§5.3 The Riemann Mapping Theorem

5.3 The Riemann Mapping Theorem

The Riemann Mapping Theorem is one of the most profound results in complex analysis; in the case of one dimension, it establishes sufficient conditions for the biholomorphic equivalence between two open subsets of the complex plane.

If there exists a biholomorphism $𝑓$ between two regions, then the two regions are said to be conformally equivalent, holomorphically equivalent, or biholomorphically equivalent. As a required intermediate for the proof, we first introduce:

Definition 5.3.1 (Holomorphic Logarithms).

Suppose $Φ : 𝑈 \to ℂ^{*} = ℂ ∖ {0}$ is holomorphic, where $𝑈$ is simply connected. Define the holomorphic logarithm of $Φ (𝑧)$ to be a branch of

log (Φ (𝑧)) = \int_{𝛾} \frac{Φ^{'} (𝜁)}{Φ (𝜁)} d 𝜁 + log (Φ (𝑧_{0}))

for any $𝑧_{0} \in 𝑈$ , where the integral is path-independent $𝛾 \subset 𝑈$ is any piecewise $𝐶^{1}$ curve from $𝑧_{0}$ to $𝑧$ .

Similarly, define the holomorphic powers of $Φ (𝑧)$ to be branches of $(Φ^{𝛼}) (𝑧) = e^{𝛼 log (Φ (𝑧))}$ , where $log (Φ (𝑧))$ is the holomorphic logarithm.

The path independence of the definition is provided by the simple connectivity of $𝑈$ . The result is the heuristic concatenation of several different branches of the complex logarithm, unique up to an additive factor of $2 π i 𝑘$ , where this additive factor is the same throughout.

Theorem 5.3.2 (Riemann Mapping Theorem).

Let

𝑈 \subset ℂ

(a proper subset, in other words,

𝑈 \neq ℂ

) be a simply connected (nonempty) open region. Let

𝑧_{0} \in 𝑈

be arbitrary. Then there exists a unique biholomorphism

𝑓 : 𝑈 \to 𝔻

such that

𝑓 (𝑧_{0}) = 0

and

𝑓^{'} (𝑧_{0}) \in ℝ_{> 0}

Proof.

First consider the case for when $𝑈$ is a bounded region. In other words, $\exists 𝑅 > 0$ such that $𝑈 \subseteq 𝐷 (0, 𝑅)$ .

Define $ℱ︀$ to be the family of all univalent functions $𝛼 : 𝑈 \to 𝔻$ (not necessarily surjective) such that $𝛼 (𝑧_{0}) = 0$ . This family is well-defined and nonempty.

To prove this assertion, observe that since $𝑧_{0} \in 𝐷 (0, 𝑅)$ , it follows that $\forall 𝑧 \in 𝑈 \subseteq 𝐷 (0, 𝑅)$ , $| 𝑧 - 𝑧_{0} | < 2 𝑅$ , and consequently, $| \frac{𝑧 - 𝑧_{0}}{2 𝑅} | < 1$ . Therefore,

𝛼 (𝑧) = \frac{𝑧 - 𝑧_{0}}{2 𝑅}

maps $𝑈$ to $𝔻$ , and it is complex affine and univalent. This shows that $𝛼 \in ℱ︀$ . It is easy to prove that $ℱ︀$ is infinite; any function in the form of $𝑧 \mapsto \frac{𝑧 - 𝑧_{0}}{𝜁}$ for $𝜁 \geq 2 𝑅$ also lies in $ℱ︀$ .

