Polar of revolution cone

Question

Let $s\in\mathbb{R}^n, \|s\|=1$ and $\theta\in[0,\frac{\pi}{2}]$. Consider the cone $$ K=\{x\in\mathbb{R}^n: \langle s, x\rangle\geq \|x\|\cos\theta\}. $$ and $$ K^*=\{x\in\mathbb{R}^n:\langle x, y\rangle\leq 0\quad \forall y\in K\}. $$ Find the closed form of $K^*$

Guess. By using geometrical illustration we guess that $$ K=\{x\in\mathbb{R}^n: \langle -s, x\rangle\geq \|x\|\cos(\frac{\pi}{2}-\theta)\}. $$ But I can not prove this assertion.

I hope someone helps me to give the full solution for this problem.

Answer 1

Let $$ K^{0}=\{x\in\mathbb{R}^n: \langle-s, x\rangle\geq\|x\|\sin\theta\}. $$ It is easy to check that $K^0$ is a cone, i.e., $\lambda x\in K^0$ for all $\lambda>0$. We will prove that $K^0=K^*$.

• $K^0\subset K^*$

Let $x_1\in K, x_2\in K^0$. We have $$ \langle s, x_1\rangle\geq\|x_1\|\cos\theta, \quad \langle-s, x_2\rangle\geq\|x_2\|\sin\theta. $$ It follows that $$ \|x_1-x_2\|=\|x_1-x_2\|\|s\|\geq\langle s, x_1-x_2\rangle\geq \|x_1\|\cos\theta+\|x_2\|\sin\theta. $$ Taking the square both sides of the above inequality $$ \|x_1-x_2\|^2\geq\left(\|x_1\|\cos\left(\theta\right)+\|x_2\|\sin\left(\theta\right)\right)^2. $$ Expanding and reducing the latter inequality we get $$ \langle x_1, x_2\rangle \leq\frac{1}{2}\left(\|x_1\|\sin\theta-\|x_2\|\cos\theta\right)^2. $$ Since $K, K^0$ are cone, it follows that $\langle x_1, x_2\rangle\leq 0$. Hence $K^0\subset K^*$.

• $K^*\subset K^0$

Let $x\in K^*$. We first prove that $s+\frac{x}{\|x\|}\sin\theta\in K$. Indeed, we have $$ \langle s, s+\frac{x}{\|x\|}\sin\theta\rangle=1+\langle s, \frac{x}{\|x\|}\sin\theta\rangle\geq 1-\sin\theta\geq 0 $$ and \begin{eqnarray*} \left(\langle s, s+\frac{x}{\|x\|}\sin\theta\rangle\right)^2&=&\left\|s+\frac{x}{\|x\|}\sin\theta\right\|^2\cos\theta^2+\left(\sin\theta+\frac{x}{\|x\|\sin\theta}\right)^2\sin^2\theta. \end{eqnarray*} Hence $$ \langle s, s+\frac{x}{\|x\|}\sin\theta\rangle\geq \left\|s+\frac{x}{\|x\|}\sin\theta\right\|\cos\theta. $$ It follows that $s+\frac{x}{\|x\|}\sin\theta\in K$. Since $x\in K^*$, $$ 0\geq \langle x, s+\frac{x}{\|x\|}\sin\theta\rangle $$ or equivalently $$ \langle-s, x\rangle\geq \|x\|\sin\theta. $$ Hence $x\in K^0$.