Timeline for Can Hölder's Inequality be strengthened for smooth functions?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| S Sep 27, 2014 at 17:18 | history | suggested | Wolfgang | CC BY-SA 3.0 |
updated broken link
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| Sep 27, 2014 at 16:45 | review | Suggested edits | |||
| S Sep 27, 2014 at 17:18 | |||||
| Sep 21, 2011 at 8:30 | answer | added | Seva | timeline score: 2 | |
| Apr 7, 2011 at 13:41 | history | edited | Denis Serre | CC BY-SA 2.5 |
added 1 characters in body
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| Apr 1, 2011 at 6:01 | answer | added | Ryan O'Donnell | timeline score: 3 | |
| Apr 1, 2011 at 5:43 | history | edited | Kevin O'Bryant | CC BY-SA 2.5 |
Added link to Vinuesa/Matolcsi paper
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| Mar 31, 2011 at 17:19 | answer | added | Terry Tao | timeline score: 12 | |
| Mar 31, 2011 at 12:42 | comment | added | Kevin O'Bryant | Mark, it's the spike at 0 that leads me to believe that the extremal functions are not symmetric. | |
| Mar 31, 2011 at 12:40 | history | edited | Kevin O'Bryant | CC BY-SA 2.5 |
added examples and umlauts; added 12 characters in body
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| Mar 31, 2011 at 3:47 | comment | added | Mark Lewko | Kevin, Is that troubling for any reason (other than that I'm trying to solve a different problem than the one you asked)? It seems that $||f*f||_{\infty}$ isn't any bigger at 0 (under my new definition of convolution than the old one) for symmetric functions which I expect to include the extremals. | |
| Mar 31, 2011 at 3:28 | comment | added | Kevin O'Bryant | @Mark: with your redefinition we end up with $\|f\ast f\|_\infty$ being enormous at $x=0$. | |
| Mar 31, 2011 at 1:21 | comment | added | Mark Lewko | It seems that redefining convolution as $\int f(x)f(c+x)dx$ could make the problem easier since we'd then have (trivially) that $||f*f||_{\infty} \geq ||f||_{2}^2$ and so $||f*f||_{\infty}$ is non-increasing when moving to the rearrangement. | |
| Mar 30, 2011 at 23:04 | comment | added | Mark Lewko | It would be nice to prove that the quantity does not increase if we replace $f$ by its decreasing rearrangement (then you could hope to show that the extremals are Gaussian via a tensorization argument). I think the denominator will not decrease by Riesz's rearrangement theorem and $||f*f||_{1} = ||f||_{1}^2$ so this term will not increase. However, I'm not sure about $||f*f||_{\infty}$ term. | |
| Mar 30, 2011 at 22:27 | comment | added | Willie Wong | I was going to mention Sidon sets, but realized that this question is probably motivated by something similar to it. :) | |
| Mar 30, 2011 at 22:06 | history | edited | Kevin O'Bryant | CC BY-SA 2.5 |
edited body
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| Mar 30, 2011 at 21:54 | comment | added | Piero D'Ancona | You might add gaussians as another example, with a slight better ratio ($\sqrt{2}$( | |
| Mar 30, 2011 at 21:22 | history | edited | Kevin O'Bryant | CC BY-SA 2.5 |
added example and definition of $\ast$
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| Mar 30, 2011 at 20:44 | comment | added | Suvrit | sorry, am not that familiar with the notation. what happens when $f$ is the constant function? | |
| Mar 30, 2011 at 20:28 | comment | added | Kevin O'Bryant | If $f$ is the indicator function of an interval of length $I$, then $\|f \ast f\|_\infty = I$, $\|f\ast f\|_1 = I^2$ and $\|f \ast f\|_2^2 = 2/3 I^3$. The ratio in question, then, is always 3/2, independent of the length of the interval | |
| Mar 30, 2011 at 19:15 | comment | added | Helge | What happens for intervals? I ask this, because I expect the answer to be obtainable by an easy computation. In particular the limit "interval length to 0" should be relevant. | |
| Mar 30, 2011 at 16:55 | history | asked | Kevin O'Bryant | CC BY-SA 2.5 |