From e8cb253fb27b0482478e3eada0732a44c4b5f225 Mon Sep 17 00:00:00 2001
From: Ian Jauslin <ian@jauslin.org>
Date: Tue, 24 Nov 2020 23:43:37 -0500
Subject: Update to v1.0:

  Fixed: Missing factor of 1/2 in expansion of f.

  Fixed: mean is not necessary in Theorem 3.

  Added: Lemma on Central Limit Theorem with infinite Variance.

  Miscellaneous changes and fixes.
---
 Carlen_Jauslin_Lieb_Loss_2020.tex | 191 +++++++++++++++++++++++---------------
 1 file changed, 114 insertions(+), 77 deletions(-)

(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 34fe424..24d7283 100644
--- a/Carlen_Jauslin_Lieb_Loss_2020.tex
+++ b/Carlen_Jauslin_Lieb_Loss_2020.tex
@@ -34,11 +34,11 @@ On the convolution inequality $f\geqslant f\star f$\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 $\int_{\mathbb R^d}f(x){\rm d}x \leqslant \textstyle\frac12$.  Moreover, if 
-$\int_{\mathbb R^d}f(x){\rm d}x = \textstyle\frac12$, then $f$ must decay fairly slowly: $\int_{\mathbb R^d}|x| f(x){\rm d}x = \infty$, and this is sharp since for all 
-$r< 1$, there are   solutions with  $\int_{\mathbb R^d}f(x){\rm d}x = \textstyle\frac12$ and $\int_{\mathbb R^d}|x|^r f(x){\rm d}x <\infty$.  However, if 
-$\int_{\mathbb R^d}f(x){\rm d}x  = : a < \textstyle\frac12$, the decay at infinity can be much more rapid: we show that for all $a<\textstyle\frac12$, there are solutions such that for some $\epsilon>0$, 
-$\int_{\mathbb R^d}e^{\epsilon|x|}f(x){\rm d}x < \infty$. 
+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.}
@@ -55,71 +55,72 @@ $\int_{\mathbb R^d}e^{\epsilon|x|}f(x){\rm d}x < \infty$.
   ,\quad\forall x\in\mathbb R^d\ ,
   \label{ineq}
 \end{equation}
-where $f\star f(x)$ denotes the convolution $f\star f(x) = \int_{\mathbb R^d} f(x-y) f(y){\rm d}y$, which by Young's inequality \cite[Theorem 4.2]{LL96} is well defined as an element of 
-$L^{p/(2-p)}(\mathbb R^d)$ for all $1 \leqslant p\leqslant 2$.  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 $\int_{\mathbb R^d} f(x){\rm d}x \leqslant 1$. 
-We prove that, in fact, all integrable solutions satisfy $\int_{\mathbb R^d} f(x){\rm d}x \leqslant \textstyle\frac12$, and this upper bound is sharp.  
+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 $\widehat{f}(k) = \int_{\mathbb R^d}e^{-i2\pi k\cdot x} f(x){\rm d}x$ is well defined as an element of $L^{p/(p-1)}(\mathbb R^d)$.  If $f$ solves the equation $f = f\star f$,
+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 
-$f = f*f$ 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 
+$f = f\star f$ 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){\rm d}k =  \frac{ \sin 2\pi xa}{\pi x}\ ,
+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},
 $$
-f_{a,t}(x) =  \int_{\mathbb R^d} e^{-i2\pi k x} g_{a,t}(k){\rm d}k = a\Gamma((d+1)/2) \pi^{-(d+1)/2} \frac{t}{(t^2 + x^2)^{(d+1)/2}}\ .
+f_{a,t}(x) =  \int_{\mathbb R^d} e^{-i2\pi k x} g_{a,t}(k)dk = a\Gamma((d+1)/2) \pi^{-(d+1)/2} \frac{t}{(t^2 + x^2)^{(d+1)/2}}\ .
 $$
 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  $\int_{\mathbb R^d}f_{a,t}(x){\rm d}x =a$, this provides a class of solutions of (\ref{ineq}) that are non-negative and satisfy 
+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){\rm d}x \leqslant \frac12\ ,
+\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){\rm d}x =\infty \ .
+ \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 $\int_{\mathbb R^d}e^{\epsilon|x|}f(x){\rm d}x < \infty$ for some $\epsilon> 0$.  The conjecture that integrable solutions of  (\ref{ineq})  
-are necessarily positive was motivated by recent work \cite{CJL19} 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{CJL19}. 
+$f$ 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}
-If $f$ is a real valued function in $L^1(\mathbb R^d)$ such that
+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 convergent series
+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 $\int_{\mathbb R^d}u(x){\rm d}x \leqslant \textstyle\frac14$, then the sum on the right in (\ref{fun}) defines an integrable function $f$ that satisfies (\ref{uineq}).
+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 $a := \int_{\mathbb R^d}f(x){\rm d}x$ and $b := \int_{\mathbb R^d}u(x){\rm d}x \geqslant 0$. Fourier transforming,
+\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)\ .
@@ -139,8 +140,8 @@ Hence (\ref{hatf}) is valid for all $k$, including $k=0$, again by continuity.
 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 $\int_{\mathbb R^d} 4^n\star^n u(x){\rm d}x \leqslant 1$, 
- 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, (\ref{hatf}) is 
+ \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
@@ -152,76 +153,75 @@ $\int_{\mathbb R^d}|x| f(x)\ dx=\infty$.
 \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 $f(x) = \sum_{n=0}^\infty \star^n w$.  Suppose that $|x|f(x)$ is integrable. Then $|x|w(x)$ is integrable.  Let $m:= \int_{\mathbb R^d}xw(x){\rm d} x$.  Since first moments add under convolution, the trivial inequality $|m||x| \geqslant m\cdot x$ yields 
-$$|m|\int_{\mathbb R^d} |x| \star^nw(x){\rm d}x \geqslant  \int_{\mathbb R^d} m\cdot x \star^nw(x){\rm d}x = n|m|^2\ .$$
- It follows that  $\int_{\mathbb R^d} |x| f(x){\rm d}x \geqslant |m|\sum_{n=1}^\infty nc_n = \infty$. Hence $m=0$. 
+\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., $\sigma^2 = \int_{\mathbb R^d}|x|^2w(x){\rm d}x\ .$
+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 
 $$
-\int_{\mathbb R^d}|x| \star^n w(x){\rm d}x =  \int_{\mathbb R^d}|n^{1/2}x| \star^n w(n^{1/2}x)n^{d/2}{\rm d}x  \geqslant  n^{1/2} \int_{\mathbb R^d}\varphi(x)\star^n w(n^{1/2}x)n^{d/2}{\rm d}x.
+\int_{\mathbb R^d}|x| \star^n w(x)dx =  \int_{\mathbb R^d}|n^{1/2}x| \star^n w(n^{1/2}x)n^{d/2}dx  \geqslant  n^{1/2} \int_{\mathbb R^d}\varphi(x)\star^n w(n^{1/2}x)n^{d/2}dx.
 $$
 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}{\rm d}x =  \int_{\mathbb R^d}\varphi(x) \gamma(x){\rm d}x =: C > 0
+\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  $\delta> 0$  for all sufficiently large $n$, $\int_{\mathbb R^d}|x| \star^n w(x){\rm d}x \geqslant \sqrt{n}\delta$, and then since $c_n\sim n^{3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x){\rm d}x= \infty$. 
+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 $R>0$, 
+ 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}{\rm d}x = 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 proved in \cite[Corollary 1]{CGR08}.
+\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
 
-\delimtitle{\bf Remark}
-For the convenience of the reader, we sketch, at the end of this paper, the part of of the argument in \cite{CGR08} that proves (\ref{CLT2}), since we know of no reference for this simple statement, and the proof in \cite{CGR08} deals with a more complicated variant of this problem.
-\endtheo
-\bigskip
-
-\theo{Theorem}
-  If $f$ satisfies\-~(\ref{ineq}), and
-$\int |x|^2[f(x)-f\star f(x)]\ dx<\infty$, then, for all $0\leqslant p<1$,
+\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 |x|^pf(x)\ dx<\infty.
+    \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 $u = f - f\star f$ as above.  Let  $t := 4\int_{\mathbb R^d}u(x){\rm d}x \leqslant 1$.  Then $t> 0$.  Define $w := t^{-1}4u$; $w$ is a probability density and 
-$$
-f(x) = \sum_{n=1}^\infty c_n t^n \star^n w(x)\ . 
-$$
-By  hypothesis, $w$ has a  mean $m := \int_{\mathbb R^d} x w(x){\rm d}x$ and  variance  $\sigma^2 =   \int_{\mathbb R^d} |x-m|^2 w(x){\rm d}x < \infty$. Since variance is additive under convolution,
+\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,
 $$
-\int_{\mathbb R^d} |x-m|^2 \star^n w(x){\rm d}x   = n\sigma^2\ .
+\int_{\mathbb R^d} |x|^2 \star^n w(x)dx   = n\sigma^2\ .
 $$
 By H\"older's inequality,  for all $0 < p < 2$,
-$\int_{\mathbb R^d} |x-m|^p \star^n w(x){\rm d}x   \leqslant  (n\sigma^2)^{p/2}$.
+$\int_{\mathbb R^d} |x|^p \star^n w(x)dx   \leqslant  (n\sigma^2)^{p/2}$.
 It follows that for $0 < p < 1$,
 $$
-\int_{\mathbb R^d} |x-m|^p f(x){\rm d}x   \leqslant  (\sigma^2)^{p/2} \sum_{n=1}^\infty n^{p/2} c_n  < \infty\ ,
+\int_{\mathbb R^d} |x|^p f(x)dx   \leqslant  \frac12(\sigma^2)^{p/2} \sum_{n=1}^\infty n^{p/2} c_n  < \infty\ ,
 $$
 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 $0<t<1$, $\sum_{n=1}^\infty n^2 c_n t^n < \infty$ 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 $\int_{\mathbb R^d}f(x){\rm d}x < \textstyle\frac12$
-can decay quite rapidly, even having finite exponential moments,  as we now show.
+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 non-negative, integrable function $u$, which integrates to $r<\frac14$, and satisfies
 \begin{equation}
-  \int_{\mathbb R^d} u(x)e^{\lambda|x|}{\rm d}x < \infty
+  \int_{\mathbb R^d} u(x)e^{\lambda|x|}dx < \infty
 \end{equation}
 for some $\lambda>0$.
 The Laplace transform of $u$ is
-$ \widetilde u(p):=\int e^{-px}u(x)\ {\rm d} x$ which is analytic for $|p|<\lambda$, and $\widetilde u(0) < \textstyle\frac14$.
+$ \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},
@@ -235,8 +235,10 @@ is an integrable solution of (\ref{ineq}).  For
 which is analytic for $|p|\leqslant \lambda_0$.
 Note that
 ${\displaystyle e^{s|x|} \leqslant \prod_{j=1}^d e^{|sx_j|} \leqslant \frac{1}{d}\sum_{j=1}^d e^{d|sx_j|} \leqslant \frac{2}{d}\sum_{j=1}^d \cosh(dsx_j)}$.
-What has just been shown, yields a $\delta>0$ so that for $|s|< \delta$, $\int_{\mathbb R^d}  \cosh(dsx_j)f(x){\rm d}x < \infty$ for each $j$. It follows that for 
-$0 < s < \delta$, $\int_{\mathbb R^d} e^{s|x|}f(x){\rm d}x < \infty$. 
+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 
 
