Ian Jauslin 70e18cbb54 Update to v1.1:
Fixed: Typo: c_n\sim n^{-3/2} not n^{3/2}
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 \documentclass[no_section_in_all]{ian}    \begin{document}    \hbox{}  \hfil{\bf\LARGE  On the convolution inequality $f\geqslant f\star f$\par  }  \vfill    \hfil{\bf\large Eric Carlen}\par  \hfil{\it Department of Mathematics, Rutgers University}\par  \hfil{\color{blue}\tt\href{mailto:carlen@rutgers.edu}{carlen@rutgers.edu}}\par  \vskip20pt    \hfil{\bf\large Ian Jauslin}\par  \hfil{\it Department of Physics, Princeton University}\par  \hfil{\color{blue}\tt\href{mailto:ijauslin@princeton.edu}{ijauslin@princeton.edu}}\par  \vskip20pt    \hfil{\bf\large Elliott H. Lieb}\par  \hfil{\it Departments of Mathematics and Physics, Princeton University}\par  \hfil{\color{blue}\tt\href{mailto:lieb@princeton.edu}{lieb@princeton.edu}}\par  \vskip20pt    \hfil{\bf\large Michael Loss}\par  \hfil{\it School of Mathematics, Georgia Institute of Technology}\par  \hfil{\color{blue}\tt\href{mailto:loss@math.gatech.edu}{loss@math.gatech.edu}}\par    \vfill      \hfil {\bf Abstract}\par  \medskip    We consider the inequality $f \geqslant f\star f$ for real functions in $L^1(\mathbb R^d)$ where $f\star f$ denotes the convolution of $f$ with itself. We show that all such functions $f$ are non-negative,  which is not the case for the same inequality in $L^p$ for any $1 < p \leqslant 2$, for which the convolution is defined. We also show that all solutions in $L^1(\mathbb R^d)$ satisfy $\int_{\mathbb R^d}f(x)dx \leqslant \textstyle\frac12$. Moreover, if  $\int_{\mathbb R^d}f(x)dx = \textstyle\frac12$, then $f$ must decay fairly slowly: $\int_{\mathbb R^d}|x| f(x)dx = \infty$, and this is sharp since for all  $r< 1$, there are solutions with $\int_{\mathbb R^d}f(x)dx = \textstyle\frac12$ and $\int_{\mathbb R^d}|x|^r f(x)dx <\infty$. However, if  $\int_{\mathbb R^d}f(x)dx = : a < \textstyle\frac12$, the decay at infinity can be much more rapid: we show that for all $a<\textstyle\frac12$, there are solutions such that for some $\epsilon>0$,  $\int_{\mathbb R^d}e^{\epsilon|x|}f(x)dx < \infty$.  \vfill    \hfil{\footnotesize\copyright\, 2020 by the authors. This paper may be reproduced, in its entirety, for non-commercial purposes.}    \eject    \setcounter{page}1  \pagestyle{plain}      \indent Our subject is the set of real, integrable solutions of the inequality  \begin{equation}   f(x)\geqslant f\star f(x)   ,\quad\forall x\in\mathbb R^d\ ,   \label{ineq}  \end{equation}  where $f\star f(x)$ denotes the convolution $f\star f(x) = \int_{\mathbb R^d} f(x-y) f(y)dy$. By Young's inequality \cite[Theorem 4.2]{LL96}, for all $1\leqslant p\leqslant 2$ and all $f\in L^p(\mathbb R^d)$, $f\star f$ is well defined as an element of  $L^{p/(2-p)}(\mathbb R^d)$. Thus, one may consider this inequality in $L^p(\mathbb R^d)$ for all $1 \leqslant p \leqslant 2$, but the case $p=1$ is special:  the solution set of (\ref{ineq}) is restricted in a number of surprising ways. Integrating both sides of (\ref{ineq}), one sees immediately that $\int_{\mathbb R^d} f(x)dx \leqslant 1$.  We prove that, in fact, all integrable solutions satisfy $\int_{\mathbb R^d} f(x)dx \leqslant \textstyle\frac12$, and this upper bound is sharp.      Perhaps even more surprising, we prove that all integrable solutions of (\ref{ineq}) are non-negative. This is {\em not true} for solutions in $L^p(\mathbb R^d)$, $1 < p \leqslant 2$.  