Weak Riemann Hypothesis?

Question

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?

Answer 1

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 $Re(s) = 1$.

Let

$$h(x) = x-\sum_{k=K}^\infty \frac{x^{1-1/k+ik^2}}{1-1/k+ik^2}$$ 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 $1-\frac{1}{k}+ik$