# What are some consequences of zero free strip of the Riemann zeta function?

A weaker version of the Riemann hypothesis is the claim that if $$\zeta(s) = 0$$ then $$Re(s) \leq 1 - h$$ for some constant $$h> 0$$. What would the consequences be of a result of this type?

• Comments are not for extended discussion; this conversation has been moved to chat. – Ben Webster Jun 13 at 0:29

If this were proven, it would be a huge breakthrough in number theory. The most direct improvement would of course be a power savings in the error term of $$\lvert \pi (x) - \mathrm{Li} (x) \rvert$$, but there are many more applications for such things. For example, this would imply that $$\zeta \left( \sigma + i t \right) = \mathcal{O} \left( \lvert t \rvert^{\varepsilon} \right)$$ for $$\sigma \geq 1 - 2 h$$, which is a significant improvement on Heath-Brown's bound $$\zeta \left( \sigma + i t \right) = \mathcal{O} \left( \lvert t \rvert^{\frac{1}{2} (1 - \sigma)^{\frac{3}{2}} + \varepsilon} \right)$$, which (if I recall correctly) is the currently best known bound for values close to $$\sigma = 1$$. Bounds like this are very useful in all types of applications, where integrals containing the zeta function appear.
Many number-theoretic functions have error bounds contingent on the supremum of the real parts of zeroes of the Riemann zeta function. Call this $$\Theta$$. If we write $$f(x)=\Omega_{\pm}(g(x))$$ for the notion that \begin{align*} \limsup_{x\to\infty}\frac{f(x)}{g(x)}&>0,\text{ and}\\ \liminf_{x\to\infty}\frac{f(x)}{g(x)}&<0, \end{align*} then (e.g. in Montgomery and Vaughan, Section 15.1), for any $$\epsilon>0$$, \begin{align*} \psi(x)-x&=\Omega_{\pm}(x^{\Theta-\epsilon}),\\ \pi(x)-\operatorname{li}(x)&=\Omega_{\pm}(x^{\Theta-\epsilon}),\\ M(x)&=\Omega_{\pm}(x^{\Theta-\epsilon}). \end{align*} Here, $$\psi(x)=\log\operatorname{lcm}(1,2,\dots,n)$$ is the Chebyshev Psi function, $$\pi(x)$$ is the prime counting function, $$\operatorname{li}(x)$$ is the logarithmic integral, and $$M(x)$$ is the Mertens function. In other words, the maximal error of these summatory functions from the standard "simple" approximations is (somewhat) precisely determined by how far the zeroes of $$\zeta$$ may stray from the line $$\Re s=1/2$$. (In fact, sharper results involving $$\Omega_{\pm}(x^\Theta)$$ directly can be shown if there is a zero $$\rho$$ with $$\Re\rho=\Theta$$). So, knowing that $$\Theta<1$$ gives quite strong information about the distribution of prime numbers, and as $$\Theta\to1/2$$ such information approximates the best possible bounds, at least as far as functions of the form $$x^\alpha$$ are concerned.
• The $\Omega_\pm$ bounds give you lower bounds on the oscillation of the error terms, not upper bounds. Your answer would make more sense to me if you were citing the $O(x^{\Theta+\epsilon})$ bounds. – Wojowu Jun 13 at 1:04