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As an undergrad, I sometimes toy with the Riemann Hypothesis (RH) to A) humble myself and B) better myself in complex analysis and proof writing. Given its infamy and my relative mathematical immaturity, I obviously don't expect to make any significant results any time soon. Hence, I apologize for fulfilling the often annoying trope of "young ambitious undergrad attempting millennium prize problems."

That said, after some playing around with the zeta function for the past few years, I believe I have come across a proof that $$\left|\frac{\zeta(s)}{\zeta(1-s)}\right|=1\implies \Re(s)=\frac{1}{2}$$ for all $s$ within the critical strip.

Assuming this is true, I am curious if this result is meaningful to RH. Is this result already known, or a triviality of the functional identity? If not, is it meaningful / significant to the RH?

Any insight as to its potential (non)significance, how this may be false, or how this is already known would be greatly appreciated.

Edit: As pointed out, this 'result' (or conjecture, if you will) fails near the origin. I'll have to overlook my proof again, but fortunately I believe it should still hold for $|\Im(s)|\ggg 0$, at the least.

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    $\begingroup$ No which I illustrated in my question at math.stackexchange.com/questions/3008597/… . $\endgroup$ Aug 5 at 19:37
  • $\begingroup$ @StevenClark: maybe you have already thought about this. It should not be too difficult to prove that for any $\sigma=\text{Re}(s)\in\left(\frac{1}{2},1\right)$ there are exactly two opposite $\tau$s such that $|\zeta(s)|/|\zeta(1-s)|=1$ at $s=\sigma\pm i\tau$. You have already noticed that $\sigma=f(\tau)$ is almost constant for $\sigma\in(0,1)$. $\endgroup$ Aug 5 at 20:22
  • $\begingroup$ @StevenClark: are non-trivial zeroes of the $\zeta$ function forced to fulfill $\sigma=\frac{1}{2}$ or $\tau=\pm f(\sigma)$? If so, this can actually lead to a proof of RH. Indeed by the integral representation the non-trivial zero closest to the origin has to fulfill $\tau > 10$, while the values of $f$ are in a tight neighbourhood of $2\pi\ll 10$. $\endgroup$ Aug 5 at 20:25
  • $\begingroup$ @JackD'Aurizio I hadn't thought about proving the conjecture in your first comment, but both of your comments seem related to my question (5) in the link above which was of primary interest to me. The comment by user reuns at the question I linked above seemed to imply there was no connection between my investigation and the RH. $\endgroup$ Aug 5 at 20:53
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    $\begingroup$ I would propose to merge this question and math.stackexchange.com/questions/266948/… $\endgroup$ Aug 6 at 23:13

1 Answer 1

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Intro: I am sorry for the following lines since your result would have been an outstanding achievement towards RH.
However...

By the reflection formula $$ \rho(s)=\frac{\zeta(s)}{\zeta(1-s)} = 2^s \pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s) $$ and $$ R(\nu)=\rho\left(\frac{1}{2}+\nu\right) =(2\pi)^{\nu} \sqrt{\frac{2}{\pi}}\sin\left(\frac{\pi \nu}{2}+\frac{\pi}{4}\right)\Gamma\left(\frac{1}{2}-\nu\right)$$ is a real analytic function on $(-1/2,1/2)$ which is also increasing on such interval and equal to $1$ at the origin. For any $\nu\in(-1/2,1/2)$ $$\|R(\nu+i\tau)\|^2 = R(\nu+i\tau)R(\nu-i\tau) = (2\pi)^{2\nu}\left(1+\frac{\sin(\pi\nu)}{\cosh(\pi \tau)}\right)\left\|\frac{\Gamma\left(\tfrac{1}{2}-\nu-i\tau\right)}{\Gamma\left(\tfrac{1}{2}-i\tau\right)}\right\|^2$$ is an even function of the $\tau$ variable. Assuming $\nu > 0$ it is also a Schwartz function.
On the other hand, by just considering some concrete instances, we may realize that your conjecture does not hold. For $\nu=\frac{1}{10}$ we have $\|R(\nu)\|^2>1.72$ and $\|R(\nu+i\tau)\|=1$ for $|\tau|\approx 6.29 $. In particular there are points on the line $\text{Re}(s)=\frac{3}{5}$ such that $|\zeta(s)|=|\zeta(1-s)|$. But exactly two of them.


On a positive note, let us try to salvage things.

Even if your conjecture does not hold, it leaves some room for an interesting attack.
Let us set $s=\sigma+i\tau$ for $\sigma\in (0,1)$.

