Let \(D=\{z\in\mathbb C:|z|<1\}\). What is the maximum possible value of \(|f'(1/2)|\) for a holomorphic function \(f:D\to D\) satisfying \(f(1/2)=3/4\)? Find all such functions that attain this maximum.
Solution
The maximum is \(\boxed{7/12}\). Every extremizer, and no other function, has the form \[ \boxed{f_\lambda(z)= \frac{\frac34+\lambda\frac{z-1/2}{1-z/2}} {1+\frac34\lambda\frac{z-1/2}{1-z/2}}, \qquad |\lambda|=1.} \]
1. Disk automorphisms reduce the derivative estimate to the origin.
For \(c\in D\), set \[ \phi_c(z)=\frac{z-c}{1-\overline c z}, \qquad \phi_c^{-1}(w)=\frac{w+c}{1+\overline c w}. \] Their denominators are nonzero on \(D\), substitution shows they are inverses, and \[ 1-|\phi_c(z)|^2= \frac{(1-|c|^2)(1-|z|^2)}{|1-\overline c z|^2}>0. \] The inverse is \(\phi_{-c}\), so it also maps \(D\) into itself. Thus these are disk automorphisms.
Put \(a=1/2\), \(b=3/4\), and \(G=\phi_b\circ f\circ\phi_a^{-1}\). Then \(G:D\to D\) is holomorphic with \(G(0)=0\). Direct differentiation gives \[ \phi_b'(b)=\frac1{1-b^2},\qquad (\phi_a^{-1})'(0)=1-a^2, \] and hence \[ G'(0)=\frac{1-a^2}{1-b^2}f'(a)=\frac{12}{7}f'(1/2). \] Schwarz’s lemma gives \(|G'(0)|\leq1\) [@SS03]. Therefore \(|f'(1/2)|\leq7/12\).
2. Equality holds exactly for the displayed functions.
Schwarz’s lemma also gives \(|G(w)|\leq|w|\). Thus \(H(w)=G(w)/w\) extends holomorphically across zero, with \(H(0)=G'(0)\), and satisfies \(|H|\leq1\) on \(D\). If \(|f'(1/2)|=7/12\), then \(|H(0)|=1\). The maximum modulus principle forces \(H\equiv\lambda\) for some \(|\lambda|=1\) [@SS03]. Thus \(G(w)=\lambda w\), and undoing the two automorphisms gives exactly \(f=\phi_b^{-1}\circ(\lambda\phi_a)\), the stated formula.
Conversely, for every \(|\lambda|=1\) this composition is a holomorphic disk automorphism, sends \(a\) to \(b\), and has \(G'(0)=\lambda\). The derivative calculation in step 1 therefore gives \(|f_\lambda'(1/2)|=7/12\). The explicit denominator cannot vanish since \(|\frac34\lambda\phi_a(z)|<3/4\) on \(D\). This proves attainment and exhausts all equality cases.