§9.3 Boundary Continuity of Biholomorphisms

9.3 Boundary Continuity of Biholomorphisms

Suppose $Ω_{1}$ and $Ω_{2}$ are two regions in the complex plane such that there is a biholomorphism $𝜑$ from $Ω_{1}$ to $Ω_{2}$ . Naturally, we are concerned about the existence of a continuous extension of $𝜑$ to $\overline{Ω_{1}}$ .

In fact, it is almost always true that such an extension exists. We will give three examples of this phenomenon, each with increasing regularity assumptions on the boundaries $𝜕 Ω_{1}$ and $𝜕 Ω_{2}$ :

If $𝜕 Ω_{1}$ and $𝜕 Ω_{2}$ are two Jordan curves, then $𝜑$ (whose existence is given by the Riemann Mapping Theorem or Theorem 5.3.2 $Theorem 5.3.2 Theorem 5.3.2$ ) extends homeomorphically to $𝜕 Ω_{1}$ (Theorem 9.3.3 $Theorem 9.3.3 Theorem 9.3.3$ ).
If $𝜕 Ω_{1}$ and $Ω_{2}$ are $𝐶^{\infty}$ Jordan curves, then $𝜑$ extends continuously and injectively to $𝜕 Ω_{1}$ and is $𝐶^{\infty}$ on $\overline{Ω_{1}}$ .
If $𝜕 Ω_{1}$ and $𝜕 Ω_{2}$ are real-analytic Jordan curves (the boundary is parameterizable by functions such that at every point, there is a power series expansion that converge to the function on a neighborhood), then $𝜑$ extends analytically past $𝜕 Ω_{1}$ (Theorem 9.3.14 $Theorem 9.3.14 Theorem 9.3.14$ ).

Figure 9.3: The region

𝑈

with a comb-shaped boundary.

Example 9.3.1.

Let

𝑈 = {𝑥 + i 𝑦 : 0 < 𝑥, 𝑦 < 2} ∖ ⋃_{𝑛 = 1}^{\infty} {i 𝑡 + \frac{1}{𝑛} : 𝑡 \in [0, 1]}

be an open, simply connected region with a comb-shaped boundary (Figure 9.3

Figure 9.3 Figure 9.3

). By the Riemann Mapping Theorem, it is conformally equivalent to

𝔻

via some biholomorphism

𝑓

. Then there is no homeomorphism

\tilde{𝑓} : \overline{𝑈} \to \overline{𝔻}

whose restriction to

𝑈

𝑓

Proof.

Suppose that such a homeomorphism $\tilde{𝑓}$ exists. Since $\tilde{𝑓} (𝑈) = 𝔻$ and $\tilde{𝑓}$ is bijective, it restricts to a homeomorphism from $𝜕 𝑈 = \overline{𝑈} ∖ 𝑈$ onto $𝜕 𝔻 = \overline{𝔻} ∖ 𝔻$ . Here

\begin{aligned} 𝜕 𝑈 & = {𝑥 + i 𝑦 : 𝑥 \in {0, 2}, 𝑦 \in [0, 2]} \\ \begin{matrix}  \end{matrix} \cup {𝑥 + i 𝑦 : 𝑥 \in [0, 2], 𝑦 \in {0, 2}} \\ \begin{matrix}  \end{matrix} \cup ⋃_{𝑛 = 1}^{\infty} {i 𝑡 + \frac{1}{𝑛} : 𝑡 \in [0, 1]} \end{aligned}

Consider the comb “tooth” $𝐿 = {1 + i 𝑡 : 𝑡 \in [0, 1]}$ . If its base point $1$ is removed from $𝜕 𝑈$ , there is no path in the remaining boundary from $𝐿 ∖ {1}$ to the rest of $𝜕 𝑈$ , since $𝐿$ meets the rest of the boundary only at $1$ (topologically, this is known as a “cut point”). Hence $𝜕 𝑈 ∖ {1}$ is not path-connected. However, removing $\tilde{𝑓} (1)$ from the circle $𝜕 𝔻$ leaves an open, path-connected arc.

Then ${\tilde{𝑓} |}_{𝜕 𝑈 ∖ {1}}$ is a homeomorphism. Let $𝑧_{1} \in 𝐿 ∖ {1}$ , $𝑧_{2} \in 𝜕 𝑈 ∖ 𝐿$ be two arbitrary points. Then there exists an arc in $𝜕 𝔻 ∖ \tilde{𝑓} (1)$ joining $\tilde{𝑓} (𝑧_{1})$ and $\tilde{𝑓} (𝑧_{2})$ . Then the image of this arc under ${\tilde{𝑓}}^{- 1}$ is a path joining $𝑧_{1}$ and $𝑧_{2}$ , a contradiction. Thus no such homeomorphism exists. ⁠ $□$

Example 9.3.2.

The biholomorphism $𝜑 : 𝔻 \to 𝔻 \cap ℍ^{+}$ defined by

𝜑 : 𝑧 \mapsto \frac{1 + i \sqrt{\frac{i (1 - 𝑧)}{1 + 𝑧}}}{1 - i \sqrt{\frac{i (1 - 𝑧)}{1 + 𝑧}}}

extends continuously but not differentiably to

\pm 1

. The boundary of

𝔻 \cap ℍ^{+}

is piecewise

𝐶^{\infty}

Theorem 9.3.3 (Osgood–Taylor–Carathéodory).

Suppose that

Ω_{1}

and

Ω_{2}

are two bounded regions in

ℂ

such that

𝜕 Ω_{1}

and

𝜕 Ω_{2}

each comprises a single Jordan curve. If

𝜑 : Ω_{1} \to Ω_{2}

is a biholomorphism (provided by the Riemann Mapping Theorem in Theorem 5.3.2

Theorem 5.3.2 Theorem 5.3.2

), then

\exists \tilde{𝜑} : \overline{Ω_{1}} \to \overline{Ω_{2}}

homeomorphic such that

\tilde{𝜑} |_{Ω_{1}} \equiv 𝜑

(the restriction of

\tilde{𝜑}

Ω_{1}

agrees with

𝜑

Proof.

The theorem will be first proven in the case that $Ω_{1}$ is the unit disk $𝔻$ . The general case will then be reduced to this special case.

Let $𝜇_{1}, 𝜇_{2} : [0, 1] \to \overline{𝔻}$ be two curves such that $𝜇_{1} ([0, 1)), 𝜇_{2} ([0, 1)) \subset 𝔻$ and $𝜇_{1} (1) = 1$ , $𝜇_{2} (1) = 1$ (as in Figure 9.4 $Figure 9.4 Figure 9.4$ ). We now aim to show that ${lim}_{𝑡 \to 1^{-}} 𝜑 (𝜇_{1} (𝑡)) = {lim}_{𝑡 \to 1^{-}} 𝜑 (𝜇_{2} (𝑡))$ (existence and equality).

