From 9c9f888bb4839dee7b34d18181d2cec50dd4b743 Mon Sep 17 00:00:00 2001
From: Ian Jauslin <ian@jauslin.org>
Date: Wed, 12 Feb 2020 22:58:32 -0500
Subject: Initial commit

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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 $\int_{\mathbb R^d}f(x){\rm d}x \leqslant \textstyle\frac12$.  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 $\epsilon>0$, 
+$\int_{\mathbb R^d}e^{\epsilon|x|}f(x){\rm d}x < \infty$. 
+\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 $f\star f(x) = \int_{\mathbb R^d} f(x-y) f(y){\rm d}y$, 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.  
+
+
+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 $f = f\star f$,
+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 
+\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}\ ,
+ \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},
+$$
+f_{a,t}(x) =  \int_{\mathbb R^d} e^{-i2\pi k x} g_{a,t}(k){\rm d}k = a\Gamma((d+1)/2) \pi^{-(d+1)/2} \frac{t}{(t^2 + x^2)^{(d+1)/2}}\ .
+$$
+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 
+\begin{equation}\label{half}
+\int_{\mathbb R^d}f(x){\rm d}x \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 \ .
+ \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}. 
+
+
+\theo{Theorem}\label{theo:positivity}
+If $f$ is 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
+\begin{equation}
+f(x) = \frac{1}{2} \sum_{n=1}^\infty  c_n 4^n (\star^n u)(x)
+\label{fun}
+\end{equation} 
+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}).
+\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,
+(\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 $\int_{\mathbb R^d} 4^n\star^n u(x){\rm d}x \leqslant 1$, 
+ 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 
+ 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 $f(x) = \sum_{n=0}^\infty \star^n w$.  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 $m=0$. 
+
+
+Suppose temporarily that in addition, $|x|^2w(x)$ is integrable. Let $\sigma^2$ be the variance of $w$; i.e., $\sigma^2 = \int_{\mathbb R^d}|x|^2w(x){\rm d}x\ .$
+Define the function $\varphi(x) = \min\{1,|x|\}$.  Then 
+$$
+\int_{\mathbb R^d}|x| \star^n w(x){\rm d}x =  \int_{\mathbb R^d}|n^{1/2}x| \star^n w(n^{1/2}x)n^{d/2}{\rm d}x  \geqslant  n^{1/2} \int_{\mathbb R^d}\varphi(x)\star^n w(n^{1/2}x)n^{d/2}{\rm d}x.
+$$
+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
+\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 $c_n\sim n^{3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x){\rm d}x= \infty$. 
+
+ 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}.
+\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 $0\leqslant p<1$,
+  \begin{equation}
+    \int |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,
+$$
+\int_{\mathbb R^d} |x-m|^2 \star^n w(x){\rm d}x   = n\sigma^2\ .
+$$
+By H\"older's inequality,  for all $0 < p < 2$,
+$\int_{\mathbb R^d} |x-m|^p \star^n w(x){\rm d}x   \leqslant  (n\sigma^2)^{p/2}$.
+It follows that for $0 < p < 1$,
+$$
+\int_{\mathbb R^d} |x-m|^p f(x){\rm d}x   \leqslant  (\sigma^2)^{p/2} \sum_{n=1}^\infty n^{p/2} c_n  < \infty\ ,
+$$
+again using the fact that $c_n\sim n^{-3/2}$. 
+\qed
+
+
+
+
+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.
+
+
+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
+\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 $\widetilde u(0) < \textstyle\frac14$.
+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
+${\displaystyle e^{s|x|} \leqslant \prod_{j=1}^d e^{|sx_j|} \leqslant \frac{1}{d}\sum_{j=1}^d e^{d|sx_j|} \leqslant \frac{2}{d}\sum_{j=1}^d \cosh(dsx_j)}$.
+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 
+$0 < s < \delta$, $\int_{\mathbb R^d} e^{s|x|}f(x){\rm d}x < \infty$. 
+\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 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 $R$. 
+
+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.
+\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 infintte initial energy}, J. Appl. Prob. {\bf 45} (2008), 95-106
+
+\bibitem[CJL19]{CJL19} E.A. Carlen, I. Jauslin and E.H. Lieb, {Analysis of a simple equation for the ground state energy of the Bose gas}, arXiv preprint arXiv:1912.04987.
+
+\bibitem[E42]{E42} C.-G. Esseen, {\em A moment inequality with an application to the central limit theorem}. Skand. Aktuarietidskr. {\bf 39} 160-170.
+
+\bibitem[LL96]{LL96} E.H. Lieb and M. Loss, {\em Analysis}, Graduate Studies in Mathematics {\bf 14}, A.M.S., Providence RI, 1996.
+
+\bibitem[SW71]{SW71} E. Stein and G. Weiss, {\em Introduction to Fourier analysis on Euclidean spaces}, Princeton University Press, Princeton NJ, 1971.
+
+
+
+
+
+\end{thebibliography}
+
+\end{document}
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