332 lines
22 KiB
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332 lines
22 KiB
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\hfil{\bf\LARGE


On the convolution inequality $f\geqslant f\star f$\par


}


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\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 nonnegative,


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 $\epsilon>0$,


$\int_{\mathbb R^d}e^{\epsilonx}f(x)dx < \infty$.


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\hfil{\footnotesize\copyright\, 2020 by the authors. This paper may be reproduced, in its entirety, for noncommercial purposes.}




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\setcounter{page}1


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\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 $f\star f(x) = \int_{\mathbb R^d} f(xy) f(y)dy$. 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


$L^{p/(2p)}(\mathbb R^d)$. 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 nonnegative. 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 $L^{p/(p1)}(\mathbb R^d)$. 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 RiemannLebesgue 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 nonnegative 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}xf(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 $\int_{\mathbb R^d}e^{\epsilonx}f(x)dx < \infty$ 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 $\sqrt{1x}$


\begin{equation}\label{3half}


\sqrt{1x}=1\sum_{n=1}^\infty c_n x^n


,\quad


c_n=\frac{(2n3)!!}{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 $k\neq 0$, $\sqrt{14\widehat u(k)} \neq 0$. By the RiemannLebesgue 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{14\widehat u(k)}


\label{hatf}


\end{equation}


for all sufficiently large $k$, and in any case $\widehat f(k)= \frac12\pm\frac12\sqrt{14\widehat u(k)}$. But by continuity and the fact that $\sqrt{14\widehat u(k)} \neq 0$ for any $k\neq 0$, the sign cannot switch.


Hence (\ref{hatf}) is valid for all $k$, including $k=0$, again by continuity. At $k=0$, $a = \textstyle\frac12  \sqrt{14b}$, 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 $\sqrt{1z}$ converges absolutely and


uniformly everywhere on the closed unit disc. Since $4 \widehat u(k) \leqslant 1$,


${\displaystyle


\sqrt{14 \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 $xf(x)$ is integrable. Then $xw(x)$ is integrable. Let $m:= \int_{\mathbb R^d}xw(x)d x$. Since first moments add under convolution, the trivial inequality $mx \geqslant m\cdot x$ yields


$$m\int_{\mathbb R^d} x \star^nw(x)dx \geqslant \int_{\mathbb R^d} m\cdot x \star^nw(x)dx = nm^2\ .$$


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 $w$,


$\int_{\mathbb R^d} x^2 \star^n w(x)dx =n^2m^2+ n\sigma^2$. For $0<t<1$, $\sum_{n=1}^\infty n^2 c_n t^n < \infty$ and we conclude that $\int_{\mathbb R^d} x^2 f(x)dx < \infty$. Finally, the final statement of Theorem~\ref{theo:positivity} shows that critical case functions $f$ satisfying the hypotheses of Theorem~\ref{theo:decay} are readily constructed.


\enddelim








Theorem \ref{theo:decay} implies that when $\int f=\frac12$, $f$ cannot decay faster than $x^{(d+1)}$. However, integrable solutions $f$ of (\ref{ineq}) such that $\int_{\mathbb R^d}f(x)dx < \textstyle\frac12$


can decay more rapidly, as indicated in the previous remark. In fact, they may even have finite exponential moments, as we now show.






Consider a nonnegative, integrable function $u$, which integrates to $r<\frac14$, and satisfies


\begin{equation}


\int_{\mathbb R^d} u(x)e^{\lambdax}dx < \infty


\end{equation}


for some $\lambda>0$.


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 welldefined Laplace transform $ \widetilde f(p)$ given by


\begin{equation}


\widetilde f(p)=\int e^{px}f(x)\ dx=\frac12(1\sqrt{14\widetilde u(p)})


\end{equation}


which is analytic for $p\leqslant \lambda_0$.


Note that


${\displaystyle e^{sx} \leqslant \prod_{j=1}^d e^{sx_j} \leqslant \frac{1}{d}\sum_{j=1}^d e^{dsx_j} \leqslant \frac{2}{d}\sum_{j=1}^d \cosh(dsx_j)}$.


Thus, for $s< \delta := \lambda_0/d$, $\int_{\mathbb R^d} \cosh(dsx_j)f(x)dx < \infty$ for each $j$, and hence


$s < \delta$, $\int_{\mathbb R^d} e^{sx}f(x)dx < \infty$.




However, there are no integrable solutions of (\ref{ineq}) that have compact support: We have seen that all solutions of (\ref{ineq}) are nonnegative, and if $A$ is the support of a nonnegative 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 $(2\pi \sigma_0^2)^{d/2}R^dB < \epsilon/3$, 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 BerryEsseen 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 DMS1764254 (E.A.C.), DMS1802170 (I.J.) and DMS1856645 (M.P.L) are gratefully acknowledged.


\vskip20pt




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\end{thebibliography}




\end{document}
