Lowener Theory on Analytic Universal Covering Maps

TL;DR

This paper extends Loewner theory to chains of universal covering maps, providing a decomposition, PDE generalization, and geometric characterizations, linking classical theory with hyperbolic geometry and Fuchsian groups.

Contribution

It introduces a new Loewner framework for universal covering maps, including a factorization, PDE, and evolution equations for deck transformations, unifying classical and geometric aspects.

Findings

Decomposition of Loewner chains into analytic and univalent parts.

Generalization of Loewner--Kufarev PDE for non-univalent chains.

Characterization of universal covering map chains via domain monotonicity.

Abstract

We study Loewner chains in $\mathcal{H}_0(\mathbb{D})$ without assuming univalence of each element. We prove a decomposition: every chain admits a factorization $f_t=F\circ g_t$, where $F$ is analytic on $\mathbb{D}(0,r)$ with $r=\lim_{t \nearrow \sup I} f_t'(0)$, and $\{g_t\}$ is a classical Loewner chain of univalent functions. Under a mild regularity assumption on $t \mapsto f_t'(0)$, we derive a partial differential equation that generalizes the Loewner--Kufarev equation. We then develop a Loewner theory for chains of universal covering maps. We characterize such chains in terms of domain families $\{\Omega_t\}$: continuity and monotonicity of $\{f_t\}$ are equivalent to kernel continuity and monotonicity of $\{\Omega_t\}$. We further show that the connectivity $C(\Omega_t)=\#(\hat{\mathbb{C}}\setminus \Omega_t)$ is a left-continuous nondecreasing function of $t$. Building on these results, we formulate a Loewner theory on Fuchsian groups and obtain evolution equations for deck transformations. As an application, we study hyperbolic metrics and establish a formula for the logarithmic derivative of the hyperbolic density along the chain. Our results provide a unified framework linking classical Loewner theory, covering maps, and the geometry of hyperbolic domains.

Equations

$$f_s\prec f_t\text{for all }(s,t)\in I_{+}^2,$$

$$\frac{\partial k}{\partial a(t)}(z,t):=$$

$$P(z,t)=\underset{\begin{array}{c}{t}_1\le t\le t_2\\ t_2-t_1\downarrow 0\end{array}}{\lim }\frac{\frac{\omega (z,t_1,t_2)}{z}-1}{\frac{a_{t_1}}{a_{t_2}}-1},z\in \mathbb{D}$$

$$\frac{\partial \omega }{\partial a(t)}(z,t)=$$

$$\frac{\partial \omega }{\partial a(t)}(z,t,t_0)=$$

$$\frac{\partial \omega }{\partial a(t)}(z,t_0,t)=$$

$$\frac{\partial f}{\partial a(t)}(z,t)=$$

$$\frac{dw}{da(t)}(t)=-\frac{1}{a(t)}wP(w,t),s\le t\in I$$

$${\Omega }_0\subset {\Omega }_1\text{if and only if}{f}_0\prec f_1.$$

$$\mathbb{D}$$

$${\Omega }_n\to \Omega \text{ in the sense of kernel with respect to a}$$

$$\text{ if and only if }{f}_n\to f\text{ locally uniformly on }\mathbb{D}.$$

$$C\cap \bigcap _{t\in I}{E}_t\ne \varnothing .$$

$${\sigma }_{t_1{t}_2}\circ {\sigma }_{t_0{t}_1}={\sigma }_{t_0{t}_2}$$

$$\frac{\partial \gamma }{\partial a(t)}(z,t)=$$

$$\frac{\partial \gamma }{\partial t}(z,t)=\frac{\dot{a}(t)}{a(t)}\left\{zP(z,t){\gamma }^{\prime }(z,t)-\gamma (z,t)P(\gamma (z,t),t)\right\}$$

$${\omega }_{t,t}={\mathrm{id}}_{\mathbb{D}}\text{and}{\omega }_{t_1,t_2}\circ {\omega }_{t_0,t_1}={\omega }_{t_0,t_2}$$

$$a_{s,t}={\omega }_{s,t}^{\prime }(0)\in (0,1],(s,t)\in I_{+}^2.$$

$$a_{t,t}=1\text{and}{a}_{t_1,t_2}{a}_{t_0,t_1}=a_{t_0,t_2}$$

$$a(t)=\{\begin{array}{ll}c{a}_{t{t}_0},& t\le t_0,\\ \frac{c}{a_{t_{0}t}},& t>t_0.\end{array}$$

$$a_{st}=\frac{a(s)}{a(t)},(s,t)\in I_{+}^2.$$

$$\left|\frac{{\omega }_{s,t}(z)}{z}-a_{s,t}\right|\le |z|\left|1-a_{s,t}\frac{{\omega }_{s,t}(z)}{z}\right|,$$

$$\left|\frac{{\omega }_{s,t}(z)}{z}-\frac{a_{s,t}(1-|z{|}^2)}{1-a_{s,t}^2|z{|}^2}\right|\le \frac{\left(1-a_{s,t}^2\right)|z|}{1-a_{s,t}^2|z{|}^2},$$

$$\left|{\omega }_{s,t}(z)-z\right|\le (1-a_{s,t})\frac{|z|(1+|z|)}{\left(1-a_{s,t}|z|\right)},$$

$$\left|{\omega }_{s,t}(z)\right|\le |z|\frac{|z|+a_{s,t}}{1+a_{s,t}|z|},$$

$$\left|{\omega }_{t_0,t_2}(z)-{\omega }_{t_0,t_1}(z)\right|\le (1-a_{t_1,t_2})\frac{|z|(1+|z|)}{\left(1-a_{t_1,t_2}|z|\right)}.$$

$$\left|\frac{g(z)-c}{1-cg(z)}\right|\le |z|,z\in \mathbb{D}$$

$$\left|g(z)-\frac{(1-|z{|}^2)c}{1-c^2|z{|}^2}\right|\le \frac{(1-c^2)|z|}{1-c^2|z{|}^2}.$$

$$1-\frac{(1-|z{|}^2)c}{1-c^2|z{|}^2}=\frac{(1-c)\left(1+c|z{|}^2\right)}{1-c^2|z{|}^2}$$

$$\left|g(z)-1\right|$$

$$\le \frac{(1-c^2)|z|}{1-c^2|z{|}^2}+\frac{(1-c)(1+c|z{|}^2)}{1-c^2|z{|}^2}\le \frac{(1-c)|(1+|z|)}{1-c|z|},$$

$$\left|{\omega }_{t_1,t_2}({\omega }_{t_0,t_1}(z))-{\omega }_{t_0,t_1}(z)\right|\le (1-a_{t_1,t_2})\frac{|{\omega }_{t_0,t_1}(z)|(1+|{\omega }_{t_0,t_1}(z)|)}{\left(1-a_{t_1,t_2}|{\omega }_{t_0,t_1}(z)|\right)}.$$

$$|g(z)|\le \frac{(1-|z{|}^2)c}{1-c^2|z{|}^2}+\frac{(1-c^2)|z|}{1-c^2|z{|}^2}=\frac{|z|+c}{1+c|z|}.$$

$$1-\epsilon <a_{s,t},\text{for all s,t with s≤t0≤t and 0<t−s<δ.}$$

$$a_{s,t}\ge a_{s^{\ast },t^{\ast }}>1-\epsilon$$

Full text

1. 1 Introduction
2. 2 Basic Estimates and the Structure Theorem
3. 3 Loewner-Kufarev Differential Equations
4. 4 Solutions to Loewner-Kufarev Ordinary Differential Equations
5. 5 Schlicht subordination and connecting chain
6. 6 Kernel convergence of domains and locally uniform convergence of covering maps
7. 7 Kernel convergence and connectivity of domains
8. 8 Loewner theory on Fuchsian Groups
9. 9 Separation Lemma

1. Introduction

Let ${\mathbb{C}}$ be the complex plane, $\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$ the extended complex plane and $\mathbb{D}(c,r)=\{z\in\mathbb{C}:|z-c|<r\}$ for $c\in\mathbb{C}$ and $r>0$. In particular we denote the unit disc $\mathbb{D}(0,1)$ by $\mathbb{D}$. Let $\mathcal{H}(\mathbb{D})$ be the the linear space of all analytic functions $f$ in $\mathbb{D}$, endowed with the topology of locally uniform convergence on $\mathbb{D}$. Set $\mathcal{H}_{0}(\mathbb{D})=\{f\in\mathcal{H}(\mathbb{D}):f(0)=0\text{ and }f^{\prime}(0)>0\}$ and $\mathfrak{B}=\{\omega\in\mathcal{H}_{0}(\mathbb{D}):|\omega(z)|\leq 1\}$.

Let $\Omega$ be a proper subdomain of $\mathbb{C}$. Then, by the Riemann mapping theorem, $\Omega$ is simply connected if and only if there exists a conformal mapping (i.e., a bijective and biholomorphic mapping) $f:\mathbb{D}\rightarrow\Omega$. Now consider a domain $\Omega$ in $\mathbb{C}$ which is not necessary simply connected. Then, by the Koebe uniformization theorem which is “the single most important theorem in the whole theory of analytic functions of one variable” (cf. Ahlfors [2, Chap.10]), the complement $\mathbb{C}\backslash\Omega$ contains at least two points if and only if there exists an analytic universal covering map $f$ of $\mathbb{D}$ onto $\Omega$, i.e, $\Omega$ is hyperbolic. Therefore analytic universal covering maps of $\mathbb{D}$ is a natural generalization of conformal mappings of $\mathbb{D}$, i.e., univalent analytic functions in $\mathbb{D}$.

The theory of univalent functions has a long history and is still an active field. The geometric theory of analytic universal covering maps has also a long history and well studied concerning the theory of Fuchsian groups. Furthermore since the hyperbolic metric on a hyperbolic domain is induced by projecting the Poincaré metric on $\mathbb{D}$ by means of any analytic universal covering maps, we can translate results on hyperbolic metrics as theorems on universal covering maps.

In the present article we shall show the Loewner theory, a powerful method in the theory of univalent functions, is applicable to study analytic universal covering maps. In 1923 Loewner [17] found that for any bounded slit mapping $f$ of $\mathbb{D}$ has a parametric representation satisfying a differential equation known as the Loewner differential equation. The parametric representation method was extensively developed and generalized by a number of researchers. We only mention the fundamental contribution by Kufarev [15], [16] and Pommerenke [24], [25, Chapter 6]. See [5] for details on the history of the theory and further references.

We focus on Loewner chains of analytic functions in $\mathbb{D}$ introduced by Pommerenke [24] in 1965. A function $f_{0}\in\mathcal{H}(\mathbb{D})$ is said to be subordinate to $f_{1}\in\mathcal{H}(\mathbb{D})$ ($f_{0}\prec f_{1}$ in short), if there exists an analytic function $\omega:\mathbb{D}\rightarrow\mathbb{D}$ with $\omega(0)=0$ and $f_{0}=f_{1}\circ\omega$. Notice that when $f_{0},f_{1}\in\mathcal{H}_{0}(\mathbb{D})$, $f_{0}\prec f_{1}$ implies $\omega\in\mathfrak{B}$.

Definition 1.1**.**

Let $I\subset[-\infty,\infty]$ and $\{f_{t}\}_{t\in I}$ a family of analytic functions in $\mathcal{H}_{0}(\mathbb{D})$. Then $\{f_{t}\}_{t\in I}$ is said to be a Loewner chain if

[TABLE]

where $I_{+}^{2}=\{(s,t)\in I^{2}:s\leq t\}$. For $(s,t)\in I_{+}^{2}$ let $\omega_{s,t}\in\mathfrak{B}$ be the unique function satisfying $f_{s}=f_{t}\circ\omega_{s,t}$. We call $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ the associated transition family of $\{f_{t}\}_{t\in I}$. By the Schwarz lemma, the function $f_{t}^{\prime}(0)$ is nondecreasing and positive on $I$ by definition. We say that a Loewner chain $\{f_{t}\}_{t\in I}$ is strictly increasing if $f_{t}^{\prime}(0)$ is strictly increasing in $t\in I$, i.e., $f_{s}^{\prime}(0)<f_{t}^{\prime}(0)$ whenever $s,t\in I$ with $s<t$. A Loewner chain $\{f_{t}\}_{t\in I}$ is called normalized if $f_{t}^{\prime}(0)=e^{t}$, $t\in I$. Also we say that a Loewner chain $\{f_{t}\}_{t\in I}$ is continuous if the mapping $I\ni t\mapsto\mathcal{H}(\mathbb{D})$ is continuous. In other words at any $t_{0}\in I$, $f_{t}\rightarrow f$ locally uniformly on $\mathbb{D}$ as $t\rightarrow t_{0}$ in $I$, which is equivalent to say that $f(z,t)$ is continuous on $\mathbb{D}\times I$ as a function of two variables. Here we follow the usual convention and write $f(z,t):=f_{t}(z)$, which will be used without mention in the sequel.

It should be noticed that in many books and papers each $f_{t}$ is assumed to be univalent on $\mathbb{D}$ in the definition of a Loewner chain, however we do not assume the univalence of each $f_{t}$. Without assuming the univalence Pommerenke ([24]) already proved that if $\{f_{t}\}_{t\in I}$ is a normalized Loewner chain, then at almost every $t$ in the interior of $I$, $\{f_{t}\}_{t\in I}$ satisfies a partial differential equation called the Loewner-Kufarev partial differential equation, which is a generalization of the original Loewner equation.

Since we shall mainly concern with continuous Loewner chains, hereafter we assume $I$ is connected, i.e., $I$ is an interval in $[-\infty,\infty]$. We consider the following three classes.

• (I)

The class of all Loewner chains $\{f_{t}\}_{t\in I}$ such that each $f_{t}\in\mathcal{H}_{0}(\mathbb{D})$ is univalent for all $t\in I$.

• (II)

The class of all Loewner chain $\{f_{t}\}_{t\in I}$ such that each $f_{t}\in\mathcal{H}_{0}(\mathbb{D})$ is the universal covering map of $\mathbb{D}$ onto $\Omega_{t}:=f_{t}(\mathbb{D})$ for all $t\in I$.

• (III)

The class of all Loewner chain $\{f_{t}\}_{t\in I}$.

In the first half of the present article (§2–5) we shall study the class (III). Particularly we shall deal with properties which are shared with (I) and (III) in §2-4 and not shared in §5. The class (III) was introduced by Pommerenke (cf. [24]). Notice that Pommerenke did not assume the connectivity of $I$ and mainly concerned with normalized Loewner chains. After ten years Pommerenke (cf. [25]) introduced the class (I) and gave a detailed study particularly on normalized Loewner chains of univalent functions. In the latter half (§6–9) we shall study the class (II) which is a geometrically natural generalization of (I).

In §2 we introduce basic estimates on transition families and prove that a Loewner chain $\{f_{t}\}_{t\in I}$ is continuous if and only if the function $a(t):=f_{t}^{\prime}(0)$ is continuous in $t\in I$. The main result of this section is the following.

Theorem 1.2** (Structure Theorem).**

Let $I\subset[-\infty,\infty]$ be an interval with $\beta=\sup I\not\in I$ and $\{f_{t}\}_{t\in I}$ a continuous Loewner chain with $a(\beta)=\lim_{t\uparrow\beta}a(t)\in(0,\infty]$.

• (i)

The locally uniform limit $f_{\beta}=\lim_{t\uparrow\beta}f_{t}$ exists if and only if $a(\beta)<\infty$. In this case there exists uniquely an analytic function $F:\mathbb{D}(0,a(\beta))\rightarrow\mathbb{C}$ with $F(0)=F^{\prime}(0)-1=0$, and a continuous Loewner chain $\{g_{t}\}_{I\cup\{\beta\}}$ of univalent functions with $\bigcup_{t\in I}g_{t}(\mathbb{D})=g_{\beta}(\mathbb{D})=\mathbb{D}(0,a(\beta))$ such that $f_{t}=F\circ g_{t}$ and $g_{t}^{\prime}(0)=a(t)$ for $t\in I\cup\{\beta\}$.

• (ii)

If $a(\beta)=\infty$, then there exists uniquely an entire function $F:\mathbb{C}\rightarrow\mathbb{C}$ with $F(0)=F^{\prime}(0)-1=0$, and a continuous Loewner chain $\{g_{t}\}_{t\in I}$ of univalent functions with $\bigcup_{t\in I}g_{t}(\mathbb{D})=\mathbb{C}$ such that $f_{t}=F\circ g_{t}$ and $g_{t}^{\prime}(0)=f_{t}^{\prime}(0)$ for $t\in I$.

In §3 we shall show, without assuming normalization, that a strictly increasing and continuous Loewner chain $\{f_{t}\}_{t\in I}$ and the associated transition family $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ satisfies a partial and ordinary differential equations with respect to the strictly increasing function $a(t):=f_{t}^{\prime}(0)$. We denote the partial derivatives of a function $k(z,t)$ with respect to $a(t)$ by

[TABLE]

and denote the Lebesgue-Stieltjes measure with respect to $a(t)$ by $\mu_{a}$.

Theorem 1.3**.**

Let $\{f_{t}\}_{t\in I}$ be a strictly increasing and continuous Loewner chain with the associated transition family $\{\omega(\cdot,s,t)\}_{(s,t)\in I_{+}^{2}}$ and $a(t)=f_{t}^{\prime}(0)$. Then there exists a $G_{\delta}$ set $N(\subset I)$ with $\mu_{a}(N)=0$ such that for all $z\in\mathbb{D}$ and $t\in I\backslash N$ the limit

[TABLE]

exists and the convergence is locally uniform on $\mathbb{D}$ for each fixed $t\in I\backslash N$. Extend $P(z,t)$ by $P(z,t)=1$ for $(z,t)\in\mathbb{D}\times N$. Then $P$ is Borel measurable on $\mathbb{D}\times I$, analytic in $z$ and satisfies $\operatorname{Re}P(z,t)>0$ and $P(0,t)=1$. Furthermore

[TABLE]

Here $\omega^{\prime}(z,t_{1},t_{2})$ and $f^{\prime}(z,t)$ mean the complex derivatives with respect to $z$. These differential equations are generalizations of the usual Loewner-Kufarev equations.

Let $\psi:I\rightarrow\mathbb{R}$ be a strictly increasing and continuous function. Let $\mu_{\psi}$ and ${\mathcal{F}}_{\psi}$ be the Lebesgue-Stieltjes measure and the associated $\sigma$-algebra on $I$ with respect to $\psi$, respectively. Notice that $(I,{\mathcal{F}}_{\psi},\mu_{\psi})$ is a complete measure space with $\mathcal{B}(I)\subset{\mathcal{F}}_{\psi}$. Here $\mathcal{B}(I)$ is the Borel $\sigma$-algebra in $I$. A family $\{P(\cdot,t)\}_{t\in I}$ in $\mathcal{H}(\mathbb{D})$ is called a Herglotz family if $\operatorname{Re}P(z,t)>0$ in $\mathbb{D}$ and $P(0,t)=1$ for each $t\in I$. Let $\mathcal{F}$ be a $\sigma$-algebra in $I$. Then we say that a Herglotz family $\{P(\cdot,t)\}_{t\in I}$ is $\mathcal{F}$-measurable if for each fixed $z\in\mathbb{D}$, $P(z,t)$ is $\mathcal{F}$-measurable in $t\in I$.

In §4 we shall show that the ordinary differential equation (1.5) always has unique solutions. Precisely for a given strictly increasing, continuous and positive function $a(t)$ on $I$, $\mathcal{F}_{a}$-measurable Herglotz family $\{P(\cdot,t)\}_{t\in I}$ and fixed $s\in I$, the ordinary differential equation

[TABLE]

with an initial condition $w(s)=z$ with $z\in\mathbb{D}$ has the unique solution $\omega_{s,t}(z)$ on $I\cap[t_{0},\infty)$. Then $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ forms a transition family. We shall also give a representation formula for $g_{t}(z)$ in Theorem 1.2.

In §5 we shall treat properties which hold for Loewner chains in the class (I) and do not necessarily hold for Loewner chains in (III). We say that $f\in\mathcal{H}_{0}(\mathbb{D})$ is maximal in the sense of continuous Loewner chain if there does not exist $\varepsilon>0$ and a continuous Loewner chain $\{f_{t}\}_{0\leq\leq\varepsilon}$ with $f_{0}=f$ and $f_{\varepsilon}^{\prime}(0)>f^{\prime}(0)$. Pommerenke ([25]) showed for any univalent $f\in\mathcal{H}_{0}(\mathbb{D})$ there exists a continuous Loewner chain $\{f_{t}\}_{0\leq t<\infty}$ of univalent functions satisfying $f_{0}=f$ and $\lim_{t\rightarrow\infty}f_{t}^{\prime}(0)=\infty$. Thus a univalent $f\in\mathcal{H}_{0}(\mathbb{D})$ is never maximal.

Theorem 1.4**.**

If $f\in\mathcal{H}_{0}(\mathbb{D})$ does not have nontangential limit at almost every $\zeta\in\partial\mathbb{D}$, then $f$ is maximal in the sense of continuous Loewner chain.

Loewner chains of covering maps, which is the main subject in the latter half of the present article, is a geometrically natural generalization of Loewner chains of univalent functions. We point out two properties shared between univalent functions and universal covering maps. For $j=1,2$ let $\Omega_{j}$ be a hyperbolic domain in $\mathbb{C}$ and $f_{j}:\mathbb{D}\rightarrow\Omega_{j}$ a universal covering map with $f_{0}(0)=f_{1}(0)$. If $\Omega_{1}$ and $\Omega_{2}$ are simply connected, then since $f_{1}$ and $f_{2}$ are univalent, it is easy to see that

[TABLE]

The above equivalence also holds for the general case. Indeed, if $f_{0}\prec f_{1}$, then it is clear that $\Omega_{0}\subset\Omega_{1}$. Assume that $\Omega_{0}\subset\Omega_{1}$. Then by the general lifting lemma (for an example see [20, Lemma 97.1]), the mapping $f_{0}:\mathbb{D}\rightarrow\Omega_{0}(\subset\Omega_{1})$ can be lifted to the unique continuous map $\omega:\mathbb{D}\rightarrow\mathbb{D}$ such that $f_{0}=f_{1}\circ\omega$ with $\omega(0)=0$,

[TABLE]

Since $f_{0}$ and $f_{1}$ are analytic and locally univalent, $\omega$ is also analytic and hence $\omega\in\mathfrak{B}$. Thus $f_{0}\prec f_{1}$.

Next let $\Omega,\Omega_{1},\Omega_{2},\ldots$ be hyperbolic domains in $\mathbb{C}$ with $a\in\Omega\cap\bigcap_{j=1}^{\infty}\Omega_{j}$ for some $a\in\mathbb{C}$, and let $f,f_{1},f_{2},\ldots\in\mathcal{H}(\mathbb{D})$ be the corresponding unique universal covering maps of $\mathbb{D}$ onto $\Omega,\Omega_{1},\Omega_{2},\ldots$ with $f(0)=f_{1}(0)=f_{2}(0)=\cdots=a$ and $f^{\prime}(0)>0,\;f_{1}^{\prime}(0)>0,\;f_{2}^{\prime}(0)>0,\;\ldots$, respectively. Let $\operatorname{Ker}(\{\Omega_{n}\}_{n=1}^{\infty},a)$ be the set consists of $a$ and all points $w$ such that there exists a domain $H$ and $N\in\mathbb{N}$ with $a,w\in H\subset\Omega_{n}$ for $n\geq N$. By definition $\operatorname{Ker}(\{\Omega_{n}\}_{n=1}^{\infty},a)$ is a domain containing $a$ or coincides with $\{a\}$. A sequence $\{\Omega_{n}\}_{n=1}^{\infty}$ is said to converge to a domain $\Omega$ in the sense of kernel with respect to $a$ if $\operatorname{Ker}(\{\Omega_{n_{k}}\}_{k=1}^{\infty},a)=\Omega$ for all subsequence $\{\Omega_{n_{k}}\}_{k=1}^{\infty}$ of $\{\Omega_{n}\}_{n=1}^{\infty}$.

If $\Omega,\Omega_{1},\Omega_{2},\ldots$ are simply connected, then since $f,f_{1},f_{2},\ldots$ are univalent, the Carathéodry kernel convergence theorem assures the equivalence relation

[TABLE]

In case that $\Omega,\Omega_{1},\Omega_{2},\ldots$ are not necessary simply connected, the Carathéodry theorem, which plays a fundamental role in the classical Loewner theory, is not applicable. However, instead of the Carathéodry theorem, we have a cerebrated generalization by Hejhal [13], and (1.8) still holds in this case. In §6 we shall introduce Pommerenke’s criterion for kernel convergence and slightly generalize the Hejhal’s theorem.

On the basis of equivalence relations (1.7) and (cf. 1.8), Pommerenke ([25]) developed his theory on Loewner chains of univalent functions in 1975. In the same way we have the following.

Theorem 1.5**.**

Let $\{\Omega_{t}\}_{t\in I}$ be a family of hyperbolic domains in $\mathbb{C}$ with $0\in\Omega_{t}$, $t\in I$. For each $t\in I$ let $f_{t}$ be the unique universal covering map of $\mathbb{D}$ onto $\Omega_{t}$ with $f_{t}(0)=0$ and $f_{t}^{\prime}(0)>0$. Then $\{f_{t}\}_{t\in I}$ is a continuous Loewner chain of universal covering maps if and only if $\{\Omega_{t}\}_{t\in I}$ is nondecreasing and continuous.

Here by “$\{\Omega_{t}\}_{t\in I}$ is continuous” we mean that for all $t_{0}\in I$ and $\{t_{n}\}_{n=1}^{\infty}\subset I$ with $t_{0}\not=t_{n}\rightarrow t_{0}$, $\{\Omega_{t_{n}}\}_{n=1}^{\infty}$ converges to $\Omega_{t_{0}}$ in the sense of kernel with respect to [math].

