The Dirichlet product (or Dirichlet convolution) of two arithmetic functions $f$ and $g$ is defined as
\[
(f * g)(n) = \sum_{d | n} f(d)g(n/d).
\] The Dirichlet product arises when multiplying two Dirichlet series, that is, if two Dirichlet series
\[
\sum_{n = 1}^{\infty} f(n) n^{-s} \quad \text{and} \quad \sum_{n = 1}^{\infty} g(n) n^{-s}
\] converge absolutely, then their product is the Dirichlet series
\[
\sum_{n = 1}^{\infty} (f * g)(n) n^{-s}.
\] It is easily verified that Dirichlet product is both commutative and associative and that $e * f = f * e =f$ for every arithmetic function $f$. Thus the set of all arithmetic functions forms a commutative monoid. The next result allows us to characterize arithmetic functions that are invertible under Dirichlet product.
Theorem. If $f$ is an arithmetic function with $f(1) \neq 0$, then there is a unique arithmetic function $f^{-1}$ such that $f^{-1} * f = f * f^{-1} = e$. The function $f^{-1}$ is given by
\[
f^{-1}(1) = \frac{1}{f(1)}, \qquad f^{-1}(n) = -\frac{1}{f(1)} \sum_{\scriptstyle d |
n \atop \scriptstyle d < n} f^{-1}(d) f(n/d) \quad \text{for } n > 1.
\] The above result show that the set of all arithmetic functions $f$ satisfying $f(1) \neq 0$ form an abelian group under Dirichlet product.
The Dirichlet multiplication provides a convenient notation to write relations among different arithmetic functions in a compact fashion
\[
\mu * 1 = e, \quad \varphi * 1 = I, \quad \varphi = \mu * I.
\]
Theorem. (Möbius inversion formula) Let $f$ and $g$ be arithmetic functions. Then
\[
f(n) = \sum_{d | n} g(d) \quad \text{for } n \in \mathbb{N}
\] if and only if
\[
g(n) = \sum_{d | n} f(d) \mu (n/d) \quad \text{for } n \in \mathbb{N}.
\]
Corollary. If $n \geq 1$, then
\[
\Lambda(n) = -\sum_{d | n} \mu(d) \log d.
\]
Proof. By Möbius inversion we have
\[
\Lambda(n) = \sum_{d | n} \mu(d) \log \frac{n}{d} = \log n \sum_{d | n} \mu(d)- \sum_{d | n} \mu(d) \log d = e(n) \log n -\sum_{d | n} \mu(d) \log d.
\] This completes the proof as $e(n) \log n = 0$ for every $n$.
An arithmetic function $f$ is called multiplicative if $f$ is not identically zero and $f(mn) = f(m) f(n)$ whenever $(m, n) = 1$. A multiplicative function $f$ is called completely multiplicative (or totally multiplicative) if $f$ is not identically zero and $f(mn) = f(m) f(n)$ for all $m, n$. We note some common examples of multiplicative functions.
- The power function $I_{\alpha}$ is completely multiplicative.
- The identity function $e$ is completely multiplicative.
- The Möbius function $\mu$ is multiplicative but not completely multiplicative as $\mu(4) = 0 \neq 1 = \mu(2)^2$.
- The Euler totient function $\varphi$ is multiplicative but not completely multiplicative as $\varphi(4) = 2 \neq 1 = \varphi(2)^2$.
- The Liouville function $\lambda$ is completely multiplicative.
Note that if $f$ is multiplicative, then $f(1) = 1$. From this it immediately follows that $\Lambda$ is not multiplicative as $\Lambda(1) = 0$.
Theorem. Let $f$ be an arithmetic function with $f(1) = 1$.
(i) $f$ is multiplicative if and only if
\[
f(p_1^{a_1} \cdots p_k^{a_k}) = f(p_1^{a_1}) \cdots f(p_k^{a_k}),
\] where $p_1, \dots, p_k$ are distinct primes.
(ii) If $f$ is multiplicative, then $f$ is completely multiplicative if and only if
\[
f(p^a) = f(p)^a
\] for all primes $p$ and all integers $a \geq 1$.
The above result shows that a multiplicative function is uniquely determined by its values on prime powers, and a completely multiplicative function is uniquely determined by its values on primes.
Theorem. If $f$ and $g$ are multiplicative, then so is their Dirichlet product $f * g$.
Proof. Let $m$ and $n$ be relatively prime integers. Then observe that
\[\begin{align*}
(f * g)(mn) &= \sum_{d | mn} f(d) g\left( \frac{mn}{d} \right) = \sum_{\scriptstyle a | m \atop \scriptstyle b | n} f(ab) g \left( \frac{mn}{ab} \right)
\end{align*}
\] as every divisor of $mn$ can be uniquely written as $ab$, where $a | m$ and $b | n$. Using the multiplicativity of $f$ and $g$ we obtain
\[\begin{align*}
(f * g)(mn) &= \sum_{\scriptstyle a | m \atop \scriptstyle b | n}f(a)f(b) g \left( \frac{m}{a} \right) g \left( \frac{n}{b} \right) = \sum_{a | m} \sum_{b | n} f(a) f(b) g \left( \frac{m}{a} \right) g \left( \frac{n}{b} \right) \\
&= \sum_{a | m} f(a) g \left( \frac{m}{a} \right) \sum_{b | n} f(b) g \left( \frac{n}{b} \right) = (f * g)(m) (f * g)(n).
