Proof. This is just a calculation, making use of the fact that $M_j - M_{j+1} = - M_j$ and $2M_{j} = M_{j+1}$ for all $j \geq 0$ , along with the definition of $c_k$ given by equation (A.15). When $k =1$ , we verify equation (3.2) by checking that $r_1^{M_1} = c_1 \cdot r_1^{M_0 + 1}$ . For the case of $k \geq 2$ , we verify equation (3.2) by computing

$$ \begin{align*} r_k^{M_k} \cdot c_k &= r_k^{M_k} \cdot \prod_{j=2}^{k} r_{j-1}^{M_{j-1} -M_j} \\&= r_k^{M_k} \cdot \prod_{j=2}^{k} r_{j-1}^{-M_{j-1}} \\&\geq r_k^{M_k} \cdot \prod_{j=2}^{k} r_{k}^{-M_{j-2}} \quad (\sqrt{r_k} \geq r_j) \\&= r_k^{M_k} \cdot r_k^{-\sum_{j=0}^{k-2} M_j} \\&= r_k^{M_k - M_{k-1} + 1} = r_k^{M_{k-1} +1}. \end{align*} $$

In the last line, we used the fact that $\sum _{j=0}^{k-2} M_j = M_{k-1} -1$ .

Proof. The claim $(1)$ is just a calculation:

(3.3) $$ \begin{align} r_2 = c_1\bigg(\frac{r_1}{2}\bigg)^{M_1} = 8^2 = 64> 16 = r_1. \end{align} $$

We will prove the second claim by induction. First, we have

(3.4) $$ \begin{align} r_3 = c_2\bigg(\frac{r_2}{2}\bigg)^{M_2} = 2^{-8}(2^5)^4= 2^{12} = 64^2. \end{align} $$

Therefore, $\sqrt {r_3} \geq r_2> r_1$ .

Suppose that for some $k \geq 3$ , we have $\sqrt {r_k} \geq r_{j}$ for all $j = 1,\ldots ,k-1$ . Then, by Lemma 3.2,

(3.5) $$ \begin{align} r_{k+1} = c_k r_k^{M_k} 2^{-M_k} \geq r_k^{M_{k-1} + 1} 2^{-M_k} = r_k^{M_{k-2} +1} r_k^{M_{k-2}} 16^{-M_{k-2}} \geq r_k^{M_{k-2}+1}. \end{align} $$

Since $k \geq 3$ , we have $M_{k-2} \geq M_1 = 2$ , so that

(3.6) $$ \begin{align} r_{k+1} \geq r_k \cdot r_k^2. \end{align} $$

Therefore, by the inductive hypothesis, we must have $\sqrt {r_{k+1}} \geq r_k \geq r_j$ for all ${j = 1,\ldots , k-1}$ . This proves the claim.

Proof. Most of the work has already been done in the proof of Lemma 3.4. We first prove equation (3.7). When $k \geq 2$ , By Lemma 3.4, we have $\sqrt {r_{k+1}} \geq r_j$ for all $j =1,\ldots k$ . Therefore, by Lemma 3.2, we obtain equation (3.7).

Equation (3.8) follows immediately. Indeed, the first two lines of equation (3.5) yields

$$ \begin{align*}r_{k+1} \geq r_k^{M_{k-1}+1} 2^{-M_k},\end{align*} $$

when $k \geq 3$ .

When $k \geq 5$ , we can refine the estimate $r_{k+1} \geq r_k^{M_{k-2}+1}$ from equation (3.5). We note that by Lemma 3.4, we have $r_k> 16$ for all $k \geq 5$ . Therefore,

$$ \begin{align*}r_{k+1} \geq r_k^{M_{k-2}+1}> 16^{M_{k-2}} = (2^4)^{M_{k-2}} = 2^{M_k}.\end{align*} $$

Finally, since $r_k> 16$ for all $k \geq 1$ , we certainly obtain equation (3.10) from equation (3.6).

Proof. This follows immediately from [Reference Burkart and LazebnikBL23, Proposition 3.21].

Proof. We abbreviate $\eta (z):=\eta _{r_k}(z)$ . We use a similar strategy as in the proof of [Reference Fagella, Jarque and LazebnikFJL19, Lemma 3.1], and the initial steps of the proof are exactly the same. We have

$$ \begin{align*} (g_k)_z(z)=M_kc_kz^{M_k-1}+r_k\eta(z)+r_kz\eta_z(z)\quad\text{and}\quad (g_k)_{\overline{z}}(z)=r_kz\eta_{\overline{z}}(z). \end{align*} $$

Solving $b"(x)=0$ , one sees that $|b'(x)|$ has a maximum at $x_0=(1/3)^{1/4}$ with $|b'(x_0)|<e$ , so that $|b'(x)|\leq e$ for $x\in [0,1]$ . Thus, $|(\widehat {\eta })'(x)|\leq e$ for all $x>0$ . Using the chain rule again, we have

$$ \begin{align*} \bigg| \frac{\partial\eta}{\partial z}(z) \bigg| = | (\widehat{\eta})'(|z|) |\cdot\bigg| \frac{\partial|z|}{\partial z} \bigg| \leq \frac{e}{2}\quad \text{and} \quad \bigg| \frac{\partial\eta}{\partial \overline{z}}(z) \bigg| = | (\widehat{\eta})'(|z|) |\cdot\bigg| \frac{\partial|z|}{\partial \overline{z}} \bigg| \leq \frac{e}{2},\end{align*} $$

where we have used the fact that

$$ \begin{align*}\frac{\partial|z|}{\partial z}=\frac{\overline{z}}{2|z|} \quad \text{and} \quad \frac{\partial|z|}{\partial \overline{z}}=\frac{z}{2|z|}.\end{align*} $$

Hence,

(3.14) $$ \begin{align} \bigg|\frac{(g_k)_{\overline{z}}(z)}{(g_k)_z(z)}\bigg| &\leq \frac{r_k|z|{e}/{2}}{| M_kc_k|z|^{M_k-1} - r_k|\eta(z)| - r_k|z|{e}/{2} | }\nonumber\\ &= \frac{{e}/{2}}{| {M_kc_k|z|^{M_k-2}}/{r_k} - {|\eta(z)|}/{|z|} - {e}/{2} | }. \end{align} $$

Let us consider the right-hand side of equation (3.14) for $|z|=r_k - 1$ , recalling $M_k:=2^k$ . We have that

$$ \begin{align*} \frac{c_k|z|^{M_k-2}}{r_k}&:=\frac{1}{r_k(r_k-1)^2}\bigg(\prod_{j=2}^k\frac{1}{r_{j-1}^{M_j-M_{j-1}}}\bigg)(r_k-1)^{2^k} \\ &= \frac{1}{r_k(r_k-1)^2}\cdot\frac{(r_k-1)^2\cdot (r_k-1)^{2^2}\cdot\cdots\cdot (r_k-1)^{2^{k-1}}}{r_1^2\cdot r_2^{2^2}\cdot \cdots \cdot r_{k-1}^{2^{k-1}}} \\ &= \frac{(r_k-1)^{2^2}}{r_k} \cdot \frac{(r_k-1)^{2^3}}{r_1^2\cdot r_2^{2^2}\cdot r_3^{2^3}} \cdot \frac{(r_k-1)^{2^4}\cdot\cdots\cdot (r_k-1)^{2^{k-1}}}{r_4^{2^4} \cdot \cdots \cdot r_{k-1}^{2^{k-1}}}. \end{align*} $$

By Lemmas 3.4 and 3.5, we may deduce for all $k \geq 5$ ,

(3.15) $$ \begin{align} r_k-1>2r_{k-1}, \end{align} $$

(3.16) $$ \begin{align} \frac{(r_k-1)^{2^3}}{r_1^2\cdot r_2^{2^2} \cdot r_3^{2^3}} \geq 1, \quad\text{and } \end{align} $$

(3.17) $$ \begin{align} (r_k-1)^2> r_k. \end{align} $$

Combining the above inequalities, it follows that when $|z| \geq r_k -1$ , we have

(3.18) $$ \begin{align} \frac{c_k|z|^{M_k-2}}{r_k} \geq r_{k-1}. \end{align} $$

Thus, it follows from equation (3.14) that in fact $|(g_k)_{\overline {z}}/(g_k)_z|\rightarrow 0$ as $k\rightarrow \infty $ .

Proof. By Proposition 3.8, we have $\sup _NK(\phi _N|_{\{|z|\leq r_N\}})<\infty $ and, by [Reference Burkart and LazebnikBL23, Proposition 4.6], we have $\sup _N K(\phi _N|_{\{|z|\geq r_N\}})<\infty $ .

