Consider the over damped Langevin dynamics: $d X_{t} = d B_{t} - \nabla U(X_{t}) dt $ on $\mathbb{R}^{d}$ where $B_t$ is a standard Brownian motion. On pages 29 and 30 of the following book

Royer, Gilles, An introduction to logarithmic Sobolev inequalities, Cours Spécialisés (Paris). 5. Paris: Société Mathématique de France. 114 p. (1999). ZBL0927.60006.

Royer, Gilles, An initiation to logarithmic Sobolev inequalities. Transl. from the French by Donald Babbitt, SMF/AMS Texts and Monographs 14; Cours Spécialisés (Paris) 5. Providence, RI: American Mathematical Society (AMS); Paris: Société Mathématique de France (ISBN 978-0-8218-4401-4/pbk). viii, 119 p. (2007). ZBL1138.60007.

the author says that if

  1. $U$ is $C^{2}$,
  2. the corresponding SDE doesn't explode in finite time almost surely, and
  3. $\exp(-U)$ is integrable on $\mathbb{R}^{d}$.

then the corresponding Boltzmann-Gibbs measure defined by $d\mu(x)\propto \exp(-U(x)) dx$ is reversible for the process (and consequently the stationary distribution for the process).

Now, consider $U$ which doesn't satisfy the third assumption (integrability). Then, according to the theory above, we can't deduce stationarity of the Boltzmann-Gibbs measure written above.

In this context, my question is:

What can we say now about the stationary distribution of the process? Can this SDE have a stationary distribution of some other form (i.e. other than the Boltzmann-Gibbs)? OR does the SDE have no stationary distribution at all (implying that if the SDE has a stationary distribution then it has to be of the Boltzmann-Gibbs form)?

I would really appreciate some help here because I am confused.


1 Answer 1


As explained here Convergence of a continuous time stochastic gradient descent algorithm even for weak regularity for $U$ (Lipschitz and L2 integrability), they indeed converge to Brownian motion as $t\to +\infty$.

here "An SDE perspective on stochastic convex optimization" they study these type of SDEs (Stochastic gradient descent) and basically ask bounded and an L2 condition for gamma_t in theorem 3.1.

Here in (H0) they also have extra constraints to $f$ that might be sharp.

As shown here a function can be convex and C1 but its gradient not be Lipschitz

The map $f:\Bbb R\to \Bbb R$, $f(x)=\frac23\lvert x\rvert^{3/2}=\int_0^x \lvert t\rvert^{1/2}\operatorname{sgn}t\,dt$ is convex and $C^1$, but $f'(x)=\lvert x\rvert^{1/2}\operatorname{sgn}x$ is not Lipschitz continuous in any neighbourhood of $0$. More generally, integrate your favourite monotone increasing continuous function which is not locally Lipschitz and you'll obtain a counterexample in $\Bbb R$.

And as shown here without this Lipschitz condition, we don't even have convergence for the deterministic gradient descent.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .