Let \(f\) be a multiplicative arithmetic function, and, for \(n \in \mathbb{Z}\) with \(n \gt 0\), let:
\[ F(n) = \sum_{d|n, d \gt 0} f(d) \]
We need to prove that \(F(n)\) would also be multiplicative. Let \(h\) and \(k\) be two relatively positive prime integer. \(F\) being multiplicative means \(F(hk) = F(h)F(k)\).
According to this lemma, for each divisor \(d\) of \(hk\), there should be integers \(d_1\) and \(d_2\) such that \(d=d_1d_2\), \(d_1|h\), \(d_2|k\) and \((d_1,d_2)= 1\).
\[\begin{gather} F(hk) = \sum_{d|n, d \gt 0} f(d) \\ F(hk) = \sum_{d_1|h,d_2|k, d_1 \gt 0,d_2 \gt 0} f(d_1d_2) \end{gather}\]
Since \(f\) is multiplicative:
\[\begin{align} F(hk) &= \sum_{d_1|h,d_2|k, d_1 \gt 0,d_2 \gt 0} f(d_1)f(d_2) \\ &= \sum_{d_1|h, d_1 \gt 0} \sum_{d_2|k,d_2 \gt 0} f(d_1) f(d_2) \end{align}\]
According to the distributive property of double summation:
\[\begin{gather} F(hk) = \sum_{d_1|h, d_1 \gt 0} f(d_1)\sum_{d_2|k,d_2 \gt 0} f(d_2) \\ \therefore F(hk)=F(h) * F(k) \end{gather}\]
For example, let \(h=3\) and \(k=4\), so \(hk=12\):
\[\begin{align} F(12) &= \sum_{d|12, d \gt 0} f(d) \\ &= f(1)+f(2)+f(3)+f(4)+f(6)+f(12) \end{align}\]
Each \(d\) can be written as \(d_1d_2\) where \(d_1|3\), \(d_2|4\), and \((d_1,d_2)=1\):
\[F(12) = f(1*1)+f(1*2)+f(3*1)+f(1*4)+f(3*2)+f(3*4)\]
Since \(f\) is multiplicative:
\[\begin{align} F(12) &= f(1)f(1)+f(1)f(2)+f(3)f(1)+f(1)f(4)+f(3)f(2)+f(3)f(4) \\ &= [f(1)+f(3)][f(1)+f(2)+f(4)]\end{align}\]