Proof. This follows immediately from the normality of ${\mathcal S}$ (Corollary A.3), combined with [Reference ConwayCon78, Proposition VII.1.16].

Proof. Item (1) follows immediately from the above definition for external hyperbolic radius. For item (2), if we temporarily let $R:= \sup \{r: \Delta _V(v, r) \subset X\}$ , then from the definition of internal hyperbolic radius, it follows easily that $V \setminus X \subset V \setminus \Delta _V(v, R^{\mathrm {int}}_{(V,v)}X)$ so that $\Delta _V(v, R^{\mathrm {int}}_{(V,v)}X) \subset X$ whence we have that $R^{\mathrm {int}}_{(V,v)}X \le R$ . Note that since $R^{\mathrm {int}}_{(V,v)}X> 0$ , this means that $R> 0$ and the set of which we take the supremum to find R must be non-empty. However, if we let $z \in V \setminus X$ (note that the requirement that $ R^{\mathrm {int}}_{(V,v)}X < \infty $ ensures that we can always find such a point), then we must have that $\rho _V(v,x) \ge R$ , and on taking an infimum over all such x, we have $R^{\mathrm {int}}_{(V,v)}X \ge R> 0$ from which we obtain item (2). Item (3) then follows from items (1) and (2) (the result being trivial in the cases where the external hyperbolic radius is infinite or the internal hyperbolic radius is zero) which completes the proof.

Proof. To prove item (1), we first observe that the result is trivial if $R^{\mathrm {int}}_{(V,v)}X = \infty $ which happens if and only if $X = V$ . So suppose now that $X \subsetneq V$ . Note that, in this case, $\partial X \cap V \ne \emptyset $ , since otherwise $\mathrm {int\,} X$ and $V \setminus \overline X$ would give a separation of the connected set V.

Now let $z \in \partial X \cap V$ and pick $\varepsilon> 0$ . Since $z \in \partial X$ , there exists $w \in V \setminus X$ with $\rho _V(z, w) < \varepsilon $ . By the triangle inequality $\rho _V(v, w) < \rho _V(v, z) + \varepsilon $ and, on taking the infimum on the left-hand side,

$$ \begin{align*} R^{\mathrm{int}}_{(V,v)}X \le \rho_V(v, z) + \varepsilon. \end{align*} $$

If we then take the infimum over all $z \in \partial X \cap V$ on the right-hand side and let $\varepsilon $ tend to $0$ , we then obtain that

$$ \begin{align*} R^{\mathrm{int}}_{(V,v)}X \le \inf_{z \in (\partial X) \cap V} \rho_{V}(v,z). \end{align*} $$

Now we show $R^{\mathrm {int}}_{(V,v)}X \geq \inf _{z \in (\partial X) \cap V} \rho _{V}(v,z)$ . Take a point $w \in V \setminus X$ and connect v to w with a geodesic segment $\gamma $ in V. Then $\gamma $ must meet $\partial X$ since otherwise, $\mathrm {int\,} X$ and $V \setminus \overline X$ would give a separation of the connected set $[\gamma ]$ (the track of $\gamma $ ). So let $z_0 \in \partial X \cap [\gamma ]$ . Clearly,

$$ \begin{align*} \rho_V(v,z_0)\leq \rho_V(v,w), \end{align*} $$

so

$$ \begin{align*} \inf_{z \in (\partial X) \cap V}\rho_{V}(v,z) \leq \rho_V(v,w), \end{align*} $$

and thus,

$$ \begin{align*} \inf_{z \in (\partial X) \cap V}\rho_{V}(v,z) \leq R^{\mathrm{int}}_{(V,v)}X. \end{align*} $$

This completes the proof of item (1).

To prove item (2), we first consider the case when $\sup _{z \in (\partial X) \cap V} \rho _{V}(v,z)=\infty $ . Note that, since the supremum of the empty set is minus infinity, this in particular implies that $(\partial X) \cap V \neq \emptyset $ . Thus, we can find a sequence $\{z_n \}\in (\partial X)\cap V$ such that $\rho _{V}(v,z_n)\rightarrow \infty $ . For each $z_n$ , choose $x_n\in X$ such that $\rho _{V}(z_n,x_n)\leq 1$ . Then, $\rho _{V}(v,x_n)\rightarrow \infty $ by the reverse triangle inequality, which shows $R^{\mathrm {ext}}_{(V,v)}X=\infty $ so that we have equality.

Now consider the case when $\sup _{z \in (\partial X) \cap V} \rho _{V}(v,z)<\infty $ . The result is trivial if this supremum is minus infinity, so again we can assume that $(\partial X) \cap V \neq \emptyset $ . Similarly to above, we can take a sequence $\{z_n \}\in (\partial X)\cap V$ for which $\rho _V(v,z_n)\rightarrow \sup _{z \in (\partial X) \cap V} \rho _{V}(v,z)$ . Then take a sequence $\{x_n \}\in X$ such that $\rho _V(x_n,z_n)< {1}/{n}$ . By definition of the external hyperbolic radius, we must have

$$ \begin{align*} \rho_V(v,x_n)\leq R^{\mathrm{ext}}_{(V,u)}X, \end{align*} $$

and since $\rho _V(x_n,z_n)< {1}/{n}$ , by the reverse triangle inequality, on letting n tend to infinity,

$$ \begin{align*} \sup_{z \in (\partial X) \cap V} \rho_{V}(v,z) \leq R^{\mathrm{ext}}_{(V,v)}X, \end{align*} $$

which proves item (2) as desired.

Now we show that, under the additional assumptions that $R^{\mathrm {ext}}_{(V,v)}X < \infty $ or $X \subsetneq V$ and $\hat {\mathbb C} \setminus X$ is connected, $\sup _{z \in (\partial X) \cap V} \rho _{V}(v,z) \geq R^{\mathrm {ext}}_{(V,v)}X$ from which item (3) follows. Assume first that $R^{\mathrm {ext}}_{(V,v)}X < \infty $ and let $\{x_n \}\in X$ be a sequence in X such that $\rho _V(v, x_n) \to R^{\mathrm {ext}}_{(V,v)}X$ as n tends to infinity (recall that we have assumed $X \neq \emptyset $ so that the set over which we are taking our supremum to obtain the external hyperbolic radius is non-empty). Note also that $R^{\mathrm {ext}}_{(V,v)}X = 0$ if and only if $X = \partial X = \{v\}$ in which case, the result is trivial, so we can assume that $\rho _V(v, x_n)> 0$ for each n. For each n, let $\gamma _n$ be the unique hyperbolic geodesic in V which passes through v and $x_n$ . Then there must be a point $z_n$ (which may possibly be $x_n$ itself) on $\gamma _n \cap \partial X$ which does not lie on the same side of $x_n$ as v since otherwise, the portion of $\gamma _n$ on the same side of v as $x_n$ and which runs from $x_n$ to $\partial V$ would be separated by the open sets $\mathrm {int\,} X$ and $V \setminus \overline X$ . However, this is impossible since $x_n \in X$ while $R^{\mathrm {ext}}_{(V,v)}X < \infty $ which forces $\gamma _n$ to eventually leave X (in both directions). It then follows that for each n, we have that

$$ \begin{align*} \rho_V(v, z_n) \ge \rho_V(v, x_n) \end{align*} $$

so that

$$ \begin{align*} \sup_{z \in (\partial X) \cap V} \rho_V(v, z) \ge \rho_V(v, x_n),\end{align*} $$

and the desired conclusion then follows on letting n tend to infinity.

Now suppose that $X \subsetneq V$ and $\hat {\mathbb C} \setminus X$ is connected. We observe that, since V is connected, we must have $(\partial X) \cap V \ne \emptyset $ similarly to in the proof of item (1) above, while if $X = U$ is a simply connected domain, then $\hat {\mathbb C} \setminus X$ is automatically connected (e.g. [Reference NewmanNew51, VI.4.1] or [Reference ConwayCon78, Theorem VIII.2.2]).

