Orthogonality de Möbius

Question

Does anyone know how prove that $$\sum_{n\leqslant x}\mu(n)\xi(n) =o(x)$$ when $\xi(n)$ is a multiplicative functions? I found one commentary that exist a connection of this problem with the Theory of L-function. I studied this theory but still did not get to solve this problem. Any suggestions?

Answer 1

Someone knows how prove that $\sum_{n\leqslant x}\mu(n)\xi(n) =o(x)$ when $\xi(n)$ is a multiplicative functions?

It is false when $\xi$ is large (such as $\xi(n)=n$) or correlated with $\mu$ (such as $\xi(n)=\mu(n)$).

Answer 2

I am going to reformulate this problem slightly.

Question: Let $\xi:\mathbb{N}\rightarrow\mathbb{C}$ be a multiplicative function with $|\xi(n)|\leq 1$ for all $n$. Can we classify all such $\xi$ for which $$\sum_{n\leq x}\xi(n)\mu(n) \neq o(x)?$$

The answer to this problem is yes. A theorem of Halasz states that this is essentially only possible when $\xi(n)$ looks like $\mu(n)n^{it}$ for some real number $t$. To state this result in full generality, we need to define a metric on the space of multiplicative functions which map into the unit disk, and in this case, we can show that if the distance between $f$ and $n^{it}$ converges to zero for every $t$, then $$\sum_{n\leq x}f(n)=o(x).$$ This is in fact an if and only if statement. For more details, see chapter three of this paper of Andrew Granville.