Let $𝑉 = 𝔻 \cap 𝐷 (1, \frac{1}{2})$ , which has a finite area. Since $𝜑$ is a biholomorphism, $𝜑 (𝑉) \subset Ω_{2}$ also has a finite area (call this $𝑀$ ) and is simply connected since its complement in $Ω_{2}$ , $𝜑 (𝔻 ∖ 𝑉)$ , is connected.

Figure 9.4: Configuration of

𝑉

, and possible

𝜇_{1}, 𝜇_{2}

Figure 9.5: Possible positioning of

𝜑 (𝑉)

relative to

Ω_{2}

Let $𝐾 \subset Ω_{2}$ be compact. Since $𝜑^{- 1}$ is continuous, it maps $𝐾$ to a compact set $𝜑^{- 1} (𝐾)$ (a function providing the compactness of the pre-image of every compact set is more generally known as proper). This implies that for any sequence ${𝑧_{𝑗}}_{𝑗 \in ℕ} \subset 𝔻$ with accumulation points only in $𝜕 𝔻$ , ${𝜑 (𝑧_{𝑗})}_{𝑗 \in ℕ}$ will have its accumulation points in $𝜕 Ω_{2}$ (because otherwise, there would be a corresponding subsequence of ${𝑧_{𝑗}}_{𝑗 \in ℕ}$ that converges to a point in $𝔻$ ). Therefore, $𝜑 (𝑉)$ does not have relatively compact closure in $Ω_{2}$ (a representation of a possible configuration in Figure 9.5 $Figure 9.5 Figure 9.5$ ).

Figure 9.6: Possible configuration of

𝛾_{𝑟}

Figure 9.7: Concept configuration of

𝜑 (𝛾_{𝑟})

For $0 < 𝑟 < \frac{1}{2}$ , we now introduce polar coordinates $(𝑟, 𝜃)$ centered at $1$ with angles offset by $π$ . Let $𝜃_{0} (𝑟)$ be the supremum of $𝜃$ for fixed $𝑟$ such that $1 - 𝑟 e^{i 𝜃} \in 𝑉$ . Define $𝛾_{𝑟} : (- 𝜃_{0} (𝑟), 𝜃_{0} (𝑟)) \to 𝑉$ to be the curve with $𝜃 \mapsto 1 - 𝑟 e^{i 𝜃}$ (as in Figure 9.6 $Figure 9.6 Figure 9.6$ ).

Lemma 9.3.4.

Let $𝑙_{𝑟}$ be the length of $𝜑 \circ 𝛾_{𝑟}$ . Then

\int_{0}^{\frac{1}{2}} \frac{𝑙_{𝑟}^{2}}{π 𝑟} d 𝑟

is finite.

Proof.

Notice that

\begin{aligned} 𝑙_{𝑟} & = \int_{𝜑 \circ 𝛾_{𝑟} ([- 𝜃_{0} (𝑟), 𝜃_{0} (𝑟)])} | d 𝑧 | = \int_{- 𝜃_{0} (𝑟)}^{𝜃_{0} (𝑟)} | \frac{d 𝜑 (1 - 𝑟 e^{i 𝜃})}{d 𝜃} | d 𝜃 \\ = \int_{- 𝜃_{0} (𝑟)}^{𝜃_{0} (𝑟)} 𝑟 | 𝜑^{'} (1 - 𝑟 e^{i 𝜃}) | d 𝜃 . \end{aligned}

Hence,

\begin{aligned} \int_{0}^{\frac{1}{2}} \frac{𝑙_{𝑟}^{2}}{π 𝑟} d 𝑟 & = \int_{0}^{\frac{1}{2}} \frac{1}{π 𝑟} {[\int_{- 𝜃_{0} (𝑟)}^{𝜃_{0} (𝑟)} | 𝜑^{'} (1 - 𝑟 e^{i 𝜃}) | 𝑟 d 𝜃]}^{2} d 𝑟 \\ = \int_{0}^{\frac{1}{2}} \frac{1}{π 𝑟} {[\int_{- 𝜃_{0} (𝑟)}^{𝜃_{0} (𝑟)} | 𝜑^{'} (1 - 𝑟 e^{i 𝜃}) | \sqrt{𝑟} \sqrt{𝑟} d 𝜃]}^{2} d 𝑟 \\ \leq \int_{0}^{\frac{1}{2}} \frac{1}{π 𝑟} [\int_{- 𝜃_{0} (𝑟)}^{𝜃_{0} (𝑟)} {| 𝜑^{'} (1 - 𝑟 e^{i 𝜃}) |}^{2} 𝑟 d 𝜃] [\int_{- 𝜃_{0} (𝑟)}^{𝜃_{0} (𝑟)} 𝑟 d 𝜃] d 𝑟 \\ \leq \int_{0}^{\frac{1}{2}} \int_{- 𝜃_{0} (𝑟)}^{𝜃_{0} (𝑟)} {| 𝜑^{'} (1 - 𝑟 e^{i 𝜃}) |}^{2} 𝑟 d 𝜃 d 𝑟 \\ = \iint_{𝑉} {| 𝜑^{'} (𝑧) |}^{2} d 𝑥 d 𝑦 = area (𝜑 (𝑉)) \leq area (Ω_{2}) < \infty, \end{aligned}

where the last step uses the Lusin Area Theorem (see Theorem 3.1.1

Theorem 3.1.1 Theorem 3.1.1

). ⁠

□

Assume that $\exists 𝑐 > 0, 𝜀 > 0$ such that $𝑙_{𝑟} \geq 𝑐$ for all $0 < 𝑟 < 𝜀$ . Then, this implies that

\int_{0}^{\frac{1}{2}} \frac{𝑙_{𝑟}^{2}}{π 𝑟} d 𝑟 \geq \int_{0}^{𝜀} \frac{𝑙_{𝑟}^{2}}{π 𝑟} d 𝑟 \geq \int_{0}^{𝜀} \frac{𝑐^{2}}{π 𝑟} d 𝑟 = \frac{𝑐^{2}}{π} \int_{0}^{𝜀} \frac{d 𝑟}{𝑟},

which is a divergent integral. This contradicts Lemma 9.3.4 $Lemma 9.3.4 Lemma 9.3.4$ . Therefore, we must have ${lim inf}_{𝑟 \to 0^{+}} 𝑙_{𝑟} = 0$ . Thus, there exists a sequence ${𝑟_{𝑗}}_{𝑗 \in ℕ} \subset (0, \frac{1}{2})$ such that $𝑟_{𝑗} \to 0$ and $𝑙_{𝑟_{𝑗}} \to 0$ as $𝑗 \to \infty$ .

Proposition 9.3.5.

For each such $𝑟_{𝑗}$ as above, the limits

lim_{𝜃 \to 𝜃_{0} {(𝑟_{𝑗})}^{-}} 𝜑 (1 - 𝑟_{𝑗} e^{i 𝜃}) and lim_{𝜃 \to - 𝜃_{0} {(𝑟_{𝑗})}^{+}} 𝜑 (1 - 𝑟_{𝑗} e^{i 𝜃})

exist.

