Proof. The desired bound can be proved by a recursive estimate. Note that we can write

(2.2) $$ \begin{align} S_{n+1} = \sum_{\mathcal{C}_{n+1}\ni J\subset \Delta_{0,n}\setminus \Delta_{0,n+1}} \frac{|J|}{d(c,J)} + \sum_{\mathcal{C}_n\ni \Delta\neq \Delta_{0,n}}\bigg( \sum_{\mathcal{C}_{n+1}\ni J\subset \Delta} \frac{|J|}{d(c,J)} \bigg). \end{align} $$

Now, since $d(c,J)> \tfrac 12|\Delta _{0,n+1}|$ for each $J\in \mathcal {C}_{n+1}$ , we certainly have

(2.3) $$ \begin{align} \sum_{\mathcal{C}_{n+1}\ni J\subset \Delta_{0,n}\setminus \Delta_{0,n+1}} \frac{|J|}{d(c,J)} \leq 2\frac{|\Delta_{0,n}|}{|\Delta_{0,n+1}|}. \end{align} $$

From the real bounds, Theorem 2.1, we know that there exists a constant $0<\alpha =\alpha (N)<1$ such that $|\Delta _{0,n}|\leq \alpha ^{-1}|\Delta _{0,n+1}|$ for all sufficiently large n. For each $\Delta \in \mathcal {C}_n$ , let $J_1,J_2,\ldots , J_{a_n}\in \mathcal {C}_{n+1}$ be all the intervals at level $n+1$ which are contained in $\Delta $ . Then, again from the real bounds, we have $\sum _{i=1}^{a_n}|J_i|\leq \beta |\Delta |$ , where $0<\beta =\beta (N)<1$ , provided the renormalization level n is sufficiently large. Moreover, $d(c,J_i)\geq d(c,\Delta )$ for all i. Hence, we have for all n sufficiently large,

(2.4) $$ \begin{align} \sum_{\mathcal{C}_n\ni \Delta\neq \Delta_{0,n}}\bigg( \sum_{\mathcal{C}_{n+1}\ni J\subset \Delta} \frac{|J|}{d(c,J)} \bigg) &\leq \sum_{\mathcal{C}_n\ni \Delta\neq \Delta_{0,n}}\bigg(\frac{\sum_{\mathcal{C}_{n+1}\ni J\subset \Delta}|J|}{d(c,\Delta)}\bigg) \nonumber\\ &\leq \beta \sum_{\mathcal{C}_n\ni \Delta\neq \Delta_{0,n}}\frac{|\Delta|}{d(c,\Delta)} = \beta S_n. \end{align} $$

Putting equations (2.3) and (2.4) back into equation (2.2), we deduce that there exists $n_0=n_0(f)$ such that $S_{n+1}\leq \beta S_n +2\alpha ^{-1}$ for all $n\geq n_0$ . By induction, it follows that $S_{n_0+k}\leq \beta ^k S_{n_0} + 2\alpha ^{-1}(1+\beta +\cdots + \beta ^{k-1})$ for all $k\geq 0$ . Since $\beta <1$ , this shows that the sequence $(S_n)_{n\geq 1}$ is bounded, and eventually universally so.

Proof. Since f has a critical point of order d at c, we have $|f'(x)|\geq C_0|x-c|^{d-1}$ for all $x\in I$ , for some $C_0=C_0(f)>0$ . Replacing, if necessary, f by $R^kf$ for sufficiently large k, we can assume that $C_0$ depends in fact only on N. Now, for each i, we can write $|\Delta _{i+1,n}|/|\Delta _{i,n}|=|f'(x_{i,n})|$ for some $x_{i,n}\in \Delta _{i,n}$ by the mean-value theorem. Hence, using that $|x_{i,n}-c|\geq d(c,\Delta _{i,n})$ , we have

$$ \begin{align*} &\frac{|\Delta_{i,n}|^2}{|\Delta_{i+1,n}|} [d(c,\Delta_{i,n})]^{d-2} = \frac{|\Delta_{i,n}|}{|f'(x_{i,n})|} [d(c,\Delta_{i,n})]^{d-2}\\ &\quad\leq C_0^{-1} \frac{|\Delta_{i,n}|}{|x_{i,n}-c|} \leq C_0^{-1} \frac{|\Delta_{i,n}|}{d(c,\Delta_{i,n})}. \end{align*} $$

This shows that $S_n^*\leq C_0^{-1}S_n$ for all (sufficiently large) n, and the desired result follows from Lemma 2.3.

Proof. This easily established from equation (3.1) by induction (write $\phi ^{k+1}=\phi \circ \phi ^k$ for the induction step).

Proof. Look at the inclusions $D(w,|\mathrm {Im}\,w|)\subseteq Y\subseteq \mathbb {C}\setminus \mathbb {R}$ and use equation (4.1) with $z_0=w$ and $R=|\mathrm {Im}\,w|$ .

Proof. Cover W with a finite number m of non-overlapping closed squares $Q_j$ , $1\leq j\leq m$ , with each $Q_j$ having the same side $\ell =(2\sqrt {2} M)^{-1}$ , and take m to be the smallest possible. Then , the diameter of $Q_j$ is $(2M)^{-1}$ and $\mathrm {dist}(Q_j, \partial V)\geq (2M)^{-1}$ for each $1\leq j\leq m$ . Since the total area of these squares cannot exceed the area of V, which is less than $\pi M^2$ , we see that $m<8\pi M^4$ . Moreover, from Koebe’s estimate in equation (4.7), we have for each j,

$$ \begin{align*} N_\varphi(Q_j)\leq (2M)^{-1}\cdot \frac{4}{(2M)^{-1}} = 4. \end{align*} $$

Now, since W is connected, given any pair of points $z,w\in W$ , we can join them by a chain of pairwise distinct squares $Q_{j_1}, Q_{j_2}, \ldots , Q_{j_n}$ such that , with $z\in Q_{j_1}$ and $w\in Q_{j_n}$ , say. Choose $z_k\in Q_{j_k}\cap Q_{j_{k+1}}$ for $k=1,2,\ldots , n-1$ , and set $z_0=z, z_n=w$ . Use equation (4.8) to get

$$ \begin{align*} \bigg|\frac{\varphi'(z)}{\varphi'(w)}\bigg| &= \prod_{k=0}^{n-1} \bigg|\frac{\varphi'(z_k)}{\varphi'(z_{k+1})}\bigg| \\ & \leq \exp{\bigg(\sum_{k=1}^{n} N_{\varphi}(Q_{j_k})\bigg)} \leq e^{4m}. \end{align*} $$

This establishes the upper bound in equation (4.9); the lower bound is obtained in the same way, or simply by interchanging z and w. Hence, assertion (i) is proved. Assertion (ii) follows from assertion (i) and the inequality in equation (4.6).

