Let \(f:\DD\to \DD\) with \(f(0) = f'(0) = 0\). Show that \(\abs{f''(0)} \leq 2\) and describe all \(f\) for which this is an equality.
Solution
By Schwarz, \(\abs{f(z)}\leq \abs z\). Write \(g(z) \da f(z)/z\), which is holomorphic on \(\DD\) since \(f\) has a zero of order at least one at \(0\), and \(\abs{g(z)}\leq 1\). Write \(f(z) = \sum_k c_kz^k\); then \(c_0 = 0\) since \(f(0) = 0\) and \(c_1 = 0\) since \(f'(0) = 0\), so \(f(z)=c_2z^2 + \bigo(z^3)\). Thus \(g(z) = c_2z + \bigo(z^2)\) and \(g(0) = 0\). By the maximum modulus principle, the nonconstant function \(g\) with \(\abs g\le1\) satisfies \(\abs g<1\) on \(\DD\) (and \(g\equiv0\) if \(g\) is constant), so \(g:\DD\to\DD\) and Schwarz applies:
- \(\abs{g(z)}\leq \abs{z}\)
- \(\abs{g'(0)} \leq 1\)
We have \(g'(z) = c_2 + \bigo(z)\) so \(g'(0) = c_2 = {f^{(2)}(0) \over 2!}\). \[ 1 \geq \abs{g'(0)} = \abs{c_2} = \abs{f^{(2)}(0) \over 2} \implies \abs{ f^{(2)}(0) } \leq 2! .\]
Suppose this is an equality – then Schwarz on \(g\) shows \(g\) is a rotation, so \[ {f(z)\over z} = g(z) = \lambda z \implies f(z) = \lambda z^2 \qquad \lambda \in S^1 .\] Conversely, \(f(z)=\lambda z^2\) with \(\abs\lambda=1\) satisfies the hypotheses and \(\abs{f''(0)}=2\).