Exercise 4.3.16* (The Dirichlet product of multiplicative functions is multiplicative)

Question

Let $f, g$ and $h$ be arithmetic functions such that $h(n) = \sum_{d \mid n } f(d) g(n/d)$ for all $n$. Show that if $f$ and $g$ are multiplicative then $h$ is also multiplicative.

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
If $n$ is a positive integer, let 
$$A_n = \{d \in \N^* :  d \mid n\}$$
denote the set of positive divisors of $n$. Then
$$h(n) = \sum_{d \in A_n}f(d) g(n/d).$$

Suppose that $n\wedge m = 1$. We know (see the proof of Theorem 4.4), that
$$\varphi
\left\{
\begin{array}{ccl}
A_n 	imes A_m & 	o & A_{nm}\
(d_1,d_2) & \mapsto & d_1d_2
\end{array}
ight.
$$
is a bijection, such that $\varphi^{-1}(d) = (d \wedge n, d \wedge m)$ for every $d \in A_{mn}$.

Then the change of index $d = \varphi(d_1,d_2)$ gives
\begin{align*}
h(nm) &= \sum_{d \in A_{nm}}  f(d) g\left(\frac{n}{d}ight)\
&=\sum_{(d_1,d_2) \in A_n 	imes A_m} f(d_1d_2) g\left(\frac{n}{d_1 d_2}ight).
\end{align*}
Since $d_1 \mid n$ and $d_2 \mid m$, where $n\wedge m = 1$, then $d_1 \wedge d_2 = 1$ and $(n/d_1) \wedge (n/d_2) = 1$, thus
\begin{align*}
h(nm) &= \sum_{(d_1,d_2) \in A_n 	imes A_m} f(d_1)f(d_2) g\left(\frac{n}{d_1} ight) g\left(\frac{n}{ d_2}ight)\
&=\sum_{d_1 \in A_n} f(d_1) g\left(\frac{n}{d_1} ight) \sum_{d_2 \in A_n}f(d_2) g\left(\frac{n}{ d_2}ight)\
&= h(n) h(m).
\end{align*}
So $h$ is multiplicative.
\end{proof}

Note: $h$ is called the Dirichlet product of $f$ and $g$. We write then $h = f \star g$.

Exercise 4.3.16* (The Dirichlet product of multiplicative functions is multiplicative)

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Exercise 4.3.16* (The Dirichlet product of multiplicative functions is multiplicative)

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