# if $f$ is an entire function satisfying $|f(z)|\leq |f(z^2)|$ for all $z\in \Bbb C$ then $f$ is constant

How do I have to show that if $$f$$ is an entire function satisfying $$|f(z)|\leq |f(z^2)|$$ for all $$z\in \Bbb C$$ then $$f$$ is constant?

I know that to show an entire function is constant, Liouville's theorem, or Cauchy's inequality(in order to show that $$f'=0$$) is useful, but in this question I don't see what theorem should I have to use.

• Hint: let $|z|=r<1$ and keep iterating the inequality to get $|f(z)| \le |f(z^{2^N})|$; let $N \to \infty$ Mar 2, 2020 at 17:15
Let $$|z|<1$$. Then, as suggested by @Conrad in the comments:
$$|f(z)|\le|f(z^2)|\le \lim_{n\to \infty}|f(z^{2^N})|=|f(0)|$$
Thus $$0$$ is a maximum for the modulus of the function inside the unitary disk. Thanks to the maximum modulus principle, the function is constant.
The result is extended to $$\mathbb{C}$$ thanks to the identity theorem.