In Fourier transform theory (on $\mathbb{R}$), three vector spaces play a very important role: $L^1(\Bbb R)$, $L^2(\Bbb R)$ and the Schwartz space $\mathcal{S}(\Bbb R)$. Arguably the nicer spaces of the three are the Schwartz space and $L^2(\Bbb R)$. To prove that the Fourier transform is unitary on $L^2(\Bbb R)$, it is sometimes shown that it is an $L^2$ isometry on $\mathcal{S}(\Bbb R)$ and that it is also invertible on $\mathcal{S}(\Bbb R)$. Since $\mathcal{S}(\Bbb R)$ is dense in $L^2(\Bbb R)$, the results extend nicely.

The Schwartz space is extremely nice but it is a little specialized for the Fourier transform. As such, for more general integral operators, we won't have such a nice space to work with. However, as we often do in measure theory, we do have characteristic functions to work with. Particularly, it's not hard to see that the set $V = \{\chi_{[a,b]}:a<b\in\Bbb R\}$ is linearly dense in $L^2(\Bbb R)$. For any reasonable (non-singular) kernel, we would expect that its corresponding integral transform is "nice" on $V$ (i.e. bounded).

Taking these cues, if one were to do Fourier theory on $V$, one would find that the Fourier transform is an $L^2$ isometry. $V$ here and $\mathcal{S}(\Bbb R)$ of course have trivial intersection. This leads me into my question: could we somehow conclude that because the Fourier transform is an $L^2$ isometry on $V$ that it is an $L^2$ isometry on $\mathcal{S}(\Bbb R)$ as an integral transform? Not as an extension of the integral transform on $V$ to $L^2(\Bbb R)$ and then restricted to $\mathcal{S}(\Bbb R)$ (which would trivially be an isometry) but as an integral transform.

I guess a further addendum would be that: since the Fourier transform (as an integral transform) is an $L^2$ isometry on $\mathcal{S}(\Bbb R)$, could we argue that it is also (as an integral transform) an $L^2$ isometry on $V$?

Of course this idea can be made much more general by considering an operator which is naturally defined (and has nice properties) on two distinct subspaces however I'm mostly curious about this particular example.

  • $\begingroup$ What do you mean by "the Fourier transform is an isometry on $\mathcal{S}$"? The Schwartz space is not a normed space. But the Fourier transform is a homeomorphism + automorphism of the Schwartz space. Or do you mean an $L^2$-isometry, when restricted to the Schwartz space? $\endgroup$ – PhoemueX Mar 19 '15 at 21:45
  • $\begingroup$ Sorry, yes I meant an $L^2$ isometry. I made the appropriate changes. Thanks for catching that. I knew what I meant but looking back, it definitely was not $100\%$ clear. $\endgroup$ – Cameron Williams Mar 19 '15 at 21:52

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