# Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if the Riemann hypothesis is false?

The wiki page on Mertens conjecture and the Connection to the Riemann hypothesis says

Using the Mellin inversion theorem we now can express $M$ in terms of 1/ζ as $$M(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{x^s}{s \zeta(s)}\, ds$$ which is valid for $\color{blue}{1} < σ < 2$, and valid for $\color{red}{1/2} < σ < 2$ on the Riemann hypothesis. ... From this it follows that $$M(x) = O(x^{\color{red}{1/2}+\epsilon})$$ for all positive ε is equivalent to the Riemann hypothesis, ...

The $\color{red}{\text{red}}$ color indicates the question My question changed, due to anon's comment's, to:

If Riemann was false, would this imply a bound of $M(x)=O(x^{\color{blue}{1}+\epsilon})\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! ---------\;\;$? $\phantom{------------------------------------------------}$ Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if RH is false?

A look at Mertens function, makes me think that it should be easy to prove this.

But I still don't have a clue...

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According to the quote, $M(x)=O(x^{1+\epsilon})$ is true regardless of RH. Perhaps you're asking if $M(x)=O(x^\sigma)$ is possible with $\sigma\le 1$ even if RH is false? –  anon Jul 12 '12 at 20:31
Ah, so my color correlation is not correct? Where do they say that it is regardless true of RH? –  draks ... Jul 12 '12 at 20:45
The point of separating the two thoughts, and qualifying the second with "on the assumption of RH," while not qualifying the first, is that the first needs no such qualification. (English! Huzzah!) –  anon Jul 12 '12 at 20:52
Even without references, the bound $|M(x)| \le x$ is trivial, which is already sharper than $O(x^{1+\epsilon})$. The point is that we don't even know a bound of the form $|M(x)| \le x^{0.999}$. The best known bound is roughly $M(x) = O(x \exp(-c \log^{0.6+\epsilon} x)$, so at least we know $M(x) = o_A(x/\log^A x)$ for all $A>0$. –  Erick Wong Jul 12 '12 at 21:11
@draks Certainly $M(x) = O(x^\sigma)$ is possible, provided RH is false in a specific way. For instance, if all the non-trivial zeros of $\zeta(s)$ are confined to the region $\{ 1/4 < \Re(z) < 3/4 \}$, then we would still have $M(x) = O(x^{3/4+\epsilon})$. –  Erick Wong Jul 12 '12 at 21:14

There is a trivial bound of $|M(x)| \le x$ for all $x\ge 0$, because the Möbius function is bounded by $1$. So we already have $M(x) = O(x^1)$ regardless of whether RH is true or false.

This turns out to be rather close to the best known unconditional bound on $M(x)$, which looks like $$M(x) = O(x \exp( -c\, \log^{0.6} x \log\log^{-0.2} x))$$ (for instance see this paper of Nathan Ng). In other words, we do not even have a proven upper bound of the form $O(x^{0.999})$.

Because the Dirichlet series for $\mu(x)$ is just $1/\zeta(s)$, bounds on $M(x)$ can be obtained by Perron's formula using knowledge of the poles of $1/\zeta(s)$, in other words the zeros of $\zeta(s)$. (It is worth noting that Granville and Soundararajan have recently developed a new approach to many such problems without intimate knowledge of $\zeta(s)$ in the critical strip.)

The fact that we are still rather ignorant about the zeros of $\zeta(s)$ is the reason we don't know a significantly better bound for $M(x)$ than the trivial one. At the same time, RH being false is not a very strong statement: it just means there is some zero of $\zeta(s)$ with $\Re(s) > 1/2$. While this does preclude a bound of $M(x) = O_\epsilon(x^{1/2+\epsilon})$, it does not rule out $M(x) = O(x^{3/4+\epsilon})$, if all the zeros of $\zeta(s)$ happen to lie to the left of $\Re(s) = 3/4$.

One last comment: it is somewhat naive to use the observed differences as evidence of how easy it is to prove a bound. For a more striking example, try graphing the prime gaps function $d(n) = p_{n+1} - p_n$. You will find that $d(n)$ appears to be $O(\log^2 n)$ (with a pretty small constant), but even assuming RH we don't know how to prove $d(n) = O(\sqrt{n}).$

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+1 Great answer. Vote up, people! –  draks ... Jul 12 '12 at 22:05