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.

decaysfaster than some power of $|t|$ as $|t| \rightarrow \infty$, and that's surely absurd... $\endgroup$ – Noam D. Elkies Jun 24 '12 at 17:13