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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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up vote 8 down vote accepted

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 $g=0$.

N.B. The basic heuristic of growth order of an entire function is that the faster the function grows, the more zeros it has.
A baby case is that of a polynomial: if its degree is $n$ it grows like $\mid z\mid^n $and has $n$ zeros.

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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
Dear Georges, thank you very much for your good wishes. I will indeed take a qualifying exam in the near future so I'm trying to solve exercises from old exams. I figure that since sometimes the professors who teach the basic graduate complex analysis courses use different books, these sort of problems cause more difficulty than originally expected due to one not having the "right" version of a theorem in hand. – Student Apr 27 '12 at 21:33
Dear @Student, yes, I agree with your analysis, which confirms my opinion of your maturity! Might I suggest that you consider using your real name (or at least a less bland nom de plume) ? It will add warmth to your exchange with users here. Whatever your decision, I'd be delighted to help you here if I can, like, I'm sure, many other StackExchangers . – Georges Elencwajg Apr 27 '12 at 21:50

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