Proof. We may assume that $z_0=0$ (otherwise we simply conjugate $\psi $ by the translation $z{\kern-1pt}\mapsto{\kern-1pt} z{\kern-1pt}-{\kern-1pt}z_0$ and work with the resulting map, which fixes $0$ ). For each $t{\kern-1pt}\in{\kern-1pt} \mathbb {D}$ , let $\varphi _t{\kern-1pt}:{\kern-1pt} \mathbb {C}{\kern-1pt}\to{\kern-1pt} \mathbb {C}$ be the unique solution to the Beltrami equation

normalized so that $\varphi _t$ fixes $0,1$ and $\infty $ . Define $\psi _t: \mathbb {C}\to \mathbb {C}$ by the formula

(4.11) $$ \begin{align} \psi_t(\zeta) =\bigg[ 1+ \frac{t}{k}( \psi(1)-1 ) \bigg] \varphi_t(\zeta). \end{align} $$

Note that $\psi _t(0)=0$ for all t. Also, for $t=k$ , we have $\psi _k(\zeta )=\psi (1)\varphi _k(\zeta )$ , so $\psi _k(1)=\psi (1)$ . Since the Beltrami form of $\psi _k$ is the same as the Beltrami form of $\varphi _k$ , which is $\mu _\psi $ , it follows from uniqueness of normalized solutions to the Beltrami equation that $\psi _k=\psi $ . This proves item (i).

Applying Lemma 4.5 to $\phi =\varphi _t$ , $z=0$ and $s=1$ , we see that for all $0<r<1$ ,

$$ \begin{align*} \max_{|\zeta|=r}|\varphi_t(\zeta)| \leq e^{\pi K_t}r^{1/K_t}, \end{align*} $$

where $K_t$ is the maximal dilatation of $\varphi _t$ , which satisfies

$$ \begin{align*} K_t \leq \frac{1+|t|}{1-|t|}. \end{align*} $$

In particular, since $K_t<3$ for all t with $|t|<\tfrac 12$ , we have

(4.12) $$ \begin{align} \varphi_t(D(0,r))\subseteq D(0,e^{3\pi}r^{1/3}). \end{align} $$

Let us now estimate the scaling factor multiplying $\varphi _t(\zeta )$ on the right-hand side of equation (4.11). Applying Lemma 4.5 with $\phi =\psi $ , $z=0$ , $s=r_0$ and $r=r_0^{-1}$ , and taking into account that the maximal dilatation of $\psi $ is less than $3$ , we get

$$ \begin{align*} \max_{|\zeta|=1}{|\psi(\zeta)|}&\leq e^{3\pi}\frac{1}{r_0^3}\min_{|\zeta|=r_0}{|\psi(\zeta)|} \\ &\leq e^{3\pi}\frac{1}{r_0^3}(Mr_0) = \frac{Me^{3\pi}}{r_0^2}. \end{align*} $$

In particular, $|\psi (1)-1|\leq 2Me^{3\pi }r_0^{-2}$ , and therefore

$$ \begin{align*} \bigg|1+\frac{t}{k}(\psi(1)-1)\bigg| \leq \frac{2Me^{3\pi}}{kr_0^2} \end{align*} $$

for all t with $|t|\leq \tfrac 12$ . Combining this fact with equation (4.12), it follows that for all such t, we have

$$ \begin{align*} \max_{|\zeta|\leq r}|\psi_t(\zeta)| \leq \frac{2Me^{6\pi}r^{1/3}}{kr_0^2}. \end{align*} $$

Therefore, $\psi _t(D(0,r)) \subseteq D(0,R)$ for all t with $|t|\leq \tfrac 12$ and all $0<r<1$ , where R is given by equation (4.10). This proves item (ii).

Proof. We may assume that z is real and non-negative, say $z=x\in [0,1)$ . Let $\varphi \in \mathrm {Aut}(\mathbb {D})$ be given by

$$ \begin{align*} \varphi(\zeta) = \frac{\zeta -x}{1-x\zeta} \end{align*} $$

and define

$$ \begin{align*} \alpha = \varphi(x-r) =\frac{-r}{1-x^2+rx};\quad \beta = \varphi(x+r) =\frac{r}{1-x^2-rx}. \end{align*} $$

Then $D_r'=\varphi (D(x,r))$ is a disk with diameter $(\alpha ,\beta )\subset (-1,1)$ . Since $|\alpha |\leq \beta $ , we see that $D_r'\supseteq D(0, |\alpha |)$ . Therefore,

$$ \begin{align*} \mathrm{mod}(\mathbb{D}\setminus D(x,r))&= \mathrm{mod}(\mathbb{D}\setminus D_r') \\ &\leq \mathrm{mod}(\mathbb{D}\setminus D(0,|\alpha|)) = \log{\frac{1}{|\alpha|}}\\ &=\log{\frac{1-x^2+rx}{r}}, \end{align*} $$

and this finishes the proof.

