Timeline for Geodesic comparison in Hadamard space
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2014 at 9:05 | vote | accept | user35593 | ||
| Oct 28, 2014 at 14:54 | comment | added | Teri | Ach, I overlooked the $f>0$ assumption, sorry. By 'my conjecture' I assume you mean the statement 'strict convexity requires strictly negative curvature'. | |
| Oct 28, 2014 at 7:52 | history | edited | user35593 | CC BY-SA 3.0 |
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| Oct 28, 2014 at 5:47 | comment | added | user35593 | I think you are right about your conjecture. Maybe its best to consider spaces of strictly negative curvature and CAT(0) spaces separately. | |
| Oct 28, 2014 at 5:42 | comment | added | user35593 | Yes but then $f(0)=0$. I stated the conjecture that if $f>0$ then $f$ is constant or strictly convex. | |
| Oct 28, 2014 at 1:48 | answer | added | Anton Petrunin | timeline score: 1 | |
| Oct 27, 2014 at 23:14 | comment | added | Teri | If you consider the lines $\gamma_1(t)=(t,0)$ and $\gamma_2(t)=(t,t)$ in the plane (which is a CAT(0) space) you see that your function $f$ equals $f(t)=t$ which is not strictly convex. I guess for strict convexity you need strictly negative curvature. | |
| Oct 27, 2014 at 22:56 | history | edited | user35593 | CC BY-SA 3.0 |
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| Oct 27, 2014 at 22:29 | history | asked | user35593 | CC BY-SA 3.0 |