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convolution inverses for arithmetic functions (Theorem)
Theorem   An arithmetic function $ f$ has a convolution inverse if and only if $ f(1) \neq 0$.
Proof. If $ f$ has a convolution inverse $ g$, then $ f*g=\varepsilon$, where $ \varepsilon$ denotes the convolution identity function. Thus, $ 1=\varepsilon(1)=(f*g)(1)=f(1)g(1)$, and it follows that $ f(1) \neq 0$.

Conversely, if $ f(1) \neq 0$, then an arithmetic function $ g$ must be constructed such that $ (f*g)(n)=\varepsilon(n)$ for all % latex2html id marker 280 $ n \in \mathbb{N}$. This will be done by induction on $ n$.

Since $ f(1) \neq 0$, we have that % latex2html id marker 286 $ \displaystyle \frac{1}{f(1)} \in \mathbb{C}$. Define $ \displaystyle g(1)=\frac{1}{f(1)}$.

Now let % latex2html id marker 290 $ k \in \mathbb{N}$ with $ k>1$ and $ g(1), \dots, g(k-1)$ be such that $ (f*g)(n)=\varepsilon(n)$ for all % latex2html id marker 298 $ n \in \mathbb{N}$ with $ n<k.$ Define

$\displaystyle g(k)=-\frac{1}{f(1)}\sum_{d\vert k \text{ and } d<k} f \left( \frac{k}{d} \right) g(d).$

Then

$ \displaystyle (f*g)(k)$ $ \displaystyle = \sum_{d\vert k} f \left( \frac{k}{d} \right) g(d)$
  $ \displaystyle = f(1)g(k) + \sum_{d\vert k \text{ and } d<k} f \left( \frac{k}{d} \right) g(d)$
  $ \displaystyle = f(1) \left( -\frac{1}{f(1)}\sum_{d\vert k \text{ and } d<k} f ... ...d) \right) + \sum_{d\vert k \text{ and } d<k} f \left( \frac{k}{d} \right) g(d)$
  $ \displaystyle =0$
  $ \displaystyle =\varepsilon(k).$
$ \qedsymbol$

In the entry titled arithmetic functions form a ring, it is proven that convolution is associative and commutative. Thus, % latex2html id marker 326 $ G=\{ f \colon \mathbb{N} \to \mathbb{C} \, \vert f(1) \neq 0 \}$ is an abelian group under convolution. The set of all multiplicative functions is a subgroup of $ G$.



"convolution inverses for arithmetic functions" is owned by Wkbj79.
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See Also: arithmetic function, multiplicative function, arithmetic functions form a ring, elementary results about multiplicative functions and convolution
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Cross-references: subgroup, multiplicative functions, abelian group, commutative, associative, arithmetic functions form a ring, induction, convolution identity function, convolution inverse, arithmetic function
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This is version 24 of convolution inverses for arithmetic functions, born on 2006-06-10, modified 2007-06-01.
Object id is 7992, canonical name is ConvolutionInversesForArithmeticFunctions.
Accessed 1267 times total.

Classification:
AMS MSC: 11A25 (Number theory :: Elementary number theory :: Arithmetic functions; related numbers; inversion formulas)
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frustrated by Wkbj79 on 2006-06-11 05:07:16
In "convolution inverses for arithmetic functions", an incorrect statement $(f*n)(n)=e(n)$ is displayed, even though I have the correct $(f*g)(n)=e(n)$ in the TeX code. This will not change despite all of the times that I have rerendered the object. How can I fix this???
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