This problem appears on the written exam for Complex Analysis.

Can we find an entire function f with $f(\log{k})=\frac{1}{1+\log{k}}$ for $k\in\mathbb{N}$? If the answer to previous question is positive, then can such $f$ satisfy $|f(z)|\leq c \exp(|z|^\alpha)$ for some positive constants $c$ and $\alpha$?

I have no idea how to do it.

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    $\begingroup$ For the first part: if $x_n$ and $y_n$ are sequences of complex numbers such that $x_n$ has no limit point, there is an entire function with $f(x_n) = y_n$. $\endgroup$ – Robert Israel Aug 23 '17 at 23:17
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    $\begingroup$ @RobertIsrael Oh, wow. Never heard of this result. Can you quote it? Is it related to Wierestrass theorem? $\endgroup$ – Vadim Aug 23 '17 at 23:20
  • $\begingroup$ Corollary of Mittag-Leffler's theorem and the Weierstrass factorization theorem. Or see this paper. $\endgroup$ – Robert Israel Aug 24 '17 at 0:22

For the second part, $f$ is to be an entire function of finite order $\alpha$. Now you want $f(\log k) = \dfrac{1}{1+\log k}$, i.e. $h(z) = (1 + z) f(z) - 1$, which is again an entire function of order $\alpha$, is to have zeros $\log k$. But since for all positive integers $m$, $$\sum_{k=2}^\infty |\log k|^{-m} = \infty$$ such a function can't have finite rank. Thus this would contradict the Hadamard factorization theorem.


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