\delta }/{\delta _3} )^{1 - 1/s}$ can take is $(\delta /2)^{1 - 1/s}$ . Thus, if we set $A = A(\delta ) = {\kappa }/{\mathcal {D} (\delta )} (\delta /2)^{1 - 1/s}$ , then this gives us the desired result for fixed $\delta>0$ .

Statement (2): this follows directly from Proposition 3.14 and Lemma 3.17. Indeed, as in the proof for the first statement, from Lemma 3.17, we know that because $x, T_\omega (x), \ldots , T_\omega ^{n-1}(x) \notin \tilde B(\delta )$ and $T_\omega ^n(x) \in \tilde B(2\delta )$ , then

$$ \begin{align*} DT_\omega^n(x) \ge \frac{\kappa}{\mathcal{D} (\delta)} \bigg( \frac{\delta}{\delta_3} \bigg)^{1 - 1/s} e^{\hat \eta n}. \end{align*} $$

However, we already know that $|x - T_0(0)| \le 4 \delta $ , so we therefore have

$$ \begin{align*} \bigg( \frac{\delta}{\delta_3} \bigg)^{1 - 1/s} \ge \bigg( \frac 1 4 \bigg)^{1 - 1/s}, \end{align*} $$

which can just be absorbed into the $\kappa> 0$ term, thereby giving us

(45) $$ \begin{align} DT_\omega^n(x) \ge \frac{\kappa}{\mathcal{D} (\delta)} e^{\hat \eta n}. \end{align} $$

Finally, if we set a constant $\beta (\delta )$ that satisfies the following criteria:

1. (1) $\beta (\delta ) \to 0 $ as $\delta \to 0$ ;

2. (2) $\hat \Lambda (\delta ) = \Lambda (\delta )^{\beta (\delta )}e^{1 - \beta (\delta )} \ge 2e$ for all sufficiently small $\delta>0$ ;

3. (3) $\Lambda (\delta )^{\beta (\delta )} \to \infty $ as $\delta \to 0$ ,

and if we set

$$ \begin{align*} \eta_0(\delta) = (1 - \beta(\delta)) \hat \eta(\delta), \end{align*} $$

then we have

$$ \begin{align*} DT_\omega^n(x) \ge \frac{\hat \Lambda(\delta)}{\mathcal{D} (\delta)} e^{ \eta_0 n}.\\[-41pt] \end{align*} $$

Proof of Proposition 4.4

From Lemma 4.5, we know that for $y, T_\omega ^n(y) \in \tilde B(\delta )$ , we have

$$ \begin{align*} \log \bigg( \frac{DT_\omega^n(y)|y| |T_\omega^{\nu_1}(y)| \cdots |T_\omega^{\nu_\Gamma}(y)|}{|\tilde B(\delta)|^\Gamma} \bigg) \ge \kappa n - \sum_{i=0}^{n-1} r_\delta(T_\omega^i(y)). \end{align*} $$

Choosing some small $0 < c < \kappa $ and setting ${\unicode{x3bb}} = \kappa - c$ , if our starting point x is in $\tilde B(\delta )$ , then we simply consider $y=x$ and directly use the above lemma:

$$ \begin{align*} DT_\omega^n(y) \ge \frac{|\tilde B(\delta)|}{|x|} \frac{|\tilde B(\delta)|}{|T_\omega^{\nu_1}(y)|} \cdots \frac{|\tilde B(\delta)|}{|T_\omega^{\nu_\Gamma}(y)|} \exp\bigg(\kappa n - \sum_{i=0}^{n-1} r_\delta (T_\omega^i(y))\bigg)> e^{{\unicode{x3bb}} n}. \end{align*} $$

Here, we can get rid of each $|\tilde B(\delta )|/|T_\omega ^i(x)|$ for $i \in \{0 , \nu _1, \ldots , \nu _\Gamma \}$ because $T_\omega ^i(x) \in \tilde B(\delta )$ .