For $f\in L^p(\mathbb R^d)$, $1\leqslant p \leqslant 2$, the Fourier transform $\widehat{f}(k) = \int_{\mathbb R^d}e^{-i2\pi k\cdot x} f(x)dx$ is well defined as an element of $L^{p/(p-1)}(\mathbb R^d)$. If $f$ solves the equation $f = f\star f$,  then $\widehat{f} = \widehat{f}^2$, and hence $\widehat{f}$ is the indicator function of a measurable set. By the Riemann-Lebesgue Theorem, if $f\in L^1(\mathbb R^d)$, then  $\widehat{f}$ is continuous and vanishes at infinity, and the only such indicator function is the indicator function of the empty set. Hence the only integrable solution of  $f = f\star f$ is the trivial solution $f= 0$. However, for $1 < p \leqslant 2$, solutions abound: take $d=1$ and define $g$ to be the indicator function of the interval $[-a,a]$. Define  \begin{equation}\label{exam}  f(x) = \int_{\mathbb R} e^{-i2\pi k x} g(k)dk = \frac{ \sin 2\pi xa}{\pi x}\ ,   \end{equation}  which is not integrable, but which belongs to $L^p(\mathbb R)$ for all $p> 1$. By the Fourier Inversion Theorem $\widehat{f} = g$. Taking products, one gets examples in any dimension.      To construct a family of solutions to (\ref{ineq}), fix $a,t> 0$, and define $g_{a,t}(k) = a e^{-2\pi |k| t}$. By \cite[Theorem 1.14]{SW71},  $$ f_{a,t}(x) = \int_{\mathbb R^d} e^{-i2\pi k x} g_{a,t}(k)dk = a\Gamma((d+1)/2) \pi^{-(d+1)/2} \frac{t}{(t^2 + x^2)^{(d+1)/2}}\ .  $$  Since $g_{a,t}^2(k) = g_{a^2,2t}$, $f_{a,t}\star f_{a,t} = f_{a^2,2t}$, Thus, $f_{a,t} \geqslant f_{a,t}\star f_{a,t}$ reduces to  $$ \frac{t}{(t^2 + x^2)^{(d+1)/2}} \geqslant \frac{2at}{(4t^2 + x^2)^{(d+1)/2}}  $$  which is satisfied for all $a \leqslant 1/2$. Since $\int_{\mathbb R^d}f_{a,t}(x)dx =a$, this provides a class of solutions of (\ref{ineq}) that are non-negative and satisfy  \begin{equation}\label{half}  \int_{\mathbb R^d}f(x)dx \leqslant \frac12\ ,  \end{equation}   all of which have fairly slow decay at infinity, so that in every case,  \begin{equation}\label{tail}   \int_{\mathbb R^d}|x|f(x)dx =\infty \ .   \end{equation}    Our results show that this class of examples of integrable solutions of (\ref{ineq}) is surprisingly typical of {\em all} integrable solutions: every real integrable solution  $f$ of (\ref{ineq}) is positive, satisfies (\ref{half}),  and if there is equality in (\ref{half}), $f$ also satisfies (\ref{tail}). The positivity of all real solutions of (\ref{ineq}) in $L^1(\mathbb R^d)$ may be considered surprising since it is  false in $L^p(\mathbb R^d)$ for all $p > 1$, as the example (\ref{exam}) shows. We also show that when strict inequality holds in (\ref{half}) for a solution $f$ of (\ref{ineq}), it is possible for  $f$ to have rather fast decay; we construct examples such that $\int_{\mathbb R^d}e^{\epsilon|x|}f(x)dx < \infty$ for some $\epsilon> 0$. The conjecture that integrable solutions of (\ref{ineq})  are necessarily positive was motivated by recent work \cite{CJL20,CJL20b} on a partial differential equation involving a quadratic nonlinearity of $f\star f$ type, and the result proved here is the key to  the proof of positivity for solutions of this partial differential equations; see \cite{CJL20}. Autoconvolutions $f\star f$ have been studied extensively; see \cite{MV10} and the work quoted there. However, the questions investigated by these authors are quite different from those considered here.      \theo{Theorem}\label{theo:positivity}  Let $f$ be a real valued function in $L^1(\mathbb R^d)$ such that  \begin{equation}\label{uineq}   f(x) - f\star f(x) =: u(x) \geqslant 0  \end{equation}  for all $x$. Then $\int_{\mathbb R^d} f(x)\ dx\leqslant\frac12$,  and $f$ is given by the series  \begin{equation}  f(x) = \frac{1}{2} \sum_{n=1}^\infty c_n 4^n (\star^n u)(x)  \label{fun}  \end{equation}  which converges in $L^1(\mathbb R^d)$, and  where the $c_n\geqslant0$ are the Taylor coefficients in the expansion of $\sqrt{1-x}$  \begin{equation}\label{3half}   \sqrt{1-x}=1-\sum_{n=1}^\infty c_n x^n   ,\quad   c_n=\frac{(2n-3)!!}{2^nn!} \sim n^{-3/2}  \end{equation}  In particular, $f$ is positive. Moreover, if $u\geqslant 0$ is any integrable function with $\int_{\mathbb R^d}u(x)dx \leqslant \textstyle\frac14$, then the sum on the right in (\ref{fun}) defines an integrable function $f$ that satisfies (\ref{uineq}), and $\int_{\mathbb R^d}f(x) dx = \frac12$ if and only if $\int_{\mathbb R^d}u(x)dx= \frac14$.  \endtheo  \bigskip    \indent\underline{Proof}: Note that $u$ is integrable. Let $a := \int_{\mathbb R^d}f(x)dx$ and $b := \int_{\mathbb R^d}u(x)dx \geqslant 0$. Fourier transforming,  (\ref{uineq}) becomes  \begin{equation} \label{ft}  \widehat f(k) = \widehat f(k)^2 +\widehat u(k)\ .  \end{equation}  At $k=0$, $a^2 - a = -b$, so that $\left(a - \textstyle\frac12\right)^2 = \textstyle\frac14 - b$. Thus $0 \leqslant b \leqslant\textstyle\frac14$. Furthermore, since $u \geqslant 0$,  \begin{equation}   |\widehat u(k)|\leqslant\widehat u(0) \leqslant \textstyle\frac14  \end{equation}  and the first inequality is strict for $k\neq 0$. Hence for $k\neq 0$, $\sqrt{1-4\widehat u(k)} \neq 0$. By the Riemann-Lebesgue Theorem, $\widehat{f}(k)$ and $\widehat{u}(k)$ are both  continuous and vanish at infinity, and hence we must have that  \begin{equation}   \widehat f(k)=\textstyle\frac12-\textstyle\frac12\sqrt{1-4\widehat u(k)}   \label{hatf}  \end{equation}  for all sufficiently large $k$, and in any case $\widehat f(k)= \frac12\pm\frac12\sqrt{1-4\widehat u(k)}$. But by continuity and the fact that $\sqrt{1-4\widehat u(k)} \neq 0$ for any $k\neq 0$, the sign cannot switch.  Hence (\ref{hatf}) is valid for all $k$, including $k=0$, again by continuity. At $k=0$, $a = \textstyle\frac12 - \sqrt{1-4b}$, which proves (\ref{half}).  The fact that $c_n$ as specified in (\ref{3half}) satisfies $c_n \sim n^{-3/2}$ is a simple application of Stirling's formula, and it shows that the power series for $\sqrt{1-z}$ converges absolutely and  uniformly everywhere on the closed unit disc. Since $|4 \widehat u(k)| \leqslant 1$,  ${\displaystyle   \sqrt{1-4 \widehat u(k)} = 1 -\sum_{n=1}^\infty c_n (4 \widehat u(k))^n}$. Inverting the Fourier transform, yields (\ref{fun}), and since $\int_{\mathbb R^d} 4^n\star^n u(x)dx \leqslant 1$,   the convergence of the sum in $L^1(\mathbb R^d)$ follows from the convergence of $\sum_{n=1}^\infty c_n$. The final statement follows from the fact that if $f$ is defined in terms of $u$ in this manner, then (\ref{hatf}) is   valid, and then (\ref{ft}) and (\ref{uineq}) are satisfied.  \qed  \bigskip    \theo{Theorem}\label{theo:decay}  Let $f\in L^1(\mathbb R^d)$ satisfy (\ref{ineq}) and $\int_{\mathbb R^d} f(x)\ dx=\textstyle\frac12$. Then  $\int_{\mathbb R^d}|x| f(x)\ dx=\infty$.  \endtheo  \bigskip      \indent\underline{Proof}: If $\int_{\mathbb R^d} f(x)\ dx=\textstyle\frac12$, $\int_{\mathbb R^d} 4u(x)\ dx=1$, then $w(x) = 4u(x)$ is a probability density, and we can write $f(x) = \frac12\sum_{n=0}^\infty \star^n w$. Aiming for a contradiction, suppose that $|x|f(x)$ is integrable. Then $|x|w(x)$ is integrable. Let $m:= \int_{\mathbb R^d}xw(x)d x$. Since first moments add under convolution, the trivial inequality $|m||x| \geqslant m\cdot x$ yields  $$|m|\int_{\mathbb R^d} |x| \star^nw(x)dx \geqslant \int_{\mathbb R^d} m\cdot x \star^nw(x)dx = n|m|^2\ .$$   It follows that $\int_{\mathbb R^d} |x| f(x)dx \geqslant \frac{|m|}2\sum_{n=1}^\infty nc_n = \infty$. Hence $m=0$.      Suppose temporarily that in addition, $|x|^2w(x)$ is integrable. Let $\sigma^2$ be the variance of $w$; i.e., $\sigma^2 = \int_{\mathbb R^d}|x|^2w(x)dx$  Define the function $\varphi(x) = \min\{1,|x|\}$. Then  $$ \int_{\mathbb R^d}|x| \star^n w(x)dx = \int_{\mathbb R^d}|n^{1/2}x| \star^n w(n^{1/2}x)n^{d/2}dx \geqslant n^{1/2} \int_{\mathbb R^d}\varphi(x)\star^n w(n^{1/2}x)n^{d/2}dx.  $$  By the Central Limit Theorem, since $\varphi$ is bounded and continuous,  \begin{equation}\label{CLT}  \lim_{n\to\infty} \int_{\mathbb R^d}\varphi(x)\star^n w(n^{1/2}x)n^{d/2}dx = \int_{\mathbb R^d}\varphi(x) \gamma(x)dx =: C > 0  \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 $c_n\sim n^{-3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x)dx= \infty$.     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}  \lim_{n\to\infty} \int_{|x| \leqslant R}\star^n w(n^{1/2}x)n^{d/2}dx = 0\ ,  \end{equation} so that the limit in (\ref{CLT}) has the value $1$. The proof then proceeds as above. The fact that (\ref{CLT2}) is valid is a consequence of Lemma~\ref{CLTL} below, which is closely based on the proof of \cite[Corollary 1]{CGR08}.  \qed  \bigskip    \theo{Theorem}\label{theo:decay3}   Let $f\in L^1(\mathbb R^d)$ satisfy\-~(\ref{ineq}), $\int_{\mathbf R^d} xu(x) dx=0$, and  $\int_{\mathbb R^d} |x|^2u(x)\ dx<\infty$, then, for all $0\leqslant p<1$,   \begin{equation}   \int_{\mathbb R^d} |x|^pf(x)\ dx<\infty.   \end{equation}  \endtheo  \bigskip    \indent\underline{Proof}: We may suppose that $f$ is not identically $0$. Let $t := 4\int_{\mathbb R^d}u(x)dx \leqslant 1$. Then $t> 0$. Define $w := t^{-1}4u$; $w$ is a probability density and  \begin{equation}\label{tfor}  f(x) = \frac12\sum_{n=1}^\infty c_n t^n \star^n w(x)\ .  \end{equation}  By hypothesis, $w$ has a zero mean and variance $\sigma^2 = \int_{\mathbb R^d} |x|^2 w(x)dx < \infty$. Since variance is additive under convolution,  $$ \int_{\mathbb R^d} |x|^2 \star^n w(x)dx = n\sigma^2\ .  $$  By H\"older's inequality, for all $0 < p < 2$,  $\int_{\mathbb R^d} |x|^p \star^n w(x)dx \leqslant (n\sigma^2)^{p/2}$.  It follows that for $0 < p < 1$,  $$ \int_{\mathbb R^d} |x|^p f(x)dx \leqslant \frac12(\sigma^2)^{p/2} \sum_{n=1}^\infty n^{p/2} c_n < \infty\ ,  $$  again using the fact that $c_n\sim n^{-3/2}$.  \qed    \delimtitle{\bf Remark}  In the subcritical case $\int_{\mathbb R^d}f(x)dx < \frac12$, the hypothesis that $\int_{\mathbb R^d} x u(x) dx = 0$ is superfluous, and one can conclude more. In this case the quantity $t$ in (\ref{tfor}) satisfies $0 < t < 1$, and if we let $m$ denote the mean of $w$,  $\int_{\mathbb R^d} |x|^2 \star^n w(x)dx =n^2|m|^2+ n\sigma^2$. For $00$.  The Laplace transform of $u$ is  $\widetilde u(p):=\int e^{-px}u(x)\ d x$ which is analytic for $|p|<\lambda$, and $\widetilde u(0) < \textstyle\frac14$.  