Let $f$ be an entire function such that $n^{-n/2}f^{(n)}(0)\in \mathbb{Z},n\geq 1$ and $f$ is not a polynomial. Show that $\limsup_{\lvert z\rvert\rightarrow \infty}\frac{\log\lvert f(z)\rvert}{\lvert z\rvert}\geq 2$. Can anyone give me a clue or an idea or a precise answer?


Write $f(z)=\sum_na_nn^{n/2}z^n/n!$, where infinitely many $a_n$ are non-zero integers, so $|a_n|\geq 1$ for those $n$. By Cauchy inequality, for every $n$ and for every $r$, we have $$\max_{|z|=r}|f(z)|\geq n^{n/2}r^n/n!,$$ so, using Stirling formula, $$\log\max_{|z|=r}|f(z)|\geq n-(n/2)\log n+n\log r+O(\log n).$$ Now maximize the RHS with respect to $n$. To do this, treat $n$ as a continuous variable, so that you can use Calculus. You obtain that the maximum is attained aproximately at $n=r^2$. Now forget this non-rigorous argument and just plug $n=r^2$, or more precisely, take $n$ for which $a_n\neq 0$, and choose $r=\sqrt{n}$. You obtain $$\limsup_{r\to\infty}\frac{\log|f(z)|}{|z|^2}\geq 1,$$ which is much stronger than what you wanted.

  • $\begingroup$ I am a little bit tired! I do not understand why you put $logmax$ at $|z|=1$? How can i get my inequality by your last statement ? $\endgroup$ – Far Mar 29 '15 at 22:37
  • $\begingroup$ This was a misprint. I corrected. How do you get your inequality? Using that $|z|^2>2|z|$ when $|z|$ is large. $\endgroup$ – Alexandre Eremenko Mar 30 '15 at 1:55
  • $\begingroup$ Very nice ! Did you use the formula $\log n! = n\log n + n - 1/2\log n - \log(\sqrt(2\pi))- 1/12n + o(1/n)$ ? $\endgroup$ – Far Mar 30 '15 at 13:13
  • $\begingroup$ Yes, but only the first 2 terms of this asymptotics are relevant. $\endgroup$ – Alexandre Eremenko Mar 30 '15 at 13:16

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