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Mar 31, 2022 at 7:38 comment added User Yes, I believe this is the context of lemma 8.1 in [3] (in a more general setting), but I am not familiar with the notation and some definitions used, so I can't really understand the proof at the moment.
Mar 30, 2022 at 13:23 history edited John Griesmer CC BY-SA 4.0
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Mar 30, 2022 at 13:18 comment added John Griesmer I should have read more carefully. I believe the method you describe was used in Frantzikinakis's article referenced above, and in Section 8 of the Ackelsberg, Bergelson, Best article.
Mar 30, 2022 at 7:22 comment added User Thank you for the answer. Indeed, the average you mention might be smaller than $\mu(A)^3$, but one can instead use a weighted average for the Kronecker factor which is larger than the same bound. So the question becomes; Is the Kronecker factor characteristic for the weighted averages as well?
Mar 30, 2022 at 1:11 history answered John Griesmer CC BY-SA 4.0