\end{align*}
\]
The Dirichlet product of two completely multiplicative functions need not be completely multiplicative. For instance, the divisor function $d = 1 * 1$ is not completely multiplicative as $d(4) = 3 \neq 4 = d(2)^2$ whereas $1$ clearly is.
Theorem. If $f$ is multiplicative, then so is it’s Dirichlet inverse $f^{-1}$.
First Proof. Suppose for the sake of contradiction that $f^{-1}$ is not multiplicative. Then there exist positive integers $m$ and $n$ with $(m, n) = 1$ such that
\[
f^{-1}(mn) \neq f^{-1}(m) f^{-1}(n).
\] We choose such a pair $m$ and $n$ for which the product $mn$ is the smallest. Since $f$ is multiplicative therefore $f^{-1}(1) = 1/f(1) = 1$ and hence neither $m$ nor $n$ can be $1$. In particular, $mn > 1$. By the construction of the product $mn$, $f(ab) = f(a)f(b)$ for all positive integers $a$ and $b$ with $(a, b) = 1$ and $ab < mn$. It now follows that
\[
\begin{align*} f^{-1}(mn) = -\sum_{\scriptstyle a | m \atop {\scriptstyle b | n \atop \scriptstyle ab < mn}} f^{-1}(a b) f \left( \frac{mn}{ab} \right) = -\sum_{\scriptstyle a | m \atop {\scriptstyle b | n \atop \scriptstyle ab < mn}} f^{-1}(a) f^{-1}(b) f \left( \frac{m}{a} \right) f\left( \frac{n}{b} \right)
\end{align*}\]
Splitting the sum we obtain
\[\begin{align*} f^{-1}(mn) &= -f^{-1}(n) \sum_{\scriptstyle a | m \atop \scriptstyle a < m} f^{-1}(a) f\left( \frac{m}{a} \right) -f^{-1}(m) \sum_{\scriptstyle b | n \atop \scriptstyle b < n} f^{-1}(b) f\left( \frac{n}{b} \right) \\ & -\sum_{\scriptstyle a | m \atop \scriptstyle a < m} \sum_{\scriptstyle b | n \atop \scriptstyle b < n} f^{-1}(a) f^{-1}(b) f \left( \frac{m}{a} \right) f\left( \frac{n}{b} \right) \\ &= f^{-1}(n)f^{-1}(m) + f^{-1}(m)f^{-1}(n) – f^{-1}(m)f^{-1}(n) \\ &= f^{-1}(m)f^{-1}(n). \end{align*} \] This contradiction proves the result.
Second Proof. Let $g$ be an arithmetic function defined as
\[
g(n) = \prod_{p^a || n} f^{-1}(p^a).
\] Then $g$ is a multiplicative function by definition and so it suffices to show that $f^{-1} = g$. Note that
\[
\begin{align*}
(g * f)(p^k) &= \sum_{d | p^k} g(d) f(p^k/d) = \sum_{i = 0}^{k} g(p^i)f(p^{k-i}) \\
&= \sum_{i = 0}^{k} f^{-1}(p^i) f(p^{k-i}) = \sum_{d | p^k} f^{-1}(d) f(p^k/d) = (f^{-1} * f)(p^k) = e(p^k).
\end{align*}\] Because $g * f$ and $e$ are both multiplicative functions and agree on prime powers, it follows that $g * f = e$ and so $g = f^{-1}$.
The next result allows us to characterize completely multiplicative functions.
Theorem. Let $f$ be multiplicative. Then $f$ is completely multiplicative if and only if $f^{-1} = \mu f$.
Proof. Suppose $f$ is completely multiplicative. Then observe that
\[
(f * \mu f)(n) = \sum_{d | n} \mu(d) f(d) f \left( \frac{n}{d} \right) = f(n) \sum_{d | n} \mu(d) = f(n) e(n) = e(n).
\] Conversely, assume that $f^{-1} = \mu f$. Then observe that
\[
\sum_{d | n} \mu(d) f(d) f \left( \frac{n}{d} \right) = 0
\] for $n > 1$. Let $n = p^a$, where $a \geq 1$. Then, we get
\[
\mu(1)f(1)f(p^{a}) + \mu(p)f(p)f(p^{a -1}) = 0.
\] It then follows that
\[
f(p^{a}) = f(p)f(p^{a -1}).
\] This implies that $f(p^{a}) = f(p)^{a}$. Thus $f$ is completely multiplicative.