Proof. By [Reference ShishikuraShi18, Lemma 10], we have that for any $p>2$ and $0<\rho <1$ , there exists $C'=C'(p,\rho )$ such that if $0<|z_2|<\rho ^2|z_1|$ , then

(3.22) $$ \begin{align} &\bigg| \hspace{-3pt}\int \!\!\!\!\int_{\mathbb{C}} \frac{\mu(z)\phi_{z_1, z_2}(z)}{1-|\mu(z)|^2}\,dx\,dy \bigg|\nonumber\\ &\quad\leq \frac{1}{1-\rho^2}\bigg|\hspace{-3pt} \int\!\!\!\! \int_{A(\rho^{-1}|z_2|, \rho|z_1|)} \frac{\mu(z)}{1-|\mu(z)|^2}\frac{dx\,dy}{z^2} \bigg| +C'I_{p,2}(\mu; |z_1|)^{1/p},\end{align} $$

and

(3.23) $$ \begin{align} &\int \!\!\!\!\int_{\mathbb{C}} \frac{|\mu(z)|^2|\phi_{z_1, z_2}(z)|}{1-|\mu(z)|^2}\,dx\,dy\nonumber\\ &\quad \leq \frac{1}{1-\rho^2} \int\!\!\!\! \int_{A(\rho^{-1}|z_2|, \rho|z_1|)} \frac{|\mu(z)|^2}{1-|\mu(z)|^2}\frac{dx\,dy}{|z|^2} +C'I_{p,2}(\mu; |z_1|)^{1/p},\end{align} $$

where

(3.24) $$ \begin{align} \phi_{z_1, z_2}(z)=\frac{z_1}{z(z-z_1)(z-z_2)}\quad\text{and}\quad I_{p,2}:=\int \!\!\!\!\int_{\mathbb{C}} \frac{|\mu(z)|^p}{(1-|\mu(z)|^2)^p} \frac{dx\,dy}{|z|^2(1+|z|/r)^2}. \end{align} $$

Thus, by [Reference ShishikuraShi18, Theorem 8], it suffices to show that the $\liminf _{z_2\rightarrow 0}$ of the four terms on the right-hand sides of equations (3.22) and (3.23) are $O(\omega _p(|z_1|))$ as $z_1\rightarrow 0$ . In fact, we have better estimates on the two integral terms: they are $O(\omega _1(|z_1|))$ by assumption (3.20). For the remaining two terms, we use [Reference ShishikuraShi18, Lemma 11]: there exist constants $C_2$ and $C_3$ depending only on $K(\psi )$ such that for $0<r<r'$ ,

(3.25) $$ \begin{align} I_{p,2}(\mu; r) \leq C_2 \int \!\!\!\!\int_{|z|<r'} \frac{|\mu(z)|^2}{1-|\mu(z)|^2}\frac{dx\,dy}{|z|^2} + \frac{C_3}{2}\bigg(\frac{r}{r'}\bigg)^2. \end{align} $$

Letting $r'=r^{1/2}$ , we see the first term on the right-hand side of equation (3.25) has order $O(\omega _1(r'))=O(\omega _1(r))$ , and the second term has order $O(r)$ , so that $I_{p,2}$ has order $O(\omega _1(r))$ . Thus, $I_{p,2}(\mu; |z_1|)^{1/p}$ has order $O(\omega _p(|z_1|))$ , and so the result follows.

Proof. Let $N\in \mathbb {N}$ . Let $\phi :=\phi _N: \mathbb {C}\rightarrow \mathbb {C}$ be a quasiconformal mapping such that ${h_{N}\circ \phi ^{-1}}$ is holomorphic and $\phi (0)=0$ . Consider

$$ \begin{align*} \psi(z):=1/\phi(1/z), \end{align*} $$

and define $I(r)$ as in equation (3.20). We wish to apply Theorem 3.14. To this end, we calculate

(3.27) $$ \begin{align} I(r) \leq \frac{k}{1-k^2}\int\!\!\!\! \int_{(|z|<r)\cap\text{supp}(\psi_{\overline{z}})} \frac{dx\,dy}{|z|^2} \leq \frac{k}{1-k^2} \sum_{j\geq j(r)} \int\!\!\!\! \int_{G_j}\frac{dx\,dy}{|z|^2}, \end{align} $$

where

(3.28) $$ \begin{align} j(r) \text{ is the smallest integer such that } 1/r<r_{j(r)} \cdot \exp(\pi/M_j) \end{align} $$

and

(3.29) $$ \begin{align} G_j:=\{z\in\mathbb{C} : r_j^{-1} \cdot \exp(-\pi/M_j) \leq |z| \leq (r_j-1)^{-1} \}. \end{align} $$

As in the proof of [Reference Burkart and LazebnikBL23, Theorem 4.8], we calculate

(3.30) $$ \begin{align} \frac{k}{1-k^2} \sum_{j\geq j(r)} \int\!\!\!\! \int_{G_j}\frac{dx\,dy}{|z|^2} &\lesssim \sum_{j \geq j(r)} \int\!\!\!\! \int_{G_j}\frac{dx\,dy}{r_j^{-2}\exp(-2\pi/M_j)} \nonumber \\[-1pt]&= \sum_{j\geq j(r)} \frac{\pi((r_j-1)^{-2}-r_j^{-2}\exp(-2\pi/M_j))}{r_j^{-2}\exp(-2\pi/M_j)} \nonumber \\[-1pt]&\simeq \sum_{j \geq j(r)} \bigg(\bigg(\frac{r_j}{r_j-1}\bigg)^{2}\exp(2\pi/M_j)-1\bigg) \nonumber \\[-1pt]&\lesssim \sum_{j \geq j(r)} \bigg(\bigg(\frac{r_j}{r_j-1}\bigg)^{2}-1 + \bigg(\frac{r_j}{r_j-1}\bigg)^{2}\frac{4\pi}{M_j}\bigg) \nonumber \\[-1pt]&\lesssim \sum_{j \geq j(r)}\bigg( \frac{2r_j -1}{(r_j-1)^2} + \frac{8\pi}{M_j} \bigg) \nonumber \\[-1pt]&\lesssim \sum_{j \geq j(r)} \bigg( \frac{1}{r_j} + \frac{1}{M_j} \bigg) \nonumber\\[-1pt]&\lesssim \bigg(\frac{1}{2} \bigg)^{j(r)}, \end{align} $$

where we have used the fact that $M_j = 2^j$ , Corollary 3.5, and the inequality

$$ \begin{align*}\exp(x) \leq 1 + 2x \quad\text{for all } x \leq 1.\end{align*} $$

Next, we note that

(3.31) $$ \begin{align} 2^{2^{(j+1)(j+2)/2}}>2^{2^{1+\cdots+j}+\cdots+2^j}>r_j \cdot \exp(\pi/M_j). \end{align} $$

Thus, it is readily calculated that

(3.32) $$ \begin{align} r_j \cdot \exp(\pi/M_j)> 1/r \implies (j+1)(j+2) > \log\log r^{-1}, \end{align} $$

and so since $(1/2)^j \simeq (1/2)^{\sqrt {(j+1)(j+2)}}$ , it follows that

(3.33) $$ \begin{align} r_j\cdot \exp(\pi/M_j)> \exp(-\pi/M_{j(r)})/r > 1/(2r) \implies \big(\tfrac{1}{2}\big)^j \lesssim \big(\tfrac{1}{2}\big)^{\sqrt{\log\log r^{-1}}}. \end{align} $$

Together, equations (3.27)–(3.33) imply that $I(r)$ is finite and has order $O(\omega _1(r))$ as $r\searrow 0$ , as needed. Thus, we may apply Theorem 3.14 to deduce that there exist $c>0$ and $r>0$ , so that

(3.34) $$ \begin{align} |\psi(z)/z-\psi'(0)| < c\cdot\omega_p(|z|) \quad\text{for } |z|<r. \end{align} $$

By multiplying $\psi $ by a complex constant, we may assume that $\psi '(0)=1$ . Since $1/\phi (1/z)=\psi (z)$ , the inequality (3.26) follows by taking $R=1/r$ and $C'>c$ . By Proposition 3.12 and Remark 3.15, the constants $C'$ and R do not depend on N since the above big-O estimates for $I(r)$ do not depend on N.

Proof. Let $\mu _N$ be the Beltrami coefficient of $\phi _N$ . As $N\rightarrow \infty $ , we have $\mu _N\rightarrow 0$ pointwise. Thus, we have $|\phi _N(z)- z|\rightarrow 0$ uniformly on the compact set $|z|\leq R$ .

Proof. This is just a simple calculation using Definition 3.13 and Corollary 3.5. Indeed, for all $k \geq 10$ , we have

$$ \begin{align*} \log \log \bigg(\frac{r_k}{20}\bigg ) &\geq \log \log \bigg(\frac{2^{M_{k-1}}}{20}\bigg) \\ &\geq \log \log (2^{M_{k-5}})\\ &= \log ( M_{k-5} \log 2) \\ &= \log (2^{k-5} \log 2) \\ &= (k-5) \log 2 + \log \log 2. \end{align*} $$

Therefore, there exists a value $k_0$ so that for all $k \geq k_0$ , we have $\log \log ({r_k}/{20}) \geq \tfrac 12k.$ For all such k, we verify that when $|z| \geq ({r_k}/{20})$ , we have

$$ \begin{align*} \omega_{2\sqrt{2}}\bigg(\frac{1}{|z|} \bigg) & = \bigg(\frac{1}{2}\bigg)^{({1}/({2 \sqrt{2}}))\sqrt{\log\log |z|}} \leq \bigg(\frac{1}{2}\bigg)^{({1}/({2 \sqrt{2}}))\sqrt{\log\log ({r_k}/{20})}}\\ & \leq \bigg( \frac{1}{2} \bigg)^{({1}/({2 \sqrt{2}})) \sqrt{{k}/{2}}} = \bigg( \frac{1}{2} \bigg)^{{\sqrt{k}}/{4}}. \end{align*} $$

This yields equation (4.1) as desired.

Proof. Equation (4.2) is just a rearrangement of equation (3.26), but using the estimate (4.1). For the second equation, just note that if $z \in \phi _N(\{w:|w| \geq ({r_k}/{20})\})$ , then there exists w with $|w| \geq ({r_k}/{20})$ so that $\phi _N(w) = z$ . Then, equation (4.2) holds with ${w = \phi _N^{-1}(z)}$ , so equation (4.3) is just a rearrangement of equation (4.2).

Proof. Part (1) is obvious. Part (2) is a simple calculation:

$$ \begin{align*} 2^N + \sum_{j=1}^k n_j = 2^N\bigg(1+ \sum_{j=0}^{k-1} 2^j\bigg) = 2^N \cdot 2^k = 2^{N+k} = n_{k+1}. \end{align*} $$

This proves the claim.

Proof. This follows for $f_N$ by combining equation (3.12) and Lemma 4.3.

Proof. If $z \in A(\exp ({\pi }/{n_{k}}) \cdot R_k, R_{k+1})$ , then

$$ \begin{align*}f_N \circ \phi_N(z) = C_{k+1} \cdot z^{n_{k+1}}.\end{align*} $$

Therefore, if $z \in \phi _N (A(\exp ({\pi }/{n_{k}}) \cdot R_k, R_{k+1}))$ , we must have

$$ \begin{align*}f_N(z) = C_{k+1} \cdot (\phi_N^{-1}(z))^{n_{k+1}}.\end{align*} $$

Therefore, it is sufficient to show that there is an M so that for all $N \geq M$ , and for all $k \geq 1$ , we have

$$ \begin{align*}A\bigg( \frac{5}{4}R_k, \frac{3}{4} R_{k+1} \bigg) \subset \phi_N \bigg( A \bigg( \exp\bigg(\frac{\pi}{n_{k}}\bigg) \cdot R_k, R_{k+1}\bigg)\bigg).\end{align*} $$

The existence of such an M is a simple calculation using Lemma 4.3.

Proof. This follows immediately from Lemma 4.3, Definition 4.5, and Lemma 4.12.

Proof. This is a simple but somewhat tedious application of Lemma 4.8. By the triangle inequality, we obtain for all $z \in A(({1}/{20})R_1, ({19}/{20})R_1)$ that

$$ \begin{align*} c_N |z|^{M_N} \bigg(1 - \frac{r_N}{c_N |z|^{M_{N} - 1}}\bigg) \leq |q_N(z)| \leq c_N|z|^{M_N}\bigg( 1 + \frac{r_N}{c_N |z|^{M_N - 1}}\bigg). \end{align*} $$

On the one hand, we have by Lemma 3.5 that

$$ \begin{align*} \max_{z \in A(({1}/{20})R_1, ({19}/{20})R_1)} \bigg( 1 + \frac{r_N}{c_N |z|^{M_N - 1}}\bigg) &= 1 + \frac{r_N}{c_N(({1}/{20})r_N)^{M_N - 1}} \\ &= 1 + \frac{r_N ({1}/{20})r_N}{({1}/{10})^{M_N} c_N({r_N}/{2})^{M_N}} \\ &= 1 + \frac{10^{M_N}}{20} \cdot \frac{r_N^{2}}{r_{N+1}} \\ &\leq 1 + \frac{10^{M_N}}{20} \cdot \frac{r_N^2}{2^{-M_N} r_N^{M_{(N-1)}}} \\ &\leq 1 + \frac{20^{M_N}}{20} \cdot \frac{1}{r_N^{M_{(N-2)}}} \\ &= 1 + \frac{1}{20} \bigg( \frac{20}{r_N^{{1}/{4}}} \bigg)^{M_N}. \end{align*} $$

By Lemma 4.8, there exists M so that for all $N \geq M$ , we have $r_N^{1/4} \geq 40$ and, therefore, we obtain

$$ \begin{align*} \max_{z \in A(({1}/{20})R_1, ({19}/{20})R_1)} \bigg( 1 + \frac{r_N}{c_N |z|^{M_N - 1}}\bigg) \leq 1+\frac{1}{20} \cdot \bigg(\frac{1}{2}\bigg)^{M_N} < 2. \end{align*} $$

Therefore, we obtain for all $z \in z \in A(({1}/{20})R_1, ({19}/{20})R_1)$ that

(4.18) $$ \begin{align} |q_N(z)| \leq 2 c_N|z|^{M_N}. \end{align} $$

The proof of the other inequality is similar.

Proof. By Lemma 4.3, there exists $M \in \mathbb {N}$ so that for all $N \geq M$ , we have

$$ \begin{align*} \phi^{-1}_N\bigg(A\bigg(\frac{1}{10}R_1, \frac{9}{10}R_1\bigg)\bigg) \subset A\bigg(\frac{1}{20}R_1, \frac{19}{20}R_1\bigg). \end{align*} $$

By perhaps choosing M larger, we have for all $N \geq M$ that $R_1 - 1 \geq ({19}/{20})R_1$ as well. Then, for all $z \in A(({1}/{10})R_1, ({9}/{10})R_1)$ , we have by Lemmas 4.3 and 4.14 that

(4.20) $$ \begin{align} \max_{z \in A(({1}/{10})R_1, ({9}/{10})R_1)} |f_N(z)| &= \max_{z \in A(({1}/{10})R_1, ({9}/{10})R_1)} |q_N(\phi_N^{-1}(z))|\nonumber\\ & \leq 2 c_N |\phi_N^{-1}(z)|^{M_N} \leq 2 c_N \alpha_1^{M_N} |z|^{M_N}. \end{align} $$

Similarly, we obtain

(4.21) $$ \begin{align} \min_{z \in A(({1}/{10})R_1, ({9}/{10})R_1)} |f_N(z)| &= \min_{z \in A(({1}/{10})R_1, ({9}/{10})R_1)} |q_N(\phi_N^{-1}(z))|\nonumber\\ & \geq \tfrac{1}{2} c_N |\phi_N^{-1}(z)|^{M_N} \geq \tfrac{1}{2} c_N \beta_1^{M_N}|z|^{M_N}. \end{align} $$

This proves the claim.

Proof. The first part uses the fact that holomorphic maps are open. The second part is an application of the maximum principle. A detailed proof can be found in [Reference BishopBis18, Lemma 11.1].

Proof. First, we prove the case of $k \geq 2$ . In this setting, by Lemma 4.13,

$$ \begin{align*} \max_{|z| = ({2}/{5})R_k} |f_N(z)| &\leq \max_{|z| = ({2}/{5})R_k} \alpha^{n_{k}}_k C_{k} |z|^{n_{k}} \leq \alpha^{n_{k}}_k C_k \big( \tfrac{2}{5} R_k \big)^{n_{k}} \leq \alpha^{n_{k}}_k \big( \tfrac{4}{5} \big)^{n_{k}} \cdot C_{k} \cdot \big(\tfrac{1}{2}R_k\big)^{n_{k}}. \end{align*} $$

By equation (4.9), we have $\alpha _k \cdot \tfrac 45 \leq \tfrac 78$ . Therefore, if $N \geq 5$ , since $R_{k+1}= C_{k} \cdot ({R_k}/{2})^{n_{k}}$ , we end up with

(4.23) $$ \begin{align} \max_{|z| = ({2}/{5})R_k} |f_N(z)| \leq \big(\tfrac{7}{8}\big)^{n_{k}} R_{k+1} < \tfrac{1}{4} R_{k+1}. \end{align} $$

Next, observe that by Lemma 4.13, we have

$$ \begin{align*} \min_{|z| = ({3}/{5})R_k} \!|f_N(z)| \geq \min_{|z| = ({3}/{5})R_k} \!\beta_k^{n_k} C_k |z|^{n_{k}} = \beta_k^{n_k} C_k \big(\tfrac{3}{5} R_k\big)^{n_{k}} = \beta_k^{n_k} \cdot \big(\tfrac{6}{5}\big)^{n_k} \cdot C_{k} \cdot \big(\tfrac{1}{2}R_k\big)^{n_{k}}. \end{align*} $$

By equation (4.9), we have $\beta _k\cdot \tfrac 65 \geq \tfrac 98$ . If $N \geq 5$ , we end up with

(4.24) $$ \begin{align} \min_{|z| = ({3}/{5})R_k} |f_N(z)| \geq \big(\tfrac{9}{8}\big)^{n_{k}} R_{k+1}> 4 R_{k+1}. \end{align} $$

The lemma for the case of $k \geq 2$ now follows from Lemma 4.16, part $(1)$ .

The case of $k =1$ is almost exactly the same, except we now have to use Lemma 4.15. By following the exact same steps as above, we obtain since $N \geq 5$ that

(4.25) $$ \begin{align} \max_{|z| = ({2}/{5})R_1} |f_N(z)| \leq 2 \big(\tfrac{7}{8}\big)^{n_1}R_{2} < \tfrac{1}{4} R_2 \end{align} $$

and

(4.26) $$ \begin{align} \min_{|z| = ({3}/{5})R_1} |f_N(z)| \geq \tfrac{1}{2}\big(\tfrac{9}{8}\big)^{n_{1}} R_{2}> 4 R_2. \end{align} $$

The lemma for the case of $k = 1$ now follows from Lemma 4.16, part $(1)$ .

