# [SOLVED] I do not understand one of the steps in this Dirichlet product proof.

• Oct 14th 2009, 07:49 AM
hatsoff
[SOLVED] I do not understand one of the steps in this Dirichlet product proof.
A note on notation:

If $\displaystyle f,g$ are arithmetic functions, then $\displaystyle f*g$ is the Dirichlet product thereof, such that $\displaystyle (f*g)(n)=\sum_{d|n}f(d)g\left(\frac{n}{d}\right)$.

Theorem: If both $\displaystyle g$ and $\displaystyle f*g$ are multiplicative, then $\displaystyle f$ is multiplicative.

Proof: We shall prove by contradiction. Suppose $\displaystyle f$ is not multiplicative. Let $\displaystyle h=f*g$. Since $\displaystyle f$ is not multiplicative, there exist $\displaystyle m,n,(m,n)=1$ such that $\displaystyle f(mn)\neq f(m)f(n)$. We choose $\displaystyle mn$ as small as possible. If $\displaystyle mn=1$, then $\displaystyle f(1)\neq f(1)f(1)$ so $\displaystyle f(1)\neq 1$. Since $\displaystyle h(1)=f(1)g(1)=f(1)\neq1$, $\displaystyle h$ is not multiplicative, a contradiction. If $\displaystyle mn>1$, we have $\displaystyle f(ab)=f(a)f(b)$ for all $\displaystyle ab<mn$ and $\displaystyle (a,b)=1$. Now,

$\displaystyle h(mn)=f(mn)g(1)+\sum_{a|m,b|n,ab<mn}f(ab)g\left(\f rac{mn}{ab}\right)$

$\displaystyle =f(mn)+\sum_{a|m,b|n,ab<mn}f(a)f(b)g\left(\frac{m} {a}\right)g\left(\frac{n}{b}\right)$

$\displaystyle =f(mn)-f(m)f(n)+h(m)h(n)$.

Since $\displaystyle f(mn)\neq f(m)f(n)$, $\displaystyle h(mn)\neq h(m)h(n)$. Therefore $\displaystyle h$ is not multiplicative, a contradiction. $\displaystyle \blacksquare$

Okay, here's my problem with this "proof": Why in the world do we have "$\displaystyle f(ab)=f(a)f(b)$ for all $\displaystyle ab<mn$ and $\displaystyle (a,b)=1$" if we are assuming that $\displaystyle f$ is not multiplicative? As far as I can see, such a statement might be true or false for some $\displaystyle f(mn)\neq f(m)f(n)$. Yet the author of the proof takes it to be true. Why?

Thanks!
• Oct 14th 2009, 08:17 AM
Bingk
Quote:

Originally Posted by hatsoff
theorem: if both $\displaystyle f$ and $\displaystyle f*g$ are multiplicative, then $\displaystyle f$ is multiplicative.

Is that right? Cuz if $\displaystyle f$ is multiplicative, then obivously $\displaystyle f$ is multiplicative, right?

As for your question, the reason is if it's not multiplicative, it means that there exists at least one case where $\displaystyle f(mn) \neq f(m)f(n)$. It doesn't mean this is true for all $\displaystyle mn$. The proof selects the smallest $\displaystyle mn$ where this happens, meaning that before this, $\displaystyle f(ab) = f(a)f(b)$ works. That's why they test when $\displaystyle mn=1$ and when $\displaystyle mn>1$. That's why the author did that ... at least that's how I've interpreted it :) ... I could be wrong :)
• Oct 14th 2009, 08:18 AM
hatsoff
Quote:

Originally Posted by Bingk
Is that right? Cuz if $\displaystyle f$ is multiplicative, then obivously $\displaystyle f$ is multiplicative, right?

As for your question, the reason is if it's not multiplicative, it means that there exists at least one case where $\displaystyle f(mn) \neq f(m)f(n)$. It doesn't mean this is true for all $\displaystyle mn$. The proof selects the smallest $\displaystyle mn$ where this happens, meaning that before this, $\displaystyle f(ab) = f(a)f(b)$ works. That's why they test when $\displaystyle mn=1$ and when $\displaystyle mn>1$. That's why the author did that ... at least that's how I've interpreted it :) ... I could be wrong :)

Ah, okay. I'm surprised I missed that, yes.

Thanks!