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'elementary results about multiplicative functions and convolution'
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| Title of object: |
elementary results about multiplicative functions and convolution |
| Canonical Name: |
ElementaryResultsAboutMultiplicativeFunctionsAndConvolution |
| Type: |
Theorem |
| Created on: |
2006-07-30 02:25:48 |
| Modified on: |
2006-07-30 02:25:48 |
| Classification: |
msc:11A25 |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}
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Content:
One of the most important elementary results about multiplicative functions and convolution is:
$\mathbf{Theorem:}$ 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 mutliplicative.
The above theorem will be proven in two separate parts.
$\mathbf{Lemma}$ $\mathbf{1:}$ If $f$ and $g$ are multiplicative, then so is $f*g$.
{\sl 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 $d=d_1d_2$, 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,
\begin{center}
\begin{tabular}{ll}
$(f*g)(ab)$ & $\displaystyle =\sum_{d|ab} f(d) g\left( \frac{ab}{d} \right)$ \\
& $\displaystyle =\sum_{d_1|a \text{ and } d_2|b} f(d_1d_2) g\left( \frac{ab}{d_1d_2} \right)$ \\
& $\displaystyle =\sum_{d_1|a \text{ and } d_2|b} f(d_1)f(d_2) g\left( \frac{a}{d_1} \right) g\left( \frac{b}{d_2} \right)$ \\
& $\displaystyle =\left( \sum_{d_1|a} f(d_1) g\left( \frac{a}{d_1} \right) \right) \left( \sum_{d_2|b} f(d_2) g\left( \frac{b}{d_2} \right) \right)$ \\
& $\displaystyle =(f*g)(a)(f*g)(b)$, QED. \end{tabular}
\end{center}
$\mathbf{Lemma}$ $\mathbf{2:}$ If $f$ is an arithmetic function and $g$ and $h$ are multiplicative functions with $f*g=h$, then $f$ is multiplicative.
{\sl 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 $a=b=1$. Since $f(1)=f(1) \cdot 1=f(1)g(1)=(f*g)(1)=h(1)=1$, then $f(ab)=1=1 \cdot 1=f(a)f(b)$.
Now let $ab$ be an arbitrary natural number. The induction hypothesis yields that, if $d_1$ divides $a$, $d_2$ divides $b$, and $d_1d_2<ab$, then $f(d_1d_2)=f(d_1)f(d_2)$. Thus,
\begin{center}
$\begin{array}{ll}
h(a)h(b) & =h(ab) \\
& =(f*g)(ab) \\
& \displaystyle =\sum_{d|ab} f(d)g\left( \frac{ab}{d} \right) \\
& \displaystyle =f(ab)g(1)+\sum_{d|ab \text{ and }d<ab} f(d)g\left( \frac{ab}{d} \right) \\
& \displaystyle =f(ab)\cdot 1+\sum_{d_1|a \text{, } d_2|b \text{, and } d_1d_2<ab} f(d_1d_2)g\left( \frac{ab}{d_1d_2} \right) \\
& \displaystyle =f(ab)+\sum_{d_1|a \text{, } d_2|b \text{, and } d_1d_2<ab} f(d_1)f(d_2)g\left( \frac{a}{d_1} \right) g\left( \frac{b}{d_2} \right) \\
& \displaystyle =f(ab)-f(a)f(b)g(1)g(1)+\sum_{d_1|a \text{ and } d_2|b} f(d_1)f(d_2)g\left( \frac{a}{d_1} \right) g\left( \frac{b}{d_2} \right) \\
& \displaystyle =f(ab)-f(a)f(b)+\left( \sum_{d_1|a} f(d_1)g\left( \frac{a}{d_1} \right) \right) \left( \sum_{d_2|b} f(d_2)g\left( \frac{b}{d_2} \right) \right) \\
& =f(ab)-f(a)f(b)+(f*g)(a)(f*g)(b) \\
& =f(ab)-f(a)f(b)+h(a)h(b). \end{array}$
\end{center}
It follows that $f(ab)=f(a)f(b)$, QED.
The theorem follows from these two lemmas and the fact that convolution is commutative.
The theorem has an obvious corollary.
$\mathbf{Corollary:}$ If $f$ is multiplicative, then so is its convolution inverse.
{\sl Proof:\/} Let $f$ be multiplicative. Since $f(1) \neq 0$, then $f$ has a convolution inverse $f^{-1}$. (See convolution inverses for arithmetic functions for more details.) Since $f*f^{-1}=\varepsilon$ and both $f$ and $\varepsilon$ are multiplicative, the theorem yields that $f^{-1}$ is multiplicative, QED. |
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