Proof. We will adopt a similar strategy to Lemma 4.17, using Lemma 4.16, part $(2)$ . First, we make the important observation that ${C_{k+1}}/{C_k} = R_k^{-n_k}$ for all $k \geq 1$ . Next, observe that if $|z| = 4 R_k$ , then we have by Lemma 4.13,

$$ \begin{align*} \max_{|z| = 4R_k} |f_N(z)| &\leq \max_{|z| = 4R_k} \alpha_k^{n_{k+1}} C_{k+1}|z|^{n_{k+1}} \\ &= \alpha_k^{n_{k+1}} C_{k+1} (4R_k)^{n_{k+1}} \\ &= \alpha_k^{n_{k+1}} 8^{n_{k+1}} \cdot C_{k} R_k^{-n_k}\cdot \big(\tfrac{1}{2}R_k\big)^{n_{k+1}} \\ &= \alpha_k^{n_{k+1}} 8^{n_{k+1}} C_{k} \big(\tfrac{1}{2}R_k\big)^{n_{k}} \cdot R_k^{-n_k} \big(\tfrac{1}{2}R_k\big)^{n_{k}}\\ &= \alpha_k^{n_{k+1}} 8^{n_{k+1}} 2^{-n_{k}} R_{k+1}. \\ &= (\alpha_k^2)^{n_{k}} 32^{n_{k}} R_{k+1}. \end{align*} $$

By equation (4.9), we have $\alpha _k^2 \leq 2$ for all $k \geq 1$ . By Lemma 4.8, there exists M so that for all $N \geq M$ , and for all $k \geq 1$ , we have $R_k^{1/2}> 256$ . Therefore,

(4.28) $$ \begin{align} \frac{64^{n_{k+1}} R_{k+1}}{R_{k+2}} \leq \frac{64^{n_{k+1}}R_{k+1}}{( {1}/{2})^{n_{k+1}} R_{k+1}^{n_k+1}} = \bigg(\frac{128}{R^{{1}/{2}}_{k+1}} \bigg)^{n_{k+1}} \leq \bigg(\frac{1}{2}\bigg)^{n_{k+1}}. \end{align} $$

By equation (4.28), and since $N \geq 5$ , we have for all $k \geq 1$ ,

(4.29) $$ \begin{align} \max_{|z| = 4R_k} |f_N(z)| \leq 64^{n_{k+1}} R_{k+1} \leq \big( \tfrac{1}{2}\big)^{n_{k+1}} R_{k+2} < \tfrac{1}{8} R_{k+2}. \end{align} $$

Next, observe that by Lemma 4.13, we similarly have

$$ \begin{align*} \min_{|z| = 4R_k} |f_N(z)| &\geq \min_{|z| = 4R_k} \beta_k^{n_{k+1}} C_{k+1}|z|^{n_{k+1}} \\ &= \beta_k^{n_{k+1}} C_k R_k^{-n_k} \cdot 2^{n_{k+1}} \bigg(\frac{R_k}{2}\bigg)^{n_{k}} \cdot \bigg(\frac{R_k}{2}\bigg)^{n_{k}} \\ &=(\beta^2_k)^{n_{k}} 2^{n_k} R_{k+1} = (2\beta_k^2)^{n_k} R_{k+1}. \end{align*} $$

By equation (4.9), we have $\beta _k^2 \cdot 2 \geq \tfrac 32$ . Therefore, since $N \geq 5$ , we have for all $k \geq 1$ that

(4.30) $$ \begin{align} \min_{|z| = 4R_k} |f_N(z)| \geq \big( \tfrac{3}{2} \big)^{n_{k}} R_{k+1}> 8 R_{k+1}. \end{align} $$

Therefore, by equations (4.29) and (4.30), for all $k \geq 1$ , we have

(4.31) $$ \begin{align} f_N(|z| = 4R_k) \subset A\big(8R_k,\tfrac{1}{8}R_{k+1}\big) \subset B_{k+1}. \end{align} $$

We can use similar techniques as above to analyze the behavior of $f_N$ on the outermost boundary of $B_k$ (see Figure 4). Indeed, we have

(4.32) $$ \begin{align} \max_{|z| = ({1}/{4})R_{k+1}} |f_N(z)| < \tfrac{1}{8} R_{k+2} \end{align} $$

and

(4.33) $$ \begin{align} \min_{|z| = ({1}/{4})R_{k+1}} |f_N(z)|> 8 R_{k+1}. \end{align} $$

Therefore, by equations (4.33) and (4.32), we have

(4.34) $$ \begin{align} f_N\big(|z| = \tfrac{1}{4} R_{k+1}\big) \subset A\big(8R_k,\tfrac{1}{8}R_{k+1}\big) \subset B_{k+1}. \end{align} $$

As we commented at the start, this proves the lemma.

Proof. By Proposition 3.6 and Lemma 4.3, there exists M so that for all $N \geq M$ , all critical points of $f_N$ satisfying $|z| \geq \tfrac 14 R_1$ belong to $\bigcup _{k=1}^{\infty } A_k$ . If $z \in A_k$ is a critical point, then Proposition 3.6 also says that $|f_N(z)| = C_k R_k^{n_k}$ . To see that $C_k R_k^{n_k} \in B_{k+1}$ , first notice that we have the identity

(4.35) $$ \begin{align} C_k R_k^{n_k} = 2^{n_k} C_k \big(\tfrac{1}{2}R_k\big)^{n_k} = 2^{n_k} R_{k+1}. \end{align} $$

It follows immediately from equation (4.35) that since $N \geq 5$ , we have

(4.36) $$ \begin{align} C_k R_k^{n_k} = 2^{n_k} R_{k+1}> 8R_{k+1}. \end{align} $$

However, we have by Lemma 4.8 that

$$ \begin{align*} \frac{2^{n_k}R_{k+1}}{R_{k+2}} \leq \frac{8^{n_k}}{R_{k+1}^{n_k}} \leq \bigg(\frac{1}{2}\bigg)^{n_k}. \end{align*} $$

So since $N \geq 5$ , we obtain from equation (4.35) that

(4.37) $$ \begin{align} C_k R^{n_k}_k \leq \big( \tfrac{1}{2} \big)^{n_{k}} R_{k+2} < \tfrac{1}{8} R_{k+2}. \end{align} $$

Therefore, by equations (4.36) and (4.37), we have $f_N(z) \in B_{k+1}$ .

Proof. These are all simple calculations. We only verify equation (5.4). If z is a critical point of $q_N$ , then we calculate that

$$ \begin{align*} q_N(z) &= c_N \bigg( \frac{-r_N}{c_N M_N}\bigg)^{{M_N}/{M_N-1}} + r_N \bigg( \frac{-r_N}{c_N M_N}\bigg)^{{1}/{M_N-1}} \\ &= \bigg( \frac{-r_N}{c_N M_N}\bigg)^{{1}/{M_N-1}} \cdot \bigg(r_N - \frac{r_N}{M_N}\bigg) \\ &= \bigg( \frac{-r_N}{c_N M_N}\bigg)^{{1}/{M_N-1}} \cdot r_N \cdot \bigg( 1 - \frac{1}{M_N} \bigg). \end{align*} $$

The result follows upon taking the absolute value.

Proof. Recall first that $r_N = c_{N-1} ({r_{N-1}}/{2})^{M_{N-1}}$ by definition. Therefore, we calculate

(5.7) $$ \begin{align} \frac{r_N}{c_N} = \frac{1}{2^{M_{(N-1)}}} \frac{c_{N-1} r^{M_{(N-1)}}_{N-1}}{c_N} = \frac{1}{2^{M_{(N-1)}}} \frac{r_{N-1}^{M_{(N-1)}}}{r_{N-1}^{-M_{(N-1)}}} = \frac{1}{2^{M_{(N-1)}}} r_{N-1}^{M_N}. \end{align} $$

First, we prove the upper bound for equation (5.6). First, note that we have for all $N \geq 5$ that

(5.8) $$ \begin{align} 2^{- {M_{(N-1)}}/{M_N - 1}} \leq \frac{1}{\sqrt{2}}, \end{align} $$

so that by combining equations (5.7) and (5.8), we obtain

(5.9) $$ \begin{align} \bigg( \frac{r_N}{c_N M_N} \bigg)^{{1}/{M_N - 1}} = \bigg(\frac{1}{M_N}\bigg)^{{1}/{M_N -1}} \cdot 2^{- {M_{(N-1)}}/{M_N - 1}} \cdot r_{N-1} \cdot r_{N-1}^{{1}/{M_N - 1}} \leq \frac{1}{\sqrt{2}} r_{N-1}^2. \end{align} $$

To prove the lower bound for equation (5.6), since $N \geq 5$ , note that we have

(5.10) $$ \begin{align} \bigg( \frac{1}{M_N} \bigg)^{{1}/{M_{N}-1}} \geq \frac{1}{2}\quad \text{and}\quad 2^{{-M_{N-1}}/{M_N -1}} \geq \frac{1}{2}. \end{align} $$

By two applications of Corollary 3.5, since $N \geq 10$ , we obtain

(5.11) $$ \begin{align} r_{N-1}^{{1}/{M_N-1}} \geq 2^{-{M_{(N-2)}}/{M_N -1}} \cdot r_{N-2}^{{M_{(N-3)}}/{M_N - 1}}\geq 2^{-1/2} r_{N-2}^{1/8} \geq 2^{-1/2} 2^{M_{N-3}/8} = 2^{M_{N-6} - 1/2}. \end{align} $$

Therefore, in a similar way to equation (5.9), except this time using equations (5.10) and (5.11), we have

(5.12) $$ \begin{align} \bigg( \frac{r_N}{c_N M_N} \bigg)^{{1}/{M_N - 1}} &= 2^{{-M_{(N-1)}}/{M_N -1}} \cdot \bigg(\frac{1}{M_N}\bigg)^{{1}/{M_N -1}} \cdot r_{N-1} \cdot r_{N-1}^{{1}/{M_N - 1}}\nonumber\\ &\geq 2^{M_{N-6} - 1/2 - 2} r_{N-1} \geq 2^{M_{N-7}} r_{N-1}. \end{align} $$

Equations (5.9) and (5.12) combine to prove equation (5.6).

