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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?

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  • $\begingroup$ Right track, but you need to differentiate the indices in your sums (for example, you might make them x, y, and z). Otherwise you will confuse yourself. Eventually you will be able to isolate a sum that leads to a delta function. $\endgroup$ – John Polcari Jul 15 '18 at 13:18
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We know that $(w_n)_{n\in\mathbb{Z}}$ is an orthonormal basis so

$$\langle f, g\rangle = \left\langle \sum_{m \in \mathbb{Z}} \hat{f}(m)w_m, \sum_{n \in \mathbb{Z}} \hat{g}(n)w_n\right\rangle = \sum_{m\in\mathbb{Z}}\sum_{n\in\mathbb{Z}} \hat{f}(m) \overline{\hat{g}(n)} \underbrace{\langle w_m, w_n\rangle}_{=\delta_{mn}} = \sum_{n \in \mathbb{Z}} \hat{f}(n) \overline{\hat{g}(n)} = \langle \hat{f}, \hat{g}\rangle$$

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