# Frequency of large values of the Mertens function

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 Montgomery-Vaughan, 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 $$f(X) = X + X^{2\sigma}$$.

It was proved by Kaczorowski and Pintz (Acta Math. Hungar. 48 (1986), 173-185, 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.
• 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?). Sep 1, 2021 at 21:15