Proof. If z is a critical point, by Lemmas 5.1 and 5.2, we have

(5.13) $$ \begin{align} |z| = \bigg( \frac{r_N}{c_N M_N} \bigg)^{{1}/{M_N - 1}} \leq \frac{1}{\sqrt{2}}r_{N-1}^2. \end{align} $$

Recall that by Corollary 3.5 that if $N \geq 10$ , we have $r_{N-1}^2\leq \tfrac 14r_N$ and $r_N^2 \leq \tfrac 14r_{N+1}$ . So by Lemmas 5.1 and 5.2, if z is a critical point, then

(5.14) $$ \begin{align} |q_N(z)| = \bigg( \frac{r_N}{c_N M_N} \bigg)^{{1}/{M_N - 1}} r_N \bigg(1 - \frac{1}{M_N} \bigg) \leq \frac{1}{\sqrt{2}} r_{N-1}^2 r_N \leq \frac{1}{4\sqrt{2}} r_N^2 < \frac{1}{16\sqrt{2}} r_{N+1} \end{align} $$

and

(5.15) $$ \begin{align} |q_N(z)| \geq 2^{M_{(N-7)}} r_{N-1} r_N \cdot \bigg( 1 - \frac{1}{M_N} \bigg)> 8 r_N. \end{align} $$

Therefore, by Definition 4.5, we have the critical values of $q_N$ contained in $B_1$ , as desired.

Proof. We first verify that $U' \subset D$ . Note that if $|z| = \tfrac 14R_1 = \tfrac 14 r_N$ , then by Lemma 4.14, we have

(5.16) $$ \begin{align} |q_N(z)| &\geq \frac{1}{2} c_N \bigg( \frac{1}{4}r_N \bigg)^{M_N} \nonumber\\\nonumber &= \bigg( \frac{1}{2}\bigg)^{M_N + 1} r_{N+1} \\\nonumber &\geq \bigg( \frac{1}{4} \bigg)^{M_N+{1}/{2}} r_N^{M_{(N-1)}} r_N \quad (\text{Lemma~}4.8) \\&= \frac{1}{2} \bigg( \frac{r^{{1}/{4}}_N}{4} \bigg)^{M_N} r_N^{M_{(N-2)}} r_N. \end{align} $$

Therefore, since $N \geq 10$ , we have $r_N^{1/4}/4> 8$ and we deduce that

(5.17) $$ \begin{align} |q_N(z)| \geq 16 r_N^3> 16 r^2_{N-1}r_N. \end{align} $$

Therefore, we must have $U' \subset D$ .

By equation (5.14), the critical values of $q_N$ satisfy $|z| \leq r_{N-1}^2r_N$ . Therefore, V contains all $2^N-1$ many critical values of $q_N$ , so that $U'$ contains the $2^N - 1$ many critical points of $q_N$ . It now follows from Lemma A.11 that $q_N: U' \rightarrow V$ is a proper degree $2^N$ branched covering map, and it follows from Theorem A.13 that $U'$ is a Jordan domain. Since $U'$ is contained in D, which is compactly contained in V, $q_N: U' \rightarrow V$ is a degree $2^N$ polynomial-like mapping (see Figure 5).

Proof. By Lemma 4.3, we verify that $U = \phi _N(U') \subset D$ . Therefore, by Lemma 3.18, $f_N = q_N \circ \phi _N^{-1}: U \rightarrow V$ is a proper, degree $2^N$ branched covering map, and is therefore a degree $2^N$ polynomial-like mapping.

Proof. Choose some $\widehat {\gamma _1} \in \widehat {\Gamma _1}$ , and let $z_0\in \widehat {\gamma _1}$ be a zero for $f_N$ . Such a $z_0$ exists since $f_N(\gamma _1) = \gamma $ surrounds the origin. Then, by Corollary A.6 and Remark 5.9 applied to the appropriate branch of the inverse $f^{-1}_N: B(0,4R_1) \rightarrow \widehat {\gamma _1}$ , there exists a constant $C>0$ such that

$$ \begin{align*} \gamma_1 \subset B\bigg(z_0, \frac{C(4R_1)}{|f_N'(z_0)|}\bigg).\end{align*} $$

Therefore, by Theorem 3.17 and Lemma 5.1, there exists another constant $L>0$ so that

$$ \begin{align*}\frac{\operatorname{\mathrm{diameter}}(f_N(\gamma_1))}{\operatorname{\mathrm{diameter}}(\gamma_1)} \kern1.5pt{\geq}\kern1.5pt \frac{1}{C} \cdot |f^{\prime}_N(z_0)| \kern1.5pt{=}\kern1.5pt \frac{1}{C} \cdot R_1(M_N \kern1.5pt{-}\kern1.5pt 1) \cdot|(\phi_N^{-1})'(z_0)| \geq \frac{R_1}{C}\kern1.5pt{\cdot}\kern1.5pt L \kern1.5pt{\cdot}\kern1.5pt (M_N\kern1.5pt{-}\kern1.5pt1).\end{align*} $$

This proves the estimate for the case of $n =1$ . For $n> 1$ , by Corollary A.4 and Remark 5.9, there exists a constant $c>1$ such that

$$ \begin{align*}\frac{\operatorname{\mathrm{diameter}}(f_N(\gamma_n))}{\operatorname{\mathrm{diameter}}(\gamma_n)} &\geq \frac{1}{c} \frac{\operatorname{\mathrm{diameter}}(f_N^{n-1}(f_N(\gamma_n)))}{\operatorname{\mathrm{diameter}}(f_N^{n-1}(\gamma_n))}\\ &\geq \frac{1}{c} \frac{\operatorname{\mathrm{diameter}}(\gamma)}{\operatorname{\mathrm{diameter}}(\gamma_1)} \geq \frac{R_1}{c \cdot C} \cdot L \cdot (M_N-1).\end{align*} $$

Therefore, there exists $M \in \mathbb {N}$ so that for all $N \geq M$ and for all $n \geq 1$ , we have

$$ \begin{align*}\frac{\operatorname{\mathrm{diameter}}(f_N(\gamma_n))}{\operatorname{\mathrm{diameter}}(\gamma_n)} \geq R_1.\end{align*} $$

This is exactly what we wanted to show.

Proof. For each $n \geq 1$ , $\widehat {\Gamma _n}$ is a covering of the Julia set of $(f_N,U,V)$ . Fix $t> 0$ . If ${\widehat {\gamma _n} \in \widehat {\Gamma _n}}$ , then $f_N(\widehat {\gamma _n}) =: \widehat {\gamma _{n-1}} \in \widehat {\Gamma _{n-1}}$ . Therefore, by Lemma 5.10, we have

(5.18) $$ \begin{align} \sum_{\widehat{\gamma_n} \in \widehat{\Gamma_n}} \operatorname{\mathrm{diameter}}(\widehat{\gamma_n})^t \leq R_1^{-t} \cdot 2^N \cdot \sum_{\widehat{\gamma_{n-1}} \in \widehat{\Gamma_{n-1}}} \operatorname{\mathrm{diameter}}(\widehat{\gamma_{n-1}})^t. \end{align} $$

It follows from equation (5.18) that

(5.19) $$ \begin{align} \sum_{n=1}^{\infty} \sum_{\widehat{\gamma_n} \in \widehat{\Gamma_n}} \operatorname{\mathrm{diameter}}(\widehat{\gamma_n})^t \leq \operatorname{\mathrm{diameter}}(\gamma)^t \cdot \sum_{n=1}^{\infty} 2^{Nn} R_1^{-tn}. \end{align} $$

The sum in equation (5.19) converges if and only if

$$ \begin{align*}2^{N} R_1^{-t} < 1.\end{align*} $$

Therefore, we use Corollary 3.5 to see that

(5.20) $$ \begin{align} R_1^{-t} 2^N = r_N^{-t} 2^{N} \leq 2^{N-M_{N-1}t}. \end{align} $$

Choose M so that for all $N \geq M$ , we have $N - 2^{N-1}t < 0$ , so that we obtain $2^N R_1^{-t} < 1$ . For such a choice of N, equation (5.19) converges, and for any $\eta> 0$ , there exists some value n so that

$$ \begin{align*}\sum_{\widehat{\gamma_n} \in \widehat{\Gamma_n}} \operatorname{\mathrm{diameter}}(\widehat{\gamma_n})^t < \eta.\end{align*} $$

Since $\widehat {\Gamma _n}$ is a covering of the filled Julia set of $f_N: U \rightarrow V$ , it follows that its Hausdorff dimension is bounded above by t.

