Let $C$ be a closed convex cone and $\nu$ be an outer support vector, i.e. $\langle\nu, x\rangle\le 0$ for all $x\in C$. Assume $\nu$ is maximized uniquely at some point $y\in C$. I am trying to show that the angle between $\nu$ and a unit vector $x\in C$ is minimized when $x$ is on an extreme ray of $C$. Let $\theta$ denote the angle between $\nu$ and $x$, then $$ \cos \theta = \frac{\langle \nu, x\rangle}{\|\nu\|} $$ Since $\nu$ is fixed, minimizing $\theta$ is the same as maximizing $\langle\nu, x\rangle$. But $\nu$ is maximized uniquely at $y$. So all I need is to show that $y$ is on the extreme ray of $C$. But I don't know how exactly I can do that. The choice of $y$ seems arbitrary. Can someone give me a hint? Thanks a lot in advance!
EDIT: I made some progress. Now, suppose $\langle\nu, x\rangle$ is maximized at some unit vector $x \in C$ and $x \in y + C \subseteq C$ for some $y$. Then $x - y \in C \Rightarrow \langle\nu, x-y\rangle \le 0 \Rightarrow \langle\nu, x\rangle \le \langle\nu, y\rangle \le 0$. Let $\hat{y} = \frac{y}{\|y\|}$ and write $y = \|y\|\hat{y}$, we have $\langle \nu, x\rangle \le \|y\|\langle\nu, \hat{y}\rangle \le 0$. Since $\|y\| \ge 0$ and $\langle\nu,x\rangle \ge \langle\nu, \hat{y}\rangle$ because $\hat{y}$ is a unit vector, in order for the inequality to be satisfied, it must be that $\|y\| \le 1$. Now suppose $\langle \nu,x\rangle > \langle\nu, \hat{y}\rangle$, which implies that $\langle\nu, x - \hat{y}\rangle > 0$. But $x - \hat{y} \in C$, which means $\langle\nu, x - \hat{y}\rangle \le 0$, a contradiction. So $x = \hat{y}$, i.e. $y \in \mathbb{R}_{\geq 0}x$. So $x$ lies on an extreme ray of $C$. But it's not very clear to me why $x - \hat{y}$ must be in $C$. Can someone point it out? Thanks!