Is the argument of the Riemann zeta function bounded in the root free region on RH?

Question

Let

• $\Re(s)=\sigma$,

• $\Im(s)=t$,

• $\zeta(s)=R(\sigma,t)e^{i\arg\zeta(\sigma,t)}$,

• $R(\sigma,t)\in\mathbb{R}$.

If the Riemann hypothesis (RH) is true, then the root free region of the Riemann zeta function $\zeta(s)$ is given by $\sigma>1/2$.

$\bf{Question}$: Assume RH is true. Is $\arg\zeta(\sigma_0,t)$ bounded in the root free region then, with $\sigma_0$ fixed?

Any proof or counter-proof or a reference would be most welcome!

Answer 1

Found it! Being referenced in Titchmarsh, The theory of the Riemann zeta-function, Oxford University Press, 2nd edition, 1951, section 8.13, p.209. The reference there is:

H.L. Montgomery, Extreme values of the Riemann zeta function, Comment. Math. Helvetici 52 (1977) 511-518

Corollary of the main result:

$Let$ $\sigma$ $be$ $fixed$, $1/2<\sigma<1$. $Then$ $as$ $t\to\infty$,

$\log\mid\zeta(s)\mid =\Omega_+ \left( \frac{(\log t)^{1-\sigma}}{(\log\log t)^\sigma} \right) $

$\arg\zeta(s) =\Omega_\pm \left( \frac{(\log t)^{1-\sigma}}{(\log\log t)^\sigma} \right) $