Timeline for Does a log-concave function on a convex set extend continuously to the boundary?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Nov 25, 2010 at 9:15 | comment | added | Suvrit | ok, i see that you meant "extension", and $e^{-1/x}$ can be extended as per your requirements. | |
| Nov 25, 2010 at 7:45 | answer | added | Theo Buehler | timeline score: 3 | |
| Nov 24, 2010 at 22:54 | comment | added | Tom LaGatta | No problem, Deane. It was a good suggestion in intent, and will perhaps help somebody answer the question. | |
| Nov 24, 2010 at 22:49 | comment | added | Deane Yang | Tom, sorry for the dyslexic comment. | |
| Nov 24, 2010 at 22:26 | history | edited | Tom LaGatta | CC BY-SA 2.5 |
added the word bounded to clarify, though the image of $f$ was already $[0,1]$
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| Nov 24, 2010 at 22:17 | comment | added | Tom LaGatta | Define $f(0) = 0$ and $f(1) = \mathrm e^{-1}$. I don't understand the point of your example; perhaps you misunderstood my question? | |
| Nov 24, 2010 at 20:32 | comment | added | Tom LaGatta | Thanks for the thought, Deane. Here it's that $-\log f$ is convex, not $\mathrm e^{-f}$. Originally, I had phrased the question in terms of convex functions instead. Since $-\log 0 = \infty$, though, I figured the question would be clearer if I just asked the log-concave version. You're right: this question translates to one about convex functions, which I also do not know the answer to. | |
| Nov 24, 2010 at 18:55 | comment | added | Deane Yang | My simple-minded reaction to this is to study $F = e^{-f}$, which is convex and bounded (if you assume the image of $f$ is in $[0,1]$). The properties of convex functions are rather well known. | |
| Nov 24, 2010 at 16:16 | history | asked | Tom LaGatta | CC BY-SA 2.5 |