Proof. We have

(3.2) $$ \begin{align} \partial_a f^{n+1}_{a,b}(x)=1+f_{a,b}'(f^n_{a,b}(x))\cdot\partial_a f^n_{a,b}(x) \end{align} $$

(note that $\partial _a f^0_{a,b}(x)=0$ and $\partial _a f^1_{a,b}(x)=1$ ). Using this formula, we prove by induction that

(3.3) $$ \begin{align} \partial_a f^n_{a,b}(x)=\sum_{k=0}^{n-1}(f^k_{a,b})'(f^{n-k}_{a,b}(x)). \end{align} $$

If $n=0$ , then both sides of (3.3) are 0. Assume now that (3.3) holds for some n and prove it for $n+1$ instead:

$$ \begin{align*} \partial_a f^{n+1}_{a,b}(x)&=1+f_{a,b}'(f^n_{a,b}(x))\cdot\sum_{k=0}^{n-1} (f^k_{a,b})'(f^{n-k}_{a,b}(x)) =1+\sum_{k=0}^{n-1}(f^{k+1}_{a,b})'(f^{n-k}_{a,b}(x))\\ &=1+\sum_{k=1}^n(f^k_{a,b})'(f^{n-(k-1)}_{a,b}(x))=\sum_{k=0}^n (f^k_{a,b})'(f^{(n+1)-k}_{a,b}(x)). \end{align*} $$

Thus, by induction, (3.3) holds for every n.

Now we have

$$ \begin{align*} (f^{n-k-1}_{a,b})'(f_{a,b}(x))\cdot (f^k_{a,b})'(f^{n-k}_{a,b}(a))=(f^{n-1}_{a,b})'(f_{a,b}(x)). \end{align*} $$

From this and (3.3) we get

$$ \begin{align*} \partial_a f^n_{a,b}(x)&=(f^{n-1}_{a,b})'(f_{a,b}(x))\sum_{k=0}^{n-1} \frac1{(f^{n-k-1}_{a,b})'(f_{a,b}(x))}\\ &=(f^{n-1}_{a,b})'(f_{a,b}(x)) \sum_{k=0}^{n-1}\frac1{(f^k_{a,b})'(f_{a,b}(x))}.\\[-3.7pc] \end{align*} $$

Proof. By Corollary 3.2,

(3.9) $$ \begin{align} |\xi_{\mu}(\omega,b)|=\int_\omega \partial_a\xi_{\mu}(t,b)\,dt\ge \int_\omega (f_{t,b}^{\mu-1})'(f_{t,b}(c))\,dt. \end{align} $$

However, by (3.8) we have

$$ \begin{align*} (f_{t,b}^{\mu-1})'(f_{t,b}(c))&=(f_{t,b}^{\mu-\nu})'(f_{t,b}^\nu(c)) \cdot (f_{t,b}^{\nu-1})'(f_{t,b}(c))\\ &\ge\inf_{a\in\omega} (f_{a,b}^{\mu-\nu})'(f_{a,b}^\nu(c))\cdot \frac1{q'}\partial_a\xi_\nu(t,b). \end{align*} $$

Together with (3.9), we get

$$ \begin{align*} |\xi_{\mu}(\omega,b)|&\ge \frac1{q'}\inf_{a\in\omega} (f_{a,b}^{\mu-\nu})'(f_{a,b}^\nu(c))\cdot \int_\omega\partial_a\xi_\nu(t,b)\,dt\\&= \frac1{q'}\inf_{a\in\omega} (f_{a,b}^{\mu-\nu})'(f_{a,b}^\nu(c))\cdot|\xi_\nu(\omega,b)|.\\[-3.5pc] \end{align*} $$

Proof. By Theorem 1.3 of [Reference Misiurewicz17] (or a result of Mañé [Reference Mañé16]), there exist $L>0$ and

such that if $x, f_{a_0} (x),\ldots , f_{a_0}^{L-1}(x) \notin I$ , then

. Therefore, if $\mathcal {N}$ is a sufficiently small neighborhood of $(a_0,1)$ , then for all $(a,b)\in \mathcal {N}$ and x such that $x, f_{a,b} (x),\ldots , f_{a,b}^{L-1}(x) \notin I$ , we have

. Since the infimum of $(f_{a,b}^i)'(x)$ over $(a,b)\in \mathcal {N}$ , $x\notin I$ and $i=0,1,\ldots ,L-1$ is positive, there exists a positive constant

such that (i) holds with

. Thus it also holds with

, and then (ii) holds with any

Proof. Assume that x is bound to c up to time p. Now choose $p_1=\frac {1}{10}\log (1/\delta _1)$ . Then by (4.3) we can estimate

$$ \begin{align*} |f_{a_0}^j(x)-f_{a_0}^j(c)|\le \delta_1 4^j\le \delta_1 4^{p_1} \end{align*} $$

if $j\le p_1$ and

$$ \begin{align*} |f_{a_0}^j(x)-f_{a_0}^j(c)|\le e^{-\beta j}\le e^{-\beta p_1} \end{align*} $$

if $p_1<j\le p$ . Thus, if $\delta _1$ is sufficiently small then $|f_{a_0}^j(x)-f_{a_0}^j(c)|\le {\overline d}/2$ and therefore

(4.6) $$ \begin{align} |f_{a_0}^j(x)-c|\ge \frac{\overline d}2 \end{align} $$

for all $j\le p$ .

This also proves the last statement of the lemma.

We have

(4.7)

The last sum in the exponential may be empty.

Similarly, using (4.6), we get

(4.8)

The sums in (4.7) and (4.8) are bounded by a constant, which only depends on $\delta _1$ , so we get (4.4). Moreover, by (4.7) and (4.8), converges to $1$ as $\delta _1\to 0$ .

Proof. By Lemma 4.3, we have

Taking into account (4.9) and (4.10), we get

If $\delta _1$ is small, then

, so taking logarithms gives us

If $\delta _1$ is small, then $-\log |x-c|$ is large, so we get

, and therefore $\tilde {\kappa}\hspace{-1pt}p<-4\log |x-c|$ .

