Find the maximum possible value of \(\abs{f'(1)}\) given that \(f\) is holomorphic on an open neighborhood of \(\{z\in\CC:\abs{z}\leq2\}\) and satisfies \(\abs{f(z)}\leq1\) when \(\abs{z}=2\).
Solution
We will use a fractional linear transformation to change the problem to one where the derivative is evaluated at the center of a disk.
The function \(z\mapsto\frac2z\left(\frac{z-1}{\bar z-1}\right)\) on \(\abs{z}=2\) has absolute value \(1\), and it extends to a fractional linear transformation \(g(z)=2\left(\frac{z-1}{4-z}\right)\). Since it also maps \(z=1\) to the interior of the unit disk, it must map the region \(\abs{z}\leq2\) bijectively onto the unit disk. We calculate \(\abs{g'(1)}=2/3\).
Now, for any other \(f\) mapping the circle \(\abs{z}=2\) into \(\abs{z}\leq1\), the composition \(h\coloneqq f\circ g^{-1}\) is holomorphic on a neighborhood of \(\abs{z}\leq1\), and maps \(\abs{z}=1\) into \(\abs{z}\leq1\). Taking absolute values in
\[ h'(0)=\frac{1}{2\pi i}\int_{\abs{z}=1}\frac{h(z)}{z^2}\,dz \]
gives \(\abs{h'(0)}\leq1\). Since \(g^{-1}(0)=1\), the chain rule gives \(h'(0)=f'(1)g'(1)^{-1}\). Thus \(\abs{f'(1)}=\abs{h'(0)}\abs{g'(1)}\leq\abs{g'(1)}=2/3\). Thus \(2/3\) is the maximum possible value of \(\abs{f'(1)}\).