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 $\prod_{p\leq x} (1-p^{-1})\sim e^{-\gamma}(\log x)^{-1} ?$

  • $\begingroup$ Do you have a reference? $\endgroup$ – András Bátkai Mar 10 '19 at 18:41
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    $\begingroup$ @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 $\endgroup$ – Will Jagy Mar 10 '19 at 22:31
  • $\begingroup$ @WillJagy: thanks, I also found this. I meant a bit more precise reference for the result recalled above. $\endgroup$ – András Bátkai Mar 11 '19 at 8:07
  • $\begingroup$ @AndrásBátkai, I see what you mean. Maybe the OP will loh at th Nicolas page and indicate which publication is relevant $\endgroup$ – Will Jagy Mar 11 '19 at 17:06
  • $\begingroup$ 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 $\endgroup$ – reuns Mar 11 '19 at 21:08

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

  • $\begingroup$ 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 $\Re(s)=1/2 + \beta =1$. $\endgroup$ – user83236 Apr 29 '19 at 18:19
  • $\begingroup$ 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 $\lim sup_{n\rightarrow \infty} \frac{\sigma(n)}{n\log \log n} =e^{\gamma}$. $\endgroup$ – user83236 Apr 29 '19 at 18:43
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    $\begingroup$ 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 $\endgroup$ – Conrad Apr 29 '19 at 21:37
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    $\begingroup$ 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 $\endgroup$ – Conrad Apr 29 '19 at 22:03
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    $\begingroup$ I am not the one making false extraordinary claims, just pointing out that the claims you are making are incorrect and based on a fundamental misunderstanding of those papers; the Nicolas and Robin papers are clear and the answer above explains precisely what Nicolas' paper proves and how it is related with classical results on RH and its possible failure; @EGME's answer below is actually included in my last paragraph with more information regarding the limit points. $\endgroup$ – Conrad Apr 30 '19 at 16:02

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).

  • $\begingroup$ i think it's because the result seems to have some highly non-trivial implications, see mathoverflow.net/q/330258/83236 $\endgroup$ – user83236 Apr 29 '19 at 6:52
  • $\begingroup$ The link you provide is not available any more. $\endgroup$ – EGME Apr 29 '19 at 11:31
  • $\begingroup$ 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 $\Re(s)>1/2$. $\endgroup$ – user83236 Apr 29 '19 at 18:35
  • $\begingroup$ ...where $b$ can be taken to have any value in $(a-1/2, 1/2]$. $\endgroup$ – user83236 Apr 29 '19 at 18:46
  • $\begingroup$ $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. $\endgroup$ – EGME Apr 29 '19 at 19:48

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