Timeline for When can we integrate a distribution over a smooth domain?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| S Oct 27, 2017 at 11:11 | history | suggested | Amir Sagiv | CC BY-SA 3.0 |
Latex + English + Format
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| Oct 27, 2017 at 7:14 | review | Suggested edits | |||
| S Oct 27, 2017 at 11:11 | |||||
| Oct 27, 2017 at 5:06 | answer | added | molendinar | timeline score: 1 | |
| Oct 27, 2017 at 2:27 | comment | added | reuns | Looking at $\lim_{\epsilon \to 0}\langle T, \psi \ast\phi_\epsilon \rangle$ for some mollifier $\phi_\epsilon$, then integrating with respect to $\psi$ a compactly supported piecewise $C^\infty$ function makes sense iff around the boundary the distribution $T$ is of order $0$. If you want continuity wrt to deformation of the mollifier and the boundary then you need a little more, see $\langle \delta, 1_{x > a}\rangle$ which is a good example. If $S$ is a compactly supported distribution then $S\ast T$ makes sense as a distribution $\langle S\ast T,\varphi\rangle=\langle T,S\ast\varphi\rangle$ | |
| Oct 27, 2017 at 2:21 | comment | added | Max Reinhold Jahnke | In general, distributions are not integrable. You can inject the space of all locally integrable functions on the space of distributions and these are all integrable distributions you are going to get. You can interpret some distributions as measures and integrate things against them. | |
| Oct 27, 2017 at 2:14 | review | First posts | |||
| Oct 27, 2017 at 3:44 | |||||
| Oct 27, 2017 at 2:10 | history | asked | Hausdorff | CC BY-SA 3.0 |