Dual of dual cone of nonconvex closed cone

Question

let $K$ be a nonconvex closed cone, then

$K^{**}=conv(K)$

should this hold? I am not quite sure about it. Thanks.

Answer 1

Yes, this is true. Recall that $\operatorname{conv} K$ is the intersection of all closed half-spaces containing $K$. Also, the dual cone $K^\circ$ consists of inward normal vectors of all closed half-spaces containing $K$. It follows that $K^\circ = (\operatorname{conv} K)^\circ$. At which point we can apply the duality theorem for closed convex cones (e.g. Theorem 14.1 in Convex Analysis by Rockafellar) and conclude that $K^{\circ\circ} = \operatorname{conv} K$.

In the degenerate case, when $K$ is not contained in any closed half-space, we have $\operatorname{conv} K = \mathbb R^n$ and $K^\circ = \{0\}$, so the statement still holds.