# Reflecting Brownian motions on horns are not Feller?

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), 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?

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, $$\mathbb{E}^{(x,y)}(\mathbb{1}_K(X_1)) \geqslant \mathbb{E}^{(x,y)}(c \mathbb{1}_{\{T_K \leqslant 1/2\}}) \geqslant \tfrac{c}{2} .$$

• Thank you for your reply. By using Skorohod type SDE for $X$ and Ito's forumula, I was able to obtain $P_{(x,y)}(T_K<1) \ge 3/4$. However, how can I use strong Markov property of $X$ to get lower estimate of $E_{(x,y)}[1_{K}(X_1)]$? – sharpe Mar 22 '18 at 6:52
• @sharpe: I expanded the last part of my answer (and fixed a typo: what I wanted to write is $T_K < \tfrac{1}{2}$, sorry). – Mateusz Kwaśnicki Mar 22 '18 at 9:01
• Thank you for your reply. This is a minute thing, but $P^{(x,y)}(T_{K} \ge 1/2) \le 1/2$, right? According to my calculations, for large $x$, $$0 \le E^{(x,y)}[u(X_{T_K})]<u(x)-(1/2)E^{(x,y)}[T_K].$$ This implies $E^{(x,y)}[T_K] <1/4$ and $P^{(x,y)}(T_{K} \ge 1/2) \le 1/2$. – sharpe Mar 22 '18 at 9:56
• @sharpe: Oh, yes, you are right! (I am used to the convention that a BM is generated by $\Delta$, not $\tfrac{1}{2} \Delta$). I updated the definition of $u$ to compensate this $1/2$. – Mateusz Kwaśnicki Mar 22 '18 at 10:17
• I understood. Thank you very much for teaching me carefully! – sharpe Mar 22 '18 at 10:21