2
$\begingroup$

Say I have a good estimate for the $L^2$ mean of the Riemann zeta function $\zeta(s)$ for $\Re s = 1/2$, $|t|\leq T$:

$$\int_0^T |\zeta(1/2+i t)|^2 = T \log T - T (1 + \log 2 \pi - 2\gamma) + O(T^\alpha).$$

for some $\alpha\in (0,1/2]$. Does a good estimate on $$\int_0^T |\zeta(\sigma+it)|^2$$ follow, for $\sigma<1$, $\sigma\ne 1/2$? What would its error term be?

$\endgroup$
3
  • $\begingroup$ If by "a good estimate" you mean a main term with a smaller error term, I don't think it is possible to deduce that from the mean value on the 1/2-line. If $\sigma > 1/2$ then the main term is $T$, not $T\log T$. The change in the shape of the main term suggests that something special is going on. If you have a bound on $\zeta(\sigma_0 + it)$ as $t\to\infty$, then by convexity you get a bound on $\zeta(\sigma + it)$ as $t\to\infty$, for any fixed $\sigma > \sigma_0$. That is sort-of like what is being asked. $\endgroup$ Feb 21, 2021 at 15:09
  • $\begingroup$ If you go to the function field / polynomial model, i.e. pretend zeta is a polynomial in $q^{-s}$, $\sum_{n=0}^{\log T} a_n q^{-ns}$, then the second moment on the $\sigma+it$ line is $\sum_{n=0}^{\log T} |a_n|^2 q^{ -2n \sigma}$. If this is $T \log T$ for $\sigma=1/2$ then this implies an upper bound of $T \log T$ for $\sigma > 1/2$, but not an asymptotic, and an upper bound of $T \log T q^{ \log T (1 - 2 \sigma)} = T^{2 -2\sigma} \log T$ for $\sigma <1/2$, but again not an asymptotic. $\endgroup$
    – Will Sawin
    Feb 21, 2021 at 19:57
  • $\begingroup$ Back to the real world, this suggests that a purely analytic convexity-like argument should exist that gives a strong upper bound on your moment, but I concur with David Farmer that an asymptotic probably isn't possible just from this. (Although maybe the fact that you know the lower-order terms of the zeta function explicitly and can bound the higher-order terms using your knowledge of $\sigma=1/2$ combines to give something useful, although I imagine your interest is not exactly in $\zeta$, which complicates things.) $\endgroup$
    – Will Sawin
    Feb 21, 2021 at 19:59

0

Your Answer

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