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The Riemann hypothesis is equivalent to the assertion that the De Bruijn-Newman constant $ \Lambda $ , as defined in https://www.sciencedirect.com/science/article/pii/S0001870809001133/pdf?md5=d2b0cbb38f79b80de06d8b9c99836fab&pid=1-s2.0-S0001870809001133-main.pdf&_valck=1, fulfills $ \Lambda\leq 0 $. On the other hand, Newman conjectured that $ \Lambda\geq 0 $, so that RH, if true, would be only barely so. So far it has been proven that $ -1.1\times 10^{-12}\lt\Lambda\lt 1/2 $.

Another assertion equivalent to RH is $ M(x)\ll_{\varepsilon}x^{1/2+\varepsilon} $ where $ M $ is the Mertens function defined as the summatory function of the Möbius function, and it has been proven that the stronger statement $ M(x)=O(\sqrt{x}) $ is incompatible with the great linear independence hypothesis, stating that, under RH, the imaginary parts of the standard L-functions are linearly independent over $ \mathbb{Q} $, the latter being widely believed to be true.

That way, the possibility that $M(x)=O(\sqrt{x}) $ may be too good to be true sounds similar to Newman's conjecture. So my question is : is $ M(x)=O(\sqrt{x}) $ actually equivalent to $ \Lambda\lt 0 $?

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The de Bruijn-Newman constant is nonnegative, as proved in this brand new preprint by Rodgers and Tao. It is also conjectured, but not proven yet, that $M(x)=O(\sqrt{x})$ is false, in which case it is actually equivalent to $\Lambda<0$. Time will tell.

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  • $\begingroup$ Wonderful ! I would not have thought that Newman's conjecture would be proven in my lifetime. Maybe a proof of RH will follow...Thank you very much for your answer. $\endgroup$
    – Sylvain JULIEN
    Commented Jan 19, 2018 at 10:00
  • $\begingroup$ @SylvainJULIEN: You are welcome. Actually, I was (or rather: we are) lucky as Rogders and Tao just proved the conjecture. There is a lot of activity and amazing progress in analytic number theory these days, thanks to great mathematicians captivated by the subject. $\endgroup$
    – GH from MO
    Commented Jan 20, 2018 at 2:26
  • $\begingroup$ @GHfromMO Hello GH, could you please elaborate on equivalence of $M(x)=O(\sqrt x)$ and $\Lambda<0$, or include a reference? $\endgroup$
    – Sungjin Kim
    Commented Jan 22, 2018 at 19:47
  • $\begingroup$ I did not say that these two statements are equivalent. I said that if $M(x)=\sqrt{x}$ is false (as expected) then it is equivalent to $\Lambda<0$ (which is false by the preprint quoted). My claim is trivial, because any two false statements are equivalent. $\endgroup$
    – GH from MO
    Commented Jan 22, 2018 at 20:24

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