User anonnn - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:31:29Z http://mathoverflow.net/feeds/user/26301 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106503/sharpening-of-lindelof-hypothesis/106643#106643 Answer by anonnn for Sharpening of Lindelöf hypothesis anonnn 2012-09-08T03:27:25Z 2012-09-08T03:37:44Z <p>There are reason to doubt that the size of $\zeta(s)$ alone could be responsible for the truth of the Riemann Hypothesis, however bounds for $\zeta(s)$ could be equivalent to a statement of the form "RH does not fail massively", as I will explain further.</p> <p>Let me present one argument to convince you that the size of $\zeta(s)$ should be independent of the truth of RH: Suppose that the following (unlikely, but currently not ruled out) configurations of zeros occur in infinitely many intervals $[T; T + 1]$: we have roughly $\asymp \log T / \log\log T$ clusters of $\log\log T$ zeros, then in such interval $\zeta(s)$ should be of size $\exp(c \log T / \log\log T)$ (in particular contradicting the conjecture of Farmer, Gonek and Hughes). And then imagine that there are a few (say $4$) zeros of $\zeta(s)$ lying off the critical line. The two behaviors are envisage-able to occur simultaneously, unless of course we prove the falsehood/truth of each statement independently. </p> <p><strong>To wit</strong>: The size of $\zeta(s)$ is in a sense a local behavior, having $O(\log T)$ badly placed zeros in a $O(1)$ vicinity of a point $1/2 + it$ is enough to produce a very large (if not super large) value of $\zeta(s)$ at that point. Therefore the size of $\zeta(s)$ will not be affected by the truth or a small failure of the Riemann Hypothesis. However good bounds for $\zeta(s)$ can prevent the Riemann Hypothesis from failing badly. For example a result of Turan and Halasz asserts that if the Lindelof Hypothesis is true then there are at most $O(T^{\varepsilon})$ zeros in the half-plane $\sigma > \tfrac 34$. </p>