formula for the convolution inverse of a completely multiplicative functionFormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction2007-04-14 01:49:132007-04-15 07:59:03Corollarypointwise multiplication of a completely multiplicative function distibutes over convolution\usepackage{amssymb}
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\begin{cor*}
If $f$ is a completely multiplicative function, then its convolution inverse is $f\mu$, where $\mu$ denotes the M\"{o}bius function.
\end{cor*}
\begin{proof}
Recall the M\"{o}bius inversion formula $1*\mu = \varepsilon$, where $\varepsilon$ denotes the convolution identity function. Thus, $f(1*\mu) = f\varepsilon$. Since \PMlinkname{pointwise multiplication of a completely multiplicative function distributes over convolution}{PropertyOfCompletelyMultiplicativeFunctions}, $(f \cdot 1)*(f\mu)=f\varepsilon$. Note that, for all natural numbers $n$, $f(n)1(n)=f(n) \cdot 1=f(n)$ and $f(n)\varepsilon(n)=\varepsilon(n)$. Thus, $f*(f \mu)=\varepsilon$. It follows that $f\mu$ is the convolution inverse of $f$.
\end{proof}