Therefore, there exists $0<\lambda_0\leqslant \lambda$ such that, for all $|p|\leqslant\lambda_0$,   $\widetilde u(p)<\textstyle\frac14$.  By Theorem \ref{theo:positivity},  ${\displaystyle   f(x):=\frac12\sum_{n=1}^\infty 4^nc_n(\star^n u)(x)}$  is an integrable solution of (\ref{ineq}). For   $|p|\leqslant\lambda_0$, it has a well-defined Laplace transform $\widetilde f(p)$ given by  \begin{equation}   \widetilde f(p)=\int e^{-px}f(x)\ dx=\frac12(1-\sqrt{1-4\widetilde u(p)})  \end{equation}  which is analytic for $|p|\leqslant \lambda_0$.  Note that  ${\displaystyle e^{s|x|} \leqslant \prod_{j=1}^d e^{|sx_j|} \leqslant \frac{1}{d}\sum_{j=1}^d e^{d|sx_j|} \leqslant \frac{2}{d}\sum_{j=1}^d \cosh(dsx_j)}$.  Thus, for $|s|< \delta := \lambda_0/d$, $\int_{\mathbb R^d} \cosh(dsx_j)f(x)dx < \infty$ for each $j$, and hence  $|s| < \delta$, $\int_{\mathbb R^d} e^{s|x|}f(x)dx < \infty$.    However, there are no integrable solutions of (\ref{ineq}) that have compact support: We have seen that all solutions of (\ref{ineq}) are non-negative, and if $A$ is the support of a non-negative integrable function, the Minkowski sum $A+A$ is the support of $f\star f$.  \bigskip      \delimtitle{\bf Remark}  One might also consider the inequality $f \leqslant f \star f$ in $L^1(\mathbb R^d)$, but it is simple to construct solutions that have both signs. Consider any radial Gaussian probability density $g$,  Then $g\star g(x) \geqslant g(x)$ for all sufficiently large $|x|$, and taking $f:= ag$ for $a$ sufficiently large, we obtain $f< f\star f$ everywhere. Now on a small neighborhood of the origin, replace the value of  $f$ by $-1$. If the region is taken small enough, the new function $f$ will still satisfy $f < f\star f$ everywhere.  \endtheo  \bigskip    We close with a lemma validating (\ref{CLT2}) that is closely based on a construction in \cite{CGR08}.    \theo{Lemma}\label{CLTL}  Let $w$ be a mean zero, infinite variance probability density on $\mathbb R^d$. Then for all $R>0$, (\ref{CLT2}) is valid.  \endtheo    \indent\underline{Proof}: Let $X_1,\dots,X_n$ be $n$ independent samples from the density $w$, and let $B_R$ denote the centered ball of radius $R$. The quantity in (\ref{CLT2}) is $p_{n,R} := \mathbb{P}(n^{-1/2}\sum_{j=1}^n X_j\in B_R)$.  Let $\widetilde X_1,\dots,\widetilde X_n$ be another $n$ independent samples from the density $w$, independent of the first $n$. Then also $p_{n,R} := \mathbb{P}(-n^{-1/2}\sum_{j=1}^n \widetilde X_j\in B_R)$. By the independence and the triangle inequality,  $$ p_{n,R}^2 \leqslant \mathbb{P}(n^{-1/2}\sum_{j=1}^n (X_j -\widetilde X_j)\in B_{2R})\ .  $$  The random variable $X_1 - \widetilde X_1$ has zero mean and infinite variance and an even density. Therefore, without loss of generality, we may assume that $w(x) = w(-x)$ for all $x$.    Pick $\epsilon>0$, and choose a large value $\sigma_0$ such that $(2\pi \sigma_0^2)^{-d/2}R^d|B| < \epsilon/3$, where $|B|$ denotes the volume of the unit ball $B$. The point of this is that if  $G$ is a centered Gaussian random variable with variance {\em at least} $\sigma_0^2$, the probability that $G$ lies in {\em any} particular translate $B_R+y$ of the ball of radius $R$ is no more than $\epsilon/3$. Let $A\subset \mathbb R^d$ be a centered cube such that  $$  \int_{A}|x|^2w(x)dx =: \sigma^2 \geqslant 2\sigma_0^2 \quad{\rm and}\quad \int_{A}w(x)dx > \frac34\ ,  $$   and note that since $A$ and $w$ are even, ${\displaystyle \int_{A} x w(x)dx = 0}$.       