TITLE: The relationship between two definitions of star-shaped domain
QUESTION [6 upvotes]: There are two definitions of star-shaped domain. One is given in wikipedia as follows. 
Def1: A set $S$ in the Euclidean space $\mathbb{R}^n$ is called a star domain (or star-convex set, star-shaped or radially convex set) if there exists $x_0$ in $S$ such that for all $x$ in $S$ the line segment from $x_0$ to $x$ is in $S$.
Another is defined in Pohozaev indentity(Pohozaev, S.I.: On the eigenfunctions of the equation $u + \lambda f (u) = 0$. Dokl. Akad.Nauk SSSR 165, 1408–1411 (1965)) as follows.
Def2: Let $\Omega\subset\mathbb{R}^N$ be a bounded and $C^{0,1}$ domain. We say that $\Omega$ is Star-shaped if, for some $z_0\in\mathbb{R}^N$,
$$(x-z_0)\cdot \nu\geq 0,\quad \forall x\in\partial\Omega$$ where $\nu$ is the unit outward normal to $\partial\Omega$ at $x$.
My question:

Are two definitions equivalence? 
If they are equivalence, how to prove?
If they are not equivalence, what relationship between them?  any examples for Def2?

Thank you!

REPLY [3 votes]: Let's clarify the geometric intuition of the two definitions by looking at the case of $\mathbb{R}^2$.
For definition 1, the definition asserts that there is some reference point $x_0$ from which can 'see' every other point $x\in S$; this will fail if there is an 'obstruction' in the way. (In other words, we can draw a line segment between $x$ and $x_0$; if some points on that segment are not in $S$, then the set isn't star-convex.) This is clearly the case in the diagram: $x_1$ is consistent with the condition for it to be part of a star-domain with respect to $x_0$, but not for $x_2$. (Note that it's sufficient to consider points on the boundary.)

What about definition 2? In that case, we instead consider points on the edge compare the angle between the vectors $x-x_0$ and $\nu$, i.e. between the vector that 'sights' a point on the boundary, and the local normal vector. For $S$ to be a star-domain with respect to $x_0$ requires that their angle never be obtuse. And this is again exactly reflected by the diagram: $(x_1-x_0)\cdot\nu_1\geq0$ since the angle between the two vectors is acute, whereas the opposite is true for $x_2$. Therefore we again conclude that $x_1$ can be part of a star-domain wrt $x_0$ but not $x_2$.
To sum up the geometry: If there is an obstruction to seeing a boundary-point $x$ from $x_0$, that means that this boundary must be 'facing towards' the point $x_0$. Conversely, if all boundaries are 'facing away' from $x_0$, then there's no way for any obstructions to exist. Hence both senses of being a star-domain are equivalent.