Arithmetic functions encode numerical data of integers. Dirichlet convolution turns divisor sums into an algebra in which multiplicativity, inversion, and Euler products become transparent.
Multiplicative functions
Definition. An arithmetic function is \(f:\mathbb N\to\mathbb C\). It is multiplicative if \[ f(mn)=f(m)f(n)\quad\text{when }\gcd(m,n)=1, \] and completely multiplicative if this holds for all \(m,n\).
Examples include \(1(n)=1\), the identity \(\operatorname{id}(n)=n\), Euler's \(\varphi\), the divisor functions \[ \tau(n)=\sum_{d\mid n}1,\qquad \sigma_k(n)=\sum_{d\mid n}d^k, \] and the M?bius function \(\mu\).
Dirichlet convolution
Definition. \[ (f*g)(n)=\sum_{d\mid n}f(d)g(n/d). \] The identity is \(\varepsilon(1)=1\), \(\varepsilon(n)=0\) for \(n>1\).
Proposition. Dirichlet convolution is commutative and associative. If \(f(1)\ne0\), then \(f\) has a unique convolution inverse. The convolution of multiplicative functions is multiplicative.
Proof. Commutativity follows from \(d\leftrightarrow n/d\). Associativity follows by rewriting both sides as \[ \sum_{abc=n}f(a)g(b)h(c). \] For inversion, solve recursively from \[ f(1)g(n)+\sum_{\substack{d\mid n\\d>1}}f(d)g(n/d)=0. \] If \(\gcd(m,n)=1\), every divisor of \(mn\) uniquely splits as \(ab\) with \(a\mid m,b\mid n\); this proves multiplicativity of \(f*g\).
Prime-power reduction
A multiplicative function is determined by its values on prime powers: \[ f(n)=\prod_{p^\alpha\parallel n}f(p^\alpha). \] Consequently, \[ \tau(n)=\prod_{p^\alpha\parallel n}(\alpha+1),\qquad \sigma_k(n)=\prod_{p^\alpha\parallel n} \frac{p^{k(\alpha+1)}-1}{p^k-1}. \]
Proposition. \[ \varphi=\operatorname{id}*\mu,\qquad \operatorname{id}=\varphi*1. \] Equivalently, \[ \sum_{d\mid n}\varphi(d)=n. \]
Proof. Partition the integers \(1\le k\le n\) according to \(d=\gcd(k,n)\). For a fixed divisor \(d\), writing \(k=d\ell\) leaves \(\varphi(n/d)\) choices. Hence \(\sum_{e\mid n}\varphi(e)=n\). Convolution with \(\mu=1^{-1}\) yields the first identity.
Average order
Example (divisor summatory function). Changing the order of summation gives \[ \sum_{n\le x}\tau(n) =\sum_{ab\le x}1 =2\sum_{a\le\sqrt x}\left\lfloor\frac xa\right\rfloor -\lfloor\sqrt x\rfloor^2. \] Using the harmonic-sum asymptotic yields \[ \sum_{n\le x}\tau(n) =x\log x+(2\gamma-1)x+O(\sqrt x). \] This Dirichlet hyperbola method turns a divisor convolution into a lattice point count.