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Suppose that $f:\mathbb C \to \mathbb C$ is holomorphic and not identically zero, and that $f$ has a zero at every lattice point (point with integer coordinates) except for $(0,0)$. Show that there is a constant $c>0$ such that $|F(z_i)|>e^{c|z_i|^2}$ for a sequence $z_1,z_2,\ldots$ of complex numbers tending to infinity.

If we take the supremum of $F$ over each circle of radius $R$, this should give a correct sequence (taking $R=1,2,\ldots$). I am not sure how to come up with a lower bound though.

user85798
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1 Answers1

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Jensen's formula gives $$ \log |f(0)| = \sum_{k=1}^n \log \frac{|a_k|}{r} + \frac{1}{2\pi} \int_0^{2\pi} \log |f(r e^{it})| \, dt, $$ for all $r>0$, where $a_1, \ldots, a_n$ are the zeros of $f$ in $|z|<r$ (repeated according to multiplicity, and assuming that there is no zero on the boundary $|z|=r$.) Assuming now that we have $|f(z)| \le e^{c|z|^2}$ for all large enough $|z|$, and writing $b = \log|f(0)|$, we get $$ \sum_{k=1}^n \log \frac{r}{|a_k|} \le cr^2-b. $$ If we denote the number of lattice points (except for 0) inside the disk $|z|<r$ by $N(r)$, we know that $\lim\limits_{r \to \infty} \frac{N(r)}{r^2} = \pi$. Now fix $\lambda \in (0,1)$ and observer that for every zero $|a_k|$ with $|a_k| < \lambda r$ we have that $\log \frac{r}{|a_k|} \ge \log \frac{1}{\lambda}$, and all the terms in the sum above are positive, so $$ N(\lambda r) \log \frac1\lambda \le cr^2 - b. $$ Dividing by $r^2$ and passing to the limit $r \to \infty$ we get $$ \pi \lambda^2 \log \frac{1}{\lambda} \le c. $$ Any particular value of $\lambda$ gives a lower bound for $c$ having the property described above. I.e., if $0<c< \pi \lambda^2 \log \frac1\lambda$, then there exists a sequence $z_k \to \infty$ with $|f(z_k)| > e^{c|z_k|^2}$. (The optimal bound using this argument is achieved for $\lambda = e^{-1/2}$ and comes out to $\frac{\pi}{2e}$, but I suspect that more sophisticated arguments would give better bounds for $c$.)

Lukas Geyer
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