My question is on whether or not there exists some monotone strictly decreasing sequence of positive numbers $c_1>c_2>\ldots$ such that given any $f$ which is a uniformly bounded holomorphic function in the right half of the complex plane with $$ f(k)\leq c_k \quad \forall\, k\in \mathbb N,$$ there holds $ f(z)=0$ on the right half plane.

2$\begingroup$ The answer is certainly "yes" (by the standard compactness argument) but I suspect that much more is known to the experts, so I'll leave it to them to answer your question properly. $\endgroup$– fedjaApr 8, 2022 at 22:43

3$\begingroup$ For example $c_k=\exp(k^{1+\epsilon})$ will do. $\endgroup$– Alexandre EremenkoApr 8, 2022 at 23:29
1 Answer
Condition $$\lim_{n\to\infty}\frac{\logc_n}{n}=\infty$$ is sufficient for $f=0$.
Since $f(z)=e^{cz}$ and $c_n=e^{cn}$ satisfy all conditions, we see that this is best possible in certain sense.
This follows for example from a (much more general) theorem of N. Levinson, Gap and density theorems, AMS, 1940, page 121. Levinson's theorem allows some growth of $F$, and much more general class of sequences instead of integers.
Remark. In fact Levinson generalizes a theorem of Vladimir Bernstein 1932 (Theorem 32 in Levinson's book), which also implies the result that I stated.

$\begingroup$ Do you know if it is possible to deduce a (possibly weaker) version of this result by the fact that the integers are a uniqueness set for bounded holomorphic functions in the halfplane? $\endgroup$ Apr 9, 2022 at 17:56

$\begingroup$ See @fedja's comment to the question. Quntitave estimates are quite different for the condition that $f(n)$ is small, vs the condition $f(n)=0$. $\endgroup$ Apr 9, 2022 at 18:58

2$\begingroup$ Thank you, I did not notice the comment of @fedja. However, I do not see how to use compactness to prove the existence of a sequence as in the statement. $\endgroup$ Apr 9, 2022 at 22:29