# Weak Riemann Hypothesis?

The Riemann Hypothesis says that all non-trivial zeros of the Riemann zeta function lie on $Re(z)=\frac{1}{2}$ line instead this region $Re(z) \in (0,1)$.

It seems a natural question to be that instead of proving that $Re(z)=\frac{1}{2}$ one could try to prove that $Re(z) \in (\epsilon,1-\epsilon)$.

I am familiar with a zero-free region that is used in the analysis of Prime-number theorem but that is too weak to conclude some like this.

My question is this question as hard as proving Riemann Hypothesis ?? Are there some developments toward proving this ?? Or a result that proving this would imply proving Riemann Hypothesis?

• It is not proven yet that $\zeta(s)$ doesn't have a sequence of zeros whose real part converges to $1$. GHFromMO on mathoverflow told me proving it would be probably worth a fields medal. It is less hard than the RH in the sense that it doesn't imply the RH. Oct 1, 2016 at 18:01
• @user1952009 Can you give a reference for this ?
– xyz
Oct 1, 2016 at 18:03
• The reference is that it is not proven in any book on $\zeta(s)$, and it is not mentionned on en.wikipedia.org/wiki/Riemann_zeta_function or en.wikipedia.org/wiki/Prime_number_theorem :) Oct 1, 2016 at 18:03
• And I think the Hurwitz zeta or the counter-example to RH constructed in Titchmarsh's book (something like $e^{i\theta} L(s,\chi_5)+e^{-i\theta}L(s,\overline{\chi_5})$) -both having a functional equation but no Euler product- have a sequence of zeros converging to $Re(s) = 1$. You can also construct easily an Euler product having a sequence of zeros converging to $Re(s) = 1$ Oct 1, 2016 at 18:08
• Read some books ? And a Siegel zeros (if it exists) is a zero at $s \in (1/2,1)$ of $L(s,\chi)$ where $\chi$ is a real character, and hence is a counter-example to the generalized Riemann hypothesis. Dec 26, 2016 at 14:56

We don't know yet if $\zeta(s)$ has a sequence of zeros converging to $Re(s)=1$.
There is an Euler product (but having no functional equation) having a sequence of zeros converging to $Re(s) = 1$.
$$h(x) = x-\sum_{k=K}^\infty \frac{x^{1-1/k+ik^2}}{1-1/k+ik^2}$$ and iteratively for every prime $q$ : $$a_{q} = h(q) -\sum_{p < q} a_{p}$$ Finally let $a_n = \prod_{p | n} a_p$ and set $$F(s) = \sum_{n=1}^\infty a_n n^{-s} = \prod_p (1+\sum_{k \ge 1} a_{p^k}p^{-sk} )= \prod_p \left( 1+ \frac{a_p}{p^s-1}\right)$$ So that $$\log F(s) = \sum_p \log (1+ \frac{a_p}{p^s-1}), \qquad \frac{F'(s)}{F(s)} = \sum_p \frac{\frac{a_pp^{s}\ln(p)}{(p^s-1)^2)} }{1+ \frac{a_p}{p^s-1}} = \sum_p a_p p^{-s} + \sum_p \sum_{k \ge 2} b_{p^k}p^{-sk}$$
By the Abel summation formula you have $$\frac{F'(s)}{F(s)} = s \int_1^\infty g(x) x^{-s-1}dx, \qquad g(x) = \sum_{p < x} a_p+\sum_{p^k < x} b_{p^k}$$ And the prime gap shows that $$g(x) = \sum_{p < x} a_p + \mathcal{O}(x^{1/2+\epsilon}) = h(x) + \mathcal{O}(x^{1/2+\epsilon})$$ i.e. $$\frac{F'(s)}{F(s)}+\frac{1}{s-1}- \sum_{k=K}^\infty \frac{1}{s-1+1/k-ik^2} = s\int_1^\infty \left(g(x)-h(x)\right) x^{-s-1}dx$$ is analytic for $Re(s) > 1/2$, and hence $F(s)$ is meromorphic there, with one pole at $s=1$ and its zeros at $1-\frac{1}{k}+ik$