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Let $X_t$ represent a continuous-time Markov process on $\mathbb{R}^d$, say a diffusion with locally Lipschitz coefficients. Suppose that there exists a unique invariant measure $\mu$ on the space, and $X_0$ is distributed as a point mass at $x_0 \in \mathbb{R}^d$, how might one compute or bound above the total variation between the measure $\mu_t$ defined by $\mu_t(A) = P(X_t \in A)$ and the invariant measure, $\mu$?

For a concrete example, if $dX_t = -X_t dt + dW_t$ (that is, $X_t$ is an Ornstein-Uhlenbeck process) one finds that the unique invariant distribution is absolutely continuous with respect to Lebesgue measure and coincides with the distribution of a standard normal variable: $$\mu(dx) = {\pi}^{-1/2}e^{-x^2} dx. $$ If $X_0 = 1$ a.s. one 'knows' that $\mu_t$ will obey the forward equation and will rapidly approach $\mu$, but I haven't found many useful tools to prove this or convergence in the general case.

Looking at contraction semigroup $T_t$, for example, one knows $$ \| T_t \| \leq Me^{\beta t} ~~ {\rm for } ~~ \beta \in \mathbb{R}, M > 0. $$ But I cannot compute $\beta$ (to show $\beta < 0$) explicitly. As $T_t$ is strongly continuous we have a uniform bound, but I cannot rule out $\beta = 0$. Some very particular results have been found, for Hamiltonian systems for example, but I imagine that general methods must exist for such an elementary question.

Thanks in advance.

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    $\begingroup$ Of course you do have $\beta = 0$ since $T_t 1 = 1$. You do not expect $T_t f$ to converge to $0$ but to $\int f\,d\mu$. In a nice symmetric case like Ornstein-Uhlenbeck you can use a spectral gap: if $N$ is the Ornstein-Uhlenbeck generator on $L^2(\mu)$, you can show its kernel is the constants and its spectrum is discrete. So if $\lambda$ is the least positive eigenvalue, by the spectral theorem you get $\|T_t f - \int f\,d\mu\|_{L^2(\mu)} \le e^{-\lambda t} \|f - \int f\,d\mu\|_{L^2(\mu)}$, which shows exponentially fast convergence. $\endgroup$
    – Nate Eldredge
    Dec 12, 2014 at 4:38
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    $\begingroup$ @ Nate Eldredge I don't see how to deduce discreteness of the spectrum from the mere fact that the constants form the kernel of N, but anyway this property has been proved (for a large class of Ornstein-Uhlenbeck operators) by Metafune, Pallara and Priola in their 2002 JFA article "Spectrum of Ornstein–Uhlenbeck operators in $L^p$ spaces with respect to invariant measures". Now, you only need to apply the spectral theorem to get the conclusion, as already suggested by Nate Eldredge. $\endgroup$
    – Delio Mugnolo
    Dec 12, 2014 at 9:49
  • $\begingroup$ @DelioMugnolo: "Kernel is the constants" and "spectrum is discrete" are two different things to show. $\endgroup$
    – Nate Eldredge
    Dec 12, 2014 at 17:27
  • $\begingroup$ Thank you both. Your comment cleared up my thinking Nate. The OU operator is a good base case to study, but I know some interesting cases do converge in spite of having no spectral gap. I take it that in general there is no way to guarantee that measures of compact support will approach the invariant measure. $\endgroup$
    – Titus
    Dec 12, 2014 at 18:32
  • $\begingroup$ A question and a note - what is the relevance of $N$ having the constant functions as its kernel? Surely the (nonconstant) invariant measure is also annihilated by $N$. $\endgroup$
    – Titus
    Dec 13, 2014 at 3:26

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