# Proving the Fourier inversion formula in finite abelian groups

The following exercise is exercise 2.2.3 from these lecture notes by Daniel Bump. If $G$ is a finite abelian group, then denote by $G^{\ast}$ the set of its characters, further if $x \in G$ let $\check x$ be the function on $G^{\ast}$ given by $\check x(\chi) = \chi(x)$.

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 $$f(x) = \frac 1{\sqrt{|G^\ast|}} \sum_{\chi \in G^\ast} \overline{\check x(\chi)}\hat f(\chi).$$ (I recently asked a question because the formula is messed up in the original document, see here).

Now I tried to prove it, using that the characters form an orthonormal set we can write $$f = \sum_i c_i \chi_i$$ where $G^{\ast} = \{ \chi_i : i = 1,\ldots, n \}$ are the characters in some order. And pluggin in the given formula for $\hat f$ with $n = |G| = |G^{\ast}|$: \begin{align*} \frac{1}{\sqrt n} \sum_{\chi \in G^{\ast}} \overline{\check x(x)} \hat f(x) & = \frac{1}{\sqrt n} \sum_{\chi \in G^{\ast}} \overline{\check x(x)} \left( \frac{1}{\sqrt n} \sum_{g\in G} \chi(g) f(g) \right) \\ & = \frac{1}{n} \sum_{\chi \in G^{\ast}} \sum_{g\in G} \overline{\check x(x)} \chi(g) f(g) \\ & = \sum_{\chi \in G} \overline{\check x(x)} \frac{1}{n}\left( \sum_{g\in G} \chi(g) f(g) \right) \\ & = \sum_{\chi \in G} \overline{\check x(x)} \langle \chi, f \rangle. \end{align*} Now with the above $\langle \chi, f \rangle = \sum_i c_i \langle \chi, \chi_i \rangle = c_k$ with $\chi = \chi_k$. So all I got is $$\frac{1}{\sqrt n} \sum_{\chi \in G^{\ast}} \overline{\check x(x)} \hat f(x) = \sum_{\chi \in G} \overline{\check x(x)} c_k = \sum_{\chi \in G} \overline{\chi(x)} c_k$$ with $\check x(x) = \chi(x)$, but I have to establish that this equals $f(x)$, so how to proceed?

I think it's not so good to use natural number indices for the representation of $f$ as a linear combination of characters. I believe it's better to write

$$f = \sum_{\psi \in G^\ast} c_{\psi}\cdot \psi.\tag{1}$$

And then compute $\hat{f}(\chi)$ using $(1)$:

\begin{align} \hat{f}(\chi) &= \frac{1}{\sqrt{\lvert G\rvert}}\sum_{g\in G} \chi(g)f(g)\\ &= \frac{1}{\sqrt{\lvert G\rvert}} \sum_{g\in G} \chi(g)\Biggl(\sum_{\psi\in G^\ast} c_\psi \psi(g)\Biggr)\\ &= \frac{1}{\sqrt{\lvert G\rvert}} \sum_{\psi \in G^\ast} c_\psi \Biggl(\sum_{g\in G} \chi(g)\psi(g)\Biggr), \end{align}

noting

$$\sum_{g\in G} \chi(g)\psi(g) = \begin{cases} \lvert G\rvert &, \psi = \overline{\chi} \\ 0 &, \psi \neq \overline{\chi}\end{cases}$$

to conclude

$$\hat{f}(\chi) = \sqrt{\lvert G\rvert}\cdot c_{\overline{\chi}}.\tag{2}$$

We note in passing that it would have been more convenient to define the Fourier coefficients with $\overline{\chi(g)}$ rather than $\chi(g)$.

Now we compute

\begin{align} \frac{1}{\sqrt{\lvert G^\ast\rvert}} \sum_{\chi \in G^\ast} \overline{\check{x}(\chi)}\hat{f}(\chi) &= \frac{1}{\sqrt{\lvert G^\ast\rvert}} \sum_{\chi \in G^\ast} \overline{\chi(x)} \cdot \sqrt{\lvert G\rvert} c_{\overline{\chi}}\\ &= \frac{\sqrt{\lvert G\rvert}}{\sqrt{\lvert G^\ast\rvert}} \sum_{\chi \in G^\ast} c_{\overline{\chi}}\cdot \overline{\chi}(x)\\ &= \frac{\sqrt{\lvert G\rvert}}{\sqrt{\lvert G^\ast\rvert}} \sum_{\psi \in G^\ast} c_\psi\cdot \psi(x)\\ &= \frac{\sqrt{\lvert G\rvert}}{\sqrt{\lvert G^\ast\rvert}} \cdot f(x). \end{align}

And now we need the fact that for finite groups we have $\lvert G\rvert = \lvert G^\ast\rvert$.