Proof. On each side of x there are the two closest preimages of c of order less than n: $\eta _1<x$ and $\eta _2>x$ . Then $f_{a_0}^n$ has positive derivative on $(\eta _1,\eta _2)$ and has negative Schwarzian derivative on that interval. Therefore on one of the intervals $[\eta _1,x]$ and $[x,\eta _2]$ the maximum of the derivative $(f_{a_0}^n)'$ is attained at x. We may assume that this is the interval $[\eta _1,x]$ . Then $f_{a_0}^n(\eta _1)=f_{a_0}^k(c)$ for some $k>0$ , so

$$ \begin{align*} |f_{a_0}^n(\eta_1)-f_{a_0}^n(x)|\ge\overline{d} -\delta_1> \overline{d} /2. \end{align*} $$

By the mean value theorem,

$$ \begin{align*} |f_{a_0}^n(\eta_1)-f_{a_0}^n(x)|=(f_{a_0}^n)'(t) |\eta_1-x| \le (f_{a_0}^n)'(t) \end{align*} $$

for some $t\in (\eta _1,x)$ , and thus,

$$ \begin{align*} (f^n_{a_0})'(x)\ge (f_{a_0}^n)'(t)> \overline{d} /2.\\[-34pt] \end{align*} $$

Proof. By Lemma 4.1, . For a general q write $q=kM_1+\ell $ , $0\leq \ell <M_1$ . Choose so that .

Since $\ell <M_1$ , we have $(f_{a,b}^{\ell })'(f_{a,b}^{kM_1}(x))> \overline {d}/4$ by Lemma 4.5 (here we can extend the estimate to a neighborhood of ${\mathcal N}_2$ of $(a_0,1)$ because we consider only finitely many iterates of the map). Then for $(a,b)\in {\mathcal N}= {\mathcal N}_1\cap {\mathcal N}_2$ ,

so (4.17) follows with $C^*=\overline {d}/8$ and

. Note that, as required, $C^*$ only depends on the unperturbed map $f_{a_0}$ , while

depends on $\delta _1$ .

Proof. By the mean value theorem, there is a y between $f_{a_0}(x)$ and $f_{a_0}(c)$ such that

$$ \begin{align*} |f_{a_0}^p(x)-f_{a_0}^p(c)|=(f_{a_0}^{p-1})'(y)|f_{a_0}(x)-f_{a_0}(c)|. \end{align*} $$

By this, Lemma 4.3 and (4.10), there exists a constant

such that if $\delta _1$ is sufficiently small then

Similarly, since $(f_{a_0}^p)'(x)=f_{a_0}'(x)\cdot (f_{a_0}^{p-1})' (f_{a_0}(x))$ , we get by Lemma 4.3 and (4.11) that there exists a constant

such that if $\delta _1$ is sufficiently small then

By the definition of p we have

$$ \begin{align*} |f_{a_0}^{p+1}(x)-f_{a_0}^{p+1}(c)|>e^{-\beta (p+1)}, \end{align*} $$

and therefore for some constant

,

Thus,

so

Together with an earlier estimate, this gives us

Since $9\beta \le \tilde {\kappa }$ , we have

$$ \begin{align*} \frac13(\tilde{\kappa}-2\beta)>\frac7{27}\tilde{\kappa}, \end{align*} $$

and therefore (4.18) holds if

(4.19)

where $p(\delta _1)$ is the bound period associated with $\delta _1$ . Since $p(\delta _1)\to \infty $ as $\delta _1\to 0$ , and $C^*$ is independent of $\delta _1$ , the above inequality holds if $\delta _1$ is sufficiently small.

Proof. Let $0=t_0<t_1<t_2<\cdots <t_S<t_{S+1}=n$ , where $t_i$ for $i\in \{1,2,\ldots ,S\}$ are the times when $f_{a,b}^{t_i}(x)\in I^{**}\setminus I$ . We want to estimate

$$ \begin{align*} (f_{a,b}^n)'(x)=\prod_{j=0}^S (f_{a,b}^{t_{j+1}-t_j})'(f_{a,b}^{t_j}(x)). \end{align*} $$

The times from $[t_0,t_1)$ form a free period; let $t_1-t_0=q_0$ . Hence, by Lemma 4.6,

Consider now times from $[t_j,t_{j+1})$ , where $j>0$ . We can write this interval as a union of a bound period $[t_j,t_j+p_j)$ and a free period $[t_j+p_j,t_{j+1})$ , and we write its length as $t_{j+1}-t_j=p_j+q_j$ . For the bound periods $[t_j,t_j+p_j)$ we can use the estimate from Lemma 4.7 if $\mathcal {N}$ is sufficiently small, because by Lemma 4.4 we work only with the finite number of iterates (for I fixed; this is why $\mathcal {N}$ depends on I). Although p may depend on the map that we are using, if $\mathcal {N}$ is sufficiently small, it may only change to $p\pm 1$ , and then by Remark 4.8 we can still use Lemma 4.7.

Thus, for the bound periods $[t_j,t_j+p_j)$ we get

(4.20) $$ \begin{align} (f_{a,b}^{p_j})'(f_{a,b}^{t_j}(x))\geq \frac1{C^*}e^{({\tilde{\kappa}}/{4}) p_j} \end{align} $$

and for the free period, as before, the estimate from Lemma 4.6 gives us

Combining these estimates, we get

with

. This completes the proof of (i).

Proof. Set $x_k=f^{\nu +k}_{a',b}(c)$ and $y_k=f^{\nu +k}_{a^{\prime \prime },b}(c)$ . Note that $f_{a,b}'(x)$ is independent of a. Therefore,

$$ \begin{align*} \log\frac{(f_{a',b}^{n-\nu})'(x_0)}{(f_{a^{\prime\prime},b}^{n-\nu})'(y_0)} & = \sum_{k=0}^{n-\nu-1}(\log f^{\prime}_{a',b}(x_k)- \log f^{\prime}_{a',b}(y_k))\\& \leq \sum_{k=0}^{n-\nu-1}\bigg(\frac{|f_{a',b}^{\prime\prime}(\eta_k)|} {f_{a',b}'(\eta_k)}\cdot |x_k-y_k| \bigg) \end{align*} $$

for some $\eta _k$ between $x_k$ and $y_k$ . Since $\eta _k\notin I^*$ for $k=0,\ldots ,n-\nu -1$ , we get, by (4.13) and (4.14),

(4.28)

Therefore,

(4.29)

We have $x_k=\xi _{\nu +k}(a',b)$ and $y_k=\xi _{\nu +k}(a^{\prime \prime },b)$ . Therefore, by Remark 4.11,

Thus,

Together with (4.29), we get

and we have proved (4.27) with

Proof. We will not give the proof since it is virtually word for word the same as that of Lemma 4.12. The only difference that the upper bound in (4.28) is replaced by $N({\mathcal N},U)$ .

