Show that $|f(z)+f(-z)+f(iz)+f(-iz)| \le 4|z|^4$

Question

Let $f:\mathbb{D}\rightarrow \mathbb{D}$ be a holomorphic function, with $f(0)=0$. Prove that $$ |f(z)+f(-z)+f(iz)+f(-iz)| \le 4|z|^4$$ for all $z\in \mathbb{D}$.

I tried Schwarz Lemma but i get $4|z|$, which is missing the power. Any hints?

Answer 1

Applying the Schwarz Lemma to each term separately gives not the intended result because the special symmetry of the problem is lost.

Instead, define $$ g(z) = \frac 14 (f(z)+f(-z)+f(iz)+f(-iz)) $$ and observe that the power series of $g$ contains only powers of $z^4$, so that $$ g(z) = h(z^4) $$ for some holomorphic function $h: \Bbb D \to \Bbb D$ with $h(0) = 0$. Now apply the Schwarz Lemma to $h$.

This can be generalized to $$ |f(z) + f(\omega z) + \ldots + f(\omega^{N-1}z)| \le N |z|^N $$ where $N \in \Bbb N$ and $\omega = e^{2\pi i/N}$ is the $N^{\text{th}}$ root of unity.