Let $\mu(x)$ be a probability measure on $\mathbb{R}^d$ which we assume to be absolutely continous with respect to Lebesgue measure, $\mbox{d}\mu(x)=\omega(x)\mbox{d}x$. I am interested in tractable sufficient conditions for $\mu$ to support a Poincare inequality, which means that there exists a finite constant $C_P$ such that $$ Var_\mu(f) \le C_P \int_{\mathbb{R}^d} |\nabla f|^2 \mbox{d}\mu $$ for all $f\in C_b^\infty(\mathbb{R}^d)$.
A necessary condition for $\mu$ to support a Poincare constant is that $\mu$ enjoys an exponential concentration property (cf. Theorem 2 in https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjmpen49f-EAxU52gIHHRZkA1M4PBAWegQIAxAB&url=https%3A%2F%2Fwww.stat.cmu.edu%2F~arinaldo%2F36788%2Fberestycki_nickl_notes.pdf&usg=AOvVaw3OZinEJYwN_mecR7WNYWnG&opi=89978449)
It is not difficult to see that exponential concentration is not sufficient for $\mu$ to support a Poincare inequality as one can construct weights with disconnected support such that the resulting measure has an exponential concentration property.
Let us now exclude that the density $\omega$ vanishes at any point and suppose we know that $\omega(x)=e^{-V(x)}$ is such that for $|x|$ sufficiently large $$ V(x) \ge \varepsilon |x| \qquad (\ast) $$ where $\varepsilon>0$.
If we have $=$ in $(\ast)$ then it is known that a Poincare inequality holds (by condition 1) in the answer to the post https://mathoverflow.net/questions/123924/sub-exponential-tail-implies-poincare-inequality). What about the inexact case though? My intuition would be that if $V(x)$ is somewhat well-behaved(e.g., not too oscillatory) a Poincare inequality should still hold. Are there any tractable conditions which guarantee this to be the case?