Proof. Write $a=|z|$ and $b=|w|$ so that $0\leq a,b<1$ . We have $1-a^2=\rho _{\mathbb {D}}(z)^{-1}$ and $1-b^2=\rho _{\mathbb {D}}(w)^{-1}$ , so equation (4.15) tells us that

(4.16) $$ \begin{align} \alpha^{-1} \leq \frac{1-a^2}{1-b^2} \leq \alpha. \end{align} $$

Let $\varphi \in \mathrm {Aut}(\mathbb {D})$ be the hyperbolic translation with axis $(-1,1)\subset \mathbb {D}$ such that $\varphi (a)=b$ . Then,

$$ \begin{align*} \varphi(\zeta) = \frac{\zeta-c}{1-c\zeta} , \end{align*} $$

where $c=(a-b)/(1-ab)\in (-1,1)$ , as a simple calculation shows. Moreover, we have

(4.17) $$ \begin{align} \varphi'(\zeta) =\frac{1-c^2}{(1-c\zeta)^2} \end{align} $$

as well as

(4.18) $$ \begin{align} \varphi"(\zeta) =\frac{2c(1-c^2)}{(1-c\zeta)^3}. \end{align} $$

Since $1-c^2 = (1-a^2)(1-b^2)/(1-ab)^2$ and since $\min \{1-a^2,1-b^2\}\leq 1-ab\leq \max \{1-a^2,1-b^2\}$ , it follows from equation (4.16) that

(4.19) $$ \begin{align} \alpha^{-1} \leq 1-c^2 \leq 1. \end{align} $$

Now, if $\zeta \in \Delta _a$ , then $|\zeta |\leq (1+a)/2$ . Hence,

$$ \begin{align*} |1-c\zeta| \geq 1-|c|\bigg(\frac{1+a}{2}\bigg) = \frac{1-|c|}{2}+\frac{1-|c|a}{2}> \frac{1-|c|a}{2}. \end{align*} $$

Here, there are two cases to consider. If $a{\kern-1pt}\geq{\kern-1pt} b$ , then $c{\kern-1pt}\geq{\kern-1pt} 0$ and $1{\kern-1pt}-{\kern-1pt}|c|a{\kern-1pt}={\kern-1pt}1{\kern-1pt}-{\kern-1pt}ca{\kern-1pt}={\kern-1pt}(1{\kern-1pt}-{\kern-1pt}a^2)/ (1-ab)$ , so from equation (4.16), we deduce that $1-|c|a\geq {\alpha ^{-1}}$ . If however $a< b$ , then $c< 0$ , and in this case, we see that

$$ \begin{align*} 1-|c|a = \frac{1-b^2+(b-a)^2}{1-ab}> \frac{1-b^2}{1-a^2} \geq \alpha^{-1}, \end{align*} $$

where once again we have used equation (4.16). Thus, in either case, we have

(4.20) $$ \begin{align} \frac{1}{2\alpha} \leq|1-c\zeta|<2 \quad\textrm{for all } \zeta\in \Delta_a. \end{align} $$

Using both equations (4.19) and (4.20) in equations (4.17) and(4.18), we easily arrive at inequalities (i) and (ii) with $\varphi $ replacing $\psi $ (and $\Delta _a$ replacing $\Delta _z$ ). Finally, we define $\psi =R_b\circ \varphi \circ R_a$ , where $R_a$ is the rigid rotation around $0$ with $R_a(z)=a$ , and $R_b$ is the rigid rotation around $0$ with $R_b(b)=w$ . Then $\psi (z)=w$ , and since $R_a,R_b$ are Euclidean isometries and $R_a(\Delta _z)=\Delta _a$ , the inequalities (i) and (ii) for $\psi $ follow from the corresponding inequalities for $\varphi $ .

Proof. The basic geometric idea behind the proof is to use macroscopic estimates on the moduli of certain annuli to bound a microscopic quantity, namely the hyperbolic Jacobian at z. Rotating the coordinate axes if necessary, we may also assume that $D\phi (z)=S\cdot T$ , where $S=\rho I=(\begin {smallmatrix} \rho & 0\\0 & \rho \end {smallmatrix})$ , for some $\rho>0$ , and $T=(\begin {smallmatrix} \unicode{x3bb} & b\\ 0 & \unicode{x3bb} ^{-1}\end {smallmatrix})$ , where $\unicode{x3bb} \geq 1$ and $b\in \mathbb {R}$ . Here, we obviously have $\rho ^2=\det {D\phi (z)}=J_\phi (z)=J_\phi ^h(z)$ . We shall prove the lemma only in the case when $b=0$ and $\unicode{x3bb}>1$ . The cases when $b\neq 0$ and/or $\unicode{x3bb} =1$ are similarly handled. Note that the linear map $D\phi (z)$ maps the circle of radius $1$ about the origin onto an ellipse with major axis $\rho \unicode{x3bb} $ and minor axis $\rho /\unicode{x3bb} $ . Since $\phi $ is $(1+\epsilon )$ -qc, we have $\unicode{x3bb} ^2\leq 1+\epsilon $ . In what follows, we assume that $\rho>\unicode{x3bb} +\epsilon $ , as otherwise, $\rho ^2\leq (\unicode{x3bb} +\epsilon )^2\leq 1+6\epsilon $ and there is nothing to prove.

If $\zeta $ is such that $|\zeta -z|\leq \delta r_z$ , we can write, using Taylor’s formula and the fact that $\phi (z)=z$ ,

(4.22) $$ \begin{align} \phi(\zeta) =z+D\phi(z)\cdot (\zeta -z) + R_\phi(\zeta), \end{align} $$

where the remainder $R_\phi (\zeta )$ satisfies $|R_\phi (\zeta )|\leq C|\zeta -z|^2$ , with $C=C_0 m_\phi (z,\delta )>0$ (and $C_0>0$ an absolute constant). Let us choose $0< r\leq \delta r_z$ so small that

(4.23) $$ \begin{align} \frac{\rho}{\unicode{x3bb}}r -Cr^2> \frac{\rho}{\unicode{x3bb}+\epsilon}r. \end{align} $$

For definiteness, we take

(4.24) $$ \begin{align} r=\min\bigg\{\delta r_z,\frac{\rho\epsilon}{C\unicode{x3bb}^2(\unicode{x3bb}+\epsilon)}\bigg\}. \end{align} $$