In view of item (2) above, item (3) holds if $\sup _{z \in (\partial X) \cap V} \rho _{V}(v,z) = \infty $ , so assume from now on that $\sup _{z \in (\partial X) \cap V} \rho _{V}(v,z) < \infty $ and note that $(\partial X) \cap V \ne \emptyset $ implies that this supremum will be non-negative so that we can set $\rho :=\sup _{z \in (\partial X) \cap V} \rho _{V}(v,z)$ . Note that, if $\rho = 0$ , then $V \setminus \{v\} \cap \partial X = \emptyset $ and, since $V \setminus \{v\}$ is connected, then either $V \setminus \{v\} \subset \mathrm {int\,} X \subset X$ or $V \setminus \{v\} \subset V \setminus \overline X \subset V \setminus X$ . In the first case, $X = V \setminus \{v\}$ , in which case, one checks easily that item (3) fails. However, we can rule out this case since $\hat {\mathbb C} \setminus X = \{v\} \cup \hat {\mathbb C} \setminus V$ is disconnected, which violates our hypothesis that $\hat {\mathbb C} \setminus X$ be connected. In the second case, we have $X = \{v\}$ , in which case one easily checks that item (3) holds. Thus, we can assume from now on that $\rho> 0$ .

Immediately from the above claim, we see that $\rho = \sup _{z \in (\partial X) \cap V} \rho _{V}(v,z) \geq R^{\mathrm {ext}}_{(V,v)}X$ , and thus, in the case where $X \subsetneq V$ and $\hat {\mathbb C} \setminus X$ is connected,

$$ \begin{align*} \sup_{z \in (\partial X) \cap V} \rho_{V}(v,z) = R^{\mathrm{ext}}_{(V,v)}X, \end{align*} $$

which proves item (3) as desired.

Proof. Let a, b be two points in $\Delta _V(z, R)$ . Using conformal invariance, we can apply a suitably chosen Riemann map from V to the unit disc ${\mathbb D}$ so that, without loss of generality, we can assume that $a=0$ while b is on the positive real axis whence the shortest geodesic segment from a to b is the line segment $[0,b]$ on the positive real axis. However, the disc $\Delta _{\mathbb D}(z, R)$ is a round disc ${\mathrm D}(w, r)$ for some $w \in D$ and $r \in (0,1)$ which is therefore convex (with respect to the Euclidean metric) and the result follows.

Proof. For the first part, if $\gamma : [a,b] \to R$ , we calculate

$$ \begin{align*} \ell_S(f(\gamma)) &=\int_{f(\gamma)}\,{d}\rho_S \\ &=\int_{a}^{b} \sigma_S(f(\gamma(t))) \cdot |f'(\gamma(t))| \cdot |\gamma'(t)|\, {d}t \\ &=\int_{a}^{b}|f^{\natural}(\gamma(t))|\cdot \sigma_R(\gamma(t)) \cdot |\gamma'(t)|\, {d}t \\ &=\int_{\gamma}|f^{\natural}|\, {d}\rho_R \\ &\leq M \int_{\gamma}\,{d}\rho_R \\ &= M \ell_R(\gamma). \end{align*} $$

The second part then follows immediately from this and the facts that by [Reference Keen and LakicKL07, Theorems 7.1.2 and 7.2.3], $\rho _R(z,w)$ is equal to the length of the shortest geodesic segment in R joining z and w, while $f(\gamma )$ is at least as long in S as the distance between $f(z)$ and $f(w)$ .

Proof. To apply Lehto’s lemma above, we need to verify two things: first that f (and the identity) can be extended as quasiconformal mappings from $\hat {\mathbb C}$ to itself and second that we have an admissible pair of mappings on $\partial A = \partial (\Omega ' \setminus \overline \Omega )$ according to Lehto’s definition given above.

First note that, in view of the argument principle, f, being univalent, is an orientation-preserving mapping on a neighborhood of $\overline \Omega $ . Using [Reference Lehto and VirtanenLV65, Satz II.8.1], f (and trivially the identity) can be extended as a quasiconformal mapping of $\mathbb C$ to itself. Using either [Reference NewmanNew51, Theorem VII.11.1] or the Orientierungssatz in [Reference Lehto and VirtanenLV65, p. 9], the above extension can be easily extended to an orientation-preserving homeomorphism of $\hat {\mathbb C}$ , which is then readily seen to be a quasiconformal mapping of $\hat {\mathbb C}$ to itself as follows easily from [Reference Lehto and VirtanenLV65, Satz I.8.1].

Both f and the identity preserve the positive orientations of $\gamma $ and $\Gamma $ , respectively. In addition, since $f(\gamma )$ lies inside $\Gamma $ , it follows that the orientations of $\gamma $ and $\Gamma $ with respect to A are the same as those of $\tilde A$ . To be precise, let $\gamma $ be positively oriented with respect to A and let ${1}/{A}$ , ${1}/({\tilde A - f(0)})$ denote the images of A, $\tilde A$ , respectively, under ${1}/{z}$ , ${1}/({z - f(0)})$ , respectively. Since A lies in the unbounded complementary component of $\gamma $ , it follows from the above definition of the orientation of a boundary curve for a domain that the winding number of ${1}/{\gamma }$ about points of ${1}/{A}$ is $1$ so that the winding number of $\gamma $ about $0$ (which lies inside $\gamma $ ) is $-1$ . By the argument principle, $f(0)$ lies inside $f(\gamma )$ and, since f is orientation-preserving, the winding number of $f(\gamma )$ about $f(0)$ is also $-1$ .

A simple calculation then shows that the winding number of ${1}/({f(\gamma ) - f(0)})$ about $0$ and thus also about points in ${1}/({\tilde A - f(0)})$ is also $1$ . This shows that f and thus F preserve the positive orientations of $\gamma $ , $f(\gamma )$ with respect to A and $\tilde A$ , respectively. Since F is the identity on $[\Gamma ]$ , it trivially preserves the positive orientation of $\Gamma $ with respect to A and $\tilde A$ (both of which lie inside $\Gamma $ ) and, with this, we have shown the hypotheses of Lemma 3.1 above from [Reference LehtoLeh65] are met.

Lemma 3.1 now allows us to extend F to a quasiconformal mapping on the conformal annulus $A = \Omega ' \setminus \overline \Omega $ such that this extension agrees with the original values of F on the boundary and maps A to $\tilde A$ . We can then use [Reference Lehto and VirtanenLV65, Satz I.8.3] on the removeability of analytic arcs or, remembering that f is defined on a neighborhood of $\overline \Omega $ while the identity is defined on all of $\hat {\mathbb C}$ , twice invoke Rickman’s lemma (e.g. [Reference Douady and HubbardDH85, Lemma 2]) to conclude that the resulting homeomorphism of $\hat {\mathbb C}$ is quasiconformal.

Proof. By the construction in equation (3.1) above, the leading coefficient has absolute value $\kappa $ while the constant term is zero. Now, for $|z|$ sufficiently large,

$$ \begin{align*} P_m^N(z) &=\unicode{x3bb} \bigg( z+\alpha_{m-1}^N+\mathcal{O}\bigg(\frac{1}{|z|}\bigg) \bigg) \bigg (1-\kappa z-\kappa \alpha_{m-1}^N+\mathcal{O}\bigg (\frac{1}{|z|}\bigg )\bigg )\\ &\quad - \alpha_{m}^N + \mathcal{O}\bigg (\frac{1}{|P\circ (\psi_{m-1}^N)^{-1}(z))|}\bigg ), \end{align*} $$

and one sees easily that the $\mathcal {O}({1}/|P\circ (\psi _{m-1}^N)^{-1}(z))|)$ term is actually $\mathcal {O}({1}/{|z|^2})$ . Therefore, the coefficient of the linear term is $\unicode{x3bb} -2 \unicode{x3bb} \kappa \alpha _{m-1}^N$ , and thus is bounded in modulus by $1+2\cdot 1\cdot \kappa \cdot ({8}/{\kappa }) =17$ . Lastly, since $\kappa \ge 1$ , ${1}/({17 + \kappa }) < \kappa < 17 + \kappa $ and so we have indeed constructed a $17+ \kappa $ -bounded sequence of polynomials (as defined near the start of §1.1), proving the lemma as desired.

Proof. Recall that $\Gamma $ is the boundary of $\Omega '$ and that we chose $\mathcal K \subset \Omega \subset {\overline \Omega } \subset \Omega ' \subset \mathrm {D}(0, ({2}/{\kappa }))$ . Let $G(z)$ be the Green’s function for P and set $h:= \sup _{z \in \Gamma }G(z)$ . Then, for each $0 \le m < N$ , $\mathrm {supp}\, \mu _m^N \subset \{z: 0 < G(z) \leq h \cdot 2^{m-N} \}$ and so $\mathrm {supp}\, \mu _0^N \subset \{z: 0 < G(z) \leq h \cdot 2^{-N} \}$ . Thus, $\mu _0^N \rightarrow 0$ everywhere as $N \rightarrow \infty $ . By [Reference Carleson and GamelinCG93, Theorem I.7.5] (see also [Reference AhlforsAhl66, Lemma 1]), we have that $\varphi _0^N$ and $(\varphi _0^N)^{-1}$ both converge uniformly to the identity on $\mathbb C$ (recall that the unique solution for $\mu \equiv 0$ is the identity in view of the uniqueness part of the measurable Riemann mapping theorem for solving the Beltrami equation e.g. [Reference Carleson and GamelinCG93, Theorem I.7.4]). Finally, $\alpha _0^N=\varphi _{0}^N(0) \rightarrow 0$ as $N\rightarrow \infty $ , and since $\psi _0^N=\varphi _0^N(z)-\alpha _0^N$ , the result follows.

Proof. Define $\psi ^{-1}: \mathbb D \rightarrow U$ to be the unique inverse Riemann map from ${\mathbb D}$ to U satisfying $\psi ^{-1}(0)=0$ , $(\psi ^{-1})'(0)>0$ . By Lemma 3.5, $\psi _{0}^N \circ \psi ^{-1} $ converges locally uniformly to $\mathrm {Id} \circ \psi ^{-1}$ on ${\mathbb D}$ . The result then follows from Theorem A.9 in view of the fact that by the above result, $(\psi _{0}^N)'(0) \to 1$ as $N \to \infty $ (so that the argument of $(\psi _{0}^N)'(0)$ converges to $0$ as $N \to \infty $ ).

Proof. Let ${d}\rho _U=\sigma _U(z)|{d}z|$ , where the hyperbolic density $\sigma _U$ is continuous on U (e.g. [Reference Keen and LakicKL07, Theorem 7.2.2]) and bounded away from $0$ on any relatively compact subset of U. For each N, $\psi _0^N$ is analytic on a neighborhood of $\overline U$ , while by Lemma 3.6 and part (2) of the definition of convergence in the Carathéodory topology (Definition A.7), $(\psi _0^N)^{-1}$ is analytic on any relatively compact subset of U for N sufficiently large, so that by Lemma 3.5, both $(\psi _0^N)'$ and $((\psi _0^N)^{-1})'$ converge uniformly to $1$ on A. Since A is a relatively compact subset of U, there exists $\delta _0>0$ such that U contains a Euclidean $2\delta _0$ -neighborhood of A. Let $\tilde A$ denote a Euclidean $\delta _0$ -neighborhood of A, so that $\tilde A$ is still a relatively compact subset of U. By Lemma 3.5 again, we can choose $N_0$ large enough such that $\psi _0^N(A)\subset \tilde {A}$ for all $N\geq N_0$ . Then, since $\sigma _U$ is continuous on the relatively compact subset A of U, there exists $\sigma>0$ such that $|\sigma _U|\ge \sigma $ on A. Then for $z \in A$ , using the uniform continuity of $\sigma _U$ on the relatively compact subset $\tilde {A}$ of U,

$$ \begin{align*} (\psi_0^N)^{\natural}(z)&= \frac{(\psi_0^N)'(z)\sigma_U(\psi_0^N(z))}{\sigma_U(z)} \end{align*} $$

converges uniformly to 1 on A, as desired. The proof for $((\psi _0^N)^{-1})^{\natural }$ is similar.

Proof. Let $\varepsilon ,\delta $ be as above and, without loss of generality, take $\varepsilon <\min \{\delta ,1\}$ . By Lemma A.4, the Euclidean and hyperbolic metrics are equivalent on compact subsets of U, and we can then use Lemma 3.5 to pick $k_0$ sufficiently large so that for all $k_1 \ge k_0$ ,

(3.2) $$ \begin{align} \rho_U((\psi_0^{n_{k_1}})^{-1}(z),z)< \frac{\varepsilon}{3(M+1)},\quad z\in A. \end{align} $$

This also implies that if we let $\check A$ be the ${\delta }/{2}$ -neighborhood of A in U, then, since $\varepsilon < \delta $ ,

(3.3) $$ \begin{align} (\psi_0^{n_{k_1}})^{-1}(A) \subset \check{A}. \end{align} $$

Next, by Lemma 3.7, we can make $k_0$ larger if needed such that for all $k_1 \ge k_0$ ,

(3.4) $$ \begin{align} |((\psi_0^{n_{k_1}})^{-1})^{\natural}(z)-1|< \frac{\varepsilon}{3},\quad z \in A. \end{align} $$

From above, since $\{P^{\circ n_k}\}_{k=1}^{\infty }$ converges locally uniformly to the identity on U (with respect to the Euclidean metric), using Lemma A.4, we can again make $k_0$ larger if necessary to ensure for all $k_1 \ge k_0$ that

(3.5) $$ \begin{align} \rho_U(P^{\circ n_{k_1}}(z),z)<\frac{\varepsilon}{3(M+1)},\quad z \in \check{A}. \end{align} $$

This also implies

(3.6) $$ \begin{align} P^{\circ n_{k_1}}(\check{A}) \subset \hat{A}. \end{align} $$

By Lemma 3.8, we can again make $k_0$ larger if needed such that for all $k_1 \ge k_0$ ,

(3.7) $$ \begin{align} |(P^{\circ n_{k_1}})^{\natural}(z)-1|< \frac{\varepsilon}{3},\quad z \in \check{A}. \end{align} $$

We remark that this is the last of our requirements on $k_0$ and we are now in a position to establish the dependencies of $k_0$ on $\kappa $ , $\gamma $ , $\Gamma $ , f, A, $\delta $ , M, $\varepsilon $ in the statement. To be precise, the requirements on $k_0$ in equation (3.2) depend on $\kappa $ , $\gamma $ , $\Gamma $ , f, A, M, $\varepsilon $ , and $\delta $ , while those in equation (3.4) depend on $\kappa $ , $\gamma $ , $\Gamma $ , f, A, $\varepsilon $ , and $\delta $ , (but not M). Note that the dependency of these two estimates on the curves $\gamma $ , $\Gamma $ , or equivalently on the domains $\Omega $ , $\Omega '$ , (which in turn depend on the scaling factor $\kappa $ ) as well as the function f, arises from the quasiconformal interpolation performed with the aid of Lemma 3.2 which is clearly dependent on these curves and this function. Further, the requirements on $k_0$ in equation (3.5) depend on $\kappa $ , A, $\delta $ , M, and $\varepsilon $ (but not $\gamma $ , $\Gamma $ , or f) while those in equation (3.7) depend on $\kappa $ , A, $\delta $ , and $\varepsilon $ (but not $\gamma $ , $\Gamma $ , f, or M). Finally, for the remaining estimates, equation (3.3) is a direct consequence of equation (3.2), while equation (3.6) follows immediately from equation (3.5) so that none of these three introduces any further dependencies.

Now fix $k_1 \ge k_0$ arbitrarily and let the finite sequence $\{P_m^{n_{k_1}}\}_{m=1}^{k_1}$ be constructed according to the sequence $\{n_k \}_{k=1}^{\infty }$ specified at the start of this section and the prescription given in equation (3.1). Note that this sequence is then $(17+\kappa )$ -bounded in view of Lemma 3.4. By construction, $Q_{n_k}^{n_k}(0)=0$ for every k so that item (3) in the statement above will be automatically satisfied.

