# Discrete Fourier tranform on $L^2(\mathbb{R})$

I'm studying on the following construction:

For $\lambda\in\mathbb{R}$, denote

$$Y_{\lambda}=\left\{g\in L^{2,loc}(\mathbb{R})\,:\,g(t+2\pi)=e^{2\pi i\lambda}g(t)\right\}$$

We give to $Y_{\lambda}$ an Hilbert space structure by $$\Vert g\Vert_{Y_{\lambda}}=\Vert e^{-it\lambda}g\Vert_{L^2(0,2\pi)}$$

Now, for $\lambda\in\mathbb{R}/\mathbb{Z}$, consider the operator $P_{\lambda}:L^2\rightarrow Y_{\lambda}$,

$$P_{\lambda}g(t)=\sum_{m\in\mathbb{Z}}e^{-2\pi i\lambda m}g(t+2\pi m)$$

Now the author says that $P_{\lambda}$ can be understood as a discrete Fourier tranform, acting on $l_m^2(L^2([2\pi m,2\pi (m+1)]))\cong L^2(\mathbb{R})$, where $\lambda\in[0,1]$ is the dual variable to $m\in\mathbb{Z}$, and there is a corresponding Plancharel identity:

$$\int_0^1\Vert P_{\lambda}\Vert_{Y_{\lambda}}^2d\lambda=\sum_{m\in Z}\Vert g\Vert_{L^2([2\pi m,2\pi (m+1)])}^2=\Vert g\Vert^2_{L^2(\mathbb{R})}$$

Intuitively it's clear to me, but I am not able to justify rigorously these statements. Can someone give me an hint, or a reference where i can clarify my ideas. Thank you in advance.

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Just look into the proof of these things for $L^2([0,2\pi])$ and prove your statement following the same ideas. It is very straightforward. –  Matthias Ludewig Apr 7 at 8:18