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Let $f, g \in C(\mathbb{R}/ \mathbb{Z} ; \mathbb{C})$. Prove the Parseval identity $$\mathcal{R} \int_0^1 f(x)\overline{g(x)} dx = \mathcal{R} \sum_{n \in \mathbb{Z}} \hat{f}(n)\overline{\hat{g}(n)}.$$ (Hint: apply the Plancherel theorem to $f + g$ and $f- g$, and subtract the two.) Then, conclude that the real parts can be removed, thus $$\int_0^1 f(x) \overline{g(x)} dx = \sum_{n \in \mathbb{Z}} \hat{f}(n) \overline{\hat{g}(n)}.$$ (Hint: apply the first identity with $f$ replaced by $if$.

Plancherel theorem: for any $f \in C(\mathbb{R}/ \mathbb{Z} ; \mathbb{C})$, the series $\sum_{n \in \mathbb{Z}} |\hat{f}(n)|^2$ is absolutely convergent, and $$||f||_2^2 = \sum_{n \in \mathbb{Z}} |\hat{f}(n)|^2.$$

$\mathcal{R}$ denotes the real part of complex number. $C(\mathbb{R}/ \mathbb{Z} ; \mathbb{C})$ is the space of continuous $\mathbb{Z}$-periodic functions. As the hint suggests, we have $||f+g||_2^2 = \sum_{n \in \mathbb{Z}} |\hat{f}(n) + \hat{g}(n)|^2$, and$ ||f-g||_2^2 = \sum_{n \in \mathbb{Z}} |\hat{f}(n) - \hat{g}(n)|^2$, but I am not sure how the subtraction of these two helps the proof.

Could you give some help?

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    $\begingroup$ Look up the polarisation identity. In an inner product space if you can express the inner product in terms of the norm (and vice versa, of course). $\endgroup$
    – copper.hat
    Apr 1, 2020 at 3:51

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Hint: The computation they want you to do is the following: $$ |a+b|^2=(a+b)\overline{(a+b)} =|a|^2+b\overline{a}+\overline{b}a+|b|^2 $$ and replacing $b$ with $-b$ you'll find $$ |a+b|^2-|a-b|^2=2a\overline{b}+2\overline{a}b=2(a\overline{b}+\overline{a\overline{b}})=4\Re(a\overline{b}) $$ Now try to apply this to the quantities of interest, moving the $\Re$ through integrals and sums.

Also note, this is "1/2" of the more general polarization identity.

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