The Dirichlet Convolution of Two Arithmetic Functions
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The Dirichlet Convolution of Two Arithmetic Functions

 Definition: Let $f$ and $g$ be arithmetic functions. The Dirichlet Convolution of $f$ and $g$ is defined to be the function $h = \sum_{d|n} f(d) g \left ( \frac{n}{d} \right )$ and is denoted by $h = f * g$.
 Theorem 1: Let $f$ and $g$ be arithmetic functions. If $f$ and $g$ are multiplicative then $h = f * g$ is multiplicative.

Theorem 1 is significant. It says that if two arithmetic functions $f$ and $g$ are multiplicative then their Dirichlet convolution $f * g$ is also multiplicative.

• Proof: Let $(m, n) = 1$. If $d | mn$ and since $(m, n) = 1$ there exists unique integers $a, b$ such that $a | m$, $b | n$, and $d = ab$. Therefore:
(1)
\begin{align} \quad h(mn) &= \sum_{d|mn} f(d) g \left ( \frac{mn}{d} \right ) \\ &= \sum_{a|m} \sum_{b|n} f(ab) g \left ( \frac{mn}{ab} \right ) \\ \end{align}
• Since $(m, n) = 1$ we must have that $(a, b) = 1$. To see this, suppose that $(a, b) = d$. Since $a | m$ and $b | n$ we must have that $d | m$ and $d | n$. So $d = 1$. Furthermore, we must also have that $\left ( \frac{m}{a}, \frac{n}{b} \right ) = 1$. Since $f$ and $g$ are multiplicative, we have that:
(2)
\begin{align} \quad h(mn) &= \sum_{a|m} \sum_{b|n} f(a)f(b) g \left ( \frac{m}{a} \right ) g \left ( \frac{n}{b} \right ) \\ &= \left [ \sum_{a|m} f(a) g \left ( \frac{m}{a} \right ) \right ] \cdot \left [ \sum_{b|n} f(b) g \left ( \frac{n}{b} \right ) \right ] \\ &= h(m) h(n) \end{align}
• Therefore $h$ is multiplicative. $\blacksquare$
 Corollary 2: Let $f$ and $g$ be arithmetic functions. If $f$ and $g$ are completely multiplicative then the convolution of $f$ and $g$, $f * g$ is multiplicative.

It is important to note that in general, the complete multiplicativity of $f$ and $g$ does NOT imply that the convolution $f * g$ is completely multiplicative.

• Proof: If $f$ and $g$ are completely multiplicative then they are multiplicative. So by Theorem 1 their Dirichlet convolution $f * g$ is multiplicative. $\blacksquare$
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