Let \(f\) be analytic on the right half-plane \[ H=\{z\in\mathbb C:\operatorname{Re}z>0\} \] and suppose \[ |f(z)|\le1\qquad(z\in H), \qquad f(1)=0. \] What is the largest possible value of \(|f'(1)|\)?
Solution
1. The map\[ \phi(z)\coloneqq\frac{z-1}{z+1} \]maps the right half-plane \(H\) biholomorphically onto the unit disk \(\mathbb D\), with inverse\[ \psi(w)\coloneqq\frac{1+w}{1-w}. \]
For \(z\in H\), \[ \abs{z+1}^2-\abs{z-1}^2 = 4\operatorname{Re}z >0. \] Hence \[ \abs{\phi(z)}<1. \] Solving \[ w=\frac{z-1}{z+1} \] for \(z\) gives \[ z=\frac{1+w}{1-w}=\psi(w). \] Thus \(\phi\) and \(\psi\) are inverse biholomorphisms. In particular, \[ \psi(0)=1. \]
2. Define\[ g(w)\coloneqq f(\psi(w)), \qquad w\in\mathbb D. \]Then\[ g(0)=0, \qquad \abs{g(w)}\leq1, \]and therefore\[ \abs{g'(0)}\leq1. \]
By step 1, \(\psi\) maps \(\mathbb D\) into \(H\), so \(g\) is analytic on \(\mathbb D\). The hypotheses on \(f\) give \[ g(0)=f(1)=0 \] and \(\abs{g(w)}\leq1\). Schwarz’s lemma therefore yields \(\abs{g'(0)}\leq1\).
3. Every admissible \(f\) satisfies\[ \abs{f'(1)}\leq\frac12. \]
Since \[ \psi'(w)=\frac{2}{(1-w)^2}, \] one has \[ \psi'(0)=2. \] By the chain rule, \[ g'(0) = f'(\psi(0))\psi'(0) = 2f'(1). \] Step 2 gives \(\abs{2f'(1)}\leq1\), proving the bound.
4. The bound \(1/2\) is attained.
Take \[ f_0(z)=\frac{z-1}{z+1}. \] By step 1, \(f_0\) maps \(H\) into \(\mathbb D\), so \(\abs{f_0(z)}<1\) on \(H\), and \[ f_0(1)=0. \] Moreover, \[ f_0'(z)=\frac{2}{(z+1)^2}, \] so \[ \abs{f_0'(1)}=\frac12. \]
5. The largest possible value is\[ \boxed{\frac12}. \]
Step 3 supplies the universal upper bound, and step 4 attains it.
6. Q.E.D.
Step 5 is the required extremal value.