Okay, firstly a bit of background to set the scene. My question comes from the approaches made by R. Spira in his paper, "An inequality for the riemann zeta function," regarding the initial steps he took in deriving and proving his inequality. A similar approach was also taken and expanded upon by F. Saidak and P. D. Zvengrowski in their paper, "On the modulus of the Riemann Zeta Function in the critical strip."

They essentially prove that for $s=\sigma + it$, with $t \geq 2\pi + 1$, and $\frac{1}{2} \leq \sigma \leq 1$,

$$ \left|\zeta(1-s) \right| \geq \left| \zeta(s)\right|\ \tag{1} $$

In this proof they all start with the functional equation for the Riemann Zeta Function, say: $$ \zeta(1-s)=(2\pi)^{-2s}\cos\left(\frac{\pi s}{2}\right)\Gamma(s)\zeta(s) \tag{2} $$ And go on to show that

$$ \left|(2\pi)^{-2s}\cos\left(\frac{\pi s}{2}\right)\Gamma(s)\right| \geq 1 \tag{3} $$

by looking at how the elementary functions $\left|(2\pi)^{-2s}\right|$ and $\left|\cos\left(\frac{\pi s}{2}\right)\right|$ change with $\sigma$ compared to $\left|\Gamma(s)\right|$. The elementary functions decrease as $\sigma$ increases, " leaving the whole burden of increase upon the $\Gamma$-function," as Spira puts it. The next bit confuses me however, as they say they require,

$$ \frac{\partial}{\partial\sigma}\left( \left|(2\pi)^{-2s}\cos\left(\frac{\pi s}{2}\right)\Gamma(s)\right|\right) \gt 0 \tag{4} $$

Where differentiating the elementary functions is somewhat trivial compared to the $\Gamma$-function, which requires use of the Stirling Approximation, which makes things a lot more complicated and introduces an error factor. Therefore, my question is whether for inequality $(3)$, it is simpler and valid to firstly rearrange for the $\Gamma$-function:

$$ \left|\Gamma(s)\right| \geq \frac{1}{\left|(2\pi)^{-2s}\cos\left(\frac{\pi s}{2}\right)\right|} $$

Square it,

$$ \left|\Gamma(s)\right|^{2} \geq \frac{1}{\left|(2\pi)^{-2s}\cos\left(\frac{\pi s}{2}\right)\right|^{2}} $$

Which can be simplified to:

$$ \left|\Gamma(s)\right|^{2} \geq \frac{\left(2\pi\right)^{2\sigma}}{2 \left( \cosh(\pi t) + \cos(\pi \sigma)\right)} \tag{5} $$

Now clearly for $\sigma = \frac{1}{2}$ this becomes:

$$ \left|\Gamma\left(\frac{1}{2}+it\right)\right|^{2} = \frac{\pi}{\cosh(\pi t)} \tag{6} $$

And so the proof boils down to showing that for $\frac{1}{2} \lt \sigma \leq 1$, that:

$$ \left|\Gamma(s)\right|^{2} \gt \frac{\left(2\pi\right)^{2\sigma}}{2 \left( \cosh(\pi t) + \cos(\pi \sigma)\right)} \tag{7} $$

Which excludes the need for differentiation as well as the more complicated Stirling Approximation, and the error factor that comes with it, that inevitably limits the proof by "always exceed[ing] the margin of safety of the inequality," as Spira says.

The reason I ask if this is a valid approach, is because then $\left| \Gamma(s) \right|^{2}$ is a much nicer term to work with due to the functional equation for the $\Gamma$-function:

$$ \Gamma(1-s)\Gamma(s) = \frac{\pi}{\sinh(\pi s)} $$

As from it we have equation $(6)$, but also:

$$ \left|\Gamma\left(1+it\right)\right|^{2} = \frac{\pi t}{\sinh(\pi t)} $$

It is also known that $\left| \Gamma(s) \right|^{2} \leq \left| \Gamma(\sigma) \right|^{2}$ where the equality is for $t=0$. This combined with the fact that the minimum of $\left|\Gamma(\sigma)\right|^{2}$ occurs at $\sigma \gt 1$ suggests that the inequality $(7)$ can be taken to its extreme at $\sigma = 1$ with $\left| \Gamma(s) \right|^{2}$ being at its minimum point for $\frac{1}{2} \lt \sigma \leq 1$ and the RHS being at its maximum, i.e $(7)$ would become:

$$ \frac{\pi t}{\sinh(\pi t)} \gt \frac{4 \pi^{2}}{2 \left( \cosh(\pi t) - 1\right)} $$

