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Let $\gamma$ denote the imaginary part of a non-trivial zero of the Riemann zeta function. Do there exist some function $f$ such that $\gamma_{n+1} - \gamma_n > f(n)>0$ for all large $n$? To be more explicit, do there exist some $c > 1$ such that $\gamma_{n+1}-\gamma_n > n^{-c}$ for all $n>n_0$ ?

Higgs Boson
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    If you do not want answers like "$f(n)=\gamma_{n+1}-\gamma_n$" or "$f(n)=0$", you must make an effort when asking. What are your conditions on $f$? – მამუკა ჯიბლაძე Aug 03 '23 at 10:44
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    Not enough effort. You still can get answers like $(\gamma_{n+1}-\gamma_n)/2$, for example – მამუკა ჯიბლაძე Aug 03 '23 at 10:54
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    It is also not expected that others will formulate your question for you. The way it is now, your question is ambiguous. Depending on restrictions on $f$ that you need it can have trivial answers, or can be equivalent to a well known open problem, or can have many other kinds of answers in between. – მამუკა ჯიბლაძე Aug 03 '23 at 11:01
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    Maybe an example will help me to explain. Here is an example showing one way to make your question unambiguous, with an answer. I have no idea however whether this is what you need, maybe you need something different. – მამუკა ჯიბლაძე Aug 03 '23 at 11:11
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    I believe it must be mentioned that another user kindly did what had to be done by you. – მამუკა ჯიბლაძე Aug 03 '23 at 12:27
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    The addendum does not figure. Since there are $\sim\frac T{2\pi}\log T$ roots with imaginary part in $[0,T]$, we have $\gamma_n\sim\frac{2\pi n}{\log n}$, and there are infinitely many $n$ such that $\gamma_{n+1}-\gamma_n\le\frac{2\pi+\epsilon}{\log n}$. That is, not only $\mu=0$ holds unconditionally, we have in fact $\liminf_{n\to\infty}(\gamma_{n+1}-\gamma_n)\log n\le2\pi$. Or did you just intended to write the latter? – Emil Jeřábek Aug 03 '23 at 13:31
  • @EmilJeřábek, see my comment below GH's answer. – Higgs Boson Aug 03 '23 at 14:35
  • I can't make much sense of the comment. It seems to be asking for something completely different from the question. – Emil Jeřábek Aug 03 '23 at 14:48
  • @EmilJeřábek, , like I said, maybe the original question is unclear as I'm an experimental physicist fascinated by number theory. But what I wrote below GH's answer is basically what I intended to ask. I saw a claim somehwere that if $\rho_T$ is a zero of $\zeta$ with $\Im(\rho_{T}) \approx T$ and $\varepsilon_{T} >0$ is small "enough", then $\zeta(1-\rho_{T} - \varepsilon_{T})$ is uniformly $o(1)$ for $T \geq T_0$. I want to understand why this should be true. – Higgs Boson Aug 03 '23 at 15:37
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    I can’t say I understand what you mean by “uniformly” when $\varepsilon_T$ depends on $T$. But anyway, it seems you want upper bounds on $\zeta'$ rather than lower bounds on gaps between zeros. – Emil Jeřábek Aug 03 '23 at 15:59
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    @EmilJeřábek I agree. I answered the OP's new question in a comment below my original answer. – GH from MO Aug 03 '23 at 17:36

1 Answers1

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Your first question can be reformulated as follows: Are the nontrivial zeros of the Riemann zeta simple? We don't know the answer to that question, even under the Riemann hypothesis.

GH from MO
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  • Maybe the question is somewhat unclear (as I'm an experimental physicist by specialty). However, it's more of a follow up to https://mathoverflow.net/q/450518/507786. That is, i'm basically asking that if $\beta+i\gamma_T$ is a zero of $\zeta$ (of whatever multiplicity) where $\gamma_T \approx T$, how small should $\varepsilon_T >0$ be for $\zeta(\beta + i\gamma_T +\varepsilon_T)$ to be uniformly $o(1)$ for all $T \geq T_0$ ? – Higgs Boson Aug 03 '23 at 12:36
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    @HiggsBoson What you are asking in your comment is very different from what you asking in your post above. First of all, if we don't assume the Riemann hypothesis, then $\beta$ should be $\beta_T$. And then, the convexity bound for $\zeta(s)$ implies that we can take $\varepsilon_T=T^{(\beta_T-\delta)/2}$ for any fixed $\delta>1$. This can be improved by using a subconvexity bound for $\zeta(s)$, the current record being due to Bourgain (2017). If we assume the Riemann hypothesis, then a much larger $\varepsilon_T$ is admissible. – GH from MO Aug 03 '23 at 17:23
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    Helpful as always, thanks GH. By the way, sorry for my unclear first question. I'm more used to analyzing data from particle accelerators! – Higgs Boson Aug 03 '23 at 18:37