Then equations (4.22) and (4.23) tell us that $\phi $ maps the disk $D(z,r)$ onto a Jordan domain $V_r$ which contains that disk and also the round annulus . Setting , we have , and so

(4.25)

Consider the images of under the forward iterates of $\phi $ , i.e. , $n\geq 0$ . The annuli are pairwise disjoint, and . By sub-additivity of the modulus, we have

(4.26)

Now, since $\phi $ is $(1+\epsilon )$ -qc, we know that $\phi ^n$ is $(1+\epsilon )^n$ -qc, and therefore,

(4.27)

Putting together equations (4.25), (4.26) and (4.27), we get

(4.28) $$ \begin{align} \log{\bigg( \frac{\rho}{\unicode{x3bb}+\epsilon} \bigg)} \sum_{n=0}^{\infty} \frac{1}{(1+\epsilon)^n} \leq\mu_r. \end{align} $$

Applying Lemma 4.10 and Remark 4.11 to our r, as defined in equation (4.24), we see that

(4.29) $$ \begin{align} \mu_r \leq\begin{cases} \log{\bigg(\dfrac{5}{\delta} \bigg)} & {\textrm{ when }} r=\delta r_z;\\[8pt] \log{\bigg(\dfrac{2C\unicode{x3bb}^2(\unicode{x3bb}+\epsilon)}{\rho\epsilon} \bigg)} & {\textrm{ when }} r=\dfrac{\rho\epsilon}{C\unicode{x3bb}^2(\unicode{x3bb}+\epsilon)}. \end{cases} \end{align} $$

Regardless of which of the two cases occur, we certainly have

(4.30) $$ \begin{align} \mu_r \leq \log{\bigg(\frac{10C\unicode{x3bb}^2(\unicode{x3bb}+\epsilon)}{\delta\rho\epsilon}\bigg)}< \log{\bigg(\frac{60C}{\delta\epsilon}\bigg)}, \end{align} $$

where in the last step, we have used that $\unicode{x3bb} ^2(\unicode{x3bb} +\epsilon )<6$ and $\rho>1$ . Combining equations (4.28) and (4.30), we deduce that

(4.31) $$ \begin{align} \log{\bigg( \frac{\rho}{\unicode{x3bb}+\epsilon} \bigg)}&\leq \frac{\epsilon}{1+\epsilon}\log{\bigg(\frac{60C}{\delta\epsilon}\bigg)} \nonumber \\ & < \epsilon \log{\bigg(\frac{60C}{\delta}\bigg)} + \epsilon \log{\frac{1}{\epsilon}}. \end{align} $$

Since $0<\epsilon <1$ , we have $\epsilon <\epsilon ^{1-\theta }$ and $\epsilon ^{\theta }\log {({1}/{\epsilon })}\leq (\theta e)^{-1}$ . Using these facts in equation (4.31), we get

(4.32) $$ \begin{align} \rho&\leq (\unicode{x3bb}+\epsilon)\exp\bigg\{\bigg(\frac{1}{\theta e}+ \log{\frac{60C}{\delta}}\bigg)\epsilon^{1-\theta}\bigg\} \end{align} $$

(4.33) $$ \begin{align} &\leq 1+ \bigg(2+180e^{{1}/{\theta e}}\frac{C}{\delta}\bigg)\epsilon^{1-\theta}, \end{align} $$

where we have used that $\unicode{x3bb} +\epsilon \leq 1+2\epsilon $ . From this, and the fact that $C=C_0 m_\phi (z,\delta )$ , it readily follows that

$$ \begin{align*} J_\phi^h(z) = \rho^2\leq 1+ 3\bigg(2+180e^{{1}/{\theta e}}C_0\frac{m_\phi(z,\delta)}{\delta}\bigg)^2 \epsilon^{1-\theta}. \end{align*} $$

This proves equation (4.21), provided we take $B_\theta (t)=3(2+ 180e^{{1}/{\theta e}}C_0t)^2$ .

Proof of Proposition 4.9

The idea, of course, is to reduce the required estimate to the case treated in Lemma 4.13. Let $\psi \in \mathrm {Aut}(\mathbb {D})$ be the conformal automorphism given by Lemma 4.12, with $\psi (z)=w=\phi (z)$ . Then the diffeomorphism $F=\psi ^{-1}\circ \phi :\mathbb {D}\to \mathbb {D}$ has a fixed point at z. Since $\psi ^{-1}$ is an isometry of the hyperbolic metric, we certainly have $J_F^h(z)=J_\phi ^h(z)$ . We would like to estimate $J_F^h(z)$ using Lemma 4.13. For this, we need an estimate on the $C^2$ norm of the composition $\psi ^{-1}\circ \phi $ in a suitable disk around z. By Koebe’s one-quarter theorem, $\psi (\Delta _z)$ contains the disk

$$ \begin{align*} D =\big\{ \zeta: |\zeta -w|<\tfrac{1}{4}|\psi'(z)|\cdot r_z \big\}. \end{align*} $$

Since we know from Lemma 4.12(i) that $|\psi '(z)|\geq (2\alpha )^{-1}$ , it follows that $\psi (\Delta _z)\supset D(w,R)$ , where $R=r_z/8\alpha $ . Now let us define

$$ \begin{align*}\delta= \frac{1}{8\alpha m_\phi(z)}\quad \textrm{and}\quad M=\sup_{\zeta\in \Delta_z}|D\phi(\zeta)| \leq m_\phi(z).\end{align*} $$

Then we have $\phi (D(z,\delta r_z))\subset D(w,M\delta r_z)\subseteq D(w,R)\subset \psi (\Delta _z)$ . We can now estimate the $C^2$ norm of F restricted to the disk $D(z,\delta r_z)$ , i.e. we can estimate $m_F(z,\delta )$ , with the help of Lemma 4.12. We do this by means of the following two steps.

