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In $\mathbb R$, the Riesz-Fischer theorem states that any square-summable series is the Fourier series of a square integrable function. Is this true in $\mathbb R^d$?

Thanks a lot in advance, Sebastien

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Yes, assuming that you are speaking of functions periodic with respect to some lattice $\Lambda \subset \mathbb{R}^d$. Then the Fourier series development yields an isometric isomorphism $L^2(\mathbb{R}^{d}/\Lambda) \to \ell^2(\Lambda)$ when normalized correctly. This is Plancherel's theorem. –  t.b. Jul 13 '11 at 9:59
If $H$ is a Hilbert space and if $\{e_{\alpha}\}_{\alpha\in A}$ is an orthonormal set in $H$, then for any $\phi\in {\cal l}^2(A)$, there exists $x\in H$ such that $\langle x,e_{\alpha} \rangle = \phi(\alpha)$ for all $\alpha\in A$. Note that $L^2(\mathbb{R}^d)$ is a Hilbert space for all positive integers $d$. –  Amitesh Datta Jul 13 '11 at 10:27
@Amitesh: Yes, but you're implicitly assuming Plancherel's theorem. Note also that in the case of $L^2(\mathbb{R}^d)$ Plancherel's theorem is something different than what you mention as $L^2(\mathbb{R}^d)$ does not contain the "Fourier basis" $t \mapsto e^{i\alpha t}$, $\alpha \in \mathbb{R}$ (and the Fourier transform is an isomorphism $L^2(\mathbb{R}^d) \to L^2(\mathbb{R}^d)$ by Plancherel (I should more accurately write $L^2(\widehat{\mathbb{R}^d})$ for the range). –  t.b. Jul 13 '11 at 10:34
@Amitesh: I was nitpicking the second sentence of your comment (reading it again, $L^2(\mathbb{R}^d)$ probably was a typo for $L^2(\mathbb{R}^d/\Lambda)$ -- you need periodicity in order to have a Fourier series). Plancherel's theorem says: If $G$ is a locally compact abelian group and $\widehat{G}$ is its dual group then there exists a normalization of the Haar measure on $\widehat{G}$ such that the Fourier transform $\mathcal{F}: L^1(G) \cap L^2(G) \to L^2(\widehat{G})$ is an isometry and extends to an isometric isomorphism $L^2(G) \to L^2(\widehat{G})$. –  t.b. Jul 13 '11 at 10:48
Thanks a lot guys! It's great to see such a discussion. I'm relieved this theorem works, because it leads to a very elegant proof of a result I had "proved" by interchanging $\int$ and $\sum$. Well, turns out the proof wasn't one, since I didn't check that interchange was indeed possible. But with this neat theorem + Parseval, all is well that ends well... Thanks again –  Sebastien Jul 13 '11 at 11:17

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