Timeline for How "should" I define "absolutely continuous" functions on e.g. n-spheres?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 7, 2012 at 23:46 | comment | added | Yemon Choi | Thanks for the comments - I will try to edit the post to clarify my own muddled thinking and come up with something better-posed. | |
| Mar 7, 2012 at 21:57 | comment | added | Bill Johnson | To me BV consists of absolutely integrable functions on a finite dimensional domain whose distributional partial derivatives exist and are measures with total bounded variation. BV spaces are interesting but difficult to work with. Pelczynski and various collaborators (including me one time) have studied BV this millennium--just do an author search on mathscinet for Pelczynski, A*. | |
| Mar 7, 2012 at 13:43 | comment | added | Tom Leinster | Yemon, you talk about metric and topological notions, but what about measure-theoretic, à la Radon-Nikodym? | |
| Mar 7, 2012 at 10:40 | answer | added | Liviu Nicolaescu | timeline score: 4 | |
| Mar 7, 2012 at 10:12 | comment | added | Willie Wong | I tend to prefer to think of AC as a fact about measures and BV as a fact about functions and by stretching the definitions a bit you get them to coincide on intervals. Reading your question it feels like you are asking about something different still. A motivation perhaps will be very much in order. (Also, I guess since you are tagging metric geometry, the naive way of BV along all geodesics is out.) Lastly, intuitively I would think that using triangulations, the scaling would be wrong. (If you take differences between adjacent vertices and add them up, the sup would be $\infty$ always.) | |
| Mar 7, 2012 at 8:58 | history | asked | Yemon Choi | CC BY-SA 3.0 |