Timeline for Interpolation between $L_1^0$ and $L_2^0$
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 18, 2016 at 15:19 | vote | accept | Bill Johnson | ||
| Jun 16, 2016 at 5:39 | history | edited | Bill Johnson |
edited tags
|
|
| Jun 15, 2016 at 17:31 | history | edited | Bill Johnson |
edited tags
|
|
| Jun 15, 2016 at 9:30 | answer | added | Uri Bader | timeline score: 7 | |
| Jun 15, 2016 at 9:02 | comment | added | Bill Johnson | "Finitely supported measures" is the case considered in the papers to which I alluded, but there were no indications that this has anything to do with getting the $L_p^0$ estimate. | |
| Jun 15, 2016 at 8:58 | comment | added | Mikael de la Salle | On the other hand, the answer to Q2 is (positive and) easier for convolution by finitely supported measures. I guess that this will be clear in Uri's forthcoming answer. | |
| Jun 15, 2016 at 8:52 | comment | added | Bill Johnson | Right. If you look at just convolution by finitely supported probabilities, I think you need the group to fail the mean-zero weak containment property in order to get the norm on $L_2^0$ less than one. | |
| Jun 15, 2016 at 8:38 | comment | added | Uri Bader | Yes, Mikael, you are right of course. I am writing things properly as an answer now. | |
| Jun 15, 2016 at 8:32 | comment | added | Mikael de la Salle | Uri: you probably have to strengthen your notion of "generating probability measure". For example, for the circle $G=\mathbf R/\mathbf Z$, the convolution by the uniform probability measure on $\{\pm \sqrt 2\}$ (and actually on any finite set) has norm $1$ on $L_2^0$. I see an argument for what Drutu and Nowak call "admissible measure". | |
| Jun 15, 2016 at 8:12 | comment | added | Bill Johnson | No, Mikael, but that case is interesting. | |
| Jun 15, 2016 at 8:11 | comment | added | Bill Johnson | @UriBader. Both comments are interesting for me. | |
| Jun 15, 2016 at 8:09 | comment | added | Mikael de la Salle | Bill, in Q1 do you assume that $T$ maps the constant function $1$ to itself? | |
| Jun 15, 2016 at 8:04 | comment | added | Mikael de la Salle | @UriBader: well, you first comment does answer Q2, no? | |
| Jun 15, 2016 at 7:43 | comment | added | Uri Bader | By "generating probability measure" I mean "not supported on a proper closed subgroup". This is necessary, otherwise $\|T\|=1$ on $L^0_p$ for every $p$. | |
| Jun 15, 2016 at 7:37 | comment | added | Uri Bader | I am aware of a direct argument that explains why the convolution operator (wrt to a generating probability measure) has norm less than 1 on each $L_p^0(G)$, $1<p<\infty$ ("direct"="not relying on interpolation" in this context). I can elaborate on that, but it does not answer the question. | |
| Jun 15, 2016 at 7:06 | history | asked | Bill Johnson | CC BY-SA 3.0 |