On a core for Neumann Laplacian on $C(\overline{D})$ Let $D \subset \mathbb{R}^d$ be a bounded $C^1$ domain. We consider a reflected Brownian motion $X=(\{X_t\}_{t \ge 0},\{P_x\}_{x \in \overline{D}})$ on $\overline{D}$. Let $\{p_t\}_{t>0}$ denote the semigroup of $X$ (in other words, $\{p_t\}_{t>0}$ is the Neumann semigroup). It is known that $\{p_t\}_{t>0}$ is a feller semigroup. That is, we have $p_t f \in C(\overline{D})$ for any $t>0$ and $f \in C(\overline{D})$. Here, $C(\overline{D})$ is the space of continuous functions on $\overline{D}.$ We always equip $C(\overline{D})$ with the sup-norm so that it becomes a real Banach space.
Let $(L,D(L))$ be the Neumann Laplacian on $C(\overline{D})$. That is,
\begin{align*}
D(L)&=\left\{ f \in C(\overline{D})\mid \lim_{t \to 0}\frac{p_t f-f}{t} \text{ exists in }C(\overline{D})\right\},\\
Lf&=\lim_{t \to 0}\frac{p_t f-f}{t}.
\end{align*}
Let $\nu$ denote the inward unit normal vector on $\partial D$, and define \begin{align*}
\mathcal{C}=\{f \in C^2(\mathbb{R}^d)|_{\overline{D}} \mid \langle \nabla f(x),\nu(x) \rangle=0\text{ for every $x \in \partial D$}\}.
\end{align*}
Can we prove that $\mathcal{C}$ is a core for $(L,D(L))$?
 A: I think $\mathcal C$ is not even dense in the space of continuous functions!
To be specific: consider a 2-D domain $D$ lying above the graph of a $C^1$ function $\phi : \mathbb R \to \mathbb R$, and assume that $\phi'$ is continuous, but nowhere differentiable — say, a generic sample path of the Wiener process.
Let $f$ be a $C^2$ function on $\mathbb R^2$ such that $\nabla f$ is orthogonal to the normal vector at each boundary point. This means that $$\partial_y f(x, \phi(x)) = \phi'(x) \partial_x f(x, \phi(x)).$$ On every interval where $\partial_x f(x, \phi(x)) \ne 0$, we find that $$\phi'(x) = \frac{\partial_y f(x, \phi(x))}{\partial_x f(x, \phi(x))}$$ is a $C^1$ function, a contradiction. Thus, $\partial_x f(x, \phi(x)) = 0$ for all $x$, and consequently also $\partial_y f(x, \phi(x)) = 0$ for all $x$. In particular, $f(x, \phi(x))$ is constant.
But that means that $\mathcal C$ only contains functions $f$ which are constant on the boundary, so $\mathcal C$ is not even dense in the space $C(\overline D)$ of continuous functions.
The same argument works for $C^{1,\alpha}$ domains, at least when $\alpha < 1$.