It seems natural to infer that Pommerenke’s theory can be generalized to Loewner chains $\{f_{t}\}_{t\in I}$ of universal covering maps. For an example, needless to say, $\{f_{t}\}_{t\in I}$ and the associated transition family satisfy the generalized Loewner-Kufarev equations (1.3)-(1.6). However we encounter some phenomena which never occur in the univalent case.

For a domain $G$ in $\hat{\mathbb{C}}$ we denote the connectivity of $G$ by $C(G)(\in\mathbb{N}\cup\{\infty\})$, i.e., $C(G)$ is the number of the connected components of $\hat{\mathbb{C}}\backslash G$. Let $\Omega,\Omega_{1},\Omega_{2},\ldots$ be hyperbolic domains in $\mathbb{C}$ and assume that $\Omega_{n}\rightarrow\Omega$ ($n\rightarrow\infty$) in the sense of kernel. If $\Omega_{1},\Omega_{2},\ldots$ are simply connected, then it is easy to see that $\Omega$ is also simply connected. Thus $C(\Omega)=1=\lim_{n\rightarrow}C(\Omega_{n})$ holds. However in general one can only show the lower semicontinuity $C(\Omega)\leq\liminf_{n\rightarrow\infty}C(\Omega_{n})$.

Example 1.6**.**

*Let $E_{0}$ be the closed line segment connecting $1$ and $2$ in the complex plane $\mathbb{C}$. For $0<t\leq 1$ let $E_{t}$ be the set obtained from $E_{0}$ by removing the concentric open line segment of length $3^{-1}t$. Then $E_{1}$ consists of two closed line segments. Next for $1<t\leq 2$ let $E_{t}$ be the set obtained from each of two closed line segments by removing the concentric open line segments of length $3^{-2}(t-1)$. This process can be continued indefinitely and we obtain $\{E_{t}\}_{t\in[0,\infty)}$. Set $E_{\infty}=\bigcap_{t\geq 0}E_{t}$. Then $E_{\infty}$ is a translation of the Cantor ternary set. Consider the universal covering maps $f_{t}$ of $\mathbb{D}$ onto $\Omega_{t}:=\mathbb{C}\backslash E_{t}$ with $f_{t}(0)=0$ and $f_{t}^{\prime}(0)$. It is not difficult to see that $\{\Omega_{t}\}_{t\in[0,\infty]}$ is strictly increasing and continuous in the sense of kernel. This implies the family $\{f_{t}\}_{t\in[0,\infty]}$ is a strictly increasing and continuous Loewner chain of universal covering maps.

E_{0}$$E_{1}$$E_{2}$$E_{t} *

Now let $\{\Omega_{t}\}_{t\in I}$ be a continuous and nondecreasing family of domains in $\hat{\mathbb{C}}$ with $E_{t}=\hat{\mathbb{C}}\backslash\Omega_{t}$, $t\in I$. Example 1.6 shows that if $C$ be a connected component of $E_{t_{0}}$ at some $t_{0}\in I$, then $C\cap E_{t}$, $t\in I\cap[t_{0},\infty)$, shrink and may split into many pieces with time $t$. In §7 we show the following.

Theorem 1.7**.**

Let $\{\Omega_{t}\}_{t\in I}$ be a continuous and nondecreasing family of domains in $\hat{\mathbb{C}}$ with $E_{t}=\hat{\mathbb{C}}\backslash\Omega_{t}$. If $C$ be a connected component of $\hat{\mathbb{C}}\backslash\Omega_{t_{0}}$ at some $t_{0}\in I$, then

[TABLE]

Particularly $C(\Omega_{t})$ is left continuous and nondecreasing in $t\in I$.

We notice that the theorem has a trivial counterpart for Loewner chains of universal covering maps.

A subset $A$ of $\hat{\mathbb{C}}$ is said to be totally disconnected if each connected component of $A$ is a one-point set. We say that a domain $\Omega$ in $\hat{\mathbb{C}}$ is maximal in the sense of kernel if there are no continuous family of domains $\{\Omega_{t}\}_{0\leq t<\varepsilon}$ with $\Omega_{0}=\Omega$ and $\Omega\subsetneq\Omega_{\varepsilon}$.

Corollary 1.8**.**

Let $\Omega$ be a domain in $\hat{\mathbb{C}}$. If the complement of $\hat{\mathbb{C}}\backslash\Omega$ is totally disconnected, then $\Omega$ is maximal in the sense of kernel.

It follows from the corollary that $\Omega_{\infty}$ in Example 1.6 is maximal. We shall give an example of a maximal domain whose complement is not totally disconnected.

In order to prove the lower semicontinuity of connectivity of domains and Theorem 1.7 we introduce a simple topological separation lemma. Let $\alpha:\partial\mathbb{D}\rightarrow\hat{\mathbb{C}}$ be a simple closed curve. By the Jordan curve theorem $\hat{\mathbb{C}}\backslash\alpha(\partial\mathbb{D})$ consists of two domains $D_{1}$ and $D_{2}$ satisfying $\partial D_{1}=\partial D_{2}=\alpha(\partial\mathbb{D})$. We say that $\alpha$ separates two sets $B_{1}$ and $B_{2}$ if $B_{1}$ and $B_{2}$ are contained in different components of $\hat{\mathbb{C}}\backslash\alpha(\partial\mathbb{D})$ respectively.

Lemma 1.9**.**

Let $\Omega$ be a domain in $\hat{\mathbb{C}}$, $E$ a connected component of $\hat{\mathbb{C}}\backslash\Omega$ and $F$ a nonempty closed subset of $\hat{\mathbb{C}}\backslash\Omega$ with $E\cap F=\emptyset$. Then there exists a simple closed curve in $\Omega$, which separates $E$ and $F$.

See Newman [22, Theorem 3.3 in Chapter VI] for a proof in the case that $F$ is also a component of $\hat{\mathbb{C}}\backslash\Omega$. The lemma seems quite natural and elementary, however we could not find a reference for the general case. We shall give a proof of Lemma 1.9 in the final section §9.

In §8 we shall study Loewner theory on Fuchsian groups. Let $\{\Omega_{t}\}_{t\in I}$ be a continuous and nondecreasing family of hyperbolic domains in $\mathbb{C}$ with $0\in\Omega_{t}$, $t\in I$, and $\{f_{t}\}_{t\in I}\subset\mathcal{H}_{0}(\mathbb{D})$ be the corresponding Loewner chain of universal covering maps. For $t\in I$ let $\Gamma_{t}$ be the covering transformation group of the universal covering map $f_{t}:\mathbb{D}\rightarrow\Omega_{t}$, i.e., $\Gamma_{t}$ is the group consisting of all $\gamma\in\text{M\"{o}b}(\mathbb{D})$ satisfying $f_{t}\circ\gamma=f_{t}$. Here $\text{M\"{o}b}(\mathbb{D})$ is the group of all automorphisms of $\mathbb{D}$.

Now we introduce $\sigma_{s,t}:\Gamma_{s}\rightarrow\Gamma_{t}$ for $(s,t)\in I_{+}^{2}$ as follows. For $\gamma\in\Gamma_{s}$ take $\alpha:[0,1]\rightarrow\mathbb{D}$ be a path from [math] to $\gamma(0)$ arbitrarily. Since $f_{s}\circ\gamma(0)=f_{s}(0)=0$, $f_{s}\circ\alpha$ is a loop ($=$ closed path) based at [math] in $\Omega_{s}(\subset\Omega_{t})$. Let $\tilde{\alpha}:[0,1]\rightarrow\mathbb{D}$ be the unique lift of $f_{s}\circ\alpha$ with $\tilde{\alpha}(0)=0$ with respect to $f_{t}:\mathbb{D}\rightarrow\Omega_{t}$. Then there exists uniquely $\tilde{\gamma}\in\Gamma_{t}$ with $\tilde{\gamma}(0)=\tilde{\alpha}(1)$. Since $\mathbb{D}$ is simply connected, the end point $\tilde{\alpha}(1)$ does not depend on the choice of $\alpha$ and hence $\tilde{\gamma}$ is uniquely determined by $\gamma$. We define $\sigma_{s,t}:\Gamma_{s}\rightarrow\Gamma_{t}$ by $\sigma_{t_{0}t_{1}}(\gamma)=\tilde{\gamma}$. It is easy to see that $\sigma_{tt}$ be the identity mapping of $\Gamma_{t}$. Notice that by definition the semigroup relation

[TABLE]

holds for $t_{0},t_{1},t_{2}\in I$ with $t_{0}\leq t_{1}\leq t_{2}$.

Theorem 1.10**.**

For $(s,t)\in I_{+}^{2}$ the map $\sigma_{s,t}:\Gamma_{s}\rightarrow\Gamma_{t}$ is an injective homomorphism and satisfies $\omega_{s,t}\circ\gamma=\sigma_{s,t}(\gamma)\circ\omega_{s,t}$.

Theorem 1.11**.**

Let $\{f_{t}\}_{t\in I}$ be Loewner chain of universal covering maps with the associated Herglotz family $\{P(\cdot,t)\}_{t\in I}$ and negligible $G_{\delta}$-set $N\subset I$. Let $t_{0}\in I$ and $\gamma\in\Gamma_{t_{0}}$ and set $\gamma_{t}=\sigma_{t_{0}t}(\gamma)\in\Gamma_{t}$ for $t\in I\cap[t_{0},\infty)$. Then the map $I\cap[t_{0},\infty)\ni t\mapsto\gamma_{t}\in\text{M\"{o}b}(\mathbb{D})$ is continuous and

[TABLE]

holds $\mu_{a}$-a.e.in $I\cap[t_{0},\infty)$. Here $\gamma(z,t):=\gamma_{t}(z)$. Particularly if $a(t)$ is locally absolutely continuous on $I$ and $\dot{a}(t):=\frac{da}{dt}(t)>0$ a.e., then $\gamma_{t}(z)$ is absolutely continuous in $t$ for each fixed $z\in\mathbb{D}$ and

[TABLE]

*holds a.e. in $I\cap[t_{0},\infty)$. *

2. Basic Estimates and the Structure Theorem

Definition 2.1**.**

For an interval $I\subset[-\infty,\infty]$ let $I_{+}^{2}=\{(s,t):s,t\in I\text{ with }s\leq t\}$. Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be a family of functions in $\mathfrak{B}$. Then we say that $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is a transition family if

[TABLE]

for all $t,t_{0},t_{1},t_{2}\in I$ with $t_{0}\leq t_{1}\leq t_{2}$. Here $\operatorname{id}_{\mathbb{D}}$ is the identity mapping of $\mathbb{D}$.

For a transition family $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ let

[TABLE]

The function $a_{s,t}$ is nonincreasing in $t\in I\cap[s,\infty]$ for each fixed $s$ and nondecreasing in $s\in I\cap[-\infty,t]$ for each fixed $t$. By (2.1) we have

[TABLE]

for all $t,t_{0},t_{1},t_{2}\in I$ with $t_{0}\leq t_{1}\leq t_{2}$. Fix $t_{0}\in I$ and $c>0$ and let

[TABLE]

Then $a(t)$ is nondecreasing and satisfies

[TABLE]

Conversely if $a(t)$, $t\in I$ is a positive and nondecreasing function, then $a_{st}$ defined by (2.4) satisfies (2.2)

Let $\{f_{t}\}_{t\in I}$ be a Loewner chain. Then for each $(s,t)\in I_{+}^{2}$ there uniquely exists $\omega_{s,t}\in\mathfrak{B}$ satisfying $f_{s}=f_{t}\circ\omega_{s,t}$. It is easy to see that $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is a transition family. We call $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ the transition family associated with $\{f_{t}\}_{t\in I}$. In this case we have $a_{s,t}=a(s)/a(t)$, where $a(t)=f_{t}^{\prime}(0)>0$, $t\in I$.

Proposition 2.2**.**

Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be a transition family. Then for $(s,t)\in I_{+}^{2}$ and $t_{0},t_{1},t_{2}\in I$ with $t_{0}<t_{1}<t_{2}$ we have

[TABLE]

Proof.

We may assume $0<a_{s,t}<1$, since otherwise $a_{s,t}=1$, and hence by the Schwarz lemma we have $\omega_{s,t}=\operatorname{id}_{\mathbb{D}}$ and (2.5)-(2.8) trivially hold. Applying the Schwarz-Pick inequality to the analytic function $g(z)=\omega_{s,t}(z)/z$ in $\mathbb{D}$ we have

[TABLE]

where $c:=g(0)=a_{s,t}\in(0,1)$. A simple calculation shows

[TABLE]

Inequalities (2.10) and (2.11) are equivalent to (2.5) and (2.6) respectively. Next, by

[TABLE]

we have

[TABLE]

which shows (2.7). Replacing $s$ and $t$ by $t_{1}$ and $t_{2}$ respectively in (2.7) and then replacing $z$ by $\omega_{t_{0},t_{1}}(z)$ we have

[TABLE]

Since $\omega_{t_{1},t_{2}}(\omega_{t_{0},t_{1}}(z))=\omega_{t_{0},t_{2}}(z)$, (2.9) easily follows from the above inequality and $|\omega_{t_{0},t_{1}}(z)|\leq|z|$.

On the other hand by (2.11) we have

[TABLE]

This implies (2.8). ∎

As simple applications of the above inequalities we give criteria for the continuities of a transition family and a Loewner chain.

Definition 2.3**.**

Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be a transition family. We say that $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is continuous if the mapping $I_{+}^{2}\ni(s,t)\mapsto\omega_{s,t}\in\mathcal{H}(\mathbb{D})$ is continuous on $I_{+}^{2}$, i.e., $\omega_{s,t}\rightarrow\omega_{s_{0},t_{0}}$ locally uniformly on $\mathbb{D}$ as $(s,t)\rightarrow(s_{0},t_{0})$ in $I_{+}^{2}$ at every $(s_{0},t_{0})\in I_{+}^{2}$. Also we say that $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is strictly decreasing if $a_{st}<1$ for $(s,t)\in I_{+}^{2}$ with $s\not=t$. This is equivalent to that $a(t)$ is strictly increasing, where $a(t)$ is defined by (2.3).

Theorem 2.4**.**

Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be a transition family with $a_{s,t}=\omega_{s,t}^{\prime}(0)$, $(s,t)\in I_{+}^{2}$ and let $a(t)$, $t\in I$ be defined by (2.3) for some $c>0$. Then the following four conditions are equivalent.

• (i)

The function $a(t)$ is continuous on $I$.

• (ii)

For all $t_{0}\in I$, $a_{s,t}\rightarrow a_{t_{0},t_{0}}=1$ as $t-s\downarrow 0$ with $s\leq t_{0}\leq t$.

• (iii)

The mapping $I_{+}^{2}\ni(s,t)\mapsto a_{s,t}\in(0,1]$ is continuous on $I_{+}^{2}$.

• (iv)

The mapping $I_{+}^{2}\ni(s,t)\mapsto\omega_{s,t}\in\mathcal{H}(\mathbb{D})$ is continuous on $I_{+}^{2}$.

Proof.

It is easy to see that (i) and (ii) are equivalent. Assume (ii) and let $(s_{0},t_{0})\in I_{+}^{2}$. We use notation $\alpha\vee\beta=\max\{\alpha,\beta\}$ and $\alpha\wedge\beta=\min\{\alpha,\beta\}$ for $\alpha,\beta\in\mathbb{R}$.

Assume (i). First we consider the case that $s_{0}=t_{0}$. Let $\varepsilon>0$ and take $\delta>0$ such that

[TABLE]

For $(s,t)\in I_{+}^{2}$ with $\max\{|s-t_{0}|,|t-t_{0}|\}<\delta/2$ put $s^{*}=\min\{s,t,t_{0}\}=\min\{s,t_{0}\}$ and $t^{*}=\max\{s,t,t_{0}\}=\max\{t,t_{0}\}$. Then $s^{*}\leq t_{0}\leq t^{*}$, $s^{*}\leq s\leq t\leq t^{*}$ and $0\leq t^{*}-s^{*}=t^{*}-t_{0}+t_{0}-s^{*}\leq|t-t_{0}|+|s-t_{0}|<\delta$. Thus we have

[TABLE]

and hence $a_{st}\rightarrow 1$ as $I_{+}^{2}\ni(s,t)\rightarrow(t_{0},t_{0})$.

Suppose that $s_{0}<t_{0}$. Let $\varepsilon>0$ and take $\delta>0$ with $0<\delta<(t_{0}-s_{0})/2$ such that

[TABLE]

for all $s_{1},s_{2},t_{1},t_{2}$ with $s_{1}\leq s_{0}\leq s_{2}$, $t_{1}\leq t_{0}\leq t_{2}$, $0<s_{2}-s_{1}<\delta$ and $0<t_{2}-t_{1}<\delta$. Then for $|s-s_{0}|<\delta$ and $|t-t_{0}|<\delta$

[TABLE]

Therefore $a_{s,t}$ is continuous at $(s_{0},t_{0})$ and (iii) holds.

Now assume (iii). Since by (2.7) we have for $(s,t)\in I_{+}^{2}$

[TABLE]

it is clear $\omega_{s,t}(z)\rightarrow\omega_{t_{0},t_{0}}(z)=z$ locally uniformly on $\mathbb{D}$ as $I_{+}^{2}\ni(s,t)\rightarrow(t_{0},t_{0})$. Suppose that $s_{0}<t_{0}$. Then by making use of

[TABLE]

for $\omega\in\mathfrak{B}$ and $|z_{0}|,|z_{1}|\leq r$ we have for $(s,t)\in I_{+}^{2}$ with $s<t_{0}$ and $s_{0}<t$ and $|z|\leq r$

[TABLE]

This implies $\omega_{s,t}(z)\rightarrow\omega_{s_{0},t_{0}}(z)$ as $I_{+}^{2}\ni(s,t)\rightarrow(s_{0},t_{0})$. Thus (iv) holds.

Finally assume (iv). Then since the mapping $I_{+}^{2}\ni(s,t)\mapsto\omega_{s,t}^{\prime}\in\mathcal{H}(\mathbb{D})$ is also continuous, (ii) holds. ∎

By slightly generalizing the original proof in [24] we show that if $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is continuous, then each $\omega_{s,t}$ is univalent. We need a lemma by Landau.

Lemma 2.5** (Landau).**

Let $\omega\in\mathfrak{B}$ with $\omega^{\prime}(0)=\sigma\in(0,1)$. Then $\omega$ is univalent in $\mathbb{D}(0,\rho)$ with $\rho=\rho(\sigma)=\sigma/(1+\sqrt{1-\sigma^{2}})$.

For a proof see [12, Theorem 10.1]. We notice that $\lim_{\sigma\uparrow 1}\rho(\sigma)=1$.

Theorem 2.6**.**

Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be a transition family. If $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is continuous, then $\omega_{s,t}$ is univalent on $\mathbb{D}$ for every $(s,t)\in I_{+}^{2}$.

Proof.

Fix $(s_{0},t_{0})\in I_{+}^{2}$. We may assume $s_{0}<t_{0}$, since otherwise the univalence is trivial.

For any $r\in(0,1)$ take $\sigma\in(0,1)$ with $\rho(\sigma)>r$. Since $a_{s,t}$ is continuous on $I_{+}^{2}$ and $a_{t,t}=1$ for $t\in I$, there exists a sequence $s_{0}<s_{1}<\cdots<s_{n}=t_{0}$ such that $a_{s_{k-1},s_{k}}>\sigma$, $k=1,\ldots,n$. Then $\omega_{s_{k-1},s_{k}}$ is univalent in $\mathbb{D}(0,r)$. From this and $\omega_{s_{k-1},s_{k}}(\mathbb{D}(0,r))\subset\mathbb{D}(0,r)$ it follows that the composition mapping $\omega_{s_{0},t_{0}}=\omega_{s_{n-1},s_{n}}\circ\cdots\circ\omega_{s_{0},s_{1}}$ is also univalent in $\mathbb{D}(0,r)$. Since $r\in(0,1)$ is arbitrary, $\omega_{s_{0},t_{0}}$ is univalent in $\mathbb{D}$. ∎

Theorem 2.7**.**

Let $\{f_{t}\}_{t\in I}\subset\mathcal{H}_{0}(\mathbb{D})$ be a Loewner chain with $a(t)=f_{t}^{\prime}(0)$, $t\in I$. Then $\{f_{t}\}_{t\in I}$ is continuous if and only if $a(t)$, $t\in I$, is continuous. Furthermore in this case the associated transition family $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is also continuous and each $\omega_{s,t}$, $(s,t)\in I_{+}^{2}$ is univalent on $\mathbb{D}$.

Proof.

The latter statement easily follows from Theorem 2.4 and 2.6.

Take $r\in(0,1)$ and consider

[TABLE]

If $\{f_{t}\}_{t\in I}$ is continuous at $t_{0}\in I$, then $f_{t}(z)\rightarrow f_{t_{0}}(z)$ uniformly on $\partial\mathbb{D}(0,r)$ as $I\ni t\rightarrow t_{0}$. Hence $a(t)\rightarrow a(t_{0})$.

To prove the converse let $t_{0}\in I$. Take $t^{*}\in I$ satisfying $t_{0}<t^{*}$ when $t_{0}<\sup I$, or put $t^{*}=t_{0}$ when $t_{0}=\sup I$. For each fixed $r\in(0,1)$ it suffices to show $|f_{t_{2}}(z)-f_{t_{1}}(z)|\rightarrow 0$ uniformly on $\overline{\mathbb{D}}(0,r)$ as $t_{2}-t_{1}\rightarrow+0$ with $t_{1}\leq t_{0}\leq t_{2}\leq t^{*}$. Let

[TABLE]

Then for any $t\in I$ with $t\leq t^{*}$ and $|z|\leq r$ we have by $|\omega_{t,t^{*}}(z)|\leq|z|$

[TABLE]

and hence

[TABLE]

Therefore by (2.7) we have

[TABLE]

as $t_{2}-t_{1}\rightarrow+0$ with $t_{1}\leq t_{0}\leq t_{2}\leq t^{*}$. ∎

Following the argument in Pommerenke [23] we shall derive a simple extendability property of transition families which leads to the structure theorem on Loewner chains. We need the Vitali convergence theorem. See [27] Chap. 7 for a proof and details.

Lemma 2.8** (the Vitali theorem).**

Let $\{g_{n}\}_{n=1}^{\infty}$ be a locally uniformly bounded sequence of analytic functions in $\mathbb{D}$. Suppose that $\lim_{n\rightarrow\infty}g_{n}(z)$ exists on a subset $A$ of $\mathbb{D}$ having at least an accumulation point in $\mathbb{D}$. Then $\{g_{n}\}_{n=1}^{\infty}$ converges locally uniformly to some analytic function.

Lemma 2.9**.**

Let $h:\mathbb{D}\rightarrow\mathbb{C}$ be a univalent analytic function with $h(0)=0$ and $h^{\prime}(0)=a>0$. Then

[TABLE]

Proof.

Notice that by the Koebe one-quarter theorem $\mathbb{D}(0,a/4)\subset f(\mathbb{D})$. Let $h^{-1}(w)=\sum_{n=1}^{\infty}b_{n}w^{n}$. Then for $R<a/4$

[TABLE]

Thus

[TABLE]

Letting $R\uparrow a/4$ we obtain the required inequality. ∎

Theorem 2.10**.**

Let $I\subset[-\infty,\infty]$ be an interval with $\beta:=\sup I\not\in I$ and $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be a transition family. Then for any $s\in I$ locally uniform limits $\omega_{s,\beta}=\lim_{t\uparrow\beta}\omega_{s,t}$ exist on $\mathbb{D}$ and the followings hold.

• (i)

If $a_{s_{0},\beta}:=\lim_{t\uparrow\beta}a_{s_{0},t}>0$ for some $s_{0}\in I$, then $a_{s,\beta}>0$ and $\omega_{s,\beta}\in\mathfrak{B}$ for all $s\in I$ and the extended family $\{\omega_{s,t}\}_{(s,t)\in(I\cup\{\beta\})_{+}^{2}}$ is a transition family on $I\cup\{\beta\}$. Here we set $\omega_{\beta,\beta}=\operatorname{id}_{\mathbb{D}}$ and $a_{\beta\beta}=1$.

• (ii)

If $a_{s_{0},\beta}=0$ for some $s_{0}\in I$, then $a_{s,\beta}=0$ and $\omega_{s,\beta}=0$ for all $s\in I$. Furthermore if $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is continuous, then for any fixed $t_{0}\in I$ and $c>0$ the locally uniform limit $g_{t}=\lim_{\tau\uparrow\beta}\frac{c}{a_{t_{0},\tau}}\omega_{t,\tau}$ exist and univalent on $\mathbb{D}$ for all $t\in I$, and $\{g_{t}\}_{t\in I}$ is a continuous Loewner chain of univalent functions having $\{\omega_{s,t}\}_{s\in I\cup\{\beta\}}$ as the associated transition family.

By a similar argument we can prove the counterpart theorem for transition families on $I$ with $\alpha=\inf I\not\in I$. To avoid complication we omit the statement.

Proof of Theorem 2.10.

Suppose $a_{s_{0},\beta}=\lim_{t\uparrow\beta}a_{s_{0},t}>0$. Then by (2.2) we have $a_{s,\beta}=\lim_{t\uparrow\beta}a_{s,t}>0$ for all $s\in I$. Fix $s\in I$ arbitrarily. Then since the family $\{\omega_{s,t}\}_{t\geq s}$ is uniformly bounded on $\mathbb{D}$, there exists a sequence $\{t_{n}\}_{n=1}^{\infty}\subset I$ with $s<t_{1}<t_{2}<\cdots<t_{n}\uparrow\beta$ such that $\{\omega_{s,t_{n}}\}_{n=1}^{\infty}$ converges to an analytic function $\varphi$ locally uniformly on $\mathbb{D}$.

