Sum Over Divisors of Multiplicative Function

Theorem
Let $$f: \Z^*_+ \to \Z^*_+$$ be a multiplicative function.

Let $$n \in \Z^*_+$$.

Let $$\sum_{d \backslash n} f \left({d}\right)$$ be the sum over the divisors of $$n$$.

Then $$F \left({n}\right) = \sum_{d \backslash n} f \left({d}\right)$$ is also a multiplicative function.

Proof
Let $$F \left({n}\right) = \sum_{d \backslash n} f \left({d}\right)$$.

Let $$m, n \in \Z^*_+: m \perp n$$.

Then by definition:
 * $$F \left({m n}\right) = \sum_{d \backslash m n} f \left({d}\right)$$

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.

Therefore:
 * $$F \left({m n}\right) = \sum_{r \backslash m, s \backslash n} f \left({r s}\right)$$

(Note of course that $$r \perp s$$ otherwise any common divisor of $$r$$ and $$s$$ would be a common divisor of $$m$$ and $$n$$.)

So, as $$f$$ is multiplicative:
 * $$F \left({m n}\right) = \sum_{r \backslash m, s \backslash n} f \left({r}\right) f \left({s}\right)$$.

But at the same time:
 * $$F \left({m}\right) F \left({n}\right) = \left({\sum_{r \backslash m} f \left({r}\right)}\right) \left({\sum_{s \backslash n} f \left({s}\right)}\right)$$.

Multiplying out the product on the right, $$F \left({m n}\right)$$ and $$F \left({m}\right) F \left({n}\right)$$ are seen to be the same.