For the statement of Carlson's theorem please see,
https://en.wikipedia.org/wiki/Carlson%27s_theorem.
There is an extension of Carlson's theorem that says that the condition that $f$ needs to vanish on integers can be replaced with $f$ vanishing on a subset A of integers provided that $A$ has upper density 1. This is a necessary and sufficient condition.
Now my question is as follows: Assuming that the condition on the growth of $f$ on the $y$-axis is more stringent, say for example that $f$ is uniformly bounded on the entire $y$-axis, can one obtain an extension of Carlson's theorem with $f$ vanishing on a monotone divergent sequence $a_1<a_2<\ldots$ with upper density strictly less than one?
Best Answer
If $f$ is bounded on the imaginary line, (and has exponential type) then $f$ has completely regular growth in the sense of Levin-Pfluger, with indicator $c|\cos\theta|$. This implies that density of zeros on the positive ray must be zero. Moreover, density of zeros in any angle $|\arg z|<\pi/2-\epsilon$ and in the vertical angle is zero. Boundedness on the imaginary line condition can be relaxed to $$\int_{-\infty}^{\infty}\frac{\log^+|f(it)|}{1+t^2}dt<\infty.$$ See any of the two books of Levin (Distribution of zeros of entire functions, or Lectures on entire functions) for terminology and proofs.
For weaker conditions on the imaginary line, implying some upper estimates on the density of real zeros, see, for example Theorem 7 (Chap IV, $\S$ 2 of the first book of Levin): the lower density of zeros is at most $$\frac{1}{2\pi}\left(h\left(\frac{\pi}{2}\right)+h\left(-\frac{\pi}{2}\right)\right),$$ where $h$ is the indicator. See also MR0207986 and https://arxiv.org/abs/0807.2054