Given a probability measure $\mu$ on $\mathbb{R}^n$, its Poincare constant is the least number $C$ such that:
$$
\int f^2 d\mu \leq C\int \|\nabla f\|^2 d\mu
$$
for all zero mean function $f$.


Is there a version of Cheeger's inequality that works in this setting? I would assume so but was only able to find references for closed Riemannian manifolds and graphs. 



Cheeger inequality for measures
Yes, the Cheeger's inequality is even known in the general framework of a Dirichlet space $(X,d,\mathcal{E},\mu)$, where $\mu$ is a probability measure. Indeed, assume that $\mathcal{E}$ is strictly local with a carre du champ $\Gamma$, that Lipschitz functions are in the domain of $\mathcal{E}$ and that $\sqrt{\Gamma(f)}$ is an upper gradient in the sense that for any Lipchitz function $f$,


$
\sqrt{\Gamma(f)}(x) =\lim \sup\_{d(x,y)\to 0} \frac{ |f(x)-f(y)|}{d(x,y)}.
$


In that case, if $A$ is a closed set of $X$, one defines its Minkowski exterior boundary measure by


$
\mu\_+(A)=\lim \inf\_{\varepsilon \to 0} \frac{1}{\varepsilon} \left( \mu(A\_\varepsilon) -\mu(A) \right),
$


where $A\_\varepsilon=\{ x \in X, d(x,A) <\varepsilon \}$. We can then define the Cheeger's constant of $(X,d,\mathcal{E})$ by


$
h=\inf \frac{\mu\_+(E)}{\mu(E)}
$
where the infimum runs over all closed sets $E$ such that $\mu(E)\le \frac{1}{2}$. Then, according to Theorem 8.5.2 in the book by [Bakry-Gentil-Ledoux](https://www.springer.com/us/book/9783319002262) , one has


$
\lambda\_1 \ge \frac{h^2}{4},
$


where $\lambda\_1$ is the spectral gap of $\mathcal{E}$. Note that the Buser's inequality for the Cheeger constant is also known in a very general framework (see Theorem 3.6 in this [paper](https://arxiv.org/pdf/1604.02520.pdf))


