It is known that with $M(x) = \sum_{n\le x}\mu(n)$, there are infinitely many $x$ s.t. $M(x)\ge x^{\frac{1}{2}  \varepsilon}$ (see Chapter 15 of MontgomeryVaughan, for example). Is there any way to make this result effective in the following sense: show that for large $X$, there exists some $x\in [X, f(X)]$ s.t. $M(x)\ge x^{\frac{1}{2}  \varepsilon}$ (or $x^{\sigma  \varepsilon}$ if $\zeta(\sigma + it) = 0$ for some $\sigma > 1/2$), where $f(X)$ is some explicit function in $X$? (Potentially naive) heuristics suggest one can take $f(X) = X + X^{2\sigma}$.
1 Answer
It was proved by Kaczorowski and Pintz (Acta Math. Hungar. 48 (1986), 173185, doi: 10.1007/BF01949062) that there exists $x\in[X,X^{1+o(1)}]$ such that $M(x)\geq x^{\sigma\varepsilon}$, and there also exists $x\in[X,X^{1+o(1)}]$ such that $M(x)\leq x^{\sigma\varepsilon}$. See Corollary 1 in their paper.

$\begingroup$ Does the result or method quickly lead to any uniformity in the height of the zero (so if there is a zero at $\sigma + it$, can one make the dependence of the result on $t$ explicit?). $\endgroup$ Sep 1, 2021 at 21:15

$\begingroup$ @MayankPandey: I don't know, I have not studied the details. $\endgroup$ Sep 2, 2021 at 1:00