Since $ℱ︀$ is uniformly bounded on $𝑈$ , by Montel’s Theorem (Theorem 5.2.6 $Theorem 5.2.6 Theorem 5.2.6$ ), $ℱ︀$ is a normal family. Let $𝑟 > 0$ satisfy $\overline{𝐷 (𝑧_{0}, 𝑟)} \subset 𝑈$ . Then by Cauchy’s Estimate (Theorem 3.2.3 $Theorem 3.2.3 Theorem 3.2.3$ ), $\forall 𝛼 \in ℱ︀$ , $| 𝛼^{'} | \leq \frac{1}{𝑟}$ on $\overline{𝐷 (𝑧_{0}, 𝑟)}$ . Hence, we have

0 < 𝑀 = sup_{𝛼 \in ℱ︀} | 𝛼^{'} (𝑧_{0}) | \leq \frac{1}{𝑟}, (5.3 . 1)

where we can assure that $𝑀$ is positive since each $𝛼 \in ℱ︀$ is univalent at $𝑧_{0}$ and by Lemma 5.1.2 $Lemma 5.1.2 Lemma 5.1.2$ .

If $𝑀$ is an accumulation point of ${| 𝛼^{'} (𝑧_{0}) |}_{𝛼 \in ℱ︀}$ , there exists a sequence ${𝛼_{𝑛}}_{𝑛 \in ℕ} \subseteq ℱ︀$ such that ${| 𝛼_{𝑛}^{'} (𝑧_{0}) |}_{𝑛 \in ℕ}$ converges to $𝑀$ . If $𝑀$ is attained as a maximum or that $| 𝛼^{'} (𝑧_{0}) | = 𝑀$ for some $𝛼 \in ℱ︀$ , we may let each $𝛼_{𝑛} \equiv 𝛼$ .

By the normality of $ℱ︀$ , there exists a subsequence ${𝛼_{𝑛_{𝑘}} (𝑧)}_{𝑘 \in ℕ} \subseteq {𝛼_{𝑛} (𝑧)}_{𝑛 \in ℕ}$ such that ${𝛼_{𝑛_{𝑘}} (𝑧)}_{𝑘 \in ℕ}$ is locally uniformly convergent in $𝑈$ to a function $\tilde{\tilde{𝛼}} (𝑧)$ (holomorphy of which is provided by Theorem 4.1.1 $Theorem 4.1.1 Theorem 4.1.1$ ). By definition, $| {\tilde{\tilde{𝛼}}}^{'} (𝑧_{0}) | = 𝑀$ , and define a function sequence with ${\tilde{𝛼}}_{𝑛_{𝑘}} = 𝛼_{𝑛_{𝑘}} \frac{| {\tilde{\tilde{𝛼}}}^{'} (𝑧_{0}) |}{{\tilde{\tilde{𝛼}}}^{'} (𝑧_{0})} \in ℱ︀$ , whose locally uniform limit is $𝑓$ . It follows that $𝑓$ is a rotation of $\tilde{\tilde{𝛼}}$ such that $𝑓^{'} (𝑧_{0}) = 𝑀$ .

Let $𝜁_{1}, 𝜁_{2} \in 𝑈$ be arbitrary and different. Choose $𝑟^{'} > 0$ to satisfy $0 < 𝑟^{'} < | 𝜁_{1} - 𝜁_{2} |$ , and let $𝜓_{𝑘} (𝑧) = {\tilde{𝛼}}_{𝑛_{𝑘}} (𝑧) - {\tilde{𝛼}}_{𝑛_{𝑘}} (𝜁_{2})$ . Since each ${\tilde{𝛼}}_{𝑛_{𝑘}}$ is univalent in $𝑈$ , it follows that each $𝜓_{𝑘}$ is non-vanishing in $𝑈 ∖ {𝜁_{2}}$ and consequently, in $\overline{𝐷 (𝜁_{1}, 𝑟^{'})}$ . By Theorem 3.3.5 $Theorem 3.3.5 Theorem 3.3.5$ , it follows that the locally uniform limit of $𝜓_{𝑘}$ , or $𝜓 = 𝑓 (𝑧) - 𝑓 (𝜁_{2})$ , is either non-vanishing or is identically zero in $\overline{𝐷 (𝜁_{1}, 𝑟^{'})}$ . The latter is an impossibility since $𝜓^{'} (𝑧_{0}) = 𝑀 > 0$ . Hence, $𝑓 (𝑧) = 𝑓 (𝜁_{2})$ has no solutions for $𝑧 \in \overline{𝐷 (𝜁_{1}, 𝑟^{'})}$ . In particular, $𝑓 (𝜁_{1}) \neq 𝑓 (𝜁_{2})$ . By the arbitrariness of $𝜁_{1}$ and $𝜁_{2}$ , the univalence of $𝑓$ follows.

Additionally, since $\forall 𝑘 \in ℕ$ , $| {\tilde{𝛼}}_{𝑛_{𝑘}} | < 1$ in $𝑈$ , it follows that $𝑓 (𝑈) \subseteq \overline{𝔻}$ . By the Open Mapping Theorem (Theorem 5.1.1 $Theorem 5.1.1 Theorem 5.1.1$ ), the condition becomes $𝑓 (𝑈) \subseteq 𝔻$ . Since ${\tilde{𝛼}}_{𝑛_{𝑘}} (𝑧_{0}) = 0$ for all $𝑘 \in ℕ$ and ${\tilde{𝛼}}_{𝑛_{𝑘}} (𝑧_{0}) \to 0 = 𝑓 (𝑧_{0})$ , it follows that $𝑓 \in ℱ︀$ .

Lastly, we aim to prove that $𝑓$ maps $𝑈$ to $𝔻$ surjectively. For the sake of contradiction, assume that $\exists 𝜉 \in 𝔻^{*}$ such that $𝜉 \notin 𝑓 (𝑈)$ . Consider the unit disk automorphism $𝜑_{𝜉} (𝑧) = \frac{𝑧 - 𝜉}{1 - 𝑧 \overline{𝜉}}$ . Since $𝜑_{𝜉} (𝑧)$ vanishes when $𝑧 = 𝜉$ , and since $𝑓 (𝑧) = 𝜉$ has no solutions in $𝑈$ , there exists a holomorphic square root

𝜇 (𝑧) = \sqrt{𝜑_{𝜉} \circ 𝑓 (𝑧)} \in 𝔻

for $𝑧 \in 𝑈$ . Let $𝜏 = 𝜇 (𝑧_{0}) = \sqrt{- 𝜉}$ , and let

𝜂 (𝑧) = 𝜑_{𝜏} \circ 𝜇 (𝑧),

where $𝜑_{𝜏} = \frac{𝑧 - 𝜏}{1 - 𝑧 \overline{𝜏}}$ . Since $𝜂 (𝑧_{0}) = 𝜑_{𝜏} (𝜏) = 0$ , it follows that $𝜂 \in ℱ︀$ . Let $\tilde{𝜂} = \frac{| 𝜂^{'} (𝑧_{0}) |}{𝜂^{'} (𝑧_{0})} 𝜂$ , which is also in $ℱ︀$ . However, since ${\tilde{𝜂}}^{'} = \frac{| 𝜂^{'} (𝑧_{0}) |}{𝜂^{'} (𝑧_{0})} 𝜂^{'}$ , we have

\begin{aligned} {\tilde{𝜂}}^{'} (𝑧_{0}) & = | \frac{𝑓^{'} (𝑧_{0}) (𝜑_{𝜏}^{'} \circ 𝜏) (𝜑_{𝜉}^{'} \circ 0)}{2 \sqrt{𝜑_{𝜉} \circ 0}} | \\ = \frac{𝑀}{2 \sqrt{| 𝜉 |}} \frac{1 - \overline{𝜏} 𝜏}{{(1 - 𝜏 \overline{𝜏})}^{2}} \frac{1 - \overline{𝜉} 𝜉}{{(1 - 0 \overline{𝜉})}^{2}} \\ = \frac{𝑀}{2 \sqrt{| 𝜉 |}} \frac{1 - {| 𝜉 |}^{2}}{1 - | 𝜉 |} = \frac{𝑀 (1 + | 𝜉 |)}{2 \sqrt{| 𝜉 |}} . \end{aligned}

Additionally, since

{(\sqrt{| 𝜉 |} - 1)}^{2} > 0 ⟺ 1 + | 𝜉 | > 2 \sqrt{| 𝜉 |} ⟺ \frac{1 + | 𝜉 |}{2 \sqrt{| 𝜉 |}} > 1,

it follows that ${\tilde{𝜂}}^{'} (𝑧_{0}) > 𝑀$ , which is a contradiction of (5.3.1) $(5.3.1) (5.3.1)$ .