Now, if $x \notin \tilde B(\delta )$ , then we instead consider the starting point $y=T_\omega ^{\nu _1}(x)$ in the context of the above lemma. Clearly, we have

$$ \begin{align*} \sum_{i=0}^{n-\nu_1-1} r_\delta(T_{\sigma^{\nu_1} \omega}^{i}(y)) = \sum_{i=0}^{n-1} r_\delta( T_\omega^i(x)) < cn, \end{align*} $$

thus, we can write

$$ \begin{align*} &DT_{\sigma^{\nu_1} \omega}^{n-\nu_1}(y) \ge \exp \bigg({\kappa(n-\nu_1) - \sum_{i=0}^{n-\nu_1-1} r_\delta(T_{ \omega}^i(y))}\bigg)\\ &\quad> \exp(\kappa(n-\nu_1) - cn ) = \exp({\unicode{x3bb}} n - \kappa \nu_1). \end{align*} $$

Furthermore, since we have assumed $x \notin \tilde B(\delta )$ and, by definition, $T_\omega ^{\nu _1}(x) \in \tilde B(\delta )$ , we can apply inequality (6) to obtain

$$ \begin{align*} DT_\omega^{\nu_1}(x) \ge A \exp(\kappa \nu_1). \end{align*} $$

Combining these together, we obtain

$$ \begin{align*} DT_\omega^n(x) = DT_{\sigma^{\nu_1} \omega}^{n-\nu_1}(y) \cdot DT_\omega^{\nu_1}(x)= \exp({\unicode{x3bb}} n - \kappa \nu_1) \cdot A \exp(\kappa \nu_1) = A e^{{\unicode{x3bb}} n}.\\[-37pt] \end{align*} $$

Proof of Lemma 4.7

First, for any $\omega \in \Omega $ , consider the set

$$ \begin{align*} Z_\delta^{\omega}(r) = \{ x \in \tilde B(\delta) | r_\delta(\omega, x) \ge r \}. \end{align*} $$

Using the definition of $r_\delta $ and condition (C2), there exists some constant $K_2>0$ such that

$$ \begin{align*} |Z_\delta^{\omega}(r)| \le K_2 e^{-r/s}|\tilde B(\delta) |, \end{align*} $$

and furthermore, there exists some constant $r^*>0$ such that $Z^{\omega }(r) \subset \tilde B(\delta /e)$ for all ${r < r^*}$ .

Now, for all $r < r^*$ , inequality (48) holds for some sufficiently large $K_1$ . For the case where $r \ge r^*$ , for any $y \in Y_\delta ^\omega (r) $ , we set $n = \hat R_\omega (y)$ and set $J= J(y)$ as the component of $(T_\omega ^n)^{-1}(\tilde B(\delta ))$ which contains y. We wish to prove that there exists a constant $K_3>0 $ such that

(50) $$ \begin{align} |J \cap Y_\delta(\omega, n)| \le K_3 e^{-r/s}|J|. \end{align} $$

To prove this, recall that since $r \ge r^*$ , we have $T_\omega ^n(y) \in \tilde B(\delta /e)$ . Thus, at least one component of $\tilde B(\delta ) \setminus \tilde B(\delta / e)$ is contained within $T_\omega ^n(J)$ , thereby implying that there exists some $\gamma>0$ such that $|T_\omega ^n(J)|/|\tilde B(\delta )| \ge \gamma> 0$ . Using Lemma 4.6, we know that $\mathcal {N}(T_\omega ^n | J) \le 1$ .

To prove equation (48), we need to show that

(51) $$ \begin{align} T_\omega^n(J \cap Y_\delta^\omega(r)) \subset Z_\delta^{\sigma^n \omega}. \end{align} $$

Now, we prove by contradiction that $\hat R_\omega (y') = \hat R_\omega (y)$ for all $y' \in J \cap Y_\delta ^\omega (q)$ . Indeed, suppose instead we have $1 \le n' < n$ , where $n' = \hat R_\omega (y')$ . We know that $T_\omega ^{n'}(y') \in \tilde B( \delta / e)$ , which, combined with Proposition 3.1, implies $T_\omega ^{n'}(J(y')) \subset \tilde B(\delta )$ . However, this contradicts our assumption that $n' $ is the first return.

We now conclude the proof by once again using Besicovitch’s covering lemma: the intervals in the set $\{J(y) | y \in Y_\delta (r)\}$ form an open cover of $Y_{\delta }(r)$ and, thus, a sub-covering can be found that has bounded multiplicity of intersection. From Lemma 4.6, we already now that $|J| \le \delta $ , which implies that $J \subset B(0, 3 \delta )$ . Thus, inequality (48) holds.

Proof of Proposition 4.3

Take the constants $K_0 , \rho _0>0 $ from Lemma 4.9, and set $\rho = \min \{ \rho _0/5, 1/2s \}$ . Using Stirling’s approximation, we know that there will exist some constant $\kappa> 1$ such that, for all constants $n, m \in {\mathbb N}$ with $n> \kappa m/2$ , we have ${{ n + m -1 \choose m-1 } \le e^{\rho n}}$ .

Let us set

$$ \begin{align*} \Xi = \{ (\omega, x) \in \Omega \times I | x \in \tilde B(\delta), r_\delta (x) \ge cn/2 - \kappa \} \end{align*} $$

as well as

$$ \begin{align*} Q_{n_1}^{n_2} (\omega, x) = \sum_{j = n_1}^{n_2} r_\delta(T_\omega^j(x)). \end{align*} $$

Furthermore, for any $m', n \ge 0$ , we set

$$ \begin{align*} \Xi_\ell^m = \{ (\omega, x) \in \Omega_\varepsilon \times I | \text{ there exists } N \ge 1 \text{ such that } \Gamma(\omega, x,N) = \ell, Q_1^N (\omega, x) = m \} \end{align*} $$

and define $\mathcal {I} = \{ (\ell , m) \in {\mathbb N}^2 | 2 \ell \ge \max \{ cn, \kappa m \}> 0 \}$ . We want to prove that

(55) $$ \begin{align} E_n \subset \Xi \cup \bigg( \bigcup_{(m', n) \in \mathcal{I} } \Xi_\ell^{ m} \bigg) \end{align} $$

and then show both of these sets decay exponentially.

To show $E_n$ is a subset, we must first show that for every $(\omega , x) \in E_n$ , there exists some $n' \ge 0$ such that

(56) $$ \begin{align} Q_0^{n'}(\omega, x)> \max \{ cn - \kappa, \kappa \Gamma_0^{n'}(\omega, x) \}. \end{align} $$

To see this, set $n_0$ as the first iterate where $Q_0^{n_0}(\omega , x) \ge cn$ . In the case where $Q_0^{n_0}(\omega , x)> \kappa \Gamma ^{n_0}(\omega , x)$ , then we can simply set $n' = n_0$ . If this is not the case, then we instead take $n' < n_0$ to be the largest iterate where $T_\omega ^{n'}(x) \in \tilde B (\delta )$ . By the definition of $n_0$ , we have $Q_0^{n'}(\omega , x) < m$ and, thus, we must have $Q_0^{n'}(\omega , x)> \kappa \Gamma ^{n'}(\omega , x)$ instead since ${(\omega , x) \in E_n }$ . Furthermore, clearly $\Gamma ^{n_0}(\omega , x) = \Gamma ^{n'}(\omega , x) + 1$ and, thus, we have

$$ \begin{align*} Q_0^n(\omega, x)>\kappa \Gamma_0^{n_0} - \kappa \ge Q_0^{n_0}(\omega, x) - \kappa \ge cn - \kappa. \end{align*} $$

This confirms inequality (56). Now, for every $(\omega , x) \in E_n \setminus \Xi $ , if we set $\ell = Q_1^n(\omega , x)$ and $m = \Gamma _1^n(\omega , x)$ , then this implies that $(\omega , x) \in \Xi $ . It is now simply a question of showing that $(\ell , m) \in \mathcal {I}$ . Indeed, using the fact that

$$ \begin{align*} 2 \ell = 2(Q^{n'}(\omega, x) - r_\delta(\omega, x)) \ge 2 ((cn - \kappa) - (cn/2 - \kappa)) = cn, \end{align*} $$

implies that $m>0$ . Furthermore, using the fact we have $2r_\delta (\omega , x) \le n - \kappa < Q_0^n(\omega , x)$ , we know that $2 \ell \ge Q_0^n(\omega , x)> \kappa \Gamma _0^n(\omega ,x) \ge \kappa n$ , thereby implying that $(\ell , m) \in \mathcal {I}$ and therefore that $E_n$ is a subset.

