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I know that the $\mu$ totient function have this two important properties:

The first one is that, supposing that $f$ is a multiplicative arithmetic function, I have that $g=f\star u$ if and only if $f=\mu \star g$, where $\star$ is the convolution of Dirichlet, $u(n)=1, \forall n\in \mathbb{N}^{\times}$, and $\mu$ is the Möbius function.

The second one is that if I have the function $I$ such that $I(n)=1$ if $n=1$ and $I(n)=0$ if $n\neq 1$, then I have that $\mu \star u=I$, being $u$ defined as the last paragraph.

These two properties have easy proofs, but I don't know how to prove that $\mu$ is the ONLY ONE function that satisfies those two properties, i.e. I have to prove the uniqueness from the $\mu$ function for the properties.

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up vote 4 down vote accepted

Note that the set $D$ of functions $f: \mathbb N\to \mathbb C$ forms a commutative ring under $+$ and $\star$. Its unit is $I$ since $f \star I = I \star f = f$ for all $f \in D$. Then $\mu \star u = I$ defines $\mu$ uniquely as the inverse element of $u$. If $\mu'$ is another such function then $$\mu' = \mu' \star I = \mu' \star (u \star \mu) = (\mu' \star u) \star \mu = I \star \mu = \mu.$$

The other condition on $\mu$ is not needed to characterize it uniquely. In fact, the two properties are easily seen to be equivalent: Given $\mu \star u = I$, multiplying both sides of $g = f \star u$ by $\mu$ gives $f = \mu \star g$, and multiplying both sides of the latter equation by $u$ gives the former. Conversely, assume $g = f \star u \Leftrightarrow f = \mu \star g$ for all $f,g \in D$. Take $f = I$ and $g = u$ to see that $I = \mu \star u$.

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Thanks @marlu but I don't see why the two properties are equivalent, I mean, I really don't know how to prove the first property, I supposed that exists another function $\mu '$ that satisfies the proposition, but there is can't continue, I don't know what I need to use... – Elmo goya Nov 12 '12 at 22:15
I just added to my answer an argument how $g = f\star u \Leftrightarrow f = \mu\star g \;\forall f,g \in D$ is equivalent to $\mu \star u = I$. I hope that helps. – marlu Nov 12 '12 at 22:33

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