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If $f$ is completely multiplicative, prove that $$(f \cdot g)^{-1}=f \cdot g^{-1},$$ for every arithmetical function $g$ with $g(1) \neq 0$.

My main problem with this is that I can't even see why it is true! As far as I understand it, we have something like $$(f \cdot g)^{-1}=f^{-1} \cdot g^{-1}= \mu f \cdot g^{-1},$$ since $f$ is completely multiplicative. But this can't be right, I don't see how we could get $\mu f$ to be $f$. I'm almost certain that I have confused something with something else, but luckily I have this wonderful community to set me straight! So my question is, what am I doing wrong?

EDIT: Copied the question verbatim from the book (Apostol, Introduction to Analytic Number Theory, p.49)

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Does the dot stand for composition or convolution ? – Sasha Oct 27 '11 at 13:37
Also is the inverse you mean a Dirichlet inverse? – Sasha Oct 27 '11 at 13:39
Sasha: see edit. And yes, the Dirichlet inverse – Carolus Oct 27 '11 at 13:41
How did you get from $f^{-1}\cdot g^{-1}$ to $\mu f\cdot g^{-1}$? – Henning Makholm Oct 27 '11 at 14:10
$(fg)^{-1}$ is not necessarily equal to $f^{-1}g^{-1}$. Take $f=g=u$. Then $fg=u$, $(fg)^{-1} = f^{-1}=g^{-1}=\mu$, but $f^{-1}g^{-1}=\mu^2\neq\mu$, since they disagree at all products of an odd number of distinct primes. – Arturo Magidin Oct 27 '11 at 16:51
up vote 2 down vote accepted

What is unclear to me is how you go from $(fg)^{-1}$ to $f^{-1}g^{-1}$. Given that we are dealing with Dirichlet inverses, this is not obvious to me. In fact, it's not true. For example, if $f=g=u$, then $(fg)^{-1} = u^{-1} = \mu$, but $f^{-1}g^{-1} = \mu\mu\neq \mu$.

So that is where the error in your reasoning comes, I think.

Remember that since ${}^{-1}$ represents the Dirichlet inverse, which is the inverse of convolution, you know that $$(f*g)^{-1} = g^{-1}*f^{-1} = f^{-1}*g^{-1},$$ but you are trying to apply this to the pointwise product, which is incorrect.

As to proving the desired identity, if $h$ is an arithmetic function with $h(1)\neq 0$, then $h^{-1}$ is given by: $$ h^{-1}(1) = \frac{1}{h(1)},\qquad h^{-1}(n) = \frac{-1}{h(1)}\sum_{{d|n},\,{d\lt n}}h\left(\frac{n}{d}\right)h^{-1}(d),\quad n\gt 1.$$

If $h$ is completely multiplicative then you have $f^{-1}(n)=\mu(n)f(n)$ for all $n\geq 1$.

So we verify that the identity holds for $f$ completely multiplicative and $g$ an arithmetic function with $g(1)\neq 0$.

At $n=1$, we have $$(f\cdot g)^{-1}(1) = \frac{1}{f(1)g(1)} = \frac{1}{g(1)}$$ (since $f(1)=1$ must hold); and $$(f\cdot g^{-1})(1) = f(1)g^{-1}(1) = \frac{1}{g(1)}.$$

If $n\gt 1$ and the result hold for all integers smaller than $n$, then we have: $$\begin{align*} (f\cdot g)^{-1}(n) &= \frac{-1}{(f\cdot g)(1)} \sum_{d|n, d\lt n}(f\cdot g)\left(\frac{n}{d}\right)(f\cdot g)^{-1}(d)\\ &= \frac{-1}{g(1)} \sum_{d|n\,d\lt n} f\left(\frac{n}{d}\right)g\left(\frac{n}{d}\right) f(d)g^{-1}(d)\\ &= \frac{-1}{g(1)}\sum_{d|n\,d\lt n}\frac{f(n)}{f(d)}g\left(\frac{n}{d}\right)f(d)g^{-1}(d)\\ &= \frac{-f(n)}{g(1)}\sum_{d|n\,d\lt n}g\left(\frac {n}{d}\right)g^{-1}(d)\\ &= f(n)\left(-\frac{1}{g(1)}\sum_{d|n\,d\lt n}g\left(\frac{n}{d}\right)g^{-1}(d)\right)\\ &= f(n)g^{-1}(n)\\ &= (f\cdot g^{-1})(n). \end{align*}$$ We used the fact that $f$ is completely multiplicative to get that $f\left(\frac{n}{d}\right) = \frac{f(n)}{f(d)}$. Equivalently, you can note that $f\left(\frac{n}{d}\right)f(d) = f(n)$.

This proves the first part of the exercise.

The second part of the exercise asks you to show that if $f$ is multiplicative and $$(f\cdot \mu^{-1})^{-1} = f\cdot \mu$$ holds, then $f$ is completely multiplicative.

Since $f$ is completely multiplicative if and only if $f^{-1} = \mu\cdot f$, you are being asked to show that $f\cdot \mu^{-1} = f$. That is, you need to show that $\mu^{-1}(n) = 1 = u(n)$.

But we know this, since $\sum_{d|n}\mu(d) = I(n)$ (the identity of the convolution), so $u$ and $\mu$ are inverses of each other.

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Fantastic answer! – Carolus Oct 27 '11 at 18:22
A minor point, isn't the sum $\sum_{d|n}\mu(d) = I(n)$? – Carolus Oct 27 '11 at 18:23
@Carolus: And what is $I(n)$ equal to? $I(n) = n$. So $\sum_{d|n}\mu(d) = I(n) = n$, which is what I wrote. – Arturo Magidin Oct 27 '11 at 19:22
But isn't $I(n)=n$ only in the case $I(1)=1$, and $I(n)=0$ for all $n>1$? – Carolus Oct 27 '11 at 20:05
@Carolus: Oh, sorry; you're right. I got my "identities" confused. It's the identity of the convolution, not the identity function. – Arturo Magidin Oct 27 '11 at 20:19

Möbius function $\mu(n)$ is defined as follows:

  • $\mu(n) = 1$ if $n$ is a square-free positive integer with an even number of prime factors.

  • $\mu(n) = −1$ if $n$ is a square-free positive integer with an odd number of prime factors.

  • $\mu(n) = 0$ if $n$ is not square-free

This means that your statement is true only if $n$ is a square-free positive integer with an even number of prime factors.

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Which statement do you mean? Do you mean: $(f \cdot g)^{-1}=f \cdot g^{-1}$? – Carolus Oct 27 '11 at 14:38
@Carolus,yes..that statement – pedja Oct 27 '11 at 14:40
Well, the book (Apostol) has an exercise where one is supposed to prove it for any completely multiplicative $f$ and any multiplicative $g$. And, I suppose any $n$... – Carolus Oct 27 '11 at 14:42
@Carolus,it is true only if $\mu(n)=1$ – pedja Oct 27 '11 at 14:44
That's what I thought... but isn't the point to prove it in the general case, for all $n$? – Carolus Oct 27 '11 at 14:48

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