Sum involving the Möbius function

Question

I have two multiplicative functions $f$ and $g$ and the expression

$$\sum_{d\mid n} \mu(d) f(d)g(n/d).$$

In case $f=1$ this is just the Möbius inversion. But what can we say about it in this more general case?

Answer 1

For two arithmetic functions $f,g $ we define the Dirichlet's convoluction as $$\left(f*g\right)\left(n\right)=\sum_{d\mid n}f\left(d\right)g\left(n/d\right) $$ so in your case, if you consider the aritmethic function $F=\mu f $ we have $$\sum_{d\mid n}f\left(d\right)\mu\left(d\right)g\left(n/d\right)=\sum_{d\mid n}F\left(d\right)g\left(n/d\right)=\left(F*g\right)\left(n\right).$$ We may also note that if $f\mu $ is multiplicative, then $F*g$ si multiplicative.