We define $U : [0, +\infty) \to [0, +\infty)$ by $U(0) := 0$ and $U (s) := s \log s$ for $s >0$. Then $U$ is strictly convex. Let $D$ be the set of all bounded non-negative continuous functions $\rho : \mathbb R^d \to \mathbb R$ such that

  • $\int_{\mathbb R^d} \rho=1$.
  • $M (\rho) :=\int_{\mathbb R^d} |x|^2 \rho (x) \, \mathrm d x < +\infty$
  • $H(\rho) := \int_{\mathbb R^d} U ( \rho (x)) \, \mathrm d x< +\infty$.

If $\rho \in D$ then $\rho$ is a probability density function whose induced measure has finite second moment and finite Boltzmann entropy.

Assume that $\rho, \rho_n \in D$ such that $\|\rho_n - \rho\|_{\infty} \to 0$ as $n \to \infty$. Is it true that $|H(\rho_n) - H(\rho)| \to 0$ as $n \to \infty$?

Thank you so much for your elaboration!


1 Answer 1


The answer is no. For instance, let $d=1$, $\rho(x):=e^{-x}\,1(x>0)$, $$\rho_n(x):=c_n\big(e^{-x}\,1(0<x\le n)+p_n\,1(n<x\le2n)\big),$$ where $c_n:=1/(1-e^{-n}+np_n)$, $p_n\in(0,\infty)$ for all $n$, and $p_n\sim1/(n\ln\ln n)$ (as $n\to\infty$).

Then $H(\rho)=-1$, $c_n\to1$, $\|\rho_n-\rho\|_\infty\to0$, $$np_n\to0,\quad np_n\ln p_n\to-\infty,$$ $$H(\rho_n)=I_n+J_n,$$ $$I_n:=\int_0^n c_n e^{-x}(\ln c_n-x)\,dx\to-1,$$ $$J_n:=\int_n^{2n} c_n p_n (\ln c_n+\ln p_n) \sim np_n \ln c_n+n p_n \ln p_n\to0-\infty,$$ so that $H(\rho_n)\to-\infty\ne-1=H(\rho)$. $\quad\Box$

  • 1
    $\begingroup$ It seems your $\rho$ is not continuous. $\endgroup$
    – Akira
    Commented Nov 22, 2023 at 15:56
  • 2
    $\begingroup$ @Akira : As in all such problems, the continuity plays here no role at all. Indeed, you can smooth out $\rho$ and $\rho_n$ in neighborhoods of the points of discontinuity so that $H(\rho)$ and $H(\rho_n)$ be approximated however closely. $\endgroup$ Commented Nov 22, 2023 at 16:12

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