Sum Over Divisors of Multiplicative Function

Theorem

Let $f: \Z_{>0} \to \Z_{>0}$ be a multiplicative function.

Let $n \in \Z_{>0}$.

Let $\ds \sum_{d \mathop \divides n} \map f d$ be the sum over the divisors of $n$.

Then $\ds \map F n = \sum_{d \mathop \divides n} \map f d$ is also a multiplicative function.

Let $\ds \map F n = \sum_{d \mathop \divides n} \map f d$.

Let $m, n \in \Z_{>0}: m \perp n$.

Then by definition:

$\ds \map F {m n} = \sum_{d \mathop \divides m n} \map f d$

The divisors of $m n$ are of the form $d = r s$ where $r$ and $s$ are divisors of $m$ and $n$ respectively, from Divisors of Product of Coprime Integers.

It is noted that $r \perp s$, otherwise any common divisor of $r$ and $s$ would be a common divisor of $m$ and $n$.

Therefore:

$\ds \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f {r s}$

So, as $f$ is multiplicative:

$\ds \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f r \map f s$

But at the same time:

$\ds \map F m \map F n = \paren {\sum_{r \mathop \divides m} \map f r} \paren {\sum_{s \mathop \divides n} \map f s}$

Multiplying out the product on the right hand side, $\map F {m n}$ and $\map F m \map F n$ are seen to be the same.

$\blacksquare$