TITLE: What are some consequences of zero free strip of the Riemann zeta function?
QUESTION [27 upvotes]: 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?

REPLY [14 votes]: 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.