elementary results about multiplicative functions and convolution


One of the most important elementary results about multiplicative functions and convolution (http://planetmath.org/DirichletConvolution) is:

Theorem.

If f, g, and h are arithmetic functionsMathworldPlanetmath with f*g=h and at least two of them are multiplicative, then all of them are multiplicative.

The above theoremMathworldPlanetmath will be proven in two separate parts.

Lemma 1.

If f and g are multiplicative, then so is f*g.

Proof.

Note that (f*g)(1)=f(1)g(1)=11=1 since f and g are multiplicative.

Let a,b with gcd(a,b)=1. Then any divisorMathworldPlanetmathPlanetmath d of ab can be uniquely factored as d=d1d2, where d1 divides a and d2 divides b. When a divisor d of ab is factored in this manner, the fact that gcd(a,b)=1 implies that gcd(d1,d2)=1 and gcd(ad1,bd2)=1. Thus,

(f*g)(ab) =d|abf(d)g(abd)
=d1|ad2|bf(d1d2)g(abd1d2)
=d1|ad2|bf(d1)f(d2)g(ad1)g(bd2)
=(d1|af(d1)g(ad1))(d2|bf(d2)g(bd2))
=(f*g)(a)(f*g)(b).

Lemma 2.

If f is an arithmetic function and g and h are multiplicative functions with f*g=h, then f is multiplicative.

Proof.

Let a,b with gcd(a,b)=1. InductionMathworldPlanetmath will be used on ab to establish that f(ab)=f(a)f(b).

If ab=1, then a=b=1. Note that f(1)=f(1)1=f(1)g(1)=(f*g)(1)=h(1)=1. Thus, f(ab)=f(1)=1=11=f(a)f(b).

Now let ab be an arbitrary natural numberMathworldPlanetmath. The induction hypothesis yields that, if d1 divides a, d2 divides b, and d1d2<ab, then f(d1d2)=f(d1)f(d2). Thus,

h(a)h(b)=h(ab)=(f*g)(ab)=d|abf(d)g(abd)=f(ab)g(1)+d|abd<abf(d)g(abd)=f(ab)1+d1|ad2|bd1d2<abf(d1d2)g(abd1d2)=f(ab)+d1|ad2|bd1d2<abf(d1)f(d2)g(ad1)g(bd2)=f(ab)-f(a)f(b)g(1)g(1)+d1|ad2|bf(d1)f(d2)g(ad1)g(bd2)=f(ab)-f(a)f(b)11+(d1|af(d1)g(ad1))(d2|bf(d2)g(bd2))=f(ab)-f(a)f(b)+(f*g)(a)(f*g)(b)=f(ab)-f(a)f(b)+h(a)h(b).

It follows that f(ab)=f(a)f(b). ∎

The theorem follows from these two lemmas and the fact that convolution is commutative.

The theorem has an obvious corollary.

Corollary.

If f is multiplicative, then so is its convolution inverse.

Proof.

Let f be multiplicative. Since f(1)0, f has a convolution inverse f-1. (See convolution inverses for arithmetic functions for more details.) Since f*f-1=ε, where ε denotes the convolution identity function, and both f and ε are multiplicative, the theorem yields that f-1 is multiplicative. ∎

Title elementary results about multiplicative functions and convolution
Canonical name ElementaryResultsAboutMultiplicativeFunctionsAndConvolution
Date of creation 2013-03-22 16:07:37
Last modified on 2013-03-22 16:07:37
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 16
Author Wkbj79 (1863)
Entry type Theorem
Classification msc 11A25
Related topic ConvolutionInversesForArithmeticFunctions