Hence, $𝑓 : 𝑈 \to 𝔻$ is biholomorphic. To prove the uniqueness of $𝑓$ , suppose $𝑔 : 𝑈 \to 𝔻$ is an arbitrary biholomorphism such that $𝑔 (𝑧_{0}) = 0$ and $𝑔^{'} (𝑧_{0}) > 0$ . Then, $𝜑 = 𝑓 \circ 𝑔^{- 1} \in Aut (𝔻)$ , and by Theorem 3.5.3 $Theorem 3.5.3 Theorem 3.5.3$ , $𝜑 (𝑧) = 𝜑_{𝑎} (𝑧 e^{i 𝜃})$ for some $𝑎 \in 𝔻$ and $0 \leq 𝜃 < 2 π$ . Since $𝜑 (0) = 0$ , it follows that $𝑎 = 0$ . Since $𝜑^{'} (0) = 𝑓^{'} (𝑧_{0}) {(𝑔^{- 1})}^{'} (0) = 𝑓^{'} \frac{𝑧_{0}}{𝑔^{'} (𝑧_{0})} > 0$ , and $𝜑^{'} (0) = 𝜑_{0}^{'} (0) e^{i 𝜃} = e^{i 𝜃} > 0$ , it follows that $𝜃 = 0$ . Hence, we have $𝜑 (𝑧) = 𝑧$ and $𝑓 \equiv 𝑔$ .

Next, assume that $𝑈$ is unbounded. It is easy to show that the boundary $𝜕 𝑈$ contains at least two points. Indeed, if $𝜕 𝑈 = ⌀$ , $𝑈$ would be closed because $𝜕 𝑈 \subseteq 𝑈$ and open by assumption. By Theorem 3.2.1.19 $Theorem 3.2.1.19 Theorem 3.2.1.19$ , $𝑈$ would either be equal to $⌀$ or $ℂ$ , both of which are impossibilities. Additionally, if $𝜕 𝑈$ comprises exactly one point $𝑎 \in ℂ$ , then in subspace defined by $𝑋 = ℂ ∖ {𝑎}$ , $𝑈$ is clopen (by the same reason as before, open by assumption and closed because $𝑋 ∖ 𝑈 = ℂ ∖ \overline{𝑈}$ is open). It follows that $𝑈 = 𝑋 = ℂ ∖ {𝑎}$ , which is not simply connected.

Suppose $𝜉_{1}$ and $𝜉_{2}$ are two distinct points in $𝜕 𝑈$ . Let us apply the linear transformation $𝜌 (𝑧) = \frac{𝑧 - 𝜉_{1}}{𝜉_{2} - 𝜉_{1}}$ to $𝑈$ , and denote the resulting region by $𝑈^{'} = 𝜌 (𝑈)$ . It follows that $0, 1 \in 𝜕 𝑈^{'}$ . Consider a branch $𝜓 (𝑧)$ of the holomorphic square root $𝑧 \mapsto \sqrt{𝑧 - 1}$ (existent by the holomorphic logarithm from simple connectivity and the fact that $1 \notin 𝑈^{'}$ ). Trivially, $𝜓$ is univalent in $𝑈^{'}$ .