Proof. We partition $(a_0-\varepsilon ,a_0-\varepsilon ^2)$ into subintervals $\eta _j=(a_0-2^{-j},a_0-2^{-j-1})=(a_j',a_j^{\prime \prime })$ , $j=J_0,\ldots ,2J_0-1$ . The critical point c of unperturbed map $f_{a_0}$ is mapped to a repelling periodic point P in m iterates. Let $U_0$ be a symmetric interval contained in the linearization domain of P such that

$$ \begin{align*} (f_{a,b}^\ell)'(x)\geq \unicode{x3bb}_1^\ell=\Lambda_1>1\quad \text{for } x\in U_0. \end{align*} $$

Let $\tilde {\eta }_j=(a_0-2^{-j},a_0)$ . Then there is a constant

so that

as long as $\xi _{m+i}(\tilde {\eta }_j,b)\subset U_0$ . We now state a version of Lemma 4.12 which will be used in the startup construction.

Lemma 5.2. Suppose that for $a\in \omega $ there is a constant $\tilde {C}=\tfrac 12$ , say, such that

(5.1) $$ \begin{align} |\xi_{m}(\omega',b)|\geq \frac1{q*}\inf_{a\in\omega'} (f_{a,b}^{m-\nu})'(f_{a,b}^\nu(c))\cdot|\xi_\nu(\omega',b)| \end{align} $$

and

(5.2) $$ \begin{align} (f^j_{a,b})'(f_{a,b}(c))\geq \tilde{C} e^{{\hat{\kappa}}/{4},},\quad 1\ \leq j\leq \nu-1. \end{align} $$

Then there is a constant such that

(5.3)

We will not give the proof of this lemma since it is identical to that of Lemma 4.12. We conclude that

and $q_*$ has a uniform control by Corollary 3.2.

We also get a uniform distortion control of $\partial _af_{a,b}^\nu (\xi _j(a,b))$ , that is, there is a constant $\tilde {C}$ depending only on $a_0$ such that for all a, $a'$ in $\tilde {\eta }_j$ ,

(5.4) $$ \begin{align} \frac{1}{\tilde{C}}\leq \frac{\partial_af_{a,b}^\nu(\xi_j(a,b))}{\partial_af_{a',b}^\nu(\xi_j(a',b))}\leq \tilde{C},\quad \nu=1,2,\ldots, \end{align} $$

as long as $\xi _{m+\nu }(\tilde {\eta }_j,b)\subset U_0$ .

It follows that there is a first time L such that $\xi _{m+L+1}(\tilde {\eta }_j,b)\not \subset U_0$ . We write $\tilde {\eta }_j$ as the disjoint union (except for an endpoint)

$$ \begin{align*} \tilde{\eta}_j={\eta}_j\cup \eta_j'. \end{align*} $$

By (5.4) it follows that $\xi _{m+L }(\eta _j,b)$ and $\xi _{m+L }(\eta _j',b)$ are comparable within a fixed constant , which only depends on $f_{a_0}$ .

We continue to iterate $\xi _{m+L+i}(\eta _j,b)$ for $i=1,2,\ldots .$ By Lemmas 4.9, 3.4 and the control of the constant $q_*$ it follows that at the first time J such that $\xi _{m+L+J}(\eta _j,b)\cap I^*\neq \emptyset $ ,

Then

is the required Lyapunov exponent in (a). It follows by Lemmas 4.9(ii), 3.4 and the control of $q_*$ that the time J will be finite. At time $N_0=m+L+J$ , we partition

$$ \begin{align*} (c-\delta,c+\delta)\cap \xi_{N_0}(\eta_j,b) \end{align*} $$

into preimages $\{\omega \}$ under the map $a\mapsto \xi _{N_0}(a,b)$ of the partition ${\mathcal Q}=\{I_{r,l}\}$ and define $n_0=N_0$ for these $\omega $ s. In the special case when $\xi _{N_0}(\eta _j,b)$ only intersects partially an end interval of ${\mathcal Q}=\{I_{r,l}\}$ , we just keep iterating until we cover complete intervals of ${\mathcal Q}$ . In other special case when $\xi _{N_0}(\eta _j,b)$ only partially covers an $I_{r,l}$ interval we adjoin the corresponding preimage to the adjacent interval. Simultaneously we delete the part of $ \eta _j$ that is mapped to $(c-e^{-\sqrt {n_0}},c+e^{-\sqrt {n_0}})$ . By the uniform distortion of both the x-derivative and a-derivative which follows from Lemmas 4.9, 4.12 and Corollary 3.2 a proportion of at most

of the piece of $\eta _j$ is mapped into $(c-e^{-\sqrt {n_0}},c+e^{-\sqrt {n_0}})$ . Here

is a constant only depending on $f_{a_0}$ . We continue to iterate $\xi _{N_0}(\eta _j,b)\setminus (c-\delta ,c+\delta ) $ , still using Lemmas 4.9, 4.12 and Corollary 3.2. For the new returning interval $\omega $ formed in this way $n_0(\omega )>N_0$ and still only a quantity proportional to $e^{-\sqrt {n_0}}/\delta $ is deleted. Conclusions (a)–(e) of Lemma 5.1 are immediately verified.

