# 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 $g(p^{k})=0$ for all primes $p$ and for all $k\in\mathbb{N}$ with $k>1$.

###### Proof.

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

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

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

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

$\begin{array}[]{ll}0&=\varepsilon(p^{2})\\ &=(f*g)(p^{2})\\ &=f(1)g(p^{2})+f(p)g(p)+f(p^{2})g(1)\\ &=g(p^{2})+f(p)(-f(p))+(f(p))^{2}\\ &=g(p^{2})\end{array}$

Let $m\in\mathbb{N}$ with $m>2$ such that, for all $k\in\mathbb{N}$ with $1, $g(p^{k})=0$. Then:

$\begin{array}[]{ll}0&=\varepsilon(p^{m})\\ &=(f*g)(p^{m})\\ &=f(1)g(p^{m})+f(p^{m-1})g(p)+f(p^{m})g(1)\\ &=g(p^{m})+(f(p))^{m-1}(-f(p))+(f(p))^{m}\\ &=g(p^{m})\end{array}$

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:

$\begin{array}[]{ll}0&=\varepsilon(p^{m})\\ &=(f*g)(p^{m})\\ &=f(p^{m-1})g(p)+f(p^{m})g(1)\\ &=(f(p))^{m-1}(-f(p))+f(p^{m})\\ &=-(f(p))^{m}+f(p^{m})\end{array}$

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

Title criterion for a multiplicative function to be completely multiplicative CriterionForAMultiplicativeFunctionToBeCompletelyMultiplicative 2013-03-22 15:58:44 2013-03-22 15:58:44 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Theorem msc 11A25 FormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction