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We have following definition of the Fourier transform $\mathcal{F} f$ of a function $f \in L_1(\mathbb{R})$: $(\mathcal{F}f)(y) := \dfrac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} f(x) e^{-ixy} dx , y\in \mathbb{R}$.

Furthermore we have:

1) For $y \in \mathbb{R}$ define the function $e_y:\mathbb{R} \rightarrow \mathbb{C}$ by $e_y(x) := \dfrac{1}{\sqrt{2 \pi}} e^{ixy}$. Then $\{ e_k: k \in \mathbb{Z}$} is a orthomal system in $L_2(-\pi, \pi)$.

2) $||\mathcal{F}f||_2^2 = \int_\mathbb{R} | \langle f, e_y \rangle |^2 dy$.

Now let space $L_2(-\pi, \pi)$ be a closed subspace of $L_2(\mathbb{R})$ (by setting $f \in L_2(\mathbb{R})$ to zero on $\mathbb{R}$ \ $(-\pi, \pi)$.

We know that for $f \in L_2(-\pi, \pi)$ it is $||\mathcal{F}f||_2^2 = ||f||_2^2$.

I need to show that for each $n \in \mathbb{N}$ and $f \in L_2 (-n \pi, n\pi)$ equation $||\mathcal{F}f||_2 = ||f||_2$ holds. To show this i should use isometric isomorphism between $L_2 (-n \pi, n\pi)$ and $L_2 (\pi, \pi)$.

Futhermore we need to show that $span \{L_2(-n \pi, n \pi): n \in \mathbb{N} \}$ is dense in $L_2({\mathbb{R}})$ and we should use it to extend $\mathcal{F} $ by continuity to an isometric operator $L_2(\mathbb{R})$

I would be grateful for any advice :-)

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@tomasz So far, we have shown that for $f\in L_2 (-\pi,\pi)$. Using that ${e_k : k \in \mathbb{Z}}$ is a complete orthonormal system in $L_2 (\mathbb{R})$, we have shown that $\left\Vert \mathcal{F}f\right\Vert _{2}^{2}=\int\limits _{\mathbb{R}}\left|\left\langle f,e_{y}\right\rangle \right|^{2}dy$. Then, by showing that $<e_k,e_y>=sinc((k-y) \pi )$ which is again an orthonormal system in $L_2 (\mathbb{R})$, we show that $\left\Vert \mathcal{F}f\right\Vert _{2}^{2}=\left\Vert f\right\Vert _{2}^{2}$. Our problem is to find this isometric isomorphism from $L_2(-\pi,\pi)$ to $L_2 (-n\pi,n\pi)$. – franckysnow Jul 2 '12 at 20:11

An explicit isometric isomorphism between $L_2(-\pi,\pi)$ and $L_2(-n\pi,n\pi)$ is $f(x)\mapsto f(x/n)/n$, which you should be able to verify easily.

To see that $\mathrm{span} \{L_2(-n\pi,n\pi):n\in\mathbb N\}$ is dense in $L_2(\mathbb R)$, note that for any $f\in L_2(\mathbb R)$ and $\epsilon>0$ we have some $r>0$ siuch that $$\int_{\mathbb R} |f(x)|^2dx - \int_{-r}^r |f(x)|^2dx<\epsilon$$ thus if we choose $n\geq r$ and let $g(x)=\chi_{(-n,n)}(x)f(x)$ we get $g\in L_2(-n\pi,n\pi)$ and $\|f-g\|_2<\epsilon$.

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Thank you for your answer. How do you define $\chi_{(-n,n)}$ ? Which properties does it have? – franckysnow Jul 3 '12 at 0:19
@FranckStudiesCommEng $\chi_{(-n,n)}$ is called the indicator function. It is $1$ if $x\in (-n,n)$ and $0$ otherwise. Really, I use it only because its more convenient than defining a function case-by-case. – Alex Becker Jul 3 '12 at 1:58

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