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I'm teaching myself Möbius inversion.

From Wikipedia it appears if $F$ and $G$ are complex-valued then

$G(x)=\sum\limits_{1 \le n \le x} F(x/n)$


$F(x)=\sum\limits_{1 \le n \le x} \mu(n) G(x/n)$

I was wondering if the implication works the other way around? That is, can we replace the "implies" with "iff"? If not, are there any simple counterexamples?

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@RobertIsrael I think the OP is referring to this:… – M Turgeon Aug 13 '12 at 16:19

Yes, it is an iff clause.

In fact, one can say something stronger:

If $\alpha(n)$ is completely multiplicative, then we have
$$ G(x) = \sum_{n \leq x} \alpha(n)F\left(\frac{x}{n}\right) \iff F(x) = \sum_{n \leq x} \mu(n)\alpha(n)G\left( \frac{x}{n}\right)$$

This is the strongest statement that I know, and is proven by considering convolutions and Dirichlet inverses. (For example, I think it's proven in Apostol's Intro to Analytic Number Theory)

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