Would like to build once more on this question.

Take $s=\sigma + ti, s \in \mathbb{C}, 0<\Re(\sigma)<1$.

Let's assume it is proven that:

$$\zeta(1-s) - \zeta(s)$$

has all its zeros on the critical line when:

$$\chi(s)=2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \phantom. \Gamma(1-s) = \pm 1$$

The other zeros are the $\rho$'s and they simply emerge when $0 - 0 = 0$, and obviously no further information can be derived about the validity of the RH. Searching for ways around this (unsuccesfully), I stumbled on another question.

From:

$$\zeta_H(s,a) = \zeta_H(s,a+1) + a^{-s}$$

it follows that:

$$\zeta(1-s) - \zeta(s) = \zeta_H(1-s,2) - \zeta_H(s,2)$$

And the latter term can be factored into:

$$ \left( \sqrt{(\zeta_H(1-s,2)} - \sqrt {\zeta_H(s,2)} \right) \left( \sqrt{(\zeta_H(1-s,2)} + \sqrt {\zeta_H(s,2)} \right)$$

A plot of both factors for $\sigma = \frac12$ revealed:

that all $\rho$ are produced as zeros of the second factor only.

the first factor shows discontinuities at the $\rho$'s, but also for a few additional values.

the discontinuities vanish when taking the absolute value (this only works when $\sigma = \frac12$).

this absolute function has zeros, but all $\rho$'s will now be residing on the line $y=2$.

Assume: $f(t) = |\left( \sqrt{(\zeta_H(\frac12-t i,2)} - \sqrt {\zeta_H(\frac12+t i,2)} \right)|$.

Note that $f(t)$ shares all its zeros with $|\chi(\frac12+t i)-1|$, but not the other way around. There appear to be 'missing zeros' in $f(t)$ that emerge at random spots. Given the similarities and overlaps between the two plots, I wondered if $f(t)$ might be a 'distorted' version of $|\chi(\frac12+t i)-1|$. A distortion potentially caused by randomly missing zeros in a "Hadamard"-type infinite product of the zeros of $|\chi(\frac12+t i)-1|$?

Question:

The function $\chi(s)-1$ is a meromorphic function, could the Weierstrass/Hadamard factorisation theorem be used to express $\chi(s)-1$ as an infinite product of its zeros?