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The inner product of the two-dimensional sequences $f(x,y)$ and $g(x,y)$ is equal to the inner product of their Fourier transforms, that is:


I am trying use a Fourier transform inverse and follow re-arranged the integrals and use the Dirac function. But I don't know Why the integrals have limits $(-\pi,\pi)$.

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Have you tried with Bessel-Parseval's inequality? – Davide Giraudo Nov 12 '12 at 15:04
I know Parseval's theorem and Bessel inequality but Bessel-Parseval's where I will be able to found? – juaninf Nov 12 '12 at 15:15
I meant the result which gives you an isometry between $L^2((-\pi,\pi)^2)$ and $\ell^2(\Bbb Z^2)$ (I actually don't know exactly the name). Then we polarize. – Davide Giraudo Nov 12 '12 at 15:17

The set $\{e^{imx}e^{iny},m,n\in\Bbb Z\}$ forms a Hilbert basis of the space $H_1:=L^2((-\pi,\pi)^2)$ with the canonical inner product. Denote $e_{m,n}$ the sequence whose all terms are $0$, except the $(m,n)$-th which is $1$. Then $\{e_{m,n},m,n\in\Bbb Z\}$ form a Hibert basis for $H_2:=\ell^2(\Bbb Z^2)$.

The two mentioned Hilbert spaces are isometric (say $\iota\colon H_1\to H_2$, with $\iota(e^{imx}e^{iny})=e_{m,n}$), so for each $x\in H_1$, $\lVert \iota(x)\rVert_{H_2}²=\lVert x\rVert_{H_1}^2$ . Such an equality is true for $x\pm iy$ and $x\pm y$, which gives $$\langle \iota(x),\iota(y)\rangle_{H_2}=\langle x,y\rangle_{H_1},$$ what is wanted.

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thanks by your reply. I don't understand your comment "Such an equality is true for $x\pm iy$ and $x\pm y$" to obtain $\langle \iota(x),\iota(y)\rangle_{H_2}=\langle x,y\rangle_{H_1},$ from $\lVert \iota(x)\rVert_{H_2}^2=\lVert x\rVert_{H_1}^2$. Which are the intermediary steps? – juaninf Nov 12 '12 at 18:03
You have a Hermitian form $q$ such that $\lVert q(u)\rVert=\lVert u\rVert$ for all $u$ (that is, it conserves the norm) and you want to show it also conserves the inner product. What does $\lVert x+iy\rVert^2+\lVert x-iy\rVert^2$ give? – Davide Giraudo Nov 12 '12 at 20:27

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