Proof.

Since $𝑙_{𝑟_{𝑗}}$ is finite, $\forall 𝜀 > 0$ , $\exists 𝛿 > 0$ such that for $𝜃 > 𝜃_{0} (𝑟_{𝑗}) - 𝛿$ ,

\int_{𝜃}^{𝜃_{0} (𝑟_{𝑗})} 𝑟 | 𝜑^{'} (1 - 𝑟 e^{i 𝜗}) | d 𝜗 < 𝜀 .

Then for any $𝜃^{'}, 𝜃^{″} \in (𝜃_{0} (𝑟) - 𝛿, 𝜃_{0} (𝑟))$ ,

\begin{matrix} | 𝜑 (1 - 𝑟 e^{i 𝜃^{'}}) - 𝜑 (1 - 𝑟 e^{i 𝜃^{″}}) | \leq length (𝜑 (𝛾_{𝑟} ([𝜃^{'}, 𝜃^{″}]))) \\ = \int_{𝜃^{'}}^{𝜃^{″}} 𝑟 | 𝜑^{'} (1 - 𝑟 e^{i 𝜗}) | d 𝜗 < 𝜀, \end{matrix}

implying that

{𝜑 (1 - 𝑟 e^{i 𝜃})}_{𝜃 \to 𝜃_{0} {(𝑟_{𝑗})}^{-}}

is Cauchy and hence convergent. The proof for the other limit follows symmetrically. ⁠

□

Lemma 9.3.6.

Let

𝜏 = 𝜕 Ω_{2}

. Then there is a function

𝜂 (𝛿)

defined for sufficiently small

𝛿

, such that

𝜂 (𝛿) \to 0

𝛿 \to 0^{+}

, such that for all

𝑎, 𝑏 \in 𝜏

satisfying

| 𝑎 - 𝑏 | \leq 𝛿

, then there is exactly one curve in

𝜏

with endpoints

𝑎

and

𝑏

with diameter

\leq 𝜂 (𝛿)

Proof.

Since $𝜏$ is a Jordan curve, it admits a homeomorphic parameterization $𝜓 : 𝑆^{1} \to 𝜏$ , and $𝜓$ is a bicontinuous bijection (onto its image) between compact Hausdorff spaces, so $𝜓^{- 1}$ is continuous.

Since $𝜓^{- 1}$ is uniformly continuous on the compact set $𝜏$ , for every $𝜀 > 0$ there exists $𝛿^{'} > 0$ such that if $| 𝑝 - 𝑞 | < 𝛿^{'}$ with $𝑝, 𝑞 \in 𝜏$ , then $| 𝜓^{- 1} (𝑝) - 𝜓^{- 1} (𝑞) | < 𝜀$ .

In particular, choose $𝛿_{0} > 0$ small enough so that whenever $| 𝜓 (𝜁) - 𝜓 (𝜁^{'}) | \leq 𝛿_{0}$ , then $| 𝜁 - 𝜁^{'} | < 2$ .

Now define

𝜂 (𝛿) = sup {diam 𝜌 : | 𝜓 (𝜁) - 𝜓 (𝜁^{'}) | \leq 𝛿}

for $0 < 𝛿 \leq 𝛿_{0}$ , where $𝜌 = 𝜓 (𝜎)$ and $𝜎 \subseteq 𝑆^{1}$ is the shorter arc connecting $𝜁, 𝜁^{'} \in 𝑆^{1}$ .

First, $𝜂 (𝛿) \to 0$ as $𝛿 \to 0^{+}$ ; if not, there exists $𝜀 > 0$ and a sequence $𝛿_{𝑛} \to 0^{+}$ with $𝜂 (𝛿_{𝑛}) \geq 𝜀$ . Then for each $𝑛 \in ℕ$ , there are $𝜁_{𝑛}, 𝜁_{𝑛}^{'} \in 𝑆^{1}$ with $| 𝜓 (𝜁_{𝑛}) - 𝜓 (𝜁_{𝑛}^{'}) | \leq 𝛿_{𝑛}$ such that the shorter arc on $𝑆^{1}$ between $𝜁_{𝑛}$ and $𝜁_{𝑛}^{'}$ has image $𝜌_{𝑛}$ with $diam 𝜌_{𝑛} \geq 𝜀$ . By compactness of $𝑆^{1}$ , we may assume (by passing to a subsequence if necessary) that $𝜁_{𝑛} \to 𝜁$ and $𝜁_{𝑛}^{'} \to 𝜁^{'}$ . Then $𝜓 (𝜁) = {lim}_{𝑛 \to \infty} 𝜓 (𝜁_{𝑛}) = {lim}_{𝑛 \to \infty} 𝜓 (𝜁_{𝑛}^{'}) = 𝜓 (𝜁^{'})$ , so $𝜁 = 𝜁^{'}$ . But then the images of the shorter arcs between each $𝜁_{𝑛}$ and $𝜁_{𝑛}^{'}$ must have diameters tending to zero (by continuity of $𝜓$ ), contradicting that $diam 𝜌_{𝑛} \geq 𝜀$ for each $𝑛$ .

To finish, choose $0 < 𝛿_{1} < 𝛿_{0}$ small enough that $𝜂 (𝛿_{1}) < \frac{1}{2} diam 𝜏$ (possible since $𝜂 (𝛿) \to 0$ as $𝛿 \to 0$ and $diam 𝜏 > 0$ ).

Then the assertion holds for all $𝛿 \leq 𝛿_{1}$ with this $𝜂$ : for $𝑎, 𝑏 \in 𝜏$ with $| 𝑎 - 𝑏 | \leq 𝛿 \leq 𝛿_{1}$ , let $𝜁, 𝜁^{'} \in 𝑆^{1}$ with $𝜓 (𝜁) = 𝑎$ , $𝜓 (𝜁^{'}) = 𝑏$ . The shorter arc on $𝑆^{1}$ between $𝜁, 𝜁^{'}$ has image $𝜌$ with $diam 𝜌 \leq 𝜂 (𝛿)$ . The image of the other arc on $𝜏$ between $𝑎, 𝑏$ , $\tilde{𝜌}$ , satisfies

diam \tilde{𝜌} + diam 𝜌 \geq diam 𝜏,

which is sufficient to imply that

diam \tilde{𝜌} \geq diam 𝜏 - diam 𝜌 > diam 𝜏 - \frac{1}{2} diam 𝜏 = \frac{1}{2} diam 𝜏 > 𝜂 (𝛿) .

Thus, exactly one arc’s image (specifically, the image of the shorter arc on $𝑆^{1}$ ) has diameter $\leq 𝜂 (𝛿)$ . ⁠ $□$

By the uniqueness as described by Lemma 9.3.6 $Lemma 9.3.6 Lemma 9.3.6$ , for any two points $𝑎, 𝑏$ sufficiently close together, we now identify the unique curve in $𝜏$ (joining $𝑎$ and $𝑏$ ) with diameter less than $𝜂 (| 𝑎 - 𝑏 |)$ as the smaller arc of $𝜏$ joining $𝑎$ and $𝑏$ , denoted by $𝜏_{𝑎, 𝑏}$ .

