# formula for the convolution inverse of a completely multiplicative function

###### Corollary 1.

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

###### Proof.

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

Title formula for the convolution inverse of a completely multiplicative function FormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction 2013-03-22 16:55:09 2013-03-22 16:55:09 Wkbj79 (1863) Wkbj79 (1863) 5 Wkbj79 (1863) Corollary msc 11A25 CriterionForAMultiplicativeFunctionToBeCompletelyMultiplicative