# Prove that Fourier transform on finite Abelian group is an isometry

Let $$G$$ be a finite abelian group,$$\hat G$$ be the dual group, $$\chi \in \hat G$$ is the character , where $$\chi: G\to \mathbb{C}^*$$ is a homomorphism. Let $$\mathbb{C}(G)$$ be the space of Complex-valued functions on G. The inner product on $$\mathbb{C}(G)$$ is defined by $$\qquad(f, g)=\sum_{a \in G} f(a) \overline{g(a)}.$$ Similarly, $$\mathbb{C}(\hat{G})$$ has an inner product given by $$\qquad (\phi, \psi)=\sum_{\chi \in \hat{G}} \phi(\chi) \overline{\psi(\chi)}$$

We define the Fourier transform $$\mathcal{F}: \mathbb{C}(G) \rightarrow \mathbb{C}(\hat{G}) \text { by } \\ \qquad(\mathcal{F} f)(\chi)=|G|^{-\frac{1}{2}} \sum_{a \in G} f(a) \chi(a)$$ and the dual Fourier transform $$\mathcal{F}^{\prime}: \mathbb{C}(\hat{G}) \rightarrow \mathbb{C}(G)$$by $$\qquad\left(\mathcal{F}^{\prime} \phi\right)(a)=|G|^{-\frac{1}{2}} \sum_{\chi \in G} \phi(\chi) \chi(a)$$

I want to show

1. $$\mathcal{F}$$ is an isometry
2. $$\left(\mathcal{F}^{\prime} \mathcal{F} f\right)(x)=f\left(x^{-1}\right)$$

Here is my work: For 1., my goal is to show $$((\mathcal{F} f)(\chi),(\mathcal{F} f)(\chi))=(\chi,\chi)$$. In fact, it is easy to show $$(\chi,\chi)=1$$. But I can only get \begin{align*} ((\mathcal{F}f)(\chi),(\mathcal{F}f)(\chi))&=\sum_{\chi \in \hat G}|(\mathcal{F}f)(\chi))|^2\\ % \left(|G|^{-\frac{1}{2}}\sum_{a\in G}f(a)\chi(a), |G|^{-\frac{1}{2}}\sum_{a\in G}f(a)\chi(a)\right)\\ &=|G|^{-1}\sum_{\chi \in \hat G}\left | \sum_{a \in G} f(a)\chi(a) \right|^2 \end{align*}

For 2., \begin{align*} (\mathcal{F'}\mathcal{F}f)(\chi)&=|G|^{-\frac{1}{2}}\sum_{\chi\in \hat G}(\mathcal{F'}\mathcal{F})(\chi) \chi(x)\\ &=|G|^{-\frac{1}{2}}\sum_{\chi\in \hat G} \left( |G|^{-\frac{1}{2}}\sum_{a\in G}f(a)\chi(a)\right)\chi(x)\\ &=|G|^{-\frac{1}{2}}\sum_{\chi\in \hat G}\left( |G|^{-\frac{1}{2}}\sum_{a\in G}f(ax^{-1})\chi(ax^{-1})\chi(x)\right)\\ &=|G|^{-\frac{1}{2}}\sum_{\chi\in \hat G}\left( |G|^{-\frac{1}{2}}\sum_{a\in G}f(ax^{-1})\chi(a)\right) \end{align*} I have used the trick write $$a$$ ranges in the whole group $$G$$ as $$ab$$ ranges in the whole group $$G$$. But I cannot move further....

You can use the following basic properties of characters: $$\forall a \in G,\ \sum_{\chi \in \hat{G}} \chi(a) = \begin{cases} 0 & \mbox{if } a \not = 1 \\ |G| & \mbox{otherwise} \end{cases}\ \mbox{ and }\ \bar{\chi} = \frac{1}{\chi}$$

Try to solve the problems before reading the solutions.

1. You're wrong. Being an isometry for $$\mathcal{F}$$ means $$\forall f, g \in \mathbb{C}(G),\ (\mathcal{F}(f), \mathcal{F}(g)) = (f, g)$$ Develop the left-hand side, commute the two sums and apply the above properties:

\begin{align} (\mathcal{F}(f), \mathcal{F}(g)) &= \sum_{\chi \in \hat{G}} \mathcal{F}(f)(\chi)\overline{\mathcal{F}(g)(\chi)} = \frac{1}{|G|}\sum_{\chi \in \hat{G}} \sum_{a, b \in G} f(a) \bar{g}(b) \chi(a) \bar{\chi}(b) \\ &= \frac{1}{|G|} \sum_{a, b \in G} f(a)\bar{g}(b) \sum_{\chi \in \hat{G}}\chi(ab^{-1}) = \frac{1}{|G|} \sum_{a \in G} |G| f(a) \bar{g}(a) = (f, g).\end{align}

1. Again develop everything, commute the sums and apply the first property above:

\begin{align} (\mathcal{F}'\mathcal{F}f)(x) &= \frac{1}{|G|^{\frac{1}{2}}} \sum_{\chi \in \hat{G}} \mathcal{F}f(\chi) \chi(x) = \frac{1}{|G|} \sum_{\chi \in \hat{G}} \sum_{a \in G} f(a)\chi(a) \chi(x) \\ &=\frac{1}{|G|} \sum_{\chi \in \hat{G}} \sum_{a \in G} f(ax^{-1})\chi(a) = \frac{1}{|G|} \sum_{a \in G} f(ax^{-1})\sum_{\chi \in \hat{G}} \chi(a) \\ &=\frac{1}{|G|} |G| f(x^{-1}) = f(x^{-1}). \end{align}

• Oh I got the wrong definition of isometry, maybe this is the reason of why I can't prove it. Anyway, thank you so much! Oct 25, 2020 at 13:26