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# 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 $f(g*h)=(fg)*(fh)$.

###### Proof.

$\displaystyle(f(g*h))(n)$ | $\displaystyle=f(n)(g*h)(n)$ |

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

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

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

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

$\displaystyle=\sum_{{d|n}}(fg)(d)(fh)\left(\frac{n}{d}\right)$ | |

$\displaystyle=((fg)*(fh))(n)$. |

∎

Related:

ArithmeticFunction, CompletelyMultiplicative, MultiplicativeFunction

Type of Math Object:

Theorem

Major Section:

Reference

## Mathematics Subject Classification

11A25*no label found*

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## Comments

## name

I feel that the name of this entry is too vague. Does anyone have suggestions for improvement?

## Re: name

something like "convolution commutes with composition"?

## Re: name

something like "composition disributes over convolution"?

## Re: name

oops, ignore that one --- "distributes over" is more accurate.

I wish we could edit postings on the PM message board...

## Re: name

It would be more like "pointwise multiplication distributes over convolution". (There is no function composition going on.) That name doesn't make me entirely happy either, because completely multiplicative functions aren't mentioned. Maybe "pointwise multiplication of a completely multiplicative function distibutes over convolution"? It's most certainly not vague, but it's long and somewhat cumbersome. I am definitely open to opinions and suggestions.

I think we need to have a way to edit our own posts. (Possibe PM project?)