1. (i) By the chain rule for first derivatives, we have $DF=D\psi ^{-1}\circ \phi \cdot D\phi $ . Since $\psi ^{-1}$ is holomorphic, for each $\zeta \in D(z,\delta r_z)$ , we have

   (4.34) $$ \begin{align} \|D\psi^{-1}(\phi(\zeta))\|\leq |(\psi^{-1})'(\phi(\zeta))| = |\psi'(\psi^{-1}\circ\phi(\zeta))|^{-1} \leq 2\alpha. \end{align} $$
   Hence, the $C^0$ norm of $DF$ in $D(z,\delta r_z)$ is bounded by $2\alpha m_\phi (z)$ .

2. (ii) By the chain rule for second derivatives, we have

   (4.35)

   
   Again, since $\psi ^{-1}$ is holomorphic, a simple calculation shows that
   $$ \begin{align*} (\psi^{-1})" = -\frac{\psi"\circ \psi^{-1}}{(\psi'\circ \psi^{-1})^3}. \end{align*} $$
   Therefore, for each $\zeta \in D(z,\delta r_z)$ , we have, with the help of Lemma 4.12,
   (4.36) $$ \begin{align} \|D^2\psi^{-1}(\phi(\zeta))\| \leq |(\psi^{-1})"(\phi(z))| \leq 128\alpha^6. \end{align} $$
   Using equations (4.34), (4.36) and the fact that in equation (4.35), we deduce that the $C^0$ norm of $D^2F$ in the disk $D(z,\delta r_z)$ is bounded by $(128\alpha ^6+2\alpha )m_\phi (z)<130\alpha ^6m_\phi (z)$ .

From steps (i) and (ii) above, we deduce that $m_F(z,\delta )\leq 130\alpha ^6m_\phi (z)$ . Therefore, applying Lemma 4.13 for F yields

$$ \begin{align*} J_\phi^h(z) =J_F^h(z) \leq 1+ B_{\theta}\bigg(\frac{m_F(z,\delta)}{\delta}\bigg)\epsilon^{1-\theta} \leq 1+ B_{\theta}\bigg(1040\alpha^7(m_\phi(z))^2\bigg)\epsilon^{1-\theta}. \end{align*} $$

This completes the proof of our theorem, provided we take $A_{\theta }(s,t)=B_{\theta }(1040s^7t^2)$ .

Proof. We present the proof of the required estimate under the additional assumption that z is a fixed-point of $\phi $ . The general case can be reduced to this one by post-composing $\phi $ with a suitable conformal automorphism of the unit disk, and proceeding just as in the proof of Proposition 4.9, mutatis mutandis. For the sake of clarity of exposition, we divide the proof into a series of steps.

1. (i) First, we introduce some notation. Throughout the proof, we denote by $c_0,c_1,\ldots $ positive constants that are either absolute or depend on the given constants $\alpha , \beta , b_0, M$ , where $M=m_\phi (z)$ . Let us write $\epsilon =b_0(1-|z|)^{\,\beta }= (b_02^{\,\beta })r_z^{\,\beta }$ . Also, let $k_0=\sup _{\zeta \in \Delta _z}|\mu _\phi (\zeta )|\leq \epsilon $ , and set $r_0=\epsilon r_z$ . We may assume without loss of generality that $\epsilon $ is small, say $\epsilon <1/{32}$ .

2. (ii) The restricted map $\phi |_{\Delta _z}: \Delta _z\to \mathbb {D}$ is a $({1+k_0})/({1-k_0})$ -quasiconformal embedding. By Lemma 4.6, the further restriction $\phi |_{D(z,r_0)}$ can be extended to a global quasiconformal homeomorphism $\psi :\mathbb {C}\to \mathbb {C}$ with $k=\|\mu _\psi \|_{\infty }$ satisfying

   $$ \begin{align*} \frac{1+k}{1-k}\leq \frac{1+\epsilon}{1-\epsilon}\cdot \frac{1+k_0}{1-k_0} \leq \bigg(\frac{1+\epsilon}{1-\epsilon}\bigg)^2. \end{align*} $$

3. (iii) In particular, $k\leq 16\epsilon <\tfrac 12$ (by our assumption on $\epsilon $ in step (i)). We may assume that $k\neq 0$ (if this is not the case, it is easy to perturb $\psi $ slightly in a neighbourhood of infinity). By Lemma 4.8(i), there exists a global holomorphic motion $\psi _t : \mathbb {C}\to \mathbb {C}$ with $\psi _k=\psi $ and $\psi _t(z)=z$ for all $t\in \mathbb {D}$ . Now choose $r_1>0$ so small that

   $$ \begin{align*} R = \frac{2Me^{6\pi}}{k_0r_0^2}\cdot r_1^{1/3}< r_z. \end{align*} $$
   For definiteness, take $r_1=c_1k^3r_z^{6\,\beta +9}$ , where $c_1=b_0^6/(M^3e^{18\pi })$ . Then, by Lemma 4.8(ii), we have $\psi _t(D(z,r_1))\subset D(z,R)$ for all t with $|t|<\tfrac 12$ (note that this includes the time $t=k$ ).

4. (iv) We may now define, for each $t{\kern-1pt}\in{\kern-1pt} D(0,\tfrac 12)$ , the map $\widetilde {\psi _t}{\kern-1pt}:{\kern-1pt} D(z,r_1)\cup (\mathbb {C}{\kern-1pt}\setminus{\kern-1pt} \mathbb {D})\to \mathbb {C}$ by

   $$ \begin{align*} \widetilde{\psi_t}(\zeta) = \begin{cases} {\psi_t(\zeta)} &\textrm{for } \zeta\in D(z,r_1), \\ {\zeta} &\textrm{for } \zeta\in\mathbb{C}\setminus \mathbb{D}. \end{cases} \end{align*} $$
   Since $D(z,R)\subset \mathbb {D}$ , we have from step (iii) that . Hence, $\widetilde {\psi _t}$ , $|t|<\tfrac 12$ , is a holomorphic family of injections, that is, a holomorphic motion of the set $D(z,r_1)\cup (\mathbb {C}\setminus \mathbb {D})$ .

