This paper p.4 claims:

Corollary C. Assume RH. For all large $t$ we have

$$|\zeta(\frac12 +it)| \le \exp\left(\frac38 \frac{\log{t}}{\log{\log{t}}}\right) \qquad (1) $$

$t$ a Gram points often appears counterexample to (1) according to mpmath, pari and maple. E.g. for $t=2381374874120.4, \, 352.4788 \not \le 24.2954 $

Other possible counterexamples $t$ are $$4992394.753, 42653554.76625,3293531640.5520, 29538618461.012969578$$ $$267653395649.1305498,2445999556058.1,2381374874120.4$$

The last one is $13$ digits.

How is "large" defined?

Added 2013-06-28 According to Computations of the Riemann zeta function the inequality fails for $t \sim 3.925 \cdot 10^{31}$.

For $t=39246764589894309155251169284104.05199$, $\zeta(1/2 + it)=15837.8712 + 3604.9344i, |\zeta(1/2 + it)|=16242.95904$, while the bound from the inequality is $580.2737$ and $\log\log{t}=4.28$.

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    $\begingroup$ "Large" $t$ in practice, for this example probably means large enough for $\log\log t$ to be non-negligible. $\endgroup$ – v08ltu Jun 19 '13 at 11:53
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    $\begingroup$ Since you are asking about a definition, “for all large $t$ ...” is defined as “there exists $t_0$ such that for all $t>t_0$ ...”. There is a priori no telling what $t_0$ is, though chances are you could extract it from the proof if it is constructive enough. $\endgroup$ – Emil Jeřábek Jun 19 '13 at 11:58
  • $\begingroup$ This is equivalent to the best bound in the Riemann von Mangoldt law ( $O( \log(T) / \log \log(T) )$ assuming RH, $O(\log T)$ unconditionally, $o( \log T)$ assuming Lindelöff). $\endgroup$ – Marc Palm Jun 19 '13 at 11:59
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    $\begingroup$ To get a feel how 'large' t ought to be in this case, I think the first is to note that the 3/8 arises as a further simplification of (1+c)/4 + o(1) where c is quite close to 1/2, namely 0.491... So the first thing to do would be check what this o(1) is more precisely (from the proof). Since after all it must be only quite small numerically, so that it not working for 13 digits would not surprise me at all. $\endgroup$ – user9072 Jun 19 '13 at 13:15
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    $\begingroup$ FYI, this inequality has been sharpened: blms.oxfordjournals.org/content/43/2/243.abstract $\endgroup$ – Micah Milinovich Jun 19 '13 at 14:43

You could look at Chandee's paper (in Proc AMS) and on arxiv at http://arxiv.org/pdf/0906.4177v1.pdf where explicit bounds are worked out for zeta and L-function, and it is specified when they start to hold. For example see Corollary 1.4 there which holds for $t$ at least $\exp(\exp(10))$ which is about $10^{9500}$.

  • $\begingroup$ Thank you. Not sure if current implementations can compute zeta or Z(t) in this range (sage, pari and lcalc fail significantly earlier). $\endgroup$ – joro Aug 18 '13 at 9:19

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