In addition, we assert that $𝜓 (𝑈^{'}) \cap (- 𝜓 (𝑈^{'})) = ⌀$ . If not, then $\exists 𝜉 \in 𝜓 (𝑈^{'})$ such that $- 𝜉 \in 𝜓 (𝑈^{'})$ . By definition, $\exists 𝑧_{1}, 𝑧_{2} \in 𝑈^{'}$ such that $𝜓 (𝑧_{1}) = 𝜉$ and $𝜓 (𝑧_{2}) = - 𝜉$ . It would then follow that $\sqrt{𝑧_{1} - 1} = - \sqrt{𝑧_{2} - 1}$ and $𝑧_{1} = 𝑧_{2}$ . It follows that $𝜉 = 0$ , which is obtained when $𝑧_{1} = 𝑧_{2} = 𝜓^{- 1} (𝜉) = 1$ . Since $1 \in 𝜕 𝑈^{'}$ and $𝑈^{'}$ is open, $1 \notin 𝑈^{'}$ , which is an impossibility.

Fix $𝜉 \in 𝜓 (𝑈^{'})$ to be arbitrary. By the Open Mapping Theorem (Theorem 5.1.1 $Theorem 5.1.1 Theorem 5.1.1$ ), there exists an open neighborhood $𝐷 (𝜉, 𝜀) \subseteq 𝜓 (𝑈^{'})$ . It follows that $𝐷 (- 𝜉, 𝜀) \cap 𝜓 (𝑈^{'}) = ⌀$ . Therefore, $\forall 𝑧 \in 𝑈^{'}$ , $| 𝜓 (𝑧) + 𝜉 | \geq 𝜀$ , and consequently, $| \frac{1}{𝜓 (𝑧) + 𝜉} | \leq \frac{1}{𝜀}$ . Hence, the function $𝜑 (𝑧) = \frac{1}{𝑧 + 𝜉}$ maps $𝑈^{'}$ to a bounded region that lies within the compact disk $\overline{𝐷 (0, \frac{1}{𝜀})}$ . Denote $𝜑 \circ 𝜓 (𝑈^{'})$ by $\tilde{𝑈}$ .

It is easy to see that $\tilde{𝑈}$ is simply connected. Let $\tilde{𝑈} = 𝜑 \circ 𝜓 \circ 𝜌 (𝑈)$ . To prove this, it suffices to show that the line integral of any holomorphic function over any closed curve in $\tilde{𝑈}$ vanishes. Let $𝑔 : \tilde{𝑈} \to ℂ$ be holomorphic, and let $Γ \subset \tilde{𝑈}$ be a closed piecewise $𝐶^{1}$ curve. Then

\oint_{Γ} 𝑔 (𝑤) d 𝑤 = \oint_{𝜌^{- 1} \circ 𝜓^{- 1} \circ 𝜑^{- 1} (Γ)} 𝑔 \circ 𝜑 \circ 𝜓 \circ 𝜌 (𝑧) \cdot {(𝜑 \circ 𝜓 \circ 𝜌)}^{'} (𝑧) d 𝑧 = 0,

by Cauchy–Goursat (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ), since $𝑈$ is simply connected by assumption, $𝜌^{- 1} \circ 𝜓^{- 1} \circ 𝜑^{- 1} (Γ)$ is a closed piecewise $𝐶^{1}$ curve in $𝑈$ , and the integrand is holomorphic on $𝑈$ . Therefore, $\tilde{𝑈}$ is simply connected.

Hence, we may use our previous result and establish a biholomorphism $\tilde{𝑓} : \tilde{𝑈} \to 𝔻$ , unique up to a transformation in $Aut (𝔻)$ . Let $𝑓 = \tilde{𝑓} \circ 𝜑 \circ 𝜓 \circ 𝜌$ , which is a biholomorphism from $𝑈$ to $𝔻$ . Similarly, it is unique up to a transformation in $Aut (𝔻)$ , and the same assertion follows. ⁠ $□$

Remark.

It is natural that we require $𝑈 \neq ℂ$ ; if there exists a univalent function $𝑓 : ℂ \to 𝔻$ , then by Liouville’s Theorem (Theorem 3.2.5 $Theorem 3.2.5 Theorem 3.2.5$ ), $𝑓$ would be a constant function.

As we will see in Section 11 $Section 11 Section 11$ , this theorem and many other properties of one-variable holomorphic functions do not extend to functions of several complex variables.