Timeline for Equivalence among these functions

Current License: CC BY-SA 4.0

12 events

when toggle format	what		by	license	comment
Jun 12 at 18:21	comment	added	Iosif Pinelis		@Fawen90 : Thank you for your appreciation. Haagerup's result is very impressive.
Jun 12 at 18:05	vote	accept	Fawen90
Jun 12 at 18:05	comment	added	Fawen90		Thank you Iosif. Now I fully understand your reasoning. This proof is so tricky and so elegant. I show the result for $p=1$ by computing explicitly the integral, while for general $p$ it's harder. Really beautiful
Jun 12 at 18:05	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 14 characters in body
Jun 12 at 17:59	comment	added	Iosif Pinelis		@Fawen90 : We do not have to deal here with individual values of $Z$. Rather, we use the fact that, by the central limit theorem, $Z$ is a limit in distribution as $n\to\infty$ of $Z_n:=\sum_{i=1}^n \frac1{\sqrt n}\,R_i$. (Of course, we also need to use here uniform integrability of the $\|Z_n\|^p$'s.)
Jun 12 at 17:54	comment	added	Fawen90		Dear Iosif, I really like this fancy alternative expression of $F_p$. There is only one step that is not perfectly clear to me : The "equivalence" between $G_p$ and $G_2$ is identified by a constant only depending on $p$ (not on the deterministic coefficients $a_i$). Does this include the case $(\sigma-1)z + mR$? Here $z$ is an arbitrary realisation of $Z$. My understanding is that $(\sigma-1)z$ cannot written as the product of a real number and a Rademacher (unless $\sigma=1$)
Jun 12 at 17:38	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 44 characters in body
Jun 12 at 17:30	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 44 characters in body
Jun 12 at 17:23	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 44 characters in body
Jun 12 at 17:17	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 44 characters in body
Jun 12 at 17:11	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 182 characters in body
Jun 12 at 17:02	history	answered	Iosif Pinelis	CC BY-SA 4.0