-1
$\begingroup$

Let $N(T)$ be the number of zeros of Riemann zeta function upto height $T$ in the critical strip and $N_0(T)$ be the number of zeros on the critical line.

What will be the significance of proving that there is an $H$ such that for $T\geq H$, $$N_0(T+1)-N_0(T)\sim \frac{1}{2\pi}\log \frac{T}{2\pi}$$ and $$N(T+1)-N(T)\sim \frac{1}{2\pi}\log \frac{T}{2\pi}$$ Are the above results (especially the first one) known? Please explain the importance of the above results.

edit So if the above two results are proved we can conclude that $$\liminf_{T\to \infty}\frac{N_0(T+1)-N_0(T)}{N(T+1)-N(T)}=1$$ Thank you.

$\endgroup$
5
  • $\begingroup$ Why the downvote? Anything wrong in the above question? $\endgroup$
    – user485483
    Jul 16, 2022 at 13:58
  • 3
    $\begingroup$ I didn't downvote, but see this question: mathoverflow.net/questions/161442/… $\endgroup$
    – Stopple
    Jul 16, 2022 at 15:28
  • $\begingroup$ @Stopple Thanks. I saw the question which you suggested and its answer also. But in that question we have the 100% hypothesis as $$\liminf_{T\to\infty}\frac{N_0(T)}{N(T)}=1$$ whereas in my question we have $$\liminf_{T\to\infty}\frac{N_0(T+1)-N_0(T)}{N(T+1)-N(T)}=1$$ Are these one and the same? $\endgroup$
    – user485483
    Jul 16, 2022 at 16:59
  • $\begingroup$ @Stopple Thanks. So what is the correct asymptotic for $N(T+1)-N(T)$? $\endgroup$
    – user485483
    Jul 16, 2022 at 18:13
  • $\begingroup$ @Stopple Thanks. What will be the Big O term? $\endgroup$
    – user485483
    Jul 16, 2022 at 19:43

1 Answer 1

2
$\begingroup$

The first two displays (together) are in between the Lindelöf hypothesis and the Riemann hypothesis. That is, they imply the Lindelöf hypothesis, while they follow from the Riemann hypothesis. They are not known unconditionally. See Sections 13.5-13.6 in Titchmarsh: The theory of the Riemann zeta-function.

It is straightforward that the first two displays imply the third display in the stronger form $$\lim_{T\to \infty}\frac{N_0(T+1)-N_0(T)}{N(T+1)-N(T)}=1.$$ To see this, divide both the numerator and the denominator by $\frac{1}{2\pi}\log \frac{T}{2\pi}$, and observe that the new numerator and denominator tend to $1$ by assumption.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.