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By use of the Riemann functional equation, it can be shown (see corollary 10.5 of Montgomery-Vaughan) that

$$|\zeta(\sigma + it)| \asymp |t|^{\sigma-1/2}|\zeta(1-\sigma - it)|$$.

where $\zeta$ denotes the Riemann zeta function and $t\in \mathbb{R}$. My questions are :

1.) Is this result also true for all Dirichlet $L-$functions and also zeta functions that do not satisfy the Riemann Hypothesis (RH) ?

2.) Denote by $\rho$ a zero of $\zeta$. Since $\lim_{s \rightarrow \rho}\Big| \frac{\zeta(s)}{\zeta(1-s)}\Big|=1$, why shouldn't this result entail the RH for large enough $|t|$ ?

3.) Is there a result of the form

$$|\zeta(\sigma + it)| \geq c |\zeta(1-\sigma - it)|$$ for all $|t|\geq t_0$, where $c$ is a positive constant ?

EDIT: I've just learnt from corollary 10.10 of Montgomery -Vaughan that a similar result holds for all Dirichlet $L-$ functions. That is, one has

$$|L(s, \chi)|\asymp (qt)^{\sigma-1/2}| L(1-s, \chi)|$$ where $\sigma=Re(s), t=\Im(s)$ and $\chi$ is a primitive character modulo $q$.

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    $\begingroup$ What do you mean by "zeta functions that do not satisfy the Riemann Hypothesis"? $\endgroup$ – Gerry Myerson Dec 2 '18 at 11:17
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This is an answer to question $2$.

Since all Dirichlet $L-$ functions satisfy a functional equation of the $\zeta$-type, note that any argument that uses the stated result (which is a consequence of the functional equation) to prove the RH would also work for any linear combination of $\zeta(s)$ and $L(s, \chi)$ whose analogue of the RH is false.

So no, the result can't be used to prove the RH.

The gap in the logic is the assumption that $$\lim_{s\rightarrow \rho} \Bigg|\frac{\zeta(s)}{\zeta(1-s)}\Bigg|=1,$$ which is not justified unless $\Re(s)=1/2$. Indeed, suppose that $\zeta(s)=s(1-s)e^s$, Then $\zeta(s)=0$ whenever $\zeta(1-s)=0$, which occurs at $s=0$ and $s=1$. However, $\lim_{s \rightarrow 1} \Bigg(\frac{s(1-s)e^s}{s(1-s)e^{1-s}}\Bigg) = \lim_{s \rightarrow 1} e^{2s-1}\neq 1$.

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  • $\begingroup$ Nevertheless it is true that for $s=\frac12+it$ with $t$ real, $\zeta(1-s)=\overline{\zeta(s)}$, so that $\left|\frac{\zeta(s)}{\zeta(1-s)}\right|=1$, no? $\endgroup$ – მამუკა ჯიბლაძე Dec 2 '18 at 12:50
  • $\begingroup$ It is true, but we can only be sure of this if $t\in \mathbb{R}$. I've edited the answer to cater for that. $\endgroup$ – 10101 Dec 2 '18 at 13:00
  • $\begingroup$ Well then the ratio, being continuous, cannot go very far from 1 in the vicinity of the critical line, so this should suffice for the limit? $\endgroup$ – მამუკა ჯიბლაძე Dec 2 '18 at 13:03
  • $\begingroup$ I mean that given a real $t$, for any $\varepsilon>0$ there is a $\delta>0$ such that whenever $\left|\frac12+it-s\right|<\delta$, one will have $\left|\left|\frac{\zeta(s)}{\zeta(1-s)}\right|-1\right|<\varepsilon$ $\endgroup$ – მამუკა ჯიბლაძე Dec 2 '18 at 13:08
  • $\begingroup$ Yes, that's true, but i don't think anything interesting can be derived from that. $\endgroup$ – 10101 Dec 2 '18 at 13:09

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