Multiplicative Function that Converges to Zero on Prime Powers

Theorem
Let $f$ be a multiplicative function such that:
 * $\displaystyle \lim_{p^k \mathop \to \infty} \map f {p^k} = 0$

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

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} \map f n = 0$

where $n$ runs through the integers.

Proof
By hypothesis, there exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > 1$.

Let $\displaystyle A = \prod_{\size {\map f {p^k} } \mathop > 1} \size {\map f {p^k} }$.

Thus $A \ge 1$.

Let $0 < \dfrac \epsilon A$.

There exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > \dfrac \epsilon A$.

Therefore there are only finitely many integers $n$ such that:
 * $\size {\map f {p^k} } > \dfrac \epsilon A$

for every prime power $p^k$ that divides $n$.

Therefore if $n$ is sufficiently large there exists a prime power $p^k$ that divides $n$ such that:
 * $\size {\map f {p^k} } < \dfrac \epsilon A$

Therefore $n$ can be written as:
 * $\displaystyle n = \prod_{i \mathop = 1}^ r p_i^{k_i} \prod_{i \mathop = r + 1}^{r + s} p_i^{k_i} \prod_{i \mathop = r + s + 1}^{r + s + t} p_i^{k_i}$

where $t \ge 1$ and:

Therefore:

This shows that $\map f n$ can be made arbitrarily small for sufficiently large $n$.