MathOverflow Asked on December 21, 2021
Here is the question.
Does the Riemann Xi function possess the universality property, or something similar to Voronin’s universality property?
Here is why the answer to this question is important.
Up to arbitrary precision, any function can be approximated by some vertical translate of the the Riemann Zeta function. The universality property is also possessed by many other classes of functions (see below).
Voronin’s universality theorem
Does the Riemann Xi function possess the universality property?
The universality property can be formulated in terms of the modulus of the Riemann Zeta function. Up to arbitrary precision, the modulus of any function can be approximated by some vertical translate of the the modulus of Riemann Zeta function (Voronin universality implies modulus universality).
We assume that the Riemann Xi function has the universality property (the modulus formulation, on the right half of the critical strip).
Now we consider the following result:
Sondow and Dumitrescu – A monotonicity property of Riemann’s xi function and a reformulation of the Riemann hypothesis (arXiv).
Now, based on the reformulation of the Riemann Hypothesis, from the paper above "A monotonicity property of the Riemann ‘s Xi function and a reformulation of the Riemann Hypothesis", we can conclude that the Riemann Hypothesis is false (the truth of the reformulation is not compatible with universality), even though we have not effectively found a zero that is not on the critical line, we proved that it must exist.
So the question is, does the Riemann Xi function possess the universality property, or something similar to Voronin’s universality property (on the right half of the critical strip)?
To me this seems like a true statement (based on the universality of Riemann’s Zeta function and the relation defining Xi), but I could be missing something.
I remember I read somewhere that Alan Turing believed that the Riemann Hypothesis was false, and he tried to find some counterexample through computation, but with the state of the art computers at that time (and in the present it is not much different, in relation to this problem), it was impossible to effectively find a zero that is not on the critical line, if its imaginary part is extremely large. Also Littlewood seems to have been convinced that Riemann’s Hypothesis was false.
The Riemann xi-function decreases exponentially as $ttoinfty$, so it can't be universal.
The decay comes from the fact that $$ xi(s) = frac12 s (s-1) pi^{-s/2} Gamma(s/2) zeta(s) . $$ If $s = sigma + i t$ with $sigma$ bounded and $tto infty$, then $frac12 s (s-1)$ has polynomial growth, $pi^{-s/2}$ is bounded, $Gamma(s/2)$ decays exponentially, and $zeta(s)$ has polynomial growth. So, the exponential decay wins.
Answered by David Farmer on December 21, 2021
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