Assume we have a probability measure $\mu$ on $\mathbb R^n$. Assume it satisfies
$$
\mu(||x|| > u) \le Ce^{-au} \ \ \forall u > 0
$$
In other words, its tail is dominated by an exponential function.
Then, how to prove Poincare inequality?
$$
Var\_{\mu}f \le K\int||\nabla f||^2d\mu
$$



Sub-exponential tail implies Poincare inequality
As Mark Meckes already pointed out, this is not enough to proof a Poncaré inequality. 


If you are interested in the case where $\mu$ is absolutely continuous, then there are conditions formulated in terms of the log-density of $\mu$, i.e. $\mu$ is of the form
$$
 \mu(\mathrm{d}x) = \exp(-H(x)) 
$$
The minus sign is just convention. 


There is a complete characterisation of the constant $K$ for the real line via the [*Muckenhoupt functional*](http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-smv44i1p4bwm). Therefore let $m$ be the median of $\mu$, i.e. $\mu((-\infty,m))=\mu((m,\infty))=1/2$, then define
$$ \begin{align}
B\_+ &= \sup\_{x\geq m} \left( \int\_{m}^x \exp(H(y))\;\mathrm{d}y\ \int\_{x}^\infty \exp(-H(y)) \; \mathrm{d} y \right) \\\\
B\_- &= \sup\_{x\leq m} \left( \int\_{x}^m \exp(H(y))\;\mathrm{d}y \ \int\_{-\infty}^x \exp(-H(y)) \; \mathrm{d}{y} \right) 
\end{align}$$
Then, one has
$$
 \frac{1}{2} \max\{B\_-,B\_+\} \leq K \leq 4 \max\{ B\_, B\_+\}
$$
I.e. the Poncaré constant can be recovered upto a factor of $8$. 


On $\mathbb{R}^n$ there is the theory of Lyapunov functions, which give sufficient conditions for Poincaré inequalities. See [A simple proof of the Poincaré inequality](http://ecp.ejpecp.org/article/view/1352) by Bakry et. al. There, two prominent assumptions on $H$ which both individually imply a Poincaré inequality are


1. For some $\alpha>0$ and all $|x|\geq R>0$ holds
$$ \langle x, \nabla H\rangle \geq \alpha |x| $$
2. For some $\alpha>0 , c>0$ and all $|x|\geq R>0$ holds
$$ \alpha |\nabla H(x)|^2 - \Delta H(x) \geq 0 $$


Then $\mu$ satisfies a Poincaré inequality with an upper bound on the constant $K$. 


