1
$\begingroup$

Consider the following Ito diffusion $X_t$ satisfying

$$dX_t=b(X_t)dt+\sigma(X_t)dB_t,\quad X_0=x\in \mathbb{R}^n,$$

with Lipschitz coefficients $b,\sigma$.

It can be shown that if $g$ is bounded and continuous, then $u(x)=E^x[g(X_t)]$ is continuous. So any Ito diffusion is $C_b$-Feller continuous.

However, some books define Feller property to be if $g$ is continuous and vanishes at infinity, then $u$ is continuous and vanishes at infinity.

It seems that Brownian Motion satisfies this property. I have also shown that one dimensional Ito diffusion has Feller property using a sufficient condition given in Lorenzi's book.

May I know whether in general the Ito diffusion $X_t$ (SDE with global Lipschitz coefficients) satisfying this property?

$\endgroup$
1
$\begingroup$

Under the hypotheses:

  1. $b$ and $\sigma$ are globally Lipschitz continuous;
  2. $\sigma(x) \sigma(x)^T$ is positive definite for all $x \in \mathbb{R}^n$;

one can prove that the semigroup $P_t$ associated to $X$ is, in fact, strong Feller, meaning that for any $t \ge 0$ the semigroup $P_t$ maps $B_b(\mathbb{R}^n)$ (bounded functions) to $C_b(\mathbb{R}^n)$ ( bounded and continuous functions). This result is typical of SDEs with non-degenerate noise. See, e.g., Chapter 12 of Giuseppe Da Prato (2014). Introduction to Stochastic Analysis and Malliavin Calculus. The Feller property is weaker, and as you say, just requires that for any $t \ge 0$ the semigroup $P_t$ maps $C_b(\mathbb{R}^n)$ into itself.

Some books discuss the related, but different, concept of Feller-Dynkin processes where $C_b(\mathbb{R}^n)$ is replaced with $C_0(\mathbb{R}^n)$, the set of continuous functions vanishing at infinity. This overloading of terminology is unfortunate. Since we have the inclusion $ C_0(\mathbb{R}^n) \subset C_b(\mathbb{R}^n)$, the notion of a Feller process is a bit more general than a Feller-Dynkin process.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.