Exercise 4.3.20 (Generalization of Möbius inversion formula)

Proof. We recall that by Theorem 4.7,

(Note that $\delta$ is the neutral element for the Dirichlet product : $F\star \delta =\delta \star f=f$.)

• We suppose that $G(x)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n\le x}F(x∕n)$ for all $x\in [1,+\infty [$. Then

  $$\begin{array}{llllll}\hfill \underset{n\le x}{\sum }\mu (n)G\left(\frac{x}{n}\right)& =\underset{n\le x}{\sum }\mu (n)\underset{m\le \frac{x}{n}}{\sum }F\left(\frac{x}{\mathit{nm}}\right)& \hfill & & \hfill & \\ \hfill & =\underset{n\le x}{\sum }\underset{m\le \frac{x}{n}}{\sum }\mu (n)F\left(\frac{x}{\mathit{nm}}\right)& \hfill & & \hfill & \\ \hfill & =\underset{\mathit{nm}\le x}{\sum }\mu (n)F\left(\frac{x}{\mathit{nm}}\right)& \hfill & & \hfill & \\ \hfill & =\underset{k\le x}{\sum }\underset{\mathit{nm}=k}{\sum }\mu (n)F\left(\frac{x}{k}\right)& \hfill & & \hfill & \\ \hfill & =\underset{k\le x}{\sum }F\left(\frac{x}{k}\right)\underset{n\mid k}{\sum }\mu (n)& \hfill & & \hfill & \\ \hfill & =\underset{k\le x}{\sum }F\left(\frac{x}{k}\right)\delta (k)& \hfill & & \hfill & \\ \hfill & =F(x)& \hfill (\mathit{by}(1)).& & \hfill & & \hfill \end{array}$$

• Conversely, suppose that $F(x)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n\le x}\mu (n)G(x∕n)$ for all $x\in [1,+\infty [$. Then, with the same steps,

  $$\begin{array}{llll}\hfill \underset{n\le x}{\sum }F\left(\frac{x}{n}\right)& =\underset{n\le x}{\sum }\underset{m\le \frac{x}{n}}{\sum }\mu (m)G\left(\frac{x}{\mathit{nm}}\right)& \hfill & \\ \hfill & =\underset{\mathit{nm}\le x}{\sum }\mu (m)G\left(\frac{x}{\mathit{nm}}\right)& \hfill & \\ \hfill & =\underset{k\le x}{\sum }\underset{\mathit{nm}=k}{\sum }\mu (m)G\left(\frac{x}{k}\right)& \hfill & \\ \hfill & =\underset{k\le x}{\sum }G\left(\frac{x}{k}\right)\underset{m\mid k}{\sum }\mu (m)& \hfill & \\ \hfill & =\underset{k\le x}{\sum }G\left(\frac{x}{k}\right)\delta (k)& \hfill & \\ \hfill & =G(x).& \hfill & \end{array}$$

So

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