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Corollary   If $f$ is a completely multiplicative function, then its convolution inverse is $f\mu$ , where $\mu$ denotes the Möbius function.

Proof. Recall the Möbius inversion formula $1*\mu = \varepsilon$ , where $\varepsilon$ denotes the convolution identity function. Thus, $f(1*\mu) = f\varepsilon$ . Since pointwise multiplication of a completely multiplicative function distributes over convolution, $(f \cdot 1)*(f\mu)=f\varepsilon$ . 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 $f$ .

formula for the convolution inverse of a completely multiplicative function is owned by Warren Buck.