Timeline for Fourier Transforms restricted to mass shell
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 11, 2011 at 20:12 | vote | accept | Jan S | ||
| Sep 11, 2011 at 20:12 | comment | added | Jan S | Yes, sorry I didn't read carefully enough. Thanks for your answer. | |
| Sep 11, 2011 at 9:59 | comment | added | Marcel Bischoff | Exactly like Pedro wrote... | |
| Sep 9, 2011 at 14:33 | comment | added | Pedro Lauridsen Ribeiro | Actually what Marcel shows goes the other way round, that is, any $v$ orthogonal to $E(\mathscr{D}(\mathscr{O}))$ is also orthogonal to $E(\mathscr{S}(\mathbb{R}^2))$. As the latter subspace is dense, we must have $v=0$. | |
| Sep 9, 2011 at 8:45 | comment | added | Jan S | @Marcel: There is one thing troubling me. Obviously $\mathcal{D}(\mathcal{O})\subset \mathcal{S}(\mathbb{R}^2)$. Now you have shown that all vectors orthogonal to $E(\mathcal{S}(\mathbb{R}^2))$ are also orthogonal to $E(\mathcal{D}(\mathcal{O}))$, but not that there are no nontrivial vectors orthogonal to $E(\mathcal{D}(\mathcal{O}))$. Or otherwise stated, why does your proof not show that any linear subspace of $\mathcal{S}(\mathbb{R}^2)$ is mapped into a dense st in $H$, which is obvioulsy false for the space spanned by only one function? | |
| Sep 8, 2011 at 14:47 | comment | added | Pedro Lauridsen Ribeiro | Ah yes, this is a direct path to show that $E(\mathscr{D}(O))$ is dense. My answer above actually suggests that an argument similar to the one used for Schwartz functions (which is essentially equivalent to showing that Hermite polynomials are an orthonormal basis in the weighted $L^2$ space whose weight is given by a Gaussian) cannot possibly work for $\mathscr{D}$. | |
| Sep 8, 2011 at 9:46 | history | edited | Marcel Bischoff | CC BY-SA 3.0 |
added 28 characters in body
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| Sep 8, 2011 at 9:40 | history | answered | Marcel Bischoff | CC BY-SA 3.0 |