Product form of a Dirichlet convolution

Question

I have derived (by a little unpure way) the product form for a Dirichlet convolution
$$f*g\left(n\right)=\sum_{d|n}f\left(d\right)g\left(\frac nd \right)=\prod_{p|n} \left( \sum_{m=0}^{k_p} f \left( p^m \right)g\left(p^{k_p -m}\right) \right)$$
where $f$ and $g$ are multiplicative functions, $p$ is a prime and
$$n=\prod_{p|n}p^{k_p} .$$
The product forms for the multiplicative function $f$ are
$$F\left( n \right) =\sum_{d|n}f\left(d\right)=\prod_{p|n}\left(\sum_{m=0}^{k_p} f \left( p^m \right) \right)$$
and
$$f\left( n \right) = \prod_{p|n} \left( F\left(p^{k_p} \right)-F\left(p^{k_{p}-1} \right)\right) $$
I get the correct result for all multiplicative function I know, but I am not able to prove it generally. Could someone to help me?

Answer 1

Since the Dirichlet convolution of multiplicative functions is multiplicative, we only need to understand $(f * g)(p^k)$ where $p$ is prime and $k \geq 0$. In particular, $$ (f * g)(p^k) = \sum_{d \mid p^k} f(d) g(p^k / d) = \sum_{m = 0}^k f(p^m) g(p^{k - m}) $$ since the only divisors of a prime power are powers of the same prime.

Now by multiplicativity, if $n = \prod p^{k_p}$, we have $$ (f * g)(n) = \prod_{p \mid n} (f * g)(p^{k_p}) = \prod_{p \mid n} \sum_{m = 0}^{k_p} f(p^m) g(p^{k_p - m}) $$ since powers of distinct primes are relatively prime.