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The functional equation for the zeta function $ζ(s)$ is given by

$ζ(s)=f(s)ζ(1-s)$

(a) We know that if $Re(s)=1/2$, then $|f(s)|=1$.

My question is about the case where $|f(s)|=1$ outside the critical line. Is this case possible, i.e.,

(b) Is this implication "if $|f(s)|=1$ then $Re(s)=1/2$" correct?

  • 0
    (b) is not correct. Please see my answer below.2014-12-12

2 Answers 2

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  • (a) if $Re(s)=1/2$, then $|f(s)|=1$ : this is true and allows us to find a simple expression of the phase of the Riemann zeta function on the critical line.

  • (b) the function $f(s)$ itself doesn't contain $\zeta$ since $f(s)$ is, from the functional equation : $\tag{1}f(s)=2^s\pi^{s-1}\sin\left(\frac {\pi s}2\right)\Gamma(1-s)$ but $|f(s)|$ may be $1$ out of the critical line as you may see on this picture of $|f(x+i y)]-1$ :

    picture The vertical line at $x=\frac 12$ gives the solution $(a)$ but two solutions to $|f(x+i y)]=1$ exist for $y$ near $2\pi$ and $-2\pi$ (for $x$ near $\frac 12$ see too this discussion concerning Siegel-$\theta$). You'll have to exclude these two solutions to make your implication correct!

$\qquad$Another perspective view :

perspective

And with this picture for $x$ between $-30$ and $30$ a kind of symbol for your 'superquest' :-) superquest


APPROXIMATIONS:
In the remaining we will always suppose that $x\in[0,1]$ and will only consider the case $y>0$.

Let's remember $6.1.30$ from A&S : $\tag{2}\left|\Gamma\left(\frac 12+iy\right)\right|^2=\frac {\pi}{\cosh(\pi y)}$ An excellent approximation for other values of $x$ (when $y>2$) is given by : $\tag{3}\left|\Gamma\left(x+iy\right)\right|^2\approx\frac {\pi\;y^{2x-1}}{\cosh(\pi y)}$ (the error is less than $0.4\%$ and quickly decreasing with $y$)

Since $\ \left|\sin\left(\frac {\pi (x+iy)}2\right)\right|^2=\cosh^2\bigl(\frac{\pi\;y}2\bigr)-\cos^2\bigl(\frac {\pi\;x}2\bigr)\ $ and $\ \left|(2\pi)^{x+iy}\right|^2=(2\pi)^{2x}\;$ we get :

\begin{align} |f(x+iy)|^2&\approx\left|\frac{(2\pi)^{x+iy}}{\pi}\sin\left(\frac {\pi (x+iy)}2\right)\right|^2\frac {\pi\;y^{1-2x}}{\cosh(\pi y)}\\ &\approx\frac{(2\pi)^{2x}}{\pi}\left(\cosh^2\left(\frac{\pi\;y}2\right)-\cos^2\left(\frac {\pi\;x}2\right)\right)\frac {\;y^{1-2x}}{\cosh(\pi y)}\\ \tag{4}&\approx \left(\frac{2\pi}y\right)^{2x}\frac y{\pi}\frac{\cosh^2\left(\frac{\pi\;y}2\right)-\cos^2\left(\frac {\pi\;x}2\right)}{\cosh(\pi y)}\\ \end{align}

Now for $y\gg 1$ the fraction at the right will converge to $\frac 12$ (since $|\cos|\le 1$) and we will have : $\tag{5}|f(x+iy)|\sim \left(\frac{2\pi}y\right)^{x-1/2}\quad\text{for}\ y\gg 1$ which shows clearly all the things of interest for us :

  • for $x=\frac 12$ we get $|f(x+iy)|=1$ independently of $y$
  • there is only one other solution : $y\approx 2\pi$ (considering $y>0$)
  • the visual aspect of the approximation shows no real difference : approximation

Of course this is not a complete proof : the error term in $(3)$ to next order should be found (possibly using A&S' expansions and propagated) proving that $x\mapsto |f(x+iy)|$ is decreasing for $y>K$ and increasing for $y (with $K$ some constant near $2\pi$). $|f(x+iy)|$ should be studied with more care for $y$ near $0$ and so on. But, at least in principle, it appears ok to me so : Fine continuation!

  • 0
    Thanks to take care of this old thread @mike and for the references. I don't have the 1965 Spira paper (I got the others) but found this [recent one by Trudgian](http://arxiv.org/pdf/1312.1409.pdf) (I got earlier too the papers from Nazardonyavi and Yakubovich). I won't hold my breath to prove RH this way but wish you fun trying it this way anyway!2014-12-12
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I plotted this $f(s)$ before. It might be easier to visualize the roots of $|f(s)|=1$ from the contour plot below. enter image description here

Milgram (arxiv:0911.1332v2) noticed the roots of $|f(s)|=1$ with $\Re(s)\not=1/2$.

Spira (1965,AN INEQUALITY FOR THE RIEMANN ZETA FUNCTION) proved that:

THEOREM 1. For $y\geq 10, 1/2 1$.

The lower bound for $y$ has been reduced to a number around 7 by others (cf. the papers that cited Spira's paper. It is also mentioned somewhere in these papers that people are wondering if the Riemann hypothesis can be proved only using the knowledge of the functional equation $ζ(s)=f(s)ζ(1-s)$).

But Spira mentioned that his theorem can not be directly used to prove that $\zeta(s)$ does not have zeros off the critical line ($\Re(s)=1/2$) for $y\geq 10$.

Albeverio and Cebulla (2007,Müntz formula and zero free regions for the Riemann zeta function, Bull. Sci. math. 131 (2007) 12–38) proved that

Corollary 4.10. There are no zeros of the $\zeta(s)$ function of the form $s = x + iy$, $x \in (0, 1)$, $0< |y|< 2\pi/\log(2)\approx9.06$.

If one can use a variation of Spira's theorem 1 to prove that $\zeta(s)$ does not have zeros off the critical line for $y\geq 9$, then, in view of Albeverio and Cebulla's Corollary 4.10., one can prove that $\zeta(s)$ does not have zeros off the critical line ($\Re(s)=1/2$) for $0<|y|$.