📂Number Theory

Dirichlet Product and Multiplicative Properties

Theorem ¹

• [1]: If $f$ and $g$ are multiplicative functions, then $f \ast\ g$ is also a multiplicative function.
• [2]: If $g$ and $f \ast g$ are multiplicative functions, then $f$ is also a multiplicative function.

Description

These properties can be used immediately when discussing the algebraic properties of multiplicative functions:

• Theorem [1] means, in other words, that multiplicative functions are closed under the convolution $\ast$.
• Theorem [2] allows proving that the inverse of a multiplicative function is also a multiplicative function by combining it with $g$ and $I = g\ast g^{-1}$.

Proof

[1]

Let’s say $h := f \ast\ g$ and $\gcd ( m , n ) = 1$ $$h(mn) = \sum_{c \mid mn} f(c) g \left( {{ mn } \over { c }} \right)$$ Now if we let $c$ to satisfy $a \mid m$ and $b \mid n$ as $c = ab$, then since $\gcd ( m, n) = 1$, it follows that $\gcd (a,b) = 1$, and $\gcd (m/a, n/b)$. Therefore $$\begin{align*} h(mn) =& \sum_{a \mid m \\ b \mid n} f (ab) g \left( {{ mn } \over { ab }} \right) \\ =& \sum_{a \mid m \\ b \mid n} f (a) f(b) g \left( {{ m } \over { a }} \right) g \left( {{ n } \over { b }} \right) \\ =& \sum_{a \mid m } f (a) g \left( {{ m } \over { a }} \right) \sum_{b \mid n} f(b) g \left( {{ n } \over { b }} \right) \\ =& h(m) h(n) \end{align*}$$

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[2]

Suppose $g$ and $h := f \ast\ g$ are multiplicative functions.

Assume for the sake of contradiction that $f$ is not a multiplicative function. If $f$ were not multiplicative, then there must exist $m, n$ that satisfies $f(mn) \ne f(m) f(n)$ and $\gcd (m,n) = 1$. Let’s, for convenience, choose $m,n$ such that $mn$ is minimized among all satisfying numbers.

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Case 1. $mn = 1$

Since $f(1) \ne f(1) f(1)$, it follows that $f(1) \ne 1$. But $$\begin{align*} 1 =& h(1) \\ =& f(1) g(1) \\ =& f(1) \cdot 1 \ne 1 \end{align*}$$ thus, $h = f \ast g$ becomes non-multiplicative, which is a contradiction.

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Case 2. $mn > 1$

Given the assumption of $m,n$, if $ab < mn$ and all $a,b$ satisfying $\gcd ( a,b) = 1$ must satisfy $f(ab) = f(a) f(b)$. Then, since $g$ is a multiplicative function, $g(1) = 1$; thus $$\begin{align*} h(mn) =& \sum_{a \mid m \\ b \mid n \\ ab < mn} f (ab) g \left( {{ mn } \over { ab }} \right) + f(mn) g(1) \\ =& \sum_{a \mid m \\ b \mid n \\ ab < mn} f (a) f(b) g \left( {{ m } \over { a }} \right) g \left( {{ n } \over { b }} \right) + f(mn) \cdot 1 \\ =& \sum_{a \mid m } f (a) g \left( {{ m } \over { a }} \right) \sum_{b \mid n} f(b) g \left( {{ n } \over { b }} \right) - f(m)f(n) + f(mn) \\ =& h(m)h(n) - f(m)f(n) + f(mn) \end{align*}$$ Summarizing $$h(mn) - h(m)h(n) = f(mn) - f(m) f(n)$$ But by the assumption of $f$, $f(mn) \ne f(m)f(n)$; thus $$h(mn) - h(m) h(n) \ne 0$$ Therefore, $h = f \ast g$ becomes non-multiplicative, which is a contradiction.

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