# Riemann zeta function and modulus

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?

• (b) is not correct. Please see my answer below. – mike Dec 12 '14 at 16:37

• (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 $|f(x+i y)]-1$ :

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 :

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

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 $y<K$ (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!

• Thank you very much. But still a mathematical proof is not available in the current literature. – Germany Dec 29 '12 at 12:52
• @RH1: the idea is original (and I don't know research about this particular aspect). Using some approximations we may probably prove that no other solution exists in $(0,1)$ than the two showed (avoiding the 'weird' solutions for $|x|>17$). – Raymond Manzoni Dec 29 '12 at 13:06
• Yes, I am praticularly interested in $(0,1)$. What the approximations we may use to prove that. – Germany Dec 29 '12 at 13:30
• @RH1: I don't know yet (every component of $f$ is acting in an opposite direction ;-)) but I'll try to update my answer in some hours... – Raymond Manzoni Dec 29 '12 at 13:59
• @ Raymond Manzoni: Can you see this link for a possible proof: math.stackexchange.com/questions/267179/… – Germany Dec 29 '12 at 18:00

I plotted this $f(s)$ before. It might be easier to visualize the roots of $|f(s)|=1$ from the contour plot below.

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 $y\geq 10, 1/2 <x< 1, |f(s)|> 1$.

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 $0<|y|$.