PlanetMath: pointwise multiplication of a completely multiplicative function distibutes over convolution

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pointwise multiplication of a completely multiplicative function distibutes over convolution

(Theorem)

Theorem Let $f$ be a completely multiplicative function and $g$ and $h$ be arithmetic functions. Then $f(g*h)=(fg)*(fh)$

Proof. Let $n$ be a positive integer. Then

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

$\qedsymbol$

"pointwise multiplication of a completely multiplicative function distibutes over convolution" is owned by Wkbj79.

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Attachments:: formula for the convolution inverse of a completely multiplicative function (Corollary) by Wkbj79

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11A25 (Number theory :: Elementary number theory :: Arithmetic functions; related numbers; inversion formulas)

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