Multiplicative Function that Converges to Zero on Prime Powers

Theorem
Let $f$ be a multiplicative function such that


 * $\displaystyle \lim_{p^k \to \infty} f(p^k) = 0$

where $p^k$ runs though all prime powers. Then


 * $\displaystyle \lim_{n \to \infty} f(n) = 0$

where $n$ runs through the integers.

Proof
By hypothesis, there exist only finitely many prime powers $p^k$ such that $|f(p^k)| > 1$.

Let $\displaystyle A = \prod_{|f(p^k)| > 1}|f(p^k)|$, so $A \geq 1$.

Let $0 < \epsilon < A$.

There exist only finitely many prime powers $p^k$ such that $|f(p^k)| > \epsilon / A$.

Therefore there are only finitely many integers $n$ such that $|f(p^k)| > \epsilon/A$ for every prime power $p^k$ that divides $n$.

Therefore if $n$ is sufficiently large there exists a prime power $p^k$ that divides $n$ and $|f(p^k)| < \epsilon / A$.

Therefore $n$ can be written as


 * $\displaystyle n = \prod_{i=1}^{r}p_i^{k_i}\prod_{i=r+1}^{r+s}p_i^{k_i}\prod_{i=r+s+1}^{r+s+t}p_i^{k_i}$

where $t \geq 1$ and

Therefore

This shows that $f(n)$ is arbitrarily small for sufficiently large $n$.