Problem Statement
Let $f(z)=\sum_{k\geq 1}a_k z^{n_k}$ be an entire function of finite order such that $\lim n_k/k=\infty$. Let $M(r)=\max_{\lvert z\rvert=r}\lvert f(z)\rvert$ and $m(r)=\min_{\lvert z\rvert=r}\lvert f(z)\rvert$. Is it true that\[\limsup\frac{\log m(r)}{\log M(r)}=1?\]
Categories:
Analysis
Progress
A problem of Pólya [Po29]. Results of Wiman [Wi14] imply that if $(n_{k+1}-n_k)^2>n_k$ then $\limsup \frac{m(r)}{M(r)}=1$. Erdős and Macintyre [ErMa54] proved this under the assumption that\[\sum_{k\geq 2}\frac{1}{n_{k+1}-n_k}<\infty.\]This was solved in the affirmative by Fuchs [Fu63], who proved that in fact for any $\epsilon>0$\[\log m(r)> (1-\epsilon)\log M(r)\]holds outside a set of logarithmic density $0$.Kovari [Ko65] has shown that the $\limsup$ is also $1$ for an arbitrary entire function given the stronger assumption that $n_k>k(\log k)^{2+c}$ for some $c>0$. It is conjectured that this condition can be replaced with $\sum \frac{1}{n_k}<\infty$. This would be best possible, as Macintyre [Ma52] has shown that, given any $n_k$ with $\sum \frac{1}{n_k}=\infty$, there is a corresponding entire function which tends to zero along the positive real axis.
In [Er61] this is asked with $m(r)=\max_n \lvert a_nr^n\rvert$, but with this definition the desired equality is a simple consequence of [Wi14] (see the comment by Quanyu Tang).
Source: erdosproblems.com/516 | Last verified: January 15, 2026