0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.