From 70e18cbb54eaff71149481d44724bb0e3e79113e Mon Sep 17 00:00:00 2001
From: Ian Jauslin <ian@jauslin.org>
Date: Wed, 19 May 2021 19:55:02 -0400
Subject: Update to v1.1:

Fixed: Typo: c_n\sim n^{-3/2} not n^{3/2}
---
 Carlen_Jauslin_Lieb_Loss_2020.tex | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

(limited to 'Carlen_Jauslin_Lieb_Loss_2020.tex')

diff --git a/Carlen_Jauslin_Lieb_Loss_2020.tex b/Carlen_Jauslin_Lieb_Loss_2020.tex
index 24d7283..fa8d279 100644
--- a/Carlen_Jauslin_Lieb_Loss_2020.tex
+++ b/Carlen_Jauslin_Lieb_Loss_2020.tex
@@ -169,7 +169,7 @@ By the Central Limit Theorem,  since $\varphi$ is bounded and continuous,
 \end{equation}
 where $\gamma(x)$ is a centered Gaussian probability density with variance $\sigma^2$. 
 
-This shows that  there is a  $\delta> 0$  such that for all sufficiently large $n$, $\int_{\mathbb R^d}|x| \star^n w(x)dx \geqslant \sqrt{n}\delta$, and then since $c_n\sim n^{3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x)dx= \infty$. 
+This shows that  there is a  $\delta> 0$  such that for all sufficiently large $n$, $\int_{\mathbb R^d}|x| \star^n w(x)dx \geqslant \sqrt{n}\delta$, and then since $c_n\sim n^{-3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x)dx= \infty$. 
 
  To remove the hypothesis that $w$ has finite variance, note that if $w$ is a probability density with zero mean and infinite variance, $\star^n w(n^{1/2}x)n^{d/2}$ is ``trying'' to converge to a ``Gaussian of infinite variance''. In particular, one would expect that for all $R>0$, 
 \begin{equation}\label{CLT2}
-- 
cgit v1.2.3-70-g09d2

