Let $D$ be an open domain in $\mathbb{R}^{2}$ whose boundary $\partial D$ is a 1-dimensional C${}^{\infty}$-submanifold of $\mathbb{R}^{2}$. Given a boundary point $x$ in $\partial D$, let $\nu(x)$ denote the outward normal vector of $D$ at $x$ and let $H_{x}$ be the closed half plane in $\mathbb{R}^{2}$ with $\nu(x)$ as outward normal vector at $x$. Thus
$$H_{x}=\{v\in\mathbb{R}^{2} \ | \ \langle \nu(x),v-x\rangle\leq0\},$$
where $\langle\cdot,\cdot\rangle$ is the notation for the inner product on $\mathbb{R}^{2}$.
Is it true that $\overline{D}$ is convex if and only if $\overline{D}\subset H_{x}$ for all $x\in \partial D$?
Intuitively this is true. However, I can't find a reference for this, nor prove this. Any references or suggestions would be greatly appreciated.
Remark: The above easily generalizes to open domains $D$ in $\mathbb{R}^{n}$.
Best Answer
Since $D \subset H_x$ for all $x \in \partial D$, we see that $\overline{D} \subset H_x$ for all $x$ and so $\overline{D} \subset \cap_{x \in \partial D} H_x$.
Now suppose $y \notin \overline{D}$, then let $x \in \partial D$ be the closest point to $x$ in $\overline{D}$, then $y-x$ is an outward normal at $x$ and hence $y \notin H_x$. In particular, $\overline{D} = \cap_{x \in \partial D} H_x$.
Note: The intersection of an arbitrary number of convex sets is a convex set.