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 $\mathbb R^d$.


  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!

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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


$\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 $\rho_n(x):=\rho(x-n)$.

$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$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 $\tau$. $\quad\Box$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.