Now equations (3.1), (3.3), and (3.6), the univalence of P on U, and the univalence of f on a neighborhood of ${\mathcal K}$ imply that $Q_{n_{k_1}}^{n_{k_1}}$ is univalent on A. Now let $z \in A$ and, using equation (3.2), consider a geodesic segment $\gamma $ connecting z to ${(\psi _{0}^{n_{k_1}})}^{-1}(z)$ which, since $\varepsilon < \delta $ , has length smaller than ${\varepsilon }/{3}$ . Since $\varepsilon < \min \{\delta , 1\}$ , ${\varepsilon }/{3}$ is in turn smaller than ${\delta }/{2}$ and so, by the definition of $\check A$ , we have $[\gamma ] \subset \check A$ . This allows us to apply equations (3.2) and (3.7), and the hyperbolic M-L estimates (Lemma 2.9) for $P^{\circ n_{k_1}}$ to conclude that the length of $P^{\circ n_{k_1}}(\gamma )$ is at most $(1 + ({\varepsilon }/{3})) {\varepsilon }/{3(M +1)}$ , which is smaller than ${\delta }/{2}$ since $\varepsilon < \min \{\delta , 1\}$ . As $[\gamma ] \subset \check A$ , by equation (3.6), $[P^{\circ n_{k_1}}(\gamma )] \subset \hat A$ and we are then able to apply the hyperbolic M-L estimates for f since by hypothesis, we have $|f^{\natural }(z)|\leq M$ on $\hat {A}$ .

In a similar manner, if instead we consider a geodesic segment connecting z to $P^{\circ n_{k_1}}(z)$ , then, since $\varepsilon < \delta $ , by equation (3.5), this segment again has length less than ${\delta }/{2}$ and starts at $z \in A$ , whence it lies inside $\check A \subset \hat A$ and we are again able to apply the hyperbolic M-L estimates for f to this segment. Recall that $\psi _{n_{k_1}}^{n_{k_1}} = f$ on U in view of the definition of this function using quasiconformal interpolation in Lemma 3.2 and also the fact that the hyperbolic distance between any two points of U is less than or equal to the hyperbolic length of any curve connecting them. Using the triangle inequality and applying the estimates in equations (3.2)–(3.7) (except equation (3.4)) as well as $|f^{\natural }(z)|\leq M$ on $\hat {A}$ from the statement, for each $z \in A$ , since $P^{\circ n_{k_1}} \circ (\psi _{0}^{n_{k_1}})^{-1}(z) \in \hat A$ and $f(\hat A) \subset U$ by hypothesis, we then have

$$ \begin{align*} \rho_U(Q_{n_{k_1}}^{n_{k_1}}(z),f(z)) &= \rho_U(\psi_{n_{k_1}}^{n_{k_1}}\circ P^{\circ n_{k_1}} \circ (\psi_{0}^{n_{k_1}})^{-1}(z),f(z)) \\ &\leq \rho_U(f\circ P^{\circ n_{k_1}} \circ (\psi_{0}^{n_{k_1}})^{-1}(z),f\circ P^{\circ n_{k_1}}(z))\\ &\quad + \rho_U(f\circ P^{\circ n_{k_1}}(z),f(z)) \\ &<M \bigg(1+\frac{\varepsilon}{3}\bigg) \bigg(\frac{\varepsilon}{3(M+1)}\bigg)+ M \bigg(\frac{\varepsilon}{3(M+1)}\bigg ) \\ &<\varepsilon \end{align*} $$

(recall that we assumed $\varepsilon <1$ ), which proves item (1). Also, using the chain rule in equation (2.2) for the hyperbolic derivative, the estimate $|f^{\natural }(z)|\leq M$ on $\hat {A}$ , and equations (3.3), (3.4), (3.6), and (3.7), for each $z \in A$ ,

$$ \begin{align*} |((Q_{n_{k_1}}^{n_{k_1}})^{\natural})(z)| &= |f^{\natural}(P^{\circ n_{k_1}}\circ (\psi_0^{n_{k_1}})^{-1}(z))\cdot (P^{\circ n_{k_1}})^{\natural}((\psi_0^{n_{k_1}})^{-1}(z))\cdot ((\psi_{0}^{n_{k_1}})^{-1})^{\natural}(z)| \\ &\leq M \bigg(1+\frac{\varepsilon}{3}\bigg) \bigg(1+\frac{\varepsilon}{3}\bigg ) \\ &<M(1+\varepsilon), \end{align*} $$

again using $\varepsilon < 1$ at the end, which proves item (2) as desired.

Proof. Let $f,g \in {\mathcal S}$ . By the Koebe one-quarter theorem (Theorem A.1) we have $\mathrm {D}(0,\tfrac 14) \subset g(\mathbb D)$ so $g^{-1}$ is defined on $\mathrm {D}(0,\tfrac 14)$ . Then if $h(w):=4g^{-1}({w}/{4})$ for $w \in \mathbb D$ , we have that $h\in {\mathcal S}$ and $g^{-1}(z) = \tfrac 14h(4z)$ for $z \in \mathrm {D}(0,\tfrac 14)$ , where $z= {w}/{4}$ . Thus, if $|z|\leq {1}/{12}$ , we have $|w| \leq \tfrac 13$ and by the distortion theorems (Theorem A.2), we have that $|h(w)|\leq \tfrac 34$ and $|g^{-1}(z)|\leq ({3}/{16}) < 1$ so that, in particular, $f \circ g^{-1}(z)$ exists. Then, using the distortion theorems again, if $z \in \mathrm {D}(0, ({1}/{12}))$ , we have that $(f \circ g^{-1})(z) \leq ({48}/{169}) < \tfrac 13$ . Thus, $f \circ g^{-1}$ is defined on $\mathrm {D}(0, ({1}/{12}))$ for all $f,g \in {\mathcal S}$ and maps $\mathrm {D}(0,({1}/{12}))$ into $\mathrm {D}(0,\tfrac 13)$ as required.

Proof. Let $f,g \in {\mathcal S}$ . By Lemma 4.1, the function $f\circ g^{-1}$ is defined on $\mathrm {D}(0, ({1}/{12}))$ . Let $w\in \mathbb D$ , $z = ({1}/{12}) w$ , so that $z \in \mathrm {D}(0,({1}/{12}))$ , and define $h(w)=12(f\circ g^{-1})({w}/{12})$ so that $h \in {\mathcal S}$ . Then, letting $w + \sum _{n=2}^{\infty } {a_n w^n}$ denote the Taylor series about $0$ for h and setting $K_0 = e \sum _{n=2}^{\infty }n^3 ({1}/{2^{n-2}})$ , if $|w|\leq \tfrac 12$ , we have

$$ \begin{align*} |h'(w)-1|&=\bigg|w\sum_{n=2}^{\infty}na_nw^{n-2} \bigg| \\ &\leq |w|\sum_{n=2}^{\infty}n|a_n||w|^{n-2} \\ &\leq |w| e\sum_{n=2}^{\infty}n^3\frac{1}{2^{n-2}} \\ &=K_0|w|, \end{align*} $$

where we used that $|a_n|\leq en^2$ as $h\in {\mathcal S}$ (see e.g. [Reference Carleson and GamelinCG93, Theorem I.1.8]). Let $\gamma =[0,w]$ be the radial line segment from $0$ to w. Then, if $|w|\leq \tfrac 12$ ,

$$ \begin{align*} |h(w)-w|&=\bigg|\int_{\gamma} [ h'(\zeta)-1 ]\,{d}\zeta\bigg| \\ &\leq K_0|w|\int_{\gamma}|\,{d}\zeta| \\ &=K_0|w|^2. \end{align*} $$

Then, if $|z|\leq {1}/{24}$ (so that $|w|\leq \tfrac 12$ ), a straightforward calculation shows

$$ \begin{align*} |(f\circ g^{-1})(z)-z|&\leq 12K_0|z|^2, \end{align*} $$

from which the lemma follows on setting $K_1=12K_0$ .

