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Let $E$ be a locally compact Hausdorff space with countable base and $X_t$ be a stochastic process taking values in the one-point compactification of $E$ (with the Borel $\sigma$-algebra). Let $f$ be a continuous function vanishing at infinity. I'm wondering under what conditions it is true that $x \mapsto \mathbf{E}^{x}[f(X_t)]$ is a continuous function?

If $X_t$ is a Brownian motion on $\mathbb{R}$, it is straightforward to verify this is true and, in fact, it is true whenever $X_t$ is an Ito diffusion.

An example where this fails is if we let $E=\mathbf{R}$ and let $X_t$ be a reflected Brownian motion on $\{ x: x\geq 0 \}$ and be the negative of the absolute value of a Brownian motion in $\mathbb{R}^3$ on $\{x : x<0 \}$.

More broadly, I'm curious as to what kind of conditions on the sample paths of a Markov process $X$ with continuous paths force it to be a Feller process. The condition I'm asking about seems to be the one that does not come for free if you start with such an $X$.

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I guess this wasn't really a well-formed question. Perhaps the answer to this is just "it does when it does" and this is why we have a name for this property. – ShawnD Feb 22 '12 at 23:40

UPDATE: This is at best a partial answer.

If $X$ is a Markov process with continuous paths, then the additional condition for it to be a diffusion it be strong Markov (see the book by Ito-McKean, Section 3.1). For example, your example is not strong Markov (stop it when it hits 0). In that case, however, it is still not necessarily Feller (this was stated previously).

If you only require $X$ to be a Markov process with not necessarily continuous paths, then your condition is (almost, see this thread) equivalent to requiring that 1) $S_t:C_0(E) \rightarrow C_0(E)$, where $(S_t)_{t\ge0}$ is the semigroup of $X$ and $C_0(E)$ is the space of continuous functions vanishing at infinity. But a Feller process requires moreover that 2) $S_tf(x) \rightarrow f(x)$ for each $x$, as $t\rightarrow 0$ (see for example Section III.3 in the book by Revuz-Yor). For example, take Brownian motion on $\mathbb R^+$ which jumps to $1$ when it hits $0$. This process satisfies 1) but not 2)

If you take for $X$ any stochastic process, I doubt that one can tell more than what you wrote in your comment.

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No, that's not right. The example given is strong Markov, but not Feller. Even simpler, take $X_t=X_0$ when $X_0\ge0$ and $X_t=X_0-t$ otherwise. – George Lowther Mar 4 '12 at 19:16
Also, minor quibble, but if your process has continuous paths you get $S_t f(x) \rightarrow f(x)$ for free (and if you have the Feller property, then you can show that you actually have $S_t f \rightarrow f$). Perhaps the theorem is that an Ito process with the strong Markov property is a diffusion? – ShawnD Mar 4 '12 at 23:15
@George: Do you mean the example given by ShawnD? Maybe I misunderstood it, probably he meant 0 to be a shunt (in which case it would be a diffusion, however). I understood that when the process starts at x < 0 it gets reflected at 0, such that it always stays below 0. In this case, I think that the process is not strong Markov, because when stopping it at 0, one has to know whether it came from below or from above in order to continue. As for your example, you are right, this is a strong Markov process with continuous paths but not a Feller process. – Pascal Maillard Mar 5 '12 at 9:32
contd. Indeed, the Ito-McKean definition of a diffusion does not imply Feller. This explains why Rogers-Williams speak of "Feller-Dynkin diffusions" and assume the Feller property. I presume that if the diffusion coefficient does not vanish, then the diffusion is Feller, but I could not find a reference to that – Pascal Maillard Mar 5 '12 at 14:37
@ShawnD: Maybe you are right, because an Ito process is a time-inhomogeneous diffusion, so the Markov property should already be enough, I suppose. However, this does not answer your question about the Feller property. I think that George's example of a process is a diffusion according to Ito-McKean or Revuz-Yor (with generator $Lf(x) = -f'(x)1_{x<0}$), but it is not Feller. – Pascal Maillard Mar 5 '12 at 14:43

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