Multiplicative Function that Converges to Zero on Prime Powers

Theorem

Let $f$ be a multiplicative function such that:

$\ds \lim_{p^k \mathop \to \infty} \map f {p^k} = 0$

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

Then:

$\ds \lim_{n \mathop \to \infty} \map f n = 0$

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

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:

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

where $t \ge 1$ and:

\(\ds 1\)

\(\le\)

\(\, \ds \size {\map f {p_i^{k_i} } } \, \)

\(\ds \)

for $i = 1, \ldots, r$

\(\ds \frac \epsilon A\)

\(\le\)

\(\, \ds \size {\map f {p_i^{k_i} } } \, \)

\(\, \ds < \, \)

\(\ds 1\)

for $i = r + 1, \ldots, r + s$

\(\ds \)

\(\)

\(\, \ds \size {\map f {p_i^{k_i} } } \, \)

\(\, \ds < \, \)

\(\ds \frac \epsilon A\)

for $i = r + s + 1, \ldots, r + s + t$

Therefore:

\(\ds \size {\map f n}\)

\(=\)

\(\ds \prod_{i \mathop = 1}^r \map f {p_i^{k_i} } \prod_{i \mathop = r + 1}^{r + s} \map f {p_i^{k_i} } \prod_{i \mathop = r + s + 1}^{r + s + t} \map f {p_i^{k_i} }\)

because $f$ is multiplicative

\(\ds \)

\(<\)

\(\ds A \paren {\frac \epsilon A}^t\)

\(\ds \)

\(<\)

\(\ds \epsilon\)

because $t \ge 1$

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

$\blacksquare$