1
$\begingroup$

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!

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $H(\rho_n)\to-\infty\ne-1=H(\rho)$. $\quad\Box$

$\endgroup$
2
  • 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

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.