Let \(f\in \Hol(\DD)\). Show that if \(f\) has a fixed point \(a\) then \(\abs{f'(a)} \leq 1\), and that \[ \abs{f(0)}^2 + \abs{f'(0)}^2 \leq 1 .\]
Both bounds fail for holomorphic maps that do not send \(\DD\) into \(\DD\): \(f(z)=2z\) fixes \(0\) and has \(f'(0)=2\).
Solution
Assume \(f(\DD)\subseteq\DD\), which the Schwarz–Pick lemma requires. Set \(f(a) = a\) in Schwarz-Pick: \[ \left|f^{\prime}(a)\right| \leq \frac{1-|f(a)|^{2}}{1-|a|^{2}} \implies \abs{f'(a)} \leq {1 - \abs{a}^2 \over 1 - \abs{a}^2} = 1 .\] Set \(a=0\): \[ \left|f^{\prime}(0)\right| \leq \frac{1-|f(0)|^{2}}{1-|0|^{2}} = 1-\abs{f(0)}^2 .\] Since the right side is at most \(1\), \(\abs{f'(0)}^2\le\abs{f'(0)}\le1-\abs{f(0)}^2\).