1
$\begingroup$

this is well-known that the Fourier transform is an automorphism of $L^2(\mathbb R)$ and also of $\mathcal S(\mathbb R)$ (Schwartz space). Is there any other spaces of functions of one real variable for which the Fourier transform is an automorphism ?

Thanks !

Improve this question
$\endgroup$
5
  • 5
    $\begingroup$ Take any subspace $V$ of either of the spaces and let $F$ denote the Fourier-transform. Then it will be an automorphism on $V+F(V)+F^2(V)+F^3(V)$. $\endgroup$
    – user1688
    Commented Mar 10, 2016 at 20:22
  • $\begingroup$ I think @ADe's observation shows that your question is not quite well-posed, in the sense you intend. I could also mention $L^1({\bf R})\cap A({\bf R})$, sometimes known as the Lebesgue-Fourier algebra $\endgroup$
    – Yemon Choi
    Commented Mar 10, 2016 at 20:49
  • $\begingroup$ Of course I mean interesting examples that I'm not aware of, not some trivial examples made from the spaces I talked about. Thanks for your answer about Lebesgue-Fourier algebra. $\endgroup$
    – Héhéhé
    Commented Mar 10, 2016 at 21:29
  • 1
    $\begingroup$ what if you write something like $S(\mathbb{R}) = P[L^2(\mathbb{R})]$ where $P[f](x) = [(f(y)e^{-\epsilon^2 y^2}) \ast (\epsilon e^{-y^2/\epsilon^2})](x)$ ? $\endgroup$
    – reuns
    Commented Mar 10, 2016 at 22:49
  • 2
    $\begingroup$ Unfortunately, "interesting examples that I'm not aware of, not some trivial examples made from the spaces I talked about" does not seem well-defined to me. The Lebesgue-Fourier algebra is just as trivial or non-trivial as @ADe's examples, but because I named the spaces maybe it looks better. "Spaces of functions" can be very very varied and I think your question needs to say something about what kinds of function spaces you want (Lorentz spaces? Banach lattices?) rather than waiting for people to suggest things and for you to say "oh, no that's not what I meant" $\endgroup$
    – Yemon Choi
    Commented Mar 11, 2016 at 1:26

1 Answer 1

Reset to default
3
$\begingroup$

I am not sure if this is in the spirit of what you are asking but it is a natural explanation of why the Schwartz space has this property. The setting is the following. Suppose that $A$ is an unbounded, positive self-adjoint operator on Hilbert space. Then the spaces $D(A^\alpha)$ and $\bigcap D(A^n)$ are in a natural way Hilbert resp. Fréchet spaces (the symbol $D$ denotes domain of definition). The relevance to your question is that if one chooses for $A$ the standard one-dimensional Schrödinger operator, then one obtains the Schwartz spaces and associated Sobolev-type spaces. One can also dualise the construction to get suitable spaces of distributions, with the tempered distributions as the limit case.

The connection with the Fourier transform comes from the fact that this operator commutes with the Schrödinger operator and this implies that it is a self-mapping of these spaces. The same thing will happen with any $A$ which commutes with the F.T. and this will provide a plethora of spaces of test functions, distribution spaces and Sobolev spaces with the property you desire.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .