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?



Asymptotics of the Liouville sum at the primes
This should be true. By a Corollary II of a [result of Pintz](http://matwbn.icm.edu.pl/ksiazki/aa/aa42/aa4214.pdf) (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 $c\ge\frac{1}{2}$.


