Timeline for Hyperfunctions supported at a point
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2015 at 12:55 | history | edited | asv | CC BY-SA 3.0 |
added 1 character in body
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| Aug 25, 2015 at 9:43 | vote | accept | asv | ||
| Aug 25, 2015 at 6:14 | comment | added | asv | @DeaneYang: Right! I missed it. | |
| Aug 24, 2015 at 23:28 | answer | added | Alexandre Eremenko | timeline score: 4 | |
| Aug 24, 2015 at 18:51 | comment | added | Deane Yang | Look at the last bullet under Examples. | |
| Aug 24, 2015 at 13:33 | comment | added | asv | @DeaneYang: I have seen this article. It is generally useful, but I could not find at this link any mentioning of hyperfunctions supported at a point. Apparently Corbennick is right. But in that case I would be interested to get a more explicit description of hyperfunctions supported at a point, if such a description exists. | |
| Aug 24, 2015 at 13:16 | comment | added | Deane Yang | You can also look at the wikipedia article: en.wikipedia.org/wiki/Hyperfunction | |
| Aug 24, 2015 at 13:10 | comment | added | Dirk | Ha! Page 5 is not available in google books for me, this explains my ignorance… | |
| Aug 24, 2015 at 11:14 | comment | added | asv | @Dirk: Distributions are canonically imbedded into hyperfunctions, see p. 5 in Schlichtkrull's book. | |
| Aug 24, 2015 at 10:36 | comment | added | Dirk | I don't know much about hyperfunctions but I am not sure what "coincide" should mean here. A distribution is an element in the dual of smooth functions with compact support while a hyperfunction is an equivalence class of holomorphic functionals. I do not see a canonical way to map one set into the other… E.g. how should a hyperfunction act on a smooth function and how to make such an identification that respects all wanted rules (such as the rules for the derivative…)? | |
| Aug 24, 2015 at 9:52 | comment | added | user1688 | Yes. See Schlichtkrull's book. | |
| Aug 24, 2015 at 9:33 | comment | added | asv | Well, what I mean when saying that the support is 0 is that the restriction of the hyperfunction to $\mathbb{R}^n\backslash\{0\}$ vanishes. Is it satisfied for $e^{1/x}$? | |
| Aug 24, 2015 at 9:32 | comment | added | user1688 | Because it is holomorphic outside zero. See the first pages of Schlichtkrull's book on how hyper functions are described as boundary values of holomorphic functions. | |
| Aug 24, 2015 at 9:22 | comment | added | asv | I don't understand why $e^{1/x}$ is supported at zero. It seems to me that its support is equal to $\mathbb{R}$. | |
| Aug 24, 2015 at 9:11 | comment | added | user1688 | See Schlichtkrull's book on page 6, where he gives the example $e^{1/x}$ as a hyper function (clearly supported at zero) which is not a distribution. | |
| Aug 24, 2015 at 8:08 | history | asked | asv | CC BY-SA 3.0 |