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

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.

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 $$K$$!