Proof. The proof will proceed by induction on k. Using (4.3), we get, in the same way as in the proof of Lemma 4.3,

(5.15) $$ \begin{align} \frac{(f_{a,b}^k)'(f_{a,b}(y))}{(f_{a,b}^k)'(f_{a,b}(c))} \leq \exp \bigg( \sum_{j=1}^{k} \dfrac{13 |f_{a,b}^j(y)-f_{a,b}^j(c)|} {f_{a,b}'(f_{a,b}^j(c))} \bigg). \end{align} $$

Furthermore,

(5.16) $$ \begin{align} |f_{a,b}^j(y)-f_{a,b}^j(c)| \le |f_{a,b}^j(x)-f_{a,b}^j(c)| \leq |f_{a,b}^j(x)-f_{a',b}^j(c)| + |f_{a',b}^j(c)-f_{a,b}^j(c)|, \end{align} $$

and by (5.11) we have

(5.17) $$ \begin{align} |f_{a,b}^j(x)-f_{a',b}^j(c)| \leq e^{-4 \sqrt{j}}. \end{align} $$

Thus, we need to estimate $|f_{a',b}^j(c)-f_{a,b}^j(c)|$ . Note that by the mean value theorem there is $a^{\prime \prime }$ between a and $a'$ such that

(5.18) $$ \begin{align} |f_{a',b}^j(c)-f_{a,b}^j(c)|=\partial_a f^j_{a^{\prime\prime},b}(c)\cdot |a-a'|. \end{align} $$

Note that $|a-a'|$ can be interpreted as $|\xi _1(a,b)-\xi _1(a',b)|$ . By Lemma 3.4,

$$ \begin{align*} |\xi_n(a,b)-\xi_n(a',b)|\geq \frac{1}{q_*}\inf_{\tilde{a}\in[a,a']}(f_{\tilde{a},b}^{n-1})'(f_{\tilde{a},b}(c))\cdot |\xi_1(a,b)-\xi_1(a',b)|. \end{align*} $$

By induction statement (iii), . We may therefore conclude that . But by the mean value theorem,

(5.19) $$ \begin{align} |f_{a,b}^j(x)-f_{a',b}^j(c)|&=|f_{a,b}^{j-1}(f_{a,b}(x))-f_{a,b}^{j-1}(f_{a,b}(c))|\nonumber\\ &=(f^{j-1}_{a,b})'(f_{a,b}(y))\cdot |f_{a,b}(c)-f_{a,b}(x)|. \end{align} $$

However, since $|\xi _n(a,b)-\xi _n(a',b)|\leq e^{-r}$ , we have

(5.20)

By the basic assumption . Note also that by Corollary 3.2, $\partial _af^j_{a^{\prime \prime },b}(c)$ is comparable within the multiplicative constant $q_*$ to $(f^{j-1}_{a^{\prime \prime },b})'(f_{a^{\prime \prime },b}(c))$ . But this quantity is in itself by induction comparable within a multiplicative constant to $\inf _{\tilde {a}\in [a,a']}(f_{\tilde {a},b}^{n-1})'(f_{\tilde {a},b}(c))$ . We use the statement of our result for $k=j-1$ .

We use (5.18) and note that $|a-a'|$ can also be written as $|\xi (a,b)-\xi _1(a',b)|$ . By Lemma 3.4,

(5.21) $$ \begin{align} |\xi_n(a,b)-\xi_n(a'b)|\geq \frac{1}{q_*}\inf_{\tilde{a}\in[a,a']}(f^{n-1})'(f_{\tilde{a},b}(c)). \end{align} $$

By combining (5.19), (5.20) and (5.21), we obtain

$$ \begin{align*} &|f^j_{a,b}(x)-f^j_{a,b}(c)|\\&\quad=|f_{a,b}^{j-1}(f_{a,b}(x))-f_{a,b}^{j-1}(f_{a,b}(c))|\geq(f^{j-1})'(f_{a,b}(y))|f_{a,b}(x)-f_{a,b}(c)| \\ &\quad\geq\inf_{y\in I_r}(f^{j-1}_{a,b})'(f_{a,b}(y))\,|f_{a,b}(x)-f_{a,b}(c)|\\ &\quad\geq C_7 e^{-3r}\inf_{y\in I_r}(f^{j-1}_{a,b})'(f_{a,b}(y))\\ &\quad\geq C_7 e^{-2r}\inf_{y\in I_r}(f^{j-1}_{a,b})'(f_{a,b}(y))|\xi_n(a,b)-\xi_n(a',b)|\\ &\quad\geq C_7e^{-2r} \inf_{y\in I_r}(f^{j-1}_{a,b})'(f_{a,b}(y))(\inf_{\tilde{a}\in[a,a']}\partial_af^{n}_{a^{\prime\prime},b}(c))|\xi_1(a,b)-\xi_1(a',b)|\\ &\quad\geq C_7 e^{-2r} \frac{ \inf_{y\in I_r}(f^{j-1}_{a,b})'(f_{a,b}(y))}{ \sup_{z\in I_r}(f^{j-1}_{a,b})'(f_{a,b}(z))} (\inf_{\tilde{a}\in[a,a']}\partial_af^{n}_{a^{\prime\prime},b}(c))|)\,|\xi_j(a,b)-\xi_j(a',b)|. \end{align*} $$

Now $\sup _{z\in I_r}(f^{j-1}_{a,b})'(f_{a,b}(z))$ and $\inf _{y\in I_r}(f^{j-1}_{a,b})'(f_{a,b}(y))$ are comparable within constant , by the statement of Lemma 5.7 with $k=j-1$ . This is where the inductive step is used. Moreover, by Corollary 3.2 and induction statement (i), we have $\partial _af_{a,b}^n(c)\geq q_*^{-1} e^{2(n-1)^{2/3}}$ . Furthermore $e^{-2r}\geq e^{-2\sqrt {n}}$ , by induction statement (v). Combining these estimates, we get

$$ \begin{align*} |\xi_j(a,b)-\xi_j(a',b)|\leq \tfrac{1}{2}|f_{a,b}^j(x)-f_{a,b}^j(c)| \end{align*} $$

since $n\geq n_0(a)$ and $n_0(a)$ can be chosen arbitrarily large.

When inserting this estimate in (5.16) we conclude that

(5.22) $$ \begin{align} |f_{a,b}^j(y)-f_{a,b}^j(c)|\leq 2e^{-4 \sqrt{j}}. \end{align} $$

To estimate $f_{a,b}'(f_{a,b}^j(c))$ from below, we use induction statement (vi) and (4.13). We get

(5.23)

Putting together (5.15), (5.22) and (5.23), we obtain

(5.24) $$ \begin{align} \frac{(f_{a,b}^k)'(f_{a,b}(y))}{(f_{a,b}^k)'(f_{a,b}(c))}< \exp\bigg(\frac{13\cdot 2}{C_9\cdot C_1^2}\sum_{j=1}^k \frac{e^{-4\sqrt{j}}}{e^{-2\sqrt {j}}}\bigg). \end{align} $$

For the lower bound, we obtain, in a similar way to (5.15),

$$ \begin{align*} \frac{(f_{a,b}^k)'(f_{a,b}(c))}{(f_{a,b}^k)'(f_{a,b}(y))} \leq \exp \bigg( \sum_{j=1}^{k} \dfrac{13|f_{a,b}^j(y)-f_{a,b}^j(c)|} {(f_{a,b})'(f_{a,b}^j(y))} \bigg). \end{align*} $$

Note, however, that

and, using (5.9) and (5.22), we get

whenever

. Now

is fixed, and thus there is a positive integer $\tilde N$ such that if $j\ge \tilde N$ then

, and then

By making $\mathcal {N}$ and $I^*$ sufficiently small, we can make $|f_{a,b}^j(c) - f_{a,b}^j(y)|$ smaller than

instead of $2e^{-4\sqrt {j}}$ , and then we get

for some constant

depending only on $a_0$ . Thus, in the same way as we obtained (5.24), we get a similar estimate for the reciprocal ratio, but with a different constant. We choose the larger of those constants as

and we get (5.13). The proof of (5.14) is completely analogous and will be omitted. As for the statement that

can be chosen arbitrarily close to $1$ (but larger than $1$ ), we refer to the argument in Lemma 4.3.

Proof. We claim that $p\leq 8r^{3/2}$ . Note that $8 r^{3/2}\leq 8 n^{3/4} < n$ . We argue by contradiction. Assume that there is $k>8r^{3/2}$ such that $f^j_{a,b}(f_{a.b}(z)$ is still bound to $f^j_{a,b}(f_{a.b}(c)$ for all $x\in I_{r,l}$ and all z between x and c. By the mean value theorem, there is a point y between x and c such that

$$ \begin{align*} |f_{a,b}^k(x)-f_{a,b}^k(c)|=|f_{a,b}(x)-f_{a,b}(c)|\cdot (f_{a,b}^{k-1})'(f_{a,b}(y)). \end{align*} $$

By Lemma 5.7 and induction statement (iii),

Since $|x-c|\ge e^{-r-1}$ , by (4.13) we get

Putting the last three inequalities together, we get

Taking into account (5.22) (which is valid also for $y=x$ ), we get

Therefore,

where

is a constant only depending on $a_0$ .

If $\delta $ is sufficiently small, then , and we get $ k^{2/3}<(k-1)^{2/3}+\tfrac 23k^{-1/3} <4r$ . We conclude that $k\leq 8r^{3/2}$ and this gives a contradiction.

Proof. We have $c=\frac 12$ and $f_{a,b}^{\prime \prime }(t)=-4\pi b\sin (2\pi t)$ . Therefore $f_{a,b}$ is strictly convex in $(c,c+\frac 12)$ , and thus for $x\in (c,c+\frac 12)$ we get (5.25). Similarly, in $(c-\frac 12,c)$ the function $f_{a,b}$ is strictly concave, and (5.25) follows.

Proof. By Definition 5.5, there exists $a\in \omega $ such that

(5.27) $$ \begin{align} |f_{a,b}^{p+1}(x)-f_{a',b}^{p+1}(c)|\geq e^{-4 \sqrt{p+1}}. \end{align} $$

We have

(5.28) $$ \begin{align} |f_{a,b}^{p+1}(x)-f_{a',b}^{p+1}(c)| \leq |f_{a,b}^{p+1}(x)-f_{a,b}^{p+1}(c)| + |f_{a,b}^{p+1}(c)-f_{a',b}^{p+1}(c)|. \end{align} $$

By the mean value theorem, there is a point y between x and c such that

(5.29) $$ \begin{align} |f_{a,b}^{p+1}(x)-f_{a,b}^{p+1}(c)| =|f_{a,b}(x)-f_{a,b}(c)| \cdot (f_{a,b}^p)'(f_{a,b}(y)). \end{align} $$

Now we estimate the second summand in (5.28). As in the proof of Lemma 5.7, we can prove that

$$ \begin{align*} |f_{a,b}^{p+1}(c)-f_{a',b}^{p+1}(c)|\leq\tfrac12|f^{p+1}_{a',b}(c) -f^{p+1}_{a,b}(x)|. \end{align*} $$

Therefore

$$ \begin{align*} |f_{a,b}^{p+1}(x)-f_{a,b}^{p+1}(c)|\geq \tfrac12|f_{a,b}^{p+1}(x)-f_{a',b}^{p+1}(c)| \geq \tfrac12e^{-4\sqrt{p+1}}. \end{align*} $$

From Lemma 5.7 we get

so

By the chain rule and Lemma 5.9 we get, from this inequality,

Since $x\in I_r$ , we have $|x-c|\le e^{-r}$ , and we get (5.26) with a suitable choice of

.

Proof. Let $\eta $ be the distance from $\xi _n(\omega ,b)$ to c. Therefore, by induction statement (i) and (4.13),

However, by (v),

, so we get

(5.30)

After time n there follows the bound period p, and by Lemma 5.7 and (iii) we get

(5.31)

for all $k\le p$ . Combining (5.30) and (5.31), we conclude that

(5.32)

For $k\leq p \leq 8r^{3/2}\leq 8 n^{3/4}$ , we conclude that

$$ \begin{align*} (f^{n+k}_{a,b})'(f_{a,b}(c))\geq e^{(n+k)^{2/3}} \end{align*} $$

At time $n+p+1$ the bound period has expired and we have for all $a\in \omega $ ,

(5.33)

where is a constant only depending on $f_{a_0}$ .

For the total derivative we obtain

(5.34)

After raising (5.33) to the power $\tfrac 23$ we obtain

(5.35)

Looking at the exponents, we get, using (5.7), the lower bound

$$ \begin{align*} 2n^{2/3}+ \frac{1}{3}p^{2/3}-\frac{8}{3}\sqrt{p+1}\geq 2(n+p)^{2/3}+\frac{1}{10}p^{2/3}. \end{align*} $$

Here we have used the information from Lemma 5.8, $p\leq 8n^{3/4}$ and that if $p\leq {1}/{100}n$ then

$$ \begin{align*} 2n^{2/3}+\tfrac{1}{3}p^{2/3}\geq 2(n+p)^{2/3}+\tfrac{1}{5}p^{2/3}. \end{align*} $$

Now, if $k=p+s$ and $0<s\le s_0$ , then we can use Lemma 4.9(ii). If $s\ge M$ then

If $s<M$ we use Lemma 4.5, which allows a perturbation to $(a,b)\in {\mathcal N}$ with a worse constant $\overline {d}/4$ instead of $\overline {d}/2$ and we get

$$ \begin{align*} (f_{a,b}^s)'(f_{a,b}^{n+p+1}(c))\ge \frac{\overline{d}}{4}. \end{align*} $$

Thus, independently of whether $s\ge M$ or $s<M$ , we have

Together with (5.32) (where we substitute $k=p+s$ ), we get

Note that since the constants , , , are absolute constants, p can be made arbitrarily large by making $\delta $ sufficiently small. Doing this, we conclude that induction statements (ii) and (iii) hold for $\nu $ satisfying $n+p<\nu < n'$ where $n'$ is the next free return time.

