# Plancherel theorem and Parseval identity

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?

• 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). Apr 1, 2020 at 3:51

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.