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 $a b$ 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 $a b$ 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,

∎

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

$\\begin{array}[]{ll}h(a)h(b)&=h(ab)\\\\\\\\&=(f*g)(ab)\\\\\\\\&\\displaystyle=\\sum_{d|ab}f(d)g\\left(\\frac{ab}{d}\\right)\\\\\\\\&\\displaystyle=f(ab)g(1)+\\sum_{\\begin{subarray}{c}d|ab\\\\d<ab\\end{subarray}}f(d)g\\left(\\frac{ab}{d}\\right)\\\\\\\\&\\displaystyle=f(ab)\\cdot 1+\\sum_{\\begin{subarray}{c}d_{1}|a\\\\d_{2}|b\\\\d_{1}d_{2}<ab\\end{subarray}}f(d_{1}d_{2})g\\left(\\frac{ab}{d_{1}d_{2}}\\right)\\\\\\\\&\\displaystyle=f(ab)+\\sum_{\\begin{subarray}{c}d_{1}|a\\\\d_{2}|b\\\\d_{1}d_{2}<ab\\end{subarray}}f(d_{1})f(d_{2})g\\left(\\frac{a}{d_{1}}\\right)g%\\left(\\frac{b}{d_{2}}\\right)\\\\\\\\&\\displaystyle=f(ab)-f(a)f(b)g(1)g(1)+\\sum_{\\begin{subarray}{c}d_{1}|a\\\\d_{2}|b\\end{subarray}}f(d_{1})f(d_{2})g\\left(\\frac{a}{d_{1}}\\right)g\\left(%\\frac{b}{d_{2}}\\right)\\\\\\\\&\\displaystyle=f(ab)-f(a)f(b)\\cdot 1\\cdot 1+\\left(\\sum_{d_{1}|a}f(d_{1})g\\left%(\\frac{a}{d_{1}}\\right)\\right)\\left(\\sum_{d_{2}|b}f(d_{2})g\\left(\\frac{b}{d_{2%}}\\right)\\right)\\\\\\\\&=f(ab)-f(a)f(b)+(f*g)(a)(f*g)(b)\\\\\\\\&=f(ab)-f(a)f(b)+h(a)h(b).\\end{array}$

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 $f(1)\eq 0$ , $f$ 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.∎

单词	ElementaryResultsAboutMultiplicativeFunctionsAndConvolution
释义	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 $fg=h$ 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 $fg$ . Proof. Note that $\\displaystyle(fg)(1)=f(1)g(1)=1\\cdot 1=1$ since $f$ and $g$ are multiplicative. Let $a,b\\in\\mathbb{N}$ with $\\gcd(a,b)=1$ . Then any divisor $d$ of $a b$ 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 $a b$ 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, ∎ Lemma 2. If $f$ is an arithmetic function and $g$ and $h$ are multiplicative functions with $fg=h$ , then $f$ is multiplicative. Proof. Let $a,b\\in\\mathbb{N}$ with $\\gcd(a,b)=1$ . Induction will be used on $a b$ to establish that $f(ab)=f(a)f(b)$ . If $ab=1$ , then $a=b=1$ . Note that $f(1)=f(1)\\cdot 1=f(1)g(1)=(fg)(1)=h(1)=1$ . Thus, $f(ab)=f(1)=1=1\\cdot 1=f(a)f(b)$ . Now let $a b$ be an arbitrary natural number. The induction hypothesis yields that, if $d_{1}$ divides $a$ , $d_{2}$ divides $b$ , and $d_{1}d_{2}<ab$ , then $f(d_{1}d_{2})=f(d_{1})f(d_{2})$ . Thus, $\\begin{array}[]{ll}h(a)h(b)&=h(ab)\\\\\\\\&=(fg)(ab)\\\\\\\\&\\displaystyle=\\sum_{d\|ab}f(d)g\\left(\\frac{ab}{d}\\right)\\\\\\\\&\\displaystyle=f(ab)g(1)+\\sum_{\\begin{subarray}{c}d\|ab\\\\d<ab\\end{subarray}}f(d)g\\left(\\frac{ab}{d}\\right)\\\\\\\\&\\displaystyle=f(ab)\\cdot 1+\\sum_{\\begin{subarray}{c}d_{1}\|a\\\\d_{2}\|b\\\\d_{1}d_{2}<ab\\end{subarray}}f(d_{1}d_{2})g\\left(\\frac{ab}{d_{1}d_{2}}\\right)\\\\\\\\&\\displaystyle=f(ab)+\\sum_{\\begin{subarray}{c}d_{1}\|a\\\\d_{2}\|b\\\\d_{1}d_{2}<ab\\end{subarray}}f(d_{1})f(d_{2})g\\left(\\frac{a}{d_{1}}\\right)g%\\left(\\frac{b}{d_{2}}\\right)\\\\\\\\&\\displaystyle=f(ab)-f(a)f(b)g(1)g(1)+\\sum_{\\begin{subarray}{c}d_{1}\|a\\\\d_{2}\|b\\end{subarray}}f(d_{1})f(d_{2})g\\left(\\frac{a}{d_{1}}\\right)g\\left(%\\frac{b}{d_{2}}\\right)\\\\\\\\&\\displaystyle=f(ab)-f(a)f(b)\\cdot 1\\cdot 1+\\left(\\sum_{d_{1}\|a}f(d_{1})g\\left%(\\frac{a}{d_{1}}\\right)\\right)\\left(\\sum_{d_{2}\|b}f(d_{2})g\\left(\\frac{b}{d_{2%}}\\right)\\right)\\\\\\\\&=f(ab)-f(a)f(b)+(fg)(a)(fg)(b)\\\\\\\\&=f(ab)-f(a)f(b)+h(a)h(b).\\end{array}$ 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 $f(1)\eq 0$ , $f$ 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.∎
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