Can we lower bound this entropy by $\int_{\mathbb R^d} \rho^k (x) \, \mathrm d x$ and $\int_{\mathbb R^d} |x|^2\rho (x) \, \mathrm d x$?

Question

We define $U : [0, \infty) \to [0, \infty)$ by $U(0) := 1$ and $U (s) := s \log s + (1-s)$ for $s >0$. Then $U$ is strictly convex. The minimum of $U$ is $0$ and is attained at $s=1$. Let $\mathcal P_2^a$ be the set of absolutely continuous Borel probability measures with finite second moment on $\mathbb R^d$. We also denote by $\rho$ the probability density function of $\rho \in \mathcal P_2^a$. 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 $\rho \in \mathcal P_2^a$.

Can we lower bound $\mathcal H (\rho)$ by some quantity depending on $\int_{\mathbb R^d} |x|^2\rho (x) \, \mathrm d x$ and $\int_{\mathbb R^d} \rho^k (x) \, \mathrm d x$ for some $k \in \mathbb N$?

Thank you so much for your elaboration!

Answer 1

$\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}$Note that $U(s)\ge-1/e+1-s$ for all real $s\ge0$ (assuming that your $\log$ is $\ln$). So, for any probability density function $\rho$, $$\H(\rho)\ge\int_{\R^d}(-1/e+1-\rho)=\int_{\R^d}(-1/e+1)-1=\infty-1=\infty.$$ So, any lower bound on $\H(\rho)$ will do.