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 functions with f*g=h and at least two of them are multiplicative, then all of them are multiplicative.
The above theorem 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)=1⋅1=1 since f and g are multiplicative.
Let a,b∈ℕ with gcd(a,b)=1. Then any divisor 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∈ℕ with gcd(a,b)=1. Induction 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=1⋅1=f(a)f(b).
Now let ab be an arbitrary natural number. 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)⋅1⋅1+(∑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 |