In my studies of cones and convexity I have recently come across the following problem:
We consider a Euclidean space $ R^d $ for $ d \geq 1 $ we look at the circular cone for a unit vector $ u \in R^d $ and an angle $ 0 < \theta < \frac{\pi}{2} $ we define the circular cone as usual $ K = K_{u,\theta}=\{x \in R^d | \langle x,u \rangle \geq \|x\|\cos{\theta} \} $. We define the extreme vectors of a convex cone as follows: $ x \in K $ is an extreme vector if $ x=y+z $ for $ y,z \in K $ implies both y,z are nonnegative multiples of $x$.
We are asked to show that if $x$ forms an angle less than $ \theta $ with $ u $ then $ x $ is not an extreme vector of $ K_{u,\theta} $.
Now I can see this intuitively but I cannot seem to be able to write out a formal proof, I am asking here in the hopes of getting some help on this. I thank all helpers.
Since $\angle (x,u)<\theta$ so there is $\theta_1$ s.t. $$ K_{\frac{x}{|x|},\theta_1} \subset K_{u,\theta } $$ where $$0<\theta_1,\ \angle (x,u) +\theta_1 < \theta $$
If $P$ is any hyperplane in $\mathbb{R}^d$ containing a line segment $Ox$ where $O$ is origin, then let $$ v\neq 0 \in P\cap\partial K_{\frac{x}{|x|},\theta_1} $$
Consider a two-dimensional plane $P_1$ containing a triangle $ \Delta Ovx$ Then we have $$ w\neq 0 \in P_1\cap \partial K_{\frac{x}{|x|},\theta_1} $$ s.t. $$ w\neq Cv $$ for all $C>0$ Then $$ x= |x|\bigg( \frac{v}{|v|} + \frac{w}{|w|} \bigg) $$
By construction of $v,\ w$ we have that $v,\ w$ are not multiple of $x$ So we complete the proof
Another : $x$ is interior pt so that there is $\delta>0$ s.t. $f(t):=tx+(1-t)u $ is interior pt for $|t-1|\leq \delta$ Hence $x= \frac{1}{2}\{f(1+\delta)+f(1-\delta)\}$