Figure 9.8: Two possibilities for the endpoints of

𝜑 (𝛾_{𝑟_{𝑗}})

Lemma 9.3.7.

Let ${𝑟_{𝑗}}_{𝑗 \in ℕ}$ be the sequence such that $𝑙_{𝑟_{𝑗}} \to 0$ as $𝑗 \to \infty$ (in the discussion preceding Proposition 9.3.5 $Proposition 9.3.5 Proposition 9.3.5$ ). Then, for each $𝑗 \in ℕ$ , either

\begin{aligned} Ω_{2} ∋ 𝑎_{𝑗} & = lim_{𝜃 \to 𝜃_{0} {(𝑟_{𝑗})}^{-}} 𝜑 (1 - 𝑟_{𝑗} e^{i 𝜃}) \\ \neq lim_{𝜃 \to - 𝜃_{0} {(𝑟_{𝑗})}^{+}} 𝜑 (1 - 𝑟_{𝑗} e^{i 𝜃}) = 𝑏_{𝑗} \in 𝜕 Ω_{2} (9.3 . 1) \end{aligned}

lim_{𝜃 \to 𝜃_{0} {(𝑟_{𝑗})}^{-}} 𝜑 (1 - 𝑟_{𝑗} e^{i 𝜃}) = lim_{𝜃 \to - 𝜃_{0} {(𝑟_{𝑗})}^{+}} 𝜑 (1 - 𝑟_{𝑗} e^{i 𝜃}) = 𝑝_{𝑗} \in 𝜕 Ω_{2} (9.3 . 2)

is satisfied. The two situations are represented in Figure 9.8

Figure 9.8 Figure 9.8

Proof.

By virtue of Proposition 9.3.5

Proposition 9.3.5 Proposition 9.3.5

, both described limits exist. Similarly, since any sequences

{1 - 𝑟_{𝑗} e^{i 𝜃}}_{𝜃 \to 𝜃_{0} {(𝑟_{𝑗})}^{-}}

and

{1 - 𝑟_{𝑗} e^{i 𝜃}}_{𝜃 \to - 𝜃_{0} {(𝑟_{𝑗})}^{+}}

approach

𝜕 𝔻

, their images under

𝜑

approach

𝜕 Ω_{2}

by properness (refer to the discussions preceding Lemma 9.3.4

Lemma 9.3.4 Lemma 9.3.4

). Therefore, both limits must lie in

𝜕 Ω_{2}

, and the conclusion follows. ⁠

□

For fixed $𝑟_{𝑗}$ , in the case of (9.3.1) $(9.3.1) (9.3.1)$ , let $𝜑 (𝛾_{𝑟_{𝑗}})$ be the described curve joining $𝑎_{𝑗}$ and $𝑏_{𝑗}$ (where $𝛾_{𝑟_{𝑗}}$ is represented by the arc as labeled in Figure 9.7 $Figure 9.7 Figure 9.7$ ). Except for the two boundary points, the curve $𝜑 (𝛾_{𝑟_{𝑗}})$ lies entirely in $Ω_{2}$ since $𝛾_{𝑟_{𝑗}} \subset 𝑉 \subset 𝔻$ . Let the shorter arc on $Ω_{2}$ joining the two points $𝑎_{𝑗}$ and $𝑏_{𝑗}$ be denoted by $𝜏_{𝑗}$ . Therefore, the curve $𝜑 (𝛾_{𝑟_{𝑗}}) \cup 𝜏_{𝑗}$ forms a Jordan curve (simplicity by injectivity).

In the event that (9.3.2) $(9.3.2) (9.3.2)$ is assumed, let $𝜏_{𝑗} = {𝑝_{𝑗}}$ , and $𝜑 (𝛾_{𝑟_{𝑗}}) (\cup 𝜏_{𝑗})$ forms a Jordan curve. By the Jordan Curve Theorem (Theorem 1.2.5 $Theorem 1.2.5 Theorem 1.2.5$ ), the selected curve bounds a region, which we will denote by $𝑊_{𝑗} \subseteq Ω_{2}$ .

For each $𝑗$ , let $𝑉_{𝑗} = {1 - 𝑟 e^{i 𝜃} : 0 < 𝑟 < 𝑟_{𝑗}, - 𝜃_{0} (𝑟) < 𝜃 < 𝜃_{0} (𝑟)}$ . Since $𝜑 (𝛾_{𝑟_{𝑗}})$ is a curve with endpoints $𝑎_{𝑗}$ and $𝑏_{𝑗}$ (or $𝑝_{𝑗}$ ) and lies in $Ω_{2}$ except for its endpoints, it splits $Ω_{2}$ into two disconnected regions, $𝑊_{𝑗}$ or $Ω_{2} ∖ \overline{𝑊_{𝑗}}$ . Since $𝛾_{𝑟_{𝑗}}$ splits $𝔻$ into two regions, one of which being $𝑉_{𝑗}$ , the image $𝜑 (𝑉_{𝑗})$ is either $𝑊_{𝑗}$ or $Ω_{2} ∖ \overline{𝑊_{𝑗}}$ .

Proposition 9.3.8.

As $𝑗 \to \infty$ ,

diam 𝑊_{𝑗} \to 0 and area 𝑊_{𝑗} \to 0 .

Proof.

Lemma 9.3.6 $Lemma 9.3.6 Lemma 9.3.6$ provides the existence of $𝜂$ such that $diam 𝜏_{𝑗} \leq 𝜂 (𝑙_{𝑟_{𝑗}})$ for sufficiently large $𝑗$ (since $| 𝑎_{𝑗} - 𝑏_{𝑗} | \leq 𝑙_{𝑟_{𝑗}}$ or $0$ for the $𝑝_{𝑗}$ case) such that $𝜂 (𝑙_{𝑟_{𝑗}}) \to 0$ as $𝑗 \to \infty$ .

Let $𝐷_{𝑗} = 𝐷 (𝑎_{𝑗}, 𝑙_{𝑟_{𝑗}} + 𝜂 (𝑙_{𝑟_{𝑗}}))$ (or $𝐷_{𝑗} = 𝐷 (𝑝_{𝑗}, 𝑙_{𝑟_{𝑗}} + 𝜂 (𝑙_{𝑟_{𝑗}}))$ in the case of the same point) for each $𝑗$ . By the previous diameter estimate, the entire Jordan curve $𝜑 (𝛾_{𝑟_{𝑗}}) \cup 𝜏_{𝑗}$ lies in $𝐷_{𝑗}$ , and hence $𝑊_{𝑗} \subset 𝐷_{𝑗}$ . This implies that

area (𝑊_{𝑗}) \leq area (𝐷_{𝑗}) = π {(𝑙_{𝑟_{𝑗}} + 𝜂 (𝑙_{𝑟_{𝑗}}))}^{2} \to 0 as 𝑗 \to \infty . □ (9.3 . 3)

Lemma 9.3.9.

For sufficiently large

𝑗 \in ℕ

𝜑 (𝑉_{𝑗}) = 𝑊_{𝑗}

Proof.

For fixed $𝑗$ , let $𝑇_{𝑗}$ be the complementary region of $𝑉_{𝑗}$ in $𝔻$ (as $𝔻 ∖ \overline{𝑉_{𝑗}}$ ). Then for an arbitrary $𝑤_{0} \in 𝑊_{𝑗}$ , there exists a point $𝑧_{0} \in 𝔻$ (specifically, either in $𝑇_{𝑗}$ or $𝑉_{𝑗}$ ) such that $𝜑 (𝑧_{0}) = 𝑤_{0}$ .

If $𝑧_{0} \in 𝑉_{𝑗}$ , then $𝜑 (𝑧_{0}) \in 𝑊_{𝑗}$ and connectivity suffices to ensure that $𝜑 (𝑉_{𝑗}) = 𝑊_{𝑗}$ .
If $𝑧_{0} \in 𝑇_{𝑗}$ , then from $𝜑 (𝑧_{0}) \in 𝑊_{𝑗}$ , connectivity implies that $𝜑 (𝑇_{𝑗}) \subseteq 𝑊_{𝑗}$ . Now by the Lusin Area Theorem (Theorem 3.1.1 $Theorem 3.1.1 Theorem 3.1.1$ ),
$area (𝜑 (𝑇_{𝑗})) = area (Ω_{2}) - area (𝜑 (𝑉_{𝑗})) = area (Ω_{2}) - \iint_{𝑉_{𝑗}} | 𝜑^{'} (𝑧) | d 𝑥 d 𝑦 .$
Since $𝑉_{𝑗}$ shrinks to the empty set, $\iint_{𝑉_{𝑗}} | 𝜑^{'} (𝑧) | d 𝑥 d 𝑦 \to 0$ as $𝑗 \to \infty$ , and hence
$area (𝜑 (𝑇_{𝑗})) \to area (Ω_{2}) . (9.3 . 4)$
Proposition 9.3.8 $Proposition 9.3.8 Proposition 9.3.8$ implies that
$area (𝑊_{𝑗}) \to 0$
as $𝑗 \to \infty$ . Hence, for sufficiently large $𝑗$ , we cannot have $𝜑 (𝑇_{𝑗}) \subseteq 𝑊_{𝑗}$ (by the induced contradiction with (9.3.4) $(9.3.4) (9.3.4)$ ). Therefore, $𝜑 (𝑇_{𝑗}) = Ω_{2} ∖ \overline{𝑊_{𝑗}}$ and $𝜑 (𝑉_{𝑗}) = 𝑊_{𝑗}$ . $□$

Lemma 9.3.10.

If the two complex curves $𝜇_{1}, 𝜇_{2} : [0, 1] \to \overline{𝔻}$ map $1$ to $1$ such that $𝜇_{1} ([0, 1)), 𝜇_{2} ([0, 1)) \subset 𝔻$ , then

lim_{𝑡 \to 1^{-}} 𝜑 (𝜇_{1} (𝑡)) = lim_{𝑡 \to 1^{-}} 𝜑 (𝜇_{2} (𝑡))

exist and are equal.

Proof.

By Lemma 9.3.9 $Lemma 9.3.9 Lemma 9.3.9$ and Proposition 9.3.8 $Proposition 9.3.8 Proposition 9.3.8$ , $\exists 𝑗 \in ℕ$ such that $diam 𝑊_{𝑗} < 𝜀$ and $𝜑 (𝑉_{𝑗}) = 𝑊_{𝑗}$ . Since $𝜇_{1} (𝑡), 𝜇_{2} (𝑡) \to 1$ as $𝑡 \to 1^{-}$ , $\exists 𝑆 \in (0, 1)$ such that $\forall 𝑡 : 𝑆 < 𝑡 < 1$ , $| 𝜇_{1} (𝑡) - 1 | < 𝑟_{𝑗}$ and $| 𝜇_{2} (𝑡) - 1 | < 𝑟_{𝑗}$ . Recall the definition of $𝑉_{𝑗}$ as the set of points ${1 - 𝑟 e^{i 𝜃} : 0 < 𝑟 < 𝑟_{𝑗}, - 𝜃_{0} (𝑟) < 𝜃 < 𝜃_{0} (𝑟)}$ ; then $𝜇_{1} (𝑡), 𝜇_{2} (𝑡) \in 𝑉_{𝑗}$ for all $𝑆 < 𝑡 < 1$ . Therefore,

𝜑 (𝜇_{1} (𝑡)), 𝜑 (𝜇_{2} (𝑡)) \in 𝑊_{𝑗}

for all $𝑆 < 𝑡 < 1$ . Then

| 𝜑 (𝜇_{1} (𝑡)) - 𝜑 (𝜇_{2} (𝑡)) | \leq diam 𝑊_{𝑗} < 𝜀

for all

𝑆 < 𝑡 < 1

, thus equality is derived. Moreover,

⋂_{𝑗} \overline{𝑊_{𝑗}}

shrinks to a single point, which is precisely the prescribed limit. ⁠

□

Corollary 9.3.10.1.

There is a continuous extension of

𝜑

𝜕 𝔻

Proof.

Let $𝜁 \in 𝜕 𝔻$ . Let $𝛾 : [0, 1] \to \overline{𝔻}$ satisfy $𝜑 ([0, 1)) \subseteq 𝔻$ and $𝛾 (1) = 1$ . Suppose $𝜑_{𝜁} : 𝑧 \mapsto 𝜑 \circ (𝑧 𝜁)$ . Then by Lemma 9.3.10 $Lemma 9.3.10 Lemma 9.3.10$ , ${lim}_{𝑡 \to 1^{-}} 𝜑_{𝜁} (𝛾 (𝑡))$ exists and is independent of the choice of $𝛾$ . Now define $\tilde{𝜑} (𝜁)$ to be the unique value of this limit and let $\tilde{𝜑} |_{𝔻} = 𝜑$ .

Fix $𝜀 > 0$ and suppose ${𝑧_{𝑛}}_{𝑛 \in ℕ} \subset \overline{𝔻}$ be a sequence such that $𝑧_{𝑛} \to 1$ as $𝑛 \to \infty$ . By Lemma 9.3.9 $Lemma 9.3.9 Lemma 9.3.9$ and Proposition 9.3.8 $Proposition 9.3.8 Proposition 9.3.8$ , $\exists 𝑗 \in ℕ$ such that $diam 𝑊_{𝑗} < 𝜀$ and $𝜑 (𝑉_{𝑗}) = 𝑊_{𝑗}$ . By the geometry of $𝑉_{𝑗}$ , $\exists 𝑁 \in ℕ$ such that $\forall 𝑛 > 𝑁$ , there exists some $0 < 𝑡_{0} = 𝑡_{0} (𝑛) < 1$ such that the line segment $(𝑡_{0} 𝑧_{𝑛}, 𝑧_{𝑛})$ lies in $𝑉_{𝑗}$ . By the previous result, we have

𝜑 (𝑡 𝑧_{𝑛}) \in 𝑊_{𝑗} ⟹ \tilde{𝜑} (𝑧_{𝑛}) = lim_{𝑡 \to 1^{-}} 𝜑 (𝑡 𝑧_{𝑛}) \in \overline{𝑊_{𝑗}} .

