0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $K$!
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.