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2 votes
1 answer
223 views

Is Boltzmann entropy well-defined for arbitrary probability density function?

As such, Boltzmann entropy is well-defined (in extended real number line) for every $f \in D_2$. …
3 votes
1 answer
130 views

Let $\mu : [0, T] \to \mathcal P_2^a (\mathbb R^d), t \mapsto \mu_t$ be absolutely continuou...

Let $\mathcal H (\mu)$ be the relative entropy of $\mu$ w.r.t. Lebesgue measure. It is well-known that $\mathcal H$ is not continuous but only lower semi-continuous w.r.t. $W_2$. …
1 vote
1 answer
143 views

Is the Boltzmann entropy continuous in the supremum norm?

If $\rho \in D$ then $\rho$ is a probability density function whose induced measure has finite second moment and finite Boltzmann entropy. …
0 votes
1 answer
87 views

Can we lower bound this entropy by $\int_{\mathbb R^d} \rho^k (x) \, \mathrm d x$ and $\int_...

We define the entropy $\mathcal H : \mathcal P_2^a \to [0, \infty]$ by $$ \mathcal H (\rho) := \int_{\mathbb R^d} U \circ \rho. $$ Then $\mathcal H (\rho) > 0$ for all $\rho \in \mathcal P_2^a$. …
0 votes
2 answers
262 views

Is the Boltzmann entropy lower semi-continuous in the weak topology induced by $C_b (\mathbb...

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