# Asymptotics of the Liouville sum at the primes

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?

• 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. Apr 11, 2023 at 18:06
• For your information, RH predicts the prime gap to be $O(x^{1/2}\log x)$ Apr 16, 2023 at 16:26

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 $$\#\mathcal S\gg Y^{1-2\varepsilon}.$$ 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 $$c\ge\frac{1}{2}$$.