Proof. To show that $F$ is a multiplicative function, we must show that if $m$ and $n$ are relatively prime positive integers, then $F(m n)=F(m) F(n)$. So let us assume that $(m, n)=1$. We have
$$F(m n)=\sum_{d \mid m n} f(d)$$

By Lemma 2.5 , since $(m, n)=1$, each divisor of $m n$ can be written uniquely as the product of relatively prime divisors $d_{1}$ of $m$ and $d_{2}$ of $n$, and each pair of divisors $d_{1}$ of $m$ and $d_{2}$ of $n$ corresponds to a divisor $d=d_{1} d_{2}$ of $m n$. Hence, we can write
$$F(m n)=\sum_{\substack{d_{1}\left|m \\ d_{2}\right| n}} f\left(d_{1} d_{2}\right)$$

Since $f$ is multiplicative and since $\left(d_{1}, d_{2}\right)=1$, we see that
$$\begin{aligned} F(m n) & =\sum_{\substack{d_{1}\left|n \\ d_{2}\right| n}} f\left(d_{1}\right) f\left(d_{2}\right) \\ & =\sum_{d_{1} \mid m} f\left(d_{1}\right) \sum_{d_{2} \mid n} f\left(d_{2}\right) \\ & =F(m) F(n) \end{aligned}$$