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Let $T\geq 0$. Assume RH(T+100), that is, assume that all non-trivial zeros $\rho$ of the Riemann zeta function with $|\Im(\rho)|\leq T+100$ satisfy $\Re(\rho)=1/2$. Can we then give a good upper bound on $|1/\zeta(s)|$ (that is, a good lower bound on $|\zeta(s)|$) for $s=\sigma+it$, $|t|\leq T$, $\sigma=3/4$ or $\sigma=1$, say?

(There are well-known analogous bounds on $\zeta'(s)/\zeta(s)$.)

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    $\begingroup$ Why call it GRH? Isn't this just (restricted) RH? $\endgroup$
    – Wojowu
    Commented Nov 4, 2019 at 22:15
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    $\begingroup$ Theorem 14.8 of Titchmarsh gives a bound for $\log\zeta(1+it)$, and 14.5 gives a bound further to the left. I guess you can truncate the integral (cf. top page 342, top page 343) in the approximate functional equation method when you only have a finite RH assumption. $\endgroup$
    – MyNinthAccount
    Commented Nov 4, 2019 at 22:56
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    $\begingroup$ I think you are going to need to assume RH up to $T+(\log T)^c$ to get a bound on $1/\zeta(s)$ up to $T$. Not sure what $c$ is optimally, but likely 2 by Titchmarsh's presentation. $\endgroup$
    – MyNinthAccount
    Commented Nov 4, 2019 at 23:08
  • $\begingroup$ Actually, why can't you just integrate the well-known $\zeta'/\zeta$ bounds to get $\log\zeta$? Also, I think that the Caratheodory/Hadamard methodology (14.2) will reduce the needed zero-free region to $T+100$ as you indicated. Maybe one could do the same with the approximate functional equation if you really wanted to. $\endgroup$
    – MyNinthAccount
    Commented Nov 5, 2019 at 10:07
  • $\begingroup$ I may be missing something obvious, but wouldn't a brutal attempt in that direction result in multiplying the bounds on $\zeta'/\zeta$ by the length of the path of integration? $\endgroup$
    – H A Helfgott
    Commented Nov 5, 2019 at 11:14

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