completely multiplicative functions whose convolution inverses are completely multiplicative

Corollary 1.

The only completely multiplicative function whose convolution inverse is also completely multiplicative is $\varepsilon$ , the convolution identity function.

Proof.

Let $f$ be a completely multiplicative function whose convolution inverse is completely multiplicative. By this entry (http://planetmath.org/FormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction), $f\mu$ is the convolution inverse of $f$ , where $\mu$ denotes the Möbius function. Thus, $f\mu$ is completely multiplicative.

Let $p$ be any prime. Then

$\begin{array}[]{rl}(f(p))^{2}&=(f(p))^{2}(-1)^{2}\\ \\ &=(f(p))^{2}(\mu(p))^{2}\\ \\ &=(f(p)\mu(p))^{2}\\ \\ &=f(p^{2})\mu(p^{2})\\ \\ &=f(p^{2})\cdot 0\\ \\ &=0.\end{array}$

Thus, $f(p)=0$ for every prime $p$ . Since $f$ is completely multiplicative,

$f(n)=\begin{cases}1&\text{if }n=1\\ 0&\text{if }n\neq 1.\end{cases}$

Hence, $f=\varepsilon$ . ∎

Title	completely multiplicative functions whose convolution inverses are completely multiplicative
Canonical name	CompletelyMultiplicativeFunctionsWhoseConvolutionInversesAreCompletelyMultiplicative
Date of creation	2013-03-22 16:55:12
Last modified on	2013-03-22 16:55:12
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	4
Author	Wkbj79 (1863)
Entry type	Corollary
Classification	msc 11A25