formula for the convolution inverse of a completely multiplicative function

Corollary 1.

If $f$ is a completely multiplicative function, then its convolution inverse is $f\mathit{}\mu$, where $\mu$ denotes the Möbius function.

Proof.

Recall the Möbius inversion formula $1*\mu =\epsilon$, where $\epsilon$ denotes the convolution identity function. Thus, $f(1*\mu )=f\epsilon$. Since pointwise multiplication of a completely multiplicative function distributes over convolution (http://planetmath.org/PropertyOfCompletelyMultiplicativeFunctions), $(f\cdot 1)*(f\mu )=f\epsilon$. Note that, for all natural numbers $n$, $f(n)1(n)=f(n)\cdot 1=f(n)$ and $f(n)\epsilon (n)=\epsilon (n)$. Thus, $f*(f\mu )=\epsilon$. It follows that $f\mu$ is the convolution inverse of $f$. ∎

Title	formula for the convolution inverse of a completely multiplicative function
Canonical name	FormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction
Date of creation	2013-03-22 16:55:09
Last modified on	2013-03-22 16:55:09
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	5
Author	Wkbj79 (1863)
Entry type	Corollary
Classification	msc 11A25
Related topic	CriterionForAMultiplicativeFunctionToBeCompletelyMultiplicative