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I was wondering if there are spaces (function spaces) where the functions have an integral representation, i.e. can be written as an integral involving Fourier coefficients and basis functions, akin to the Fourier transform. If so, what are they called and where may I find more information about them?

Edit: I will try to make this more specific: It is well known that a Hilbert space has an orthonormal basis such that every vector can be written as a series (which is always well defined even if the orthonormal basis is uncountable since only countably many of the Fourier coefficients are nonzero). What I'm looking for is a function space that behaves just like a Hilbert space except that instead of a series you have an integral, e.g. $L^1(\mathbf R)$ with the "basis" $\{\phi_x \mid x \in \mathbf R, \phi_x(t) = e^{itx}\}$ as used in the inverse Fourier transform.

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I'm not sure if this question can be answered as it stands. Could you be a bit more specific as to what exactly you're looking for? For instance, $L^1(S^1)$ or $L^2(S^1)$ or more generally, $L^1(G)$ or $L^2(G)$ where $G$ is a locally compact abelian group (see e.g. here) and also the Hardy spaces would be viable answers. – t.b. Jun 18 '11 at 3:22

It would be nice to make the question more specific. If you mean just integral transforms which have inverse, then the answer is there are many of them: sine and cosine Fourier transform, Laplace transform, Bessel transform etc. And the reason is simple. Consider non-degenerate linear transform $A:\mathbb R^n\to \mathbb R^n$ and for $f\in \mathbb R^n$ put $\tilde f=Af$. Then we can recover $f$ with inverse transform using "transform coefficients" $\tilde f$, $f=A^{-1}\tilde f$. In infinite dimensional spaces like $L_2(\mathbb R)$ there are plenty of non-degenerate bounded linear integral operators with bounded inverse. So Fourier transform in this sense is far from unique. Here a question may arise as to why inverse transform also will be an integral one. General answer is given by the Schwartz kernel theorem.

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+1 for mentioning the Schwartz kernel theorem, Sir. – Jonas Teuwen Aug 17 '11 at 22:01

I am not really sure what you are looking for. But one possibility is that you are looking for representation theory of compact groups, or locally compact abelian (LCA) groups.

The motivating example (which is both compact and abelian) is standard Fourier series. Consider integrable periodic complex functions on the real line. These can be interpreted as functions on the circle understood as a group $SO(2)$. Now, every LCA group $G$ carries Haar measure and has associated with it its Pontryagin dual $\widehat G$ (which consists of irreducible representations endowed with group and topological structure in a natural way) $-$ here we have $\widehat {SO(2)} = {\mathbb Z}$. The conclusion: we can index periodic functions by integers and this is precisely the Fourier series. Same approach works for the Fourier transform, where we have the LCA group ${\mathbb R}^n$, and the dual is also ${\mathbb R}^n$.

Another related area is that of compact groups. Here we no longer require commutativity but this comes at the cost of losing local compactness. In any case, it is possible to form similar approach as in the above and index functions by some irreducible representations. The complication here is that because the groups are no longer required to be commutative, so the representations do no longer need to be one dimensional. Due to Peter-Weyl theorem we have $$L^2(G) \cong \widehat{\bigoplus_{V \in {\widehat G}}} {\rm End}(V)$$ E.g. one can represent functions on $SU(2)$ by operators on $(2j+1)$-dimensional spin-$j$ representations where $j$ is any non-negative half-integer. This is all part of non-commutative harmonic analysis.

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I know about this theory. However, it doesn't yield the classical Fourier transform though (where thinkings happen in $L^1(\mathbf R)$. – echoone Jun 19 '11 at 10:51
@echoone: I don't agree that thinking happens in $L^1({\mathbb R})$. There are lots of applications of Fourier transform where you only work in $L^2({\mathbb R})$ (a prominent example being quantum theory but there are many others). Also, for standard results like Plancherel one needs both $L^1$ and $L^2$. For me the basic theory arises from group theory and the rest are just extensions. Sorry, if this isn't the view you were looking for. – Marek Jun 19 '11 at 11:18

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