State the Schwarz lemma for analytic functions in the unit disc.
Let \(f: \mathbb{D} \to \mathbb{D}\) be an analytic map from the unit disc \(\mathbb{D}\) into itself. Use the Schwarz lemma to show that for each \(a\in \mathbb{D}\) we have \[\begin{align*} \dfrac{|f'(a)|}{1-|f(a)|^2} \leq \dfrac{1}{1-|a|^2} \end{align*}\]
Solution
Part 1: Statement of the Schwarz Lemma
Schwarz Lemma: Let \(\mathbb{D} = \{z \in \mathbb{C} : |z| < 1\}\). If \(g: \mathbb{D} \to \mathbb{D}\) is a holomorphic function satisfying \(g(0) = 0\), then:
\(|g(z)| \leq |z|\) for all \(z \in \mathbb{D}\).
\(|g'(0)| \leq 1\). Furthermore, if \(|g(z)| = |z|\) for some non-zero \(z \in \mathbb{D}\), or if \(|g'(0)| = 1\), then \(g(z) = e^{i\theta} z\) for some constant \(\theta \in \mathbb{R}\) (i.e. \(g\) is a rotation).
Part 2: Proof of the Schwarz-Pick Derivative Inequality
1. Define the disk automorphisms \(\phi_a\) and \(\psi_{f(a)}\).
1.1. For any \(w \in \mathbb{D}\), define the Möbius transformation \(\phi_w(z) = \frac{w - z}{1 - \bar{w} z}\).Proof: This is a definition; the denominator does not vanish for \(|z|<1\) because \(|\bar w z|<1\).
1.2. \(\phi_w\) is a biholomorphic map from \(\mathbb{D}\) onto \(\mathbb{D}\) with \(\phi_w(w) = 0\), \(\phi_w(0) = w\), and \(\phi_w^{-1} = \phi_w\).Proof: Direct substitution gives \(\phi_w(w)=0\), \(\phi_w(0)=w\), and \(\phi_w(\phi_w(z))=z\). The identity \(1-|\phi_w(z)|^2=\frac{(1-|w|^2)(1-|z|^2)}{|1-\bar wz|^2}\) shows that \(\phi_w\) maps \(\mathbb D\) into \(\mathbb D\); being its own inverse, it is a biholomorphism of \(\mathbb D\).
1.3. Compute the derivative of \(\phi_w(z)\): \[\phi_w'(z) = \frac{-(1 - \bar{w}z) - (w - z)(-\bar{w})}{(1 - \bar{w}z)^2} = \frac{-1 + \bar{w}z + |w|^2 - \bar{w}z}{(1 - \bar{w}z)^2} = \frac{-(1 - |w|^2)}{(1 - \bar{w}z)^2}.\] Proof: Quotient rule.
1.4. Evaluating \(\phi_w'\) at \(z = 0\) and \(z = w\): \[\phi_w'(0) = -(1 - |w|^2), \qquad \phi_w'(w) = \frac{-(1 - |w|^2)}{(1 - |w|^2)^2} = -\frac{1}{1 - |w|^2}.\] Proof: Substitution into derivative formula.
1.5. Q.E.D.
2. Construct the normalized map \(g: \mathbb{D} \to \mathbb{D}\).
2.1. Set \(b = f(a) \in \mathbb{D}\).Define \(g: \mathbb{D} \to \mathbb{D}\) by: \[g(z) = (\phi_b \circ f \circ \phi_a)(z) = \phi_b(f(\phi_a(z))).\] Proof: Composition of holomorphic maps.
2.2. Since \(\phi_a(\mathbb{D}) = \mathbb{D}\), \(f(\mathbb{D}) \subseteq \mathbb{D}\), and \(\phi_b(\mathbb{D}) = \mathbb{D}\), \(g\) maps \(\mathbb{D}\) into \(\mathbb{D}\).Proof: Image containment under composition.
2.3. Evaluate \(g(0)\): \[g(0) = \phi_b(f(\phi_a(0))) = \phi_b(f(a)) = \phi_b(b) = 0.\] Proof: \(\phi_a(0) = a\), \(f(a) = b\), \(\phi_b(b) = 0\).
2.4. Q.E.D.
3. Apply the Schwarz Lemma to \(g\).
3.1. By Part 1, since \(g\) is holomorphic on \(\mathbb{D}\), \(g(\mathbb{D}) \subseteq \mathbb{D}\), and \(g(0) = 0\), we have: \[|g'(0)| \leq 1.\] Proof: Schwarz Lemma derivative bound.
3.2. Q.E.D.
4. Compute \(g'(0)\) via the Chain Rule.
4.1. By the chain rule applied to \(g(z) = \phi_b(f(\phi_a(z)))\): \[g'(z) = \phi_b'(f(\phi_a(z))) \cdot f'(\phi_a(z)) \cdot \phi_a'(z).\] Proof: Chain rule for holomorphic functions.
4.2. Evaluating at \(z = 0\): \[g'(0) = \phi_b'(f(\phi_a(0))) \cdot f'(\phi_a(0)) \cdot \phi_a'(0) = \phi_b'(f(a)) \cdot f'(a) \cdot \phi_a'(0) = \phi_b'(b) \cdot f'(a) \cdot \phi_a'(0).\] Proof: \(\phi_a(0) = a\) and \(f(a) = b\).
4.3. Substitute \(\phi_b'(b) = -\frac{1}{1 - |b|^2} = -\frac{1}{1 - |f(a)|^2}\) and \(\phi_a'(0) = -(1 - |a|^2)\) from <1>1.<2>4: \[g'(0) = \left(-\frac{1}{1 - |f(a)|^2}\right) \cdot f'(a) \cdot \big(-(1 - |a|^2)\big) = \frac{1 - |a|^2}{1 - |f(a)|^2} f'(a).\] Proof: Product of the derivative values.
4.4. Taking the absolute value: \[|g'(0)| = \frac{1 - |a|^2}{1 - |f(a)|^2} |f'(a)|.\] Proof: \(1 - |a|^2 > 0\) and \(1 - |f(a)|^2 > 0\).
4.5. Q.E.D.
5. Conclusion.
5.1. From <1>3.<2>1 and <1>4.<2>4: \[\frac{1 - |a|^2}{1 - |f(a)|^2} |f'(a)| \leq 1.\] Proof: \(|g'(0)| \leq 1\).
5.2. Dividing both sides by \(1 - |a|^2 > 0\): \[\frac{|f'(a)|}{1 - |f(a)|^2} \leq \frac{1}{1 - |a|^2}.\] Proof: Division by positive real number.
5.3. Q.E.D.