# convolution inverses for arithmetic functions

###### Theorem.

An arithmetic function $f$ has a convolution inverse if and only if $f(1)\neq 0$.

###### Proof.

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

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

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

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

 $g(k)=-\frac{1}{f(1)}\sum_{d|k\text{ and }d

Then

 $\displaystyle(f*g)(k)$ $\displaystyle=\sum_{d|k}f\left(\frac{k}{d}\right)g(d)$ $\displaystyle=f(1)g(k)+\sum_{d|k\text{ and }d $\displaystyle=f(1)\left(-\frac{1}{f(1)}\sum_{d|k\text{ and }d $\displaystyle=0$ $\displaystyle=\varepsilon(k).$

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

Title convolution inverses for arithmetic functions ConvolutionInversesForArithmeticFunctions 2013-03-22 15:58:32 2013-03-22 15:58:32 Wkbj79 (1863) Wkbj79 (1863) 27 Wkbj79 (1863) Theorem msc 11A25 ArithmeticFunction MultiplicativeFunction ArithmeticFunctionsFormARing ElementaryResultsAboutMultiplicativeFunctionsAndConvolution