# Show that if $f$ is entire and $(\forall z \in \mathbb{C})$ $|f(z^2)| \leq |f(z)|$, then $f$ is constant

Show that, if $f$ is entire and $(\forall z \in \mathbb{C})$ $|f(z^2)| \leq |f(z)|$, then $f$ is constant

How should I approach this problem? Should I use the power series of $f(z)$ or Liouville's Theorem?

• You have $|f(z^{2^n})| \leq |f(z) |$, hence $|f(z) | \leq \sup_{|z| \leq 2} |f(z)|$ for all $z$, so $f$ is bounded. – pisco Apr 16 '18 at 13:49
• A different approach. Prove first (using Liouville's theorem) that $f(z^2)=c\,f(z)$ for some constant $c$. Then compare the Taylor expansion of both sides of the equality. – Julián Aguirre Apr 16 '18 at 13:54

Consider the region $|z|=R$ for $R>1$. Since $\left|f(z^2)\right|\leq\left|f(z)\right|$,
$$\max_{|z|=R^2}\left|f(z)\right| \leq \max_{|z|=R}\left|f(z)\right|$$ which contradicts the maximum modulus principle over the region $|z|\leq R^2$, unless $f$ is constant.