We shall show $\omega_{s,t}(z)\rightarrow\varphi(z)$ as $t\uparrow\beta$ locally uniformly on $\mathbb{D}$. For any $r\in(0,1)$ and $\varepsilon>0$ take $N\in\mathbb{N}$ such that for all $n\geq N$

[TABLE]

and

[TABLE]

Then for $t\in(t_{N},\beta)$ and $|z|\leq r$ we have by (2.7) and $|\omega_{s,t_{N}}(z)|\leq|z|$

[TABLE]

Therefore $\omega_{s,t}(z)\rightarrow\varphi(z)$ as $t\uparrow\beta$ locally uniformly on $\mathbb{D}$.

Hereafter we write $\varphi$ as $\omega_{s,\beta}$. Then by letting $u\uparrow\beta$ in

[TABLE]

we have

[TABLE]

Thus the extended family $\{\omega_{s,t}\}_{(s,t)\in(I\cup\{\beta\})_{+}^{2}}$ is a transition family on $I\cup\{\beta\}$.

Now suppose $a_{s_{0},\beta}=\lim_{t\uparrow\beta}a_{s_{0},t}=0$. Then by (2.2) we have $a_{s,\beta}=\lim_{t\uparrow\beta}a_{s,t}=0$ for all $s\in I$. Let $s\in I$, $r\in(0,1)$ be fixed and $\{c_{n}\}_{n=1}^{\infty}$ be a sequence of positive numbers with $1>c_{1}>c_{2}>\cdots>c_{n}\downarrow 0$. Then there exists $\{t_{n}\}_{n=1}^{\infty}\subset I$ with $t_{0}:=s<t_{1}<t_{2}<\cdots t_{n}\uparrow\beta$ such that

[TABLE]

Put

[TABLE]

Then $1>\rho_{1}>\rho_{2}>\cdots>r$ and by (2.8)

[TABLE]

Therefore for $|z|\leq r$ and $t\geq t_{n}$ we have

[TABLE]

It easily follows from this that $\omega_{s,t}(z)\rightarrow 0$ as $t\uparrow\beta$ locally uniformly on $\mathbb{D}$.

Finally suppose that $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is continuous. Then by Theorem 2.6 each $\omega_{s,t}$ is univalent on $\mathbb{D}$ for $(s,t)\in I_{+}^{2}$. Let $t_{0}\in I$ and $c>0$ be fixed and define $a(t)$ by (2.3). For fixed $\tau\in I$ consider the family $\{a(t)\omega_{\tau,t}\}_{\tau\leq t\in I}$. By the growth theorem for univalent functions we have

[TABLE]

This implies that $\{a(t)\omega_{\tau,t}\}_{\tau\leq t\in I}$ is locally uniformly bounded and forms a normal family. Thus there exists a sequence $\{t_{n}\}_{n=1}^{\infty}\subset I$ such that $s<t_{1}<\cdots<t_{n}\uparrow\beta$ and a locally uniform limit

[TABLE]

exists. Since $g_{\tau}^{\prime}(0)=a(\tau)>0$, by Hurwitz’s theorem, $g_{\tau}$ is univalent. For $t\in I$ with $t\geq\tau$ we have

[TABLE]

Hence $\lim_{n\rightarrow\infty}a(t_{n})\omega_{t,t_{n}}(\zeta)$ exists for $\zeta\in\omega_{s,t}(\mathbb{D})$. Since $\omega_{s,t}(\mathbb{D})$ is a nonempty subdomain of $\mathbb{D}$ and the family $\{a(t_{n})\omega_{t,t_{n}}\}_{t_{n}>t}$ is locally uniformly bounded on $\mathbb{D}$, by the Vitali convergence theorem, the locally uniform limit $g_{t}(z):=\lim_{n\rightarrow\infty}a(t_{n})\omega_{t,t_{n}}(z)$ exists on $\mathbb{D}$ and we obtain $g_{\tau}(z)=g_{t}(\omega_{\tau,t}(z))$, $z\in\mathbb{D}$. Since $g_{\tau}$ and $\omega_{\tau,t}$ are univalent, so is $g_{t}$. For $s\in I$ with $s<\tau$ we have

[TABLE]

Therefore for $z\in\mathbb{D}$ $g_{s}(z):=\lim_{n\rightarrow\infty}a(t_{n})\omega_{s,t_{n}}(z)$ exists and $g_{s}(z)=g_{\tau}(\omega_{s,\tau}(z))$ holds. It is easy to see that the convergence is locally uniformly on $\mathbb{D}$. Again by Hurwitz’s theorem, $g_{s}$ is univalent.

We have shown that the locally uniform limit $g_{t}=\lim_{n\rightarrow\infty}a(t_{n})\omega_{t,t_{n}}$ exists and univalent on $\mathbb{D}$ for all $t\in I$, and that $\{g_{t}\}_{t\in I}$ is a Loewner chain of univalent functions with $g_{t}^{\prime}(0)=a(t)$, $t\in I$ having $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ as the associated transition family. Particularly since $a(\tau)\rightarrow\infty$ as $\tau\uparrow\beta$, by Lemma 2.9 we have $a(\tau)\omega_{t,\tau}(z)=a(\tau)g_{\tau}^{-1}(g_{t}(z))\rightarrow g_{t}(z)$ as $\tau\uparrow\beta$. ∎

Now we prove the following slightly generalized version of structure theorem.

Theorem 2.11** (Structure Theorem).**

Let $I\subset[-\infty,\infty]$ be an interval with $\beta=\sup I\not\in I$ and $\{f_{t}\}_{t\in I}$ a Loewner chain with $a(t)=f_{t}^{\prime}(0)$, $t\in I$. Let $a(\beta)=\lim_{t\uparrow\beta}a(t)\in(0,\infty]$.

• (i)

The locally uniform limit $f_{\beta}=\lim_{t\uparrow\beta}f_{t}$ exists if and only if $a(\beta)<\infty$. In this case there exists uniquely an analytic function $F:\mathbb{D}(0,a(\beta))\rightarrow\mathbb{C}$ with $F(0)=F^{\prime}(0)-1=0$ and a Loewner chain $\{g_{t}\}_{I\cup\{\beta\}}$ with $\bigcup_{t\in I}g_{t}(\mathbb{D})=g_{\beta}(\mathbb{D})=\mathbb{D}(0,a(\beta))$ such that $f_{t}=F\circ g_{t}$, $g_{t}^{\prime}(0)=a(t)$ for $t\in I\cup\{\beta\}$. Furthermore if $\{f_{t}\}_{t\in I}$ is continuous, each $g_{t}$, $t\in I\cup\{\beta\}$ is univalent in $\mathbb{D}$.

• (ii)

If $\{f_{t}\}_{t\in I}$ is continuous and $a(\beta)=\infty$, then there exists uniquely an entire function $F:\mathbb{C}\rightarrow\mathbb{C}$ with $F(0)=F^{\prime}(0)-1=0$ and a Loewner chain $\{g_{t}\}_{t\in I}$ of univalent functions with $\bigcup_{t\in I}g_{t}(\mathbb{D})=\mathbb{C}$ such that $f_{t}=F\circ g_{t}$, $g_{t}^{\prime}(0)=f_{t}^{\prime}(0)$ for $t\in I$.

Pommerenke already studied a similar representation formula for normalized Loewner chain. See [24, Satz 5].

Notice that since $F^{\prime}(0)\not=0$, $\{g_{t}\}_{t\in I}$ shares the associated transition family with $\{f_{t}\}_{t\in I}$.

Proof.

Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be the transition family associated with $\{f_{t}\}_{t\in I}$.

We show (i). If the locally uniform limit $f_{\beta}=\lim_{t\uparrow\beta}f_{t}$ exists, then it is clear that $a(\beta)=\lim_{t\uparrow\beta}f_{t}^{\prime}(0)=f_{\beta}^{\prime}(0)<\infty$. Conversely if $a(\beta)<\infty$, then by Theorem 2.10 $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ has the extension $\{\omega_{s,t}\}_{(s,t)\in(I\cup\{\beta\})_{+}^{2}}$ with $\omega_{s,\beta}=\lim_{t\uparrow\beta}\omega_{s,t}$, $s\in I$. For $t\in I$ let

[TABLE]

Then $g_{t}^{\prime}(0)=a(t)=f_{t}^{\prime}(0)$.

Now consider the family $\{f_{t}\circ\omega_{t\beta}^{-1}\}_{t\in I}$. Since $\omega_{t,\beta}\in\mathfrak{B}$ and $\omega_{t,\beta}^{\prime}(0)=a_{t,\beta}=\frac{a(t)}{a(\beta)}$, by Lemma 2.5 the function $\omega_{t,\beta}$ is univalent on $\mathbb{D}(0,\rho(a_{t,\beta}))$. Combining this and (2.7) we conclude that $\omega_{t,\beta}$ maps the disc $\mathbb{D}(0,\rho(a_{t,\beta}))$ univalently onto $\omega_{t,\beta}(\mathbb{D}(0,\rho(a_{t,\beta})))$ which contains the disc $\mathbb{D}(0,r(a_{t,\beta}))$, where

[TABLE]

Therefore we may assume that $f_{t}\circ\omega_{t,\beta}^{-1}$ is defined on $\mathbb{D}(0,r(a_{t,\beta}))$, $t\in I$. Notice that $a_{t,\beta}$ is nondecreasing in $t$ and $a_{t,\beta}\uparrow 1$ as $t\uparrow\beta$, and that $r(\sigma)$ is strictly increasing on $(0,1)$ and $r(\sigma)\uparrow 1$ as $\sigma\uparrow 1$. From $\omega_{s,\beta}=\omega_{t,\beta}\circ\omega_{s,t}$ for $(s,t)\in I_{+}^{2}$, it follows that $f_{s}\circ\omega_{t,\beta}^{-1}$ coincides with $f_{t}\circ\omega_{t,\beta}^{-1}$ on a neighborhood of the origin. Hence by the identity theorem for analytic functions $f_{s}\circ\omega_{s,\beta}^{-1}$ coincides with $f_{t}\circ\omega_{t,\beta}^{-1}$ on $\mathbb{D}(0,r(a_{s,\beta}))$ Therefore there exists uniquely a function $f_{\beta}:\mathbb{D}\rightarrow\mathbb{C}$ such that for all $t\in I$

[TABLE]

Thus again by the identity theorem for analytic functions we have $f_{t}=f_{\beta}\circ\omega_{t,\beta}$ on $\mathbb{D}$. Now it is clear that $\{f_{t}\}_{t\in I\cup\beta}$ is a Loewner chain with the associated transition family $\{\omega_{s,t}\}_{(s,t)\in(I\cup\{\beta\})}$. Let

[TABLE]

Then the family $\{g_{t}\}_{t\in I\cup\{\beta\}}$ is also a Loewner chain having $\{\omega_{s,t}\}_{(s,t)\in(I\cup\{\beta\})}$ as the associated transition family. Since $a_{t,\beta}\rightarrow 1$, it follows from Proposition 2.2 that $g_{t}\rightarrow g_{\beta}$ locally uniformly on $\mathbb{D}$ as $t\uparrow\beta$. Let

[TABLE]

Then $F\circ g_{t}=f_{\beta}\circ\omega_{t,\beta}=f_{t}$, as required.

To see the uniqueness assume that an analytic function $\tilde{F}:\mathbb{D}\rightarrow\mathbb{D}(0,a(\beta))$ with $\tilde{F}(0)=\tilde{F}^{\prime}(0)-1=0$ and a Loewner chain $\{\tilde{g}_{t}\}_{t\in I\cup\{\beta\}}$ with $\tilde{g}_{\beta}(\mathbb{D})=\mathbb{D}(0,a(\beta))$ satisfy $F\circ g_{t}=\tilde{F}\circ\tilde{g}_{t}$, $t\in I\cup\{\beta\}$. Notice that by $F\circ g_{t}=\tilde{F}\circ\tilde{g}_{t}$ the Loewner chain $\{\tilde{g}_{t}\}_{t\in I\cup\{\beta\}}$ also has $\{\omega_{s,t}\}_{(s,t)\in(I\cup\{\beta\})_{+}^{2}}$ as the associated transition family. Since $\tilde{g}_{\beta}(\mathbb{D})=\mathbb{D}(0,a(\beta))$, $\tilde{g}_{\beta}(0)=0$ and $g_{\beta}^{\prime}(0)=a(\beta)$, by the Schwarz lemma we have $\tilde{g}_{\beta}(z)=a(\beta)z=g_{\beta}(z)$ and hence $\tilde{g}_{t}(z)=\tilde{g}_{\beta}(\omega_{t,\beta}(z))=a(\beta)\omega_{t,\beta}(z)=g_{t}(z)$, $t\in I$. This also implies $\tilde{F}=F$.

If $\{f_{t}\}_{t\in I}$ is continuous, then the function $a(t)$ is positive and continuous on $I\cup\{\beta\}$, and so is $a_{s,t}=a(s)/a(t)$ on $I_{+}^{2}$. Therefore by Theorem 2.4 $\{\omega_{s,t}\}_{(s,t)\in(I\cup\{\beta\})_{+}^{2}}$ is continuous and hence by Theorem 2.6 $\omega_{t,\beta}$ and $g_{t}=a(\beta)\omega_{t,\beta}$ are univalent on $\mathbb{D}$.

To show (ii) suppose that $\{f_{t}\}_{t\in I}$ is continuous and $a(\beta)=\infty$. Then $a_{s,\beta}=\lim_{t\uparrow\beta}\frac{a(s)}{a(t)}=0$ for $s\in I$. By applying Theorem 2.10 (ii) with $t_{0}\in I$ and $c=a(t_{0})$ the locally uniform limit $g_{t}=\lim_{\tau\uparrow\beta}a(\tau)\omega_{t,\tau}$ exists and univalent on $\mathbb{D}$ for $t\in I$ and $\{g_{t}\}_{t\in I}$ forms a Loewner chain having $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ as the associated transition family. Notice that $g_{t}^{\prime}(0)=a(t)=f_{t}^{\prime}(0)$, $t\in I$.

Now consider the family $\{f_{t}\circ g_{t}^{-1}\}_{t\in I}$. Each $f_{t}\circ g_{t}^{-1}$ is defined on the domain $g_{t}(\mathbb{D})$ and $\{g_{t}(\mathbb{D})\}_{t\in I}$ is nondecreasing in $t$. For $(s,t)\in I_{+}^{2}$ we have on $g_{s}(\mathbb{D})$

[TABLE]

By the Koebe theorem we have $\mathbb{D}(0,a(t)/4)\subset g_{t}(\mathbb{D})$. Combining this and $\lim_{t\uparrow\beta}g_{t}^{\prime}(0)=a(\beta)=\infty$ it follows that $\bigcup_{t\in I}g_{t}(\mathbb{D})=\mathbb{C}$. Therefore $\{f_{t}\circ g_{t}^{-1}\}_{t\in I}$ defines a unique entire function $F$ with $F(0)=F^{\prime}(0)-1=0$ satisfying

[TABLE]

for all $t\in I$. Thus $f_{t}=F\circ g_{t}$, as required.

Finally assume that an entire function $\tilde{F}:\mathbb{C}\rightarrow\mathbb{C}$ with $\tilde{F}(0)=\tilde{F}^{\prime}(0)-1=0$ and a Loewner chain $\{\tilde{g}_{t}\}_{t\in I}$ of univalent functions satisfy $\tilde{g}_{t}^{\prime}(0)=g_{t}^{\prime}(0)$ and $F\circ g_{t}=\tilde{F}\circ\tilde{g}_{t}$, $t\in I\cup\{\beta\}$. Then since $\{\tilde{g}_{t}\}_{t\in I}$ shares $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ as the transition family with $\{g_{t}\}_{t\in I}$, we have by Lemma 2.9

[TABLE]

This also implies $\tilde{F}=F$. ∎

Definition 2.12**.**

Let $\{f_{t}\}_{t\in I}$ be a Loewner chain with $\beta=\sup I\not\in I$. Then the domain $\Omega_{\beta}=\bigcup_{t\in I}f_{t}(\mathbb{D})$ is called the Loewner range of $\{f_{t}\}_{t\in I}$.

We study the relation between $\Omega_{\beta}$ and $a(\beta)$.

Proposition 2.13**.**

Let $\{f_{t}\}_{t\in I}$ be a Loewner chain of univalent functions with $\beta=\sup I\not\in I$ and $\Omega_{\beta}$ be the Loewner range of $\{f_{t}\}_{t\in I}$. Then $a(\beta)=\lim_{t\uparrow\beta}f_{t}^{\prime}(0)=\infty$ if and only if $\Omega_{\beta}=\mathbb{C}$.

Proof.

For $t\in I$ let $\rho_{1}(t)=\sup\{r>0:\mathbb{D}(0,r)\subset f_{t}(\mathbb{D})\}$. Then the proposition easily follows from the inequality

[TABLE]

which is is a simple application of of the Koebe one-quarter theorem and the Schwarz lemma. ∎

In general case we have the following. Here temporarily we use notion of universal covering. It will be systematically treated in §6

Theorem 2.14**.**

Let $\{f_{t}\}_{t\in I}$ be a Loewner chain with $\beta=\sup I\not\in I$ and $\Omega$ be the Loewner range of $\{f_{t}\}_{t\in I}$. If ${}^{\#}(\mathbb{C}\backslash\Omega)\geq 2$, then $a(\beta)<\infty$.

Proof.

Take distinct points $w_{0},w_{1}\in(\mathbb{C}\backslash\Omega_{\beta})$ and let $F:\mathbb{D}\rightarrow\mathbb{C}\backslash\{w_{0},w_{1}\}$ be the universal covering map with $F(0)=0$ and $F^{\prime}(0)>0$. Then for each $f_{t}:\mathbb{D}\rightarrow\mathbb{C}\backslash\{w_{0},w_{1}\}$ have a unique analytic lift $\phi_{t}:\mathbb{D}\rightarrow\mathbb{D}$ with $F\circ\phi_{t}=f_{t}$ and $\phi(0)=0$. Hence $a(t)=f_{t}^{\prime}(0)\leq F^{\prime}(0)$ by the Schwarz lemma. ∎

From the above theorem it follows that if $a(\beta)=\infty$, $\Omega_{\beta}=\mathbb{C}$ or $\Omega_{\beta}=\mathbb{C}\backslash\{w_{0}\}$ for some $w_{0}\in\mathbb{C}\backslash\{0\}$. It is easy to construct examples of Loewner chains in both cases.

The reverse of the above theorem is not always true. Actually for any surjective analytic map $F:\mathbb{D}\rightarrow\mathbb{C}$, the family $\{F(tz)\}_{0<t<1}$ is a strictly increasing and continuous Loewner chain with $a(\beta)<\infty$ and $\Omega_{\beta}=\mathbb{C}$.

3. Loewner-Kufarev Differential Equations

Intuitively (2.9) in Proposition 2.2 means that for fixed $z\in\mathbb{D}$ and $t_{0}\in I$, the function $\omega_{t_{0},t}(z)$, $t\in I\cap[t_{0},\infty)$ is Lipschitz continuous with respect to the function $a(t)$, which is strictly increasing and continuous. We now summarize basic facts concerning the Fundamental theorem of calculus with respect to a strictly increasing and continuous function. Let $\mu_{1}$ be the ($1$-dimensional) Lebesgue measure and $\psi:[\alpha,\beta]\rightarrow\mathbb{R}$, $-\infty<\alpha<\beta<\infty$, be a continuous and strictly increasing function. By $\mu_{\psi}$ we denote the Lebesgue-Stieltjes measure associated with $\psi$.

Lemma 3.1**.**

Suppose that $\psi$ is strictly increasing and continuous on $[\alpha,\beta]$. Then a subset $A$ of $[\alpha,\beta]$ is $\mu_{\psi}$-measurable if and only if $\psi(A)$ is Lebesgue measurable and in this case we have $\mu_{\psi}(A)=\mu_{1}(\psi(A))$.

For a proof see [6] p.135. We notice that every Borel subset of $[\alpha,\beta]$ is $\mu_{\psi}$-measurable, since $\psi$ is a homeomorphism of $[\alpha,\beta]$ onto $[\psi(\alpha),\psi(\beta)]$.

For a function $u:[\alpha,\beta]\rightarrow\mathbb{R}$ we define the upper and lower $\psi$-derivatives of $u$ at $t$ respectively by

[TABLE]

Then it is a simple exercise to see

[TABLE]

If the upper and lower derivatives are finite and equal at $t$, we say that $u$ is $\psi$-differentiable at $t$. Their common value is denoted by $D_{\psi}u(t)$ and is called the $\psi$-derivative of $u$ at $t$. It is easy to verify that $u$ is $\psi$-differentiable at $t$ if and only if the limit

[TABLE]

exists and the limit coincides with $D_{\psi}u(t)$. We denote the usual derivative (with respect to the identity function on $I$) by $D$.

We say that a function $u:[\alpha,\beta]\rightarrow\mathbb{R}$ is absolutely $\psi$-continuous if for each $\varepsilon>0$ there exists $\delta>0$ such that, if $\{[\alpha_{n},\beta_{n}]\}$ is any at most countable collection of non-overlapping closed intervals in $[\alpha,\beta]$ with $\sum_{k}(\psi(\beta_{k})-\psi(\alpha_{k}))<\delta$, then $\sum_{k}|u(\beta_{k})-u(\alpha_{k})|<\varepsilon$. Here we say that a collection of closed intervals is non-overlapping if their interiors are disjoint. For a complex valued function $h=u+iv:[\alpha,\beta]\rightarrow\mathbb{C}$ we say that $h$ is absolutely $\psi$-continuous (or $\psi$-differentiable) if both $u$ and $v$ are $\psi$-absolutely continuous ($\psi$-differentiable, respectively). Notice that an absolutely $\psi$-continuous function is continuous.

Lemma 3.2**.**

Suppose that a function $h$ is absolutely $\psi$-continuous on $[\alpha,\beta]$. Then for $\mu_{\psi}$-almost every $t\in[\alpha,\beta]$, $h$ is $\psi$-differentiable at $t$ and $D_{\psi}h$ is $\mu_{\psi}$-integrable. Furthermore

[TABLE]

Conversely if $k$ is an $\mu_{\psi}$-integrable function on $[\alpha,\beta]$ and

[TABLE]

then $h$ is absolutely $\mu_{\psi}$-continuous on $[\alpha,\beta]$ and $D_{\psi}h(t)=k(t)$ for $\mu_{\psi}$-almost every $t\in[\alpha,\beta]$.

Proof.

Put $\tilde{\alpha}=\psi(\alpha)$, $\tilde{\beta}=\psi(\beta)$. Then, by definition, $h\circ\psi^{-1}$ is an ordinary absolutely continuous function on $[\tilde{\alpha},\tilde{\beta}]$. Therefore there exists a set $\tilde{N}\subset[\tilde{\alpha},\tilde{\beta}]$ of Lebesgue measure zero such that for $s\in[\tilde{\alpha},\tilde{\beta}]\backslash\tilde{N}$, $u\circ\psi^{-1}$ is differentiable at $s$, i.e., the limit

[TABLE]

exists. Replacing $\tilde{N}$ by a larger set if necessary, we can assume that $\tilde{N}$ is a $G_{\delta}$ set. Let $N=\psi^{-1}(\tilde{N})$. Then by Lemma 3.1 we have $\mu_{\psi}(N)=\mu_{1}(\tilde{N})=0$ and it is easy to see that

[TABLE]

where $D(h\circ\psi^{-1})$ means the usual derivative of $h\circ\psi^{-1}(s)$ with respect to $s$. Notice that both $D_{\psi}h$ and $D(h\circ\psi^{-1})$ are Borel measurable on $[\alpha,\beta]\backslash N$ and $[\tilde{\alpha},\tilde{\beta}]\backslash\tilde{N}$, respectively and by Lemma 3.1, $\mu_{\psi}(A)=\mu_{1}(\psi(A))$ for any Borel subset $A$ of $[\alpha,\beta]$. Since $h\circ\psi^{-1}$ is absolutely continuous, by the fundamental theorem of calculus we have for $s\in[\tilde{\alpha},\tilde{\beta}]$

[TABLE]

which implies (3.1). The remaining part also follows from the corresponding part of the fundamental theorem of calculus. ∎

Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be a transition family. We write $\omega(z,s,t)$ instead of $\omega_{s,t}(z)$ for $(s,t)\in I_{+}^{2}$ and $z\in\mathbb{D}$ and denote

[TABLE]

Theorem 3.3**.**

Let $\{\omega(\cdot,s,t)\}_{(s,t)\in I_{+}^{2}}$ be a strictly decreasing and continuous transition family and let $a(t)$, $t\in I$ be a strictly increasing and positive function defined by (2.3) for some $c>0$. Then there exists a $G_{\delta}$ set $N(\subset I)$ of $\mu_{a}$-measure zero and a Herglotz family $\{P(\cdot,t)\}_{t\in I}$ such that $P(z,t)$ is Borel measurable on $\mathbb{D}\times I$, and that for each $t\in I\backslash N$

[TABLE]

and the convergence is locally uniform on $\mathbb{D}$. Furthermore for each fixed $t_{0}\in I$ and $z\in\mathbb{D}$

[TABLE]

for $t\in(I\cap[-\infty,t_{0}))\backslash N$ and

[TABLE]

for $t\in(I\cap(t_{0},\infty])\backslash N$, where each convergence is locally uniform on $\mathbb{D}$.

Proof of Theorem 3.3.

We show the theorem in the case that $\alpha=\inf I\not\in I$.

Step 1. Take a sequence $\{s_{k}\}_{j=1}^{\infty}\subset I$ with $s_{k}\downarrow\alpha$ and a distinct sequence $\{z_{j}\}_{j=1}^{\infty}\subset\mathbb{D}$ with $z_{j}\rightarrow 0$. By (2.9) we have for $t_{1},t_{2}\in I$ with $s<t_{1}<t_{2}$

[TABLE]

From this it follows that for each fixed $j,k\in\mathbb{N}$, $\omega(z_{j},s_{k},t)$ is locally Lipschitz continuous and hence locally absolutely continuous with respect to $a(t)$ on $I\cap(s_{k},\infty)$. Thus there exists a set $N_{j,k}\subset I\cap[s_{k},\infty)$ of $\mu_{a}$-measure zero such that for $t\in(I\cap[s_{k},\infty))\backslash N_{j,k}$ the limit

[TABLE]

exists. By (3.9) the family of analytic functions

[TABLE]

is locally uniformly bounded, where $(t_{1},t_{2})\in I_{+}^{2}$ with $t_{2}-t_{1}>0$. By making use of the Vitali convergence theorem, at $t\in(I\cap(s_{k},\infty))\backslash\cup_{j=1}^{\infty}N_{j,k}$ the limit

[TABLE]

exists for every $z\in\mathbb{D}$ and the convergence is locally uniform on $\mathbb{D}$.

Step 2. Let $N$ be a $G_{\delta}$-set of $\mu_{a}$-measure zero with $\cup_{k=1}^{\infty}\left(\cup_{j=1}^{\infty}N_{j,k}\right)\subset N$. Let $t\in I\backslash N$ and take $k\in\mathbb{N}$ with $s_{k}<t$. Then for $z\in\mathbb{D}$ we claim

[TABLE]

This is because we have by (3.9) and $\omega^{\prime}(z,t_{1},t_{2})\rightarrow 1$ locally uniformly on $\mathbb{D}$ as $t_{2}-t_{1}\downarrow 0$ with $t_{1}\leq t\leq t_{2}$,

[TABLE]

where $\gamma(\lambda)$ is the line segment connecting $\omega(z,s_{k},t_{1})$ and $\omega(z,s_{k},t)$, i.e.,

[TABLE]

Step 3. For $t\in I\backslash N$ we show the limit

[TABLE]

exists for every $z\in\mathbb{D}$ and the convergence is locally uniform on $\mathbb{D}$.

The Step 2 implies that the limit

[TABLE]

exists for $\zeta\in\omega(\mathbb{D},s_{k},t)$. Since $\omega(\mathbb{D},s_{k},t)$ is a nonempty domain, again by (3.9) and the Vitali convergence theorem, the above limit exists for every $\zeta\in\mathbb{D}$ and the convergence is locally uniform on $\mathbb{D}$. From this it easily follows that the limit in (3.11) exist and the convergence is locally uniform on $\mathbb{D}$.

Step 4. For $t\in I\backslash N$ we show the limit

[TABLE]

exists for every $\zeta\in\mathbb{D}$ and the convergence is locally uniform on $\mathbb{D}$. Furthermore as a function of $z\in\mathbb{D}$, $\Phi(z,t)$ is analytic in $\mathbb{D}$ and satisfies $\Phi(0,t)=0$ and $\Phi(\mathbb{D},t)\subset\mathbb{D}$.

Let

[TABLE]

Then

[TABLE]

and hence by the Schwarz’s lemma

[TABLE]

Notice that

[TABLE]

and $B(0,t_{1},t_{2})=\tilde{\omega}(0,t)+1=2$. We show the analytic function $\tilde{\omega}(z,t)+1$ has no zeros in $\mathbb{D}$. Suppose, on the contrary, that there is $z_{0}\in\mathbb{D}\backslash\{0\}$ with $\tilde{\omega}(z_{0},t)+1=0$. Then by (3.12), $\tilde{\omega}(z_{0},t)-1=0$, which is a contradiction. Therefore for $t\in I\backslash N$

[TABLE]

is analytic in $z$ satisfying $|\Phi(z,t)|\leq|z|$. Also $\Phi$ is Borel measurable on $\mathbb{D}\times(I\backslash N)$. For $t\in N$ we define $\Phi(z,t)=0$, $z\in\mathbb{D}$. Then $\Phi$ is a Borel measurable function on $\mathbb{D}\times I$.

Step 5. Let

[TABLE]

We show (3.6).

Since $\Phi(z,t)$ is Borel measurable on $\mathbb{D}\times I$, so is $P(z,t)$. Furthermore $\operatorname{Re}P(z,t)>0$ with $P(0,t)=1$, since $|\Phi(z,t)|\leq|z|$ and $\Phi(0,t)=0$. Hence $\{P(\cdot,t)\}_{t\in N}$ is a Borel measurable Herglotz family. The equation (3.6) follows from (3.11) and

[TABLE]

Step 6. We show (3.7) and (3.8).

Let $t,t_{0},t_{1},t_{2}\in I$ with $t_{1}\leq t\leq t_{2}<t_{0}$ and $t_{2}-t_{1}>0$. Put

[TABLE]

Then for $t\in I\backslash N$ letting $t_{2}-t_{1}\downarrow 0$ with $t_{1}\leq t\leq t_{2}$ we have by (3.6)

[TABLE]

Let $t_{0}\in I$ and $t\in\backslash N$ with $t_{0}<t$. Since the convergence of (3.6) is locally uniform on $\mathbb{D}$, letting $t_{2}-t_{1}\downarrow 0$ with $t_{1}\leq t\leq t_{2}$ we have

[TABLE]

Notice in (3.7) and (3.8) that by the Vitali convergence theorem the convergence is locally uniform on $\mathbb{D}$. ∎

Theorem 3.4**.**

Let $\{f_{t}\}_{t\in I}$ be a strictly increasing and continuous Loewner chain with the associated transition family $\{\omega(\cdot,t_{0},t_{1})\}$ and $a(t)=f_{t}^{\prime}(0)$. Then there exist a $G_{\delta}$ set $N(\subset I)$ with $\mu_{a}(N)=0$ and a Herglotz family $\{P(\cdot,t)\}_{t\in I}$, which is Borel measurable on $\mathbb{D}\times I$ such that

[TABLE]

Proof.

Let $N$ and $\{P(\cdot,t)\}_{t\in I}$ as in Theorem 3.3. Notice that since $\{f_{t}\}_{t\in I}$ is continuous, $f_{\tau}^{\prime}\rightarrow f_{t}^{\prime}$ locally uniformly on $\mathbb{D}$ as $\tau\rightarrow t$.

Let $t\in I\backslash N$. Then by letting $t_{2}-t_{1}\downarrow 0$ with $t_{1}\leq t\leq t_{2}$

[TABLE]

∎

We notice that Theorem 1.3 easily follows from Theorems 3.3 and 3.4.

Lemma 3.5** (Zarecki).**

Suppose that $\psi$ is a strictly increasing and continuous function on $[\alpha,\beta]$. Then the inverse function $\psi^{-1}$ is absolutely continuous on $[\psi(\alpha),\psi(\beta)]$ if and only if $\frac{d\psi}{dt}(t)>0$ $\mu_{1}$-a.e. on $[\alpha,\beta]$.

For a proof and details see [21] p.271.

Theorem 3.6**.**

Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be a strictly decreasing and continuous transition family and let $a(t)$, $t\in I$ be a strictly increasing and positive function defined by (2.3) for some $c>0$. If $a(t)$ is locally absolutely continuous on $I$ and $\dot{a}(t):=\frac{da}{dt}>0$ $\mu_{1}$-a.e., then there exist a $G_{\delta}$ set $E(\subset I)$ with $\mu_{1}(E)=0$ and a Herglotz family $\{P(\cdot,t)\}_{t\in I}$ which is Borel measurable on $\mathbb{D}\times I$, such that for $z\in\mathbb{D}$

[TABLE]

In particular if $\{\omega(\cdot,s,t)\}_{(s,t)\in I_{+}^{2}}$ is associated with a strictly increasing and continuous Loewner chain $\{f_{t}\}_{t\in I}$ satisfying $a(t)=f_{t}^{\prime}(0)$, then

[TABLE]

Notice in the case that $a(t)=e^{t}$, $t\in I$, (3.16) and (3.17) are reduced to the usual Loewner-Kufarev ordinary and partial differential equations, respectively.

Proof.

It suffices to show the theorem in the case that $I=[\alpha,\beta]$ with $-\infty<\alpha<\beta<\infty$. Take a $G_{\delta}$ set $N\subset[\alpha,\beta]$ with $\mu_{a}(N)=0$ as in Theorem 3.3. Then by Lemma 3.1 we have $\mu_{1}(a(N))=\mu_{a}(N)=0$. From Lemma 3.5 it follows that $a^{-1}$ is absolutely continuous and hence $a^{-1}$ has the Lusin (N) property, i.e., $a^{-1}$ maps a $\mu_{1}$-null set to a $\mu_{1}$-null set. Therefore $\mu_{1}(N)=\mu_{1}(a^{-1}(a(N)))=0$.

Let $E_{0}(\subset[\alpha,\beta])$ be the set of all $t\in I$ at which $a$ is not differentiable. Since $a(t)$ is absolutely continuous on $[\alpha,\beta]$, we have $\mu_{1}(E_{0})=0$. Also let $E_{1}(\subset[\alpha,\beta])$ be the set of all $t\in I$ at which $a$ is differentiable and $\dot{a}(t)=0$. Then by the assumption $\mu_{1}(E_{1})=0$. Take a $G_{\delta}$ set $E_{2}$ with $E_{0}\cup E_{1}\subset E_{2}$ and $\mu_{1}(E_{2})=0$. Let $E=N\cup E_{2}$. Then $E$ is a $G_{\delta}$ set with $\mu_{1}(E)=0$. By (3.6) we have for $t\in I\backslash E$ and $z\in\mathbb{D}$

[TABLE]

Similarly (3.15)–(3.17) follow from (3.7), (3.8) and (3.13) respectively. ∎

4. Solutions to Loewner-Kufarev Ordinary Differential Equations

Let $I$ be an interval and $a(t)$ be a strictly increasing, positive and continuous function on $I$. In this section for a given $\mathcal{F}_{a}$-measurable Herglotz family $\{P(\cdot,t)\}_{t\in I}$ we shall solve the ordinary differential equation (3.8). Precisely we study for each fixed $t_{0}\in I$ and $z\in\mathbb{D}$ a differential equation

[TABLE]

with an initial condition

[TABLE]

Let $I_{0}$ be a compact subinterval of $I$, $c_{1},c_{2}\in\mathbb{C}$ and $u$, $v$ $a$-absolutely continuous functions on $I_{0}$. Then $c_{1}u+c_{2}v$ and $uv$ are also $a$-absolutely continuous on $I_{0}$, and $D_{a}(c_{1}u+c_{2}v)(t)=c_{1}D_{a}u(t)+c_{2}D_{a}v(t)$, $D_{a}(uv)(t)=D_{a}u(t)\cdot v(t)+u(t)\cdot D_{a}v(t)$ hold $\mu_{a}$-a.e. Furthermore if $h$ is a function of class $C^{1}$ defined on an interval containing $u(I_{0})$, then $h\circ u$ is $a$-absolutely continuous on $I_{0}$ and $D_{a}(h\circ u)(t)=Dh(u(t))D_{a}u(t)$ holds $\mu_{a}$-a.e. Particularly from this and Lemma 3.2 we have for $n=1,2,\ldots$ and $[t_{0},t]\in I$

[TABLE]

Lemma 4.1**.**

Let $u$ be an absolutely $a$-continuous function on $[\alpha,\beta]\subset I$ with $-\infty<\alpha<\beta<\infty$ satisfying

[TABLE]

for some positive constant $M$. Then for $t\in[\alpha,\beta]$

[TABLE]

Proof.

Since $u$ is absolutely $a$-continuous on $[\alpha,\beta]$, so is $|u|$. Thus we have for $\mu_{a}$ almost all $t\in I$

[TABLE]

From this it follows that for $\mu_{a}$ almost all $t\in I$

[TABLE]

Thus $|u(t)|a(t)^{M}$ and $|u(t)|a(t)^{-M}$ are nondecreasing and nonincreasing, respectively and hence we have

[TABLE]

as required. ∎

We need estimates for analytic functions of positive real part. For details see [25, §2.1].

Lemma 4.2**.**

Let $p\in\mathcal{H}(\mathbb{D})$ with $\operatorname{Re}P(z)>0$ and $p(0)=1$. Then for $z\in\mathbb{D}$ we have

[TABLE]

Lemma 4.3**.**

Let $\mathcal{F}$ be a $\sigma$-algebra on $I$ containing the Borel algebra $\mathcal{B}(I)$ on $I$. Let $\{P(\cdot,t)\}_{t\in I}$ be a ${\mathcal{F}}$-measurable Herglotz family and $w(z,t)$ be a function in $\mathbb{D}\times I$ such that $w(z,t)$ is continuous in $t$ and is analytic in $z$. Then $P(w(z,t),t)$ is ${\mathcal{F}}$-measurable in $t$ and is analytic in $z$.

Proof.

For each fixed $t\in I$ it is clear that $P(w(z,t),t)$ is analytic in $z$. For $k\in\mathbb{N}$ take a sequence of disjoint Borel subsets $\{S_{j}^{(k)}\}_{j=1}^{N_{k}}$ of $\mathbb{D}$ with $\mathbb{D}=\cup_{j=1}^{N_{k}}S_{j}^{(k)}$ satisfying $\operatorname{diam}(S_{j}^{(k)})\>=\sup_{w,z\in S_{j}^{(k)}}|w-z|\leq\frac{1}{k}$. For $k\in\mathbb{N}$ and $1\leq j\leq N_{k}$ take $\zeta_{j}^{(k)}\in S_{j}^{(k)}$ arbitrarily and define $P_{k}(z,t)$, $(z,t)\in\mathbb{D}\times I$ by

[TABLE]

Then $P_{k}(z,t)\rightarrow P(z,t)$ as $k\rightarrow\infty$ for all $(z,t)\in\mathbb{D}\times I$. Therefore $P_{k}(w(z,t),t)\rightarrow P(w(z,t),t)$ as $k\rightarrow\infty$. To show the lemma it suffices to see that $P_{k}(w(z,t),t)$ is $\mathcal{F}$-measurable in $t$ for each fixed $z\in\mathbb{D}$.

For an open set $V\subset\mathbb{C}$ we have

[TABLE]

Thus for each fixed $z\in\mathbb{D}$ we have

[TABLE]

Since $\{t\in I:w(z,t)\in S_{j}^{(k)}\}\in\mathcal{B}(I)\subset\mathcal{F}$ and $\{t\in I:P(\zeta_{j}^{(k)},t)\in V\}\in\mathcal{F}$, the above set is clearly $\mathcal{F}$-measurable. ∎

Theorem 4.4**.**

Let $I$ be an interval, $a(t)$ a strictly increasing, positive and continuous function on $I$ and $\{P(\cdot,t)\}_{t\in I}$ an $\mathcal{F}_{a}$-measurable Herglotz family. Then for each fixed $s\in I$ and $z\in\mathbb{D}$ there exists a unique locally absolutely $a$-continuous function $w:I\cap[s,\infty)\rightarrow\mathbb{D}$ satisfying the differential equation

[TABLE]

with the initial condition $w(s)=z$. Furthermore for $(s,t)\in I_{+}^{2}$ and $z\in\mathbb{D}$ let $\omega_{s,t}(z)$ denote the unique solution to (4.4) with $\omega_{s,s}(z)=z$. Then $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is a transition family satisfying $\omega_{s,t}^{\prime}(0)=a(s)/a(t)$, $(s,t)\in I_{+}^{2}$. Particularly $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is continuous and each $\omega_{s,t}(z)$ is univalent in $\mathbb{D}$.

The following proof is a straight forward generalization of that of Theorem 6.3 in Pommerenke [25].

Proof.

Step 1. Let $z\in\mathbb{D}$, $t_{0},t_{1}\in I$ with $t_{0}<t_{1}$ and $w:[t_{0},t_{1}]\rightarrow\mathbb{D}$ be a function with $w(t_{0})=z$. We show that $w(t)$ is an absolutely $a$-continuous solution to (4.4) on $[t_{0},t_{1}]$ if and only if $w(t)$ is a continuous solution to the integral equation

[TABLE]

on $[t_{0},t_{1}]$.

The sufficiency follows from Lemma 3.2. Suppose that $w(t)$ is an absolutely $a$-continuous solution to (4.4) on $[t_{0},t_{1}]$ with $w(t_{0})=z$. First we assume $z\not=0$. Put $\rho=\max_{t_{0}\leq t\leq t_{1}}|w(t)|\in[0,1)$ and $M=\frac{1+\rho}{1-\rho}$. Then by Lemma 4.2 $|D_{a}w(t)|\leq Ma(t)^{-1}|w(t)|$. Hence it follows from Lemma 4.1 that $|w(t)|\geq|z|(a(t_{0})/a(t_{1}))^{M}>0$. Therefore we can choose a single-valued branch $\log w(t)$ of the logarithm. It is easy to see that $\log w(t)$ is absolutely $a$-continuous on $[t_{0},t_{1}]$. We have

[TABLE]

and by integration

[TABLE]

which is equivalent to (4.5). Furthermore by taking real parts of both sides of the equation we see that $|w(t)|$ is strictly decreasing in $t$.

In the case that $z=0$ we have to show that if $w(t)$ is an absolutely $a$-continuous solution to (4.4) with $w(t_{0})=0$, then $w(t)\equiv 0$. This easily follows from Lemma 4.1.

Step 2. We solve the integral equation (4.5). Define $w_{1}(z,t)\equiv z$ for $(z,t)\in\mathbb{D}\times(I\cap[t_{0},\infty))$. Suppose, inductively, that we obtained a function $w_{n}(z,t)$ satisfying the followings,

• (a)

$w_{n}(z,t)$ is continuous in $t\in I\cap[t_{0},\infty)$ with $w_{n}(z,t_{0})=z$,

• (b)

$|w_{n}(z,t)|\leq|z|$,

• (c)

$w_{n}(z,t)$ is analytic in $z\in\mathbb{D}$.

Then by Lemma 4.3 $P(w_{n}(z,t),t)$ is $\mathcal{F}_{a}$-measurable in $[t_{0},\infty)$. By (b) and Lemma 4.2 we have

[TABLE]

Hence $P(w_{n}(z,t),t)$ is locally $\mu_{a}$-integrable in $t$. Therefore we can inductively define

[TABLE]

and it is clear that $w_{n+1}$ satisfies (a). Since $\operatorname{Re}P(z,t)>0$ and $a(t)>0$, $w_{n+1}$ satisfies (b). Put $q(z,t)=P(w_{n}(z,t),t)$. Then $q(z,t)$ is analytic in $z$ with $\operatorname{Re}q(z,t)>0$ and $q(0,t)=1$. From these properties and Lemma 4.2 it follows that $|q^{\prime}(z,t)|\leq\frac{2}{(1-r)^{2}}$ on $\overline{\mathbb{D}}(0,r)\times I$ for each fixed $r\in(0,1)$. For fixed $t\in I$ with $t>t_{0}$ put

[TABLE]

Then by the Lebesgue dominated convergence theorem

[TABLE]

as $\Delta z\rightarrow 0$ with $z,z+\Delta z\in\mathbb{D}(0,r)$. Hence $h(z)$ is analytic in $z$ and $w_{n+1}(z,t)$ satisfies (c).

Step 3. We show $\{w_{n}(z,t)\}_{n=1}^{\infty}$ converges locally uniformly in $\mathbb{D}\times(I\cap[t_{0},\infty))$ and $w(z,t):=\lim_{n\rightarrow\infty}w_{n}(z,t)$ satisfies (a), (b) and (c) and (4.5).

For $\operatorname{Re}a\geq 0$ and $\operatorname{Re}b\geq 0$ we have

[TABLE]

By making use of this and the inequalities and Lemma 4.2 we have for $n\geq 2$

[TABLE]

and similarly by (4.3)

[TABLE]

Therefore we have inductively by using (4.3)

[TABLE]

Hence $\{w_{n}(z,t)\}_{n=1}^{\infty}$ converges locally uniformly in $\mathbb{D}\times(I\cap[t_{0},\infty))$ and $w(z,t):=\lim_{n\rightarrow\infty}w_{n}(z,t)$ satisfies (a), (b) and (c) and (4.5).

Step 4. We show the uniqueness of the solution $w(z,t)$ and the univalence in $z$. To see this assume that a function $\tilde{w}(z,t)$ is also absolutely $a$-continuous in $t\in[t_{0},t_{1}]$ with $\tilde{w}(z,t_{0})=z$ satisfying (4.4). Notice that $|\tilde{w}(z,t)|\leq|z|$ holds for $t\in I\cap[t_{0},\infty)$, since $\tilde{w}(z,\cdot)$ satisfies (4.5).

Since by Lemma 4.2 $|\{zP(z,t)\}^{\prime}|\leq|P(z,t)|+|zP^{\prime}(z,t)|\leq\frac{2}{(1-|z|)^{2}}$, we have $|zP(z,t)-\tilde{z}P(\tilde{z},t)|\leq\frac{2|z-\tilde{z}|}{(1-r)^{2}}$ for $z,\tilde{z}\in\overline{\mathbb{D}}(0,r)$. From this we have

[TABLE]

Since $w(z,t_{0})-\tilde{w}(z,t_{0})=z-z=0$, by Lemma 4.1 we have $w(z,t)=\tilde{w}(z,t)$ for all $t\in I\cap[t_{0},t_{1}]$.

Similarly since for $z_{1},z_{2}\in\overline{\mathbb{D}}(0,r)$

[TABLE]

we have by Lemma Lemma 4.1.

[TABLE]

Thus $w(z,t)$ is univalent in $z$.

Step 5. Now we write $\omega(z,t_{0},t)$ instead of $w(z,t)$ for $(t_{0},t)\in I_{+}^{2}$ and $z\in\mathbb{D}$. Finally we show that $\{\omega(\cdot,s,t)\}_{(s,t)\in I_{+}^{2}}$ forms a transition family with $\omega^{\prime}(0,s,t)=\frac{a(s)}{a(t)}$, $(s,t)\in I_{+}^{2}$.

Let $(t_{0},t_{1})\in I_{+}^{2}$. As a function of $t\in I\cap[t_{1},\infty)$, $\omega(z,t_{0},t)$ and $\omega(\omega(z,t_{0},t_{1}),t_{1},t)$ satisfy the same equation (4.4) and the initial condition $\omega(z,t_{0},t_{1})=\omega(\omega(z,t_{0},t_{1}),t_{1},t_{1})$. Therefore by the uniqueness we have $\omega(z,t_{0},t)=\omega(\omega(z,t_{0},t_{1}),t_{1},t)$ and $\{\omega(\cdot,s,t)\}_{(s,t)\in I_{+}^{2}}$ forms a transition family.

Next by $\omega(0,t_{0},t)=0$ and $P(0,t)=1$ we have

[TABLE]

∎

Suppose that $\beta:=\sup I\not\in I$. In §2 we saw that if $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ is a continuous transition family, the locally uniform limit $g_{t}(z):=\lim_{\tau\uparrow\beta}a(\tau)\omega(z,t,\tau)$ exists on $\mathbb{D}$ and $\{g_{t}\}_{t\in I}$ forms a Loewner chain having $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ as the associated transition family.

By (4.5) we have

[TABLE]

Since $\frac{a(t)}{a(s)}\omega(\cdot,s,t)$ is univalent, we have by the growth theorem

[TABLE]

Combining this, Lemma 4.2 and $|\omega(z,s,t)|\leq|z|$ we have for $|z|\leq r$

[TABLE]

Since $\frac{1}{a(t)^{2}}$ is $\mu_{a}$-integrable on $[t_{0},\beta)$, we have the expression

[TABLE]

5. Schlicht subordination and connecting chain

Let $I\subset[-\infty,\infty]$ be an interval. First we point out a few simple facts concerning differences between (I) the class of Loewner chains of univalent functions on $I$ and (III) the class of Loewner chains on $I$.

Let $\{f_{t}\}_{t\in I}$ be a Loewner chain with $\Omega_{t}=f_{t}(\mathbb{D})$, $t\in I$. Then $\{\Omega_{t}\}_{t\in I}$ is nondecreasing, i.e., $\Omega_{s}\subset\Omega_{t}$ for $(s,t)\in I_{+}^{2}$.

Definition 5.1**.**

We say that a Loewner chain $\{f_{t}\}_{t\in I}$ is strictly expanding if the corresponding family of domains $\{f_{t}(\mathbb{D})\}_{t\in I}$ is strictly increasing, i.e.,

[TABLE]

By the uniqueness part of the Schwarz lemma if a Loewner chain $\{f_{t}\}_{t\in I}$ is strictly expanding, then the function $a(t)=f_{t}^{\prime}(0)$, $t\in I$, is strictly increasing. When $\{f_{t}\}_{t\in I}$ consists of univalent functions, the reverse is also true. However when $f_{t}$ are not necessary univalent, the reverse is not always true. We have a simple counter example.

Example 5.2**.**

For $t>0$ let $g_{t}$ be the conformal mapping of $\mathbb{D}$ onto the rectangle $\{w\in\mathbb{C}:|\operatorname{Re}w|<1,|\operatorname{Im}w|<t\;\}$ with $g_{t}(0)=0$ and $g_{t}^{\prime}(0)>0$. Put $f_{t}=e^{g_{t}}$, $t\in I$. Then it is easy to see that $f_{t}^{\prime}(0)$ is strictly increasing in $t$ and hence $\{f_{t}\}_{t>0}$ is a Loewner chain. However $f_{t}(\mathbb{D})=\{w\in\mathbb{C}:e^{-1}<|w|<e\}$ for $t>\pi$. Thus $\{f_{t}\}_{t>0}$ is not strictly expanding.

