In my studies of cones and convexity I have recently come across the following unexplained piece of information presented:
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 dual of a convex cone $K$ to be defined as follows: $ K^*= \{ y | \forall x \in S : \langle y,x \rangle \geq 0 \} $ Now, the book (in my native language) presents these two facts as ordinary notes without proof:
$ {K_{u,\theta}}^*=K_{u, \frac{\pi}{2}-\theta} $
If L is a vector subspace (of the vector space the convex cones of ours are in) then we have: $ L^* = L^\perp $
I cannot seem to be able to write a formal proof for each of these two cases presented here and I would certainly appreciate help in proving these. I thank all helpers.
(1) Note that $$ K=K_{u,\theta}=\{x|\angle (u,x)\leq \theta \} $$
And for $x\in K$, $$ \angle (x,y)\leq \angle (x,u)+\angle (u,y) \leq \theta + \angle (u,y) $$ Hence $$ \angle (u,y) \leq \frac{\pi}{2}-\theta\Rightarrow x\cdot y\geq 0 $$
That is $$ K_{u,\frac{\pi}{2}-\theta} \subset K^\ast$$
In further for $\angle (u,y)>\frac{\pi}{2}-\theta $ consider a two dimensional plane $$P=(u,w),\ w=\frac{y-y\cdot uu }{|y-y\cdot uu |} $$
Then $$\cos\ \theta u-\sin\ \theta w\in K$$ $$ \angle (\cos\ \theta u-\sin\ \theta w , y) > \frac{\pi}{2} $$ So $y$ is not in $K^\ast$ so that $K_{u,\frac{\pi}{2}-\theta} = K^\ast$
(2) $L$ is a subspace so that $$L=\{v\in \mathbb{R}^d | v\cdot v_i=0,\ 1\leq i\leq m\} $$ for some unit vectors $v_i$ Note that $L^\perp =(v_1,\cdots,v_m)$ In further $L^\perp\subset L^\ast$ If $v=v_L+\sum_i c_iv_i$ for some $v_L\in L$, then $$ v\cdot (-v_L) < 0 $$ Hence $v$ is not in $L^\ast$ so that $L^\perp=L^\ast$