We now turn to verifying (i) at the next free return time $n'$ . Using the previous derivative estimates and (i) of Lemma 4.9, we get, after writing $n'=n+p+1+q$ , that

where

is an absolute constant only depending on $a_0$ . Arguing in different cases depending on the relative sizes of n and q, one can verify that induction statement (i) with n replaced by $n'$ holds.

Since

and

do not depend on $\delta $ , while by making $\delta $ sufficiently small we can make p as large as we want, we conclude using Lemma 5.8 that

This proves (iv) for $n'$ .

Proof. Let us fix $k \leq n-1$ . Set $t_0=1$ and let $\{t_j\}_{j=1}^m$ be the free return times arranged in increasing order. Here m is defined by the condition $t_{m-1} < k \leq t_m-1$ , and we can assume that $t_m=n$ . Observe that for all free returns $t_j$ there is $r_j$ such that $\xi _{t_j}(a,b), \xi _{t_j}(a',b) \in I_{r_j}$ .

Note that $f^{\prime }_{a,b}(x)=f^{\prime }_{a',b}(x)$ . Thus, using the mean value theorem, we can write the logarithm of the left-hand side of (6.1) as

$$ \begin{align*} &\log \frac{(f_{a,b}^{k})'(f_{a,b}(c))}{(f_{a',b}^{k})'(f_{a',b}(c))} = \log\frac{\prod_{j=1}^{k} f^{\prime}_{a,b}(\xi_j(a,b))} {\prod_{j=1}^{k} f^{\prime}_{a,b}(\xi_j(a',b))}\\ &\quad \leq\sum_{i=0}^{m-1}\bigg(\sum_{j=t_i}^{t_i+p_i}\bigg| \frac{f_{a,b}^{\prime\prime}(\eta_j)}{f_{a,b}'(\eta_j)}\bigg| |\xi_j(a,b)-\xi_j(a',b)| +\log\frac{(f_{a,b}^{t_{i+1}-t_i-p_i-1})'(\xi_{t_i+p_i+1}(a,b))} {(f_{a',b}^{t_{i+1}-t_i-p_i-1})'(\xi_{t_i+p_i+1}(a',b))}\bigg) \end{align*} $$

for some $\eta _j$ between $\xi _j(a,b)$ and $\xi _j(a',b)$ , where $p_i$ is the corresponding bound time and $p_0=-1$ . We will denote the first sum in the parentheses above by $S_i'$ and the second term in the parentheses by $S_i^{\prime \prime }$ . Note that the sum $S_0'$ is empty.

By (4.13) and (4.14) we get

(6.2)

Set $\sigma _i=|\xi _{t_i}(a,b)-\xi _{t_i}(a',b)|$ . We claim that the sum $S_i'$ can be estimated from above by a constant times $\sigma _i e^{r_i}$ .

First we note that by (6.2), the first term of $S_i'$ can be estimated by .

For the remaining terms we introduce the reference interval $\Omega _{t_i}=I_{r_i+1}$ and intervals $\Omega _{t_i+\nu }= f_{a,b}^\nu (\Omega _{t_i})$ , $\nu =0,1,2,\ldots ,p_i$ . We have

$$ \begin{align*} \xi_{t_i+1}(a,b)-\xi_{t_i+1}(a',b)&= (f_{a,b}(\xi_{t_i}(a,b))-f_{a,b}(\xi_{t_i}(a',b))\\ &\quad+(f_{a,b}(\xi_{t_i}(a',b)-f_{a',b}(\xi_{t_i}(a',b))\\ &=f_{a,b}'(y)(\xi_{t_i}(a,b)-\xi_{t_i}(a',b))+(a-a') \end{align*} $$

for some y between $\xi _{t_i}(a,b)$ and $\xi _{t_i}(a',b)$ . Furthermore, $|\Omega _{t_i+1}|=f_{a,b}'(y')|\Omega _{t_i}|$ for some $y'\in \Omega _{t_i}$ .

We get

(6.3) $$ \begin{align} \frac{|\xi_{t_i+1}(a,b)-\xi_{t_i+1}(a',b)|}{|\Omega_{t_i+1}|} =\frac{f_{a,b}'(y)}{f_{a,b}'(y')}\cdot\frac{\sigma_i} {|\Omega_{t_i}|}\pm\frac{|a-a'|}{f_{a,b}'(y')|\Omega_{t_i}|}. \end{align} $$

By the mean value theorem,

$$ \begin{align*} \frac{|a-a'|}{f_{a,b}'(y)\sigma_i}=\frac1{f_{a,b}'(y)\partial_a \xi_{t_i}(a^{\prime\prime},b)} \end{align*} $$

for some $a^{\prime \prime }$ between a and $a'$ . By (3.5) and induction statement (iii),

By (4.13) and induction statement (v),

Therefore, we get

Since $t_i$ can be made as large as we want (because $t_i\ge n_0$ ), we get $|a-a'|<f_{a,b}'(y)\sigma _i/2$ . Therefore, from (6.3) we get

$$ \begin{align*} \frac12\frac{f_{a,b}'(y)}{f_{a,b}'(y')}\cdot\frac{\sigma_i} {|\Omega_{t_i}|}<\frac{|\xi_{t_i+1}(a,b)-\xi_{t_i+1}(a',b)|} {|\Omega_{t_i+1}|}<2\frac{f_{a,b}'(y)}{f_{a,b}'(y')}\cdot \frac{\sigma_i}{|\Omega_{t_i}|}. \end{align*} $$

Since $y,y'\in I_{r_i}\cup I_{r_i+1}$ , we have $|y-c|\in [e^{-r_i-2},e^{-r_i}]$ , and the same holds for $y'$ . Therefore, by (4.13), we get

