Koebe uniformization of nondegenerate domains with bounded gap-ratio

Let $\Omega_{0}$ be an arbitary domain that combined $\Omega$ with some components of $\hat{\mathbb{C}}-\Omega$. For each $b\in\mathcal{C}(\Omega)$, recall that $N(\Omega_{0},b)$ is the connected component of $\mathcal{E}(\Omega_{0})-\beta(b)$ containing $b$. Take $w\in\ulcorner b\urcorner$ and $R_{0}>0$, let $C(w,R_{0})$ be the circle of radius $R_{0}$ cetered at $w$ such that $C(w,R_{0})\subset\ulcorner N(\Omega_{0},b)\urcorner$. Since $\beta(b)\in\Omega-\{\infty\}$, the complementary components intersecting $C(w,R_{0})$ will not intersect $\beta(b)$. Hence they must be contained in $\ulcorner N(\Omega_{0},b)\urcorner$. Let $\mathcal{B}_{R_{0}}\subset\mathcal{C}_{nt}(\Omega)$ be the components that intersect both $C(w,R_{0})$ and $C(w,R_{0}/2)$. Clearly, for each $a\in\mathcal{B}_{R_{0}}$, we have $\operatorname{diam}(\ulcorner a\urcorner)\geq R_{0}/2$. This implies that $area(\ulcorner a\urcorner)\geq\kappa R^{2}_{0}/4$. However, the area of $\ulcorner N(\Omega_{0},b)\urcorner$ is limited, we can conclude that $\mathcal{B}_{R_{0}}$ is a finite set. Generally, we denote $\mathcal{B}_{R_{j}}(j=1,2,\ldots)$ as the collection of complementary components of $\Omega$ that intersect both $C(w,R_{j})$ and $C(w,R_{j}/2)$, where $R_{j}=\min_{a\in\mathcal{B}_{R_{j-1}}}\operatorname{dist}(\ulcorner a\urcorner,w).$ The setting of $R_{j}$ ensures that there is no complementary component intersecting both $C(w,R_{j-1})$ and $C(w,R_{j})$.

Now we consider a subset of $\Gamma_{\Omega_{0}}(q,b)$, where $q\in\mathcal{E}(\Omega_{0})$. Suppose that $\ulcorner q\urcorner$ is inside the circle $C(w,R_{2M+1})$ for some positive integer $M$. We take $r\in(R_{2M+1},R_{1})$ and let $C(w,r)$ be the circle of radius $r$ centered at $w$. Moreover, we let $C^{\prime}(w,r)=C(w,r)-\ulcorner b\urcorner$, then $C^{\prime}(w,r)\cup\ulcorner b\urcorner$ separates $\ulcorner q\urcorner$ from $\infty$. Set $\gamma_{r}=\pi_{\Omega_{0}}(C^{\prime}(w,r))$, we have $\{\gamma_{r}:r\in(R_{M},R_{1})\}\subset\Gamma_{\Omega_{0}}(q,b)$. For any extended metric $m$ on $\Omega_{0}$ such that $l=L_{m}(\Gamma_{\Omega_{0}}(q,b))>0$, we infer from the definition of $m-$length that

where $\chi_{a}$ is defined as $\chi_{a}(A)=1$ if $\ulcorner a\urcorner\cap A\neq\emptyset$ and $\chi_{a}(A)=0$ if $\ulcorner a\urcorner\cap A=\emptyset$. Let $A_{j}$ be the annulus bounded by $C(w,R_{j})$ and $C(w,R_{j}/2)$. In order to prevent the double counting of complementary components in the subsequent deduction, we integrate (3.2) from $r=R_{2j}/2$ to $r=R_{2j}$ for $j=1,2,\cdots$. Then we have

By Cauchy’s inequality,

Since the complementary components that intersect $A_{2j}$ must be contained in $B(w,R_{2j-1})$, the sum of their area is controlled by the area of $B(w,R_{2j-1})$. Then we have

This implies

Recall that

There is at least one complementary component $a^{*}\in\mathcal{B}_{R_{j}}$ intersecting both $C(w,R_{j})$ and $C(w,R_{j+1})$, but it does not contain any inner point of the disk bounded by $C(w,R_{j+1})$, see figure 2.

