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Let $\zeta$ denote the Riemann zeta function and let denote it's complex zero. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), but not assuming Riemann Hypothesis ?

Assuming both the RH and the SZC, one can mimick the proof of Theorem 15.6 of Montgomery-Vaughan's Multiplicative Number Theory and show that $\frac{1}{\zeta'(\rho)} \ll X$, where $X$ is any real number $\geq |\rho| $. (Actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$). However, it looks like the bound $\frac{1}{\zeta'(\rho)} \ll X$ could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the $S(T)=\arg \zeta(\sigma + iT)$. On the RH, it is a classical fact that $S(T) \ll \frac{\log T}{\log \log T}$, whilst $S(T) \ll \log T$ unconditionally. The unconditional bound seems sufficient for the purposes of showing that $\frac{1}{\zeta'(\rho)} \ll X$ from the Montgomery-Vaughan argument.

Let $\zeta$ denote the Riemann zeta function and let $\rho$ denote one of its complex zeros. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), but not assuming Riemann Hypothesis ?

Assuming both the RH and the SZC, one can mimick the proof of Theorem 15.6 of Montgomery-Vaughan's Multiplicative Number Theory and show that $$ \frac{1}{\zeta'(\rho)} \ll X,\label{1}\tag{1} $$ where $X$ is any real number $\geq |\rho| $ (actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$).
However, it looks like the bound \eqref{1} could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the $S(T)=\arg \zeta(\sigma + iT)$. On the RH, it is a classical fact that $$ S(T) \ll \frac{\log T}{\log \log T} $$ whilst $S(T) \ll \log T$ unconditionally. The unconditional bound seems sufficient for the purposes of showing that \eqref{1} comes from the Montgomery-Vaughan argument.

Let $\zeta$ denote the Riemann zeta function and let denote it's complex zero. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), but not assuming Riemann Hypothesis ?

Assuming both the RH and the SZC, one can mimic the proof of Theorem 15.6 of Montgomery-Vaughan's Multiplicative Number Theory and show that $\frac{1}{\zeta'(\rho)} \ll X$, where $X$ is any real number $\geq |\rho| $. (Actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$). However, it looks like the bound $\frac{1}{\zeta'(\rho)} \ll X$ could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the Selberg $S(T)$ function. On the RH, it is a classical fact that $S(T) \ll \frac{\log T}{\log \log T}$, whilst $S(T) \ll \log T$ unconditionally. So the unconditional bound seems sufficient for the purposes of showing that $\frac{1}{\zeta'(\rho)} \ll X$ from the Montgomery-Vaughan argument.

Let $\zeta$ denote the Riemann zeta function and let denote it's complex zero. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), but not assuming Riemann Hypothesis ?

Assuming both the RH and the SZC, one can mimick the proof of Theorem 15.6 of Montgomery-Vaughan's Multiplicative Number Theory and show that $\frac{1}{\zeta'(\rho)} \ll X$, where $X$ is any real number $\geq |\rho| $. (Actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$). However, it looks like the bound $\frac{1}{\zeta'(\rho)} \ll X$ could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the $S(T)=\arg \zeta(\sigma + iT)$. On the RH, it is a classical fact that $S(T) \ll \frac{\log T}{\log \log T}$, whilst $S(T) \ll \log T$ unconditionally. The unconditional bound seems sufficient for the purposes of showing that $\frac{1}{\zeta'(\rho)} \ll X$ from the Montgomery-Vaughan argument.

Let $\zeta$ denote the Riemann zeta function and let denote it's complex zero. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), but not assuming Riemann Hypothesis ?

Assuming both the RH and the SZC, one can mimic the proof of Theorem 15.6 of Montgomery-Vaughan and show that $\frac{1}{\zeta'(\rho)} \ll X$, where $X$ is any real number $\geq |\rho| $. (Actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$). However, it looks like the bound $\frac{1}{\zeta'(\rho)} \ll X$ could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the Selberg $S(T)$ function. On the RH, it is a classical fact that $S(T) \ll \frac{\log T}{\log \log T}$, whilst $S(T) \ll \log T$ unconditionally. So the unconditional bound seems sufficient for the purposes of showing that $\frac{1}{\zeta'(\rho)} \ll X$ from the Montgomery-Vaughan argument.

Let $\zeta$ denote the Riemann zeta function and let denote it's complex zero. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), but not assuming Riemann Hypothesis ?

Assuming both the RH and the SZC, one can mimic the proof of Theorem 15.6 of Montgomery-Vaughan's Multiplicative Number Theory and show that $\frac{1}{\zeta'(\rho)} \ll X$, where $X$ is any real number $\geq |\rho| $. (Actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$). However, it looks like the bound $\frac{1}{\zeta'(\rho)} \ll X$ could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the Selberg $S(T)$ function. On the RH, it is a classical fact that $S(T) \ll \frac{\log T}{\log \log T}$, whilst $S(T) \ll \log T$ unconditionally. So the unconditional bound seems sufficient for the purposes of showing that $\frac{1}{\zeta'(\rho)} \ll X$ from the Montgomery-Vaughan argument.