Hence, any two $𝑧_{𝑛}, 𝑧_{𝑚}$ for $𝑛, 𝑚 > 𝑁$ satisfy $| \tilde{𝜑} (𝑧_{𝑛}) - \tilde{𝜑} (𝑧_{𝑚}) | \leq diam 𝑊_{𝑗} < 𝜀$ . Hence, ${\tilde{𝜑} (𝑧_{𝑛})}_{𝑛 \in ℕ}$ is Cauchy and hence convergent to $\tilde{𝜑} (1)$ .

This is sufficient to show that $\tilde{𝜑}$ is continuous at $1$ . Indeed, assuming otherwise implies $\exists 𝜀 > 0$ such that $\forall 𝛿 > 0$ , $\exists 𝑧_{𝛿} \in \overline{𝔻}$ within $𝛿$ apart from $1$ such that $| \tilde{𝜑} (𝑧_{𝛿}) - \tilde{𝜑} (1) | \geq 𝜀$ . Let $𝛿_{𝑛} = \frac{1}{𝑛}$ , and hence $𝛿_{𝑛} \to 0$ with the sequence ${𝑧_{𝛿_{𝑛}}}_{𝑛 \in ℕ}$ converging to $1$ . By the previous result, $\tilde{𝜑} (𝑧_{𝛿_{𝑛}}) \to \tilde{𝜑} (1)$ , which contradicts $| \tilde{𝜑} (𝑧_{𝛿_{𝑛}}) - \tilde{𝜑} (1) | \geq 𝜀$ for all $𝑛 \in ℕ$ . The same argument can be repeated to show that $\tilde{𝜑}$ is continuous at every point of $𝜕 𝔻$ . ⁠ $□$

Proposition 9.3.11.

Let

𝑓 \in 𝐶 (\overline{𝔻})

such that

𝑓

is holomorphic on

𝔻

. If there exists an open arc

𝛾 \subseteq 𝜕 𝔻

such that

𝑓 \equiv 𝑐

𝛾

(

𝑐

is some complex constant), then

𝑓

is constant on

𝔻

Proof.

Without loss of generality, assume $𝑐 = 0$ . If $𝛾 = 𝜕 𝔻$ , then by the continuity of $| 𝑓 |$ , it attains its maximum somewhere in $\overline{𝔻}$ (by the Extreme Value Theorem in Theorem 1.2.23 $Theorem 1.2.23 Theorem 1.2.23$ ). If $| 𝑓 |$ attains its maximum at some $𝑧_{0} \in 𝔻$ , it must be an isolated extrema (otherwise the Identity Theorem implies constancy on all of $𝔻$ ), and thus the Maximum Modulus Principle (Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ ) implies that $𝑓$ is constant and subsequently identically zero. If $| 𝑓 |$ attains its maximum on $𝜕 𝔻$ , the conclusion follows trivially.

Hence we may assume that $\exists 𝑧 \in 𝜕 𝔻$ such that $𝑓 (𝑧) \neq 0$ . Without loss of generality, we assume $𝑧 = - 1$ (generality by applying a rotation). Then the Möbius transform $𝜙 : 𝑧 \mapsto i \frac{1 - 𝑧}{1 + 𝑧}$ (in the inverse form of Example 5.1.7 $Example 5.1.7 Example 5.1.7$ ) maps $𝛾$ bijectively to a connected segment $Γ \subset ℝ$ . Then the function $𝑓 \circ 𝜙^{- 1}$ maps $Γ = 𝜙 (𝛾)$ to ${0}$ and is continuous on $ℍ^{+} \cup ℝ$ and holomorphic on $ℍ^{+}$ . Let $𝑈$ define the upper half-disk with diameter $Γ$ . By the Schwarz Reflection Principle (Theorem 5.5.2 $Theorem 5.5.2 Theorem 5.5.2$ ), $𝑓 \circ 𝜙^{- 1}$ may be analytically continued to a function $\tilde{𝑔}$ on $𝑈 \cup Γ \cup {\overline{𝑧} : 𝑧 \in 𝑈}$ . Since the zero set $Γ$ has an accumulation point in $𝑈 \cup Γ \cup {\overline{𝑧} : 𝑧 \in 𝑈}$ , the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ) implies that $𝑔$ vanishes on all of $𝑈 \cup Γ \cup {\overline{𝑧} : 𝑧 \in 𝑈} \to 𝑈 \to ℍ^{+}$ , or equivalently, that $𝑓$ vanishes on $𝔻$ . ⁠ $□$

Lemma 9.3.12.

The extension

\tilde{𝜑}

of the biholomorphism

𝜑

\overline{𝔻}

is injective.

Figure 9.9: Two line segments

𝛼

and

𝛽

mapping to a Jordan curve bounding

𝑊

Proof.

Notice that $\tilde{𝜑} (𝔻) = Ω_{2}$ , and from the earlier discussion regarding properness, $\tilde{𝜑} (𝑆^{1}) \subseteq 𝜕 Ω_{2}$ . By the biholomorphy on $𝔻$ , it thus suffices to show that $\tilde{𝜑} |_{𝜕 𝔻}$ is one-to-one, or that for any two points $𝜁, 𝜁^{'} \in 𝜕 𝔻$ such that $\tilde{𝜑} (𝜁) = \tilde{𝜑} (𝜁^{'})$ , $𝜁 = 𝜁^{'}$ . Assume, for the sake of contradiction, that $𝜁 \neq 𝜁^{'}$ . The straight line segment connecting $0$ to $𝜁$ (denoted $𝛼$ ), and the straight line segment joining $0$ and $𝜁^{'}$ (denoted $𝛽$ ) then split $𝔻$ into two domains $𝑈_{1}$ and $𝑈_{2}$ , and $\tilde{𝜑} (𝛼) \cup \tilde{𝜑} (𝛽)$ forms a Jordan curve enclosing some region $𝑊$ . See Figure 9.9 $Figure 9.9 Figure 9.9$ . By connectivity, either $𝑈_{1}$ or $𝑈_{2}$ maps to $𝑊$ . Without loss of generality, assume $\tilde{𝜑} (𝑈_{1}) = 𝑊$ . Since $\tilde{𝜑}$ is continuous on $\overline{𝑈_{1}}$ , it follows that

\tilde{𝜑} (𝜕 𝔻 \cap 𝜕 𝑈_{1}) \subseteq 𝜕 Ω_{2} \cap 𝜕 𝑊 = {\tilde{𝜑} (𝜁)} = {\tilde{𝜑} (𝜁^{'})} .

