convolution inverses for arithmetic functions

Theorem.

An arithmetic function $f$ has a convolution inverse if and only if $f\mathit{}\mathrm{(}\mathrm{1}\mathrm{)}\mathrm{\ne }\mathrm{0}$.

Proof.

If $f$ has a convolution inverse $g$, then $f*g=\epsilon$, where $\epsilon$ denotes the convolution identity function. Thus, $1=\epsilon (1)=(f*g)(1)=f(1)g(1)$, and it follows that $f(1)\ne 0$.

Conversely, if $f(1)\ne 0$, then an arithmetic function $g$ must be constructed such that $(f*g)(n)=\epsilon (n)$ for all $n\in \mathbb{N}$. This will be done by induction on $n$.

Since $f(1)\ne 0$, we have that ${\displaystyle \frac{1}{f(1)}}\in ℂ$. Define $g(1)={\displaystyle \frac{1}{f(1)}}$.

Now let $k\in \mathbb{N}$ with $k>1$ and $g(1),\mathrm{\dots },g(k-1)$ be such that $(f*g)(n)=\epsilon (n)$ for all $n\in \mathbb{N}$ with Define

Then

∎

In the entry titled arithmetic functions form a ring, it is proven that convolution is associative and commutative. Thus, $G=\{f:\mathbb{N}\to ℂ|f(1)\ne 0\}$ is an abelian group under convolution. The set of all multiplicative functions is a subgroup of $G$.