Timeline for A Balog-Szemeredi-Gowers-type question
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 27, 2013 at 6:42 | comment | added | DmitryZ | Brendan, unfortunately one cannot hope for a strict inclusion. Indeed, suppose in the additive case, that $B$ is an AP and $G$ is such that $B._GB$ contains only odd numbers. Then we cannot find $B'$ such that $B'+B' \subset B+_GB$ (it must contain even numbers), but of course BSG gives us $B'$ with comparable sumset size. However, I don't see how to control $|3B'B'-3B'B'|$ in terms of $|B'B'|$ only, since we inevitably have additional elements not in $B._GB$. | |
| Aug 27, 2013 at 0:02 | answer | added | Brendan Murphy | timeline score: 1 | |
| Aug 26, 2013 at 23:21 | comment | added | Brendan Murphy | If you could find $B'\subseteq B$ such that $|B'|>_\epsilon |B|$ and $B'.B'\subseteq B\cdot_G B$, that would prove the theorem. I don't see how to get the last condition, but Bourgain's theorem seems closest. | |
| Aug 26, 2013 at 14:17 | comment | added | DmitryZ | @BrendanMurphy Thanks for suggestions. I tried to approach the problem this way, like TGF Jones does in his thesis arxiv.org/abs/1301.4853v1 (he uses the Bourgain version of BSG), but I don't see how to handle such complex combination of multiplication and addition. | |
| Aug 23, 2013 at 17:05 | comment | added | Brendan Murphy | Also, Bourgain's version of BSG might be helpful here, but I couldn't make it work straight away (see Bourgain's paper "On the Dimension of Kakeya Sets and Related Maximal Inequalities", or the last exercise in section 6.4 of Tao and Vu's book). | |
| Aug 23, 2013 at 17:02 | comment | added | Brendan Murphy | It's still true by the usual argument (averaging over additive energies) that there exists $\lambda\in\mathbb{F}_p$ such that $|B+_G \lambda\cdot B|\geq\frac 12\min\{|G|,p\}$. After that Glibichuk and Konyagin's argument breaks down pretty badly. It's worth noting that they actually show that either $|(c-d)A+(a-b)A|\geq\frac 12\min\{|A|^2,p\}$ or $|(c-d)A+(a-b+c-d)A|\geq\frac 12\min\{|A|^2,p\}$ where $a,b,c,d$ are some elements of $A$. | |
| Aug 23, 2013 at 15:23 | history | edited | DmitryZ | CC BY-SA 3.0 |
Better notation used
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| Aug 23, 2013 at 15:13 | comment | added | DmitryZ | @AndresCaicedo Thanks for the comment, I have updated the question to clarify this. | |
| Aug 23, 2013 at 15:12 | history | edited | DmitryZ | CC BY-SA 3.0 |
added 3 characters in body
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| Aug 23, 2013 at 15:02 | history | edited | DmitryZ | CC BY-SA 3.0 |
Notation explained in more detail.
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| Aug 23, 2013 at 14:33 | comment | added | Andrés E. Caicedo | @LevBorisov $B\cdot B=\{ab\mid a,b\in B\}$, $A\pm B=\{a\pm b\mid a\in A,b\in B\}$. Also, writing $3(B\cdot B-B\cdot B)$ runs the risk of being ambiguous: Would this mean the set on the left of the display, or the set $\{3(ab-cd)\mid a,b,c,d\in B\}$? | |
| Aug 23, 2013 at 14:01 | comment | added | Lev Borisov | I have a hard time understanding the notation. Can you please clarify it? There are a bunch of B.B's with plus and minus signs that seem to formally cancel. | |
| Aug 23, 2013 at 13:25 | history | asked | DmitryZ | CC BY-SA 3.0 |