Simplifying the equation and we get:

$$ t \gt 2 \pi \coth\left(\frac{\pi t}{2}\right) $$

where $\coth(\pi t)$ quickly limits to $1$ as $t \gt 2$, and so we get:

$$ t \gt 2 \pi $$

For which the inequality $(7)$ holds, presumably excluding any kind of ambiguity. Would this then mean that the inequality in $(1)$ can be strengthened to exclude the equality over the range $\frac{1}{2} \lt \sigma \leq 1$ and be only equal for $\sigma = \frac{1}{2}$ for $t \gt 2\pi$?

Now I have a gut feeling that something (or a few things) above is incorrect and that I have misinterpreted a result in the mentioned papers and/or have done something incorrect in my maths, either explicitly or in the form of an incorrect assumption made. So, I would kindly ask for some clarification. Thanks!

Edit 06/12/18: My reasoning that the LHS of $(7)$ is at its minimum at $\sigma = 1$ is incorrect as for $t \gt \gt 1$, $\left| \Gamma(s) \right|^{2}$ is smaller for $\sigma < 1$. However, here I found an alternative approximation for $\left| \Gamma(s) \right|^{2}$ and found that the error in the approximation approaches zero as $t \rightarrow \infty$ and is valid for $t \gt 2\pi + 1$. Of course, this is assuming the maths is correct.

  • $\begingroup$ I don't understand your question. $|\Gamma(s)| > |\pi^{s-1/2}\Gamma(1-s)|$ is equivalent to $|\Gamma(s)|^2 > |\pi^{2s}\Gamma(s)\Gamma(1-s)|= |\pi^{s-1}/\sin(\pi s)|^2$. To show that last identity (for $|\Im(s)| > T_0, \Re(s) \in (1/2,1)$) we need the explicit formulas for $\Gamma(s), \pi^2/\sin(\pi s)^2$ (the infinite products or the series for their $\log$ or $\log$ derivative) files.ele-math.com/articles/jmi-07-16.pdf $\endgroup$ – reuns Dec 3 '18 at 22:16
  • $\begingroup$ My first question would be: if I take the magnitude squared of the functional equation for the Riemann Zeta function, does proving that $\left|\Gamma(s)\right|^{2} \gt (2\pi)^{2\sigma} / 2\left(\cosh(\pi t) + \cos( \pi \sigma)\right)$ also prove that $\left| \zeta(1-s)\right| \gt \left| \zeta(s) \right|$ (for $\sigma \in \left(1/2, 1\right)$? Then, if it does, are infinite products or the series for their log or log derivatives actually needed for said proof? If so why? Since $\left|\Gamma(s)\right|^{2}$ has nice elementary forms for certain values of $s$ can those not be used? $\endgroup$ – Steven Graham Dec 3 '18 at 22:32
  • $\begingroup$ Of course it proves so for the $s$ where $\zeta(s) \ne 0$ (and $t$ large enough) $\endgroup$ – reuns Dec 3 '18 at 22:48
  • $\begingroup$ Awesome, thank you. So then to take it further: Given the inequalities are true, if we can combine $(7)$ with $\left| \Gamma(\sigma) \right|^{2} \gt \left| \Gamma(s) \right|^{2}$ for all $t \neq 0$ to see that the inequality is most at risk for $t=0$. Then, using $\left| \Gamma(\sigma) \right|^{2}$ instead on the LHS of $(7)$, in the specified range for $\sigma$, $\left| \Gamma(\sigma) \right|^{2}$ is minimum at $\sigma = 1$, and the RHS of $(7)$ is maximum, yet we see even here $(7)$ is still true for $t \gt 2\pi$. Does this then prove that $(1)$ is an equality iff $\sigma = 1/2, t \gt 2\pi$? $\endgroup$ – Steven Graham Dec 3 '18 at 23:27

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