Then the arc of $𝜕 𝔻$ in $𝜕 𝑈_{1}$ maps to a constant, which by Proposition 9.3.11 $Proposition 9.3.11 Proposition 9.3.11$ , implies that $\tilde{𝜑}$ is constant, which is an impossibility. (The same argument is used for $𝑈_{2}$ )

Therefore, by the induced contradiction, we must have $𝜁 = 𝜁^{'}$ , which implies the injectivity of $\tilde{𝜑}$ . ⁠ $□$

Thus, the preceding results gives the construction of an injective, continuous extension of $𝜑$ to $\overline{𝔻}$ . Moreover, the extension is onto since a continuous function maps compact sets to compact sets.

Next, we show that the extension is a homeomorphism.

Proposition 9.3.13.

The extension

\tilde{𝜑}

of the biholomorphism

𝜑

\overline{𝔻}

is a homeomorphism.

Proof.

It suffices to show that the restriction

\tilde{𝜑} |_{𝜕 𝔻}

is a homeomorphism between

𝜕 𝔻

and

𝜕 Ω_{1}

. Since

𝜕 𝔻

is compact (and one under the subspace topology, a compact space) and

𝜕 Ω_{1}

is trivially Hausdorff (under the subspace topology), the continuity and injectivity of

\tilde{𝜑} |_{𝜕 𝔻}

suffices to show that it is a homeomorphism (by Theorem 3.2.1.16

Theorem 3.2.1.16 Theorem 3.2.1.16

). ⁠

□

Proof (Alternate).

We show that the inverse of $\tilde{𝜑} |_{𝜕 𝔻}$ is continuous, which also suffices to show that $\tilde{𝜑} |_{𝜕 𝔻}$ is a homeomorphism.

By the final argument of the proof of Corollary 9.3.10.1 $Corollary 9.3.10.1 Corollary 9.3.10.1$ , it suffices to show that any sequence ${𝑧_{𝑛}}_{𝑛 \in ℕ} \subset 𝜕 Ω_{2}$ that converges to a point $𝑧 \in 𝜕 Ω_{2}$ has a corresponding sequence ${𝑤_{𝑛}}_{𝑛 \in ℕ} = {{(\tilde{𝜑} |_{𝜕 𝔻})}^{- 1} (𝑧_{𝑛})}_{𝑛 \in ℕ}$ in $𝜕 𝔻$ which converges to $𝑤 = {(\tilde{𝜑} |_{𝜕 𝔻})}^{- 1} (𝑧)$ . Assume, for contradiction, that some sequence as labeled above does not converge to $𝑤$ . By the Bolzano–Weierstrass Theorem (Theorem 1.1.10 $Theorem 1.1.10 Theorem 1.1.10$ ), some subsequence of ${𝑤_{𝑛}}_{𝑛 \in ℕ}$ , denoted by ${𝑤_{𝑛_{𝑘}}}_{𝑘 \in ℕ}$ , converges to $𝑤_{\infty} \in 𝜕 𝔻$ (which is not equal to $𝑤$ ). Then $\tilde{𝜑} (𝑤_{𝑛_{𝑘}}) \to \tilde{𝜑} (𝑤_{\infty})$ as $𝑘 \to \infty$ by the continuity of $(\tilde{𝜑})$ . However, each $\tilde{𝜑} (𝑤_{𝑛_{𝑘}}) = 𝑧_{𝑛_{𝑘}} \to 𝑧 = \tilde{𝜑} (𝑤)$ . Hence, $\tilde{𝜑} (𝑤) = \tilde{𝜑} (𝑤_{\infty})$ which by injectivity, implies $𝑤 = 𝑤_{\infty}$ , which is a contradiction. Thus $\tilde{𝜑}$ establishes a homeomorphism between $\overline{𝔻}$ and $\overline{Ω_{1}}$ . ⁠ $□$

Now consider the general case for arbitrary $Ω_{1}$ as in the theorem statement. Let $𝜑 : Ω_{1} \to Ω_{2}$ be a biholomorphism.

By the Riemann Mapping Theorem (Theorem 5.3.2 $Theorem 5.3.2 Theorem 5.3.2$ ), there exist biholomorphisms $𝜑_{1} : 𝔻 \to Ω_{1}$ and $𝜑_{2} : 𝔻 \to Ω_{2}$ , which extend homeomorphically to ${\tilde{𝜑}}_{1}$ and ${\tilde{𝜑}}_{2}$ , respectively, by the preceding results.

Thus, $𝜓 = 𝜑_{2}^{- 1} \circ 𝜑 \circ 𝜑_{1} \in Aut (𝔻)$ , which by Theorem 3.5.3 $Theorem 3.5.3 Theorem 3.5.3$ , is the composition of a Möbius transformation and a rotation, which extends homeomorphically to some function $\tilde{𝜓}$ on $\overline{𝔻}$ . Thus, we define $\tilde{𝜑} = {\tilde{𝜑}}_{2} \circ \tilde{𝜓} \circ {\tilde{𝜑}}_{1}^{- 1}$ to be the composition of three continuous functions, defining a continuous extension of $𝜑$ , completing the proof. ⁠ $□$

The second case pertaining to $𝐶^{\infty}$ boundaries will be proved later. We now entertain the third.

Theorem 9.3.14.

Let

Ω_{1}

and

Ω_{2}

be two regions in

ℂ

each bounded by a single real-analytic Jordan curve. Then a biholomorphism

𝜑 : Ω_{1} \to Ω_{2}

can be analytically continued to a neighborhood of

\overline{Ω_{1}}

Figure 9.10: The construction of

𝑁_{2}

𝑉_{𝑞}

𝜂_{2}

, and

𝜓

, and similarly, that of

𝑁_{1}

𝑉_{𝑝}

𝜂_{1}

, and

𝜙

Proof.

By the Osgood–Taylor–Carathéodory Theorem (Theorem 9.3.3 $Theorem 9.3.3 Theorem 9.3.3$ ), $𝜑$ extends continuously to $\overline{Ω_{1}}$ with $𝜑 (𝜕 Ω_{1}) = 𝜕 Ω_{2}$ ; retain the name $𝜑$ for this extension.

Fix $𝑝 \in 𝜕 Ω_{1}$ and set $𝑞 = 𝜑 (𝑝) \in 𝜕 Ω_{2}$ . Since $𝜕 Ω_{2}$ is real-analytic, the function $𝜂_{2}$ defined by a locally injective power series on a subset of the real axis extends (by power series) to a biholomorphism $𝜂_{2} : 𝑁_{2} \to 𝑉_{𝑞}$ from a region $𝑁_{2} ∋ 0$ onto a neighborhood $𝑉_{𝑞} ∋ 𝑞$ , with $𝜂_{2} (0) = 𝑞$ and $𝜂_{2} (𝑁_{2} \cap ℝ) = 𝑉_{𝑞} \cap 𝜕 Ω_{2}$ (existence by real-analyticity and biholomorphy by Theorem 5.1.3 $Theorem 5.1.3 Theorem 5.1.3$ ). Set $𝜓 = 𝜂_{2}^{- 1} : 𝑉_{𝑞} \to 𝑁_{2}$ .

