Hi, I don't know if this question is appropriate for Math Overflow but I was wondering if there is anything known about the following: Let $$ S(\alpha) = \sum_{n \leq x}\Lambda(n)e(n\alpha). $$

Then asymptotically, how small can $$ \inf_{\alpha}\left|S(\alpha)\right| $$ be relative to $x$? Also, for each $x$, if $\alpha_x$ is a value at which the sum takes the infimum, then how are the $\alpha_x$ distributed in $(0,1)$ as $x \rightarrow \infty$?

Ok, thanks.


1 Answer 1



This is likely to be a pretty difficult question I think. For a random sequence of $\pm 1$s in place of the von Mangoldt function $\Lambda(n)$ the answer is a little surprising: the infimum is basically $1/\sqrt{x}$, a result of Konyagin and Schlag. This is available here:


I say surprising because most people, if they were given 10 seconds to guess the answer, would probably go for $\sqrt{x}$ (I certainly would have).

I'm not sure there's any real reason to suppose that the answer for the deterministic function $\Lambda(n)$ will be much different, except perhaps in logarithmic factors.

I think you have precisely no chance of saying anything useful about the $\alpha_x$, but maybe someone will prove me wrong! I would be surprised if they were not close to equidistributed, though there may be some repulsion effects away from rationals with small denominator (where $S(\alpha)$ will be large).

EDIT: Thinking about it some more, it's not obvious to me even how one would show that $S(1/x) \neq 0$, though maybe this does follow from some kind of lower bound for linear forms in $\log p$. My point is that if there is deviation from the behaviour for a random sequence I would expect that one would see it near $\alpha = 0$ (and near other rationals with small denominator).

  • $\begingroup$ Thank you very much, Professor Green. It is really interesting that the infimum goes to zero as $x \rightarrow \infty$ for the case you described! I'll look at the paper. Thanks. $\endgroup$ Oct 20, 2011 at 11:23
  • $\begingroup$ For a long time it was a conjecture of Littlewood that the infimum is $o(\sqrt{x})$. Actually it was Konyagin who got down to $x^{-1/2 + o(1)}$. $\endgroup$
    – Ben Green
    Oct 20, 2011 at 11:27

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