Proof. Let
$f$ be a completely multiplicative function whose convolution inverse is completely multiplicative. By
this entry,
$f\mu$ is the convolution inverse of
$f$ where
$\mu$ denotes the
Möbius function. Thus,
$f\mu$ is completely multiplicative.
Let $p$ be any prime. Then
$\begin{array}{rl} (f(p))^2 & =(f(p))^2(-1)^2 \\ \\ & =(f(p))^2(\mu(p))^2 \\ \\ & =(f(p)\mu(p))^2 \\ \\ & =f(p^2)\mu(p^2) \\ \\ & =f(p^2) \cdot 0 \\ \\ & =0. \end{array}$
Thus, $f(p)=0$ for every prime $p$ Since $f$ is completely multiplicative,
$$f(n)=\begin{cases} 1 & \text{if } n=1 \\ 0 & \text{if } n\neq 1. \end{cases}$$
Hence, $f=\varepsilon$ 
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