Timeline for Cyclic vectors for the translation operator
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 10, 2021 at 12:32 | comment | added | Giorgio Metafune | @yemonChoi Of course I am only considering $C_0(R)$, that is continuous functions vanishing at infinity. | |
| Jun 10, 2021 at 12:24 | comment | added | Giorgio Metafune | The core theorem is Theorem 1.9 pag. 8 of the book by B. Davies "One-parameter semigroups". | |
| Jun 10, 2021 at 12:16 | comment | added | Giorgio Metafune | @Yemon Choi No way. If he is right I am wrong and conversely. | |
| Jun 10, 2021 at 12:13 | comment | added | Yemon Choi | @GiorgioMetafune How does your comment reconcile with the one of ChristianRemling above? | |
| Jun 10, 2021 at 11:09 | comment | added | ABIM | @GiorgioMetafune Do you have a reference to the core theorem? | |
| Jun 10, 2021 at 11:07 | comment | added | Giorgio Metafune | If $f \in H^1$ and the span of its translates is dense in $L^2$, then it is also dense in $H^1$. This follows from the core theorem, since this span is invariant under the translation semigroup whose generator has domain $H^1$. In particular, the span of the translates of such a function is dense with respect to the sup-norm. | |
| Jun 10, 2021 at 9:00 | history | edited | ABIM | CC BY-SA 4.0 |
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| Jun 10, 2021 at 8:59 | comment | added | ABIM | Oh, but of course the topology on $C(\mathbb{R})$ must be the uniform on compacts one; not the uniform one (or else yes this would be essentially obvious that it won't work the reason you stated) | |
| Jun 9, 2021 at 18:01 | comment | added | Christian Remling | I don't quite know how to prove this rigorously, but I think it's in fact clear that the translates of a single function can never be uniformly dense because we won't be able to approximate functions that have asymptotics incompatible with the given function. | |
| Jun 9, 2021 at 17:19 | comment | added | ABIM | @ChristianRemling This is a good point, I made the appropriate modification to the question's formulaiton | |
| Jun 9, 2021 at 17:18 | history | edited | ABIM | CC BY-SA 4.0 |
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| Jun 9, 2021 at 16:55 | comment | added | Christian Remling | Wiener's Tauberian theorem, as usually formulated, is about the span of $\{ f(x+a): a\in\mathbb R\}$ being dense, not $\{ f(x+nb): n\in\mathbb N\}$. | |
| Jun 9, 2021 at 16:53 | history | edited | Johannes Hahn | CC BY-SA 4.0 |
fixed the link
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| Jun 9, 2021 at 15:06 | history | edited | YCor | CC BY-SA 4.0 |
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| Jun 9, 2021 at 14:42 | history | asked | ABIM | CC BY-SA 4.0 |