Is $\sum_{d | n} f(d)$ completely multiplicative if $f$ is completely multiplicative?

Question

In these excellent notes by Pete L. Clark, it is claimed that if $f(n)$ is a multiplicative function, then $F(n) = \sum_{d|n} f(d)$ is multiplicative as well. I understand the proof of this statement, however it is also said that this does not hold true for completely multiplicative functions. Why, in general, does the relation not preserve completely multiplicative functions? I tried to begin by assuming that $(m,n) = d$ and to derive a contradiction for $F(m) * F(n)$, but I was unable to.

Thanks!

Answer 1

The question is whether $F(p^{a+b})=F(p^a)F(p^b)$ holds for primes $p$, where $F(n)=\sum_{d\mid n}f(d)$ and $f$ itself is completely multiplicative. As a special case, we should have $F(p^2)=F(p)^2$. If we write out $f(1)=1$ and $f(p)=x$ (for some particular $p$) this condition reads

$$ 1+x+x^2=(1+x)^2.$$

If $x\ne0$ the above equation is false, so if any $f(p)$ is nonzero then $F$ cannot be completely multiplicative. Indeed, this shows the only completely multiplicative $f$ for which $F$ is also completely multiplicative is $f=\delta_1(n)$ (which is $1$ if $n=1$ and $0$ otherwise).

Answer 2

I think the easiest way to see why it's not always true for completely multiplicative functions is to come up with a counterexample.

Perhaps the simplest counterexample comes from one of the easiest completely multiplicative functions, the $1$ function, $f(n) \equiv 1$. Then $$ F(n) = \sum_{d \mid n} f(d) = \sum_{d \mid n} 1 = d(n) = \sigma_0(n),$$ and so $F(n)$ is the number-of-divisors function. This is multiplicative, but it is not completely multiplicative. For instance, $d(4) = 3$ while $d(2) = 2$, so $d(4) \neq d(2)^2$.