The right-hand side above is a weighted average between $(2-2b)/(2-2b)=1$ and

, so it is smaller than

. Since we can switch y and $y'$ , we get

Thus, we get the following inequality with

:

(6.4)

Now we want to estimate

$$ \begin{align*} \frac{|\xi_{t_i+{\nu}}(a,b)-\xi_{t_i+{\nu}}(a',b)|} {|\Omega_{t_i+{\nu}}|} \end{align*} $$

from above. The numerator can be estimated as follows:

(6.5) $$ \begin{align} |\xi_{t_i+{\nu}}(a,b)-\xi_{t_{i+{\nu}}}(a',b)| &\leq |f_{a,b}^{{\nu}-1}(f^{t_i+1}_{a,b}(c)) -f_{a,b}^{{\nu}-1}(f^{t_i+1}_{a',b}(c))|\\ &+|f_{a,b}^{\nu}(f^{t_i}_{a',b}(c)) -f_{a',b}^{\nu}(f^{t_i}_{a',b}(c))|. \nonumber \end{align} $$

By a similar argument to the proof of Lemma 5.8, in particular estimating $|a-a'|$ as in that lemma, we obtain that

$$ \begin{align*} |f_{a,b}^{\nu}(f^{t_i}_{a',b}(c)) -f_{a',b}^{\nu}(f^{t_i}_{a',b}(c))|\leq \tfrac{1}{2}|\xi_{t_i+{\nu}}(a,b)-\xi_{t_{i+{\nu}}}(a',b)|. \end{align*} $$

Therefore, we get the estimate

(6.6) $$ \begin{align} |\xi_{t_i+{\nu}}(a,b)-\xi_{t_{i+{\nu}}}(a',b)| &\leq 2|f_{a,b}^{{\nu}-1}(f^{t_i+1}_{a,b}(c)) -f_{a,b}^{{\nu}-1}(f^{t_i+1}_{a',b}(c))|\\ &=2(f^{{\nu}-1}_{a,b})'(y)|\xi_{t_i+1}(a,b)- \xi_{t_i+1}(a',b)|,\nonumber \end{align} $$

where y is between $\xi _{t_{i+1}}(a,b)$ and $\xi _{t_{i+1}}(a',b)$ (using the mean value theorem). Again by the same theorem there is $y^{\prime \prime }\in \Omega _{t_i+1}$ such that

$$ \begin{align*} |\Omega_{t_i+{\nu}}|= (f^{{\nu}-1}_{a,b})'(y^{\prime\prime})|\Omega_{t_i+1}|. \end{align*} $$

By Lemma 5.7, , so we get

(6.7)

Let us consider the interval $\omega _i\in \mathcal {P}_{t_i}$ containing $\omega $ and denote its midpoint by $\hat {a}_i$ . Then by (5.11),

(6.8) $$ \begin{align} |\xi_{t_i+\nu}(\tilde{a},b)-\xi_\nu(\hat{a}_i,b)|\le e^{-4\sqrt{\nu}} \end{align} $$

for all $\tilde {a}\in \omega _i$ and $\nu \le p_i$ .

We claim that for all $\tilde {a}\in \omega _i$ and $\nu \le p_i$ we have

(6.9) $$ \begin{align} |\xi_{t_i+\nu}(\tilde{a},b)-c|\geq\tfrac12|\xi_\nu(\hat{a}_i,b)-c|. \end{align} $$

There is an integer $\nu _0$ such that

(6.10)

for all $\nu \geq \nu _0$ . Note that $\nu _0$ depends only on , which is independent of $\delta $ . Therefore, we may assume that $\delta $ is so small that

(6.11)

Moreover, by induction statement (v),

(6.12)

Consider $\nu \le p_i$ . If $\nu \ge \nu _0$ , then by (6.8), (6.10) and (6.12), we get

(6.13) $$ \begin{align} |\xi_{t_i+\nu}(\tilde{a},b)-\xi_\nu(\hat{a}_i,b)|\le \tfrac12|\xi_\nu(\hat{a}_i,b)-c|. \end{align} $$

Therefore,

$$ \begin{align*} |\xi_{t_i+\nu}(\tilde{a},b)-c|\ge |\xi_\nu(\hat{a}_i,b)-c|- |\xi_{t_i+\nu}(\tilde{a},b)-\xi_\nu(\hat{a}_i,b)|\ge \tfrac12|\xi_\nu(\hat{a}_i,b)-c| \end{align*} $$

and (6.9) follows.

If $\nu <\nu _0$ then by (4.3) and (5.12),

$$ \begin{align*} |\xi_{t_i+\nu}(\tilde{a},b)-\xi_\nu(\hat{a}_i,b)| &\le |\xi_{t_i+\nu}(\tilde{a},b)-\xi_\nu(\tilde{a},b)|+ |\xi_\nu(\tilde{a},b)-\xi_\nu(\hat{a}_i,b)|\\ &\le 4^\nu|\xi_{t_i}(\tilde{a},b)-c|+4^\nu|\tilde{a}-\hat{a}_i|\le 4^\nu(\delta+|\omega_i|). \end{align*} $$

By Lemma 5.1(b) for $j=n_0$ and by making $n_0$ sufficiently large, we get $\partial _a\xi _j(a,b)\ge 1$ . Therefore,

$$ \begin{align*} |\omega|\le |\xi_{n_0}(\omega,b)|\le\delta. \end{align*} $$

Thus, $|\omega _i|\le \delta $ , and we get

$$ \begin{align*} |\xi_{t_i+\nu}(\tilde{a},b)-\xi_\nu(\hat{a}_i,b)|\le 2\delta\cdot 4^\nu. \end{align*} $$

Together with (6.11) and (6.12) in this case we also get (6.13), and (6.9) follows.

Now for each $\nu $ we choose $\tilde {a}_\nu $ between a and $a'$ , so that $\xi _{t_i+\nu }(\tilde {a}_\nu ,b)=\eta _{t_i+\nu }$ . Thus, by (6.9) and (6.12),

(6.14)

By (6.2) we have

(6.15)

By the definition of bound periods and the definition of $\Omega _{t_i+\nu }$ , we have

$$ \begin{align*} |\Omega_{t_i+\nu}|\le e^{-4\sqrt{\nu}}. \end{align*} $$

Moreover,

$$ \begin{align*} |\Omega_{t_i}|=e^{-r_i-1}-e^{-r_i-2}. \end{align*} $$

Substituting those two inequalities, (6.14), (6.7) and (6.4) into the right-hand side of (6.15), we get

for some constant

. This implies that there is a constant

such that

Together with the estimate on the first term of $S_i'$ , that we obtained long ago, we get a constant

such that

(6.16)

Note that

depends on

, but not on $\delta $ .

