# On $\zeta(s)=h^2(s)$ when we presume that the Riemann zeta function has no zeros for $\Re s>\frac{1}{2}$

By specialization of a theorem from complex analysis, one has that on assumption that the Riemann zeta function $\zeta(s)$ has no zeros with $\Re s>\frac{1}{2}$, then there exists an analytic function $h$ defined on $\Re s>\frac{1}{2}$ such that $\zeta=h^2$.

I've computed and did a construction with the purpose to understand and learn more about complex analysis techniques.

Question 1. Can you say if any of the following claims are wrongs or there are mistakes?

1) One has that $$\zeta'(s)=2\zeta(s)\cdot\frac{h'(s)}{h(s)}$$ for $\Re s>\frac{1}{2}$, and it is not necessary declare a special branch of $h^2$ to do previous computation.

2) It is possible justify for $\Re s>\frac{1}{2}$ $$(h(s))^2-2h(s)h'(s)=\frac{1}{s-1} \left( \frac{2s-1}{s-1} \right)-s^2\int_1^\infty\frac{(x-\lfloor x\rfloor)\log x}{x^{s+1}}dx,$$ and from here it is possible compute the limit as $s\to\frac{1}{2}^{+}$ as $\zeta(\frac{1}{2})-\frac{1}{2}\zeta'(\frac{1}{2})-1$.

Let a disk $\mathcal{D}$ centered in the abscissa $b$ and ordinate $t$ that is the ordinate of a non-trivial zero $\rho$ of the Riemann zeta function ( a zero with $\Im \rho>0$ and $\sigma=\Re \rho=\frac{1}{2}$), radius $r=\overrightarrow{ab}$, where $\frac{1}{2}<a<b$ and $\overrightarrow{\sigma a}=\epsilon$ thus with we take $\epsilon>0$. We consider $\xi$ as the complex number $\frac{a+b}{2}+it$, and the rectangle $\mathcal{R}$ whose vertices are $a-irt,b-irt,b+irt$ and $a+irt$.

Now we consider the sequence of previous items defined as before where $\mathcal{D}'$ will be the disk centered in the abscissa $b':=a-\frac{\epsilon}{3}$ and ordinate $t'$ that is the ordinate of the following non-trivial zero when the imaginary part increases, and radius $r'=\overrightarrow{a'b'}$, ... For this new iterated denoted by the prime $'$, I've computed using the Cauchy's formula $$\zeta(\xi')=h(\xi')^2=-\frac{1}{4\pi^2} \left( \int_{\partial \mathcal{D}'} \frac{h(z)}{z-{\xi'}}dz\right).$$ On the other hand, using the Hadamard three-lines theorem applied to the strip $a'<b'$ we can deduce that $\log M$ is convex, where $$M(x):=\sup_{y} \left| h(x+iy) \right|,$$ and we are defining now (update to makes sense the Question 2., I believe that now has the meaning that I wanted) $$\hat{M}(x):=\sup_{-\sqrt{(r')^2-(x')^2}\leq y\leq \sqrt{(r')^2-(x')^2}} \left| h(x+iy) \right|$$

Question 2. I am agree that previous is an hypothetical approach, and perhaps we can not do computations: it is possible a comparison of sizes between $\left| \zeta(\xi') \right|$ and $(\hat{M}(\Re\xi'))^2$ (thus one has $\Re \xi'$ near to $\frac{1}{2}$)? If previous computations are no feasibles or are non- comparable quantities, explain us.

Thanks in advance. If some part of previous isn't well-possed, please add a comment, and I can improve my post or delete that part, while it is possible.

• I did an update, ask me if it is conffused, I mean the comparison between values relatively larges on the chord through the abscissa $\Re\xi'$ and low values on arcs measured by Cauchy integral. – user243301 May 26 '16 at 11:07
• by the way, $\frac{1}{s-1}$ is meromorphic with a simple pole at $s=1$, as $\zeta(s)$. is there an analytic/meromorphic function $h(s)$ such that $\frac{1}{s-1} = h(s)^2$ ? – reuns May 26 '16 at 11:55
• (and searching for a function $h$ such that $f(s) = h(s)^2$ is not so different as searching for a function $g$ such that $f(s) = e^{g(s)}$ (i.e. $g(s) = \ln f(s)$) ) – reuns May 26 '16 at 12:00
• and how do you solve this problem, that $\frac{1}{s-1}$ has no square root (nor logarithm) in the case of $\zeta(s)$ which is very similar to $\frac{1}{s-1}$ ? – reuns May 26 '16 at 12:03
• the answer is : consider $F(s) = (s-1) \zeta(s)$ which is analytic on $\mathbb{C}$ and assuming the Riemann hypothesis $\log F(s)$ also is analytic on $Re(s) > 1/2$, hence there is a function $h$ analytic on $Re(s) > 1/2$ such that $F(s) = h(s)^2$ – reuns May 26 '16 at 12:08

Consider $$F(s) = (s-1) \zeta(s)$$ which is analytic on $\mathbb{C}$ and assuming the Riemann hypothesis $$\log F(s)$$ also is analytic on $Re(s) > 1/2$, hence there is a function $h$ analytic on $Re(s) > 1/2$ such that $$F(s) = h(s)^2.$$