# 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 $f*g=h$ 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 $f*g$.

###### Proof.

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

Let $a,b\in\mathbb{N}$ with $\gcd(a,b)=1$. 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 $\gcd(a,b)=1$ implies that $\gcd(d_{1},d_{2})=1$ and $\displaystyle\gcd\left(\frac{a}{d_{1}},\frac{b}{d_{2}}\right)=1$. Thus,

 $(f*g)(ab)$ $\displaystyle=\sum_{d|ab}f(d)g\left(\frac{ab}{d}\right)$ $\displaystyle=\sum_{\begin{subarray}{c}d_{1}|a\\ d_{2}|b\end{subarray}}f(d_{1}d_{2})g\left(\frac{ab}{d_{1}d_{2}}\right)$ $\displaystyle=\sum_{\begin{subarray}{c}d_{1}|a\\ d_{2}|b\end{subarray}}f(d_{1})f(d_{2})g\left(\frac{a}{d_{1}}\right)g\left(% \frac{b}{d_{2}}\right)$ $\displaystyle=\left(\sum_{d_{1}|a}f(d_{1})g\left(\frac{a}{d_{1}}\right)\right)% \left(\sum_{d_{2}|b}f(d_{2})g\left(\frac{b}{d_{2}}\right)\right)$ $\displaystyle=(f*g)(a)(f*g)(b)$.

###### Lemma 2.

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

###### Proof.

Let $a,b\in\mathbb{N}$ with $\gcd(a,b)=1$. 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 $d_{1}d_{2}, then $f(d_{1}d_{2})=f(d_{1})f(d_{2})$. Thus,

$\begin{array}[]{ll}h(a)h(b)&=h(ab)\\ \\ &=(f*g)(ab)\\ \\ &\displaystyle=\sum_{d|ab}f(d)g\left(\frac{ab}{d}\right)\\ \\ &\displaystyle=f(ab)g(1)+\sum_{\begin{subarray}{c}d|ab\\ d

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

Title elementary results about multiplicative functions and convolution ElementaryResultsAboutMultiplicativeFunctionsAndConvolution 2013-03-22 16:07:37 2013-03-22 16:07:37 Wkbj79 (1863) Wkbj79 (1863) 16 Wkbj79 (1863) Theorem msc 11A25 ConvolutionInversesForArithmeticFunctions