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For Lebesgue-absolutely continuous probability measures $\rho\ll \mathcal{L}^d$ in the whole space $\mathbb{R}^d$ with finite second moments (i-e $\rho\in \mathcal{P}^2_{ac}(\mathbb{R}^d)$), let $$ \mathcal{H}(\rho)=\int_{\mathbb{R}^d}\rho\log \rho \,dx $$ be the Boltzmann entropy. It is well-known that $\mathcal{H}$ is lower semi-continuous (l.s.c.) in the Wasserstein metric $W_2$. We denoted by $\tau$ the weak topology induced by the set of real-valued bounded continuous functions on $\mathbb R^d$.

Questions:

  1. Is $\mathcal{H}$ l.s.c. in $\tau$?
  2. Fix $C>0$. Is the sublevel set $\{\rho \in \mathcal{P}^2_{ac}(\mathbb{R}^d) : \mathcal{H}(\rho) \le C\}$ compact in $\tau$?

Thank you so much for your elaboration! Any reference is greatly approciated!

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  • $\begingroup$ According to MathOverflow guidelines, there should be only one question in one post. $\endgroup$ Commented Nov 2, 2023 at 14:09
  • $\begingroup$ @IosifPinelis I have seen you deleted your answer. I have a modified version of Boltzmann entropy. I think that your proof works for this modified version. If you don't mind, I will send you an email for that proof... $\endgroup$
    – Akira
    Commented Nov 2, 2023 at 14:33
  • $\begingroup$ What do you call "weak topology" here ? $H$ is a function on ac square-integrable pdf but you take a topology generated on $\mathbb R^d$ and not on $P=P^2_{ac}(\mathbb R^d)$ so i don't see what your question means. Do you mean the topology induced on $P$ by the "integration" linear functionals associated to bounded continuous functions ? $\endgroup$
    – plm
    Commented Nov 2, 2023 at 14:50
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    $\begingroup$ @plm : The definition of convergence here is as common in probability. It does define a topology over the set of probability measures. $\endgroup$ Commented Nov 2, 2023 at 16:55
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    $\begingroup$ @Akira : Of course, you can send me an email. On the other hand, see the new version of the answer. $\endgroup$ Commented Nov 2, 2023 at 16:56

2 Answers 2

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$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$1. The answer to Question 1 is no. Indeed, let $d=1$ and $$\rho_n:=c_n p_n,$$ where $$p_n(x):=f(x)+\frac{1(e^n<x<e^{e^n})}{x\ln^2 x}$$ for natural $n$ and real $x$, $f$ is the standard normal pdf, and $c_n:=1/\int_\R p_n\to1$.

Then $\rho_n\to\rho:=f$ pointwise (as $n\to\infty$), so that the probability measures with densities $\rho_n$ converge in total variation and hence in $\tau$ to the probability measure with density $\rho$.

On the other hand (assuming that your $\log$ is $\ln$), $$\H(\rho_n)\sim-\int_{e^n}^{e^{e^n}}\frac{dx}{x\ln x}\sim -n\to-\infty,$$ whereas $\H(\rho)\in\R$. $\quad\Box$

  1. The answer to Question 2 is no: As in the comment by Kostya_I, let $\rho_n(x):=\rho(x-n)$.
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$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$The answer to Question 1 becomes yes if, as suggested by the OP in a personal message, we use the following definition of the entropy instead of the one in the OP: $$\H(\rho):=\int_{\R^d}U(\rho(x))\,dx,$$ where $U(r):=r\ln r+1-r$ for real $r>0$, with $U(0):=1$, so that $U$ is a nonnegative convex function on $[0,\infty)$.

Indeed, suppose that a sequence of probability measures $\mu_n$ with densities $\rho_n$ converges in $\tau$ to a probability measure $\mu$ with density $\rho$. For real $\si>0$, let $\ga_\si$ be the Gaussian measure over $\R^d$ with mean $0$ and covariance matrix $\si I_d$, where $I_d$ is the identity matrix. Let $f_\si$ be the density of $\ga_\si$.

Then for each $\si>0$ the density $f_\si *\rho_n$ of the probability measure $\ga_\si *\mu_n$ converges pointwise to the density $f_\si *\rho$ of the probability measure $\ga_\si *\mu$ (this follows because $f_\si$ is a bounded continuous function). So, by the Fatou lemma, $$\liminf_n\H(f_\si *\rho_n)\ge \H(f_\si *\rho). \tag{1}\label{1}$$

By Jensen's inequality applied to the convex function $U$, $$\H(\rho_n)\ge\H(f_\si *\rho_n). \tag{2}\label{2}$$

Also, $f_\si *\rho\to\rho$ almost everywhere as $\si\downarrow0$. So, again by the Fatou lemma, $$\liminf_{\si\downarrow0}\H(f_\si *\rho)\ge \H(\rho). \tag{3}\label{3}$$

It follows by \eqref{2},\eqref{1}, and \eqref{3} that $\liminf_n\H(\rho_n)\ge \H(\rho)$. So, $\H$ is l.s.c. in $\tau$. $\quad\Box$

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