Proof. The only critical points of $f_N$ contained inside of D are of the form $\phi _N(z)$ , where z is a critical point of $q_N$ . By Lemma 5.3, the critical values of $q_N$ are contained in $B_1$ . Therefore, the critical values of $f_N$ associated to the critical points in D belong to $B_1$ .

Proof. By Lemma 4.18, there exists $M \in \mathbb {N}$ such that for all $N \geq M$ , we have ${f_N(B_k) \subset B_{k+1}}$ . This implies that each point in $B_k$ escapes locally uniformly to $\infty $ .

Proof. By Proposition 4.11 and by Lemma 4.16, it is sufficient to verify that there exists $M \in \mathbb {N}$ so that for all $N \geq M$ , we have $f_N(|z| = 4R_k) \subset B_{k+1}$ and $f_N(|z| = \tfrac 54R_k) \subset B_{k+1}$ . The former has already been verified in equation (4.31), so we only need to prove the latter. By the maximum principle for holomorphic functions, we have

(6.1) $$ \begin{align} \max_{|z| = ({5}/{4})R_k} |f_N(z)| \leq \max_{|z| = 4R_k} |f_N(z)| < \tfrac{1}{8}R_{k+2}. \end{align} $$

By Lemma 4.13, we have

$$ \begin{align*} \min_{|z| = ({5}/{4})R_k} |f_N(z)| &\geq \min_{|z| = ({5}/{4})R_k} C_{k+1} \beta_k^{n_{k+1}} |z|^{n_{k+1}} = \beta_k^{n_{k+1}} C_k \bigg( \frac{R_k}{2} \bigg)^{n_k} \bigg(\frac{5}{2}\bigg)^{n_{k+1}} \bigg(\frac{1}{2}\bigg)^{n_k}\\ & \geq \beta_k^{n_{k+1}} R_{k+1} \bigg(\frac{5}{4}\bigg)^{n_{k+1}}. \end{align*} $$

By equation (4.9), we have $\beta _k \cdot \tfrac 54\geq \tfrac 65$ . Therefore, since $N \geq 5$ , we have

(6.2) $$ \begin{align} \min_{|z| = ({5}/{4})R_k} |f_N(z)| \geq \big(\tfrac{6}{5}\big)^{n_{k+1}} R_{k+1}> 8R_{k+1}. \end{align} $$

It follows from equations (6.2) and (6.1) that

(6.3) $$ \begin{align} f_N\big(|z| = \tfrac{5}{4}R_k\big) \subset A\big(8R_{k+1}, \tfrac{1}{8}R_{k+2}\big) \subset B_{k+1}. \end{align} $$

As discussed at the beginning, this proves the claim.

Proof. We consider the cases of $k \geq 2$ and $k = 1$ separately.

When $k \geq 2$ , equation (4.34) implies that $f_N(|z| = \tfrac 14R_k) \subset B_{k}$ . By equation (4.23), we have $\max _{|z| = (2/5)R_k} |f_N(z)| \leq \tfrac 14R_{k+1}$ . Next, we observe that by Lemma 4.8,

$$ \begin{align*} \min_{|z| = ({2}/{5})R_k} \!\!|f_N(z)| \geq\! \min_{|z| = ({2}/{5})R_k} \!\beta_k^{n_k} C_k |z|^{n_{k}} = \beta_k^{n_k} C_k \big(\tfrac{4}{5}\big)^{n_k} \big(\tfrac{1}{2}R_k\big)^{n_{k}} &= \beta_k^{n_k} R_{k+1} \big(\tfrac{4}{5}\big)^{n_k} \\&\geq \beta_k^{n_k} \big(\tfrac{4}{10}\big)^{n_k} R_k^{n_{k-1}} R_k. \end{align*} $$

Since $N \geq 5$ , by equation (4.9) and Lemma 4.8, we have

(6.4) $$ \begin{align} \beta_k \tfrac{4}{10} R^{{1}/{2}}_k \geq 4. \end{align} $$

Therefore, we obtain

(6.5) $$ \begin{align} \min_{|z| = ({2}/{5})R_k} |f_N(z)| \geq 4^{n_k} R_k> 8 R_k. \end{align} $$

It follows that

(6.6) $$ \begin{align} f_N\big(|z| = \tfrac{2}{5}R_k\big) \subset A\big(8R_{k}, \tfrac{1}{8}R_{k+1}\big) \subset B_k \end{align} $$

whenever $k \geq 2$ . Therefore, the case of $k \geq 2$ follows from part (2) of Lemma 4.16 and Proposition 4.11.

For the $k = 1$ case, we use slightly different estimates. First, notice that by following a similar argument as above, but applying Lemma 4.15, we obtain

$$ \begin{align*} \min_{|z| = ({2}/{5})R_1} |f_N(z)| \geq \tfrac{1}{2} 4^{n_1} R_1> 8R_1. \end{align*} $$

This, combined with equation (4.25), allows us to conclude that

(6.7) $$ \begin{align} f_N\big(|z|= \tfrac{2}{5}R_1\big) \subset B_1. \end{align} $$

Next, by following similar reasoning as in equation (5.16), except this time applying Lemma 4.15, we have

$$ \begin{align*} \min_{|z| = ({1}/{4})R_1} |f_N(z)| \geq \frac{1}{4} \beta_1^{M_N} \bigg( \frac{r_N^{{1}/{2}}}{4}\bigg)^{M_N} r_N^{M_{N-2}} r_N = \frac{1}{4}\bigg( \frac{\beta_1 r_N^{{1}/{2}}}{4}\bigg)^{M_N} r_N^{M_{N-2}} r_N. \end{align*} $$

By equation (4.9) and Lemma 4.8, along with similar reasoning as equation (5.17), we have

(6.8) $$ \begin{align} \min_{|z| = ({1}/{4})R_1} |f_N(z)|> 16 r^2_{N-1} r_N = 16 r^2_{N-1} R_1^2 > 4 R_1. \end{align} $$

So by equation (6.8) and the maximum principle used with equation (4.25), we have

(6.9) $$ \begin{align} f_N\big(|z| = \tfrac{1}{4}R_1\big) \subset B_1. \end{align} $$

The $k=1$ case now follows from Proposition 4.11 and part (2) of Lemma 4.16.

Proof. We again must argue the $k=1$ and $k \geq 2$ cases separately.

When $k \geq 2$ , we have by Lemma 4.13 that

$$ \begin{align*} \max_{|z| = ({3}/{5})R_k} |f_N(z)| &\leq \alpha_k^{n_k} C_k \big(\tfrac{3}{5}R_k \big)^{n_k} = \alpha_k^{n_k} C_k \big( \tfrac{6}{5} \big)^{n_k} \big( \tfrac{1}{2} R_k \big)^{n_k} = \alpha_k^{n_k} \big( \tfrac{6}{5} \big)^{n_k} R_{k+1}. \end{align*} $$

By equation (4.9), we have $\alpha _k \tfrac 65 \leq \tfrac 75$ . Therefore, since $N \geq 5$ , we obtain

$$ \begin{align*} \frac{(\alpha_k ({6}/{5}))^{n_k}R_{k+1}}{R_{k+2}} \leq \frac{( {7}/{5})^{n_k}R_{k+1}}{2^{-n_{k+1}} R_{k+1}^{n_k + 1}} = \bigg( \frac{28}{5 R_{k+1}} \bigg)^{n_k} < \bigg(\frac{1}{4}\bigg)^{n_k}. \end{align*} $$

It follows that

(6.11) $$ \begin{align} \max_{|z| = ({3}/{5})R_k} |f_N(z)| \leq \big(\tfrac{1}{4}\big)^{n_k} R_{k+2} < \tfrac{1}{8}R_{k+2}. \end{align} $$

Therefore, we have by equations (4.24) and (6.11) that

(6.12) $$ \begin{align} f_N\big(|z| = \tfrac{3}{5}R_k\big)\subset A\big(8R_{k+1}, \tfrac{1}{8}R_{k+2}\big) \subset B_{k+1}. \end{align} $$

For the $k = 1$ case, the arguments are similar, and by using Lemma 4.15 and requiring $N \geq 5$ , we obtain

(6.13) $$ \begin{align} \max_{|z| = ({3}/{5})R_1} \leq 2 \big( \tfrac{1}{4} \big)^{n_k} R_{3} < \tfrac{1}{8} R_3. \end{align} $$

Therefore, by equations (4.26) and (6.13), we obtain

(6.14) $$ \begin{align} f_N\big(|z| = \tfrac{3}{5}R_1\big) \subset A\big(8R_2,\tfrac{1}{8}R_3\big) \subset B_{3}. \end{align} $$

The lemma now follows from equations (6.12) and (6.14).

