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 $d=d_{1}d_{2}$, 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,

∎

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 $d_{1}d_{2}<ab$, then $f(d_{1}d_{2})=f(d_{1})f(d_{2})$. Thus,

$\begin{array}[]{ll}h(a)h(b)&=h(ab)\\ \\ &=(f*g)(ab)\\ \\ &\displaystyle=\sum_{{d|ab}}f(d)g\left(\frac{ab}{d}\right)\\ \\ &\displaystyle=f(ab)g(1)+\sum_{{\substack{d|ab\\ d<ab}}}f(d)g\left(\frac{ab}{d}\right)\\ \\ &\displaystyle=f(ab)\cdot 1+\sum_{{\substack{d_{1}|a\\ d_{2}|b\\ d_{1}d_{2}<ab}}}f(d_{1}d_{2})g\left(\frac{ab}{d_{1}d_{2}}\right)\\ \\ &\displaystyle=f(ab)+\sum_{{\substack{d_{1}|a\\ d_{2}|b\\ d_{1}d_{2}<ab}}}f(d_{1})f(d_{2})g\left(\frac{a}{d_{1}}\right)g\left(\frac{b}{d_{2}}\right)\\ \\ &\displaystyle=f(ab)-f(a)f(b)g(1)g(1)+\sum_{{\substack{d_{1}|a\\ d_{2}|b}}}f(d_{1})f(d_{2})g\left(\frac{a}{d_{1}}\right)g\left(\frac{b}{d_{2}}\right)\\ \\ &\displaystyle=f(ab)-f(a)f(b)\cdot 1\cdot 1+\left(\sum_{{d_{1}|a}}f(d_{1})g\left(\frac{a}{d_{1}}\right)\right)\left(\sum_{{d_{2}|b}}f(d_{2})g\left(\frac{b}{d_{2}}\right)\right)\\ \\ &=f(ab)-f(a)f(b)+(f*g)(a)(f*g)(b)\\ \\ &=f(ab)-f(a)f(b)+h(a)h(b).\end{array}$

It follows that $f(ab)=f(a)f(b)$. ∎

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. ∎