Convex sets and semicontinuous functions

Question

I have this problem: Let $X$ be an open bounded subset of $\mathbb{R}^n,$ and fix $x_0\in X.$ For all $e\in S^{n-1}$ we put $$\phi(e)=\sup\{t\geq0: x_0+te\in X\}$$ $$\overline{\phi}(e)=\inf\{t\geq0: x_0+te\in \mathbb{R}^n-\overline{X}\}$$ I was able to prove that $\phi:S^{n-1}\to[0,\infty)$ is a lower semicontinuous function and $\overline{\phi}:S^{n-1}\to[0,\infty)$ is an upper semicontinuous function. Now, if $X$ is also convex, I would like to prove that these two maps coincide. If $X$ is convex the two sets $A=\{t\geq0: x_0+te\in X\}$, $B=\{t\geq0: x_0+te\in \mathbb{R}^n-\overline{X}\}$ are intervals, and $0\in A.$ Moreover, they are disjoint, since it is not possible that a point belongs to both $A$ and $B,$ and $\phi(e)=\sup{A}\leq\inf{B}=\overline{\phi}(e).$ If $\phi(e)<\overline{\phi}(e),$ I should have an interval of points $t$ such that $x_0+te\in\partial A,$ but I can't see how to conclude. How can I prove that $\phi=\overline{\phi}?$

Answer 1

Consider the set $\{ x_0 + t e: t\geq 0 \} := L $ which is a ray in $\mathbb{R}^n$. In view of boundedness of $X$ the intersection $L \cap \partial X \neq \emptyset$. Let $t_0 > 0$ be such that $p_0: = x_0 + t_0 e \in \partial X$. Thanks to the convexity of $X$ there exists a supporting hyperplane of $X$ through $p_0$, meaning that there is a hyperplane $\mathcal{H} \subset \mathbb{R}^n$ passing through the point $p_0$ and having $X$ entirely on one side of it. From this it follows that the intersection of $L$ with $\partial X$ consists of a single point, which is $p_0 = x_0 + t_0 e$.

We now show that $\phi(e) = \overline{\phi}(e) = t_0$. Indeed, due to the definition of $t_0$, for all $0<t<t_0$ we have $x_0 + t e \in X$ and $x_0 + t_0 e \notin X$, hence $\varphi (e) = t_0$. Similarly for all $t>t_0$ we have $x_0 + te \notin \overline{X}$ and $x_0 + t_0 e \in \overline{X}$, implying that $\overline{\phi}(e) = t_0$.