# convolution inverses for arithmetic functions

###### Theorem.

An arithmetic function^{} $f$ has a convolution inverse if and only if $f\mathit{}\mathrm{(}\mathrm{1}\mathrm{)}\mathrm{\ne}\mathrm{0}$.

###### Proof.

If $f$ has a convolution inverse $g$, then $f*g=\epsilon $, where $\epsilon $ denotes the convolution identity function. Thus, $1=\epsilon (1)=(f*g)(1)=f(1)g(1)$, and it follows that $f(1)\ne 0$.

Conversely, if $f(1)\ne 0$, then an arithmetic function $g$ must be constructed such that $(f*g)(n)=\epsilon (n)$ for all $n\in \mathbb{N}$. This will be done by induction^{} on $n$.

Since $f(1)\ne 0$, we have that $\frac{1}{f(1)}}\in \u2102$. Define $g(1)={\displaystyle \frac{1}{f(1)}}$.

Now let $k\in \mathbb{N}$ with $k>1$ and $g(1),\mathrm{\dots},g(k-1)$ be such that $(f*g)(n)=\epsilon (n)$ for all $n\in \mathbb{N}$ with $$ Define

$$ |

Then

$(f*g)(k)$ | $={\displaystyle \sum _{d|k}}f\left({\displaystyle \frac{k}{d}}\right)g(d)$ |

$$ | |

$$ | |

$=0$ | |

$=\epsilon (k).$ |

∎

In the entry titled arithmetic functions form a ring, it is proven that convolution is associative and commutative^{}. Thus, $G=\{f:\mathbb{N}\to \u2102|f(1)\ne 0\}$ is an abelian group under convolution. The set of all multiplicative functions is a subgroup^{} of $G$.

Title | convolution inverses for arithmetic functions |
---|---|

Canonical name | ConvolutionInversesForArithmeticFunctions |

Date of creation | 2013-03-22 15:58:32 |

Last modified on | 2013-03-22 15:58:32 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 27 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 11A25 |

Related topic | ArithmeticFunction |

Related topic | MultiplicativeFunction |

Related topic | ArithmeticFunctionsFormARing |

Related topic | ElementaryResultsAboutMultiplicativeFunctionsAndConvolution |