5. (v) Now apply Slodkowski’s Theorem 4.7 to get a global extension $\widehat {\psi }_t: \mathbb {C}\to \mathbb {C}$ of the motion $\widetilde {\psi _t}$ , with time parameter t in $D(0,\tfrac 12)$ . In particular, the map $\widehat {\psi }=\widehat {\psi }_k$ is K-quasiconformal with $K=({1+2k})/({1-2k})$ , and it maps the unit disk onto itself. Moreover, we have

   $$ \begin{align*} \widehat{\psi}|_{D(z,r_1)}=\psi|_{D(z,r_1)}=\phi|_{D(z,r_1)}. \end{align*} $$
   Thus, $\widehat {\psi }$ is the desired modification of $\phi $ away from z.

6. (vi) We are now in a position to use the same annulus trick we employed in the proof of Lemma 4.13. Let $\rho>0$ , $\unicode{x3bb}>1$ and the absolute constant $C_0>0$ be as in the proof of that Lemma. In particular, $\rho ^2=J_\phi ^h(z)=J_{\widehat {\psi }}^h(z)$ , and thus our goal is to bound $\rho $ from above. We have $\unicode{x3bb} \leq 1+\epsilon $ , and we may assume that $\rho>\unicode{x3bb} +\epsilon $ , otherwise there is nothing to prove. Now let $r_2>0$ be given by

   $$ \begin{align*} r_2 = \frac{\epsilon}{3C_0M}< \frac{\rho\epsilon}{C_0M\unicode{x3bb}^2(\unicode{x3bb}+\epsilon)}. \end{align*} $$
   Then for all $r\leq r_2$ , the inequality in equation (4.23) holds. Let us choose $r=\min \{r_1,r_2\}$ . With this choice of r, using the Taylor expansion in equation (4.22) as in the proof of Lemma 4.13, we see that is a conformal annulus, with
   (4.40)

   

7. (vii) Now define for all $n\geq 0$ , and note that

   (4.41)

   
   Since , we deduce from equations (4.40) and (4.41) that
   (4.42) $$ \begin{align} \log{\bigg(\frac{\rho}{\unicode{x3bb}+\epsilon}\bigg)} \sum_{n=0}^{\infty} \bigg(\frac{1-2k}{1+2k}\bigg)^n \leq \log{\frac{2}{r}}, \end{align} $$
   where we have used the estimate on $\mod {(\mathbb {D}\setminus D(z,r))}$ given by Lemma 4.10 (and Remark 4.11). From equation (4.42), it follows that
   (4.43) $$ \begin{align} \log{\bigg(\frac{\rho}{\unicode{x3bb}+\epsilon}\bigg)} \leq \frac{4k}{1+2k}\log{\frac{2}{r}}<4k\log{\frac{2}{r}}. \end{align} $$

8. (vii) However, from our choices of $r_1$ and $r_2$ , we see that $r=\min \{r_1,r_2\}=c_2k^3r_z^{6\,\beta +9}$ for some constant $c_2>0$ . Hence,

   $$ \begin{align*} \log{\frac{2}{r}} \leq \log{\frac{2}{c_2}} +3\log{\frac{1}{k}}+(6\,\beta+9)\log{\frac{1}{r_z}}. \end{align*} $$
   Putting this back into equation (4.43) and using that $k\leq \mathrm {(const.)}r_z^{\,\beta }$ , we deduce that for each $0<\theta <1$ ,
   $$ \begin{align*} \log{\bigg(\frac{\rho}{\unicode{x3bb}+\epsilon}\bigg)} &\leq c_3k+c_4k\log{\frac{1}{k}} + c_5k\log{\frac{1}{r_z}} \\ &\leq c_6r_z^{\,\beta(1-\theta)} + c_7r_z^{\,\beta}\log{\frac{1}{r_z}} \\ &\leq c_8 r_z^{\,\beta(1-\theta)}. \end{align*} $$
   Here, the constants $c_6, c_7, c_8$ depend on $M,\beta , b_0$ and also on $\theta $ . From this, it follows that
   $$ \begin{align*} \rho \leq 1+ c_9r_z^{\,\beta(1-\theta)}, \end{align*} $$
   and therefore,
   $$ \begin{align*} J_\phi^h(z) = \rho^2 \leq 1 + c_{10}r_z^{\,\beta(1-\theta)}, \end{align*} $$
   where the constant $c_{10}$ depends on $M,\beta , b_0$ and $\theta $ .

Hence, we have established equation (4.39), with $c_{10}$ playing the role of $C_\theta $ , in the case when z is a fixed-point of $\phi $ . As we already remarked, the general case follows from this one by post-composition of $\phi $ with a suitable automorphism of the disk, using the same procedure given in the proof of Proposition 4.9. It is here, and only here, that equation (4.37) is used. Hence, the final constant $C_\theta $ indeed depends on $M, \alpha , \beta , b_0$ and of course also on $\theta $ . This finishes the proof.

Proof. The hard work has already been done in Propositions 4.9 and 4.14, and all we have to do is to show, with the help of equation (4.44), how to reduce the present theorem to the situation in those auxiliary results. There is no loss of generality in assuming that $\phi $ preserves $Y^+=Y\cap \mathbb {C}^+$ (and therefore also $Y^{-}=Y\cap \mathbb {C}^{-}$ ). Also, it suffices to establish the upper estimates in equations (4.45) and (4.46), since the lower estimates follow by replacing $\phi $ with its inverse. Moreover, by symmetry, we only need to establish these upper estimates for points $z\in U\cap Y^+$ .

