Timeline for Proof of a simple property of real, constant functions.
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| S Feb 26, 2012 at 22:18 | history | suggested | Neil G | CC BY-SA 3.0 |
align env.
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| Feb 26, 2012 at 22:18 | review | Suggested edits | |||
| S Feb 26, 2012 at 22:18 | |||||
| Feb 26, 2012 at 19:28 | comment | added | ThisIsNotAnId | @WimC I see it. Really nice. Thanks! | |
| Feb 26, 2012 at 19:16 | comment | added | Alex Becker | @WimC Very nice concise argument, will +1 as soon as I get more votes. | |
| Feb 26, 2012 at 19:12 | comment | added | WimC | @ThisIsNotAnId: I divided the interval $[0,1]$ into $n$ equal pieces and used the inequality on all of those. I implicitly used to triangle inequality for $|x + y|$. | |
| Feb 26, 2012 at 19:12 | comment | added | Pete L. Clark | +1: this works and is a nice "discrete analogue" of my answer. Maybe one or two more lines of detail would help... | |
| Feb 26, 2012 at 19:08 | comment | added | Davide Giraudo | Yes (the last bound). Note that the same idea works if we assume that $|f(x_1)-f(x_2)|\leq |x_1-x_2|^{1+p}$ where $p>0$. | |
| Feb 26, 2012 at 19:05 | comment | added | ThisIsNotAnId | Though that looks interesting, I'm not able to see how you derived the expansion above. Also, does it make use of the hypothesis that $|f(x_1) - f(x_2)| \leq (x_1 - x_2)^2?$ | |
| Feb 26, 2012 at 18:56 | history | answered | WimC | CC BY-SA 3.0 |