Let $f : \mathbb{C} → \mathbb{C}$ be an entire function. Prove that if $|f(z^2)| ≤ 2|f(z)|$ for all z ∈ C, then f is constant. To start, I proved that $h(z) = f(z^2)/f(z)$ is entire and used Liouville's theorem to $h(z) = c$. I am stuck on showing that f is a constant.
 A: Suppose $f(z^2) = c f(z)$ where $c$ is a nonzero constant.  Then $f(z^{2^n}) = c^n f(z)$ for all positive integers $n$.  For  any $2^{n}$'th root of unity $\omega$, $(\omega z)^{2^n} = z^{2^n}$, so $f(z) = f(\omega z)$.   Since
$f$ is constant on a set that has a finite limit point, it is constant.
A: To show that $h$ is entire, you just need that $h$ is bounded in each neighborhood of the possible zeros of $f$. Therefore, each singularity of $h$ is removable. 
Now, Liouville's Theorem gives that $h$ is constant, i.e. there is some $c\in\mathbb C$ such that $f(z^2)=cf(z)$ for all $z\in\mathbb C$. If $c=0$, we are done, so assume that $c\neq 0$. Differentiating both sides yields $2zf'(z^2)=cf'(z)$. In particular, $f'(0)=0$. Again, differentiating gives us $2f'(z^2)+4z^2f''(z^2)=cf''(z)$, and again $f''(0)=0$. Continuing this process leads to $f^{(n)}(0)=0$ for all $n\geq 1$. But then (Taylor!) $f$ is constant.
A: Another approach: Let $M(R)= \sup_{|z|=R} |f(z)|.$ Use the hypotheses and induction to see
$$M(2^{2^k}) \le 2^kM(2),\,\, k= 1,2,\dots$$
This gives a sequence of radii $1<R_1 < R_2 < \cdots \to \infty$ and a constant $C>0$ such that
$$M(R_k) \le C\ln R_k.$$
This is way less than linear growth, so in the usual way, Cauchy's estimates on the Taylor coefficients of $f$  at $0$ show all coefficients are $0$ except for the constant term.