Let $(a,b)=V\cap \mathbb {R}$ , and let $\varphi :V\to \widehat {\mathbb {C}}$ be a holomorphic univalent map with $\varphi (Y^+)=\mathbb {D}$ , , normalized so that $\varphi (a)=-1$ , $\varphi (b)=+1$ . Let $W^*=\bigcup _{\zeta \in \varphi (U^+)} \Delta _\zeta \subset \mathbb {D}$ , and consider $W=\varphi ^{-1}(W^*)\subset Y^+$ . Note that $W\supset U^+$ . By Lemma 4.4(ii), the $C^2$ norms of the restrictions $\varphi |_{W}$ and $\varphi ^{-1}|_{\varphi (W^*)}$ are both bounded by a constant that depends only on $\mathrm {dist}(\partial V, \partial W)$ , and it is not difficult (albeit a bit laborious) to see that this last distance is bounded by a constant that depends only on M. These bounds also imply that there exists a constant $K_1>1$ depending only on M such that

(4.47) $$ \begin{align} \frac{1}{K_1}(1-|\varphi(z)|) \leq |\mathrm{Im}\,z| \leq K_1(1-|\varphi(z)|) \end{align} $$

for all $z\in W$ .

Now consider the $C^2$ diffeomorphism $\psi :\mathbb {D}\to \mathbb {D}$ given by $\psi =\varphi \circ \phi \circ \varphi ^{-1}$ . Note that, by the chain rule and the bounds on $\varphi $ , $\varphi ^{-1}$ stated above, the $C^2$ norm of $\psi |_{W^*}$ is also bounded by a constant that depends only on M.

Given a point $z\in Y^+$ and a vector $v\in T_zY^+\equiv T_zY$ , let $\zeta =\varphi (z)\in \mathbb {D}$ and $w=D\varphi (z)v\in T_\zeta \mathbb {D}$ . Since $\varphi $ yields an isometry between the hyperbolic metric of $Y^+$ (that is, of Y) and the hyperbolic metric of $\mathbb {D}$ , we have $|v|_Y=|w|_{\mathbb {D}}$ . Moreover, by the chain rule, we have

$$ \begin{align*} |D\phi(z)v|_Y = |D\varphi^{-1}(\psi(\zeta))\,D\psi(\zeta)w|_Y = |D\psi(\zeta)w|_{\mathbb{D}}, \end{align*} $$

where, in the last step, we have used that $\varphi ^{-1}$ yields an isometry between the hyperbolic metric of $\mathbb {D}$ and the hyperbolic metric of $Y^+$ (and therefore the derivative $D\varphi ^{-1}(\psi (\zeta ))$ is an infinitesimal isometry between corresponding tangent spaces). This shows that for each $z\in Y^+$ and each non-zero tangent vector $v\in T_zY$ , we have

(4.48) $$ \begin{align} \frac{|D\phi(z)v|_Y}{|v|_Y} = \frac{|D\psi(\zeta)w|_{\mathbb{D}}}{|w|_{\mathbb{D}}}. \end{align} $$

In addition, since $\varphi $ and $\varphi ^{-1}$ are conformal, we have that $\psi $ and $\phi $ have the same dilatation at corresponding points, i.e. $K_\psi (\zeta )=K_\phi (z)$ for all $z\in Y^+$ . Also, since $\varphi $ and $\varphi ^{-1}$ are hyperbolic isometries, the hyperbolic Jacobians of $\psi $ and $\phi $ agree on corresponding points, that is, $J_\psi ^h(\zeta )=J_\phi ^h(z)$ .

Putting these facts together, we see that the assertions (i) and (ii) in the statement (that is, the estimates in equations (4.45) and (4.46)) will be proved for $\phi $ as soon as the corresponding assertions for $\psi $ are proved. However, assertion (i) for $\psi $ follows by putting together Proposition 4.9 and equation (4.44), whereas assertion (ii) for $\psi $ follows by putting together Proposition 4.14 and equation (4.44). To see why this is so, we need to check that, in each case, the hypotheses of the corresponding propositions are satisfied by $\psi $ .

Case (i). If $\phi $ is $(1+\delta )$ -quasiconformal, as in assertion (i), then $\psi $ is $(1+\delta )$ -quasiconformal as well. The hypotheses on $\phi $ imply that there exists a constant $K_2>1$ depending only on M such that

(4.49) $$ \begin{align} \frac{1}{K_2} \leq \frac{|\mathrm{Im}\,z|}{|\mathrm{Im}\,\phi(z)|} \leq K_2 \end{align} $$

for all $z\in W$ . Applying this with $z=\varphi ^{-1}(\zeta )$ for $\zeta \in W^*$ and using equation (4.47), we deduce that there exists $K_3>1$ depending only on M such that

$$ \begin{align*} \frac{1}{K_3} \leq \frac{\rho_{\mathbb{D}}(\zeta)}{\rho_{\mathbb{D}}(\psi(\zeta))} \leq K_2 \end{align*} $$

for all $\zeta \in W^*$ . This shows that the inequality of equation (4.13) in the hypothesis of Proposition 4.9 is satisfied for $\psi $ . Moreover, for each $\zeta \in \varphi (U^+)$ , we have $\Delta _\zeta \subset W^*$ , and so, in the notation introduced before, $m_\psi (\zeta )\leq \|\psi |_{W^*}\|_{C^2}\leq K_4$ , where $K_4>0$ is a constant that depends only on M. Hence, all the hypotheses of Proposition 4.9 are satisfied by $\psi $ . It follows that for each $0<\theta <1$ , there exists a constant $K_\theta $ depending only on $\theta $ and M such that

(4.50) $$ \begin{align} J_{\psi}^h(\zeta) \leq 1+ K_\theta \delta^{1-\theta} \end{align} $$

for all $\zeta \in \varphi (U^+)$ . Combining equation (4.50) with the general upper estimate in equation (4.44) (for $\psi $ ), we see that for each $0<\theta <1$ , there exists a constant $C_\theta>0$ depending only on $\theta $ and M such that

(4.51) $$ \begin{align} \frac{|D\psi(\zeta)w|_{\mathbb{D}}}{|w|_{\mathbb{D}}}\leq 1+ C_\theta \delta^{1-\theta} \end{align} $$

for all $\zeta \in \varphi (U^+)$ and each non-zero tangent vector $w\in T_\zeta \mathbb {D}$ . Putting equation (4.51) together with equation (4.48) for $z=\varphi ^{-1}(\zeta )\in U^+$ and $v=D\varphi ^{-1}(\zeta )w\in T_zY^+$ , we deduce the upper estimate in equation (4.45), as desired.

