# Is the Boltzmann entropy continuous in the supremum norm?

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!

The answer is no. For instance, let $$d=1$$, $$\rho(x):=e^{-x}\,1(x>0)$$, $$\rho_n(x):=c_n\big(e^{-x}\,1(0 where $$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 $$H(\rho_n)\to-\infty\ne-1=H(\rho)$$. $$\quad\Box$$

• It seems your $\rho$ is not continuous. Commented Nov 22, 2023 at 15:56
• @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. Commented Nov 22, 2023 at 16:12