Some basic topological properties of dual cones

Question

For a cone $K\subseteq\mathbb{R}^n$ (not necessarily convex nor closed), we define its dual cone as $$K^*=\{y\vert x^Ty\ge 0\,\text{for all }x\in K\}.$$ I know that $K^*$ is a closed, convex cone. I would like help proving the following (coming from page 53 in Boyd and Vandenberghe):

1. If the closure of $K$ is pointed (i.e., if $x\in\text{cl} K$ and $-x\in\text{cl} K$, then $x=0$), then $K^*$ has nonempty interior.
2. $K^{**}=\text{cl}(\text{conv}K))$, i.e., $K^{**}$ is the closure of the convex hull of $K$.

First attempts:

• For 1), I began by assuming that the interior of $K^*$ is empty. It follows that since $K^*$ is nonempty and convex that it lies in a hyperplane $H=\{x\vert a^T x=b\}$ for some $a\ne 0$, $b=0$. (I believe that $b=0$ since the origin is contained in $K^*$; I know that, however, $b$ is not necessarily $0$ for arbitrary nonempty convex sets with empty interiors.) Somehow I want this to imply that the closure of $K$ is not pointed, but I can't figure out that argument. The fact that $K^*$ is contained in a hyperplane through the origin clearly gives us symmetry with respect to the origin that I would like to use to show the pointedness of $K$.

• For 2), since $K^{**}$ is closed and convex, we immediately get $\text{cl}(\text{conv}K))\subseteq K^{**}$, so I would like help proving the other direction. Perhaps proving 2) first might help with 1), but I'm not sure.

Answer 1

For a complete solution for a slight variation (K is assumed to be closed convex) of the problem 2, look here: https://math.stackexchange.com/posts/2333573

Here are my attempts to prove them: To show $K^{**} \subseteq cl(convK)$ (the other way is relatively easy to prove)

• assume $\exists y \in K^{**}~|~y \notin cl(convK)$
• $a \cdot y > b, ~ a \cdot x < b ~\forall x\in cl(convK)$ (strict separating plane between closed convex disjoint sets, one of which is compact)
• $0 \in K \implies b > 0 \implies a \cdot y b > 0$
• Using the cone property, $t y\in K^{**}$ (Note $t y \notin cl(convK)$, since $y \notin cl(convK))$ for any arbitrarily small $t > 0$, $a \cdot t y$ can be made arbitrarily small (< any given b) which is a contradiction.

Using this one can prove (1): assuming K is convex. K* has empty interior implies $a \cdot x = b ~\forall~ x \in K^*$, with origin in K*, $b=0$. cl(K) since K is convex $\implies K^{**} = K \implies a, -a \in cl(K)$. $a=0$ implies $K^* = 0$. assuming K* is non-trivial, this would be a contradiction. Possibly this should be included in the problem?

Answer 2

1- Is incorrect: take $$K=\{(x,y): \quad x=y,~ x \leq 0 \} \cup \{(x,y): \quad-x=y,~ x\leq 0 \}\cup \{(x,y): \quad x\ge 0 ,~ y=0 \}$$ Then $K^* = \{0\}$ (this example was for original question before Edit, when convexity of $K$ wasn't assumed) Edit: In 1 further if $K$ is assumed convex, then 1 is true. Proof: following OP's proof since $K\subseteq H$ then $0 \in H$ so $b=0$ and now it is enough to show that $a\in clK \cap - clK $

2-Hint: To show $K^{**} \subseteq \text{cl}(\text{conv}K))$ : Proof by contradiction: assume there exist $y \in K^{**}$ such that $y \notin \text{cl}(\text{conv}K))$, now by sparation theorem , separate $y$ and $\text{cl}(\text{conv}K))$. And see what happens....