Update to v1.1:

Fixed: Typo: c_n\sim n^{-3/2} not n^{3/2}

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}

Changelog

@@ -1,3 +1,8 @@
+v1.1:
+
+  * Fixed: Typo: c_n\sim n^{-3/2} not n^{3/2}
+
+
 v1.0:
 
   * Fixed: Missing factor of 1/2 in expansion of f.