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 $? 



Does $ M(x)=O(\sqrt{x}) $ if and only if the De Bruijn-Newman constant is negative?
The de Bruijn-Newman constant is nonnegative, as proved in this brand new [preprint](https://arxiv.org/abs/1801.05914) 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.


