The dual of a convex cone is defined as $K^{*} = \{ y : x^{T}y \geq 0 \text{ for all } x \in K \}$.
Dual cone $K^{*}$ is apparently always convex, even if original $K$ is not.
I think I can prove it by the definition of the convex set. Say $x_1, x_2 \in K^{*}$ then $ \theta x_1 + (1-\theta) x_2 \in K^{*}$
But I don't quite get what the official solution says:
$K^{∗}$ is the intersection of a set of homogeneous halfspaces (meaning, halfspaces that include the origin as a boundary point). Hence it is a closed convex cone.
Can someone explain?
Best Answer
The closed halfspaces are $H_x:=\{y\in\Bbb R^n\,:\, x^Ty\ge 0\}$ and $K^*=\bigcap_{x\in K} H_x$. Each closed halfspace is closed and convex. If it contains the origin (which these do), then it is a cone. Intersection of cones (resp. closed sets / convex sets) is a cone (resp. a closed set / a convex set).