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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.I must be missing something stupid

Looking for references.

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Ok. So one day someone says to you this.

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

My question then arises, is

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

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. I must be missing something stupid.

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