# elementary results about multiplicative functions and convolution

One of the most important elementary results about multiplicative functions and convolution (http://planetmath.org/DirichletConvolution) is:

###### Theorem.

If $f$, $g$, and $h$ are arithmetic functions^{} with $f\mathrm{*}g\mathrm{=}h$ and at least two of them are multiplicative, then all of them are multiplicative.

The above theorem^{} will be proven in two separate parts.

###### Lemma 1.

If $f$ and $g$ are multiplicative, then so is $f\mathrm{*}g$.

###### Proof.

Note that $(f*g)(1)=f(1)g(1)=1\cdot 1=1$ since $f$ and $g$ are multiplicative.

Let $a,b\in \mathbb{N}$ with $\mathrm{gcd}(a,b)=1$. Then any divisor^{} $d$ of $ab$ can be uniquely factored as $d={d}_{1}{d}_{2}$, where ${d}_{1}$ divides $a$ and ${d}_{2}$ divides $b$. When a divisor $d$ of $ab$ is factored in this manner, the fact that $\mathrm{gcd}(a,b)=1$ implies that $\mathrm{gcd}({d}_{1},{d}_{2})=1$ and $\mathrm{gcd}({\displaystyle \frac{a}{{d}_{1}}},{\displaystyle \frac{b}{{d}_{2}}})=1$. Thus,

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

$={\displaystyle \sum _{\begin{array}{c}{d}_{1}|a\\ {d}_{2}|b\end{array}}}f({d}_{1}{d}_{2})g\left({\displaystyle \frac{ab}{{d}_{1}{d}_{2}}}\right)$ | |

$={\displaystyle \sum _{\begin{array}{c}{d}_{1}|a\\ {d}_{2}|b\end{array}}}f({d}_{1})f({d}_{2})g\left({\displaystyle \frac{a}{{d}_{1}}}\right)g\left({\displaystyle \frac{b}{{d}_{2}}}\right)$ | |

$=\left({\displaystyle \sum _{{d}_{1}|a}}f({d}_{1})g\left({\displaystyle \frac{a}{{d}_{1}}}\right)\right)\left({\displaystyle \sum _{{d}_{2}|b}}f({d}_{2})g\left({\displaystyle \frac{b}{{d}_{2}}}\right)\right)$ | |

$=(f*g)(a)(f*g)(b)$. |

∎

###### Lemma 2.

If $f$ is an arithmetic function and $g$ and $h$ are multiplicative functions with $f\mathrm{*}g\mathrm{=}h$, then $f$ is multiplicative.

###### Proof.

Let $a,b\in \mathbb{N}$ with $\mathrm{gcd}(a,b)=1$. Induction^{} will be used on $ab$ to establish that $f(ab)=f(a)f(b)$.

If $ab=1$, then $a=b=1$. Note that $f(1)=f(1)\cdot 1=f(1)g(1)=(f*g)(1)=h(1)=1$. Thus, $f(ab)=f(1)=1=1\cdot 1=f(a)f(b)$.

Now let $ab$ be an arbitrary natural number^{}. The induction hypothesis yields that, if ${d}_{1}$ divides $a$, ${d}_{2}$ divides $b$, and $$, then $f({d}_{1}{d}_{2})=f({d}_{1})f({d}_{2})$. Thus,

$$

It follows that $f(ab)=f(a)f(b)$. ∎

The theorem follows from these two lemmas and the fact that convolution is commutative.

The theorem has an obvious corollary.

###### Corollary.

If $f$ is multiplicative, then so is its convolution inverse.

###### Proof.

Let $f$ be multiplicative. Since $f(1)\ne 0$, $f$ has a convolution inverse ${f}^{-1}$. (See convolution inverses for arithmetic functions for more details.) Since $f*{f}^{-1}=\epsilon $, where $\epsilon $ denotes the convolution identity function, and both $f$ and $\epsilon $ are multiplicative, the theorem yields that ${f}^{-1}$ is multiplicative. ∎

Title | elementary results about multiplicative functions and convolution |
---|---|

Canonical name | ElementaryResultsAboutMultiplicativeFunctionsAndConvolution |

Date of creation | 2013-03-22 16:07:37 |

Last modified on | 2013-03-22 16:07:37 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 16 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 11A25 |

Related topic | ConvolutionInversesForArithmeticFunctions |