I am proving the identity $\langle f, g\rangle = \langle\hat{f}, \hat{g}\rangle$, using the Discrete Fourier analysis, which is the Parseval’s identity.
I already know that if $$\langle f, g\rangle=\sum_{x\in \mathbb{Z}_n}f(x)\overline{g(x)},$$ $$w_n = \exp\left(\frac{2πix}{n}\right)$$ and $\hat{f}$ is the fourier transform of $f$. Then
\begin{eqnarray} % \nonumber to remove numbering (before each equation) \langle\hat{f}, \hat{g}\rangle&=&\sum_{x\in \mathbb{Z}_n}\hat{f(x)}\overline{\hat{g(x)}} \\ &=& \sum_{x\in\mathbb{Z}_n}\left( \left(\sum_{x\in\mathbb{Z}_n}f(x)w^{-rx}\right)\overline{\left(\sum_{x\in\mathbb{Z}_n}g(x)w^{-sx} \right)} \right)\\ &=& \end{eqnarray}
I don't know whether I am on the right track and what next follow?
We know that $(w_n)_{n\in\mathbb{Z}}$ is an orthonormal basis so
$$\langle f, g\rangle = \left\langle \sum_{m \in \mathbb{Z}} \hat{f}(m)w_m, \sum_{n \in \mathbb{Z}} \hat{g}(n)w_n\right\rangle = \sum_{m\in\mathbb{Z}}\sum_{n\in\mathbb{Z}} \hat{f}(m) \overline{\hat{g}(n)} \underbrace{\langle w_m, w_n\rangle}_{=\delta_{mn}} = \sum_{n \in \mathbb{Z}} \hat{f}(n) \overline{\hat{g}(n)} = \langle \hat{f}, \hat{g}\rangle$$