Proof. Set $C_0=C_0(R_0)=2\max _{z\in {\overline {\tilde U}_{R_0}}}|\psi _{\unicode{x3bb} } '(z)| = ({2}/{\kappa })\max _{z\in {\overline {U}_{R_0}}}| \psi '(z)|$ . Then $C_0(R_0)$ does not depend on $\kappa $ and is clearly increasing in $R_0$ , and therefore bounded on any bounded subinterval of $[0,\infty )$ . For $z \in {\mathrm D}(z_0, r_0)$ , set $\zeta = {(z-z_0)}/{r_0}$ and note that if we define $\varphi (\zeta ):= ({\psi (r_0\zeta +z_0)-\psi (z_0)})/{r_0\psi '(z_0)}$ , we have that $\varphi \in {\mathcal S}$ . Applying the distortion theorems (Theorem A.2) to $\varphi $ , we see

$$ \begin{align*} |\varphi(\zeta)|&\leq\frac{|\zeta|}{(1-|\zeta|)^2} \\ &\leq\frac{{s}/{r_0}}{(1- ({s}/{r_0}))^2}, \end{align*} $$

from which we can conclude (using $r_0 = {\tilde r_0}/{\kappa }$ and $\tilde {r}_0 \le 2$ )

$$ \begin{align*} |\psi(z)-\psi(z_0)| \le \frac{{s}/{r_0}}{(1- ({s}/{r_0}))^2}\cdot C_0, \end{align*} $$

which proves item (1). For (2), we again apply the distortion theorems to $\varphi $ and observe

$$ \begin{align*} \frac{1- ({s}/{r_0})}{(1+ ({s}/{r_0}))^3}\leq\frac{1-|\zeta|}{(1+|\zeta|)^3}\leq |\varphi'(\zeta)|\leq\frac{1+|\zeta|}{(1-|\zeta|)^3} \leq\frac{1+ ({s}/{r_0})}{(1- ({s}/{r_0}))^3}, \end{align*} $$

from which item (2) follows as $\varphi '(\zeta )= {\psi '(z)}/{\psi '(z_0)}$ .

Proof. Fix $\kappa _0\geq 48$ . By Lemma 4.2, we have, on $U\subset \mathrm {D}(0, ({2}/{\kappa }))\subset \mathrm {D}(0,({1}/{24}))$ , that $|(f\circ g^{-1})(z)-z|<K_1|z|^2$ for some $K_1>0$ (note that $f\circ g^{-1}$ is defined on U by Lemma 4.1). So $|(f\circ g^{-1})(z)-z|< {4K_1}/{\kappa ^2}$ since $|z|< {2}/{\kappa }$ . Then make $\kappa _0$ larger if necessary to ensure that ${4K_1}/{\kappa ^2} \le \eta r_0 = \eta {\tilde r_0}/{\kappa }$ for all $\kappa \geq \kappa _0$ (where we recall that $\tilde r_0 = \tilde r_0(R_0) = {d}(\partial {\tilde U}_{R_0},\partial U_{\unicode{x3bb} }) = \kappa r_0$ ). In fact, $\kappa _0=\max \{48, ({4K_1}/{\eta \tilde r_0})\}$ will suffice and since $\tilde r_0$ depends only on $R_0$ , we have the correct dependencies for $\kappa _0$ and the proof is complete.

Proof. For $R> 0$ , set $c_R:= {e^R-1}/{e^R+1}$ . Then, if we fix $z_0 \in \overline U_{R_0}$ , we have that $|\psi (z_0)|\leq c_{R_0}$ (recall that $\rho _{\mathbb D}(0,z)=\log (({1+|z|})/({1-|z|}))$ for $z \in \mathbb D$ ). Thus, $c_{R_0}<1$ and

(4.1) $$ \begin{align} 1-|\psi(z_0)|^2 \ge 1-c_{R_0}^2>0. \end{align} $$

As in the proof of Lemma 4.3, set $C_0=C_0(R_0)=2\max _{z\in {\overline {\tilde U}_{R_0}}}|{\tilde \psi _{\unicode{x3bb} }}'(z)|$ . Let $0<\eta _1=\eta _1(R_0)<\tfrac 12$ be such that

(4.2) $$ \begin{align} \frac{C_0 \eta_1}{(1-\eta_1)^2}\leq \frac{1}{2} (\log 10 - \log 9)(1-c_{R_0 + \log 3}^2) \end{align} $$

and note that $\eta _1$ depends only on $R_0$ . Using Lemma 4.4, we can pick $\kappa _1=\kappa _1(R_0, \eta _1) = \kappa _1(R_0)>0$ such that, if $\kappa \geq \kappa _1$ , then $|(f\circ g^{-1})(z)-z|<\eta _1 r_0$ on $U \supset \overline U_{R_0}$ (recall the definitions of $\tilde {r_0} = \tilde {r_0}(R_0)$ and $r_0 = r_0(\kappa , R_0)$ given before Lemma 4.3).

Now set $s:=|(f\circ g^{-1})(z_0)-z_0|$ . We have $|(f\circ g^{-1})(z_0)-z_0|=s<\eta _1 r_0 < {r_0}/{2}$ as $\eta _1<\tfrac 12$ . Then, recalling the definition of $r_0 = {d}(\partial U_{R_0}, \partial U)$ , we have $(f\circ g^{-1})(z_0) \in {\mathrm D}(z_0, ({r_0}/{2})) \subset {\mathrm D}(z_0, r_0) \subset U$ as in the statement so that, in particular, $\psi (f\circ g^{-1})(z_0)$ is well defined. Again using $\rho _{\mathbb D}(0,z)=\log (({1+|z|})/({1-|z|}))$ for $z \in \mathbb D$ combined with the Schwarz lemma for the hyperbolic metric, we must have that $(f\circ g^{-1})(z_0) \in \overline \Delta _U(z_0, \log 3)$ . By the triangle inequality for the hyperbolic metric, $(f\circ g^{-1})(z_0) \in \overline \Delta _U(0, R_0 + \log 3) = \overline U_{R_0 + \log 3}$ so that $|\psi (f\circ g^{-1})(z_0)| \le c_{R_0 + \log 3}$ . Then, similarly to equation (4.1),

(4.3) $$ \begin{align} 1-|\psi((f\circ g^{-1})(z_0))|^2 \ge 1-c_{R_0 + \log 3}^2> 0. \end{align} $$

We may then apply item (1) of Lemma 4.3 and equation (4.2) to see that

$$ \begin{align*} |\psi(z_0)-\psi((f\circ g^{-1})(z_0))|&\leq \frac{C_0 ({s}/{r_0})}{(1- ({s}/{r_0}))^2} \\ &\leq \frac{C_0\eta_1}{(1-\eta_1)^2} \\ &\leq \frac{1}{2} (\log 10 - \log 9)(1-c_{R_0 + \log 3}^2). \end{align*} $$

Thus, using the triangle inequality (and the fact that $\psi $ is a Riemann mapping to the unit disc which has radius $1$ ), we see that

$$ \begin{align*} |(1-|\psi(z_0)|^2)-(1-|\psi((f\circ g^{-1})(z_0))|^2)|< (\log 10 - \log 9)(1-c_{R_0 + \log 3}^2). \end{align*} $$

Making use of equations (4.1) and (4.3), noting that $C_R$ is an increasing function of R, and applying the mean value theorem to the logarithm function on the interval $[1-c_{R_0 + \log 3}^2,\infty )$ , we have

$$ \begin{align*} |\!\log(1-|\psi(z_0)|^2)-\log(1-|\psi((f\circ g^{-1})(z_0))|^2)|<\log 10-\log 9 \end{align*} $$

from which item (1) follows easily. For item (2), let $0<\eta _2<1$ (e.g. $\eta _2 = {1}/{35}$ ) be such that

$$ \begin{align*} \frac{1+\eta_2}{(1-\eta_2)^3}<\frac{9}{8}. \end{align*} $$

By Lemma 4.4, using the same $R_0> 0$ as above, we can pick $\kappa _2=\kappa _2(R_0, \eta _2) = \kappa _2(R_0)>48$ such that for all $\kappa \geq \kappa _2$ , if $z\in U \supset \overline U_{R_0}$ ,

$$ \begin{align*} |(f\circ g^{-1})(z)-z|<\eta_2r_0. \end{align*} $$

Using the same $z_0\in \overline U_{R_0}$ as above, in a similar way to how we used item (1) of Lemma 4.3 above, we can apply item (2) of the same result to see that

$$ \begin{align*} \frac{|\psi'((f\circ g^{-1})(z_0))|}{|\psi'(z_0)|}\leq \frac{9}{8}, \end{align*} $$

as desired. The result follows if we set $\kappa _0=\kappa _0(R_0)= \max \{\kappa _1(R_0),\kappa _2(R_0)\}$ .

