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Assume we have a reflected Brownian motion in a smooth bounded domain $D \subseteq \mathbb R^d$. It can have nonzero (but constant) drift, non-identity (but constant) covariance matrix, and oblique (=not normal) reflection.

We want to prove exponential ergodicity, that is, $$ ||P^t(\cdot, x) - \pi(\cdot)||_{TV} \le a(x)e^{-kt}, $$ for some constant $k > 0$, which is called {\it the exponent of ergodicity}. TV is the total variation distance, $\pi$ is the invariant distribution (provided it exists and is unique).

There is a following theorem (see, e.g. the article "Lyapunov implies Poincare" by Arnaud Guillin):

if we construct a function V (Lyapunov function) such that $$ LV \le -kV + c1_B, $$ where $k, c > 0$, $B$ is some {\it petite} set (in practice, it means compact set), and $L$ is the generator, then exponential ergodicity holds (with the same exponent $k$).

This works perfectly fine in domains like $\mathbb R^d_+$ (posiitve orthant). However, if $D$ is bounded, then we can take $B = D$, and the condition is trivially fulfilled with ANY k (just take any smooth $V$ in the domain of the generator and adjust $c$, taking it large enough!) So it means that exponent of ergodicity can be taken as large as possible??

Thank you in advance.

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The constant $K$ in the exponential ergodicity and the constant $k$ in the Lyapunov condition are not the same in general. The detailed dependence of the constants is rather involved, but in particular it depends on $c$ and gets small for $c$ large. In the case of the Poincaré inequality there is an explicit formula (Bakry et al. A simple proof of the Pincaré inequa.) . In the case of exponential ergodicity, you can try to follow the original proof of Meyn & Tweedie: Stability of Markovian Processes III or another very good reference is M. Hairer: Convergence of Markov Processes. –  André Schlichting Feb 27 '13 at 9:46
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In addition working on a bounded domain involves to take care of the boundary conditions of your Lyapunov function $V$. An essential tool which is used in the proofs is integration by parts for the generator $L$ wrt. to $V$. This means $V$ has to satisfies in your case Neumann boundary conditions. If your domains is convex, that is easy to accomplish. But if your domain is not convex, think of spiral like domains, this can be rather involved. –  André Schlichting Feb 27 '13 at 9:57

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