A Loewner chain $\{f_{t}\}_{t\in I}$ is bounded on $I\cap[-\infty,t_{0}]\times\overline{\mathbb{D}}(0,r)$ for every $t_{0}\in I$ and $r\in(0,1)$. In fact, let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be the associated transition family. Then for $t\in I$ with $t\leq t_{0}$ and $r\in(0,1)$ we have by the Schwarz lemma

[TABLE]

Suppose that $\{f_{t}\}_{t\in I}$ is a Loewner chain of univalent functions. Then by the growth theorem for univalent analytic functions we have

[TABLE]

Therefore for any $M>0$, the class of Loewner chains $\{f_{t}\}_{t\in I}$ of univalent functions defined on some interval $I$ satisfying $\sup_{t\in I}f_{t}^{\prime}(0)\leq M$ is uniformly bounded on $I\times\overline{\mathbb{D}}(0,r)$ for every $r\in(0,1)$.

Contrary there are no local upper bounds for the class of all Loewner chains $\{f_{t}\}_{t\in I}$ satisfying $\sup_{t\in I}f_{t}^{\prime}(0)\leq M$. For example let

[TABLE]

Then it is easy to see that $\{f_{n}(z,t)\}_{n=1}^{\infty}$ is a sequence of normalized Loewner chains, and that for any $r\in(0,1)$ and $t_{0}\in\mathbb{R}$

[TABLE]

Later we shall give a family of Loewner chains $\{f_{t}\}_{0<t<\infty}$ of universal covering maps on $I=(0,\infty)$ with $f_{t}^{\prime}(0)=t$, $0<t<\infty$ which is not uniformly bounded on $(0,t_{0}]\times\overline{\mathbb{D}}(0,r)$ for any $t_{0}\in\mathbb{R}$ and $r\in(0,1)$. See Example 7.12.

Definition 5.3**.**

A function $f\in\mathcal{H}(\mathbb{D})$ is said to be schlicht subordinate to a function $g\in\mathcal{H}(\mathbb{D})$ if there exists a univalent analytic map $\omega:\mathbb{D}\rightarrow\mathbb{D}$ with $\omega(0)=0$ and $f=g\circ\omega$. We say that $f\in\mathcal{H}_{0}(\mathbb{D})$ can be continuously connected with $g\in\mathcal{H}_{0}(\mathbb{D})$ by a Loewner chain if there exists a continuous Loewner chain $\{f_{t}\}_{\alpha\leq t\leq\beta}$ with $f_{\alpha}=f$ and $f_{\beta}=g$.

The following result is known. See Pommerenke [24, §4 Folgerung 1]. For completeness and later applications we shall give a proof.

Proposition 5.4**.**

A function $f\in\mathcal{H}_{0}(\mathbb{D})$ can be continuously connected with a function $g\in\mathcal{H}_{0}(\mathbb{D})$ by a Loewner chain if and only if $f$ is schlicht subordinate to $g$.

Proof.

The necessity follows from Theorem 2.7.

Conversely let $\omega\in\mathfrak{B}$ be the unique univalent mapping with $f=g\circ\omega$. We may assume $\omega^{\prime}(0)=f^{\prime}(0)/g^{\prime}(0)\in(0,1)$, since otherwise $f$ coincides with $g$. Take a sequence $\{r_{n}\}_{n=1}^{\infty}$ with $0<r_{1}<\cdots<r_{n}\uparrow 1$ and put $\omega_{n}(z)=\omega(r_{n}z)$, $n\in\mathbb{N}$. Then for each $n\in\mathbb{N}$, $\omega_{n}(\overline{\mathbb{D}})\subset\mathbb{D}$. Take $z_{n}\in\partial\mathbb{D}$ with $|\omega_{n}(z_{n})|=\max_{z\in\overline{\mathbb{D}}}|\omega_{n}(z)|$. Let $\gamma_{n}:[0,1]\rightarrow\overline{\mathbb{D}}$ be the curve consists of radial line segment starting at $\omega_{n}(z_{n})/|\omega_{n}(z_{n})|\in(\partial\mathbb{D})$ and ending at $\omega_{n}(z_{n})$, and the boundary curve $\omega_{n}(\partial\mathbb{D})$ begins and ends on $\omega_{n}(\partial\mathbb{D})$. We assume that $\gamma_{n}$ is injective on $[0,1)$. For each fixed $0<t\leq 1$ let $\omega_{n}(z,t)$, $z\in\mathbb{D}$ be the unique conformal mapping of $\mathbb{D}$ onto the simply connected domain $\mathbb{D}\backslash\{\gamma_{n}(s):0\leq s\leq 1-t\}$, and let $\omega_{n}(z,0)=\omega_{n}(z)$, $z\in\mathbb{D}$. Notice that $\omega_{n}(z,1)=z$, $z\in\mathbb{D}$. Then for fixed $n\in\mathbb{N}$, since a family of simply connected domains $\{\omega_{n}(\mathbb{D},t)\}_{0\leq t\leq 1}$ is strictly increasing and continuous in the sense of kernel, $\{\omega_{n}(\cdot,t)\}_{0\leq t\leq 1}$ is a Loewner chain. For details see [2, Chapter 6]. By taking a new parameter we may assume $\omega_{n}^{\prime}(0,t)=t$, $r_{n}\alpha\leq t\leq 1$, where $\alpha:=\omega^{\prime}(0)\in(0,1)$. Then, since $|\omega_{n}(z,t)|\leq 1$, the sequence $\{\omega_{n}(z,t)\}_{n=1}^{\infty}$ of functions of two variable $(z,t)\in\mathbb{D}\times[\alpha,1]$ is uniformly bounded on $\mathbb{D}\times[\alpha,1]$.

For $n\in\mathbb{N}$ let $\{\omega_{n}(\cdot,s,t)\}_{\alpha\leq s\leq t\leq 1}$ be the transition family of $\{\omega_{n}(\cdot,t)\}_{\alpha\leq t\leq 1}$. Since $|\omega_{n}^{\prime}(z,t)|\leq(1-|z|^{2})^{-1}$ and $|\omega_{n}(z,t)|\leq|z|$, by Theorem 2.2 we have for $s\leq t$ and $z_{0},z_{1}\in\overline{\mathbb{D}}(0,r)$

[TABLE]

Therefore the sequence $\{\omega_{n}(z,t)\}_{n=1}^{\infty}$ is equicontinuous on $\overline{\mathbb{D}}(0,r)\times[\alpha,1]$ for any fixed $r\in(0,1)$. Thus by the Arzelà-Ascoli theorem some subsequence $\{\omega_{n_{k}}(z,t)\}_{k=1}^{\infty}$ converges to a function $\phi(z,t)$, $(z,t)\in\mathbb{D}\times[\alpha,1]$ uniformly on $\overline{\mathbb{D}}(0,r)\times[\alpha,1]$ for every $r\in(0,1)$. For each fixed $t\in[\alpha,1]$, as a function of $z\in\mathbb{D}$, $\phi(z,t)$ is a locally uniform limit of the sequence of univalent functions $\{\omega_{n_{k}}(z,t)\}_{k=1}^{\infty}$ in $\mathbb{D}$ and $\phi^{\prime}(0,t)=\lim_{k\rightarrow\infty}\omega_{n_{k}}^{\prime}(0,t)=t\not=0$. Hence $\phi(z,t)$ is univalent analytic in $\mathbb{D}$ with $\phi^{\prime}(0,t)=t$, $\alpha\leq t\leq 1$. For $\alpha\leq s\leq t\leq 1$, since $|\omega_{n_{k}}(z,s,t)|\leq 1$, by taking a subsequence if necessary, we may suppose that $\{\omega_{n_{k}}(z,s,t)\}_{k=1}^{\infty}$ converges locally uniformly in $\mathbb{D}$ to a function in $\mathfrak{B}$. Combining this and $\omega_{n_{k}}(z,s)=\omega_{n_{k}}(\omega_{n_{k}}(z,s,t),t)$ it follows that $\phi(\cdot,s)\prec\phi(\cdot,t)$. Hence $\{\phi(z,t)\}_{\alpha\leq t\leq 1}$ is a Loewner chain. Since $\phi^{\prime}(0,t)=t$, $\alpha\leq t\leq 1$, $\{\phi(z,t)\}_{\alpha\leq t\leq 1}$ is continuous. Furthermore by (5.3) we have

[TABLE]

as $k\rightarrow\infty$. Thus

[TABLE]

Also we have

[TABLE]

Put $f_{t}(z)=g(\phi(z,t))$, $t\in[\alpha,1]$. Then $\{f_{t}\}_{\alpha\leq t\leq 1}$ is a continuous Loewner chain connecting $f_{\alpha}=g\circ\omega=f$ and $f_{1}=g\circ\operatorname{id}_{\mathbb{D}}=g$. ∎

Definition 5.5**.**

Let $f\in\mathcal{H}_{0}(\mathbb{D})$. We say that $f$ is maximal in the sense of continuous Loewner chain if there is no continuous Loewner chain $\{f_{t}\}_{0\leq t\leq\varepsilon}$ for some $\varepsilon>0$ satisfying $f_{0}=f$ and $f^{\prime}(0)<f_{\varepsilon}^{\prime}(0)$.

Pommerenke ([25, Theorem 6.1]) proved that for any univalent $f\in\mathcal{H}_{0}(\mathbb{D})$ there exists a normalized Loewner chain $\{f_{t}\}_{\alpha\leq t<\infty}$ with $\alpha=\log f^{\prime}(0)$ and $f=f_{\alpha}$. Thus a univalent function $f$ can not be maximal.

Theorem 5.6**.**

Let $f\in\mathcal{H}_{0}(\mathbb{D})$. If for almost every $\zeta\in\partial\mathbb{D}$ the nontangential limit of $f$ does not exists at $\zeta$, then $f$ is maximal in the sense of continuous Loewner chain.

Proof.

Assume that $\{f_{t}\}_{0\leq t\leq\varepsilon}$ is a Loewner chain with $f_{0}=f$ and $f_{0}^{\prime}(0)<f_{\varepsilon}^{\prime}(0)$. Take $\omega\in\mathfrak{B}$ with $f=f_{\varepsilon}\circ\omega$. Since $\omega$ is bounded, it follows from the Fatou theorem ([8, Theorem 2.2]) that for almost every $\zeta\in\partial\mathbb{D}$ the nontangential limit $\omega(\zeta)$ exists.

Claim. The Lebesgue measure of the subset $A:=\{\zeta\in\partial\mathbb{D}:|\omega(\zeta)|<1\}$ of $\partial\mathbb{D}$ is positive.

To show the claim suppose, on the contrary, that $|\omega(\zeta)|=1$ for almost all $\zeta\in\partial\mathbb{D}$. Then $\omega$ is an inner function and hence by the Frostman theorem (see [10, Theorem 2.6.4]), for all $c\in\mathbb{D}$, except possibly for a set of capacity zero, the function

[TABLE]

is a Blaschke product. Since by Proposition 5.4 $\omega$ is univalent, for such $c$, $B_{c}$ is a Blaschke product of order one. Therefore $B_{c}$ and $\omega$ is a linear fractional transformation preserving $\mathbb{D}$, and hence by $\omega(0)=0$ and $\omega^{\prime}(0)>0$, we have $\omega(z)\equiv z$. This contradicts $\omega^{\prime}(0)=f^{\prime}(0)/f_{\varepsilon}^{\prime}(0)<1$ and a proof the claim is completed.

Now for all $\zeta\in A$ we have $f(z)=f_{\varepsilon}(\omega(z))\rightarrow f_{\varepsilon}(\omega(\zeta))$ as $z\rightarrow\zeta$ nontangentially in $\mathbb{D}$, which contradicts the assumption of the theorem. ∎

In 1962 MacLane [18] constructed an analytic function $F$ in $\mathbb{D}$ satisfying for all $\zeta\in\partial\mathbb{D}$

[TABLE]

The function $F$ cannot have a nontangential limit at any point of $\partial\mathbb{D}$, since $F$ has no radial limits at any point of $\partial\mathbb{D}$. Therefore by Theorem 5.6, $F$ is maximal in the sense of continuous Loewner chain.

6. Kernel convergence of domains and locally uniform convergence of covering maps

Let $w_{0}\in\hat{\mathbb{C}}$ and $\{\Omega_{n}\}_{n=1}^{\infty}$ be a sequence of domains in $\hat{\mathbb{C}}$. Suppose that $w_{0}\in\Omega_{n}$ for all sufficiently large $n$. The kernel of $\{\Omega_{n}\}_{n=1}^{\infty}$ with respect to the reference point $w_{0}$ is defined as the set consisting of $w_{0}$ together with all points $w$ such that there exists a domain $H$ and $N\in\mathbb{N}$ with

[TABLE]

We denote the kernel by $\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$. Then $\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$ is a domain containing $w_{0}$ or coincides with $\{w_{0}\}$. A sequence $\{\Omega_{n}\}_{n=1}^{\infty}$ of domains said to converges to a domain $\Omega$ with respect to $w_{0}$ in the sense of kernel if $\ker(w_{0},\{\Omega_{n_{k}}\}_{k=1}^{\infty})=\Omega$ for every subsequence $\{\Omega_{n_{k}}\}_{k=1}^{\infty}$ of $\{\Omega_{n}\}_{n=1}^{\infty}$. Also $\{\Omega_{n}\}_{n=1}^{\infty}$ is said to degenerates to $\{w_{0}\}$ if $\ker(w_{0},\{\Omega_{n_{k}}\}_{k=1}^{\infty})=\{w_{0}\}$ for every subsequence $\{\Omega_{n_{k}}\}_{k=1}^{\infty}$.

Since $\hat{\mathbb{C}}$ is compact, it is easy to see that $\{\Omega_{n}\}_{n=1}^{\infty}$ converges to $\hat{\mathbb{C}}$ if and only if $\Omega_{n}=\hat{\mathbb{C}}$ for all sufficiently large $n$.

Suppose temporarily that $w_{0}\in\mathbb{C}$ and each $\Omega_{n}$ is a simply connected domain contained in $\mathbb{C}$. Let $f_{n}$ be the unique conformal map of $\mathbb{D}$ onto $\Omega_{n}$ with $f_{n}(0)=w_{0}$ and $f_{n}^{\prime}(0)>0$. Then the Carathéodory convergence theorem states the followings.

• (1)

If $\{\Omega_{n}\}_{n=1}^{\infty}$ converges to a domain $\Omega$ with $\Omega\subsetneq\mathbb{C}$ in the sense of kernel with respect to $w_{0}$, then $\Omega$ is simply connected and $\{f_{n}\}_{n=1}^{\infty}$ converges locally uniformly on $\mathbb{D}$ to the unique conformal map $f$ of $\mathbb{D}$ onto $\Omega$ with $f(0)=w_{0}$ and $f^{\prime}(0)>0$. If $\{\Omega_{n}\}_{n=1}^{\infty}$ degenerates to $\{w_{0}\}$, then $f_{n}\rightarrow w_{0}$ locally uniformly on $\mathbb{D}$.

• (2)

If $\{f_{n}\}_{n=1}^{\infty}$ converges locally uniformly on $\mathbb{D}$ to a nonconstant function $f$, then $f$ is analytic and univalent in $\mathbb{D}$ and $\{\Omega_{n}\}_{n=1}^{\infty}$ converges to $f(\mathbb{D})$ in the sense of kernel with respect to $w_{0}$. If $f_{n}\rightarrow w_{0}$ locally uniformly on $\mathbb{D}$, then $\{\Omega_{n}\}_{n=1}^{\infty}$ degenerates to $\{w_{0}\}$.

By replacing conformal maps with universal covering maps Hejhal ([13]) generalized the Carathéodry theorem to the case that each $\Omega_{n}$ is not necessarily simply connected. In this section we shall generalize Hejhal’s theorem and give a detailed proof. Before this we study basic properties of the notion of kernel convergences, particularly when $\Omega_{n}$, $n\in\mathbb{N}$ are multiply connected.

We start with an equivalent condition introduced by Pommerenke ([25], Problem 3, p.31 and [26], §1.8). For the reader’s convenience we shall give a proof of the equivalence. For $z,w\in\mathbb{C}$ and $E\subset\hat{\mathbb{C}}$ let

[TABLE]

and for $z\in\hat{\mathbb{C}}$ and $E\subset\hat{\mathbb{C}}$ let

[TABLE]

We set

[TABLE]

Theorem 6.1**.**

Let $\{\Omega_{n}\}_{n=1}^{\infty}$ be a sequence of domains in $\hat{\mathbb{C}}$ and $\Omega$ a domain in $\hat{\mathbb{C}}$. Then the following three conditions are equivalent.

• (i)

Both

• (a)

for every compact subset $K$ of $\Omega$ there exists $N\in\mathbb{N}$ with $K\subset\Omega_{n}$ for all $n\geq N$

and

• (b)

for every $c\in\partial\Omega$, $d^{*}(c,\partial\Omega_{n})\rightarrow 0$

are satisfied.

• (ii)

For every $w_{0}\in\Omega$, $\Omega_{n}\rightarrow\Omega$ in the sense of kernel with respect to $w_{0}$.

• (iii)

For some $w_{0}\in\Omega$, $\Omega_{n}\rightarrow\Omega$ in the sense of kernel with respect to $w_{0}$.

Proof.

Assume (i) and take $w_{0}\in\Omega$ arbitrarily. First we show $\Omega\subset\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$. To see this let $w\in\Omega$ and take a polygonal line $\ell$ connecting $w_{0}$ and $w$ contained in $\Omega$. Let $\delta_{1}\in(0,d^{*}(\ell,\partial\Omega))$ and put $H=\bigcup_{\zeta\in\ell}\mathbb{D}^{*}(\zeta,\delta_{1})$. Then $H$ is a domain with $w_{0},w\in H\subset\overline{H}\subset\Omega$. Since $\overline{H}$ is compact, by (a) we have that $\overline{H}\subset\Omega$ for all sufficiently large $n$. Hence $w\in\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$ and we have $\Omega\subset\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$.

Next we show $\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})\subset\Omega$. Suppose, on the contrary, that $\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})\backslash\Omega\not=\emptyset$. By (a), as shown above, $\Omega\subset\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$. Since $\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$ is connected, there exists $c\in\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})\cap\partial\Omega$. Then since $c\in\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$, some neighborhood of $c$ is contained in $\Omega_{n}$ for all sufficiently large $n$. However since $c\in\partial\Omega$, by (b) we have $d^{*}(c,\partial\Omega_{n})\rightarrow 0$, which is a contradiction. Thus $\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})=\Omega$.

Notice that if $\{\Omega_{n}\}_{n=1}^{\infty}$ and $\Omega$ satisfy (a) and (b), then any subsequence $\{\Omega_{n_{k}}\}_{k=1}^{\infty}$ and $\Omega$ also satisfy (a) and (b). Hence $\ker(w_{0},\{\Omega_{n_{k}}\}_{k=1}^{\infty})=\Omega$. Therefore (ii) holds.

It is clear that (ii) implies (iii). Now assume (iii) holds for some $w_{0}\in\Omega$. Let $K$ be a compact subset of $\Omega=\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$. For $w\in K$ there exist a domain $H_{w}$ contained in $\Omega$ and $n(w)\in\mathbb{N}$ such that $w_{0},w\in H\subset\Omega_{n}$ for all $n\geq n(w)$. Since $K\subset\bigcup_{w\in K}H_{w}$ is an open covering of the compact set $K$, we can choose $w_{1},\ldots,w_{j}$ such that $K\subset H_{w_{1}}\cup\cdots\cup H_{w_{j}}$. Therefore we have $K\subset H_{w_{1}}\cup\cdots\cup H_{w_{j}}\subset\Omega_{n}$ for all $n\geq\max\{n(w_{1}),\ldots,n(w_{j})\}$, and hence (a) holds.

To show (b) suppose, on the contrary, that there exist $c\in\partial\Omega$, $\delta_{2}>0$ and a subsequence $\{\Omega_{n_{k}}\}_{k=1}^{\infty}$ such that $d^{*}(c,\partial\Omega_{n_{k}})\geq\delta_{2}$ for all $k\in\mathbb{N}$. This implies $\mathbb{D}^{*}(c,\delta_{2})\subset\Omega_{n_{k}}$ or $\mathbb{D}^{*}(c,\delta_{2})\subset\hat{\mathbb{C}}\backslash\overline{\Omega_{n_{k}}}$. Since $c\in\partial\Omega$, we can take $c^{*}\in\mathbb{D}^{*}(c,\delta_{2})\cap\Omega=\mathbb{D}^{*}(c,\delta_{2})\cap\ker(w_{0},\{\Omega_{n}\}_{n=1}^{\infty})$ and a domain $H$ with $w_{0},c^{*}\subset H\subset\Omega_{n}$ for all sufficiently large $n$. Hence $\mathbb{D}^{*}(c,\delta_{2})\subset\Omega_{n_{k}}$ for all sufficiently large $k$. Since $c^{*}\in H\cap\mathbb{D}^{*}(c,\delta_{2})$, the union $H\cup\mathbb{D}^{*}(c,\delta_{2})$ is a domain containing $c$ and $w_{0}$. Therefore

[TABLE]

which contradicts $c\in\partial\Omega$. ∎

By modifying the above proof we obtain the following criterion.

Theorem 6.2**.**

Let $w_{0}\in\hat{\mathbb{C}}$ and $\{\Omega_{n}\}_{n=1}^{\infty}$ be a sequence of domains in $\hat{\mathbb{C}}$ with $w_{0}\in\Omega_{n}$ for all sufficiently large $n$. Then $\{\Omega_{n}\}_{n=1}^{\infty}$ degenerates to $\{w_{0}\}$ if and only if

• (c)

$d^{*}(w_{0},\partial\Omega_{n})\rightarrow 0$**

holds.

By making use of Theorem 6.1 we can omit a reference point $w_{0}$ from the definition of the kernel convergence.

Definition 6.3**.**

We say that a sequence of domains $\{\Omega_{n}\}_{n=1}^{\infty}$ converges to a domain $\Omega$ in the sense of kernel (indicated simply by $\Omega_{n}\rightarrow\Omega$ as $n\rightarrow\infty$), if both (a) and (b) hold.

We have an important remark. In our definition the limit domain $\Omega$ is not uniquely determined by $\{\Omega_{n}\}_{n=1}^{\infty}$.

Example 6.4**.**

*For $n\in\mathbb{N}$ let $\Omega_{n}$ be the domain obtained from the strip $\{w\in\mathbb{C}:|\operatorname{Im}w|<1\}$ by removing the line segments $\{k+is:|s|\leq 1-n^{-1}\}$, $k\in\mathbb{Z}$. Then for each $k\in\mathbb{Z}$ , $\Omega_{n}\rightarrow D_{k}:=\{w\in\mathbb{C}:k<\operatorname{Re}w<k+1,\;0<\operatorname{Im}w<1\}$ as $n\rightarrow\infty$.

\Omega_{n}$$-1[math]1$$2$$3

-1$$D_{-1}[math]D_{0}$$1$$D_{1}$$2$$D_{2}$$3$$D_{3} *

Proposition 6.5**.**

If $\Omega_{n}\rightarrow\Omega$ and $\Omega_{n}\rightarrow\Omega^{\prime}$ as $n\rightarrow\infty$ in the sense of kernel, then either $\Omega=\Omega^{\prime}$ or $\Omega\cap\Omega^{\prime}=\emptyset$ holds.

Proof.

It suffices to show that $\Omega=\Omega^{\prime}$ whenever $\Omega\cap\Omega^{\prime}\not=\emptyset$. To this end suppose that $c\in\Omega\cap\partial\Omega^{\prime}$. Take $r>0$ with $\overline{\mathbb{D}}^{*}(c,r)\subset\Omega$. Then there exists $N\in\mathbb{N}$ such that for $n\geq N$ we have $\overline{\mathbb{D}}^{*}(c,r)\subset\Omega_{n}$. Hence $d^{*}(c,\partial\Omega_{n})\geq r$ for $n\geq N$ and this contradicts $c\in\partial\Omega^{\prime}$. Therefore $\Omega\cap\partial\Omega^{\prime}=\emptyset$. Since $\Omega$ is connected, either $\Omega\subset\Omega^{\prime}$ or $\Omega\subset\hat{\mathbb{C}}\backslash\Omega^{\prime}$ holds. From $\Omega\cap\Omega^{\prime}\not=\emptyset$ it follows that $\Omega\subset\Omega^{\prime}$.

Similarly by replacing $\Omega$ with $\Omega^{\prime}$ we have $\Omega^{\prime}\subset\Omega$ and hence $\Omega=\Omega^{\prime}$. ∎

When $\{\Omega_{n}\}_{n=1}^{\infty}$ is monotone, one can easily prove the following. We denote the set of interior points of $A\subset\hat{\mathbb{C}}$ by $\operatorname{Int}A$.

Theorem 6.6**.**

Let $\{\Omega_{n}\}_{n=1}^{\infty}$ be a sequence of domains in $\mathbb{C}$.

• (i)

If $\Omega_{1}\subset\Omega_{2}\subset\cdots$, then $\Omega_{n}\rightarrow\bigcup_{k=1}^{\infty}\Omega_{k}$ as $n\rightarrow\infty$.

• (ii)

If $\Omega_{1}\supset\Omega_{2}\supset\cdots$ and $\operatorname{Int}\left(\bigcap_{k=1}^{\infty}\Omega_{k}\right)\not=\emptyset$, then $\Omega_{n}\rightarrow\Omega$ as $n\rightarrow\infty$ for every connected component $\Omega$ of $\operatorname{Int}\left(\bigcap_{k=1}^{\infty}\Omega_{k}\right)$.

Theorem 6.7**.**

Let $\{\Omega_{n}\}_{n=1}^{\infty}$ and $\{\Omega_{n}^{\prime}\}_{n=1}^{\infty}$ be sequences of domains in $\hat{\mathbb{C}}$ which converge to domains $\Omega$ and $\Omega^{\prime}$ in $\hat{\mathbb{C}}$, respectively. Suppose that $w_{0}\in\Omega\cap\Omega^{\prime}$. Then there exists $N\in\mathbb{N}$ such that $w_{0}\in\Omega_{n}\cap\Omega_{n}^{\prime}$ for all $n\geq N$ and the component of $\Omega_{n}\cap\Omega_{n}^{\prime}$ containing $w_{0}$ converges to the component of $\Omega\cap\Omega^{\prime}$ containing $w_{0}$ as $n\rightarrow\infty$.

