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.

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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....

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  • $\begingroup$ Thanks. For 1), the problem that I was having was that, as written, $K$ was not assumed to be convex. I believe 1) is correct when we add this assumption? @Ashkan $\endgroup$ – Satana Jun 4 '17 at 20:59
  • $\begingroup$ @Atsana Clearly Yes.. easy to proof $\endgroup$ – Red shoes Jun 4 '17 at 21:02

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