# The dual of a circular cone

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:

1. ${K_{u,\theta}}^*=K_{u, \frac{\pi}{2}-\theta}$

2. 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 $L^\perp=L^\ast$

• I think you have a mistake in (1) you showed both inclusions as the same direction and u for the P is already reserved for the unit vector – kroner Oct 11 '16 at 10:55
• First : Draw vectors $u,\ w$ on $\mathbb{R}^2$ which is related to orientation Second : note that $u\cdot (\cos\ \theta u-\sin\ \theta w)=\cos\ \theta$ – HK Lee Oct 11 '16 at 11:24
• thanks but what if I am in $R^d$ for $d>2$ does this still work? – kroner Oct 11 '16 at 11:24
• Yes $P$ is two-dimensional $subspace$ – HK Lee Oct 11 '16 at 11:26
• I can not understand your question $\cos\ \theta u$ is just multiplication of scalar and vector – HK Lee Oct 11 '16 at 11:30