# On an oscillation Theorem involving the Chebyshev function and the zeros of the Riemann zeta function

Define $$\theta(x)=\sum_{p\leq x} \log p$$, where $$p>1$$ denotes a prime. Nicolas proved that if the Riemann zeta function $$\zeta(s)$$ vanishes for some $$s$$ with $$\Re(s)\leq 1/2 + b$$, where $$b\in(0, 1/2)$$, then

$$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{-b}).$$

Is the reverse necessarily true ? It appears so to me, since

$$\theta(x)=x +O(\sqrt{x})-\sum_{\rho} \frac{x^{\rho}}{\rho},$$

where the sum is over the complex zeros of $$\zeta$$ and $$\prod_{p\leq x} (1-p^{-1})\sim e^{-\gamma}(\log x)^{-1} ?$$

• Do you have a reference? – András Bátkai Mar 10 at 18:41
• @AndrásBátkai Nicolas puts most of his stuff on his site. math.univ-lyon1.fr/~nicolas and publications at math.univ-lyon1.fr/~nicolas/publications.html – Will Jagy Mar 10 at 22:31
• @WillJagy: thanks, I also found this. I meant a bit more precise reference for the result recalled above. – András Bátkai Mar 11 at 8:07
• @AndrásBátkai, I see what you mean. Maybe the OP will loh at th Nicolas page and indicate which publication is relevant – Will Jagy Mar 11 at 17:06
• The explicit formula doesn't help a lot because it doesn't converge absolutely. Under the RH $\theta(x) = x+O(\sqrt{x} \log^2x)$ you can't improve it much. It is easy to show if $\theta(x)-x = o(x^{\sigma})$ then $\zeta(s)$ has no zeros for $\Re(s) \ge \sigma$, the converse is much harder – reuns Mar 11 at 21:08

## 2 Answers

The result Nicolas proved (Theorem 3, Jean-Louis Nicolas. Petites valeurs de la fonction d'Euler, J. Number Theory, 17, 1983, 375--388, paper linked HERE) is actually not quite what is claimed in this post, but:

$$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{b-\frac{1}{2}-\epsilon})$$, for all $$\epsilon > 0$$, where $$0< b < \frac{1}{2}$$ is such that RZ has a zero with real part $$\frac{1}{2} + b$$, so it is in the classical spirit as the worse RH fails (the higher the supremum of the real part of the critical zeros), the stronger the claimed $$\Omega$$.

The estimate claimed in the post above is wrong (obviously because it is stronger for a weaker claim, not to speak that since RZ always vanishes on the critical line the claim of the post would imply the estimate to be true for $$b$$ going to zero and that contradicts the later claim when RZ is true). Using the zeros symmetry around the critical line, one can rephrase the $$\Omega$$ result to:

$$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{-\beta-\epsilon})$$, for all $$\epsilon > 0$$, where $$0< \beta < \frac{1}{2}$$ is such that RZ has a zero with real part $$\beta$$

He also proved that if RH is true $$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)x^{\frac{1}{2}}\log x$$ is always negative for $$x>2$$, has inferior limit $$\log{4\pi}-4-\gamma$$ and superior limit less or equal to $$\gamma - \log{4\pi}$$ which are both close to $$-2$$, so obviously any $$\Omega$$ result as above with a power $$x^{-\beta}, 0 < \beta < \frac{1}{2}$$ dispproves RH, with the lower the $$\beta$$ the worse RH fails, while any $$O$$ result gives bounds on the supremum of the real parts of critical zeros, with the higher the $$\beta$$ the better the bounds

• You claim in the MSE post that the inequality is reversed and the PNT corresponds to $\beta=0$. Both of these claims are clearly false because 1.) The inequality sign is clearly ''>'' not otherwise, and you can find this everywhere in the literature, e.g Lagarias ''An elementary problem equivalent to the RH''. 2.) The PNT certainly doesn't correspond to $\beta=0$, because the PNT is the statement that $\zeta(s)\neq 0$ for $\Re(s)=1/2 + \beta =1$. – ABD. Apr 29 at 18:19
• Actually, the fact that the inequality is false for $\beta=0$ is independent of the PNT. It is indeed an immediate consequence of Gronwall's classical result that $\lim sup_{n\rightarrow \infty} \frac{\sigma(n)}{n\log \log n} =e^{\gamma}$. – ABD. Apr 29 at 18:43
• The mse post is a fallacy and it is based on a complete misinterpretation of Robin results. Robin proved 3 things, equivalence to RH, unconditional result that shows an O result and what happens when there are zeros with real part in $(.5,1)$ and in that case there is an omega results that gets more powerful, the higher the real parts of zeros get. The fact that there are no zeros at 1 has no relevance there, but it can be interpreted - not proved per se, just INTERPRETED to confirm that the powerful omega result cannot hold at zero, so the statement on mse – Conrad Apr 29 at 21:37
• see mathoverflow.net/q/329837/83236. Quite stranger that you've now diverted from your first comments on the subject. – ABD. Apr 29 at 21:45
• We seem to go in circles - as I noted in other posts on this theme, you ain't going to convince me that false is true whatever rhetoric or sophistry you use; the papers of Robin and Nicolas are quite clear and I looked them up earlier today to confirm details like theorem number and page to avoid misrepresentations like the one above regarding PNT; your post on mse, comments here etc are fallacies and they seem to be in a malicious intentional spirit not in an honest mistake one and as noted it is pointless to keep explaining your fallacy, but feel free to go ahead and publish your results – Conrad Apr 29 at 22:03

Actually, in the paper cited in answer one, the following is proved:

$$\log (e^\gamma\log(\theta(x))\prod_{p\le x}(1-1/p))<0$$ for $$x\ge 3$$ if and only if the Riemann Hypothesis holds ... I don't know how to link to other posts, but essentially this same question is asked again (this one came first).

• i think it's because the result seems to have some highly non-trivial implications, see mathoverflow.net/q/330258/83236 – ABD. Apr 29 at 6:52
• The link you provide is not available any more. – EGME Apr 29 at 11:31
• In the post mathoverflow.net/q/329837/83236 it seems you agreed that $a\in(1/2, 1]$ is the supremum of the real parts of the zeros of $\zeta$ if and only if $f(x)=\Omega_{\pm}(x^{-b})$, Notice that if the inequality concerned is false for some $b$, then it is also false for any $b'<b$. Thus since $a\in(1/2, 1]$ is the supremum of the real parts of the zeros of $\zeta$ if and only if $f(x)=\Omega_{\pm}(x^{-b})$, it follows that the inequality is false for $\beta=1/2$ since $\zeta(s)\neq 0$ for $\Re(s)=1/2+\beta\geq 1$, hence $\zeta(s)\neq 0$ for $\Re(s)>1/2$. – ABD. Apr 29 at 18:35
• ...where $b$ can be taken to have any value in $(a-1/2, 1/2]$. – ABD. Apr 29 at 18:46
• $ABD: Ok, I see the post again, strange. If$a>1/2$the$f$oscillates (proved in theorem 3, more specifically, for all$b$between$1-a$and$1/2$,$f(x)=\Omega_{\pm}(x^{-b})$), and if$f$oscillates, then$a>1/2$(this much follows). Hmm, now I am not sure which inequality you are referring to which may be false for some$b\$, maybe this is better discussed in that post, I can’t see two posts at the same time on my device. I hope I have no typos, I can’t see the latex in comments. – EGME Apr 29 at 19:48