Case (ii). If $\phi $ is asymptotically holomorphic (near the real axis), then so is $\psi $ (near the boundary of the unit disk). Verifying the hypotheses of Proposition 4.14 for $\psi $ in this case is similar to what was done in case (i), and hence we omit the details.

Proof. First, we give an informal description of the argument. For a suitable constant $0<\unicode{x3bb} <1$ , we partition the domain of $f=\phi \circ g$ into a sequence of scales, the nth scale being the set of points in the domain (off the real axis) whose distance to the real axis is of the order $\unicode{x3bb} ^n$ . The rough idea then is that at each level, the worst expansion of the hyperbolic metric of Y by g beats the best contraction of that metric by $\phi $ . In this, we are aided by Theorem 4.15 and Lemma 4.3. We warn the reader that in what follows, whenever invoking Theorem 4.15, we denote by $C_\theta $ the largest of the two constants with that name appearing in assertions (i) and (ii) of said theorem.

Let us now present the formal proof. Let us assume we are given a large number $\alpha>1$ . How large $\alpha $ must be will be determined in the course of the argument.

To start with, note that by equation (4.2) in Lemma 4.2, we have for all $z\in U_\alpha $ ,

(5.2) $$ \begin{align} \frac{1}{|\mathrm{Im}\,z|} \leq \rho_Y(z) \leq \frac{1}{|\mathrm{Im}\,z|}\bigg(1-\frac{1}{2\alpha}\bigg)^{-1}. \end{align} $$

Let us fix for the time being a real number $0<\unicode{x3bb} <1$ , which we will use to define the scales we mentioned above. For definiteness, we take $\unicode{x3bb} =M^{-1}$ . For each $n\geq 1$ , we define

$$ \begin{align*} W_n = \bigg\{z\in U_\alpha : \frac{\unicode{x3bb}^n}{\alpha M}\leq |\mathrm{Im}\,z| < \frac{\unicode{x3bb}^{n-1}}{\alpha M}\bigg\}. \end{align*} $$

Also, we set $W_0=U\setminus U_\alpha \subset Y$ . Then we have, of course, $U\setminus \mathbb {R}= \bigcup _{n=0}^\infty W_n$ .

Claim. There exists a sequence of numbers $\xi _n>1$ , $n\geq 0$ , with $\xi _n\to 1$ as $n\to \infty $ , having the following property: for each $z\in W_n$ and each tangent vector $v\in T_{z}Y$ , we have

(5.3) $$ \begin{align} |D(g\circ \phi)(z)v|_Y \geq \xi_n|v|_Y. \end{align} $$

With the claim at hand, we proceed to the proof of the assertions in the statement of our theorem. Let $z\in \mathcal {K}_f$ be a point whose iterates up to time $n>1$ stay off the real axis—in other words, $f^i(z)\in Y$ for all $0\leq i\leq n$ . Note that, since $f=\phi \circ g$ , we have $f^n=\phi \circ (g\circ \phi )^{n-1}\circ g$ . Write $z_1=g(z)$ and define inductively $z_{j+1}=g\circ \phi (z_j)$ for $j=1,\ldots , n-1$ . Then for each non-zero tangent vector $v\in T_zY$ , we have, by the chain rule,

(5.19) $$ \begin{align} Df^n(z)v = D\phi(z_n)\bigg[\prod_{j=1}^{n-1} Dg(\phi(z_j))D\phi(z_j)\bigg] Dg(z)v. \end{align} $$

Now, since the holomorphic map g expands the hyperbolic metric of Y, we have that $|Dg(z)v|_Y>|v|_Y$ . Moreover, the amount of possible contraction of the hyperbolic metric by the $(1+\delta )$ -quasiconformal diffeomorphism $\phi $ is bounded from below. Indeed, we have $|D\phi (\zeta )w|_Y\geq (1-C_\theta \delta ^{1-\theta })|w|_Y$ for all $\zeta \in Y$ and all $w\in T_{\zeta }Y$ . Moreover, writing $v_1=Dg(z)v\in T_{z_1}Y$ and $v_{j+1}=D(g\circ \phi )(z_j)v_j\in T_{z_{j+1}}Y$ for $j=1,\ldots , n-1$ , and applying the above claim, we get

$$ \begin{align*} |v_{j+1}|_Y = |D(g\circ \phi)(z_j)v_j|_Y \geq \xi_{k_j}|v_j|_Y, \end{align*} $$

where $k_j\geq 0$ is the unique integer such that $z_j\in W_{k_j}$ . Setting $\eta =1-C_\theta \delta ^{1-\theta }<1$ and carrying these facts back into equation (5.19), we deduce that

(5.20) $$ \begin{align} |Df^n(z)v|_Y> \eta\,\bigg[\prod_{k=1}^{\infty} \xi_k^{N_{k,n}(z)}\bigg] |v|_Y, \end{align} $$

where $N_{k,n}(z)$ is the total number of j terms in the range $1\leq j\leq n-1$ such that $z_j\in W_{k}$ (in particular, the product appearing in the right-hand side is actually finite). This proves assertion (a). Now suppose that z is such that its -limit set accumulates at a point off the real axis, say $p\in Y$ . This is the case, for instance, if z is a recurrent or periodic point for f. Then there exist $k\geq 0$ and a sequence $j_\nu \to \infty $ such that $z_{j_\nu }\to p$ as $\nu \to \infty $ and $z_{j_\nu }\in W_k$ for all $\nu $ . However, this tells us that $N_{k,n}(z)\to \infty $ as $n\to \infty $ , and therefore, from equation (5.20), we deduce at last that $|Df^n(z)v|_Y/|v|_Y\to \infty $ as $n\to \infty $ . This proves the desired expansion property stated in assertion (b), and it also proves assertion (c). Hence, it remains to prove assertion (d).