I have a question about functions satisfying a condition.

Let $D \subset \mathbb{R}^d$ be a Lipschitz domain. That is, for each $x \in \partial D$, there exists an open neighborhood $U$ of $x$ in $\mathbb{R}^d$ and a bi-Lipschitz function $\psi_{x}:B(1) \to U$ such that $\psi(0)=x$ and $\psi_{x}(B_{+}(1))=U \cap D$. Here, $B(1)=\{x \in \mathbb{R}^d \mid |x|<1\}$ and $B_{+}(1)=\{x=(x_1,\ldots,x_d) \in B(1) \mid x_d>0\}$.

We denote $H^{1}(D)$ by the 1-st order Sobolev space on $D$ with Neumann boundary condition: $$H^{1}(D)=\{f \in L^{2}(D,dx)\mid\frac{\partial f}{\partial x_i} \in L^{2}(D,dx),\ 1\le i\le d\}.$$ Here, $\frac{\partial f}{\partial x_i}$ is the distributional derivative of $f$. We also denote $(L,\mathcal{D}(L))$ be the Neumann Laplacian on $D$. We note that \begin{equation*} D(L):=\{f \in H^{1}(D) \mid H^{1}(D) \ni g \mapsto \mathcal{E}(f,g) \text{ is continuous w.r.t. } L^{2}(D,dx)\text{-topology}\}, \end{equation*} where $\mathcal{E}(f,g)=\sum_{i=1}^{d}\int_{D}\frac{\partial f}{\partial x_i}\frac{\partial g}{\partial x_i}\,dx$.

**My question**

Fix $p,q \in \partial D$ with $p \neq q$. Then, can we find an $f \in D(L)\cap C(\bar{D})$ such that $f(p) \neq f(q)$ ?

**My attempt**

Fix an open neighborhood $U$ of $p$ in $\mathbb{R}^d$ and a bi-Lipschitz function $\psi:B(1) \to U$ such that $\psi(0)=p$ and $\psi_{p}(B_{+}(1))=U \cap D$. We may assume $p,q \in U$. Since $$\psi^{-1}(p)=0\text{ and }\psi^{-1}(q) \in \{x \in B(1)\mid x_d=0\},$$ it is easy to construct a smooth function $F$ on $B(1)$ such that $\text{supp}[F] \subset B(1)$ and $$F(\psi^{-1}(p))\neq F(\psi^{-1}(q))\text{ and }\left.\frac{\partial F}{\partial x_d}\right|_{x_d=0}=0.$$

That is, $F$ satisfies the Neumann boundary condition in the sense that $\sum_{i=1}^{d}\frac{\partial F}{\partial x_i}n_i=0$. Here, $n=(n_1,\ldots,n_d)$ denotes the outward unit normal to the boundary of upper half space.

Clearly, $\tilde{F}(x)=F(\psi^{-1}(x))$ satisfies $\tilde{F}(p) \neq \tilde{F}(q)$. **Can we show the following?** $$(1)\quad(\nu,\nabla \tilde{F}):=\sum_{i=1}^{d}\nu_i\frac{\partial \tilde{F}}{\partial x_i}=0\ \mathcal{H}^{d-1} \text{-a.e. on }\partial D.$$ Here, $\mathcal{H}^{d-1}$ is the $(d-1)$-dim Hausdorff measure and $\nu$ is the outward unit normal to $\partial D$. If (1) holds, I think $\tilde{F} \in D(L)$ follows by using Green's formula:
\begin{equation*}
-\int_{D}\frac{\partial^2 \tilde{F}}{\partial x_i^2}f\,dx=\int_{D}\frac{\partial f}{\partial x_i}\frac{\partial \tilde{F}}{\partial x_i}\,dx+\int_{\partial D}f (\nu,\nabla \tilde{F})\,d\mathcal{H}^{d-1}.
\end{equation*}