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The question is about the proof of Theorem 2.17 (Page 36) of the book Introduction to Analytic Number Theory by Apostol:

Theorem 2.17. Let $f$ be multiplicative. Then $f$ is completely multiplicative if, and only if, $$f^{-1}(n)=\mu(n)f(n)\text{ for all }n\geq 1.$$ Proof. Let $g(n)=\mu(n)f(n)$. If $f$ is completely multiplicative we have $$(g\ast f)(n)=\sum_{d\mid n}\mu(d)f(d)f(\tfrac{n}{d})=f(n)\sum_{d\mid n}\mu(d)=f(n)I(n)=I(n)$$ since $f(1)=1$ and $I(n)=0$ for $n>1$. Hence $g=f^{-1}$.

Conversely...

Why is it $$(g\ast f)(n) = \cdots = f(n)I(n) = I(n)\qquad?$$

I think it should be $$(g*f)(n) = \cdots = f(n)I(n) = f(n)$$

Thanks in advance!

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Could you please provide the proof in question? All questions should be self contained. –  JavaMan Nov 26 '11 at 4:15
2  
Remember that $I(n)$ is not the constant function 1, but rather the function $I(n)=1$ if $n=1$ and $I(n)=0$ if $n\gt 0$. So $f(n)I(n) = 0$ if $n\gt 1$, and $f(1)I(1) = f(1)$. –  Arturo Magidin Nov 26 '11 at 4:19

1 Answer 1

up vote 5 down vote accepted

Read the next line:

since $f(1)=1$ and $I(n)=0$ for $n>1$.

For any $n>1$, we have that $f(n)I(n)=f(n)\cdot0=0$, and $f(1)I(1)=1\cdot1=1$. Thus, we have shown that $f(n)I(n)=I(n)$ for all $n$.

As Arturo emphasizes, the definition of $I(n)$ is $$I(n)=\begin{cases}1\text{ if }n=1\\0\text{ if }n>1\end{cases}$$ not $$I(n)=1\quad\text{for all }n$$

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