Suppose \(f:\DD\to \DD\) is analytic. Prove that \[ \forall a\in \DD, \qquad {\abs{f'(a)} \over 1 - \abs{f(a)}^2 } \leq {1 \over 1 - \abs{a}^2} .\]
Solution
Goal: Prove the Schwarz–Pick inequality: if \(f: \DD \to \DD\) is analytic, then for every \(a \in \DD\), \[\frac{\abs{f'(a)}}{1 - \abs{f(a)}^2} \leq \frac{1}{1 - \abs a^2}.\]
1. Recall the automorphism \(\phi_a(z) := \frac{z - a}{1 - \bar a z}\) of \(\DD\), which maps \(a\) to \(0\).
\(\phi_a\) is a M"obius map taking \(\DD\) onto \(\DD\) (its pole \(1/\bar a\) lies outside the unit disk) and \(\phi_a(a) = 0\); its inverse is \(\phi_a^{-1}(w) = \frac{w + a}{1 + \bar a w}\).
2. Define \(F := \phi_{f(a)} \circ f \circ \phi_a^{-1}\); then \(F: \DD \to \DD\) is analytic and \(F(0) = 0\).
\(\phi_{f(a)}(f(\phi_a^{-1}(0))) = \phi_{f(a)}(f(a)) = 0\), using \(\phi_a^{-1}(0) = a\).
3. By Schwarz’s lemma, \(\abs{F'(0)} \leq 1\).
Schwarz lemma applies to the analytic map \(F: \DD \to \DD\) with \(F(0) = 0\).
4. Compute \(F'(0)\) by the chain rule.
4.1. \(F'(0) = \phi_{f(a)}'(f(a)) \cdot f'(a) \cdot (\phi_a^{-1})'(0)\).
Chain rule for the composition \(\phi_{f(a)} \circ f \circ \phi_a^{-1}\).
4.2. \(\phi_w'(z) = \frac{1 - \abs w^2}{(1 - \bar w z)^2}\), so \(\phi_{f(a)}'(f(a)) = \frac{1}{1 - \abs{f(a)}^2}\).
Differentiate \(\phi_w\); at \(z = w\) the denominator is \(1 - \abs w^2\).
4.3. \((\phi_a^{-1})'(0) = 1 - \abs a^2\).
\(\phi_a^{-1}(w) = \frac{w + a}{1 + \bar a w}\); differentiate and evaluate at \(w = 0\): \(\frac{1 \cdot 1 - (0 + a)\bar a}{1^2} = 1 - \abs a^2\).
4.4. Hence \(F'(0) = f'(a) \cdot \frac{1 - \abs a^2}{1 - \abs{f(a)}^2}\).
<2>1–<2>3.
5. Combine with Schwarz’s lemma.
5.1. \(\abs{f'(a)} \cdot \frac{1 - \abs a^2}{1 - \abs{f(a)}^2} \leq 1\).
3 and 4.4.
5.2. Divide by the positive quantity \(1 - \abs a^2\).
\(\abs a < 1\), so \(1 - \abs a^2 > 0\).
6. Q.E.D.
5 gives \(\frac{\abs{f'(a)}}{1 - \abs{f(a)}^2} \leq \frac{1}{1 - \abs a^2}\).