A theorem of Duffin and Schaeffer says (quoting from R. P. Boas, Entire Functions, Theorem 10.5.1):
Let $f(z)$ be regular in $|\arg z| \leq \alpha \leq \pi / 2$ and let $h(\theta) \leq a |\cos \theta| + b |\sin \theta|$, [where $h$ is the indicator function of $f$,] and $b < \pi$. If $\{\lambda_n\}$ is an increasing sequence of real numbers such that $\lambda_{n + 1} - \lambda_n \geq 2 \delta > 0$, $|\lambda_n - n| \leq L$, $n = 1, 2, \cdots$, and $\{f(\lambda_n)\}$ is bounded, then $f(x)$ is bounded for $x > 0$; if $f(\lambda_n) \to 0$, $f(x) \to 0$ as $x \to \infty$.
Boas then says about the theorem:
The theorem is not true without some condition more restrictive than $|\lambda_n - n| = o(n)$, but it is an open question whether the condition $|\lambda_n - n| < L$ can be weakened to $|\lambda_n - n| < \epsilon(n)$ with some $\epsilon(n)$ which becomes infinite.
My questions are:
- Do the conclusions change if we allow $|\lambda_n - n| \to \infty$ but impose the condition that the $\lambda_n$ have larger than unit density, i.e. $\lim_{n \to \infty} n/\lambda_n = M > 1$ for some arbitrarily large $M$?
- Has there been progress towards concluding that $f(x)$ is bounded with a weaker condition than $|\lambda_n - n| < L$?
