Möbius Inversion and Fourier transform. 
Is there any relation between the Möbius Inversion formula for posets and the inversion theorem for the Fourier transform?

The two formulas have a conceptual similarity in that each says that a convolution operator is inverse to another, which looks like the former but has minus signs injected into it. Rota's musings about generating functions being harmonic analysis suggest that a relation like this exists in the folklore.
 A: The theory surrounding Dirichlet series, Dirichlet convolutions, Möbius transforms, Möbius inversion can be thought as Fourier analysis on the semigroup $\mathbb{N}$, the natural numbers. 
The characters of $\mathbb{N}$ are parametrized by complex numbers $s\in\mathbb{C}$, given by $n\mapsto n^{-s}$. Thus the Fourier transform is defined by $$f=\{a_n\}_{n\in\mathbb{N}}\mapsto \mathcal{F}(f):=\sum_{n\in\mathbb{N}}\frac{a_n}{n^{s}}$$
which is the Dirichlet series. The Dirichlet convolution is the convolution with respect to the semigroup $\mathbb{N}$: $(f*g)(n)=\sum_{ab=n}f(a)g(b)$, and it satisfies a typical property in Fourier analysis 
$$\mathcal{F}(f*g)=\mathcal{F}(f)\cdot\mathcal{F}(g).$$
The Möbius transform is $f\mapsto f*1$, and on the Fourier side, $\mathcal{F}(f)\mapsto \mathcal{F}(f)\cdot \mathcal{F}(1)$, where $\mathcal{F}(1)$ is the zeta function $\zeta(s)$. Thus the Möbius inversion is given on the Fourier side by $\mathcal{F}(g)\mapsto \mathcal{F}(g)\cdot\mathcal{F}(1)^{-1}=\mathcal{F}(g)\cdot\zeta(s)^{-1}$, and on the physical side (i.e., on the side of natural numbers),  the Möbius inversion is given by $g\mapsto g*\mathcal{F}^{-1}(\zeta(s)^{-1})=g*\mu$ where $\mu$ is the Möbius function. 
One sees that $\mu*1=\varepsilon$, where $\varepsilon(1)=1$ and $\varepsilon(n)=0$ for $n\neq 1$. And we have $f*\varepsilon=f$. In particular, this $\varepsilon$ function resembles the $\delta$ function in Fourier analysis on $\mathbb{R}^n$ and $\mathbb{T}^n$.
