# The addition of two dual closed convex cones is a closed convex cone.

Given that $C$ and $D$ are closed convex cones. I need to prove that for dual cones $C^*$ and $D^*$ that $C^*+D^*$ is a closed convex cone. I know that because $C$ and $D$ are closed convex cones, then $C^*$ and $D^*$ are also closed convex cones because of dual cone properties.

Because addition is closed then $C^*+D^*$ is closed.

To show its a cone. I take an $x \in C^*+D^*$ and let $\alpha$ be a positive scalar. Then I let $x=x_1+x_2$ for some $x_1 \in C^*$ and $x_2 \in D^*$ since $C^*$ and $D^*$ are cones $\alpha x \in C^*$ and $\alpha x \in D^*$. Since $\alpha x= \alpha(x_1+x_2)=\alpha x_1+ \alpha x_2$, $\alpha x \in C^*+D^*$.

I am just having problems showing it is convex? I was thinking about starting it similar to showing it was a cone but I get stuck?

## 1 Answer

The set $C^* + D^*$ might not be closed!

As an example, take $$C^* = \{x \in \mathbb R^3 \mid x_1 \ge \sqrt{x_2^2 + x_3^3}\}$$ and $$D^* = \{x \in \mathbb R^3 \mid x_3 = 0 \text{ and } x_1 = x_3\}.$$