Proof. By equations (6.6) and (6.7), along with equations (6.12) and (6.14), we have $f_N(|z|=\tfrac 25R_k) \subset B_k$ and $f_N(|z| = \tfrac 35R_k) \subset B_{k+1}$ for all $k \geq 1$ . Therefore, by part (2) of Lemma 4.16, we have $f_N(V_k) \subset B_k \cup A_{k+1} \cup B_{k+1}$ , as desired.

Proof. Note that by Lemmas A.19 and 4.3, there exist constants $\unicode{x3bb}>0$ and $\delta>0$ so that

(6.18) $$ \begin{align} B(0, \delta C_k R_k^{n_k}) \subset f_N(B(w^k_j, \unicode{x3bb} ( \exp(\pi/n_k) - 1 ) R_k)) \subset B\big(0, \tfrac{1}{2} C_kR_k^{n_k}\big). \end{align} $$

Moreover, Lemmas A.19 and 4.3 imply that the mapping in equation (6.15) is injective, and therefore conformal. Next, note that

(6.19) $$ \begin{align} \frac{2^{-n_k} R_k}{\unicode{x3bb}(\exp(\pi/n_k)-1)R_k} \leq \frac{2^{-n_k}}{\unicode{x3bb}\pi / n_k} = \frac{n_k}{2^{n_k} \pi \unicode{x3bb}} \xrightarrow[]{k \rightarrow \infty} 0. \end{align} $$

Therefore, there exists M so that for all $N \geq M$ , we have equation (6.15) and $B(w_j^k, R_k/2^{n_k}) \subset B(w^k_j, \unicode{x3bb} ( \exp (\pi /n_k) - 1 ) R_k)$ . It remains to verify that equation (6.17) holds.

To see this, note that by Theorem A.1 and equation (6.18), we have

$$ \begin{align*} |(f_N)'(w^k_j)| &\geq \frac{1}{4} \frac{\operatorname{\mathrm{dist}}(0, \partial B(0,\delta C_k R_k^{n_k}))}{\operatorname{\mathrm{dist}}(w^k_j,\partial B(w^k_j, \unicode{x3bb} ( \exp(\pi/n_k) - 1 ) R_k) ))} \\ &= \frac{\delta }{4 \unicode{x3bb}} \frac{C_kR_k^{n_k}}{(\exp(\pi/n_k) - 1 )R_k} \\ &\geq \frac{\delta}{8 \unicode{x3bb} \pi} n_k C_k R_k^{n_k - 1} \\ &= \frac{\delta}{8 \unicode{x3bb} \pi} 2^{n_k} n_k R_{k+1} R_k^{-1}. \end{align*} $$

Therefore, if $w^k_j$ is a zero of $f_N$ in $A_k$ , we have

(6.20) $$ \begin{align} |(f_N)'(w^k_j)| \geq \frac{\delta}{8 \unicode{x3bb} \pi} 2^{n_k} n_k R_{k+1} R_k^{-1}. \end{align} $$

Next, consider the branch of the inverse $f_N^{-1}: B(0,\delta C_k R_k^{n_k}) \rightarrow D'$ , where

$$ \begin{align*}D' \subset B(w^k_j, \unicode{x3bb} ( \exp(\pi/n_k) - 1 ) R_k).\end{align*} $$

Since $\delta>0$ is fixed, there exists a perhaps larger $M \in \mathbb {N}$ so that for all $N \geq M$ , we have $\delta 2^{n_1}> 4$ . Therefore,

(6.21) $$ \begin{align} \delta C_k R_k^{n_k} = \delta 2^{n_k} C_k\big(\tfrac{1}{2}R_k\big)^{n_k} = \delta 2^{n_k} R_{k+1}> 2^{n_{k-1}} 4R_{k+1} > 4R_{k+1}. \end{align} $$

We can further deduce from equation (6.21) that the modulus of the annulus $B(0,\delta C_k R_k^{n_k}) \setminus \overline {B(0,4R_{k+1})}$ is bounded below by some fixed constant independent of k. Denote $D = f_N^{-1}(B(0,4R_{k+1})) \subset D'$ . By applying Corollary A.6 to ${f_N^{-1}: B(0,4R_{k+1}) \rightarrow D}$ , we see that there exists a constant $L'>0$ such that

(6.22) $$ \begin{align} \frac{4R_{k+1}}{L'} \frac{1}{|f'(w^k_j)|} \leq r_{D,w^k_j} \leq R_{D,w^k_j} \leq L' \frac{4R_{k+1}}{|f'(w^k_j)|}. \end{align} $$

Since the modulus of $B(0,\delta C_k R_k^{n_k}) \setminus \overline {B(0,4R_{k+1})}$ is bounded below by some fixed constant independent of k, the constant $L'$ is independent of k and $w^k_j$ . By perhaps increasing M, we have that for all $N\geq M$ ,

(6.23) $$ \begin{align} 4 R_{k+1}L' |f'(w_j^k)|^{-1} \leq 4 R_{k+1}L'\frac{8 \unicode{x3bb} \pi}{\delta n_k} 2^{-n_k} R^{-1}_{k+1} R_k \leq \frac{R_k}{2^{n_k}}. \end{align} $$

Therefore, by equations (6.22) and (6.23), and equation (A.6), we have

$$ \begin{align*}D \subset B\bigg(w_j^k, L'\frac{4R_{k+1}}{|f'(w_j^k)|}\bigg) \subset B\bigg(w_j^k,\frac{R_k}{2^{n_k}}\bigg).\end{align*} $$

This proves equation (6.17), which is exactly what we wanted to show.

Proof. First, observe that every connected component of the boundary of $A(\tfrac 35R_k, \tfrac 54R_k) \setminus \overline {\bigcup _{j=1}^{n_k} B(w^k_j, R_k/2^{n_k})}$ is mapped inside of $B_{k+1}$ by $f_N$ . Indeed, by Lemma 6.5, $f_N(|z| = \tfrac 35R_k) \subset B_{k+1}$ , and by equation (6.3), $f_N(|z| = \tfrac 54 R_k) \subset B_{k+1}$ . The rest of the connected components of the boundary of $A(\tfrac 35R_k, \tfrac 54R_k) \setminus \overline {\bigcup _{j=1}^{n_k} (w^k_j, R_k/2^{n_k})}$ are the boundaries of the petals $P_k \subset \mathcal {P}_k$ . By equation (6.17),

(6.24) $$ \begin{align} f_N\bigg(\partial B\bigg(w_j^k, \frac{R_k}{2^{n_k}}\bigg)\bigg) \subset \{|z| \geq 4 R_{k+1}\}. \end{align} $$

By equations (4.37) and (6.18), we have

(6.25) $$ \begin{align} f_N\bigg(\partial B\bigg(w_j^k, \frac{R_k}{2^{n_k}}\bigg)\bigg) \subset B\bigg(0, \frac{1}{2}C_kR_k^{n_k}\bigg) \subset B\bigg(0, \frac{1}{4}R_{k+2}\bigg). \end{align} $$

Equations (6.24) and (6.25) imply

(6.26) $$ \begin{align} f_N\bigg(\partial B\bigg(w_j^k, \frac{R_k}{2^{n_k}}\bigg)\bigg) \subset B_{k+1}. \end{align} $$

By Proposition 4.11, $f_N$ has no additional zeros in $A(\tfrac 35R_k, \tfrac 54R_k) \setminus \overline {\bigcup _{j=1}^{n_k} B(w^k_j, R_k/2^{n_k})}$ . The result now follows from the maximum principle and minimum principle for non-zero holomorphic functions.

Proof. We will show that all other points get mapped to $B_k$ or $B_{k+1}$ , so that they belong to the Fatou set of $f_N$ by Lemma 6.1. Suppose that $z \in A_k \cap \mathcal {J}(f_N)$ , but $z \notin \mathcal {P}_k$ . Then, by Lemma 6.3, $z \notin A(\tfrac 54R_k, 4 R_k)$ , and by Lemma 6.4, $z \notin A(\tfrac 14R_k, \tfrac 25 R_k)$ . Similarly, Lemma 6.8 and the assumption that $z \notin \mathcal {P}_k$ shows that $z \notin A(\tfrac 35R_k, \tfrac 54R_k)$ . Finally, observe that we cannot have $z \in \{|z| = \tfrac 25R_k\}$ , $z \in \{|z| = \tfrac 35R_k\}$ , or $z \in \{|z| = \tfrac 54 R_k\}$ by equations (6.6) and (6.7), (6.10), and (6.3), respectively. Since $z \in A_k \cap J(f_N)$ , but $z \notin \mathcal {P}_k$ , it follows that $z \in A(\tfrac 25R_k, \tfrac 35R_k) = V_k$ , which proves the theorem.

Notation 7.2. We will use the following notation in this section.

1. (1) The polynomial-like mapping $(f_N,U,V)$ will be the one defined as in Lemma 5.6.

2. (2) The filled Julia set of $(f_N,U,V)$ will be denoted as E.

Proof. The filled Julia set E of $(f_N,U,V)$ is forward invariant and contained in D. Therefore, if $z \in E$ , it is impossible for $z \in A_k$ or $z \in B_k$ for any integer k by Definition 7.1. So we will suppose that $z \notin E$ . By Definition 4.14, if $|z|> \tfrac {1}{4}{R_1}$ , then there exists $k \geq 0$ so that z must belong to exactly one of $B_k$ or $A_k$ . Therefore, we only have to focus our attention on the case $|z| \leq \tfrac {1}{4}{R_1}$ .

Suppose first that $z \in U \subset \overline {D}$ , recalling that $U \subset \overline {D}$ by Lemma 5.6. Then, since ${z \notin E}$ , there exists a smallest integer $l \geq 1$ so that $f_N^l(z) \notin U$ . First, we consider the case that $|f_N^l(z)|> R_1/4$ . Then, by equation (6.9), $f_N(\partial D) \subset B_1$ , so by our choice of l and the maximum principle, we must have either $f_N^l(z) \in A_1$ or $f_N^l(z) \in B_1$ . It follows from Definition 7.1 that either $z \in A_{1-l}$ or $z \in B_{1-l}$ .

Next, we consider the case where $|f_N^l(z)| \leq \tfrac 14R_1$ , so that $f^l(z) \in \overline {D} \setminus U$ . Observe that by Definition 5.4, $f_N(\partial U) = \{z: |z| = 16r_{N-1}^2 R_1\} \subset B_1$ . Indeed, we certainly have $16r_{N-1}^2 R_1> 4 R_1$ , and we may argue using Lemma 4.8 that $16 r^2_{N-1} R_1 < R_2/4$ . We also have $f_N(\partial D) \subset B_1$ by equation (6.9). Therefore, by Proposition 4.11, we must have $f^{l+1}(z) \in B_1$ , so that $z \in B_{-l}$ .

Proof. By assumption, we have $f_N(z) \in A_k$ for some integer k. By Lemma 7.3, either ${z \in E}$ , $z \in A_j$ for some integer j, or $z \in B_j$ for some integer j. We cannot have $z \in E$ , because $f_N(E) \subset E$ . If $z \in B_j$ and $j \leq 0$ , then by Definition 7.1, we must have $f_N(z) \in B_{j+1}$ . If $z \in B_j$ for some $j \geq 1$ , then $f_N(z) \in B_{j+1}$ by Lemma 4.18. Therefore, we cannot have $z \in B_j$ for any integer j. The only remaining possibility is that we must have $z \in A_j$ for some integer j.

Notation 7.5. For an open set $\Omega \subset \mathbb {C}$ , we let $\widehat \Omega $ denote the union of $\Omega $ and its bounded complementary components.

Proof. By Lemma 3.18, $f_N$ has no asymptotic values. By Proposition 4.19, for all sufficiently large N, all of the critical points of $f_N$ with $|z| \geq \tfrac {1}{4}{R_1}$ are mapped into $B_{k+1}$ for some $k \geq 1$ . By Lemma 5.12, all of the critical points of $f_N$ with $|z| \leq \tfrac {1}{4}{R_1}$ are mapped into $B_1$ . Therefore, $SV(f_N) \subset \bigcup _{k \geq 1} B_k$ . It follows from Lemma 4.18 that $P(f_N) \subset \bigcup _{k\geq 1} B_k$ as well.

Proof. To prove part (1), note that by Lemma 7.8, there are no critical values of $f_N$ contained in $\overline {B(0,4R_1)}$ . Therefore, the claim follows by Lemma A.15, which further implies that each connected component of $\widehat Z_k$ for $k \leq 0$ is a Jordan domain.

Part (2) follows immediately from part (1).

To see part (3), note that Lemma A.15 implies that $Z_0$ consists of $2^N$ many connected components. Each connected component of $\widehat Z_0$ is mapped conformally onto $\widehat Z_1$ by $f_N$ . Therefore, $f_N(\widehat Z_0)$ contains the $2^N$ many connected components of $Z_0$ , and it follows that each connected component of $\widehat Z_0$ contains $2^N$ many connected components of $Z_{-1}$ . Therefore, $Z_{-1}$ consists of $2^{2N}$ many connected components. By proceeding similarly, we deduce that every connected component of $\widehat Z_{k+1}$ for $k \leq -1$ contains $2^{N}$ many connected components of $Z_{k}$ , and $Z_k$ therefore must have $2^{N(-k+1)}$ many connected components.

Proof. When $k \geq 2$ , observe that on $\phi _N^{-1}(W)$ , we have $f_N \circ \phi _N = C_{k} z^{n_{k}}$ and that the mapping $f_N \circ \phi _N: \phi ^{-1}_N(W) \rightarrow A_{k+1}$ is a degree $n_{k}$ covering map. Since $\phi _N$ is quasiconformal, it follows that $f_N: W \rightarrow A_{k}$ is a degree $n_{k}$ covering map as well. The $k = 1$ case is similar, except this time, we use the fact that on $\phi _N^{-1}(W)$ , $f_N \circ \phi _N(z) = q_N(z)$ is a degree $n_1$ covering map to conclude that $f_N \circ \phi _N: \phi _N^{-1}(W) \rightarrow A_{2}$ is a degree $n_1$ covering map.

Proof. By Lemma 4.18, we have $f_N(B_{k}) \subset B_{k+1}$ for all $k \geq 1$ . Therefore, we must have $f_N(\Omega _k) \subset \Omega _{k+1}$ for all $k \geq 1$ , and since $f_N$ has no asymptotic values by Lemma 3.18, we actually have

(7.4) $$ \begin{align} f_N(\Omega_k) = \Omega_{k+1} \end{align} $$

for all $k \geq 1$ , (see [Reference HerringHer98, Corollary 2]).

Suppose for the sake of contradiction that we have $\Omega _j = \Omega _k$ for some $k> j \geq 1$ . It follows from Definition 7.11 that $\Omega _j = \Omega _{j+1}$ . This, combined with equation (7.4), implies that $\Omega _j$ is unbounded. This contradicts [Reference BakerBak75, Theorem 1].

Proof. By Lemma 7.4, we have $f_N^{-1}(A_k) \subset A$ for each integer k. Since

$$ \begin{align*}f_N^{-1}(A) = \bigcup_{k \in \mathbb{Z}} f_N^{-1}(A_k),\end{align*} $$

the conclusion follows immediately.

Proof. This follows immediately from Definition 7.16 along with equations (A.11) and (A.12).

Proof. Since $(f_N,U,V)$ is a polynomial-like mapping and E is its filled Julia set, E coincides with the closure of the repelling periodic cycles for $(f_N,U,V)$ . These are also repelling periodic cycles for $f_N$ viewed as an entire function, so $E \subset \mathcal {J}(f_N)$ . Since $\mathcal {J}(f_N)$ is backwards invariant, it follows immediately from Definition 7.16 that $E' \subset \mathcal {J}(f_N)$ as well.

Now, suppose that $z \in \mathcal {J}(f_N)$ , but $z \notin E'$ . The set $E'$ is both forward and backward invariant. Therefore, for all $n \geq 0$ , we have $f_N^n(z) \notin E'$ and, in particular, we have ${f_N^n(z) \notin E}$ . By Lemma 6.1 and Definition 7.1, if $f_N^n(z) \in B_k$ for some $n \geq 0$ and $k \in \mathbb {Z}$ , we must have z in the Fatou set of $f_N$ . Therefore, by Lemma 7.3, we must have $f_N^n(z) \in A$ for all $n \geq 0$ so that $z \in X$ , as desired.

Proof. First, suppose that $z \in A_k$ for some $k \leq 0$ . Then, by Definition 7.1, $f_N(z) \in A_{k+1}$ , so that $j = k+1$ .

Next, suppose that $z \in A_k$ for some $k \geq 1$ . Then, by equation (4.31), we have $f_N(|z|=4R_k) \subset B_{k+1}$ . Therefore, by the maximum principle for holomorphic functions, we must have $f_N(B(0,4R_k)) \subset \widehat B_{k+1}.$ So if $f_N(z) \in A_j$ , we must have $j \leq k+1$ .

Proof. If $k \leq 0$ , we have $f_N(z) \in A_{k+1}$ by Definition 7.1, so we must have $k \geq 1$ .

By Lemma 6.10, we must have $z \in V_k$ or we must have $z \in P_k$ for some petal $P_k \subset \mathcal {P}_k$ . If $z \in V_k$ , then by Corollary 6.6, we must have $f_N(z) \in B_k \cup A_{k+1} \cup B_{k+1}$ . Since ${f_N(z) \in A_j}$ and $j \leq k$ , we must have $z \in P_k$ for some petal $P_k \subset \mathcal {P}_k.$

Proof. By Lemma 7.8, there are no singular values of $f_N$ in $\overline {U}$ . Therefore, $f_N^{-1}(U)$ is the disjoint union of Jordan domains. By Lemma 7.10, there are exactly $n_k$ many connected components of $f_N^{-1}(U)$ contained inside of $V_k$ . By Lemma 6.7, there is exactly one connected component of $f_N^{-1}(U)$ contained in each petal $P_k \subset \mathcal {P}_k$ , which yields $n_k$ many more connected components of $f_N^{-1}(U)$ . There are no other connected components contained inside of $A_k$ by Lemma 6.10. Therefore, the total number of connected components contained inside of $A_k$ is $n_k + n_k = n_{k+1}$ .