elementary results about multiplicative functions and convolution

One of the most important elementary results about multiplicative functions and convolution (http://planetmath.org/DirichletConvolution) is:

Theorem.

If $f$ , $g$ , and $h$ are arithmetic functions with $f*g=h$ and at least two of them are multiplicative, then all of them are multiplicative.

The above theorem will be proven in two separate parts.

Lemma 1.

If $f$ and $g$ are multiplicative, then so is $f*g$ .

Proof.

Note that $\displaystyle(f*g)(1)=f(1)g(1)=1\cdot 1=1$ since $f$ and $g$ are multiplicative.

Let $a,b\in\mathbb{N}$ with $\gcd(a,b)=1$ . Then any divisor $d$ of $a b$ can be uniquely factored as $d=d_{1}d_{2}$ , where $d_{1}$ divides $a$ and $d_{2}$ divides $b$ . When a divisor $d$ of $a b$ is factored in this manner, the fact that $\gcd(a,b)=1$ implies that $\gcd(d_{1},d_{2})=1$ and $\displaystyle\gcd\left(\frac{a}{d_{1}},\frac{b}{d_{2}}\right)=1$ . Thus,

$(f*g)(ab)$	$\displaystyle=\sum_{d\|ab}f(d)g\left(\frac{ab}{d}\right)$
	$\displaystyle=\sum_{\begin{subarray}{c}d_{1}\|a\\ d_{2}\|b\end{subarray}}f(d_{1}d_{2})g\left(\frac{ab}{d_{1}d_{2}}\right)$
	$\displaystyle=\sum_{\begin{subarray}{c}d_{1}\|a\\ d_{2}\|b\end{subarray}}f(d_{1})f(d_{2})g\left(\frac{a}{d_{1}}\right)g\left(% \frac{b}{d_{2}}\right)$
	$\displaystyle=\left(\sum_{d_{1}\|a}f(d_{1})g\left(\frac{a}{d_{1}}\right)\right)% \left(\sum_{d_{2}\|b}f(d_{2})g\left(\frac{b}{d_{2}}\right)\right)$
	$\displaystyle=(fg)(a)(fg)(b)$ .

∎

Lemma 2.

If $f$ is an arithmetic function and $g$ and $h$ are multiplicative functions with $f*g=h$ , then $f$ is multiplicative.

Proof.

Let $a,b\in\mathbb{N}$ with $\gcd(a,b)=1$ . Induction will be used on $a b$ to establish that $f(ab)=f(a)f(b)$ .

If $ab=1$ , then $a=b=1$ . Note that $f(1)=f(1)\cdot 1=f(1)g(1)=(f*g)(1)=h(1)=1$ . Thus, $f(ab)=f(1)=1=1\cdot 1=f(a)f(b)$ .

Now let $a b$ be an arbitrary natural number. The induction hypothesis yields that, if $d_{1}$ divides $a$ , $d_{2}$ divides $b$ , and $d_{1}d_{2}<ab$ , then $f(d_{1}d_{2})=f(d_{1})f(d_{2})$ . Thus,

$\begin{array}[]{ll}h(a)h(b)&=h(ab)\\ \\ &=(f*g)(ab)\\ \\ &\displaystyle=\sum_{d|ab}f(d)g\left(\frac{ab}{d}\right)\\ \\ &\displaystyle=f(ab)g(1)+\sum_{\begin{subarray}{c}d|ab\\ d<ab\end{subarray}}f(d)g\left(\frac{ab}{d}\right)\\ \\ &\displaystyle=f(ab)\cdot 1+\sum_{\begin{subarray}{c}d_{1}|a\\ d_{2}|b\\ d_{1}d_{2}<ab\end{subarray}}f(d_{1}d_{2})g\left(\frac{ab}{d_{1}d_{2}}\right)\\ \\ &\displaystyle=f(ab)+\sum_{\begin{subarray}{c}d_{1}|a\\ d_{2}|b\\ d_{1}d_{2}<ab\end{subarray}}f(d_{1})f(d_{2})g\left(\frac{a}{d_{1}}\right)g% \left(\frac{b}{d_{2}}\right)\\ \\ &\displaystyle=f(ab)-f(a)f(b)g(1)g(1)+\sum_{\begin{subarray}{c}d_{1}|a\\ d_{2}|b\end{subarray}}f(d_{1})f(d_{2})g\left(\frac{a}{d_{1}}\right)g\left(% \frac{b}{d_{2}}\right)\\ \\ &\displaystyle=f(ab)-f(a)f(b)\cdot 1\cdot 1+\left(\sum_{d_{1}|a}f(d_{1})g\left% (\frac{a}{d_{1}}\right)\right)\left(\sum_{d_{2}|b}f(d_{2})g\left(\frac{b}{d_{2% }}\right)\right)\\ \\ &=f(ab)-f(a)f(b)+(f*g)(a)(f*g)(b)\\ \\ &=f(ab)-f(a)f(b)+h(a)h(b).\end{array}$

It follows that $f(ab)=f(a)f(b)$ . ∎

The theorem follows from these two lemmas and the fact that convolution is commutative.

The theorem has an obvious corollary.

Corollary.

If $f$ is multiplicative, then so is its convolution inverse.

Proof.

Let $f$ be multiplicative. Since $f(1)\neq 0$ , $f$ has a convolution inverse $f^{-1}$ . (See convolution inverses for arithmetic functions for more details.) Since $f*f^{-1}=\varepsilon$ , where $\varepsilon$ denotes the convolution identity function, and both $f$ and $\varepsilon$ are multiplicative, the theorem yields that $f^{-1}$ is multiplicative. ∎

Title	elementary results about multiplicative functions and convolution
Canonical name	ElementaryResultsAboutMultiplicativeFunctionsAndConvolution
Date of creation	2013-03-22 16:07:37
Last modified on	2013-03-22 16:07:37
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	16
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 11A25
Related topic	ConvolutionInversesForArithmeticFunctions

$(f*g)(ab)$	$\displaystyle=\sum_{d\|ab}f(d)g\left(\frac{ab}{d}\right)$
	$\displaystyle=\sum_{\begin{subarray}{c}d_{1}\|a\\ d_{2}\|b\end{subarray}}f(d_{1}d_{2})g\left(\frac{ab}{d_{1}d_{2}}\right)$
	$\displaystyle=\sum_{\begin{subarray}{c}d_{1}\|a\\ d_{2}\|b\end{subarray}}f(d_{1})f(d_{2})g\left(\frac{a}{d_{1}}\right)g\left(% \frac{b}{d_{2}}\right)$
	$\displaystyle=\left(\sum_{d_{1}\|a}f(d_{1})g\left(\frac{a}{d_{1}}\right)\right)% \left(\sum_{d_{2}\|b}f(d_{2})g\left(\frac{b}{d_{2}}\right)\right)$
	$\displaystyle=(fg)(a)(fg)(b)$ .