Proof.

Let $D$ be the component of $\Omega\cap\Omega^{\prime}$ containing $w_{0}$. Since $\{w_{0}\}$ is compact and contained in $\Omega$ and $\Omega^{\prime}$, there exists $N_{0}\in\mathbb{N}$ such that $w_{0}\in\Omega_{n}\cap\Omega_{n}^{\prime}$ for all $n\geq N$. For $n\geq N$ let $D_{n}$ be the component of $\Omega_{n}\cap\Omega_{n}^{\prime}$ containing $w_{0}$.

We show $\{D_{n}\}_{n\geq N}$ and $D$ satisfy the condition (a). Let $K$ be a compact subset of $D$. We can take a compact and connected set $\tilde{K}$ with $K\cup\{w_{0}\}\subset\tilde{K}\subset D$ and take $N_{1}\geq N$ with $\tilde{K}\subset\Omega_{n}\cap\Omega_{n}^{\prime}$ for all $n\geq N_{1}$. Since $\tilde{K}$ is connected and contains $w_{0}$, we have

[TABLE]

We show $\{D_{n}\}_{n\geq N}$ and $D$ satisfy the condition (b). Let $c\in\partial D$. Since

[TABLE]

$c\in\partial\Omega$ or $c\in\partial\Omega^{\prime}$. We may assume $c\in\partial\Omega$ without loss of generality. Then for any $\varepsilon>0$, there exists $N_{2}\in\mathbb{N}$ such that $d^{*}(c,\partial\Omega_{n})<\varepsilon$ for $n\geq N$. Thus $\mathbb{D}(c,\varepsilon)\backslash\Omega_{n}\not=\emptyset$ and hence $\mathbb{D}(c,\varepsilon)\backslash D_{n}\not=\emptyset$ for all $n\geq N_{2}$. Also by $c\in\partial D$ there exists $\tilde{c}\in D\cap\mathbb{D}(c,\varepsilon)$. As shown above we can take $N_{3}\in\mathbb{N}$ such that $\tilde{c}\in D_{n}$ for all $n\geq N_{3}$. Therefore $\mathbb{D}(c,\varepsilon)\cap D_{n}\not=\emptyset$ and $\mathbb{D}(c,\varepsilon)\backslash D_{n}\not=\emptyset$ for all $n\geq N_{4}:=\max\{N_{2},N_{3}\}$. Since $D_{n}$ is connected, for $n\geq N_{4}$ this implies $\mathbb{D}(c,\varepsilon)\cap\partial D_{n}\not=\emptyset$ which is equivalent to $d^{*}(c,\partial D_{n})<\varepsilon$. We obtain $d^{*}(c,\partial D_{n})\rightarrow 0$ as $n\rightarrow\infty$. ∎

In the above theorem one can not replace intersection symbol by union symbol.

Example 6.8**.**

Let $\{\Omega\}_{n=1}^{\infty}$ and $\{D_{k}\}_{k\in\mathbb{Z}}$ as in Example 6.4. Let $\Omega^{\prime}=\Omega_{n}^{\prime}=\mathbb{D}(0,\rho)$ with $0<\rho<1$. Then although $\Omega_{n}\rightarrow D_{0}$ and $\Omega_{n}^{\prime}\rightarrow\Omega^{\prime}$ we have $\Omega_{n}\cup\Omega_{n}^{\prime}\rightarrow D_{-1}\cup D_{0}\cup\Omega^{\prime}\supsetneq D_{0}\cup\Omega^{\prime}$ as $n\rightarrow\infty$.

Now we consider a generalization of the Hejhal theorem. First we study relations between $\ker(w_{0},\{f_{n}(D)\}_{n=1}^{\infty})$ and $f(D)$ when $w_{0}\in f(D)$ and $f_{n}\in\mathcal{H}(D)$ converges to $f\in\mathcal{H}(D)$ locally uniformly on $D$. The following is fairly easy and we omit a proof which is a simple application of Rouche’s theorem.

Proposition 6.9**.**

Let $D$ be a domain in $\mathbb{C}$ and $\{f_{n}\}_{n=1}^{\infty}$ a sequence of nonconstant analytic functions in $D$ which converges to a function $f$ locally uniformly on $D$. If $f$ is nonconstant, then $\{f_{n}(D)\}_{n=1}^{\infty}$ and $f(D)$ satisfy the condition (a) and $f(D)\subset\ker(w_{0},\{f_{n}(D)\}_{n=1}^{\infty})$ for all $w_{0}\in f(D)$.

Without adding some condition, we can not conclude the revers inclusion relation $\ker(w_{0},\{f_{n}(D)\}_{n=1}^{\infty})\subset f(D)$. For example consider the sequence $\{z^{n}\}_{n=1}^{\infty}$ of functions in $\mathbb{D}$. We have another less trivial example.

Example 6.10**.**

Let $D_{n}$ be the domain obtained from the rectangle $\{w\in\mathbb{C}:|\operatorname{Re}w|<1,|\operatorname{Im}w|<2\pi\;\}$ by removing the two line segments $\{t\pm\pi i/6:-1+n^{-1}\leq t\leq 1\}$. Then $D_{n}\rightarrow D:=\{w\in\mathbb{C}:|\operatorname{Re}w|<1,|\operatorname{Im}w|<\pi/6\;\}$ in the sense of kernel. Let $g_{n}$ and $g$ be the conformal map of $\mathbb{D}$ onto $D_{n}$ and $D$ with $g_{n}(0)=g(0)=1$, $g_{n}^{\prime}(0)>0$ and $g^{\prime}(0)>0$, respectively. Then by the Carathéodry kernel convergence theorem $g_{n}\rightarrow g$ locally uniformly in $\mathbb{D}$ and hence $f_{n}:=e^{g_{n}}\rightarrow f:=e^{g}$ locally uniformly in $\mathbb{D}$. Each $f_{n}(\mathbb{D})$, $n\in\mathbb{N}$ and $\ker(w_{0},\{f_{n}(D)\}_{n=1}^{\infty})$ coincide with the annulus $\{\zeta\in\mathbb{C}:e^{-1}<|\zeta|<e\}$. However $f(\mathbb{D})=\{\zeta\in\mathbb{C}:e^{-1}<|\zeta|<e,\;|\operatorname{Arg}\zeta|<\pi/6\}$ which is a proper subset of the annulus.

Let $R$ be a Riemann surfaces. An analytic surjection $p:\tilde{R}\rightarrow R$ of a Riemann surface $\tilde{R}$ is called a covering map if for each $x\in R$ there exists a connected neighborhood $V$ of $x$ such that $p^{-1}(V)$ can be written as the union of disjoint open sets $\{\tilde{V}_{\lambda}\}_{\lambda\in\Lambda}$ and for each $\lambda\in\Lambda$, the restriction $p|_{\tilde{V}_{\lambda}}$ is a conformal map of $\tilde{V}_{\lambda}$ onto $V$. The Riemann surface $\tilde{R}$ is called a covering surface of $R$, and $V$ is called an evenly covered neighborhood of $x$. Note that each $\tilde{V}_{\lambda}$ is a component of $p^{-1}(V)$. When $\tilde{R}$ is simply connected, $p$ and $\tilde{R}$ are called a universal covering map and a universal covering surface, respectively.

Every Riemann surface has a universal covering surface. By the Koebe uniformization theorem a simply connected Riemann surface is conformally equivalent to one of the three Riemann surfaces: $\mathbb{D}$, $\mathbb{C}$ or $\hat{\mathbb{C}}$. A Riemann surface is called hyperbolic if it has $\mathbb{D}$ as a universal covering surface. Let $\Omega$ be a domain in $\hat{\mathbb{C}}$. Then $\Omega$ is a Riemann surface by definition and $\Omega$ is hyperbolic if and only if ${}^{\#}(\hat{\mathbb{C}}\backslash\Omega)\geq 3$, i.e. the complement of $\Omega$ has at least three points. When $\Omega$ is hyperbolic domain in $\mathbb{C}$, for any $z_{0}\in\mathbb{D}$, $w_{0}\in\Omega$ and $\alpha\in\mathbb{R}$ there exists the unique covering map $f:\mathbb{D}\rightarrow\Omega$ with $f(z_{0})=w_{0}$ and $\arg f^{\prime}(z_{0})=\alpha$. For details see [2], [29] or [3].

We say that an analytic function $f$ in a domain $D\subset\mathbb{C}$ is a covering if $f:D\rightarrow f(D)(\subset\mathbb{C})$ is a covering map. By definition if $f$ is univalent in a domain $D$, then $f$ is a covering.

Lemma 6.11**.**

Let $p:\tilde{R}\rightarrow R$ be an analytic covering map of a Riemann surface $\tilde{R}$ onto a Riemann surface $R$. Suppose $X$ is a simply connected Riemann surface and $h:X\rightarrow R$ is analytic. Then for any $x\in X$ and $\tilde{a}\in\tilde{R}$ with $p(\tilde{a})=h(x)$ there exists uniquely an analytic mapping $\tilde{h}:X\rightarrow\tilde{R}$ with $p\circ\tilde{h}=h$ and $\tilde{h}(x)=\tilde{a}$. Furthermore if $h$ is injective, the restriction $p|_{\tilde{h}(X)}:\tilde{h}(X)\rightarrow h(X)$ is an analytic bijection.

Proof.

The existence of $\tilde{h}$ follows from the simple connectivity of $X$ and the lifting lemma for covering spaces (see [19] Theorem 5.1 or [20] Lemma 79.1). Then since $p$ and $h$ are analytic and $p$ is a local homeomorpshim, $\tilde{h}$ is also analytic.

Assume that $h$ is injective. Then clearly $\tilde{h}$ is also injective and hence both $h:X\rightarrow h(X)$ and $\tilde{h}:X\rightarrow\tilde{h}(X)$ are bijective. Thus $p|_{\tilde{h}(X)}=h\circ\tilde{h}^{-1}:\tilde{h}(X)\rightarrow h(X)$ is a bijection. ∎

We have the following immediate implication.

Proposition 6.12**.**

Let $f\in\mathcal{H}_{0}(\mathbb{D})$ and let $g\in\mathcal{H}_{0}(\mathbb{D})$ be a universal covering. Then $f\prec g$ if and only if $f(\mathbb{D})\subset g(\mathbb{D})$.

Lemma 6.13**.**

Let $g:\mathbb{D}\rightarrow\mathbb{C}$ be analytic with $0,1\not\in g(\mathbb{D})$. Then

[TABLE]

on $\mathbb{D}$.

Here $\log^{+}y=\max\{\log y,0\}$ for $y>0$. For a proof see [2, Theorem 1-13].

Theorem 6.14**.**

Let $D$ be a hyperbolic domain in $\mathbb{C}$ and $\{f_{n}\}_{n=1}^{\infty}$ be a family of analytic functions in $D$ such that each $f_{n}$ is a covering. Suppose that $\{f_{n}\}_{n=1}^{\infty}$ converges to a nonconstant analytic function $f$ locally uniformly on $D$. Then $f$ is also a covering and $f_{n}(D)\rightarrow f(D)$ as $n\rightarrow\infty$ in the sense of kernel.

Proof.

Step 1. We show that if $a\in D$ and $V$ is a simply connected and bounded domain with $f(a)\in V\subset\overline{V}\subset\ker(f(a),\{f_{n}(D)\}_{n=1}^{\infty})$, then $V\subset f(D)$ and there exists a univalent analytic function $\varphi:V\rightarrow D$ satisfying $f(\varphi(w))\equiv w$ on $V$ and $\varphi(f(a))=a$.

Since $f_{n}(a)\rightarrow f(a)\in V$ and $\overline{V}$ is compact, by Proposition 6.9 there exists $N\in\mathbb{N}$ such that $f_{n}(a)\in V\subset\overline{V}\subset f_{n}(D)$ for all $n\geq N$.

For each $n\geq N$, applying Lemma 6.11 to covering map $f_{n}:D\rightarrow f_{n}(D)$ and the inclusion map $i_{V}:V\xhookrightarrow{}f_{n}(D)$ there exists a subdomain $\tilde{V}_{n}$ of $D$ and a conformal map $\varphi_{n}:V\rightarrow\tilde{V}_{n}$ such that $a\in\tilde{V}_{n}$ and the restriction $f_{n}|_{\tilde{V}}$ is a conformal map of $\tilde{V}_{n}$ onto $V$ with $\varphi_{n}=(f_{n}|_{\tilde{V}})^{-1}$. Notice that $\varphi_{n}(f_{n}(a))=a$.

Now we claim that the family $\{\varphi_{n}\}_{n=1}^{\infty}$ is locally uniformly bounded in $V$ and form a normal family. Indeed, since $D$ is hyperbolic we can take $\alpha,\beta\in\mathbb{C}\backslash D$ with $\alpha\not=\beta$. Let $h:\mathbb{D}\rightarrow V$ be a conformal map with $h(0)=f(a)(\in V)$ and $\zeta_{n}=h^{-1}(f_{n}(a))$, $n\in\mathbb{N}$. Then the function

[TABLE]

avoids [math] and $1$, and $H_{n}(0)=(a-\beta)/(\alpha-\beta)$. Thus by Lemma 6.13

[TABLE]

By replacing $\zeta$ with $\frac{\zeta-\zeta_{n}}{1-\overline{\zeta_{n}}\zeta}$ we have

[TABLE]

Combining this and $\zeta_{n}\rightarrow h^{-1}(f(a))=0$ it follows that $\{\varphi_{n}\}_{n=1}^{\infty}$ is locally uniformly bounded in $V$.

Let $\{\varphi_{n_{k}}\}_{k=1}^{\infty}$ be a subsequence with $\varphi_{n_{k}}\rightarrow\varphi$ locally uniformly in $V$. Then $\tilde{V}:=\varphi(V)\subset\overline{D}$. By the Hurwitz theorem and $\varphi_{n_{k}}(f_{n_{k}}(a))=a$, $\varphi$ is univalent in $V$ or $\varphi=a$. Hence $\varphi(V)\subset D$ and

[TABLE]

This implies $\varphi$ is univalent in $V$ and hence $\varphi:V\rightarrow\tilde{V}$ and $f|_{\tilde{V}}:\tilde{V}\rightarrow V$ are conformal and $\varphi=f|_{\tilde{V}}^{-1}$.

Step 2. We show $V\subset f(D)$ and $f_{n}(D)\rightarrow f(D)$ in the sense of kernel.

It follows from (6.1) that $V=f(\varphi(V))\subset f(D)$ and hence we have $\ker(f(a),\{f_{n}(D)\}_{n=1}^{\infty})\subset f(D)$. Since by Proposition 6.9 the reverse inclusion relation holds, we have $f(D)=\ker(f(a),\{f_{n}(D)\}_{n=1}^{\infty})$. The entire argument can be repeated for any subsequence $\{f_{n_{k}}\}_{k=1}^{\infty}$ and we obtain $f(D)=\ker(f(a),\{f_{n_{k}}(D)\}_{k=1}^{\infty})$. Hence $f_{n}(D)\rightarrow f(D)$ in the sense of kernel.

Step 3. For any $w_{0}\in f(D)$ let $V$ be a simply connected and bounded domain with $w_{0}\in V\subset\overline{V}\subset f(D)$. We show that $V$ is an evenly covered neighborhood of $w_{0}$ and $f$ is a covering.

For any $a\in f^{-1}(V)$, since $\overline{V}\subset f(D)=\ker(f(a),\{f_{n}(D)\}_{n=1}^{\infty})$, it follows from Step 1 that there exists a univalent function $\varphi:V\rightarrow D$ with $\varphi(f(a))=a$ and $f(\varphi(w))\equiv w$ on $V$. Let $U$ be the connected component of $f^{-1}(V)$ containing $a$. Then we claim $\tilde{V}:=\varphi(V)=U$.

Since $\tilde{V}$ is a domain with $a\in\tilde{V}\subset f^{-1}(V)$, we have $\tilde{V}\subset U$. To show the reverse inclusion relation suppose, on the contrary, that $U\backslash\tilde{V}\not=\emptyset$. Then there exists $a^{\prime}\in U\cap\partial\tilde{V}$. By $a^{\prime}\in U$ we have $f(a^{\prime})\in V$. Since $a^{\prime}\in\partial\tilde{V}$, this violates $f|_{\tilde{V}}:\tilde{V}\rightarrow V$ is univalent.

Now Let $f^{-1}(V)=\cup_{\lambda\in\Lambda}U_{\lambda}$ be the decomposition into connected components. For each lambda take $a(\lambda)\in U_{\lambda}$. Then as shown above, there exists a conformal map $\varphi_{\lambda}:V\rightarrow U_{\lambda}$ such that $f|_{U\lambda}:U_{\lambda}\rightarrow V$ is also conformal map and $f|_{U_{\lambda}}^{-1}=\varphi_{\lambda}$. Therefore $V$ is a evenly covered neighborhood of $w_{0}$ and hence $f:D\rightarrow f(D)$ is a covering map. ∎

Corollary 6.15**.**

Let $D$, $\{f_{n}\}_{n=1}^{\infty}$ and $f$ be as in Theorem 6.14, and let $a\in D$ and $V$ be a simply connected and bounded domain with $f(a)\in V\subset\overline{V}\subset D$. Then there exists $N\in\mathbb{N}$ such that $f_{n}(a)\in V\subset f_{n}(D)$ for $n\geq N$. Furthermore for $n\geq N$ let $\varphi_{n}=f_{n}^{-1}$ on $V$ with $\varphi_{n}(f_{n}(a))=a$. Then $\varphi_{n}\rightarrow\varphi$ locally uniformly on $V$, where $\varphi=f^{-1}$ on $V$ with $\varphi(f(a))=a$.

Proof.

Step 1 in the proof of the theorem shows that every subsequence of $\{\varphi_{n}\}_{n=1}^{\infty}$ has a further subsequence which converges to $\varphi=(f|_{\tilde{V}})^{-1}$ locally uniformly on $V$. Hence $\{\varphi_{n}\}_{n=1}^{\infty}$ converges to $\varphi$ locally uniformly on $V$. ∎

The following is not new, however for completeness we give a proof.

Theorem 6.16**.**

Let $\{\Omega_{n}\}_{n=1}^{\infty}$ be a sequence of hyperbolic domains in $\mathbb{C}$ which converges to a hyperbolic domain $\Omega$ in $\mathbb{C}$ in the sense of kernel. Let $f$ and $f_{n}$, $n\in\mathbb{N}$, be analytic universal covering maps of $\mathbb{D}$ onto $\Omega$ and $\Omega_{n}$, respectively. Let $\{a_{n}\}_{n=1}^{\infty}$ be a sequence in $\mathbb{D}$ which converges to $a\in\mathbb{D}$. Suppose that $f_{n}(a_{n})\rightarrow f(a)$ and $\arg f_{n}^{\prime}(a_{n})\rightarrow\arg f^{\prime}(a)$. Then $f_{n}\rightarrow f$ as $n\rightarrow\infty$ locally uniformly on $\mathbb{D}$.

Proof.

Since $\Omega$ is hyperbolic, we can take $\alpha,\beta\in\partial\Omega$ with $\alpha\not=\beta$. Since $\Omega_{n}\rightarrow\Omega$ in the sense of kernel, there exist sequences $\{\alpha_{n}\}_{n=1}^{\infty}$ and $\{\beta_{n}\}_{n=1}^{\infty}$ with $\alpha_{n},\beta_{n}\in\partial\Omega_{n}$ satisfying $\alpha_{n}\rightarrow\alpha$ and $\beta_{n}\rightarrow\beta$ as $n\rightarrow\infty$. We may assume that $\alpha_{n}\not=\beta_{n}$. Since $f_{n}$ avoid $\alpha_{n}$ and $\beta_{n}$ and $f_{n}(a_{n})\rightarrow f(a)$, similarly as in the proof of Theorem 6.14, we infer by Lemma 6.13 that $\{f_{n}\}_{n=1}^{\infty}$ forms a normal family. Thus any subsequence of $\{f_{n}\}_{n=1}^{\infty}$ has a further subsequence which converges to some analytic function $g$ locally uniformly on $\mathbb{D}$. By Theorem 6.14 $g$ is also a universal covering. Since $f_{n}(a_{n})\rightarrow f(a)$ and $\arg f_{n}^{\prime}(a_{n})\rightarrow\arg f^{\prime}(a)$, we have $g(a)=f(a)$ and $\arg g^{\prime}(a)=\arg f^{\prime}(a)$. From the uniqueness theorem for universal covering maps it follows that $g=f$. Therefore the original sequence $\{f_{n}\}_{n=1}^{\infty}$ converges to $f$ locally uniformly on $\mathbb{D}$. ∎

Definition 6.17**.**

Let $I\subset[-\infty,\infty]$ be an interval and $\{\Omega_{t}\}_{t\in I}$ be a family of domains in $\hat{\mathbb{C}}$. Then we say that $\{\Omega_{t}\}_{t\in I}$ is continuous at $t_{0}\in I$ (in the sense of kernel) if

• (a’)

for any compact subset $K$ of $\Omega$ there exists $\delta>0$ such that $K\subset\Omega_{t}$ for all $t\in I$ with $0<|t-t_{0}|<\delta$

and

• (b’)

for every $c\in\partial\Omega$, $\operatorname{dist}(c,\partial\Omega_{t})\rightarrow 0$ as $I\backslash\{t_{0}\}\ni t\rightarrow t_{0}$.

If $\{\Omega_{t}\}_{t\in I}$ is continuous at every $t_{0}\in I$, then $\{\Omega_{t}\}_{t\in I}$ is simply called continuous (in the sense of kernel).

It is easy to see that $\{\Omega_{t}\}_{t\in I}$ is continuous at $t_{0}\in I$ if and only if $\Omega_{t_{n}}\rightarrow\Omega$ as $n\rightarrow\infty$ in the sense of kernel for all sequences $\{t_{n}\}_{n=1}^{\infty}\subset I$ with $t_{0}\not=t_{n}\rightarrow t_{0}$.

Here we summarize results concerning relations between a $1$-parameter family of hyperbolic domains and the corresponding family of universal covering maps. The following theorem, containing Theorem 1.5, directly follows from Proposition 6.12, Theorems 6.14 and 6.16.

Theorem 6.18**.**

Let $\{\Omega_{t}\}_{t\in I}$ be a family of hyperbolic domains in $\mathbb{C}$ with $0\in\Omega_{t}$, $t\in I$. For each $t$ let $f_{t}:\mathbb{D}\rightarrow\Omega_{t}$ be the universal covering map with $f_{t}(0)=0$ and $f_{t}^{\prime}(0)>0$. Then

• (i)

$\{f_{t}\}_{t\in I}$* is a Loewner chain if and only if $\{\Omega_{t}\}_{t\in I}$ is nondecreasing, i.e., $\Omega_{s}\subset\Omega_{t}$ whenever $s,t\in I$ with $s\leq t$,*

• (ii)

$\{f_{t}\}_{t\in I}$* is a strictly increasing Loewner chain if and only if $\{\Omega_{t}\}_{t\in I}$ is strictly increasing, i.e., $\Omega_{s}\subsetneq\Omega_{t}$ whenever $s,t\in I$ with $s<t$,*

• (iii)

$\{f_{t}\}_{t\in I}$* is continuous if and only if $\{\Omega_{t}\}_{t\in I}$ is continuous in the sense of kernel.*

7. Kernel convergence and connectivity of domains

In this section we first pay our attention to families of domains in $\hat{\mathbb{C}}$ and state several results. Notice, unless otherwise stated, that these have counterparts for families of universal covering maps.

The connectivity of a domain $\Omega$ in $\mathbb{C}$ (or in $\hat{\mathbb{C}}$) is the number of connected components of $\hat{\mathbb{C}}\backslash\Omega$. We denote it by $C(\Omega)$. Then $C(\Omega)$ is lower semicontinuous with respect to kernel convergence.

Theorem 7.1**.**

Let $\{\Omega_{n}\}_{n=1}^{\infty}$ be a sequence of domains in $\hat{\mathbb{C}}$. Suppose $\{\Omega_{n}\}_{n=1}^{\infty}$ converges to a domain $\Omega$ in $\hat{\mathbb{C}}$ in the sense of kernel. Then

[TABLE]

Proof.

When $C(\Omega)=0$, the inequality (7.1) trivially holds. If $C(\Omega)=1$, then there exists $c\in\partial\Omega$ and hence by Definition 6.3, $d^{*}(c,\partial\Omega_{n})\rightarrow 0$ as $n\rightarrow\infty$. This implies $C(\Omega_{n})\geq 1$ for all sufficiently large $n$. and (7.1) holds.

Assume $2\leq k:=C(\Omega)<\infty$. Let $E_{1},\ldots,E_{k}$ be the component of $\hat{\mathbb{C}}\backslash\Omega$. For each $j=1,\ldots,k$, by Lemma 1.9 there exists a simple closed curve $\alpha_{j}$ which separates $E_{j}$ and $\bigcup_{\ell\not=j}E_{\ell}$ and let $D_{j}$ be the component of $\hat{\mathbb{C}}\backslash\alpha_{j}$ which contains $E_{j}$. Take $\zeta_{j}\in\partial E_{j}(\subset\partial\Omega\cap D_{j})$ arbitrarily and let $\delta_{j}=d^{*}(\zeta_{j},\alpha_{j})>0$. Then by Definition 6.3 there exists $N\in\mathbb{N}$ such that

[TABLE]

Thus for each $j=1,\ldots,k$ and $n\geq N$ there exists $\zeta_{j,n}\in\partial\Omega_{n}\cap D_{j}$ and hence there exists the unique component $E_{j,n}$ of $\hat{\mathbb{C}}\backslash\Omega_{n}$ containing $\zeta_{j,n}$. Since $\alpha_{j}$ separates $E_{j,n}$ and $\bigcup_{\ell\not=j}E_{\ell,n}$, $\hat{\mathbb{C}}\backslash\Omega_{n}$ has at least $k$ components. Therefore $C(\Omega_{n})\geq k$ for $n\geq N$.