For the estimate on $|\Xi |$ , note that by the definition of $r_\delta $ , there is a constant $C=C(\kappa )$ such that for all $\omega \in \Omega _\varepsilon $ , we have

$$ \begin{align*} |\{ x \in \tilde B(\delta) | r_\delta(x) \ge cn/ 2 - \kappa \}| \le C |\tilde B(\delta)| e^{-n/2s} \le C |\tilde B(\delta)| e^{-\rho n} \end{align*} $$

and, thus, by using Fubini’s theorem, we have

(57) $$ \begin{align} |\Xi| \le C |\tilde B(\delta)| e^{-\rho n}. \end{align} $$

As for the estimate on $|\Xi _\ell ^{ m}|$ for $(\ell , m) \in \mathcal {I}$ , for every $x \in \tilde B(\delta )$ , we set

$$ \begin{align*} \hat E_\ell^{ m }(x) = \{ \omega \in \Omega_\varepsilon | x \in \Xi_\ell^{ m} \}. \end{align*} $$

Clearly, we have

$$ \begin{align*} \hat E_\ell^{ m }(x) \subset \bigcup_{\substack{\mathbf{r} \in {\mathbb N}^m \\ |\mathbf{r}| = \ell}} \hat Y_{\delta}^m(x, \mathbf{r}). \end{align*} $$

Additionally, there are ${\ell + m - 1 \choose m - 1} $ of $\mathbf r \in {\mathbb N}^m$ with $|\mathbf r| = \ell $ . Thus, using Lemma 4.10, we have

$$ \begin{align*} \mathbb{P}_\varepsilon (\hat E_\ell^{ m}(x)) &\le {\ell + m - 1 \choose m - 1} \le K_0^m e^{- \rho_0 | \mathbf{\ell}|} \\ & \le e^{\rho \ell} e^{-\rho_0 \ell} e^{\kappa \rho m} \le e^{- \rho(4 \ell - \kappa m)} \le e^{- 2 \rho \ell}. \end{align*} $$

Thus, we have

$$ \begin{align*} \sum_{(\ell, m) \in \mathcal{I}} \mathbb{P}_\varepsilon (\Xi_\ell^{ m}) \le \sum_{\ell \ge \left\lfloor {n/2} \right\rfloor } \sum_{\substack{ m \text{ such that } \\ (\ell, m) \in \mathcal{I}}} \mathbb{P}_\varepsilon (\Xi_\ell^{ m}) \le \sum_{\ell \ge \left\lfloor {n/2} \right\rfloor } e^{- 2 \rho \ell} \frac{2 \ell}{\kappa} \le C e^{- \rho n}.\\[-55pt] \end{align*} $$

Proof of Lemma 5.2

We follow the same steps as in [Reference Du24]. First, we want to show that $(\omega ,x) \notin E_n$ implies that $h(\omega , x) \le n$ . To this end, we set $a_i = c - r_\delta (T_\omega ^i(x))$ . By definition, for any $(\omega , x) \notin E_n$ , we have $\sum _{j=0}^{n-1}a_j> 0$ . For any $c'> c$ , if we set $A = c$ , $c_2 = 0$ and $c_1 = c - c' <0$ (and thus $A \ge c_2> c_1$ ), we can apply Pliss’s lemma: there exists some integer $1 \le n_1 \le n$ such that for all $0 \le k < n_1$ , we have

$$ \begin{align*} \sum_{j=k}^{n_1-1} a_j> c_1(n_1 -k)= (c - c')(n_1 -k). \end{align*} $$

Using how we have defined $a_i$ , this can be rewritten as

$$ \begin{align*} \sum_{j=k}^{n_1-1} r_\delta(T_\omega^j(x)) < c'(n_1 -k). \end{align*} $$

Thus, $n_1$ is a $(\delta , c')$ -hyperbolic time for $(\omega , x)$ .

For the tails of first hyperbolic times, we simply use the contrapositive that $h(\omega , x)>n$ implies $(\omega , x) \in E_n$ , which decays exponentially by Proposition 4.3. Thus, the proof is complete.