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.

Thanks in advance.

  • 2
    $\begingroup$ Hint: let $|z|=r<1$ and keep iterating the inequality to get $|f(z)| \le |f(z^{2^N})|$; let $N \to \infty$ $\endgroup$
    – Conrad
    Mar 2, 2020 at 17:15

1 Answer 1


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.


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