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 $(a-1/2, 1/2]$ ?
 A: 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 $c<1/2$.
A: 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).
