Theorem   If $f$ , $g$ , and $h$ are arithmetic functions with $f*g=h$ and at least two of them are multiplicative, then all of them are multiplicative.

Lemma 1   If $f$ and $g$ are multiplicative, then so is $f*g$ .

Proof. Note that $\displaystyle (f*g)(1)=f(1)g(1)=1 \cdot 1=1$ since $f$ and $g$ are multiplicative.

Let $a,b \in \mathbb{N}$ with $\gcd(a,b)=1$ . Then any divisor $d$ of $ab$ can be uniquely factored as $d=d_1d_2$ , where $d_1$ divides $a$ and $d_2$ divides $b$ . When a divisor $d$ of $ab$ is factored in this manner, the fact that $\gcd(a,b)=1$ implies that $\gcd(d_1,d_2)=1$ and $\displaystyle \gcd \left( \frac{a}{d_1}, \frac{b}{d_2} \right)=1$ . Thus,

Lemma 2   If $f$ is an arithmetic function and $g$ and $h$ are multiplicative functions with $f*g=h$ , then $f$ is multiplicative.

Proof. Let $a,b \in \mathbb{N}$ with $\gcd(a,b)=1$ . Induction will be used on $ab$ to establish that $f(ab)=f(a)f(b)$ .

If $ab=1$ , then $a=b=1$ . Note that $f(1)=f(1) \cdot 1=f(1)g(1)=(f*g)(1)=h(1)=1$ . Thus, $f(ab)=f(1)=1=1 \cdot 1=f(a)f(b)$ .

Now let $ab$ be an arbitrary natural number. The induction hypothesis yields that, if $d_1$ divides $a$ , $d_2$ divides $b$ , and $d_1d_2<ab$ , then $f(d_1d_2)=f(d_1)f(d_2)$ . Thus,

It follows that $f(ab)=f(a)f(b)$ .

Corollary   If $f$ is multiplicative, then so is its convolution inverse.

Proof. Let $f$ be multiplicative. Since $f(1) \neq 0$ , $f$ has a convolution inverse $f^{-1}$ . (See convolution inverses for arithmetic functions for more details.) Since $f*f^{-1}=\varepsilon$ , where $\varepsilon$ denotes the convolution identity function, and both $f$ and $\varepsilon$ are multiplicative, the theorem yields that $f^{-1}$ is multiplicative.