  
@@ -247,42 +249,77 @@ $f$ by $-1$. If the region is taken small enough, the new function $f$ will stil
 \endtheo
 \bigskip
 
-We close by sketching a proof of (\ref{CLT2}) using the construction in \cite{CGR08}.  Let $w$ be a mean zero, infinite variance probability density on $\mathbb R^d$.  
-Pick $\epsilon>0$, and choose a large value $\sigma$ such that $(2\pi \sigma^2)^{-d/2}R^d|B| < \epsilon/2$, where $|B|$ denotes the volume of the ball.   The point of this is that if 
-$G$ is a centered Gaussian random variable with variance $\sigma^2$, the probability is no more than $\epsilon/2$ that $G$ lies in {\em any} particular translate $B_R+y$ of the ball of radius $R$. 
+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.
 
-Let $A \subset \mathbb R^d$ be a bounded set so that $\int_{A}xw(x){\rm d}x = 0$ and  $\int_{A}|x|^2w(x){\rm d}x = \sigma^2$.  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 greater than $\sigma^2$, $\alpha$ is a Bernoulli variable with success probability $\int_{A}w(x){\rm d}x$, which we can take arbitrarily close to $1$ by 
-increasing the size of $A$, and finally such that $\alpha X + (1-\alpha)Y $  has the probability density $w$. Taking independent i.i.d. sequences of such random variables, $w(n^{1/2}x)n^{d/2}$ is the probability density of
-${\displaystyle W_n :=  \frac{1}{\sqrt{n}}\sum_{j=1}^n \alpha_j X_j +  \frac{1}{\sqrt{n}} \sum_{j=1}^n(1-\alpha_j)Y_j}$.
-We are interested  in estimating 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. These conditional expectations have the form 
-${\mathbb E}\left[ 1_{B_R + y}\left( \frac{1}{\sqrt{n}}\sum_{j=1}^n \alpha_j X_j \right)\right]$
-for some translate $B_R +y$ of the unit Ball. Since $A$ is bounded, the $X_j$'s all have the same finite third moment, and now the Berry-Esseen Theorem \cite{B41,E42}, a version of the Central Limit Theorem 
-with rate information,  allows us to control the error in approximating this expectation by 
-$\mathbb{E}(1_{B_R +y}(G))$ where $G$ is centered Gaussian with variance $\sigma^2$. By the choice of $\sigma$, this is no greater than $\epsilon/2$, independent of $y$.  For $n$ large enough, 
-the remaining errors  -- coming from the small probability that $\sum_{n=1}^n \alpha_j$ is significantly less $n$, and the error bound provided by the Berry-Esseen Theorem, are readily absorbed into the remaining 
-$\epsilon/2$ for large $n$. Thus for all sufficiently large $n$, the integral in (\ref{CLT2}) is no more than $\epsilon$. 
 
 
 \vfill
 \eject
 {\bf Acknowledgements}:
 
-U.S.~National Science Foundation grants DMS-1764254 (E.A.C.),  DMS-1802170 (I.J.)  and  NSF grant DMS-1856645 (M.P.L)  are gratefully acknowledged.
+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
 
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+
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+
+
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