# An entire function $g$ such that $|g(z^2)| \leq e^{|z|}$ and $g(m) = 0 \quad \forall m \in \mathbb{Z}$ is identically $0$

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?

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Since $\mid g(z) \mid \leq exp(\mid z \mid^{\frac{1}{2}})$, $g$ has growth order $\leq \frac{1}{2}$ by definition, as you correctly remark.
If $g$ were not the zero function and if $(z_k)_k$ were some enumeration of its non-zero zeros, we would have for all $s\gt \frac{1}{2}$ (Stein-Shakarchi, Theorem 2.1, page 138) $$\sum_k \frac{1}{\mid z_k \mid ^s } \lt \infty$$ In particular, taking $s=1$ and realizing that the positive integers are among the zeros $z_k$ according to the statement of the exercise, we would get $$\sum_{n=1} ^\infty \frac{1}{n } \leq \sum_k \frac{1}{\mid z_k \mid } \lt \infty$$
Since the harmonic series actually diverges ( $\sum_{n=1} ^\infty \frac{1}{n }=\infty$ ) this is false and we must have $g=0$.
A baby case is that of a polynomial: if its degree is $n$ it grows like $\mid z\mid^n$and has $n$ zeros.
Thank you very much. I actually thought about this same argument at first, but the book I was looking at had that theorem but with $\sum\frac{1}{|z_k|^{\lambda + 1}} < \infty$ where $\lambda$ is the order of growth so I obviously wasn't able to use it. I see know why some people say it isn't such a great book after all ;) – Student Apr 27 '12 at 20:53
Dear @Student, yes, I had noticed that your question was very mature and that you were at $\epsilon$ distance from the solution. Are you going to take a qualification examination in order to do a Ph.D. in Analysis? In that case you have all my best wishes for success. – Georges Elencwajg Apr 27 '12 at 21:15