Proof. As in the proof of Lemma 4.2, define $h(w)=12(f\circ g^{-1})({w}/{12})$ . Note that h is defined on all of ${\mathbb D}$ by Lemma 4.1 and that $h\in {\mathcal S}$ . Let $z={w}/{12}$ . Using the distortion theorems (Theorem A.2), we have that, for $z \in {\mathrm D}(0, ({1}/{12}))$ ,

(4.4) $$ \begin{align} |(f\circ g^{-1})'(z)|\leq \frac{1+|12z|}{(1-|12z|)^3}. \end{align} $$

If $\kappa \geq \kappa _0$ , we have that $\mathrm {D}(0,({2}/{\kappa })) \subset \mathrm {D}(0,({2}/{\kappa _0}))=\mathrm {D}(0,({1}/{288}))$ . Let $z \in \overline {U}$ and, since ${\overline U}\subset \mathcal K \subset \mathrm {D}(0,({2}/{\kappa }))\subset \mathrm {D}(0,({1}/{288}))$ , we have $|z|< {1}/{288}$ for $\kappa \geq \kappa _0$ . Thus, the right-hand side of equation (4.4) is less than ${25\cdot 24^2}/{23^3}$ , which in turn is less than $\tfrac 65$ for all $\kappa \geq \kappa _0$ as desired.

Proof. Applying Lemmas 4.5 and 4.6 to the definition in equation (2.1) of the hyperbolic derivative taken with respect to the hyperbolic metric of U, and letting $\kappa _0$ be the maximum of the two lower bounds on $\kappa $ in these lemmas, we have that there exists a $\kappa _0 \ge 576$ depending on $R_0$ such that, for all $\kappa \geq \kappa _0$ , and $z \in \overline U_{R_0}$ , $(f\circ g^{-1})(z) \in U$ and

$$ \begin{align*} |(f\circ g^{-1})^{\natural}(z)|&=\frac{1-|\psi(z)|^2}{1-|\psi((f\circ g^{-1})(z))|^2}\cdot\frac{2\cdot|\psi'((f\circ g^{-1})(z))|}{2\cdot|\psi'(z)|}\cdot |(f\circ g^{-1})'(z)| \\ &\leq \frac{10}{9}\cdot\frac{9}{8}\cdot\frac{6}{5} \\ &= \frac{3}{2}, \end{align*} $$

as desired.

Proof. Step 1: Setup. Without loss of generality, make $\varepsilon $ smaller if necessary to ensure $\varepsilon <R_0$ . Let $\kappa _0= \kappa _0(R_0) \ge 576$ be as in the statement of Lemma 4.7 so that the conclusions of this lemma as well as those of Lemmas 4.5 and 4.6 also hold. Then for all $\kappa \geq \kappa _0$ , we have $U \subset \mathcal K \subset \mathrm {D}(0,({2}/{\kappa }))\subset \mathrm {D}(0,({1}/{288})) \subset \mathrm {D}(0, ({1}/{12}))$ . Note that the last inclusion implies that, if $f,g \in {\mathcal S}$ , then $f\circ g^{-1}$ is defined on U in view of Lemma 4.1.

Step 2: Application of the Polynomial Implementation Lemma. First apply Lemma 4.7 with $5R_0 + 1$ replacing $R_0$ so that, for all $\kappa \geq \kappa _0$ , if $f,g \in {\mathcal S}$ , we have $(f\circ g^{-1})(U_{5R_0 +1}) \subset U$ and

(4.5) $$ \begin{align} \| (f \circ g^{-1})^{\natural} \|_{U_{5R_0}}\leq \| (f \circ g^{-1})^{\natural} \|_{U_{5R_0+1}} \le\frac{3}{2}. \end{align} $$

Note that, by Lemmas 2.8 and 2.9, since $U_{2R_0}$ is then hyperbolically convex while $(f \circ g^{-1})(0)=0$ , this implies

(4.6) $$ \begin{align} (f \circ g^{-1})(U_{2R_0}) &\subset U_{3R_0}. \end{align} $$

We observe that since $\mathrm {Id} \in {\mathcal S}$ , in particular, we have $f(U_{2R_0})\subset U_{3R_0}$ for all $f \in {\mathcal S}$ .

Fix $\kappa \ge \kappa _0$ and for each $0 \leq i \leq N$ , using equation (4.5), apply the Polynomial Implementation Lemma (Lemma 3.9), with $\Omega =\mathrm {D}(0, ({1}/{24}))$ , $\Omega '=\mathrm {D}(0,\tfrac 12)$ , $\gamma =\mathrm {C}(0,({1}/{24}))$ , $\Gamma =\mathrm {C}(0,\tfrac 12)$ (where both of these circles are positively oriented with respect to the round annulus of which they form the boundary), $f=f_{i+1}\circ f_{i}^{-1}$ , $A=U_{5R_0}$ , $\delta = 1$ (and hence, ${\hat {A}}= U_{5R_0+1}$ ), $M = \tfrac {3}{2}$ , and $\varepsilon $ replaced with ${\varepsilon }/{3^{N}}$ . Note that $f(0)=0$ , $(f\circ g^{-1})(U_{5R_0 +1}) \subset U$ (as noted above), and that, in view of Lemma 4.1, f is analytic and injective on a neighborhood of $\overline \Omega $ and maps $\gamma $ inside $\mathrm {D}(0,\tfrac 13)$ which lies inside $\Gamma $ , so that $(f, \mathrm {Id})$ is indeed an admissible pair on $(\gamma , \Gamma )$ in the sense given in Definition 3.3 in §3 on the Polynomial Implementation Lemma, which then allows us to obtain a quasiconformal homeomorphism of $\hat {\mathbb C}$ using Lemma 3.2.

Let $M_N$ be the maximum of the integers $n_{k_0}$ in the statement of Lemma 3.9 for each of the $N+1$ applications of this lemma above. Note that each $k_0$ depends on $\kappa $ , the curves $\gamma =\mathrm {C}(0,({1}/{24}))$ , $\Gamma =\mathrm {C}(0,\tfrac 12)$ , and the individual function $f = f_{i+1}\circ f_{i}^{-1}$ being approximated, as well as A, $\delta $ , the upper bound M on the hyperbolic derivative (which, in our case, by equation (4.5) is $\tfrac {3}{2}$ for every function we are approximating) and finally $\varepsilon $ . Thus, $M_N$ , in addition to N, then also depends on $R_0$ , $\kappa $ , the finite sequence of functions $\{f_i\}_{i=0}^{N+1}$ , and, finally (recalling that here we have $\gamma =\mathrm {C}(0,({1}/{24}))$ , $\Gamma =\mathrm {C}(0,\tfrac 12)$ and $A=U_{5R_0}$ , $\delta = 1$ , $M = \tfrac {3}{2}$ ), $\varepsilon $ . From these $N+1$ applications, we also then obtain (after a suitable and obvious labeling) a finite $(17 + \kappa )$ -bounded sequence $\{P_m\}_{m=1}^{(N+1)M_N}$ such that each $Q_{iM_N,(i+1)M_N}$ is univalent on $U_{5R_0}$ , and we have, for each $0 \le i \le N$ and each $z \in U_{5R_0}$ ,

(4.7) $$ \begin{align} \rho_U(Q_{iM_N,(i+1)M_N}(z),f_{i+1}\circ f_{i}^{-1}(z) )<\frac{\varepsilon}{3^N}. \end{align} $$

It also follows from Lemma 3.9 that each $Q_{iM_N,(i+1)M_N}$ depends on N, $R_0$ , $\kappa $ , the functions $f_i$ , $f_{i+1}$ , and $\varepsilon $ so that we obtain the correct dependencies for $M_N$ and $\{P_m\}_{m=1}^{(N+1)M_N}$ in the statement. Lastly, by item (3) of Lemma 3.9, $Q_{iM_N,(i+1)M_N}(0)=0$ , for each i, proving item (1) in the statement above.

