Say $f$ is entire, $|f'(z)|\le e^{|z|}$, and $f$ vanishes on the set $\{\frac{n}{\sqrt{1+|n|}}: n\in \mathbb{Z}\}$. Why must $f$ be constantly zero?

Question

Say $f$ is entire, $|f'(z)|\le e^{|z|}$, and $f$ vanishes on the set $\{\frac{n}{\sqrt{1+|n|}}: n\in \mathbb{Z}\}$.

Why must $f$ be constantly zero?

Answer 1

I solved it. Hints for those of you who are interested:

1) Show that $|f'(z)|\le e^{|z|}$ implies that $|f(z)|\lesssim e^{|z|}$ where the implicit constant is independent of $z$.

2) If $f$ is non-zero, then, by a corollary to Poisson-Jensen, it follows that $\sum \frac{1}{|{z_k}|^s}<\infty$ for $s>1$ where the $z_k$ are the nonzero zeroes of $f$.

Answer 2

This is not an answer, I just wanted to post it to see if someone thinks it could go anywhere and a comment doesn't really provide a nice format to write it up in. I have only recently just learnt some basic complex analysis and wouldn't mind any experts pointing out any flaws in my logic too.

Suppose $f$ doesn't vanish everywhere. Then by the Residue theorem

$$\frac{1}{2 \pi i} \int_C \frac{f'(z) dz}{f(z)} \geq r^2$$ where $C$ is the circle of radius $r$, since we know a lower bound on how many zeros of $f(z)$ there are within this circle.

This implies that

$$ \int_C \left|\frac{f'(z)}{f(z)}\right| dz \geq 2 \pi r^2$$

Now I want to use the fact $|f'(z)|$ is of order $e^r$ and so the term within the integral cannot be too large. That is the part I'm not sure about. It seems likely because the only examples of things where the derivative of a function is significantly bigger than the function is for things like $e^{f(r)}$ where $f'(r) > 1$.

Edit: I just thought of something like $\sin(kt)$ as $k$ gets large. This throws the last paragraph out the window :(

Thanks

Answer 3

The Identity Theorem applies since the vanishing set for the function accumulates near $z = 1$ and $z = -1$ as $n \to \infty$ and $n \to -\infty$ respectively.