I am trying to prove the following claim:
Let $f$ be an arithmetic function and, for $n \in \mathbb{Z}$ with $n > 0$, let
$$
F(n) = \sum_{d\ \vert\ n,\ d > 0}f(d)
$$
If $F$ is multiplicative, then $f$ is multiplicative.
So far I have the following: given relatively prime integers $a$ and $b$, we can uniquely express each divisor $d$ of $ab$ as $d_1d_2$ where $d_1\ \vert\ a$ and $d_2\ \vert\ b$. This means:
$$
F(ab) = \sum_{d\ \vert\ ab}f(d) = \sum_{d_1\ \vert\ a} \sum_{d_2\ \vert\ b} f(d_1d_2)
$$
$$
F(a)F(b) = \sum_{d_1\ \vert\ a} \sum_{d_2\ \vert\ b} f(d_1)f(d_2)
$$
Since $F$ is multiplicative,
$$
\sum_{d_1\ \vert\ a} \sum_{d_2\ \vert\ b} f(d_1)f(d_2) = \sum_{d_1\ \vert\ a} \sum_{d_2\ \vert\ b} f(d_1d_2)
$$
I am not sure where to proceed from here to show that corresponding terms of the two summations are equal. Am I approaching this the right way?
Best Answer
Using the Möbius inversion formula one gets $f(n)=\sum_{d|n} F(d)\mu(n/d)$ or $f=F*\mu$ where $\mu$ denotes the Möbius function and $*$ the convolution operator. Note that $\mu$ is multiplicative and that the convolution of two multiplicative functions is multiplicative as well. Thus $f=F*\mu$ is multiplicative.