Since $\Omega$ is a domain with bounded gap-ratio, we take the first $R_{0}$ sufficiently small so that the gap-ratios of all complementary components inside $B(w,R_{0})$ to $b$ are uniformly bounded. Hence we have

where $\rho\geq 1$ is a constant. It then follows that

We indicate that the complementary components intersecting $A_{2j}$ must be contained in $B(w,R_{2j-1})\setminus B(w,R_{2j+1})$, then for each $a\in\mathcal{C}(\Omega_{0})$, there is at most one $j\in\{1,2,\cdots,M\}$ such that $\ulcorner a\urcorner\cap A_{2j}\neq\emptyset$. We add (3.3) from $j=1$ to $j=M$, to get

This implies $Ml^{2}\leq 4\pi\rho^{2}(1+\kappa^{-1})A(m)$. Hence

Given $\varepsilon>0$, we can choose sufficiently large $M$ so that the right hand of (3.4) is less than $\varepsilon$. Then we have for each $w\in\ulcorner b\urcorner$, there is a $R_{2M+1}>0$ such that $EL(\Gamma)<\varepsilon$ when $d(\ulcorner q\urcorner,w)<R_{2M+1}$. According to the compactness of $\ulcorner b\urcorner$, the proof is completed. ∎

We first normalize the sequence $f_{n}$ by requiring $\infty\in\Omega$ and $f_{n}(\infty)=\infty$ due to theorem 3.2. Without loss of generality, we assume that $B_{0}^{*}$ is the Hausdorff limit of the sequence $\ulcorner f_{n}(b)\urcorner$. Let $B^{*}$ be the complement of the connected component of $\hat{\mathbb{C}}-B_{0}^{*}$ that contains $\infty$. Our goal is to show that $B^{*}=\ulcorner f(b)\urcorner$. According to the Carathéodory kernel convergence theorem, we have $B^{*}\subset\ulcorner f(b)\urcorner$. Hence we only need to show that $\ulcorner f(b)\urcorner\subset B^{*}$.

Striving for a contradiction, we assume $\ulcorner f(b)\urcorner-B^{*}\neq\emptyset$. Let $p^{*}\in\ulcorner f(b)\urcorner-B^{*}$ as shown in figure 3. Set $\Omega^{*}_{n}=\hat{f}(\Omega_{n})$, $q^{*}_{n}=\pi_{\Omega^{*}_{n}}(p^{*})$ and $q_{n}=\hat{f}_{n}^{-1}(q_{n}^{*})$. (It is possible that $q_{n}\in\Omega_{n}$ or $q_{n}\in\mathcal{C}(\Omega_{n})$.) Since $p^{*}\in\ulcorner f(b)\urcorner$, it follows that the distance in $\mathbb{C}$ between $\ulcorner q_{n}\urcorner$ and $\ulcorner b\urcorner$ tends to zero as $n\rightarrow\infty$. (Otherwise, we consider a Jordan curve in $\Omega$ which separates $\ulcorner b\urcorner$ from $\ulcorner q_{n}\urcorner$ for infinite many $n$, then the image of this curve under $f$ will separate $\ulcorner f(b)\urcorner$ from $p^{*}$.)

Part I: Prove that $EL(\Gamma_{n})\rightarrow 0$ as $n\rightarrow\infty$.

Recall that $\beta(b)\subset\Omega-\{\infty\}$ is a Jordan curve that separates $b$ from $\infty$ in $\mathcal{E}(\Omega)$, and $N(\Omega_{n},b)$ is the connencted component of $\mathcal{E}(\Omega_{n})\setminus\beta(b)$ that contains $b$. Similarly, we let $N(\Omega^{*}_{n},b)$ be the connencted component of $\mathcal{E}(\Omega^{*}_{n})\setminus f_{n}(\beta(b))$ that contains $f_{n}(b)$. Let $\Gamma_{n}=\Gamma_{\Omega_{n}}(q,b)$ be the collection of all Jordan curves $\gamma\subset N(\Omega_{n},b)\setminus\{b,q_{n}\}$ satisfies either of the following conditions:

• •

  $\gamma$ separates $\{b,q_{n}\}$ from $\infty$;

• •

  $\bar{\gamma}=\gamma\cup\{b\}$ and $\bar{\gamma}$ separates $q_{n}$ from $\infty$.

Since $d(\ulcorner q_{n}\urcorner,\ulcorner b\urcorner)$ tends to zero as $n\rightarrow\infty$, it follows from theorem 3.1 and definition 2.1 that $EL(\Gamma_{n})$ tends to zero as $n\rightarrow\infty$.

Let $\Gamma^{*}_{n}=\{\hat{f}_{n}(\gamma):\gamma\in\Gamma_{n}\}$.

Part II: Fix an extended metric to get a positive lower bound of $EL(\Gamma^{*}_{n})$.

It follows from the normalization $f_{n}(\infty)=\infty$ that there is some radius $R$ so that $f_{n}(\beta(b))\subset B(0,R)$ for all $n$. We now define for all $z\in\Omega^{*}_{n}$:

and for $a^{*}\in\mathcal{C}(\Omega^{*}_{n})$:

It is clear that $m^{*}_{n}$ is an extended metric on $\mathcal{E}(\Omega^{*}_{n})$. In the following we will show that the ratio $L_{m^{*}_{n}}(\gamma^{*})/A(m^{*}_{n})$ is strictly greater than zero under the metric $m^{*}_{n}$ for any $n$, so that the transboundary extremal length $EL(\Gamma^{*}_{n})$ is bounded away from zero as $n\rightarrow\infty$.

Since $p^{*}\notin B^{*}$, there is a curve $\eta\subset\hat{\mathbb{C}}-B^{*}_{0}$ that connects $p^{*}$ and $\infty$ as shown in figure 3. Let $\delta^{*}>0$ be a number that smaller than the distance from $\eta$ to $B^{*}_{0}$. It is obvious that for sufficiently large $n$ we have $\operatorname{diam}(\ulcorner\gamma^{*}\urcorner)>\delta^{*}$ for all $\gamma^{*}\in\Gamma^{*}_{n}$.

Let $z_{1},z_{2}\in\ulcorner\gamma^{*}\urcorner$, and for any $z\in\mathbb{C}$, set $\phi(z)=\mid z-z_{1}\mid$. It is easy to know that $\phi(z)$ is a non-negative real valued function and $\phi(\ulcorner\gamma^{*}\urcorner)$ covers the interval $[\ 0,\mid z_{1}-z_{2}\mid\ ]$. Parametering $\gamma^{*}$ as $\gamma^{*}:I\rightarrow\mathcal{E}(\Omega^{*}_{n})$, where $I$ is an interval. It is clear that $\gamma^{*-1}(\Omega^{*}_{n})$ is a collection of connected components of $I(\gamma^{*-1}(\Omega^{*}_{n}))$. We denote the collection of connected components as $\mathcal{J}$. Then for each $J\in\mathcal{J}$, we have $\operatorname{diam}(\phi(\gamma^{*}(J)))\leq L_{m^{*}_{n}}(\gamma^{*}(J)).$

On the other hand, for each $a^{*}\in\gamma^{*}\cap\mathcal{C}(\Omega^{*}_{n})$, it follows that

Since the set of such $J$ and $a^{*}$ is countable, and the interval $[\ 0,\mid z_{1}-z_{2}\mid\ ]$ is covered by the sets

this implies

Since $z_{1},z_{2}$ are chosen arbitrarily, we conclude that

Moreover, we note that $\hat{f}_{n}(\Omega_{n})$ is $\kappa-$nondegenerate. Then for each non-trival component $a^{*}\in\mathcal{C}(\Omega^{*}_{n})$, we have

Thus $m^{*}_{n}(a^{*})^{2}\leq\kappa^{-1}area(\ulcorner a^{*}\urcorner)$ for $\ulcorner a^{*}\urcorner\subset B(0,R)$ and $m^{*}_{n}(a^{*})^{2}=0$ for $\ulcorner a^{*}\urcorner\subset\hat{\mathbb{C}}-B(0,R)$. Moreover, we have