Dual cone's dual cone is the closure of primal cone's convex hull

Question

Assume $K$ is a cone and its dual cone is $K^* = \{y:x^Ty \geq 0,\, \forall x \in K\}$. Then we have $K^{**} = \text{cl}(\text{conv}\ K)$, where cl means closure, conv means convex hull.

How to prove it? (Especially $K^{**} \subseteq \text{cl}(\text{conv}\ K)$ since the other direction's proof is trivial.)

I also found a same question here. The answer only gives a hint and I still don't know how to prove it.

Answer 1

This roughly corresponds to BV 2.31, part (f). The questions are equivalent once we show that $conv(K)$ is a convex cone.

BV 2.31 part (f) asks: Let $K^*$ be the dual cone of a convex cone $K$, as defined above, then show $K^{**} = cl(K)$ (hence if $K$ is closed, $K^{**} = K$).

To prove this, observe that, by definition, for $y \in K^*$, $y \neq 0$ is the normal vector of a (closed, homogenous) halfspace (homogenous means containing the origin). (Closed because we have $\leq$.) And also, that the closure of a convex cone is the intersection of all homogenous halfspaces containing $K$.

Applying these two:

$ cl K = \bigcap_{y \in K^*} \{x | y^Tx \geq 0\} = \{x | y^Tx \geq 0 $ for all $ y \in K^* \} = K^{**}$

Where the last equality follows by the definition of the dual of the cone, taking the dual of $K^*$

Finally, we sketch why $conv(K)$ is in fact a convex cone, and then we are done.

Clearly $C = conv(K)$ is convex. It turns out that it is also cone (and is in fact the conic hull, defined as the smallest convex cone that contains K!)

First, $C$ is a cone: Observe that if $K$ is convex then $C = K$, so wlog have that $K$ is not convex. We can then use induction. In the case of two points, $x_1, x_2$ in $K$, since $K$ is a cone, $\alpha x_1, \beta x_2$ are both also in $K$, by the definition of a cone. By definition of convex hull, $\tilde{x} = \theta(\alpha x_1) + (1-\theta) (\beta x_2) \in C$ for all $\theta \geq 0$. Now observe that $\gamma * \tilde{x}$ is also in $C$ for $\gamma \geq 0$, since $\gamma \tilde{x} = \gamma\theta(\alpha x_1) + \gamma(1-\theta) (\beta x_2) = \theta(\alpha \gamma x_1) + (1-\theta) (\beta \gamma x_2) $, which is another convex combination of two points in K, and so $\gamma * \tilde{x}$ must also be in $C = conv(K)$.

We can then proceed by induction on more than two points in $K$ and show that $C = conv(K)$ must be a cone. Hence $conv(K)$ is a convex cone and the result above can be used.

Fun note: we can also show that $C$ is in fact the conic hull of $K$ as follows:

• $C = conv(K)$ and $D = conichull(K)$. Then by definition, we have $K \subseteq C$ and $K \subseteq D$.
• Since $D$ is the smallest convex cone containing $K$, and since $C$ is a convex cone containing $K$, we know that $D \subseteq C$
• $D$, the conic hull, contains all conic combinations of points in $K$, and $C$ contains all convex combinations of points in $K$. From the definitions of these two combinations, we see that a convex combination is a conic combination (convex combinations are restricted to coefficients $\theta$ that sum to 1, whereas conic combinations do not have this restriction). So $C \subseteq D$
• Thus $C = D$ so the convex hull of a cone $K$ is also the conic hull of $K$!