Step 3: Estimates on the compositions $\{Q_{iM_N}\}_{i=1}^{N+1}$ . We use the following claim to prove items (2) and (3) in the statement (note that we do not require part (ii) of the claim below for this, but we will need it in proving item (4) later).

Claim 4.9. For each $1\leq j \leq N+1$ , we have that $Q_{jM_N}$ is univalent on $U_{2R_0}$ and, for each $z \in U_{2R_0}$ ,

$$ \begin{align*} \mathrm{(i)}& \; \rho_U(Q_{jM_N}(z),f_{j}(z))< \frac{\varepsilon}{3^{N+1-j}}, \\ \mathrm{(ii)}& \; \rho_U(Q_{jM_N}(z),0)<4 R_0. \end{align*} $$

Note that the error in this polynomial approximation for $j=N+1$ is the largest, as this error combines errors from the largest number of prior mappings.

Proof. We prove the claim by induction on j. Let $z \in U_{2R_0}$ . For the base case, we have that univalence and part (i) in the claim follow immediately from our applications of the Polynomial Implementation Lemma and in particular from equation (4.7) (with $j = i+1 = 1$ so that $i=0$ ) since $f_0= \mathrm {Id}$ . For part (ii), using part (i) (or equation (4.7)) and equation (4.6), compute

$$ \begin{align*} \rho_U(Q_{M_N}(z),0) &\leq \rho_U(Q_{M_N}(z),f_{1}(z))+\rho_U(f_{1}(z),0) \\ &<\frac{\varepsilon}{3^{N}}+3R_0 \\ &<4R_0, \end{align*} $$

which completes the proof of the base case since we had assumed $\varepsilon < R_0$ . Now suppose the claim holds for some $1\leq j <N+1$ . Then,

$$ \begin{align*} \rho_U(Q_{(j+1)M_N}(z),f_{j+1}(z) )& \leq \rho_U(Q_{jM_N,(j+1)M_N}\circ Q_{jM_N}(z),(f_{j+1}\circ f_{j}^{-1})\circ Q_{jM_N}(z)) \\ &\quad +\rho_U((f_{j+1}\circ f_{j}^{-1})\circ Q_{jM_N}(z),(f_{j+1}\circ f_{j}^{-1})\circ f_j(z) ). \end{align*} $$

Now $Q_{jM_N}(z) \in U_{4R_0} \subset U_{5R_0}$ by the induction hypothesis, so equation (4.7) implies that the first term on the right-hand side in the inequality above is less than ${\varepsilon }/{3^N}$ . Again by the induction hypothesis, $Q_{jM_N}(z) \in U_{4R_0}\subset U_{5R_0}$ , while we also have $f_j(z)\in U_{3R_0}\subset U_{5R_0}$ by equation (4.6). Thus, equation (4.5), the hyperbolic convexity of $U_{5R_0}$ from Lemmas 2.8, 2.9, and the induction hypothesis imply that the second term in the inequality is less than $\tfrac 32\cdot {\varepsilon }/{3^{N+1-j}}$ . Thus, we have $\rho _U(Q_{(j+1)M_N}(z),f_{j+1}(z) ) < {\varepsilon }/{3^{N+1-(j+1)}}$ , proving the first part of the claim.

Also, using what we just proved, equation (4.6), and our assumption that $\varepsilon < R_0$ ,

$$ \begin{align*} \rho_U(Q_{(j+1)M_N}(z),0) &\leq \rho_U(Q_{(j+1)M_N}(z),f_{j+1}(z))+\rho_U(f_{j+1}(z),0) \\ &<\frac{\varepsilon}{3^{N+1-(j+1)}}+3R_0 \\ &<4R_0, \end{align*} $$

which proves part (ii) in the claim. Univalence of $Q_{(j+1)M_N}$ follows by hypothesis as $Q_{jM_N}(U_{2R_0})\subset U_{4R_0}$ , while $Q_{(j+1)M_n,jM_N}$ is univalent on $A=U_{5R_0}\supset U_{4R_0}$ by the Polynomial Implementation Lemma as stated immediately before equation (4.7). This completes the proof of the claim, from which items (2) and (3) in the statement of Phase I follow easily.

Step 4: Proof of item (4) in the statement. To finish the proof, we need to give a bound on the size of the hyperbolic derivatives of the compositions $Q_{iM_N}$ , $1 \le i \le N+1$ . It will be of essential importance to us later that this bound not depend on the number of functions being approximated, the reason being that, in the inductive construction in Lemma 6.2, the error from the prior application of Phase II (Lemma 5.17) needs to pass through all these compositions while remaining small. This means that the estimate on the size of the hyperbolic derivative in item (2) of the statement of Lemma 3.9 is too crude for our purposes and so we have to proceed with greater care.

Let ${d}\rho _U(z)$ be the hyperbolic length element in U and write ${d}\rho _U(z)=\sigma _U(z)|{d}z|$ , where the hyperbolic density $\sigma _U$ (as introduced in the proof of Lemma 3.7) is continuous and positive on U (e.g. [Reference Keen and LakicKL07, Theorem 7.2.2]) and therefore uniformly continuous on $U_{4R_0}$ , as $U_{4R_0}$ is relatively compact in U. Let $\sigma = \sigma (R_0)> 0$ be the infimum of $\sigma _U$ on $U_{4R_0}$ so that

(4.8) $$ \begin{align} \sigma_U(z) \ge \sigma, \quad z \in U_{4R_0}. \end{align} $$

Let $z \in U_{2R_0}$ and observe that, since $\kappa \ge \kappa _0 \ge 576$ , $U \subset {\mathrm D}(0, ({1}/{288})) \subset {\mathbb D}$ . Then item (3) in the statement together with the Schwarz lemma for the hyperbolic metric (e.g. [Reference Carleson and GamelinCG93, Theorems 4.1 or 4.2]) give, for $1 \leq i \leq N+1$ , $\rho _{\mathbb D}(Q_{iM_N}(z),f_i(z))\leq \rho _{U}(Q_{iM_N}(z),f_i(z))<\varepsilon $ . If $\gamma $ is a geodesic segment in ${\mathbb D}$ from $Q_{iM_N}(z)$ to $f_i(z)$ , we see that

$$ \begin{align*} \varepsilon&>\rho_{U}(Q_{iM_N}(z),f_i(z)) \\ &\geq \rho_{\mathbb D}(Q_{iM_N}(z),f_i(z)) \\ &= \int_{\gamma}\,{d}\rho_{\mathbb D} \\ &= \int_{\gamma}\frac{2|{d}w|}{1-|w|^2} \\ &\geq \int_{\gamma}2|{d}w| \\ &=2l(\gamma) \\ &\geq 2|Q_{iM_N}(z)-f_i(z)| \end{align*} $$

and so, in particular,

(4.9) $$ \begin{align} |Q_{iM_N}(z)-f_i(z)|<\varepsilon. \end{align} $$

Now suppose further that $z \in U_{R_0}$ and set

(4.10) $$ \begin{align} \delta_0 = \delta_0(R_0) = \min_{w\in \partial U_{R_0}}{d}(w,\partial U_{({3}/{2})R_0}), \end{align} $$

where ${d}(\cdot \,, \cdot )$ denotes Euclidean distance. By [Reference NewmanNew51, Theorem VII.9.1], the winding number of $\partial U_{\tfrac 32R_0}$ (suitably oriented) around z is $1$ . Then, using [Reference ConwayCon78, Corollary IV.5.9] together with the standard distortion estimates in Theorem A.2 and equation (4.9) above, we obtain

$$ \begin{align*} |Q_{iM_N}^{\prime}(z)|&\leq |f_i^{\prime}(z)|+|Q_{iM_N}^{\prime}(z)-f_i^{\prime}(z)| \\ &= |f_i^{\prime}(z)|+\bigg| \frac{1}{2\pi i} \int_{\partial U_{({3}/{2})R_0}}\frac{Q_{iM_N}(w)-f_i(w)}{(w-z)^2}\,{d}w\bigg| \\ &\leq \frac{1+ ({1}/{288})}{(1- ({1}/{288}))^3} + \frac{\varepsilon}{2\pi\delta_0^2}l(\partial U_{({3}/{2})R_0}), \end{align*} $$

where $l(\partial U_{\tfrac 32R_0})$ is the Euclidean length of $\partial U_{\tfrac 32R_0}$ . By making $\varepsilon $ smaller if needed, we can thus ensure, for $z \in U_{R_0}$ , that

