3
$\begingroup$

Is it true that

$$|\zeta(\frac{1}{2}+it)|< \frac{1}{4}+t^2$$ for all $t\geq 0$, where $\zeta$ denotes the Riemann zeta function ?

It is known that the left hand-side is $O(t^{0.25})$, and on the the Lindelof Hypothesis, it is actually $O(t^\epsilon)$ for any positive $\epsilon$. But since these results are only valid for large $|t|$, thyey dont seem to answer my question.

Note: I have added abs value for The LHS of inequality because it is not clear if $\zeta(\frac{1}{2}+it)$ is always real number

Improve this question
$\endgroup$
2
  • $\begingroup$ In the paper: Hiary, Ghaith A. An explicit van der Corput estimate for ζ(1/2+it). Indag. Math. (N.S.) 27 (2016), no. 2, 524–533. It is proved that $|\zeta(1/2+it)|\le 0.63 t^{1/6}\log t$ for $t\ge 3$ and we have $|\zeta(1/2+it)|\le 1.461$ for $0\le t\le 3$. Of course the proof is more complicate than that in the answer of Carlo Beenakker. $\endgroup$
    – juan
    Commented Feb 26, 2018 at 16:18
  • $\begingroup$ You can find further useful bounds here: iml.univ-mrs.fr/~ramare/TME-EMT/Articles/Art06.html $\endgroup$
    – GH from MO
    Commented Feb 26, 2018 at 23:02

2 Answers 2

Reset to default
10
$\begingroup$

The inequality at the top of page 99 in Lectures on The Riemann Zeta–Function gives $$|\zeta(\tfrac{1}{2}+it)|<9t^{1/2},\;\;t\geq 1.$$ To cover the whole the interval $t\geq 0$, just add $3/2$ to the right-hand-side of the inequality.

The inequality $|\zeta(\tfrac{1}{2}+it)|<\tfrac{1}{4}+t^2$ holds for $t> 0.8$.

Improve this answer
$\endgroup$
1
$\begingroup$

Visually,

enter image description here

where black dots are values of $\zeta\left(\frac12+it\right)$ and the red ones are at the distance $\frac14+t^2$ from the origin.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .