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| -rw-r--r-- | Carlen_Jauslin_Lieb_Loss_2020.tex | 2 | ||||
| -rw-r--r-- | Changelog | 5 |
2 files changed, 6 insertions, 1 deletions
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} @@ -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. |