Let $\zeta$ denote the Riemann zeta function. By an argument of Robin http://zakuski.utsa.edu/~jagy/Robin_1984.pdf, we know that $\zeta(\rho)=0$ for some $\rho$ with $\Re(\rho) \in (1/2, 1/2 + \beta]$, where $0<\beta\leq 1/2$, if and only if there exists some positive constants $\beta'$ and $c$ such that \begin{equation} \sum_{d|N} d \geq e^{\gamma}N\log \log N + \frac{cN\log \log N}{(\log N)^{\beta'}} \end{equation} for infinitely many positive integers $N$, where $\gamma$ denotes the Euler-Mascheroni constant and $\beta'$ can be taken to have any value satisfying $1/2 -\beta <\beta'<1/2$.

Notice that if it could be shown that the above inequality is false for some $\beta'$, then it must also be false for any $\beta''<\beta'$, implying that if $\zeta(\rho)\neq 0$ for $\Re(\rho)=1/2 + \beta$, then $\zeta(\rho)\neq 0$ for $1/2<\Re(\rho)\leq 1/2 + b$, where $0<b< \beta$.

Is this a known result ?


put on hold as off-topic by Yemon Choi, Jan-Christoph Schlage-Puchta, Dima Pasechnik, András Bátkai, Alexey Ustinov 27 mins ago

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  • $\begingroup$ I don't see how this follows from Robin's result - it concerns the existence of zeros with real part in the strip $(1/2,1/2+\beta]$, so I'm not sure how you're using the assumption that $\zeta$ has no zeros on the line $\Re(\rho)=1/2+\beta$? $\endgroup$ – Thomas Bloom Mar 10 at 9:34
  • $\begingroup$ I'm voting to close this question because of the sock-puppetry (visible to users with 10k+ rep) $\endgroup$ – Yemon Choi yesterday