Inner product identity for cones

Question

Let $\emptyset \neq C\subseteq \mathbb R^n$ be a convex, open cone with the property that $\operatorname{int }C^* \neq \emptyset$, where $C^*$ denotes the dual cone defined by $$C^* = \{x \in \mathbb R^n: \langle x,y \rangle \geq 0 \quad \forall y \in C\}.$$ (always a closed and convex cone). Then we have for each $y\in C$ $$\inf_{x\in C^* \cap S^{n-1}} \langle x,y \rangle \geq c_y \lVert y \rVert$$ for some constant $c_y >0$. I was unable to show this. I know that $C^* \cap S^{n-1}$ is compact and the inner product is continuous, so this attains a maximum. But I failed to see why it is impossible for this minimum to be $0$. Any help appreciated!

Answer 1

Big Hint:

I think the key is that $C$ is a non-empty open cone.

If $C$ were a close cone, then $c_y$ would be 0 for any non-zero $y$ on the boundary of $C$. For example, suppose $C\subset R^2$ and $C=\{(x,0) | x\geq 0\}.$ Then $C^*=\{(x,y)| x\geq 0, y\in\mathbb R\}$. Notice that $C^*\cap S^{n-1} = \{(\sin t, \cos t) | 0\leq t\leq \pi\}$, so for any fixed $y\in C$, $y=(y_0, 0)$ and $$ \inf_{x\in C^*\cap S^{n-1}} \langle y,x\rangle = \langle(y_0, 0),(0,1)\rangle=0. $$ Hence $c_y$ would be 0.

On the other hand, if $C$ is a open cone and $y\in C\setminus\{0\}$, then there exist a positive number $\epsilon>0$ such that $B(y,\epsilon)\subset C$. We know that $\langle c^*, y_b\rangle \geq 0$ for all $y_b\in B(y,\epsilon)$ and $c^* \in C^*$ by the definition of $C^*$ and the fact that $B(y,\epsilon)\subset C$. That implies that $\inf_{y_b \in B(y,\epsilon) } \langle c^*, y_b\rangle \geq 0$ for all $ c^*\in C^*$. So now we can write that for any $c^*\in S^{n-1}\cap C^*$, $$\langle B(y, \epsilon), c^* \rangle \geq 0,$$ $$\langle y - \epsilon c^*, c^* \rangle \geq 0,$$ $$\langle y , c^* \rangle - \epsilon||c^*||^2\geq 0,....$$

That line of reasoning should lead you to the answer.