Suppose \(f:\DD\to \DD\) is holomorphic and \(f(0)=0\). Show that \[ \abs{f(z)+f(-z)}\le 2\abs z^2 \qquad(z\in\DD). \] If equality holds at some \(z_0\neq0\), show that \[ f(z)=\lambda z^2 \] for some \(\abs\lambda=1\).
Solution
By Schwarz’s lemma, \(|f(z)|\le |z|\). Hence \[ g(z)= \begin{cases} f(z)/z,&z\neq0,\\ f'(0),&z=0 \end{cases} \] is holomorphic on \(\DD\) and satisfies \(|g|\le1\). Set \[ h(z)=\frac{g(z)-g(-z)}2. \] Then \(h\) is holomorphic, \(h(0)=0\), and \(|h(z)|\le1\). Schwarz’s lemma gives \(|h(z)|\le|z|\). Since \[ f(z)+f(-z)=z\bigl(g(z)-g(-z)\bigr)=2zh(z), \] we obtain \[ |f(z)+f(-z)|\le2|z|^2. \]
Suppose equality holds at some \(z_0\neq0\). Then \(|h(z_0)|=|z_0|\), so the equality case of Schwarz’s lemma gives \[ h(z)=\lambda z, \qquad |\lambda|=1. \] Let \[ e(z)=\frac{g(z)+g(-z)}2. \] Then \[ g(z)=e(z)+\lambda z, \qquad g(-z)=e(z)-\lambda z. \] Because \(|g(z)|,|g(-z)|\le1\), the parallelogram identity yields \[ 2|e(z)|^2+2|z|^2 =|g(z)|^2+|g(-z)|^2 \le2, \] so \(|e(z)|\le\sqrt{1-|z|^2}\). On \(|z|=r<1\) this gives \(|e(z)|\le\sqrt{1-r^2}\); by the maximum modulus principle the same bound holds on \(|z|\le r\). Letting \(r\uparrow1\) shows \(e\equiv0\). Therefore \[ g(z)=\lambda z, \qquad f(z)=zg(z)=\lambda z^2. \] Conversely, these maps attain equality for every \(z\).