Timeline for An asymmetric quadrilinear estimate
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1 at 21:31 | vote | accept | Medo | ||
| S Mar 30 at 21:18 | history | edited | fedja | CC BY-SA 4.0 |
Only typos
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| S Mar 30 at 21:18 | history | suggested | Medo | CC BY-SA 4.0 |
Only typos
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| Mar 30 at 21:15 | comment | added | fedja | @Medo Nope to the first: $\beta$ is the denominator size, so it is $1-x^2-y^2$, indeed. Yes to the second: $\alpha$ and $\beta$ are as you said. I was just a bit sloppy in the formula (it was supposed to mean "each of $\alpha$ and $\beta$ is of the kind $2^{-k}$ for possibly different $k$") | |
| Mar 30 at 19:21 | comment | added | Medo | Thank you so much for your answer. Many nice and useful tricks here. In the proof of the sufficient condition, did you mean to put $x^2+y^2$ (rather than $1-x^2-y^2$) in the dyadic set of size about $\beta$ ? Also, I can't see why we can set $\alpha,\beta=2^{-k}$, $k\geq 0$. Do you mean $\alpha=2^{-k_{1}}$, $\beta=2^{-k_{2}}$, $k_{1},k_{2}\geq 0$ and then translate the relation between $\alpha $ and $\beta$ into restrictions on the double sum ? Thank you. | |
| Mar 30 at 18:19 | review | Suggested edits | |||
| S Mar 30 at 21:18 | |||||
| Mar 30 at 3:08 | history | answered | fedja | CC BY-SA 4.0 |