Let $\lambda$ denote the Liouville function, and let $L(x):=\sum_{n \leq x} \lambda(n)$ be the Liouville sum. Define $c$ to be the supremum of the real parts of the zeros of the Riemann zeta function. It is a straightforward exercise to show that for any $\varepsilon>0$, one has $L(x) =\Omega_{\pm }(x^{c\varepsilon})$ as $x \rightarrow \infty$. Is this also true when $x$ runs through some special set, specifically the set of primes?

3$\begingroup$ If we know that prime gaps are $O(x^d)$, then the maximal values of $L$ on primes cannot deviate from general ones by at most $O(x^d)$. So if we know $d<c$ then we get the required result. Obvious arguments give a prime gap of $O(x^c\log x)$ which isn't quite good enough, but we might be able to hammer it out. $\endgroup$– WojowuApr 11, 2023 at 18:06

$\begingroup$ For your information, RH predicts the prime gap to be $O(x^{1/2}\log x)$ $\endgroup$– TravorLZHApr 16, 2023 at 16:26
1 Answer
This should be true. By a Corollary II of a result of Pintz (with not too much work, one can get this to work for the Liouville function in place of Mobius), we have that $$\sum_{Y/(100\log Y)\le n\le Y} L(n)\gg Y^{1 + c  \varepsilon},$$ in your notation. We also have that for $n$ in this range, $L(n) \ll Y^{c + \varepsilon}$, so we get that with $$\mathcal S = \{Y/(100\log Y)\le n\le Y : L(n)\ge Y^{c  10\varepsilon}\},$$ we have that $\#\mathcal S\gg Y^{12\varepsilon}.$ At this point, we wish to show there is a prime within $Y^{c  10\varepsilon}/2$ of an element of $\mathcal S$. This should follow comfortably from results on primes in almost all short intervals as $c\ge\frac{1}{2}$.

$\begingroup$ Thanks for your answer! Do you think Pintz's result/methods could help answer this: mathoverflow.net/q/445289/501735 ? $\endgroup$ Apr 22, 2023 at 11:40