# A curious relationship betwen $|\zeta(\sigma+it)|$ and $|\zeta(1-\sigma - it)|$

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 $$q$$.

• What do you mean by "zeta functions that do not satisfy the Riemann Hypothesis"? – Gerry Myerson Dec 2 '18 at 11:17

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, $$\lim_{s \rightarrow 1} \Bigg(\frac{s(1-s)e^s}{s(1-s)e^{1-s}}\Bigg) = \lim_{s \rightarrow 1} e^{2s-1}\neq 1$$.

• 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? – მამუკა ჯიბლაძე Dec 2 '18 at 12:50
• 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. – 10101 Dec 2 '18 at 13:00
• 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? – მამუკა ჯიბლაძე Dec 2 '18 at 13:03
• 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 $\left|\left|\frac{\zeta(s)}{\zeta(1-s)}\right|-1\right|<\varepsilon$ – მამუკა ჯიბლაძე Dec 2 '18 at 13:08
• Yes, that's true, but i don't think anything interesting can be derived from that. – 10101 Dec 2 '18 at 13:09