1
$\begingroup$

Claim: Take any function $f(t) > 0$ for $t > 0$, such that $f(t) \to \infty$ as $t \to \infty$, then for $\sigma > 0$ $$|\zeta(\sigma + it)| = o(f(t))$$

Is there any already existing evidence, like papers or proofs or something that can debunk this?

As far as I know, under the Lindelof hypothesis $$|\zeta(\frac{1}{2} + it)| = o(t^\epsilon)$$ and Littlewood has already proved that under the Riemann hypothesis $$|\zeta(\frac{1}{2} + it)| = o\left(\exp\left(\frac{10\log t}{\log \log t}\right)\right)$$ both of which agree with the argument.

Also I know from this paper at http://arxiv.org/pdf/math/0612106v2.pdf that $$\int_0^T |\zeta(1/2 + it)|^{2k}dt \gg_k T (\log T)^{k^2}$$ which kind of gives me a hint that there must be an obvious lower bound which can probably show that the condition given above for $\zeta(\sigma + it)$ and $f(t)$ is invalid.

Looking for references.

$\endgroup$
1
  • $\begingroup$ If $0 < \sigma < 1/2$ then we could set $\sigma' = 1 - \sigma$ and deduce using the functional equation that $\zeta(\sigma' + it)$ decays faster than some power of $|t|$ as $|t| \rightarrow \infty$, and that's surely absurd... $\endgroup$ – Noam D. Elkies Jun 24 '12 at 17:13
11
$\begingroup$

First, your condition seems a bit strange to me. Since it seems to me it would imply that that the absolute value of $\zeta(1/2 + it)$ is bounded which would contradict the lower bound on the moments you recall.

Yet, second, here is one (unconditional) result on $|\zeta(1/2 + it)|$ that gives some information you seem to seek (Jutila 1983, Bull LMS):

There exist positve constants $a,b,c$ such that for each $T\ge 10$ one has

$$ \exp( a(\log \log T)^{1/2}) \le |\zeta(1/2 + it)| \le \exp ( b(\log \log T)^{1/2})$$
for $t$ in a subset of $[0,T]$ of measure at least $cT$.

$\endgroup$
2
4
$\begingroup$

Read Theorem 8.12 in Titchmarsh:

For $\frac12 \le \sigma <1$ take $0<\alpha <1-\sigma$. Then the inequality $|\zeta(\sigma+it)| > \exp(\log^\alpha t)$ is satisfied for indefinitely large values of $t$.

Therefore $f(t)=\exp(\log^\alpha t)$ does not satisfy your assertion for this $\sigma$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.