# Do relaxed Liouville functions violate Chowla's conjecture?

Let $\lambda$ be the Liouville function. One version of Chowla's conjecture says that for each set of distinct natural numbers $h_1 , \dots , h_k$,

$$\sum_{n\leq x} \lambda(n+h_1) \dots \lambda(n+h_k) = o(x)$$

Is this conjectural behavior stable under perturbations?

Say that a function $f: \mathbb N \to \pm 1$ is a relaxed Liouville function with error $\epsilon$ if for each $n$, $f(nm)=\lambda(n)f(m)$ for a density $1-\epsilon$ set of $m$.

Note that this definition is meaningful for $\epsilon = 0$. Then the obvious questions are:

Does there exist a relaxed Liouville function $f$ with error $0$ and a set of numbers $h_1, \dots, h_k$ such that

$$\sum_{n\leq x} f(n+h_1) \dots f(n+h_k) \neq o(x)?$$

and more vaguely:

Does there exist a relaxed Liouville function $f$ with error $\epsilon$ for $\epsilon$ "small", a set of numbers $h_1, \dots, h_k$, and a "large" constant $\delta$ such that:

$$\left|\sum_{n\leq x} f(n+h_1) \dots f(n+h_k)\right| \not \leq ( \delta +o(1) ) x?$$

For instance, can we take $\delta/\epsilon$ to be arbitrarily large with fixed $h_1, \dots, h_k$?

Clearly negative answers to these questions would imply Chowla's conjecture. So I am hoping instead for positive examples.

My motivation is that many elementary approaches that can prove some weak partial results towards Chowla's conjecture are very stable and therefore apply even for relaxed Liouville functions. I want to understand the limits of these elementary methods by explicit counterexample.

Here is an alternate definition which I think captures the same spirit. Can we find, for some $h_1, \dots, h_k$, for each $N$ a function $f: \mathbb N \to \pm 1$ with $f(pn) = - f(n)$ for $p< N$ such that

$$\left|\sum_{n\leq x} f(n+h_1) \dots f(n+h_k)\right| \not \leq ( \delta +o(1) ) x$$

for some $\delta$ independent of $N$? Or even just some $\delta$ with $\delta \log N$ arbitrarily large?

The idea being that about $1/\log N$ numbers have no prime factors less than $N$ and hence are arbitrary.

If you are willing to violate multiplicativity $f(nm)=f(n)f(m)$ about $\varepsilon$ of the time, then one can make $f$ more or less arbitrary on an interval of the form $[(1-\varepsilon) x, x]$, and this is enough to violate Chowla-type conjectures. So the answer to your question as stated is "yes".
On the other hand, if one retains multiplicativity, then it was conjectured by Elliott that the analog of Chowla's conjecture should hold whenever $\sum_p \frac{1 - \hbox{Re} f(p) \chi(p) p^{it}}{p} = \infty$ for any Dirichlet character $\chi$ and real $t$, which would imply a negative answer to your question for any $\varepsilon < 1/2$ if one insists on perfect multiplicativity. (Technically, Elliott's conjecture is false, see this recent paper of mine with Matomaki and Radziwill, but if one repairs it in the way indicated in that paper, the conclusion of the above argument still holds.) The technical counterexample aside, all other known evidence points to the truth of the (corrected) Elliott conjecture, for instance our paper establishes an averaged form of this conjecture, and a different averaged form was recently established by Frantzikinakis and Host.
• Thanks! This is helpful, but isn't quite what I'm looking for. In your example the error you are creating in Chowla's conjecture is only of size $\epsilon$. I want to know whether a small $\epsilon$ can cause a large error in Chowla's conjecture. This would demonstrate that local constructions that only use the multiplicativity for small sets of numbers are incapable of proving Chowla. (This question was actually inspired by Maksym Radziwill's talk on that paper.) – Will Sawin Jun 15 '15 at 3:52
• I think the current version of your notion of "relaxed Liouville function" may be too strong. If one has $f(nm) = \lambda(n) f(m)$ for most $n,m$, then one also has $f(nm) = \lambda(m) f(n)$ for most $n,m$, and so $f(n)/\lambda(n) = f(m)/\lambda(m)$ for most $n,m$, and so $f(n)$ is a constant multiple of $\lambda(n)$ for most $n$, and then the Chowla-type conjecture for $f$ will follow from that of $\lambda$ (with an error proportional to $\varepsilon$). – Terry Tao Jun 15 '15 at 21:18
• I meant the set of $(n,m)$ satisfying this condition to have density $1-\epsilon$ for any fixed $n$ but not necessarily as a set of pairs of natural numbers. So the intersection of the set of $(n,m)$ and $(m,n)$ covered could be empty. Sorry if that was a bit confusing! – Will Sawin Jun 15 '15 at 22:56
• I've been trying to think of different relaxations of this concept that could have an interesting answer. Another is to look at functions where we require $f(nm)=f(n)f(m)$ only if $n \leq N$ for some large constant $N$, and $f(p)=-1$ for primes $p \leq N$. Does a Chowla-type statement hold with error going to $0$ as $N$ goes to $\infty$? I can't imagine it does but I don't see a trivial argument why. – Will Sawin Jun 15 '15 at 23:07