Prove that $g(n)$ is a multiplicative function as well Suppose that $f(n)$ is any multiplicative function , and define a new function as $g(n) = f(d_1) + f(d_2) +...+f(d_r)$, where $d_1, d_2,...,d_r$ are divisors of $n$.
Prove that $g(n)$ is a multiplicative function as well. 
A little confused about this problem , but I  know what a multiplicative function is, since it is a function $f(n)$ that satisfies the multiplication formula 
$f(mn)=f(m)f(n)$  for all numbers $m$ and $n$ with $\gcd(m,n) = 1$
Any Help with this problem. Would love to see an example of a solution to this. Thanks.                      

Proof
$g(n)=\displaystyle\sum_{d \mid n}f(d),\ g(m)=\displaystyle\sum_{d \mid m}f(d)$
$g(mn)\\=\displaystyle\sum_{d \mid mn}f(d)\\=\displaystyle\sum_{d_1 \mid m, \ {d_2\mid n} }f(d_1d_2)\\=\displaystyle\sum_{d_1 \mid m, \ {d_2\mid n} }f(d_1)f(d_2)\\=\left(\displaystyle\sum_{d_1 \mid m}f(d_1)\right)\left(\displaystyle\sum_{d_2 \mid n}f(d_2)\right)\\=g(m)g(n)$
$\\$
Example
$g(24)\\=\displaystyle\sum_{d \mid 24}f(d)\\=f(1)+f(2)+f(3)+f(4)+f(6)+f(8)+f(12)+f(24)\\=f(1\cdot1)+f(2\cdot1)+f(1\cdot3)+f(4\cdot1)+f(2\cdot3)+f(8\cdot1)+f(4\cdot3)+f(8\cdot3)\\=\left(f(1)+f(2)+f(4)+f(8)\right)\left(f(1)+f(3)\right)=g(8)g(3)$
