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Robin showed that if $a\in(1/2, 1]$ is the supremum of the real parts of the zeros of the Riemann zeta function $\zeta(s)$, then $f(x)=\Omega_{\pm} (x^{-b})$, where $b$ is some number on $(a-1/2, 1/2],$ $$f(x)=\log \Big(e^{\gamma}\log \theta(x)\prod_{p\leq x} (1-p^{-1})\Big),$$ $\theta(x)=\sum_{p\leq x} \log p$, the Chebyshev sum over the primes $p\leq x$ and $\gamma=0.577\cdots$ the Euler constant.

But is it true that if $\zeta(s)\neq 0$ for $\Re(s)\in(1/2 , 1]$, then $f(x)\neq \Omega_{\pm} (x^{-c})$ for any $c\in (0, 1/2]$?

In other words, is it true that $a\in(1/2, 1]$ is the supremum of the real parts of the zeros of the Riemann zeta function if and only if $f(x)=\Omega_{\pm} (x^{-b})$, where $b$ is some number on $(a-1/2, 1/2]$ ?

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It is well-known that the Riemann hypothesis implies $$ \theta(x)=x+O(\sqrt{x}\ln^2 x). $$ Therefore, under the Riemann hypothesis we have $$ \ln\theta(x)=\ln x+O\left(\frac{\ln^2 x}{\sqrt{x}}\right). $$ Also, from the partial summation we get $$ \sum_{p\leq x}\frac{1}{p}=\int_{1.5}^x \frac{d\theta(t)}{t\ln t}=\ln\ln x+M+O\left(\frac{\ln x}{\sqrt x}\right). $$ Now, from Mertens' theorems we obtain $$ \ln(e^\gamma \prod_{p\leq x}\left(1-\frac{1}{p}\right))=-\sum_{p\leq x} \frac{1}{p}+M+O\left(\frac{1}{x}\right)=-\ln\ln x+O\left(\frac{\ln x}{\sqrt x}\right). $$ Therefore assuming RH we deduce that $$ f(x)=\ln\ln\theta(x)+\ln(e^\gamma \prod_{p\leq x}\left(1-\frac{1}{p}\right))=O\left(\frac{\ln x}{\sqrt x}\right), $$

which is certainly not $\Omega(x^{-c})$ for any $c<1/2$.

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  • $\begingroup$ Thanks, so $a\in(1/2, 1]$ is the supremum of the real parts of the zeros of the Riemann zeta function if and only if $f(x)=\Omega_{\pm} (x^{-b})$, where $b$ is some number on $(a-1/2, 1/2]$ ! Interesting... $\endgroup$ – Fourton. Apr 24 '19 at 12:09
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A bit more can be said. Assuming RH, $f(x)<0$ for $x$ sufficiently large; and conversely, if $f(x)<0$ for $x$ sufficiently large, then RH holds (you can take $x\geq 3$, this is in Nicolas, J.L., “Petits valeurs de la fonction d’Euler”, Journal of Number Theory 17 (1983) 375-388).

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    $\begingroup$ thanks ! Only if MO could allow one to accept two answers ! $\endgroup$ – Fourton. Apr 24 '19 at 18:38
  • $\begingroup$ Notice that if the inequality $\sum_{d\mid N} d \geq e^{\gamma}N\log \log N + \frac{c\log \log N}{(\log N)^{\beta}}$ is false for some $\beta$, then it is also false for any $\beta'< \beta$. And since $a\in(1/2, 1]$ is the supremum of the real parts of the complex zeros of $\zeta(s)$ iff $f(x)=\Omega_{\pm}(x^{-b})$, it follows that this inequality is false for $\beta=1/2$ since $\zeta(s)\neq 0$ for $\Re(s)=1/2 + \beta\geq 1$. Hence it must also be false for any $\beta<1/2$, thus $\zeta(s)\neq 0$ for $\Re(s)=1/2+\beta, \beta>0$. $\endgroup$ – user83236 Apr 29 '19 at 20:50
  • $\begingroup$ Note that Robin demonstrated that if $\theta=1/2 + \beta, \beta\in(0, 1/2]$ is the supremum of the zeros of $\zeta(s)$, then $\sum_{d\mid N} d \geq e^{\gamma}N\log \log N + \frac{c\log \log N}{(\log N)^{c}}$, where $c$ can be taken to have any value on $[a-1/2, 1/2)$. $\endgroup$ – user83236 Apr 29 '19 at 20:59
  • $\begingroup$ ...for infinitely many positive integers $N$... $\endgroup$ – user83236 Apr 29 '19 at 21:12
  • $\begingroup$ Robin's result is a corollary of Nicolas' result that if $a\in(1/2, 1]$ is the supremum of the real parts of the zeros of the Riemann zeta function, then $f(x)=\Omega_{\pm}(x^{-c})$. Thus if it could be shown that $\zeta(s)\neq 0$ for $\Re(s)>1/2$ entails that $f(x)\neq \Omega_{\pm}(x^{-c})$, then it would follow that $a\in(1/2, 1]$ is the sup of the zeros of $\zeta$ iff $f(x)=\Omega_{\pm}(x^{-c})$, where $c$ is some real number on $(a-1/2, 1/2]$. $\endgroup$ – user83236 Apr 29 '19 at 21:20

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