📂Number Theory

Generalized Dirichlet Product

Definition ¹

Let $F : \mathbb{R}^{+} \to \mathbb{C}$ be a function where $x \in (0,1)$, and let $F(x) = 0$ apply. For any arithmetic function $\alpha$, the following operation $\circ$ is defined as the generalized Dirichlet product. $$(\alpha \circ F)(x) := \sum_{n \le x} \alpha (n) F \left( {{ x } \over { n }} \right)$$

Basic Properties

Let $\alpha$ and $\beta$ be arithmetic functions, and let $F , G : \mathbb{R}^{+} \to \mathbb{C}$ be a function where its function value is $0$ in $x \in (0,1)$.

• [1]: $\alpha \circ \left( \beta \circ F \right) = \left( \alpha \ast\ \beta \right) \circ F$
• [2] Left unit element: $(I \circ F) = F$
• [3] Generalized inverse formula: If $\alpha$ has an inverse $\alpha^{-1}$, then $$G(x) = \sum_{n \le x} \alpha (n) F \left( {{ x } \over { n }} \right) \iff F(x) = \sum_{n \le x} \alpha^{-1} (n) G \left( {{ x } \over { n }} \right)$$
• [4] Generalized Möbius inverse formula: If $\alpha$ is completely multiplicative, then $$G(x) = \sum_{n \le x} \alpha (n) F \left( {{ x } \over { n }} \right) \iff F(x) = \sum_{n \le x} \mu (n) \alpha (n) G \left( {{ x } \over { n }} \right)$$

Explanation

The generalized Dirichlet product commonly appears throughout analytic number theory. The difference from the original convolution is that one of the two functions does not have to be an arithmetic function and that the index of $\sum$ has been changed from $d \mid n$ to $n \le x$. If $F$ is an arithmetic function where $F(x) = 0$ for all places where $x \notin \mathbb{N}$ applies, then $\circ$ precisely becomes $\ast$. In other words, for all $m \in \mathbb{N}$ $$(\alpha \circ F) (m) = (\alpha \ast\ F)(m)$$ hence, calling $\circ$ a generalization of $\ast$. Such operation $\circ$ may generally not follow the associative or commutative law. From the definition alone, there is a constraint that the left side of the operator must be an arithmetic function, which is why theorem [2] specifically mentions the property of being a left unit element.

Theorem [4] is a generalization of Möbius’s inverse formula, noting that it involves the extended functions $F$ and $G$ instead of the arithmetic function $\alpha$.

[1]

For $x > 0$, $$\begin{align*} \left[ \alpha \circ \left( \beta \circ F \right) \right] (x) =& \sum_{n \le x} \alpha (n) \sum_{m \le x/n} \beta (m) F \left( {{ x } \over { mn }} \right) \\ =& \sum_{mn \le x} \alpha (n) \beta (m) F \left( {{ x } \over { mn }} \right) \\ =& \sum_{k \le x} \left[ \sum_{n \mid k } \alpha (n) \beta \left( {{ k } \over { n }} \right) \right] F \left( {{ x } \over { k }} \right) \\ =& \sum_{k \le x} ( \alpha \ast\ \beta ) (k) F \left( {{ x } \over { k }} \right) \\ =& \left[ ( \alpha \ast\ \beta ) \circ F \right] (x) \end{align*}$$

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[2]

The identity $I$, for all $m \in \mathbb{N}$, $$(I \circ F)(x) = \sum_{n \le x} \left[ {{ 1 } \over { n }} \right] F \left( {{ x } \over { n }} \right) = F(x)$$

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[3]

If $G = \alpha \circ F$, then by Theorem [1] and Theorem [2], $$\alpha^{-1} \circ G = \alpha^{-1} \circ \left( \alpha \circ F \right) = \left( \alpha^{-1} \ast\ \alpha \right) \circ F = I \circ F = F$$ The proof in the reverse direction is similar.

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[4]

If a property of completely multiplicative functions $f$ is multiplicative, then being a completely multiplicative function $f$ is equivalent to the inverse of the Dirichlet product of $f$ $f^{-1}$ being as follows. $$f^{-1} (n) = \mu (n) f (n)$$

Since $\alpha$ is assumed to be completely multiplicative, applying Theorem [3] to $\alpha^{-1} (n) = \mu (n) \alpha (n)$ gives the result.

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See Also

• The Dirichlet product of arithmetic functions