To estimate $S_i^{\prime \prime }$ , we use Lemma 4.12, and immediately get

However, $\delta>e^{-r_{i+1}}$ , so

(6.17)

This estimate also applies to $S_0^{\prime \prime }$ .

By Lemma 3.4 applied to a subinterval $\omega '=[a,a']$ (or $[a',a]$ ) of $\omega $ (made possible by Remark 5.13) and (5.36) (see Remark 5.12), we get

As we already noticed in the proof of Lemma 5.11, by taking $\delta $ sufficiently small we can make $p_i$ as large as we need and we may assume that

and therefore we get

(6.18) $$ \begin{align} \sigma_{i+1}\ge2\sigma_i. \end{align} $$

We are now ready to estimate the logarithm of the left-hand side of (6.1), which is less then or equal to $\sum _{i=0}^{m-1}(S_i'+S_i^{\prime \prime })$ . By (6.16) and (6.17), we get

Rearrange the sum $\sum _{i=0}^m\sigma _i/e^{-r_i}$ and group it according to the values of $r_i$ . Set $W_k=\{i\in [1,m]:r_i=k\}$ . Consider k such that $W_k$ is non-empty. Then we can write $W_k=\{i_s<i_{s-1}<\cdots <i_0\}$ , and by (6.18) we have $\sigma _{i_j}\le \sigma _{i_0}/2^j$ . Thus,

$$ \begin{align*} \sum_{i\in W_k}\frac{\sigma_i}{e^{-r_i}}\le2\frac{\sigma_{\mu_k}} {e^{-k}}, \end{align*} $$

where $\mu _k$ is the largest element of $W_k$ . However, $\sigma _{\mu _k}$ is the length of an interval which is contained in the union of three subintervals of $I_k$ , and the length of each of those subintervals is $|I_k|/k^2$ . Moreover, $|I_k|<e^{-k}$ . Thus,

(6.19) $$ \begin{align} \sum_{i\in W_k}\frac{\sigma_i}{e^{-r_i}}\le\frac6{k^2}. \end{align} $$

If $W_k$ is empty, then of course (6.19) also holds. In this way we get

The right-hand side of the above inequality is finite, so we can denote its exponential by

and then (6.1) holds.

Proof. The lemma follows immediately from Lemma 6.1, Corollary 3.2 and induction statement (iii).

Outline of proof of Proposition 7.1.

The proof of Proposition 7.1 is based on induction. Note that the Cantor set construction can be stopped at a finite stage $\hat {N}$ , which is defined by the relation

After this time the term $2-2b$ of equation (4.12) dominates in the derivative and we conclude that for all $a\in \tilde {E}_b$ there is a constant $C>0$ such that

(7.7) $$ \begin{align} (f^n_{a,b})'(f_{a,b}(c))\geq Ce^{\tilde{\kappa}n}\quad \text{for all } n\geq 0. \end{align} $$

A more general result with a more detailed proof is given in Proposition 8.1.

Proof. The proof is initially the same as the proof of Proposition 7.1 .

As before, we carry out the construction only until time $\hat {N}$ . Here $\hat {N}$ is the smallest integer satisfying

$$ \begin{align*} e^{-\sqrt{ \hat{N}}}\leq d. \end{align*} $$

At time $\hat {N}$ we have a partition ${\mathcal P}_{\hat {N}}$ consisting of finitely many intervals $\{\omega _j\}_{j=1}^{M_{\hat {N}}}$ .

We now aim to prove that the hyperbolicity statement (8.1) is true.

We first recall the two outside expansion statements of Lemma 4.9. Suppose that $(a,b)\in {\mathcal N}$ and chose $I=I^*=(c-\delta ,c+\delta )$ in that lemma. Then the following assertions hold.

1. (1) If $x,\ f_{a,b}x,\ldots , f_{a,b}^{n-1}x\not \in I^*$ and $f_{a,b}^n x\in I^*$ , then

   

2. (2) There is an integer M such that if $x, f_{a,b}x,\ldots ,f_{a,b}^{M-1}x\not \in I^*$ then

   

Proof of Theorem B

Fix $b<1$ . Each tongue is open, so the set $T_b$ is open. Therefore, it is the union of countably many components, each of them an open interval. Since the points on the boundary of a tongue belong to $TN$ , and the sets T and $TN$ are disjoint, each component is contained in one tongue.

We claim that the intersection of the closures of two distinct components $A_1$ and $A_2$ is empty. Suppose it is not and that a belongs to this intersection. Then $(a,b)\in TN$ , so it has its type. This type must be the same as the type of each of the tongues containing $A_1$ and $A_2$ , so those types are the same, that is, $A_1$ and $A_2$ are contained in the same tongue. If n is the period of the neutral periodic orbit of $f_{a,b}$ , the map $f_{a,b}^n$ has an interval on which it looks like one of Cases 1, 2 or 4 of Lemma 4.1 of [Reference Misiurewicz and Rodrigues19]. By Theorem 4.1 and Lemma 2.6 of [Reference Misiurewicz and Rodrigues20], this cannot be Case 4 (a neutral periodic point repelling from both sides), and by Lemma 4.2 of [Reference Misiurewicz and Rodrigues19] it cannot be Case 1 or 2 (a neutral periodic point repelling from one side). This proves our claim.

If a parameter $a\in TN_b$ does not belong to a boundary of a component of $T_b$ , then by Lemma 4.2 of [Reference Misiurewicz and Rodrigues19] the neutral periodic orbit of $f_{a,b}$ is repelling from both sides (Case 4), so by Theorem 4.1 and Lemma 2.6 of [Reference Misiurewicz and Rodrigues20] a is isolated in the set of elements of $T_b\cup TN_b$ which have type of the same period. This proves that there are only countably many such values of a.

By the claim, the complement of $T_b$ is a closed set without isolated points. The set $TN_b$ is countable. Therefore $E_b$ (which is the complement of $T_b$ minus $TN_b$ ) is dense in the complement of $T_b$ .

The second part of the statement follows from the first and the fact that each component of $T_b$ is contained in one tongue.