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?

$\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – 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 HeathBrown'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.
Of course, this is just one of many and varied consequences.

$\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Ben Webster♦ Jun 13 at 0:29
Many numbertheoretic 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.

6$\begingroup$ 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. $\endgroup$ – Wojowu Jun 13 at 1:04