# Robin's inequality and the zeros of the Riemann zeta function

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 $$(a-1/2, 1/2]$$ ?

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 $$c<1/2$$.

• 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... – Fourton. Apr 24 '19 at 12:09

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).

• thanks ! Only if MO could allow one to accept two answers ! – Fourton. Apr 24 '19 at 18:38
• 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 $\Re(s)=1/2+\beta, \beta>0$. – user83236 Apr 29 '19 at 20:50
• 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 $[a-1/2, 1/2)$. – user83236 Apr 29 '19 at 20:59
• ...for infinitely many positive integers $N$... – user83236 Apr 29 '19 at 21:12
• 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 $(a-1/2, 1/2]$. – user83236 Apr 29 '19 at 21:20