It is then easy to find mutually independent random variables $X$, $Y$ and $\alpha$ such that  $X$ takes values in $A$ and, has zero mean and variance $\sigma^2$, $\alpha$ is a Bernoulli variable with success probability $\int_{A}w(x)dx$, and finally such that $\alpha X + (1-\alpha)Y$ has the probability density $w$. Taking independent identically distributed (i.i.d.) sequences of such random variables, $w(n^{1/2}x)n^{d/2}$ is the probability density of  ${\displaystyle W_n := n^{-1/2}\sum_{j=1}^n \alpha_j X_j + n^{-1/2} \sum_{j=1}^n(1-\alpha_j)Y_j}$, and we seek to estimate   the expectation of $1_{B_R}(W_n)$. We first take the conditional expectation, given the values of the $\alpha$'s and the $Y$'s, and we define $\hat{n} = \sum_{j=1}^n\alpha_j$. These conditional expectations have the form  ${\mathbb E}\left[ 1_{B_R + y}\left(\sum_{j=1}^n n^{-1/2}\alpha_j X_j \right)\right]$  for some translate $B_R +y$ of $B_R$, the ball of radius $R$. The sum $n^{-1/2}\sum_{j=1}^n \alpha_j X_j$ is actually the sum of $\hat{n}$ i.i.d. random variables with mean zero and variance $\sigma^2/n$. The probability that $\hat{n}$ is significantly less than $\frac34 n$ is negligible for large $n$; by classical estimates associated with the Law of Large Numbers, for all $n$ large enough, the probability that $\hat{n} < n/2$ is no more than $\epsilon/3$. Now let $Z$ be a Gaussian random variable with mean zero and variance  $\sigma^2\hat{n}/n$ which is at least $\sigma^2_0$ when $\hat{n} \geqslant n/2$. Then by the multivariate version \cite{R19} of the Berry-Esseen Theorem \cite{B41,E42}, a version of the Central Limit Theorem  with rate information, there is a constant $K_d$ depending only on $d$ such that  $${\textstyle  \left|{\mathbb E}\left[ 1_{B_R + y}\left(\sum_{j=1}^n n^{-1/2}\alpha_j X_j \right)\right] - {\mathbb P}\{Z \in B_R + y\}\right| \leqslant K_d \hat{n} \frac{{\mathbb E}|X_1|^3}{n^{3/2}} \leqslant K_d \frac{{\mathbb E}|X_1|^3}{n^{1/2}}  \ .}  $$  Since $A$ is bounded, ${\mathbb E}|X_1|^3 < \infty$, and hence for all sufficiently large $n$, when $\hat{n} \geqslant n/2$.  $${\textstyle  {\mathbb E}\left[ 1_{B_R + y}\left(\sum_{j=1}^n n^{-1/2}\alpha_j X_j \right)\right] \leqslant \frac23 \epsilon  \ .}  $$  Since this is uniform in $y$, we finally obtain ${\mathbb P}(W_n \in B_R) \leqslant \epsilon$ for all sufficiently large $n$. Since $\epsilon>0$ is arbitrary, (\ref{CLT2}) is proved.  \qed    We close by thanking the anonymous referee for useful suggestions.        \vfill  \eject  {\bf Acknowledgements}:    U.S.~National Science Foundation grants DMS-1764254 (E.A.C.), DMS-1802170 (I.J.) and DMS-1856645 (M.P.L) are gratefully acknowledged.  \vskip20pt    \begin{thebibliography}{WWW99}    \bibitem[B41]{B41} A. Berry, {\em The Accuracy of the Gaussian Approximation to the Sum of Independent Variates}. Trans. of the A.M.S. {\bf 49} (1941),122--136.    \bibitem[CGR08]{CGR08} E.A. Carlen, E. Gabetta and E. Regazzini, {\it Probabilistic investigation on explosion of solutions of the Kac equation with infinite initial energy}, J. Appl. Prob. {\bf 45} (2008), 95--106    \bibitem[CJL20]{CJL20} E.A. Carlen, I. Jauslin and E.H. Lieb, {Analysis of a simple equation for the ground state energy of the Bose gas}, Pure and Applied Analysis, 2020, in press, arXiv preprint arXiv:1912.04987.    \bibitem[CJL20b]{CJL20b} E.A. 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