Finally we assume $C(\Omega)=\infty$. For any $k\in\mathbb{N}$ take components $E_{1},\ldots,E_{k}$ of $\hat{\mathbb{C}}\backslash\Omega$ arbitrarily. Then by repeating the same argument we have $\liminf_{n\rightarrow\infty}C(\Omega_{n})\geq k$ for all $k\in\mathbb{N}$, and hence $\liminf_{n\rightarrow\infty}C(\Omega_{n})=\infty$. ∎

Example 6.4 shows one can not replace the inequality sign in (7.1) with equality.

Let $\{\Omega_{t}\}_{0\leq t\leq\infty}$ with $E_{t}=\hat{\mathbb{C}}\backslash\Omega_{t}$, $t\in I$ as in Example 1.6. Then the corresponding family of universal covering maps $\{f_{t}\}_{0\leq t\leq\infty}$ with $f_{t}(0)=0$ and $f_{t}^{\prime}(0)>0$ is a strictly increasing and continuous Loewner chain. Let $t_{0}\in I$ and $C_{0}$ be a component of $E_{t_{0}}$. Since $\{E_{t}\}_{t\in I}$ is nonincreasing with time $t$, so is $\{C_{0}\cap E_{t}\}_{t\in I}$. Intuitively the family $\{C_{0}\cap E_{t}\}_{t\in I}$ shrinks and splits into many pieces with time $t$, but it never vanishes. This holds in general. To prove this we show first a weaker result.

Proposition 7.2**.**

Let $\{\Omega_{t}\}_{t\in I}$ be a nondecreasing and continuous family of domains in $\hat{\mathbb{C}}$ with $E_{t}=\hat{\mathbb{C}}\backslash\Omega_{t}$, $t\in I$. Let $t_{0}\in I$, $F$ be a closed subset of $E_{t_{0}}$ and $V$ a domain in $\hat{\mathbb{C}}$ with $\emptyset\not=F=E_{t_{0}}\cap V$. Then for any $t\in I$

[TABLE]

Proof.

It suffices to show (7.2) for $t\in I\cap(t_{0},\infty]$. Suppose, on the contrary, that $F\cap E_{t}=\emptyset$ for some $t\in I\cap(t_{0},\infty]$. Then since $\{E_{t}\}_{t\in I}$ is nonincreasing with $t$, there exists $t_{1}\in I\cap(t_{0},\infty)$ such that

[TABLE]

Assume $F\cap E_{t_{1}}=\emptyset$. Then since $F$ is compact and $F\subset\Omega_{t_{1}}$, there exists $\delta>0$ such that $F\subset\Omega_{t}=\hat{\mathbb{C}}\backslash E_{t}$ for $|t-t_{1}|<\delta$ with $t\in I$. This contradicts (7.3).

Next we assume $F\cap E_{t_{1}}\not=\emptyset$. Then since $F\subset V$, we have

[TABLE]

Notice that $\Omega_{t_{0}}\cap V\not=\emptyset$, since otherwise $V\subset E_{t_{0}}$ and $F=E_{t_{0}}\cap V=V$, and hence $F(\not=\emptyset,\hat{\mathbb{C}})$ is open and closed, which is a contradiction. Particularly we have

[TABLE]

Therefore, since $V$ is connected, we have $V\cap\partial\Omega_{t_{1}}\not=\emptyset$. Take $c\in V\cap\partial\Omega_{t_{1}}$. Then

[TABLE]

While for $t\in I\cap(t_{1},\infty)$ we have $F\subset\Omega_{t}$ and

[TABLE]

Thus we have $V\subset\Omega_{t}$ for $t\in I\cap(t_{1},\infty)$, which contradict $c\in V$ and (7.5). ∎

Theorem 7.3**.**

Let $I\subset[-\infty,\infty]$ be an interval and $\{\Omega_{t}\}_{t\in I}$ be a nondecreasing and continuous family of domains in $\hat{\mathbb{C}}$ with $E_{t}=\hat{\mathbb{C}}\backslash\Omega_{t}$. Let $t_{0}\in I$ and $C_{0}$ be a connected component of $E_{t_{0}}$. Then

[TABLE]

Particularly $C(\Omega_{t})$ is nondecreasing and left continuous on $I$.

Proof.

Let $E_{t_{0}}=C_{0}\bigcup_{\lambda\in\Lambda}C_{\lambda}$ be the decomposition of $E_{t_{0}}$ into connected components. Fro each $\lambda\in\Lambda$ by Lemma 1.9 there exists a simple closed curve $\alpha_{\lambda}$ in $\Omega_{t_{0}}$ which separates $C_{0}$ and $C_{\lambda}$. Let $V_{\lambda}$ be the component of $\hat{\mathbb{C}}\backslash\alpha_{\lambda}$ containing $C_{0}$ and let $F_{\lambda}=V_{\lambda}\cap E_{t_{0}}$. Then since $C_{0}\subset F_{\lambda}$ and $\partial V_{\lambda}\cap E_{t_{0}}=\alpha_{\lambda}\cap(\hat{\mathbb{C}}\backslash\Omega_{t_{0}})=\emptyset$, we have $F_{\lambda}=\overline{V_{\lambda}}\cap E_{t_{0}}$ and $F$ is a nonempty closed set. Therefore by Proposition 7.2 $F_{\lambda}\cap E_{t}\not=\emptyset$ for all $t\in I$.

Claim. For any fixed $t\in I$ the family $\{F_{\lambda}\cap E_{t}\}_{\lambda\in\Lambda}$ of compact subsets in $\hat{\mathbb{C}}$ has the finite intersection property, i.e., for any $\lambda_{1},\ldots,\lambda_{n}\in\Lambda$

[TABLE]

We show Claim. Let $V$ bet the component of the open set $V_{\lambda_{1}}\cap\cdots\cap V_{\lambda_{n}}$ containing $C_{0}$ and $F=V\cap E_{t_{0}}$. Then since $C_{0}\subset F$ and

[TABLE]

$F$ is nonempty and closed. Therefore by Proposition 7.2 again, we have $F\cap E_{t}\not=\emptyset$. The claim easily follows from this and

[TABLE]

Since a family of compact sets having the finite intersection property has nonempty intersection, we have by the claim

[TABLE]

for all $t\in I$. While for any $\lambda\in\Lambda$ we have

[TABLE]

It follows that $\bigcap_{\lambda\in\Lambda}F_{\lambda}=C_{0}$ and hence

[TABLE]

Since $\{E_{t}\}_{t\in I}$ is nonincreasing with $t\in I$, the family $\{C_{0}\cap E_{t}\}_{t\in I}$ of compact sets also has the finite intersection property. Therefore $\bigcap_{t\in I}(C_{0}\cap E_{t})=C_{0}\cap\bigcap_{t\in I}E_{t}\not=\emptyset$ as required.

Finally we show $C(\Omega_{t})$ is nondecreasing in $t\in I$. Notice that the left continuity follows from this and the lower semicontinuity of $C(\Omega_{t})$ (see Theorem 7.1). Let $t_{0}\in I$. If $C(\Omega_{t_{0}})=0$, then $C(\Omega_{t})\geq C(\Omega_{t_{0}})$ trivially holds for all $t\geq t$. Assume that $C(\Omega_{t_{0}})\geq 1$ and $k\in\mathbb{N}$ satisfies $k\leq C(\Omega_{t})$. Let $C_{1},\ldots,C_{k}$ be disjoint components of $E_{t_{0}}$. Then as shown above there exists $\zeta_{j}\in C_{j}\cap\bigcap_{t\in I}E_{t}$, $j=1,\ldots,k$. For $j=1,\ldots,k$ and $t>t_{0}$ let $C_{j}(t)$ be the component of $E_{t}$ containing $\zeta_{j}$. Then since $C_{j}(t)\subset C_{j}$, $j=1,\ldots,k$, $C_{1}(t),\ldots,C_{k}(t)$ are disjoint. Therefore $C(\Omega_{t})\geq k$ and hence $C(\Omega_{t})\geq C(\Omega_{t_{0}})$ for $t>t_{0}$. ∎

Definition 7.4**.**

Let $\Omega$ be a domain in $\hat{\mathbb{C}}$. We say that $\Omega$ is maximal in the sense of kernel if there is no nondecreasing and continuous family $\{\Omega\}_{0\leq t\leq\varepsilon}$ of domains in $\hat{\mathbb{C}}$ with $\varepsilon>0$ and $\Omega=\Omega_{0}\subsetneq\Omega_{\varepsilon}$.

Assume that $\Omega$ is a hyperbolic domain $\mathbb{C}$ and $0\in\Omega$. Let $f:\mathbb{D}\rightarrow\Omega$ be the unique universal covering map with $f(0)=0$ and $f^{\prime}(0)>0$. Then $\Omega$ is maximal in the sense of kernel if and only if there is no continuous Loewner chain $\{f_{t}\}_{0\leq t\leq\varepsilon}$ of universal covering maps with $f_{0}=f$ and $f_{\varepsilon}^{\prime}(0)>f^{\prime}(0)$.

Corollary 7.5**.**

Let $\Omega$ be a domain in $\hat{\mathbb{C}}$ with $E=\hat{\mathbb{C}}\backslash\Omega$. If $E$ is totally disconnected, then $\Omega$ is maximal in the sense of kernel.

Proof.

Suppose, on the contrary, there exists a nondecreasing and continuous family $\{\Omega_{t}\}_{0\leq t\leq\varepsilon}$ of domains in $\hat{\mathbb{C}}$. Since $E$ is totally disconnected, for any $w\in E$ the component of $E$ containing $w$ coincides with $\{w\}$. Thus by Theorem 7.3 $w\in E_{\varepsilon}:=\hat{\mathbb{C}}\backslash\Omega_{\varepsilon}$ and hence $E\subset E_{\varepsilon}$. This contradicts $\Omega=\Omega_{0}\subsetneq\Omega_{\varepsilon}$. ∎

From the corollary it follows that the $\Omega_{\infty}$ in Example 1.6 is maximal. We have an example of maximal domain whose complement is not totally disconnected.

Example 7.6**.**

For $n\in\mathbb{N}$ and $k=0,\ldots,n$ let

[TABLE]

and

[TABLE]

Then $\Omega:=\hat{\mathbb{C}}\backslash E$ is maximal.

Proof.

Suppose, on the Contrary, $\{\Omega_{t}\}_{0\leq t\leq\varepsilon}$ is a nonshrinking and continuous family of domains in $\hat{\mathbb{C}}$ for some $\varepsilon>0$ with $\Omega_{0}=\Omega\subsetneq\Omega_{\varepsilon}$. Let $E_{t}=\hat{\mathbb{C}}\backslash\Omega_{t}$, $0\leq t\leq\varepsilon$. Since for each $n\in\mathbb{N}$ and $k=0,\ldots,n$, $\{a_{n,k}\}$ is a component of $E$, by Theorem 7.3 we have $a_{n,k}\in E_{\varepsilon}$. Since $E_{\varepsilon}$ is closed, this implies $[1,2]\subset E_{\varepsilon}$. We also have $\infty\in E_{\varepsilon}$. Therefore $E\subset E_{\varepsilon}$, which contradicts $E_{\varepsilon}\subsetneq E$. ∎

Definition 7.7**.**

Let $D_{0}$ and $D_{1}$ be domains in $\hat{\mathbb{C}}$ with $D_{0}\subset D_{1}\subsetneq\hat{\mathbb{C}}$. We say that $D_{0}$ is continuously connected with $D_{1}$ if there is a nondecreasing and continuous family $\{\Omega_{t}\}_{0\leq t\leq 1}$ of domains in $\hat{\mathbb{C}}$ with $\Omega_{0}=D_{0}$ and $\Omega_{1}=D_{1}$.

When $0\in D_{0}$ and $D_{1}$ is hyperbolic, $D_{0}$ is continuously connected with $D_{1}$ if and only if there exists a continuous Loewner chain of universal covering maps which connects $f_{0}$ with $f_{1}$, where $f_{j}\in\mathcal{H}_{0}(\mathbb{D})$ is the unique universal covering map of $\mathbb{D}$ onto $D_{j}$, $j=1,2$.

In the case that $C(D_{1})$ is finite we shall give a necessary and sufficient condition for the existence of a nondecreasing and continuous family of domains connecting $D_{0}$ with $D_{1}$. To show this we need an elementary topological lemma and a weaker result when $D_{0}$ and $D_{1}$ are simply connected.

Lemma 7.8**.**

Let $\Omega$ be a domain in $\hat{\mathbb{C}}$ and $C$ be a connected component of the complement $\hat{\mathbb{C}}\backslash\Omega$. Then $\hat{\mathbb{C}}\backslash C$ is connected. In particular $\hat{\mathbb{C}}\backslash C$ is a simply connected domain.

Proof.

Since $\Omega\subset\hat{\mathbb{C}}\backslash C$, there is the unique component $\tilde{\Omega}$ of $\hat{\mathbb{C}}\backslash C$ containing $\Omega$. Assume that $D$ is another component of $\hat{\mathbb{C}}\backslash C$. Then since $D\cap\Omega\subset D\cap\tilde{\Omega}=\emptyset$ and $\partial D\subset C$, $C\cup\overline{D}$ is connected and $C\subsetneq C\cup\overline{D}\subset\hat{\mathbb{C}}\backslash\Omega$. This contradicts the maximality of the component $C$ of $\hat{\mathbb{C}}\backslash\Omega$. Thus $\hat{\mathbb{C}}\backslash C=\tilde{\Omega}$ and $\hat{\mathbb{C}}\backslash C=\tilde{\Omega}$ is a domain. Furthermore since $\hat{\mathbb{C}}\backslash\tilde{\Omega}=C$ is connected, $\hat{\mathbb{C}}\backslash C$ is simply connected. ∎

We say that a set $E\subset\hat{\mathbb{C}}$ is a continuum if $E$ is a nonempty compact and connected subset of $\hat{\mathbb{C}}$. A continuum is said to be nondegenerate if it contains at least two points, and to be degenerate if it consist of a single point.

Proposition 7.9**.**

Let $D_{0}$ and $D_{1}$ be simply connected domains in $\hat{\mathbb{C}}$ with $D_{0}\subset D_{1}\subsetneq\hat{\mathbb{C}}$ . Then $D_{0}$ is continuously connected with $D_{1}$

Proof.

We may assume $0\in D_{0}\subsetneq D_{1}\subset\mathbb{C}$ after a linear fractional transformation, if necessary. Assume that $\hat{\mathbb{C}}\backslash D_{1}$ is a nondegenerate continuum. Let $g_{j}\in\mathcal{H}_{0}(\mathbb{D})$ be the unique conformal mapping of $\mathbb{D}$ onto $D_{j}$, $j=0,1$. By Proposition 5.4 there exists a continuous Loewner chain $\{f_{t}\}_{0\leq t\leq 1}$ of univalent functions with $f_{j}=g_{j}$, $j=0,1$. Let $D_{t}=f_{t}(\mathbb{D})$ for $0<t<1$. Then $\{D_{t}\}_{0\leq t\leq 1}$ is a nondecreasing and continuous family of domains in $\mathbb{C}$ connecting $D_{0}$ with $D_{1}$.

Assume $\hat{\mathbb{C}}\backslash D_{1}=\{\infty\}$, i.e., $D_{1}=\mathbb{C}$. Since $D_{0}\subsetneq D_{1}=\mathbb{C}$ and $D_{0}$ is simply connected, there is the unique conformal map $g_{0}\in\mathcal{H}_{0}(\mathbb{D})$ of $\mathbb{D}$ onto $D_{0}$. By Theorem 6.1 in [25] there exists a normalized Loewner chain $\{f_{s}\}_{\alpha\leq s<\infty}$ with $\alpha=\log g_{0}^{\prime}(0)$ and $g_{0}=f_{\alpha}$. Let $D_{t}=f_{\alpha+t(1-t)^{-1}}(\mathbb{D})$ for $0<t<1$. Then $\{D_{t}\}_{0\leq t\leq 1}$ is a nondecreasing and continuous family of domains in $\mathbb{C}$ connecting $D_{0}$ with $D_{1}=\mathbb{C}$. ∎

Theorem 7.10**.**

Let $D_{0}$ and $D_{1}$ be domains in $\hat{\mathbb{C}}$ with $D_{0}\subset D_{1}\subsetneq\hat{\mathbb{C}}$ and $C(D_{1})<\infty$. Then $D_{0}$ is continuously connected with $D_{1}$ if and only if for every component $C$ of $\hat{\mathbb{C}}\backslash D_{0}$, there are components $C^{\prime}$ of $\hat{\mathbb{C}}\backslash D_{1}$ with $C^{\prime}\subset C$.

Proof.

The necessity easily follows from Theorem 7.3.

To show sufficiency we may assume $0\in D_{0}\subsetneq D_{1}\subset\mathbb{C}$ after a linear fractional transformation, if necessary. Let $E_{i}=\hat{\mathbb{C}}\backslash D_{i}$, $i=1,2$. Then since $E_{1}\subset E_{0}$, for each component $C^{\prime}$ of $E_{1}$, there exists a unique component $C$ of $E_{0}$ satisfying $C^{\prime}\subset C$. By the assumption of the theorem $E_{0}$ and $E_{1}$ can be decomposed into connected components as follows.

[TABLE]

with $\bigcup_{k=1}^{p_{j}}C_{j,k}^{\prime}\subset C_{j}$, $j=1,\ldots,n$, where $n=C(D_{0})$ and $p_{j}\in\mathbb{N}$, $j=1,\ldots,n$.

For each $j=1,\ldots,n$ and $k=1,\ldots,p_{j}$ let

[TABLE]

We notice that $\Omega_{j}$ and $\tilde{\Omega}_{j,k}$ are simply connected domains with $\Omega_{j}\subset\tilde{\Omega}_{j,k}$. For each $j=1,\ldots,n$ and $k=1,\ldots,p_{j}$ by Proposition 7.9 we can take a nondecreasing and continuous family $\{\Omega_{t}^{j,k}\}_{0\leq t\leq 1}$ of domains with $\Omega_{0}^{j,k}=\Omega_{j}$ and $\Omega_{1}^{j,k}=\Omega_{j,k}^{\prime}$.

For $t\in[0,1]$ let $\Omega_{t}$ be the connected component of

[TABLE]

which contains [math]. By repeated use of Theorem 6.7, $\{\Omega_{t}\}_{0\leq t\leq 1}$ is a nondecreasing and continuous family of domains. Since

[TABLE]

we have $\Omega_{0}=D_{0}$ and $\Omega_{1}=D_{1}$, as required. ∎

Now we shall give a structure theorem for Loewner chains of universal covering maps.

Theorem 7.11**.**

Let $\{f_{t}\}_{t\in I}$ be a continuous Loewner chain of universal covering maps with $\beta=\sup I\not\in I$ and $\Omega_{t}=f_{t}(\mathbb{D})$, $t\in I$. Let $a(\beta)=\lim_{t\uparrow\beta}a(t)\in(0,\infty]$ and $\Omega_{\beta}=\bigcup_{t\in I}\Omega_{t}$. Then $a(\beta)<\infty$ if and only if

• (i)

$\Omega_{\beta}$* is a hyperbolic domain and $f_{t}\rightarrow f_{\beta}$ locally uniformly as $t\uparrow\beta$, where $f_{\beta}$ is the unique universal covering map of $\mathbb{D}$ onto $\Omega_{\beta}$ with $f_{\beta}(0)=0$ and $f_{\beta}^{\prime}(0)>0$. In this case $\{f_{t}\}_{t\in I\cup\{\beta\}}$ is also a continuous Loewner chain of universal covering maps.*

Furthermore $a(\beta)=\infty$ if and only if one of the following conditions holds.

• (ii)

The Loewner chain $\{f_{t}\}_{t\in I}$ consists of univalent functions and $\Omega_{\beta}=\mathbb{C}$.

• (iii)

There exists $w_{0}\in\mathbb{C}\backslash\{0\}$ and $t_{0}\in I$ such that with $\Omega_{\beta}=\mathbb{C}\backslash\{w_{0}\}$ and $C(\Omega_{t})\equiv 2$ for $t_{0}<t<\beta$. In this case we have

[TABLE]

where $C_{t}^{0}$ and $C_{t}^{\infty}$ are the components of $\hat{\mathbb{C}}\backslash\Omega_{t}$ containing $w_{0}$ and $\infty$, respectively.

Proof.

First we notice that by Theorem 6.6, $\Omega_{t}\rightarrow\Omega_{\beta}$ as $t\uparrow\beta$ in the sense of kernel. Let $E_{t}=\hat{\mathbb{C}}\backslash\Omega_{t}$ for $t\in I$ and $E_{\beta}=\hat{\mathbb{C}}\backslash\Omega_{\beta}$. Then $\infty\in E_{\beta}\subset E_{t}$, $t\in I$.

Assume that $E_{\beta}$ contains at least two distinct points other than $\infty$. Then $\Omega_{\beta}$ is hyperbolic. Therefore by Theorem 6.16, $f_{t}\rightarrow f_{\beta}$ as $t\uparrow\beta$ locally uniformly on $\mathbb{D}$ and $f_{\beta}$ is also a universal covering map. Furthermore by Theorem 2.10, the locally uniform limit $\omega_{s,\beta}=\lim_{t\uparrow\beta}\omega_{s,t}\in\mathfrak{B}$ exists. Then we have $f_{s}=\lim f_{t}\circ\omega_{s,t}=f_{\beta}\circ\omega_{s,\beta}$ for $s\in I$ and hence $\{f_{t}\}_{t\in I\cup\{\beta\}}$ is a Loewner family of universal covering maps.

Next assume that $E_{\beta}=\{\infty\}$. Then since by Theorem 7.3, the function $C(\Omega_{t})$, $t\in I\cup\{\beta\}$, is nondecreasing, $C(\Omega_{t})\equiv 1$. Therefore each $f_{t}$ is univalent and $a(\beta)=\infty$ follows from Proposition 2.13.

Finally assume that there exists $w_{0}\in\mathbb{C}\backslash\{0\}$ with $E_{\beta}=\{w_{0},\infty\}$. Then as before, since the function $C(\Omega_{t})$ is left continuous and nondecreasing, there exists $t_{0}\in I$ such that $C(\Omega_{t})\equiv 2(=C(\Omega_{\beta}))$ on $[t_{0},\beta)$. Therefore for $t_{0}\leq t<\beta$, $E_{t}$ decomposed into two components $C_{t}^{0}$ and $C_{t}^{\infty}$ containing $w_{0}$ and $\infty$, respectively. It is easy to see that (7.9) holds and $\rho_{t}=\max_{w\in C_{t}^{0}}|w-w_{0}|\rightarrow 0$ as $t\uparrow\beta$.

We show $a(\beta)=\infty$. Without loss of generality we may assume $w_{0}=-c$, $c>0$. For $R>1$ let

[TABLE]

and $g_{R}:\mathbb{D}\rightarrow A_{R}$ be the universal covering map with $g_{R}(0)=0$ and $g_{R}^{\prime}(0)>0$. Then

[TABLE]

Since $\overline{A_{R}}\subset\Omega_{\beta}$ and $\Omega_{t}\rightarrow\Omega_{\beta}$ as $t\uparrow\beta$, there exists $t_{0}\in I$ such that $\overline{A_{R}}\subset\Omega_{t}$ for $t_{0}<t<\beta$. Thus

[TABLE]

and $a(\beta)=\lim_{t\uparrow\infty}f_{t}^{\prime}(0)=\infty$. ∎

Example 7.12**.**

Let $a>0$ and

[TABLE]

Then $f_{a}(\cdot,t)$ is the unique universal covering map of $\mathbb{D}$ onto $\mathbb{C}\backslash\overline{\mathbb{D}}(-a,ae^{-\frac{t}{2a}})$ with $f_{a}(0,t)=0$ and $f_{a}^{\prime}(0,t)=t$. Since the all Maclaurin coefficients of $f_{a}(\cdot,t)$ are positive, it is not difficult to see that for fixed $t>0$ and $r\in(0,1)$

[TABLE]

Therefore a family of normalized Loewner chains $\{f_{a}(z,e^{t})\}_{a>0}$ is not uniformly bounded on $\overline{\mathbb{D}}(0,r)\times[-T,T]$ for any fixed $r\in(0,1)$ and $T>0$.

Let $\{f_{t}\}_{t\in I}$ be a Loewner chain such that every $\Omega_{t}=f_{t}(\mathbb{D})$, $t\in I$, is hyperbolic. For each $t\in I$ let $\tilde{f}_{t}$ be the unique universal covering map of $\mathbb{D}$ onto $\Omega_{t}$ with $\tilde{f}_{t}(0)=0$ and $\tilde{f}_{t}^{\prime}(0)>0$. Then $\{\tilde{f}_{t}\}_{t\in I}$ is a Loewner family of universal covering maps with $f_{t}\prec\tilde{f}_{t}$, i.e., there exist the lift $\tilde{\omega}_{t}:\mathbb{D}\rightarrow\mathbb{D}$ with $f_{t}=\tilde{f}_{t}\circ\tilde{\omega}_{t}$ and $\tilde{\omega}_{t}(0)=0$, $t\in I$. Since $\{\Omega_{t}\}_{t\in I}$ is nondecreasing, $\{\tilde{f}_{t}\}_{t\in I}$ is a Loewner family of universal covering maps. For $(s,t)\in I_{+}^{2}$ let $\tilde{\omega}_{s,t}:\mathbb{D}\rightarrow\mathbb{D}$ be the lift with $\tilde{f}_{s}=\tilde{f}_{t}\circ\tilde{\omega}_{s,t}$ and $\tilde{\omega}_{s,t}(0)=0$. Then we have the following commutative diagram.

