criterion for a multiplicative function to be completely multiplicative

Theorem.

Let $f$ be a multiplicative function with convolution inverse $g$. Then $f$ is completely multiplicative if and only if $g\mathit{}\mathrm{(}{p}^k\mathrm{)}\mathrm{=}\mathrm{0}$ for all primes $p$ and for all $k\mathrm{\in }\mathrm{N}$ with $k\mathrm{>}\mathrm{1}$.

Proof.

Note first that, since $f(1)=1$ and $f*g=\epsilon$, where $\epsilon$ denotes the convolution identity function, then $g(1)=1$. Let $p$ be any prime. Then

$$0=\epsilon (p)=(f*g)(p)=f(1)g(p)+f(p)g(1)=g(p)+f(p).$$

Thus, $g(p)=-f(p)$.

Assume that $f$ is completely multiplicative. The statement about $g$ will be proven by induction on $k$. Note that:

Let $m\in \mathbb{N}$ with $m>2$ such that, for all $k\in \mathbb{N}$ with , $g(p^k)=0$. Then:

Conversely, assume that $g(p^k)=0$ for all $k\in \mathbb{N}$ with $k>1$. The statement $f(p^k)={(f(p))}^k$ will be proven by induction on $k$. The statement is obvious for $k=1$. Let $m\in \mathbb{N}$ such that $f(p^{m-1})={(f(p))}^{m-1}$. Then:

Thus, $f(p^m)={(f(p))}^m$. It follows that $f$ is completely multiplicative. ∎

Title	criterion for a multiplicative function to be completely multiplicative
Canonical name	CriterionForAMultiplicativeFunctionToBeCompletelyMultiplicative
Date of creation	2013-03-22 15:58:44
Last modified on	2013-03-22 15:58:44
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	9
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 11A25
Related topic	FormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction