# Fourier inversion formula on finite abelian groups

The following exercise is exercise 2.2.3 from these lecture notes by Daniel Bump.

Let $\mathcal F : L^2(G) \to L^2(G^{\ast})$ be the Fourier transform, defined by $\mathcal{F}f = \hat f$, where $\hat{f} : L^2(G^{\ast}) \to \mathcal C$ $$\hat f(\chi) = \frac{1}{\sqrt{|G|}} \sum_{x\in G} \chi(x) f(x).$$ Prove that $$f(g) = \frac{1}{\sqrt{|G|}} \sum_{x\in G} \overline{\check x(\chi)} \hat f(x).$$ The formula to prove does not make much sense to me, at first the argument "$g$" of $f$ does not appear on the right hand side, also the expression $\hat f(x)$ does not make much sense, as $\hat f$ expects elements from $G^*$, not from $G$. Also where comes the $\chi$ in $\overline{\check x(\chi)}$ from?

Maybe that might be a typo, but as there is so much I do not understand about the formula I do not know how to fix it, and solve the exercise, so any hints?

There are some typos, the sum has to be over $\chi \in G^\ast$, that is where the $\chi$ comes from, and $g$ has to be replaced by $x$ (the usual name for elements of $G$ in this exercise). That is:
Prove that $$f(x) = \frac 1{\sqrt{|G^\ast|}} \sum_{\chi \in G^\ast} \overline{\check x(\chi)}\hat f(\chi)$$