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 $f(g*h)=(fg)*(fh)$.

Proof.

Let $n$ be a positive integer. Then

 $\displaystyle(f(g*h))(n)$ $\displaystyle=f(n)(g*h)(n)$ $\displaystyle=f(n)\sum_{d|n}g(d)h\left(\frac{n}{d}\right)$ $\displaystyle=\sum_{d|n}f(n)g(d)h\left(\frac{n}{d}\right)$ $\displaystyle=\sum_{d|n}f\left(d\cdot\frac{n}{d}\right)g(d)h\left(\frac{n}{d}\right)$ $\displaystyle=\sum_{d|n}f(d)f\left(\frac{n}{d}\right)g(d)h\left(\frac{n}{d}\right)$ $\displaystyle=\sum_{d|n}(fg)(d)(fh)\left(\frac{n}{d}\right)$ $\displaystyle=((fg)*(fh))(n)$.

Title pointwise multiplication of a completely multiplicative function distibutes over convolution PointwiseMultiplicationOfACompletelyMultiplicativeFunctionDistibutesOverConvolution 2013-03-22 15:59:49 2013-03-22 15:59:49 Wkbj79 (1863) Wkbj79 (1863) 8 Wkbj79 (1863) Theorem msc 11A25 ArithmeticFunction CompletelyMultiplicative MultiplicativeFunction