(4.11) $$ \begin{align} |Q_{iM_N}^{\prime}(z)|\leq \tfrac{3}{2}. \end{align} $$

We can make $\varepsilon $ smaller still if needed to guarantee that, if $z,w \in U_{4R_0}$ and $|z-w|<\varepsilon $ , then, by uniform continuity of $\sigma _U$ on $U_{4R_0}$ ,

(4.12) $$ \begin{align} |\sigma_U(z)-\sigma_U(w)|<\sigma. \end{align} $$

Note that both equations (4.11) and (4.12) required us to make $\varepsilon $ smaller, but these requirements depended only on $R_0$ and, in particular, not on the sequence of polynomials we have constructed. Although this means we may possibly need to run the earlier part of the argument again to find a new integer $M_N$ and then construct a new polynomial sequence $\{P_m \; : \; 1 \leq m \leq (N+1)M_N, \; 0 \leq i \leq N \}$ , our requirements on $\varepsilon $ above will then automatically be met. Alternatively, these requirements on $\varepsilon $ could be made before the sequence is constructed. However, we decided to make them here for the sake of convenience.

If $z \in U_{R_0}$ , we then have

$$ \begin{align*} |Q_{iM_N}^{\natural}(z)|&\leq |f_i^{\natural}(z)| + |Q_{iM_N}^{\natural}(z)-f_i^{\natural}(z)| \\ &=|f_i^{\natural}(z)| + \bigg|\frac{\sigma_U(Q_{iM_n}(z))}{\sigma_U(z)}Q_{iM_N}^{\prime}(z)-\frac{\sigma_U(f_{i}(z))}{\sigma_U(z)}f_{i}^{\prime}(z)\bigg| \\ &\leq |f_i^{\natural}(z)| + \bigg|\frac{\sigma_U(Q_{iM_n}(z))}{\sigma_U(z)}Q_{iM_N}^{\prime}(z)-\frac{\sigma_U(f_{i}(z))}{\sigma_U(z)}Q_{iM_N}^{\prime}(z)\bigg| \\ & \quad + \bigg|\frac{\sigma_U(f_i(z))}{\sigma_U(z)}Q_{iM_N}^{\prime}(z)-\frac{\sigma_U(f_{i}(z))}{\sigma_U(z)}f_{i}^{\prime}(z)\bigg|. \\ \end{align*} $$

We need to bound each of the three terms on the right-hand side of the above inequality. Recall that, as $g=\mathrm {Id} \in {\mathcal S}$ , we have that $|f_i^{\natural }(z)|\leq \tfrac 32$ by equation (4.5). For the second term, by equations (4.6), (4.8), (4.9), (4.11), and (4.12), and part (ii) in Claim 4.9, we have

$$ \begin{align*} &\bigg|\frac{\sigma_U(Q_{iM_n}(z))}{\sigma_U(z)}Q_{iM_N}^{\prime}(z)-\frac{\sigma_U(f_{i}(z))}{\sigma_U(z)}Q_{iM_N}^{\prime}(z)\bigg| \\ &\quad = \frac{1}{|\sigma_U(z)|}\cdot |Q_{iM_N}^{\prime}(z)|\cdot |\sigma_U(Q_{iM_N}(z))-\sigma_U(f_{i}(z))| \\ &\quad \leq \frac{1}{\sigma}\cdot \frac{3}{2}\cdot \sigma = \frac{3}{2}. \end{align*} $$

For the third and final term, recall that we chose $\kappa _0 = \kappa _0(R_0)$ sufficiently large to ensure that the conclusions of Lemmas 4.5 and 4.6 hold. We can then apply Lemmas 4.5 and 4.6, together with equation (4.11) to obtain that

$$ \begin{align*} \bigg|\frac{\sigma_U(f_i(z))}{\sigma_U(z)}Q_{iM_N}^{\prime}(z)-\frac{\sigma_U(f_{i}(z))}{\sigma_U(z)}f_{i}^{\prime}(z)\bigg| &\leq \bigg|\frac{\sigma_U(f_{i}(z))}{\sigma_U(z)}\bigg|\cdot (|Q_{iM_N}^{\prime}(z)|+|f_{i}^{\prime}(z)|) \\ &\leq \frac{10}{9}\cdot \frac{9}{8}\bigg(\frac{3}{2}+\frac{6}{5} \bigg) \\ &< 4. \end{align*} $$

Thus,

$$ \begin{align*} |Q_{iM_N}^{\natural}(z)|&\leq \frac{3}{2}+\frac{3}{2}+4 =7, \end{align*} $$

as desired.

Proof. The continuity of w implies item (1) of Carathéodory convergence in the sense of Definitions A.7, 5.1. For item (2), fix $h_0 \in I$ , let $K \subset W_{h_0}$ be compact, and let $z \in K$ . Set $\delta :={\mathrm d}(K,\partial W_{h_0})$ . By the uniform continuity of F on compact subsets of $\mathbb {T}\times I$ , we can find $\eta>0$ such that, for each $h \in I$ with $|h - h_0| < \eta $ ,

$$ \begin{align*} |\gamma_h(t) - \gamma_{h_0}(t)| < \frac{\delta}{2} \quad \mbox{for all } t\in \mathbb T, \end{align*} $$

and $\gamma _h$ is thus homotopic to $\gamma _{h_0}$ in $\mathbb C \setminus K$ . We observe that we have not assumed that $I \cap (h_0-\eta , h_0 + \eta )$ is an interval, so we may not be able to use the parameterization induced by $\gamma _h(z)$ to make the homotopy. However, using the above, it is a routine matter to construct the desired homotopy using convex linear combinations. By the above remark on winding numbers and Cauchy’s theorem, one then obtains

$$ \begin{align*} n(\gamma_h,w)= n(\gamma_{h_0},w) = 1 \quad \mbox{for all } w \in K. \end{align*} $$

Thus, if $|h - h_0| < \eta $ , then $K \subset W_h$ , and item (2) of Carathéodory convergence follows readily from this.

To show item (3) of Carathéodory convergence, let $\{h_n\}$ be any sequence in I which converges to $h_0$ and suppose N is an open connected set containing $w(h_0)$ such that $N\subset W_{h_n}$ for infinitely many n. Without loss of generality, we may pass to a subsequence to assume that $N\subset W_{h_n}$ for all n. Let $z \in N$ and connect z to $w(h_0)$ by a curve $\eta $ in N. As $[\eta ]$ is compact, there exists $\delta>0$ such that a Euclidean $\delta $ -neighborhood of $[\eta ]$ is contained in N and thus avoids $\gamma _{h_n}$ for all n. By the continuity of F, this neighborhood also avoids $\gamma _{h_0}$ . Since $w(h_0)$ and z are connected by $\eta $ which avoids $\gamma _{h_0}$ , they are in the same region determined by $\gamma _{h_0}$ so that $n(\gamma _{h_0},z)=n(\gamma _{h_0},w(h_0))$ . Hoewever, since by hypothesis $w(h_0) \in W_{h_0}$ , by [Reference NewmanNew51, Corollary 2 to Theorem VII.8.7 combined with Theorem VII.9.1], $n(\gamma _{h_0},w(h_0))=1$ whence $z \in W_{h_0}$ . As z is arbitrary, we have $N \subset W_{h_0}$ and item (3) of Carathéodory convergence, and the result then follows.