The interior of the set of continuously differentiable nonnegative functions Let $\Omega \subset \mathbb{R}^N$ be a bounded open set with smooth boundary. Consider the space $$C^1_0(\overline{\Omega}) := \{u \in C^{1}(\overline{\Omega}) : u = 0\text{ in }\partial \Omega\}$$
with the norm
$$
\|u\| = \|u\|_{\infty} + \|\,|\nabla u|\,\|_{\infty}
,\quad \forall u \in C^1_0(\overline{\Omega}).$$
Also consider $P = \{u \in C^1_0(\overline{\Omega}) : u \geq 0\}$. How can I show that $int(P) = A$, where
$$
A = \{ u \in C^1_0(\overline{\Omega}) : u(x) > 0, \forall x \in \Omega \text{ and }\langle \nabla u(x), \eta(x)\rangle < 0, \forall x \in \partial \Omega\} ?
$$
Given a point $u \in A$, I know that it's graphic has a specific behavior in the boundary, because the cosine of the angle between the gradient and the normal vectors in the boundary is negative. I imagine that a ball around it should be a "curved cover" involving its graphic. However, I didn't succeed in showing that there's a very small $r > 0$ such that the functions on the open ball $B_{r}(u)$ should be nonnegative.
I would appreciate any help.
 A: Here's an idea of how I think the proof can be done.
Let $u \in A$ and denote by $\partial_\eta$ the outer normal derivative. Since $\partial \Omega$ is compact, there exists $\delta>0$ such that $\sup_{x \in \partial \Omega}\partial_\eta u(x) < - \delta <0$. Now, since $u \in C^1(\overline{\Omega})$, there is a neighborhood $N$ of $\partial \Omega$ on which $\partial_\eta u \leq - \delta/2$ (use the continuity of $\partial_\eta u$ in $x \in \partial \Omega$ with $\varepsilon=\delta/2$ combined with the compactness of $\partial \Omega$). Let
$$\alpha=\inf\{u(x):x \in \Omega \setminus N\}$$ and
$$\beta=\min\{\alpha, \delta/2\}.$$
We claim that $B(u,\beta) \subset P$.
Indeed, let $v \in C_0^1(\overline{\Omega})$ such that $\|u-v\|<\beta$. Then,
$$|u(x)-v(x)|<\alpha, \hbox{ for all } x \in \Omega.$$
In particular, if $x \in \Omega\setminus N$, we obtain that
$$-\alpha<u(x)-v(x)<\alpha,$$
which implies that
$$0\leq u(x)-\alpha<v(x),$$
since $u(x) \geq \alpha$ in $\Omega \setminus N$.
If $x \in N$, we obtain that
$$\left|\partial_\eta u(x)-\partial_\eta v(x) \right|=|\langle \nabla (u-v)(x), \eta \rangle | \leq |\nabla(u-v)(x)| <\delta/2, \hbox{ for all }  x \in N.$$
Therefore,
$$\partial_\eta v(x) < \partial_\eta u(x)+\delta/2<-\delta+\delta/2,\hbox{ for all } x \in N.$$
By using this question: we obtain that
$$v(x) >0 \hbox{ for all } x \in N.$$
Consequently, $v(x) \geq 0$ for all $x \in \overline{\Omega}$, which implies that $B(u,\beta) \subset P$ and hence, $u \in \operatorname{int}(P)$.
