pointwise multiplication of a completely multiplicative function distibutes over convolution

Theorem.

Let $f$ be a completely multiplicative function and $g$ and $h$ be arithmetic functions. Then $f\mathit{}\mathrm{(}g\mathrm{*}h\mathrm{)}\mathrm{=}\mathrm{(}f\mathit{}g\mathrm{)}\mathrm{*}\mathrm{(}f\mathit{}h\mathrm{)}$.

Proof.

Let $n$ be a positive integer. Then

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