[Fall 2010, Problem \(\# 5 ]\) Let \(H : = \{ z \in \mathbb { C } : \operatorname { R e } ( z ) > 0 \}\) . Suppose \(f\) is an analytic function which maps the open unit disk \(D\) into \(H\) and satisfies \(f ( 0 ) = 2\) . Find a sharp upper bound for \(\left| f ^ { \prime } ( 0 ) \right|\), justifying your bound by a proof and its sharpness by an example.
Solution
Define \[ \Phi(w)=\frac{w-2}{w+2}. \]
1. The map \(\Phi\) sends the right half-plane \(H\) into the open unitdisk.
If \(w\in H\), then \(\operatorname{Re}w>0\). Hence \[ \begin{aligned} \abs{w+2}^2-\abs{w-2}^2 &= (w+2)(\bar w+2)-(w-2)(\bar w-2)\\ &= 8\operatorname{Re}w\\ &> 0. \end{aligned} \] Therefore \[ \abs{w-2}<\abs{w+2}, \] so \(\abs{\Phi(w)}<1\).
2. The function\[ g=\Phi\circ f \]is an analytic self-map of the unit disk satisfying \(g(0)=0\).
The denominator \(w+2\) does not vanish on \(H\), so \(\Phi\) is analytic there. Since \(f\) is analytic and maps the unit disk into \(H\), step 1 shows that \(g\) is an analytic self-map of the unit disk. Moreover, \[ g(0) = \frac{f(0)-2}{f(0)+2} = 0. \]
3. One has\[ \abs{g'(0)}\leq1. \]
This is Schwarz’s lemma applied to the function in step 2.
4. The derivatives satisfy\[ g'(0)=\frac{f'(0)}4. \]
Differentiating \[ \Phi(w)=\frac{w-2}{w+2} \] gives \[ \Phi'(w)=\frac4{(w+2)^2}. \] By the chain rule and \(f(0)=2\), \[ g'(0) = \Phi'(f(0))f'(0) = \frac4{(2+2)^2}f'(0) = \frac{f'(0)}4. \]
5. The sharp upper bound is\[ \boxed{ \abs{f'(0)}\leq4. } \]
Combining steps 3 and 4 gives \[ \frac{\abs{f'(0)}}4 = \abs{g'(0)} \leq1. \]
6. The bound in step 5 is attained by\[ f_0(z)=2\frac{1+z}{1-z}. \]
The Cayley transform \[ \frac{1+z}{1-z} \] maps the open unit disk into the right half-plane, so \(f_0\) maps the disk into \(H\). Also \[ f_0(0)=2. \] Differentiating gives \[ f_0'(z)=\frac4{(1-z)^2}, \] and hence \[ \abs{f_0'(0)}=4. \] Thus equality occurs in step 5.
7. Q.E.D.
Step 5 proves the upper bound, and step 6 proves its sharpness.