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


(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?

  • $\begingroup$ (b) is not correct. Please see my answer below. $\endgroup$
    – mike
    Commented Dec 12, 2014 at 16:37

2 Answers 2

  • (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 :


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

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 $y<K$ (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!

  • $\begingroup$ Thank you very much. But still a mathematical proof is not available in the current literature. $\endgroup$
    – Safwane
    Commented Dec 29, 2012 at 12:52
  • 1
    $\begingroup$ @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 $|x|>17$). $\endgroup$ Commented Dec 29, 2012 at 13:06
  • $\begingroup$ Yes, I am praticularly interested in $(0,1)$. What the approximations we may use to prove that. $\endgroup$
    – Safwane
    Commented Dec 29, 2012 at 13:30
  • $\begingroup$ @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... $\endgroup$ Commented Dec 29, 2012 at 13:59
  • $\begingroup$ @ Raymond Manzoni: Can you see this link for a possible proof: math.stackexchange.com/questions/267179/… $\endgroup$
    – Safwane
    Commented Dec 29, 2012 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. enter image description here

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


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 $0<|y|$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .