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I proved Parseval's identity as follows:

If $f\in L^2\left(\mathbb{R}/\mathbb{Z}\right)$, then \begin{equation*} \int\left|f\left(t\right)\right|^2\,dt=\sum_{n\in\mathbb{Z}}\left|c\left(n\right)\right|^2. \end{equation*} Proof. \begin{align*} \int\left|f\left(t\right)\right|^2\,dt&=\int\left|\sum_{n\in\mathbb{Z}}c\left(n\right)\exp\left(2\pi int\right)\right|^2\,dt\\ &=\int\sum_{n\in\mathbb{Z}}c\left(n\right)\exp\left(2\pi int\right)\overline{\sum_{n\in\mathbb{Z}}c\left(n\right)\exp\left(2\pi int\right)}\,dt\\ &=\int\sum_{n\in\mathbb{Z}}c\left(n\right)\exp\left(2\pi int\right)\sum_{n\in\mathbb{Z}}\overline{c\left(n\right)}\exp\left(-2\pi int\right)\,dt\\ &=\sum_{n,k\in\mathbb{Z}}c\left(n\right)\overline{c\left(k\right)}\int\exp\left(2\pi it\left(k-n\right)\right)\,dt\tag{$*$}\\ &=\sum_{n\in\mathbb{Z}}\left|c\left(n\right)\right|^2.\tag*{$\blacksquare$} \end{align*}

However, I took a leap of faith with equality $\left(*\right)$. How can I justify moving the infinite sum outside? I know that I need the monotone/dominated convergence theorem, but I do not know how to show that \begin{equation*} g_{n,k}\left(t\right)=c\left(n\right)\overline{c\left(k\right)}\exp\left(2\pi it\left(k-n\right)\right) \end{equation*} satisfies the requirements.

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Actually $L^{2}({\bf{R}}/{\bf{Z}})$ can be identified with $L^{2}[0,1]$, so $|g_{n,k}(t)|\leq|c(n)|\cdot|c(k)|$ and $\displaystyle\int_{0}^{1}|c(n)|\cdot|c(k)|dt=|c(n)|\cdot|c(k)|<\infty$, so Lebesgue Dominated Convergence Theorem goes through.

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  • $\begingroup$ I see! But then I need to show $(c(n))_n$ is in $\ell^2$ also! Thanks! $\endgroup$ – Josué Jan 20 '18 at 20:27

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