Timeline for Convex extensibility of combination of two lines
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2014 at 17:56 | comment | added | Jan Kyncl | If instead of the lines we have only two segments of finite length, then it seems both convex and concave continuous functions exist: the function must only go below or above the tetrahedron which is a convex hull of the two segments. | |
| Mar 22, 2014 at 7:00 | comment | added | Bjørn Kjos-Hanssen | Oh, I see... but this seems to make crucial use of the fact that we can extend lines to infinity, which actually in my intended application we can't. | |
| Mar 22, 2014 at 6:28 | comment | added | Jan Kyncl | No, such function cannot be continuous at points $(x,0)$ for $x>0$. To see this, consider (in 3D) the segments connecting $(0,0,0)$ with $(x,1,x)$. They pointwise approach the line formed by the points $(x,0,x)$, and they must lie below the graph of $f$. An analogous argument also shows that there is no convex function solving the question. | |
| Mar 22, 2014 at 6:21 | comment | added | Bjørn Kjos-Hanssen | Nice. Any thoughts on whether there is a continuous one? | |
| Mar 22, 2014 at 6:19 | vote | accept | Bjørn Kjos-Hanssen | ||
| Mar 22, 2014 at 6:12 | history | answered | Jan Kyncl | CC BY-SA 3.0 |