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Dirichlet convolution

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For two functions f, g : \mathbb{N} \rightarrow \mathbb{C}, the Dirichlet convolution (or simply convolution, when the context is clear) f * g of f and g is defined as

\sum_{d\mid n} f(d)g\left( \frac{n}{d} \right).

We usually only consider positive divisors of n. We are often interested in convolutions of weak multiplicative functions; the set of weak multiplicative functions is closed under convolution. In general, convolution is commutative and associative; it also has an identity, the function f(n) defined to be 1 if n=1, and 0 otherwise. Not all functions have inverses (e.g., the function f(n) : n \mapsto 0 has no inverse, as f*g = f, for all functions g: \mathbb{N} \rightarrow \mathbb{C}), although all functions f such that f(1) \neq 0 have inverses.

Closure of Weak Multiplicative Functions Under Convolution

Theorem. If f,g : \mathbb{N} \rightarrow \mathbb{C} are weak multiplicative functions, then so is f*g.

Proof. Let m,n be relatively prime. We wish to prove that (f*g)(mn) = (f*g)(m)\cdot (f*g)(n).

For a \in \mathbb{N}, let {\rm D}_a be the set of divisors of a. For relatively prime m,n, we claim that the function p : (d_m,d_n) \mapsto d_md_n is a bijection from {\rm D}_m \times {\rm D}_n to {\rm D}_{mn}. Indeed, for any d_m \mid m and d_n \mid n, so p({\rm D}_m \times {\rm D}_n) \subseteq {\rm D}_{mn}. Furthermore, for each d \mid mn, there exist unique d_m, d_n such that d_m \mid m, d_n \mid n, d_md_n = d. Thus p is bijective. As a result of our claim, we have the identity

\sum_{d_m \mid m, d_n \mid n}a(d_m)b(d_n) = \sum_{d_m \mid m} a(d_m) \cdot \sum_{d_n \mid n}b(d_n),

for any functions a,b mapping subsets of \mathbb{N} into \mathbb{C}. In particular, we may let the domains of a and b be {\rm D}_m, {\rm D}_n, and define a(d_m) = f(d_m)g(m/d_m) and b(d_n) = f(d_n)g(n/d_n). We then have

\sum_{d_m \mid m}f(d_m)g(m/d_m) \cdot \sum_{d_n \mid n}f(d_n)g(n/d_n) = \sum_{d_m \mid m, d_n \mid n}f(d_m)g(m/d_m)f(d_n)g(n/....

But since each divisor of m is relatively prime to every divisor of n, we have

\sum_{d_m \mid m, d_n \mid n}f(d_m)g(m/d_m)f(d_n)g(n/d_n) = \sum_{d_m \mid m, d_n \mid n}f(d_md_n)g\left(\frac{mn}{d_md_n}\ri...,

as desired.

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