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

$(f*g)(k)$	$={\displaystyle \sum _{d|k}}f\left({\displaystyle \frac{k}{d}}\right)g(d)$
	$=0$
	$=\epsilon (k).$

∎

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 $G$.

Title	convolution inverses for arithmetic functions
Canonical name	ConvolutionInversesForArithmeticFunctions
Date of creation	2013-03-22 15:58:32
Last modified on	2013-03-22 15:58:32
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	27
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 11A25
Related topic	ArithmeticFunction
Related topic	MultiplicativeFunction
Related topic	ArithmeticFunctionsFormARing
Related topic	ElementaryResultsAboutMultiplicativeFunctionsAndConvolution