the boundary of a star shaped domain has a subset for which the elements stand at an obtuse angle on the outer normal vector I'm trying to understand a section in (PUCCI, PATRIZIA, and JAMES SERRIN. “A General Variational Identity.” Indiana University Mathematics Journal, vol. 35, no. 3, 1986, pp. 681–703.):
Let $\Omega\subset\mathbb{R}^n$ be a smooth, bounded, star shaped domain with respect to the origin and let $\nu(x)$ be the outer normal at $x\in\partial\Omega$. It is stated that this implies $x\cdot\nu(x)\geq0$ on $\partial\Omega$, and there even is a subset of $\partial\Omega$ of positive measure on which $x\cdot\nu(x)>0$.
I understand the first part but why is there is a subset on which $x\cdot\nu(x)>0$?
 A: Suppose $\partial \Omega$ can be described as a smooth surface. It follows that $\nu(x)$, the outward normal, is continuous. Thus $h(x) = x\cdot \nu(x)$ is a continuous function. If we have that $h(x_0) > 0$ at some point $x_0 \in \partial \Omega$, then, since $h$ is continuous, there exists $\epsilon > 0$ such that $h(x) > 0$ for all $x \in \partial \Omega$ such that $d(x, x_0) < \epsilon$. This suffices to show that $x \cdot \nu(x) > 0$ on a subset of $\partial \Omega$ of positive measure.
All we have to do now is show that such an $x_0$ really does exist. Since we have already assumed $\partial \Omega$ is a smooth surface, there exists an open subset $U$ of $\partial \Omega$ not containing $0$ which can be described as the $x \in \mathbb{R}^n$ such that $g(x) = 0$ for some smooth function $g : U \to \mathbb{R}$. The upshot of defining such a $g$ is that now we can write
$$
\nu= \frac{\nabla g}{|\nabla g|}
$$
We now try to find the critical points of $|x|^2$ on $U$. If there are no critical points, then $|x|^2 = \text{constant}$, i.e. $U$ is a subset of a sphere. As you can check, $x \cdot \nu(x) > 0$ on a sphere, so we are done. Suppose now there is a critical point $x_c \in \partial \Omega$. Then $x_c$ is a critical point of the function $C : \mathbb{R}^n \to \mathbb{R}$ given by
$$
C(x) = |x|^2 - \lambda g(x)
$$
for some $\lambda \in \mathbb{R}$ (this is the method of Lagrange multipliers). We calculate
$$
\nabla C(x) = 2x - \lambda \nabla g(x)
$$
Thus, since $\nabla C(x_c) = 0$
$$
x_c = \frac{\lambda}{2} \nabla g(x_c)
$$
But now
$$
x_c \cdot \nu = \left( \frac{\lambda}{2} \nabla g\right) \cdot \left( \frac{\nabla g}{|\nabla g|}\right) = \frac{\lambda |\nabla g|}{2}
$$
We know that we must have $\lambda \geq 0$ since $x_c \cdot \nu \geq 0$. If $\lambda = 0$, then we must have $x_c = 0$, since the only critical point of $C(x) = |x|^2$ is $0$. But we already said that $U$ does not contain $0$. Thus, it must be the case that $\lambda > 0$. Therefore $x_c \cdot \nu > 0$, and we are done.
I have to confess this argument doesn't exactly seem airtight, but it's the best I could think of. Maybe someone else can identify some holes in it.
