# completely multiplicative functions whose convolution inverses are completely multiplicative

###### Corollary 1.

The only completely multiplicative function^{} whose convolution inverse is also completely multiplicative is $\epsilon $, the convolution identity function.

###### Proof.

Let $f$ be a completely multiplicative function whose convolution inverse is completely multiplicative. By this entry (http://planetmath.org/FormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction), $f\mu $ is the convolution inverse of $f$, where $\mu $ denotes the Möbius function^{}. Thus, $f\mu $ is completely multiplicative.

Let $p$ be any prime. Then

$\begin{array}{cc}\hfill {(f(p))}^{2}& ={(f(p))}^{2}{(-1)}^{2}\hfill \\ & \\ & ={(f(p))}^{2}{(\mu (p))}^{2}\hfill \\ & \\ & ={(f(p)\mu (p))}^{2}\hfill \\ & \\ & =f({p}^{2})\mu ({p}^{2})\hfill \\ & \\ & =f({p}^{2})\cdot 0\hfill \\ & \\ & =0.\hfill \end{array}$

Thus, $f(p)=0$ for every prime $p$. Since $f$ is completely multiplicative,

$$f(n)=\{\begin{array}{cc}1\hfill & \text{if}n=1\hfill \\ 0\hfill & \text{if}n\ne 1.\hfill \end{array}$$ |

Hence, $f=\epsilon $. ∎

Title | completely multiplicative functions whose convolution inverses are completely multiplicative |
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Canonical name | CompletelyMultiplicativeFunctionsWhoseConvolutionInversesAreCompletelyMultiplicative |

Date of creation | 2013-03-22 16:55:12 |

Last modified on | 2013-03-22 16:55:12 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 4 |

Author | Wkbj79 (1863) |

Entry type | Corollary |

Classification | msc 11A25 |