[Math] Some basic topological properties of dual cones

convex-analysisdual-coneduality-theorems

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.

Best Answer

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?

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