I have been trying to solve the following exercise from a collection of old complex analysis qualifier exams.
Suppose that $g$ is an entire function that satisfies the inequality $|g(z^2)| \leq e^{|z|}$. Also suppose that $g(m) = 0 \quad \forall m \in \mathbb{Z}$. Then prove that $g(z) \equiv 0$ (i. e. that $g$ is identically $0$).
So what I think is that the inequality by putting $z^{1/2}$ gives me $|g(z)| \leq e^{|z|^{1/2}}$ and this means that the entire function $g$ is of finite order and its order $\lambda = \lambda(g) \leq \frac{1}{2}$. Then I have been looking at the basic theorems for finite order entire functions but I don't really see if one of them would be helpful here.
So my question is how can I solve this problem? Is it really helpful to look at the theorems for finite order entire functions?