\mathbb{D}$$\mathbb{D}$$\mathbb{D}$$\mathbb{D}$$\Omega_{s}$$\Omega_{t}$$\tilde{\omega}_{s,t}$$\tilde{\omega}_{s}$$\tilde{\omega}_{t}$$\omega_{s,t}$$\tilde{f}_{s}$$\tilde{f}_{t}$$f_{s}$$f_{t}$$\operatorname{inc}

It is clear that $\{\tilde{f}_{t}\}_{t\in I}$ is expanding if and only if so is $\{f_{t}\}_{t\in I}$. However notice that continuity and strict increasing property of $\{f_{t}\}_{t\in I}$ are not inherited by $\{\tilde{f}_{t}\}_{t\in I}$. By modifying Example 6.10 one can easily obtain counter examples.

8. Loewner theory on Fuchsian Groups

In §6 (see Proposition 6.12) we see that if $f,g\in\mathcal{H}_{0}(\mathbb{D})$ and $g$ is a covering with $f(\mathbb{D})\subset g(\mathbb{D})$, then $f\prec g$. That is, there exists $\omega\in\mathfrak{B}$ with $f=g\circ\omega$. For later applications, let us recall a construction of $\omega$. For notations and terminology see [19] or [20] for examples.

For $z\in\mathbb{D}$ let $\alpha:[0,1]\rightarrow\mathbb{D}$ be a path in $\mathbb{D}$ from [math] to $z$, i.e., $\alpha$ is a continuous map with $\alpha(0)=0$ and $\alpha(1)=z$. Then $f\circ\alpha$ is a path in $f(\mathbb{D})(\subset g(\mathbb{D}))$ from $f(0)(=g(0))$ to $f(z)$. Then there exists a unique path $\tilde{\alpha}:[0,1]\rightarrow\mathbb{D}$ called the lift of $f\circ\alpha$ with respect to the covering map $g:\mathbb{D}\rightarrow g(\mathbb{D})$ such that $\tilde{\alpha}(0)=0$ and $g\circ\tilde{\alpha}=f\circ\alpha$. If we choose another path $\alpha^{\prime}:[0,1]\rightarrow\mathbb{D}$ in $\mathbb{D}$ from [math] to $z$ and obtain the lifted path $\tilde{\alpha}^{\prime}$ in $\mathbb{D}$ as above, then since $\mathbb{D}$ is simply connected, $\alpha^{\prime}$ is path homotopic to $\alpha$ and hence $\tilde{\alpha}^{\prime}$ is path homotopic to $\tilde{\alpha}^{\prime}$. Particularly the end point $\tilde{\alpha}^{\prime}(1)$ coincides with $\tilde{\alpha}(1)$. Therefore the end point $\tilde{\alpha}(1)$ does not depend on the choice of $\alpha$ and only on $z$, and we can define $\omega(z)=\tilde{\alpha}(1)\in\mathbb{D}$. By $g\circ\tilde{\alpha}(1)=g\circ\alpha(1)$ we have $g(\omega(z))=f(z)$. And $\omega(0)=0$ clearly follows from $f(0)=g(0)$. It is easy to see that $\omega$ is analytic.

The following theorem is a direct consequence of Theorems 2.6 and 2.7. However we give a completely different and topological proof.

Theorem 8.1**.**

Let $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$ be an associated transition family of a continuous Loewner family $\{f_{t}\}_{t\in I}$ of universal covering maps. Then $\omega_{s,t}$ is univalent in $\mathbb{D}$ for every $(s,t)\in I_{+}^{2}$.

Proof.

Suppose, on the contrary, that $\omega_{t_{0},s}$ is not univalent for some $t_{0},s\in I$ with $t_{0}<s$. Then since $\omega_{t_{0},t}(z)$ is continuous in $t\in I\cap[t_{0},\infty)$ for each fixed $z\in\mathbb{D}$ and $\omega_{t_{0},t_{0}}$ coincides with the identity mapping of $\mathbb{D}$, there exist distinct $z_{1},z_{2}\in\mathbb{D}$ and $t_{1}\in I\cap(t_{0},\infty)$ such that

[TABLE]

Let $\alpha,\beta:[0,1]\rightarrow\mathbb{D}$ be paths from [math] to $z_{1}$ and $z_{2}$, respectively. And let $\tilde{\alpha},\tilde{\beta}:[0,1]\rightarrow\mathbb{D}$ be the paths with $\tilde{\alpha}(0)=0$ and $\tilde{\beta}(0)=0$ such that $f_{t_{1}}\circ\tilde{\alpha}=f_{t_{0}}\circ\alpha$ and $f_{t_{1}}\circ\tilde{\beta}=f_{t_{0}}\circ\beta$. Then $\tilde{\alpha}(1)=\omega_{t_{0},t_{1}}(z_{1})=\omega_{t_{0},t_{1}}(z_{2})=\tilde{\beta}(1)$. This implies $f_{t_{0}}\circ\alpha(1)=f_{t_{1}}\circ\tilde{\alpha}(1)=f_{t_{1}}\circ\tilde{\beta}(1)=f_{t_{0}}\circ\beta(1)$. Since $\mathbb{D}$ is simply connected, there exists a path homotopy $F:[0,1]\times[0,1]\rightarrow\mathbb{D}$ between $\tilde{\alpha}$ and $\tilde{\beta}$, i.e., $F$ is a continuous map and satisfies

[TABLE]

for each $s,\lambda\in[0,1]$. Then $f_{t_{1}}\circ F$ is a path homotopy between $f_{t_{0}}\circ\alpha=f_{t_{1}}\circ\tilde{\alpha}$ and $f_{t_{0}}\circ\beta=f_{t_{1}}\circ\tilde{\beta}$ in $\Omega_{t_{1}}$. Since $\{\Omega_{t}\}_{t\in I}$ is continuous at $t_{1}$ in the sense of kernel, there exists $\delta>0$ such that for $t\in I$ with $|t-t_{1}|<\delta$, $f_{t_{1}}\circ F([0,1]\times[0,1])\subset\Omega_{t}$. Thus for each $t\in I$ with $|t-t_{1}|<\delta$, with respect to the covering map $f_{t}:\mathbb{D}\rightarrow\Omega_{t}$, the path homotopy $f_{t_{1}}\circ F$ can be lifted to a unique path homotopy $F_{t}:[0,1]\times[0,1]\rightarrow\mathbb{D}$ with $F_{t}(0,\lambda)=0$, $\lambda\in[0,1]$. Since $F_{t}$ is a path homotopy, we have $F_{t}(1,0)=F_{t}(1,1)$ and hence $\omega_{t_{0},t}(z_{1})=\omega_{t_{0},t}(z_{2})$, which contradicts (8.1). ∎

For $t\in I$ let $\Gamma_{t}$ be the covering transformation group of the universal covering map $f_{t}:\mathbb{D}\rightarrow\Omega_{t}$, i.e., $\Gamma_{t}$ is the class of analytic homeomorphisms $\gamma$ of $\mathbb{D}$ onto $\mathbb{D}$ satisfying $f_{t}\circ\gamma=f_{t}$, which forms a group under composition. Each $\Gamma_{t}$ is a subgroup of the group $\operatorname{Aut}(\mathbb{D})$ of linear fractional transformations $\gamma$ of the form

[TABLE]

which maps $\mathbb{D}$ conformally onto $\mathbb{D}$. Notice that $\Gamma_{t}$ is fixed point free, i.e., every $\gamma\in\Gamma_{t}$ with $\gamma\not=\operatorname{id}_{\mathbb{D}}$ has no fixed point in $\mathbb{D}$. From this it follows that for $\gamma_{j}\in\Gamma_{t}$, $j=1,2$,

[TABLE]

Now we introduce $\sigma_{s,t}:\Gamma_{s}\rightarrow\Gamma_{t}$ for $(s,t)\in I_{+}^{2}$ as follows. The remaining of the section is devoted to study relations between $\{f_{t}\}_{t\in I}$, $\{\omega_{s,t}\}_{(s,t)\in I_{+}^{2}}$, $\{\Gamma_{t}\}_{t\in I}$ and $\{\sigma_{s,t}\}_{(s,t)\in I_{+}^{2}}$.

For $\gamma\in\Gamma_{s}$ take $\alpha:[0,1]\rightarrow\mathbb{D}$ be a path from [math] to $\gamma(0)$ arbitrarily. Since $f_{s}\circ\gamma(0)=f_{s}(0)=0$, $f_{s}\circ\alpha$ is a loop ($=$ closed path) based at [math] in $\Omega_{s}(\subset\Omega_{t})$. Let $\tilde{\alpha}:[0,1]\rightarrow\mathbb{D}$ be the unique lift of $\alpha$ with $\tilde{\alpha}(0)=0$ satisfying $f_{s}\circ\tilde{\alpha}=f_{t}\circ\alpha$. Then there exists uniquely $\tilde{\gamma}\in\Gamma_{t}$ with $\tilde{\gamma}(0)=\tilde{\alpha}(1)$. Since $\mathbb{D}$ is simply connected, the end point $\tilde{\alpha}(1)$ does not depend on the choice of $\alpha$ and hence $\tilde{\gamma}$ is uniquely determined by $\gamma$. We define $\sigma_{s,t}:\Gamma_{s}\rightarrow\Gamma_{t}$ by $\sigma_{s,t}(\gamma)=\tilde{\gamma}$. Notice that by definition, $\sigma_{t,t}$ is the identity mapping of $\Gamma_{t}$, and that the semigroup relation

[TABLE]

holds for $t_{0},t_{1},t_{2}\in I$ with $t_{0}\leq t_{1}\leq t_{2}$.

Now we prove $\sigma_{s,t}$ is an injective homomorphism and satisfies $\omega_{s,t}\circ\gamma=\sigma_{s,t}(\gamma)\circ\omega_{s,t}$.

Proof of Theorem 1.10.

Let $\gamma\in\Gamma_{s}$ and $z\in\mathbb{D}$, and let $\alpha,\beta:[0,1]\rightarrow\mathbb{D}$ be paths from [math] to $\gamma(0)$ and from [math] to $z$, respectively. Notice that $\alpha(1)=\gamma(0)=\gamma\circ\beta(0)$. By $\alpha*(\gamma\circ\beta)$ we denote the path obtained by first traversing $\alpha$ and then traversing by $\gamma\circ\beta$. Let $\tilde{\alpha},\tilde{\beta}:[0,1]\rightarrow\mathbb{D}$ be the lifted paths from [math] with $f_{t}\circ\tilde{\alpha}=f_{s}\circ\alpha$ and $f_{t}\circ\tilde{\beta}=f_{s}\circ\beta$, respectively. Since $f_{s}(\alpha(1))=f_{s}\circ\gamma(0)=f_{s}(0)$, the path $f_{s}\circ\alpha$ is a loop. Thus there exists uniquely $\tilde{\gamma}\in\Gamma_{t}$ with $\tilde{\gamma}(0)=\tilde{\alpha}(1)$. Then the path $\tilde{\alpha}*(\tilde{\gamma}\circ\tilde{\beta})$ coincides with the unique lifted path of $f_{s}(\alpha*(\gamma\circ\beta))$ from [math]. Therefore $\omega_{s,t}$ maps the end point $\gamma(z)$ of $(\alpha*(\gamma\circ\beta))$ to the end point $\tilde{\gamma}(\tilde{\beta}(1))$ of $\tilde{\alpha}*(\tilde{\gamma}\circ\tilde{\beta})$, i.e., $\omega_{s,t}(\gamma(z))=\tilde{\gamma}(\tilde{\beta}(1))$. By $\tilde{\gamma}=\sigma_{s,t}(\gamma)$ and $\tilde{\beta}(1)=\omega_{s,t}(z)$ we have $\omega_{s,t}(\gamma(z))=\sigma_{s,t}(\gamma)(\omega_{s,t}(z))$.

O\mathbb{D}$$\gamma(0)$$\alpha$$\beta$$z$$\gamma\circ\beta$$\gamma(z)O\mathbb{D}$$\tilde{\alpha}$$\tilde{\gamma}(0)$$\tilde{\beta}$$\omega_{s,t}(z)$$\tilde{\gamma}\circ\tilde{\beta}$$\tilde{\gamma}(\omega_{s,t}(z))$$\Omega_{s}OO\Omega_{t}$$f_{s}\circ\alpha=f_{t}\circ\tilde{\alpha}$$f_{s}\circ\beta=f_{t}\circ\tilde{\beta}$$f_{s}$$f_{t}$$\omega_{s,t}

Let $\delta\in\Gamma_{s}$ and consider the case that $z=\delta(0)$. Then $\gamma(z)=\gamma\circ\delta(0)$. Furthermore we have $\omega_{s,t}(z)=\sigma_{s,t}(\delta)(0)$. Thus

[TABLE]

By (8.2), this implies $\sigma_{s,t}(\gamma\circ\delta)=\sigma_{s,t}(\gamma)\circ\sigma_{s,t}(\delta)$. Therefore $\sigma_{s,t}$ is a group homomorphism.

Finally we show $\sigma_{s,t}$ is injective. Since $\sigma_{s,t}$ is a homomorphism, it suffices to show the kernel of $\sigma_{s,t}$ is trivial. By the fixed point free properties of $\Gamma_{s}$ and $\Gamma_{t}$ it is reduced to show that if $\sigma_{s,t}(\gamma)(0)=0$, then $\gamma(0)=0$. Suppose that $\tilde{\gamma}:=\sigma_{s,t}(\gamma)$ satisfies $\tilde{\gamma}(0)=0$. Then $\omega_{s,t}(\gamma(0))=\tilde{\gamma}(0)=0$. Since $\omega_{s,t}$ is univalent, we have $\gamma(0)=0$ as required. ∎

Corollary 8.2**.**

For $(s,t)\in I_{+}^{2}$ and $\gamma\in\Gamma_{s}$, the image domain $\omega_{s,t}(\mathbb{D})$ is $\sigma_{s,t}(\gamma)$ invariant, i.e., $\sigma_{s,t}(\gamma)(\omega_{s,t}(\mathbb{D}))=\omega_{s,t}(\mathbb{D})$.

Proof.

This follows from $\omega_{s,t}\circ\gamma=\sigma_{s,t}(\gamma)\circ\omega_{s,t}$ and $\gamma(\mathbb{D})=\mathbb{D}$. ∎

Now we derive the differential equation satisfied by $\sigma_{t_{0},t}(\gamma)$.

Proof of Theorem 1.11.

It suffice to show in the case that $I=[t_{0},t_{0}^{*}]$ with $-\infty<t_{0}<t_{0}^{*}<\infty$. Since $\sigma_{t_{0},t}:\Gamma_{t_{0}}\rightarrow\Gamma_{t}$ is a homomorphism, we have $(\gamma^{-1})_{t}=\sigma_{t_{0},t}(\gamma^{-1})=(\sigma_{t_{0},t}(\gamma))^{-1}=(\gamma_{t})^{-1}$, $\gamma\in\Gamma_{t_{0}}$. Thus we may write simply $\gamma_{t}^{-1}$ without any ambiguity.

If $\gamma=\operatorname{id}_{\mathbb{D}}$, then (1.10) holds trivially. Suppose that $\gamma\in\Gamma_{t_{0}}\backslash\{\operatorname{id}_{\mathbb{D}}\}$. Then since $\sigma_{t_{0},t}$ is injective, $\gamma_{t}\not=\operatorname{id}_{\mathbb{D}}$. It follows from the fixed point free property of $\Gamma_{t}$ that $\gamma_{t}(0)\not=0$ for $t\in I\cap[t_{0},\infty)$. By Theorem 1.10 and $\omega_{t_{0},t}(0)=0$ we have $\gamma_{t}(0)=\omega_{t_{0},t}(\gamma(0))\not=0$. Particularly the mapping $[t_{0},t_{0}^{*}]\ni t\mapsto\gamma_{t}(0)\in\mathbb{D}$ is continuous and

[TABLE]

Similarly the mapping $[t_{0},t_{0}^{*}]\ni t\mapsto\gamma_{t}^{-1}(0)\in\mathbb{D}$ is also continuous and by $|\gamma_{t}^{-1}(0)|=|\gamma_{t}(0)|$ we have $\min_{t\in[t_{0},t_{0}^{*}]}|\gamma_{t}^{-1}(0)|=m>0$ and $\max_{t\in[t_{0},t_{0}^{*}]}|\gamma_{t}^{-1}(0)|=M<1$. For $t\in[t_{0},t_{0}^{*}]$ take $\zeta_{t}\in\mathbb{D}$ and $\theta_{t}\in\mathbb{R}$ such that

[TABLE]

Then we have $\gamma_{t}(0)=-e^{i\theta_{t}}\zeta_{t}$ and $\gamma_{t}^{-1}(0)=\zeta_{t}$. Hence

[TABLE]

are continuous on $[t_{0},t_{0}^{*}]$ with $0<m\leq|\zeta_{t}|\leq M<1$. We may assume $\theta_{t}$ is also continuous on $[t_{0},t_{0}^{*}]$. From these properties it follows that $\gamma_{s}\rightarrow\gamma_{t}$ locally uniformly on $\mathbb{D}$ as $[t_{0},t_{0}^{*}]\ni s\rightarrow t$, i.e., the map $[t_{0},t_{0}^{*}]\ni t\mapsto\gamma_{t}\in\mathcal{H}(\mathbb{D})$ is continuous.

By Theorem 1.10 we have

[TABLE]

Since $\gamma_{t_{2}}^{\prime}(z)\rightarrow\gamma_{t}^{\prime}$ and $\omega_{t_{1},t_{2}}(z)\rightarrow z$ locally uniformly on $\mathbb{D}$ as $t_{2}-t_{1}\downarrow 0$ with $t_{1}\leq t\leq t_{2}$, we have

[TABLE]

If $t\in[t_{0},t_{0}^{*}]\backslash N$, then by Theorem 3.3

[TABLE]

and the convergence is locally uniform on $\mathbb{D}$. Since $\gamma_{t_{1}}(z)\rightarrow\gamma_{t}(z)$, we have

[TABLE]

Combining these equalities and (8.4) we have (1.10).

Suppose that $a(t)$ is absolutely continuous and $\dot{a}(t)>0$ a.e. Let $E_{0}$ be the set of all $t\in I$ at which $a$ is not differentiable. Also let $E_{1}(\subset[\alpha,\beta])$ be the set of all $t\in I$ at which $a$ is differentiable and $\dot{a}(t)=0$. Then $E_{0}\cup E_{1}$ is the set of Lebesgue measure [math] and $t\in[t_{0},t_{0}^{*}]\backslash(N\cup E_{0}\cup E_{1})$ we have

[TABLE]

Combining this and (1.10) we obtain (1.11). ∎

9. Separation Lemma

One can prove the separation lemma (Lemma 1.9) in a purely topological manner on the basis of methods in Newman [22]. However, for the sake of simplicity we shall make use of the Riemann mapping theorem and avoid elaborate arguments.

We have repeatedly and implicitly used the criterion: a domain in $\hat{\mathbb{C}}$ is simply connected if and only if its complement is connected or empty. Combining this, the Riemann mapping theorem and the Jordan curve theorem it is not difficult to see the following classical result.

Lemma 9.1**.**

Let $\Omega$ be a domain in $\mathbb{C}$. Then $\Omega$ is simply connected if and only if the inside domain of any simple closed curve in $\Omega$ is contained in $\Omega$.

For a purely topological proof see Newman [22, Chapter VI]. From the above lemma it is easy to see the following.

Lemma 9.2**.**

Let $E$ be a nonempty compact connected set in $\hat{\mathbb{C}}$. Then each component of $\hat{\mathbb{C}}\backslash E$ is simply connected.

By a partition of a set $E\subset\hat{\mathbb{C}}$ we means two nonempty subsets $H_{1}$ and $H_{2}$ of $E$ such that $H_{1}$ and $H_{2}$ are both closed in the relative topology of $E$ with $H_{1}\cup H_{2}=E$ and $H_{1}\cap H_{2}=\emptyset$. For a proof of the following lemma see Newman [22, Theorem 5.6].

Lemma 9.3**.**

Let $E$ be a compact set in a metric space $X$, $F_{1}$, $F_{2}$ be nonempty closed subsets of $E$ such that for any component $B$ of $E$, $F_{1}\cap B=\emptyset$ or $F_{2}\cap B=\emptyset$. Then there exists a partition $H_{1}$, $H_{2}$ of $E$ with $F_{1}\subset H_{1}$ and $F_{2}\subset H_{2}$.

Now we prove the separation lemma.

Proof of Lemma 1.9.

We may assume that $\infty\in F$ and $F$ contains at least a point other than $\infty$. By Lemma 9.3 there exist closed sets $H_{1}$ and $H_{2}$ with $H_{1}\cap H_{2}=\emptyset$ and $H_{1}\cup H_{2}=\hat{\mathbb{C}}\backslash\Omega$ satisfying $C\subset H_{1}$ and $F\subset H_{2}$. Then since $\infty\not\in H_{1}$, $H_{1}$ is compact in $\mathbb{C}$ and we have $0<d(H_{1},H_{2}):=\{|z-w|:z\in H_{1},\;w\in H_{2}\}<\infty$.

Let $S$ be a square with $H_{1}\subset\operatorname{Int}S$ such that its sides are parallel to the coordinate axes. Here by a square we mean a closed solid square consisting of both boundary and interior, and denote the set of interior points of $S$ by $\operatorname{Int}S$. Let $\ell$ be the length of edges of $S$ and take $n\in\mathbb{N}$ with

[TABLE]

By means of equally spaced horizontal and vertical lines we divide $S$ into nonoverlapping small squares having the edge length $\ell/n$. We call $\hat{\mathbb{C}}\backslash\operatorname{Int}S$ the unbounded square. Let $K$ be the union of the unbounded square and those squares that intersect $H_{2}$.

Claim. $\partial K\cap(H_{1}\cup H_{2})=\emptyset$, i.e., $\partial K\subset\Omega$.

We show the Claim. Notice that $\partial K$ consists of edges $A$ of small squares. In the case that $A\subset\partial S$ there exists a unique small square $T$ having $A$ as its edge. Since $\operatorname{diam}T=\frac{\sqrt{2}\ell}{n}<d(H_{1},\partial S)$, we have $T\cap H_{1}=\emptyset$. Also we have $T\cap H_{2}=\emptyset$. Indeed if $T\cap H_{2}\not=\emptyset$, then $A\subset T\cup(\hat{\mathbb{C}}\backslash\operatorname{Int}S)\subset K$. This implies $A^{\circ}\subset\operatorname{Int}K$ which contradicts $A\subset\partial K$. Here $A^{\circ}$ is the open segment obtained from $A$ by removing two vertices of $A$.

Next we consider the case that $A\backslash\partial S\not=\emptyset$. In this case there exist uniquely two small adjacent squares $T_{1}$ and $T_{2}$ with $A\subset T_{1}\cap T_{2}$, and we may assume $T_{1}\cap H_{2}=\emptyset$ and $T_{2}\cap H_{2}\not=\emptyset$. Then $A\cap H_{2}\subset T_{1}\cap H_{2}=\emptyset$. Also we have $A\cap H_{1}=\emptyset$, since $T_{2}\cap H_{2}\not=\emptyset$ and $\operatorname{diam}(T_{1}\cup T_{2})=\frac{\sqrt{3}\ell}{n}<d(H_{1},H_{2})$. Thus we have completed the proof of the claim.

Let $K_{0}$ be the component of $K$ containing the unbounded square $\hat{\mathbb{C}}\backslash\operatorname{Int}S$ and $K=K_{0}\cup K_{1}\cup\cdots\cup K_{m}$ be the decomposition of $K$ into components. Since $\partial K_{j}\subset\partial K\subset\Omega$ for $j=0,\ldots,m$, there exists path $\alpha_{j}:[0,1]\rightarrow\Omega$ with $\alpha_{j}(0)\in\partial K_{0}$ and $\alpha_{j}(1)\in\partial K_{j}$ for $j=1,\ldots,m$. Then $\tilde{F}=K\cup\bigcup_{j=1}^{m}\alpha_{j}([0,1])$ is connected and closed in $\hat{\mathbb{C}}$ satisfying

[TABLE]

Let $\Omega_{0}$ be the component of $\hat{\mathbb{C}}\backslash\tilde{F}$ containing $C$. Since

[TABLE]

we have $\partial\Omega_{0}\cap H_{1}=\emptyset$ and hence $\Omega_{0}\cap H_{1}$ is compact. By Lemma 9.2 $\Omega_{0}$ is simply connected and hence by the Riemann mapping theorem there exists a conformal map $h:\mathbb{D}\rightarrow\Omega_{0}$. Since $\Omega_{0}\cap H_{1}$ is compact, there exists $r\in(0,1)$ with

[TABLE]

Define a simple closed curve $\alpha:\partial\mathbb{D}\rightarrow\Omega_{0}$ by

[TABLE]

Then the inside domain of $\alpha$ coincides with $h(\mathbb{D}(0,r))$ and contains $C(\subset\Omega_{0}\cap H_{1})$. Since $\alpha(\partial\mathbb{D})\cap\tilde{F}=\emptyset$, the connected set $\tilde{F}$ is contained in either the inside or outside domains of $\alpha$. By $\infty\in\tilde{F}$, $\tilde{F}$ is contained in the outside domain of $\alpha$. Therefore $\alpha$ separate $C$ and $F(\subset\tilde{F})$.

K_{0}$$K_{1}$$K_{2}$$C$$\Omega_{0}$$\alpha_{1}$$\alpha_{2}

∎

Index