criterion for a multiplicative function to be completely multiplicative


Let f be a multiplicative functionMathworldPlanetmath with convolution inverse g. Then f is completely multiplicative if and only if g(pk)=0 for all primes p and for all kN with k>1.


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


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

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


Let m with m>2 such that, for all k with 1<k<m, g(pk)=0. Then:


Conversely, assume that g(pk)=0 for all k with k>1. The statement f(pk)=(f(p))k will be proven by induction on k. The statement is obvious for k=1. Let m such that f(pm-1)=(f(p))m-1. Then:


Thus, f(pm)=(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