In your case of the Bessel function a good choice appear to be $|J_0(x)+iY_0(x)|$. This has to be proved, but the figure with Mathematica is spectacular. Perhaps this is how you get your figure. These are the Hankel functions.

The Riemann hypothesis is equivalent to the behavior of a simple power series. So this is a difficult problem. In my paper arxiv.org/abs/1505.00440 I give a solution for a particular example of the problem. This can be applied to similar cases, but is not useful in the case of Riesz and Hardy and Littlewood series related to RH.

We apply probability theory to mathematical phenomena. In this case small probabilities occur. For example R. Brent, van de Lune, and I compute the probability of $|\arg(\zeta(\sigma+it)|>\pi/2$ for $\sigma=1.165$ as $1.279\dots\times10^{-283}$ and this really happens for some $t$. We can obtain smaller probabilities for other $\sigma>1$.

By homogeneity reasons this can not be true.

@TPTW There are many zeros o f $\zeta'(s)$ with real part $>1$. For each one of these zeros there are two zeros of $\Re \zeta'(1+it)$ and $\Im\zeta'(1+it)$. For the derivative there are parallel real and imaginary lines (to the real axis) separated by $\pi/\log2$. Each one of these parallel lines gives a zero of $\Re \zeta'(1+it)$ or $\Im\zeta'(1+it)$ respectively. The x-ray of $\zeta'(s)$ show this, but I have not published this.

Hence the two first solutions of $\Re\zeta(1+it)=0$ are (with 30 decimal digits) $$t_1=682112.891338239941159556828817,\quad t_2=682112.944250491762439022676048$$

@paul garret In the case simple to state you say that $h(s)$ do not vanish in $\sigma>1/2$. But later says that may be exceptions (zeros off the critical line) when $h(s)=h(1-s)=0$. Any such exception implies a zero of $h(s)$ with $\sigma>1/2$. Can you explain what I am missing.

@paul garret Thanks for pointing the interesting work of Kim Klimger-Logan. I am really interested in that preprint.

Yes. You have reason.

I was trying to find the first zero off the line. This needs no explanation. I have only to show it. But it is not so easy. Perhaps the question is more interesting that I thought.

They are the zeros of $\sqrt{2}\sin(t\log2)-\sqrt{3}\sin(t\log3)$. I make some plots. A complex zero of this function will be "visible" in the real plot. Again this have a long justification. Each time the function decreases it cuts the real axis. each time the function increases cuts the real axis. (I have no time for English grammar).

All solutions with $ |\Im(z)|\le 7000$ are on the critical line.

This appears as (part of) Conjecture 5.6 in Zhi-Wei Sun "Two new kinds of numbers and related divisibility results", arXiv:1408.5381 v8. In Remark 5.3 Zhi-Wei Sun asserts it is divisible by n+1.