Jack's conjecture 1: The locus of points of the critical strip such that $|\zeta(s)|=|\zeta(1-s)|$ is made by the line $\sigma=\frac{1}{2}$ and by the union of two graphs, symmetric with respect to the real line, given by $\tau=\pm f(\sigma)$.

Jack's conjecture 2: In order to prove RH it is sufficient to show that there are no zeroes of the $\zeta$ function on the curve $\tau=f(\sigma)$.

Numerically it seems that $f$ is almost constant, taking values between $6.28$ and $6.3$.
And the non-trivial zero closest to the origin is well-known to fulfill $\tau > 10$ (this just follows from integral representations for $\zeta(s)$).

If this actually works we'll share the prize, deal? :)


Update. As shown in the comments, there is no hope to prove RH through this naive approach, but I am still convinced that something (like improving the shape of the zero-free region) can be done. Let us assume that all the exceptional zeroes in the critical strip are simple. Then for any one of them the limit of $\frac{\zeta(s)}{\zeta(1-s)}$ equals $-\frac{\zeta'(s)}{\zeta'(1-s)}$, so they have to lie on the region defined by the equality between $f(s)=\left|\frac{\zeta(s)}{\zeta(1-s)}\right|$ and $g(s)=\left|\frac{\zeta'(s)}{\zeta'(1-s)}\right|$. $f(s)$ essentially is an elementary function and its behavior for a fixed $\sigma\in\left(\frac{1}{2},1\right)$ and a varying $\tau\in\mathbb{R}$ is Gaussian-like. $\zeta'(s)$ and $\zeta'(1-s)$ are still related via the derivative of the reflection formula, but the ratio $\frac{\zeta'(s)}{\zeta'(1-s)}$ is less elementary, depending on bounds for $|\zeta(s)|$. Numerical experiments reveal that the behaviour of $g(s)$ for a fixed $\sigma$ and a varying $\tau$ still is essentially Gaussian, but for $\tau$ close to zero $g(s)$ is much larger than $f(s)$. If we prove that the curve given by $f(s)=g(s),\tau > 0$ lies in the currently known zero-free region, we have proven RH under the assumption that all the exceptional zeroes are simple.

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    $\begingroup$ According to my version of Mathematica $f(0)$ is a tad larger than $2\pi$, but pretty close indeed :) $\endgroup$ Aug 5 at 20:12
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    $\begingroup$ Really grateful for the counterexample. I'll review my work with your points in mind and see where I went wrong. Fortunately, as you also suggested, It may be salvageable, as I had a hint that my conjecture may fail close to the origin. I believe the "proof" should still work for some significant subset of the critical strip, at the least. (also, deal ;) $\endgroup$
    – Graviton
    Aug 5 at 20:42
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    $\begingroup$ note that if $\rho$ is a non critical line zero (counterexample to RH), there is no particular reason to have $|f(\rho)| =1$ as we can easily give counterxamples of (entire, conjugate invariant) functions $g$ satisfying $g(s)=f(s)g(1-s)$ with $f$ having no zero, but for which $|f(\rho)|$ is arbitrary (non zero) at a zero of $g$ so where $g(\rho)=g(1-\rho)=0$; similarly there is no particular reason to have any relation between a zero of $g$ and some specific behavior of $f$ there $\endgroup$
    – Conrad
    Aug 5 at 20:54
  • $\begingroup$ @Conrad: of course you are right, if at some $s_0=\sigma+i\tau$ with $\sigma\in(0,1)\setminus\{1/2\}$ we have $\zeta(s_0)=0$, then $\lim_{s\to s_0}\left|\frac{\zeta(s)}{\zeta(1-s)}\right|$ simply equals the modulus of $\zeta'(s_0)/\zeta'(1-s_0)$. On the other hand $\zeta'(s)$ and $\zeta'(1-s)$ are still related via the (derivative of the) reflection formula, so it does not sound completely insane that we can get something by studying the ratio $\zeta(s)/\zeta(1-s)$. $\endgroup$ Aug 5 at 23:14
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    $\begingroup$ @JackD'Aurizio truly, your optimism and ardent analysis of my question is inspiring. If you'd like, (although I lack as much of a toolset as you do) I'd be delighted to share more of my work / ideas on RH as per the email in my profile. I've taken a very geometric approach of the partial sums of the zeta function which would be nice to flesh out with more experienced eyes. $\endgroup$
    – Graviton
    Aug 6 at 0:43

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