I have a question about Feller property of reflecting Brownian motions. 


Let $D \subset \mathbb{R}^2$ be a domain. Assume $D$ is represented as
\begin{equation\*}
D=\{(x,y) \in \mathbb{R} \times \mathbb{R} \mid |y|<H(|x|)\}.
\end{equation\*}
Here, $H$ denotes a smooth function on $[0,\infty)$.


Let $X\_t$ be the reflecting Brownian motion on $\overline{D}$ and $X\_t^1$ be the first coordinate of $X\_t$. According to Pinsky's heuristic argument (see p.4 of [1](https://projecteuclid.org/download/pdfview_1/euclid.aop/1241099925)), the behavior of $\rho(t)=|X\_t^1|$ over the long run should be like the behavior of the one dimensional diffusion generated by
\begin{equation\*}
\frac{1}{2}\frac{d^2}{d \rho^2}+\frac{C H'(\rho)}{H(\rho)(H'(\rho)^2+1)}\frac{d}{d \rho},
\end{equation\*}
where $C$ is some positive constant. Hence, if $H'(\rho) \to 0$ as $\rho \to \infty$, the above diffusion is not much different from the one generated by 
\begin{equation\*}
\frac{1}{2}\frac{d^2}{d \rho^2}+\frac{C H'(\rho)}{H(\rho)}\frac{d}{d \rho}.
\end{equation\*}


**My question** 


Let $H=\exp(-x^4)$. If we believe the above heuristic argument, the inner drift of $|X\_t^1|$ becomes stronger as $X\_t$ goes to infinity. Hence, I think $X\_t$ is not Feller process by means of $p\_{t}(C\_{\infty}(\overline{D})) \subset C\_{\infty}(\overline{D})$. But I couldn't prove this claim which is seemingly correct.


Do you know how to prove this?



Reflecting Brownian motions on horns are not Feller?
You are right: $X\_t$ is not a Feller process. The following argument is somewhat sketchy, but it should not be too difficult to make it complete.




---


Let us consider the function $$u(x,y) = \tfrac{1}{4} (1 - x^{-2} - 4 y^2) $$ Observe that:


* the normal derivative of $u$ on the boundary of $D = \{(x, y) : |y| < e^{-x^4}\}$ is zero (except, of course, at $(0, 1)$ and at $(0, -1)$);
* $-\tfrac{1}{2} \Delta u(x, y) = 1 + \tfrac{3}{4} x^{-4}$ is everywhere greater than $1$;
* $u(x, y) > 0$ if $(x, y) \in D$ and $|x| > 2$.


The above properties imply that $u(x, y)$ provides an upper bound for the mean hitting time $T\_K$ of $K = \{(x, y) \in D : |x| \leqslant 2\}$ by $X\_t$ (the reflected Brownian motion in $D$) started at $(x, y) \in D$, with $|x| > 2$.


Since $u$ is bounded above by $\tfrac{1}{4}$, the probability that $T\_K \geqslant \tfrac{1}{2}$ is at most $\tfrac{1}{2}$. It follows that with probability at least $\tfrac{1}{2}$, $T\_K$ is less than $\tfrac{1}{2}$.


The above observation and the strong Markov property imply that $\mathbb{E}^{(x,y)}(\mathbb{1}\_K(X\_1))$ is bounded below by a positive constant in $D$, and so $X\_t$ is not a Feller process. To see this, denote $\phi(t, x) = \mathbb{E}^{(x,y)}(\mathbb{1}\_K(X\_t))$, and write
$$
 \begin{aligned} \mathbb{E}^{(x,y)}(\mathbb{1}\_K(X\_1)) & \geqslant \mathbb{E}^{(x,y)}(\mathbb{1}\_K(X\_1) \mathbb{1}\_{\{T\_K \leqslant 1/2\}}) \\
 & = \mathbb{E}^{(x,y)}(\phi(1 - T\_K, X\_{T\_K}) \mathbb{1}\_{\{T\_K \leqslant 1/2\}}) . \end{aligned}
$$
Now $\phi(t, x)$ is jointly continuous and positive, and hence bounded below by a positive constant, say $c$, on a compact set $\{(t, x) : t \in [\tfrac{1}{2}, 1], x \in K\}$. Therefore,
$$
 \mathbb{E}^{(x,y)}(\mathbb{1}\_K(X\_1)) \geqslant \mathbb{E}^{(x,y)}(c \mathbb{1}\_{\{T\_K \leqslant 1/2\}}) \geqslant \tfrac{c}{2} .
$$


