Let \(\HH_R\da\{w\in\CC:\Re w>0\}\). Suppose \(f:\DD\to\HH_R\) is analytic and satisfies \(f(0)=2\). Find a sharp upper bound for \(|f'(0)|\), and prove sharpness by example.
Solution
Let \[ r(w)=\frac{i}{2}w:\HH_R\to\HH, \qquad C(w)=\frac{w-i}{w+i}:\HH\to\DD. \] Then \(r(2)=i\) and \(C(i)=0\), so \[ F\da C\circ r\circ f:\DD\to\DD \] satisfies \(F(0)=0\). Schwarz’s lemma gives \(|F'(0)|\le1\).
By the chain rule, \[ F'(0)=C'(r(f(0)))\,r'(f(0))\,f'(0) =C'(i)\frac{i}{2}f'(0). \] Since \[ C'(w)=\frac{2i}{(w+i)^2}, \qquad |C'(i)|=\frac12, \] we obtain \[ 1\ge |F'(0)|=\frac14|f'(0)|, \qquad |f'(0)|\le4. \]
The bound is sharp. For \(|\lambda|=1\), take \(F(z)=\lambda z\) and unwind the conjugation: \[ f(z)=r^{-1}\!\left(C^{-1}(\lambda z)\right) =2\frac{1+\lambda z}{1-\lambda z}. \] This maps \(\DD\) biholomorphically onto \(\HH_R\), satisfies \(f(0)=2\), and \[ f'(0)=4\lambda, \] so \(|f'(0)|=4\).