The relationship between two definitions of star-shaped domain

Question

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:

1. Are two definitions equivalence?

2. If they are equivalence, how to prove?

3. If they are not equivalence, what relationship between them? any examples for Def2?

Thank you!

Answer 1

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.

Answer 2

Let $\Omega\subset \mathbb{R}^n$ be a bounded Lipschitz domain and let $\nu:\partial\Omega\to\mathbb{R}^n$ be the exterior normal vector field, which is defined a.e..

We will prove the equivalence for a.e. such point and we will assume without loss of generality that $z_0=0$ (so $\Omega$ is starshaped with respect to the origin).

By definition, if $\nu(y)$ is well defined then, there is a ball $B(y,r)$ such that $B(y,r)\cap \partial \Omega$, can be seen as the graph of a Lipschitz function $f:B(0,1)\to B(y,r)$, such that $f$ is differentiable in $f(0)=y$, $f'(0)(e_n)=\nu(y)$ is normal to the hyperplane $f'(0)(\mathbb{R}^n)$.

Let $d\in\mathbb{R}^n$ be such that for small $t>0$, $y+td\in \Omega$. Because of the choosen orientation of $\nu$, we have that $$\nu(y)\cdot (td)\le 0, \forall t \tag{1}.$$

Thus, by taking $d=-y$, we conclude from $(1)$ that $\nu(y)\cdot (-t)y\le 0$ for all small $t\in [0,1]$. Letting $t\to 1$ we have that $\nu(y)\cdot y\ge 0$.

On the other hand, assume that $\nu(y)\cdot y\ge 0$, a.e. $y\in \partial \Omega$. Suppose that there is $y\in \partial \Omega$ and $t\in (0,1)$ such that $ty\notin\Omega$. If $\nu(y)$ exist (if $\nu(y)$ does not exist, we take $y'$ close to $y$ such that $\nu(y')$ is well defined), we can consider two cases:

Case 1: $ty\notin \Omega$ for all $t\in (0,1)$.

In this case we must have $\nu(y)\cdot y=0$. Choose $\eta\in\mathbb{R}^n$ such that $\eta\cdot y<0$ and $\eta\cdot y$ is small. As $\nu$ is oriented outward, there is $y_1$ close to $y$ and $t_1\in (0,1)$, such that $t_1y_1\in \partial \Omega$ and $ty_1\in \Omega$ for $t\in (t_1,1)$. If $\nu (y_1)$ is well defined, we reach na absurd, because $t_1y_1$ points inward $\Omega$. If $\nu(y_1)$ is not defined, we take $y_2$ close to it with the same properties.

Case 2: There is $t_0\in (0,1)$ such that $t_0y\in \Omega$.

Let $t_1$ be the first number in $(0,1)$, for which $(1-t_1)y\in \partial \Omega$ and now apply the same reasoning of case 1.

We conclude from cases 1 and 2 that for each $y\in \partial \Omega$, $ty\in \Omega$ for all $t\in (0,1)$, therefore, $\Omega$ is star shaped with respect to the origin.

Answer 3

If we don't assume the set $S$ to be a $C^{0,1}$ domain to begin with, the unit disk with a cut, i.e. \begin{align*} S = \{ x \in \mathbb{R}^2 \mid |x|<1 \} \setminus [0,1]\times\{0\} \end{align*} is an example satisfying Definition 1 with $x_0 = (-\frac{1}{2},0)$, but not Definition 2.