State the Riemann mapping theorem and prove the uniqueness part.
Solution
Statement of the Riemann Mapping Theorem
Theorem (Riemann Mapping Theorem): Let \(\Omega \subsetneq \mathbb{C}\) be a non-empty simply connected domain (open connected subset), and let \(z_0 \in \Omega\) be an arbitrary fixed point. Then there exists a unique biholomorphic (conformal bijection) mapping \(f: \Omega \to \mathbb{D}\) (where \(\mathbb{D} = \{z \in \mathbb{C} : |z| < 1\}\) is the open unit disk) such that: \[f(z_0) = 0 \quad \text{and} \quad f'(z_0) > 0.\]
Proof of the Uniqueness Part
1. Set up the comparison between two such Riemann mappings.
1.1. Suppose \(f_1, f_2: \Omega \to \mathbb{D}\) are two biholomorphic mappings satisfying the normalization conditions: \[f_1(z_0) = 0, \quad f_1'(z_0) > 0, \qquad f_2(z_0) = 0, \quad f_2'(z_0) > 0.\] Proof: This is the hypothesis of the uniqueness assertion.
1.2. Define \(g: \mathbb{D} \to \mathbb{D}\) by \(g = f_1 \circ f_2^{-1}\).Proof: Well-defined because \(f_2: \Omega \to \mathbb{D}\) is biholomorphic, so its inverse \(f_2^{-1}: \mathbb{D} \to \Omega\) exists and is biholomorphic.
1.3. Q.E.D.
2. \(g\) is an automorphism of the unit disk fixing the origin.
2.1. \(g\) is a composition of two biholomorphisms (\(f_1\) and \(f_2^{-1}\)), hence \(g \in \text{Aut}(\mathbb{D})\).Proof: The set of biholomorphic self-maps \(\text{Aut}(\mathbb{D})\) is a group under composition.
2.2. \(g(0) = f_1(f_2^{-1}(0)) = f_1(z_0) = 0\).Proof: \(f_2(z_0) = 0 \implies f_2^{-1}(0) = z_0\), and \(f_1(z_0) = 0\).
2.3. Q.E.D.
3. By the Schwarz Lemma, \(g\) is a pure rotation.
3.1. Any holomorphic map \(g: \mathbb{D} \to \mathbb{D}\) with \(g(0) = 0\) that is bijective (an automorphism) is of the form \(g(w) = e^{i\theta} w\) for some constant \(\theta \in \mathbb{R}\).Proof: By the Schwarz Lemma, \(|g'(0)| \leq 1\) and \(|(g^{-1})'(0)| \leq 1\).By the chain rule, \(g'(0) \cdot (g^{-1})'(0) = 1\), so \(|g'(0)| = 1\).The equality case of the Schwarz Lemma implies \(g(w) = e^{i\theta} w\).
3.2. Therefore, \(g(w) = e^{i\theta} w\) for all \(w \in \mathbb{D}\).Proof: Follows from
3.1.
3.3. Q.E.D.
4. Determine the angle \(\theta\) using the derivative normalization.
4.1. Differentiating \(g(w) = e^{i\theta} w\) gives \(g'(0) = e^{i\theta}\).Proof: Derivative of linear function.
4.2. By the chain rule on \(g(w) = f_1(f_2^{-1}(w))\) at \(w=0\): \[g'(0) = f_1'(f_2^{-1}(0)) \cdot (f_2^{-1})'(0) = f_1'(z_0) \cdot \frac{1}{f_2'(z_0)} = \frac{f_1'(z_0)}{f_2'(z_0)}.\] Proof: Inverse function theorem derivative rule \((f_2^{-1})'(0) = \frac{1}{f_2'(z_0)}\).
4.3. Since \(f_1'(z_0) > 0\) and \(f_2'(z_0) > 0\), the ratio \(\frac{f_1'(z_0)}{f_2'(z_0)}\) is a strictly positive real number: \[\frac{f_1'(z_0)}{f_2'(z_0)} > 0.\] Proof: Quotient of two positive real numbers.
4.4. From 1 and 2, \(e^{i\theta} = \frac{f_1'(z_0)}{f_2'(z_0)} \in (0, \infty)\).Since \(|e^{i\theta}| = 1\), we must have \(e^{i\theta} = 1\).Proof: The only positive real number on the unit circle is \(1\).
4.5. Q.E.D.
5. Conclusion: \(f_1 = f_2\).
5.1. Since \(e^{i\theta} = 1\), \(g(w) = w\) for all \(w \in \mathbb{D}\), which means \(g = \text{id}_{\mathbb{D}}\).Proof: Substitution of \(e^{i\theta} = 1\) into \(g(w) = e^{i\theta} w\).
5.2. \(f_1 \circ f_2^{-1} = \text{id}_{\mathbb{D}} \implies f_1(z) = f_2(z)\) for all \(z \in \Omega\).Proof: Composing both sides with \(f_2\) on the right.
5.3. Q.E.D.