# Trying to debunk a claim

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.

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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... – Noam D. Elkies Jun 24 '12 at 17:13

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 $cT$.

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Awesome... Let me probe on that a bit! :) – Roupam Ghosh Jun 23 '12 at 15:38
This is also a corollary of Selberg's central limit theorem: terrytao.wordpress.com/2009/07/12/… – Terry Tao Jun 24 '12 at 3:22

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 $\sigma$ .