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 $$u(x,y) = \tfrac{1}{8} (1 - x^{-2} - 4 y^2) $$ 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 $(0, -1)$);

• $-\Delta u(x, y) = 1 + \tfrac{3}{4} x^{-4}$ 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 $\tfrac{1}{8}$, the probability that $T_K \geqslant \tfrac{1}{2}$ is at most $\tfrac{1}{4}$. It follows that with probability at least $\tfrac{3}{4}$, $T_K$ 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, $$ \mathbb{E}^{(x,y)}(\mathbb{1}_K(X_1)) \geqslant \mathbb{E}^{(x,y)}(c \mathbb{1}_{\{T_K \leqslant 1/2\}}) \geqslant \tfrac{3 c}{4} . $$

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

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 $$u(x,y) = \tfrac{1}{8} (1 - x^{-2} - 4 y^2) $$ 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 $(0, -1)$);

• $-\Delta u(x, y) = 1 + \tfrac{3}{4} x^{-4}$ 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 $\tfrac{1}{8}$, the probability that $T_K \geqslant 1$ is at most $\tfrac{1}{4}$. It follows that with probability at least $\tfrac{3}{4}$, $T_K$ is less than one.

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.

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 $$u(x,y) = \tfrac{1}{8} (1 - x^{-2} - 4 y^2) $$ 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 $(0, -1)$);

• $-\Delta u(x, y) = 1 + \tfrac{3}{4} x^{-4}$ 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 $\tfrac{1}{8}$, the probability that $T_K \geqslant \tfrac{1}{2}$ is at most $\tfrac{1}{4}$. It follows that with probability at least $\tfrac{3}{4}$, $T_K$ 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, $$ \mathbb{E}^{(x,y)}(\mathbb{1}_K(X_1)) \geqslant \mathbb{E}^{(x,y)}(c \mathbb{1}_{\{T_K \leqslant 1/2\}}) \geqslant \tfrac{3 c}{4} . $$