pointwise multiplication of a completely multiplicative function distibutes over convolution

Theorem.

Let $f$ be a completely multiplicative function and $g$ and $h$ be arithmetic functions. Then $f\mathit{}\mathrm{(}g\mathrm{*}h\mathrm{)}\mathrm{=}\mathrm{(}f\mathit{}g\mathrm{)}\mathrm{*}\mathrm{(}f\mathit{}h\mathrm{)}$.

Proof.

Let $n$ be a positive integer. Then

$(f(g*h))(n)$	$=f(n)(g*h)(n)$
	$=f(n){\displaystyle \sum _{d|n}}g(d)h\left({\displaystyle \frac{n}{d}}\right)$
	$={\displaystyle \sum _{d|n}}f(n)g(d)h\left({\displaystyle \frac{n}{d}}\right)$
	$={\displaystyle \sum _{d|n}}f\left(d\cdot {\displaystyle \frac{n}{d}}\right)g(d)h\left({\displaystyle \frac{n}{d}}\right)$
	$={\displaystyle \sum _{d|n}}f(d)f\left({\displaystyle \frac{n}{d}}\right)g(d)h\left({\displaystyle \frac{n}{d}}\right)$
	$={\displaystyle \sum _{d|n}}(fg)(d)(fh)\left({\displaystyle \frac{n}{d}}\right)$
	$=((fg)*(fh))(n)$.

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Title	pointwise multiplication of a completely multiplicative function distibutes over convolution
Canonical name	PointwiseMultiplicationOfACompletelyMultiplicativeFunctionDistibutesOverConvolution
Date of creation	2013-03-22 15:59:49
Last modified on	2013-03-22 15:59:49
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	8
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 11A25
Related topic	ArithmeticFunction
Related topic	CompletelyMultiplicative
Related topic	MultiplicativeFunction