From e8cb253fb27b0482478e3eada0732a44c4b5f225 Mon Sep 17 00:00:00 2001 From: Ian Jauslin Date: Tue, 24 Nov 2020 23:43:37 -0500 Subject: [PATCH] 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 ++++++++++++++++++------------ Changelog | 10 ++ 2 files changed, 124 insertions(+), 77 deletions(-) create mode 100644 Changelog 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 +\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, $$ -f(x) = \sum_{n=1}^\infty c_n t^n \star^n w(x)\ . -$$ -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, -$$ -\int_{\mathbb R^d} |x-m|^2 \star^n w(x){\rm d}x = n\sigma^2\ . +\int_{\mathbb R^d} |x|^2 \star^n w(x)dx = n\sigma^2\ . $$ 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 $00$. 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 \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[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[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[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[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[E42]{E42} C.-G. Esseen, {\em A moment inequality with an application to the central limit theorem}. Skand. Aktuarietidskr. {\bf 39} 160-170. +\bibitem[CJL20b]{CJL20b} E.A. Carlen, I. Jauslin and E.H. Lieb, {Analysis of a simple equation for the ground state of the Bose gas II: Monotonicity, Convexity and Condensate Fraction}, arXiv preprint arXiv:2010.13882. + +\bibitem[E42]{E42} C.-G. Esseen, {\em A moment inequality with an application to the central limit theorem}. Skand. Aktuarietidskr. {\bf 39} (1942) 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[MV10]{MV10} M. Matolcsi, C. Vinuesa, {\it Improved bounds on the supremum of autoconvolutions}, J. Math. Anal. Appl. {\bf 372} (2010) 439--447. + +\bibitem[R19]{R19} M. Rai\v c, {\em A multivariate Berry-Esseen Theorem with explicit constants}, Bernoulli {\bf 25} (2019) 2824--2853 + + \bibitem[SW71]{SW71} E. Stein and G. Weiss, {\em Introduction to Fourier analysis on Euclidean spaces}, Princeton University Press, Princeton NJ, 1971. diff --git a/Changelog b/Changelog new file mode 100644 index 0000000..50e06cb --- /dev/null +++ b/Changelog @@ -0,0 +1,10 @@ +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. +