How to define the Fourier transform on arbitrary Hilbert spaces?

Question

The Fourier transform is a unitary operator from $L^2$ to $L^2$. But all infinite-dimensional Hilbert spaces are isometrically isomorphic to $L^2$. So that means we can define the Fourier transform on arbitrary infinite-dimensional Hilbert spaces.

So my question is, what does the Fourier transform look like on an arbitrary infinite-dimensional Hilbert space? Can it still be defined in terms of an integral?

Your question doesn't make sense. For any separable Hilbert space $H$, any unitary operator $U$ with eigenvalues $1,i,-1,-i$ each of them occurring infinitely many times is of the form $U = P^* \mathcal{F} P$ with $P$ an isomorphism $H \to L^2$. — reuns, Apr 08 '19 at 06:07
@reuns I didn’t know that. In any case is it possible to write it as an integral? — Keshav Srinivasan, Apr 08 '19 at 06:09
What do you mean by "integral" on an arbitrary Hilbert space? — Eric Wofsey, Apr 08 '19 at 06:14
@EricWofsey Well, there’s the Bochner integral. In any case, I should say that my question is related to my other question here. — Keshav Srinivasan, Apr 08 '19 at 06:18
You need to make clear fixing an orthonormal basis of $H$ then any linear map becomes some kind of infinite dimensional matrix. The Fourier transform is a very special linear map defined only on $L^2(G)$. — reuns, Apr 08 '19 at 06:33

score 0 · Answer 1 · answered Apr 08 '19 at 06:12

0

There is nothing as the Fourier transform in arbitrary Hilbert space. The Fourier transform is the unitary operator diagonalizing the shift operator $S_a f(x) = f(x+a)$ and every convolution operator.

That is it applies to $L^2(G)$ for $G$ an abelian group, and it is unitary $L^2(G) \to L^2(\widehat{G})$ the Pontryagin dual.

With $G = \Bbb{R}$ then $\widehat{G} = \Bbb{R}$.

With $G = \Bbb{R/Z}$ then $\widehat{G} = \Bbb{Z}$, the corresponding Fourier transform is called Fourier series.

answered Apr 08 '19 at 06:12

reuns

77,999

What do you mean by “There is nothing as the Fourier transform in arbitrary Hilbert space.”? Didn’t you say any Unitary operator with those four eigenvalues occurring infinitely many times corresponds to the Fourier transform with respect to some isomorphism with $L^2$? – Keshav Srinivasan Apr 08 '19 at 06:20

How to define the Fourier transform on arbitrary Hilbert spaces?

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