Let \(f\) be analytic from the open unit disk into itself. Prove that \[ |f'(z)|\le\frac1{1-|z|^2} \] for every \(|z|<1\).
Solution
Let \[ \DD=\{z\in\CC:\abs z<1\}. \]
1. For \(a\in\DD\), define\[ \phi_a(w) \coloneqq \frac{w-a}{1-\overline a w}. \]Then \(\phi_a\) is an automorphism of \(\DD\) satisfying\[ \phi_a(a)=0, \qquad \phi_a'(a)=\frac{1}{1-\abs a^2}. \]
For \(w\in\DD\), \[ \abs{1-\overline a w}^2-\abs{w-a}^2 = (1-\abs a^2)(1-\abs w^2)>0, \] so \(\abs{\phi_a(w)}<1\). Solving \[ \zeta=\frac{w-a}{1-\overline a w} \] for \(w\) gives \[ w=\frac{\zeta+a}{1+\overline a\zeta}, \] which also maps \(\DD\) to itself by the same calculation. Hence \(\phi_a\) is a biholomorphic self-map of \(\DD\). Differentiating, \[ \phi_a'(w) = \frac{1-\abs a^2}{(1-\overline a w)^2}, \] and evaluation at \(w=a\) gives the stated derivative.
2. Fix \(z\in\DD\) and set\[ g \coloneqq \phi_{f(z)}\circ f\circ\phi_z^{-1}. \]Then \(g:\DD\to\DD\) is analytic and\[ g(0)=0. \]
By step 1, both disk automorphisms are analytic self-maps of \(\DD\), and \(f\) maps \(\DD\) into itself. Since \[ \phi_z^{-1}(0)=z, \] we obtain \[ g(0) = \phi_{f(z)}(f(z)) = 0. \]
3. One has\[ \abs{g'(0)}\leq1. \]
This is Schwarz’s lemma applied to the analytic self-map \(g\) of the unit disk with \(g(0)=0\).
4. The derivative in step 3 is\[ g'(0) = \frac{1-\abs z^2}{1-\abs{f(z)}^2}\,f'(z). \]
By the chain rule, \[ g'(0) = \phi_{f(z)}'(f(z))\,f'(z)\,(\phi_z^{-1})'(0). \] Step 1 gives \[ \phi_{f(z)}'(f(z)) = \frac1{1-\abs{f(z)}^2}. \] Since \[ \phi_z'(z)=\frac1{1-\abs z^2}, \] the inverse-function derivative formula gives \[ (\phi_z^{-1})'(0) = 1-\abs z^2. \] Substitution yields the formula.
5. Therefore\[ \abs{f'(z)} \leq \frac{1-\abs{f(z)}^2}{1-\abs z^2}. \]
Take absolute values in step 4 and apply step 3.
6. Hence, for every \(z\in\DD\),\[ \boxed{ \abs{f'(z)} \leq \frac1{1-\abs z^2}. } \]
Because \(f(z)\in\DD\), \[ 0<1-\abs{f(z)}^2\leq1. \] Apply this inequality to the numerator in step 5.
7. Q.E.D.
Step 6 is the required estimate.