Since $𝜑$ is continuous at $𝑝$ and $𝜑 (𝑝) = 𝑞 \in 𝑉_{𝑞}$ , the set $𝜑^{- 1} (𝑉_{𝑞})$ is an open neighborhood of $𝑝$ . Since $𝜕 Ω_{1}$ is real-analytic, choose a biholomorphism $𝜂_{1} : 𝑁_{1} \to 𝑉_{𝑝}$ from a region $𝑁_{1} ∋ 0$ onto a neighborhood $𝑉_{𝑝} ∋ 𝑝$ contained in $𝜑^{- 1} (𝑉_{𝑞})$ , with $𝜂_{1} (0) = 𝑝$ and $𝜂_{1} (𝑁_{1} \cap ℝ) = 𝑉_{𝑝} \cap 𝜕 Ω_{1}$ . Set $𝜙 = 𝜂_{1}^{- 1} : 𝑉_{𝑝} \to 𝑁_{1}$ . Denote by $𝑁_{1}^{+}$ the component of $𝑁_{1} ∖ ℝ$ whose image under $𝜂_{1}$ is $𝑉_{𝑝} \cap Ω_{1}$ . See Figure 9.10 $Figure 9.10 Figure 9.10$ .

By construction $𝜑 (𝑉_{𝑝} \cap \overline{Ω_{1}}) \subseteq 𝑉_{𝑞}$ , so the composition $𝐹 = 𝜓 \circ 𝜑 \circ 𝜂_{1}$ is defined on

𝑁_{1}^{+} \cup (𝑁_{1} \cap ℝ) . (the striped region with the thick line in Figure 9.10 Figure 9.10 Figure 9.10)

On $𝑁_{1}^{+}$ , $𝜂_{1}$ maps to $𝑉_{𝑝} \cap Ω_{1}$ on which $𝜑$ is holomorphic, and $𝜑 \circ 𝜂_{1}$ maps to a subset of $𝑉_{𝑞} \cap Ω_{2}$ , on which $𝜓$ is holomorphic, so $𝐹$ is holomorphic on $𝑁_{1}^{+}$ (the regions mapped from here are visualized in Figure 9.10 $Figure 9.10 Figure 9.10$ with striped lines). By Osgood–Taylor–Carathéodory (Theorem 9.3.3 $Theorem 9.3.3 Theorem 9.3.3$ ), $𝐹$ is continuous up to $𝑁_{1} \cap ℝ$ with

𝐹 (𝑁_{1} \cap ℝ) = 𝜓 (𝜑 (𝜂_{1} (𝑁_{1} \cap ℝ))) \subseteq 𝜓 (𝑉_{𝑞} \cap 𝜕 Ω_{2}) = 𝑁_{2} \cap ℝ .

Since $𝜓 (𝑉_{𝑞} \cap Ω_{2})$ is connected and disjoint from $ℝ$ , the image $𝐹 (𝑁_{1}^{+}) \subset 𝜓 (𝑉_{𝑞} \cap Ω_{2})$ lies entirely in one component of $ℂ ∖ ℝ$ ; without loss of generality $𝐹 (𝑁_{1}^{+}) \subset ℍ^{+}$ .

Thus $𝐹 : 𝑁_{1}^{+} \to ℍ^{+}$ is univalent, continuous up to $𝑁_{1} \cap ℝ$ , and real-valued there. The Schwarz Reflection Principle (Theorem 5.5.2 $Theorem 5.5.2 Theorem 5.5.2$ ) extends $𝐹$ biholomorphically to

𝑈 = 𝑁_{1}^{+} \cup (𝑁_{1} \cap ℝ) \cup {\overline{𝑧} : 𝑧 \in 𝑁_{1}^{+}}

an open neighborhood of $0$ , and maps to

𝑉 = 𝐹 (𝑁_{1}^{+}) \cup 𝐹 (𝑁_{1} \cap ℝ) \cup {\overline{𝑧} : 𝑧 \in 𝐹 (𝑁_{1}^{+})} ∋ 0 .

Setting

𝑊_{𝑝} = 𝜂_{1} (𝑈 \cap 𝐹^{- 1} (𝑉 \cap 𝑁_{2}) \cap 𝑁_{1})

(the $\cdot \cap 𝑁_{1}$ is added here to ensure that the region in the parenthesis is in the domain of $𝜂_{1}$ ), an open neighborhood of $𝑝$ , the map

𝜑_{𝑝} = 𝜂_{2} \circ 𝐹 \circ 𝜙 : 𝑊_{𝑝} \to ℂ

is holomorphic and agrees with $𝜑$ on $𝑊_{𝑝} \cap Ω_{1}$ . To justify the domain restrictions, observe that applying $𝜙$ on $𝑊_{𝑝}$ gives $𝑈 \cap 𝐹^{- 1} (𝑉 \cap 𝑁_{2}) \cap 𝑁_{1}$ , and $𝐹$ is defined thereon (since it is a subset of $𝑈$ ), and maps to a subset of $𝑉 \cap 𝑁_{2}$ , on which $𝜂$ is defined.

The main observation here is that each composition will map open sets to open sets and vice versa (although we are being a bit pedantic).

Applying this construction at every $𝑧 \in 𝜕 Ω_{1}$ yields an open neighborhood $𝑊_{𝑧} ∋ 𝑧$ and a holomorphic map $𝜑_{𝑧}$ extending $𝜑 |_{\overline{Ω_{1}}}$ .

The collection of all $𝑊_{𝑧}$ ’s form an open cover of $𝜕 Ω_{1}$ ; thus the Heine–Borel Theorem (Theorem 1.1.9 $Theorem 1.1.9 Theorem 1.1.9$ ) gives the existence of a finite subcover $𝒲︀ = {𝑊_{𝑧_{𝑘}}}_{𝑘}$ where each $𝑧_{𝑘} \in 𝜕 Ω_{1}$ .

For two sets $𝑊_{1}, 𝑊_{2} \in 𝒲︀$ where $𝑊_{1} \cap 𝑊_{2} \neq ⌀$ , both $𝜑_{𝑧_{1}}$ and $𝜑_{𝑧_{2}}$ agree with $𝜑$ on the nonempty open connected set $𝑊_{𝑧_{1}} \cap 𝑊_{𝑧_{2}} \cap Ω_{1}$ , so the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ) gives $𝜑_{𝑧_{1}} \equiv 𝜑_{𝑧_{2}}$ on $𝑊_{𝑧_{1}} \cap 𝑊_{𝑧_{2}}$ .

The local extensions are therefore mutually consistent and, together with $𝜑 |_{Ω_{1}}$ , define a holomorphic extension of $𝜑$ to the open neighborhood $Ω_{1} \cup ⋃_{𝑊 \in 𝒲︀} 𝑊$ . ⁠ $□$