# Proof of $(K_1+K_2)^* = K_1^*\cap K_2^*$: the dual of sum of convex cones is same to the intersection of duals of convex cones

Let $K\subset R^n$ be a cone: $\lambda x \in K$ if $x \in K$ and $\lambda>0$. Let $K^*$ be the conjugate cone of $K$, defined as $K^* = \{x^* | \langle x, x^*\rangle \ge 0, \forall x \in K\}$. I am trying to prove the next theorem and do not understand why convexity is needed for $K_1$ and $K_2$.

Let $K_1$ and $K_2$ be convex cones in $R^n$. Then \begin{equation} (K_1+K_2)^* = K_1^*\cap K_2^*. \end{equation}

My trial is as follows:

If $x^* \in K_1^*\cap K_2^*$, then $\langle x_1+x_2, x^*\rangle = \langle x_1,x^*\rangle + \langle x_2,x^*\rangle \ge 0$ for every $x_1 \in K_1$ and $x_2 \in K_2$, so $x^* \in (K_1+K_2)^*$ and $K_1^* \cap K_2^* \subset (K_1 + K_2)^*$. Suppose $x^* \in (K_1+K_2)^*$ but $x^* \notin K_1^*$. Then, there exists $x_1 \in K_1$ such that $\langle x_1, x^*\rangle < 0$. Since $K_2$ is a cone, $\lambda x_2 \in K_2$ for every $x_2 \in K_2$ and $\lambda>0$. Thus, $\langle x_2,x^*\rangle$ can be arbitrarily small and there exists $x_2 \in K_2$ such that $\langle x_1,x^*\rangle + \langle x_2,x^*\rangle < 0$. It contradicts the assumption that $x^* \in (K_1+K_2)^*$. Therefore, if $x^* \in (K_1+K_2)^*$, then $x^*\in K_1^*\cap K_2^*$ and $(K_1+K_2)^* \subset K_1^*\cap K_2^*$. It completes the proof.

Where should I use convexity of $K_1$ and $K_2$ or can I remove convexity from the Theorem?

## 1 Answer

The convexity of $K_1$ and $K_2$ is not needed (as shown by your proof).

This can also be seen from the calculus rules for the conjugate cone: For an arbitrary set $K$, you have $K^* = \operatorname{clcone}(K)^*$, where $\operatorname{clcone}(K)$ is the closed conic hull of $K$, i.e., the smallest closed, convex cone containing $K$. Using this relation, one can also remove the convexity assumption on $K_i$ (if you would have a proof which requires convexity).

• Thanks for your reply but I am not sure about the closed conic hull part. If $K$ is a singleton with a non-zero point, then $K^* = \{x^*|\langle x, x^*\rangle \ge 0, \forall x \in K\}$ will be a halfspace, while the smallest closed convex cone containing $K$ seems a ray that passes through $x$. – flyingwith Aug 27 '16 at 12:34
• But the conjugate cone of that ray is again the half-space. – gerw Aug 27 '16 at 19:57
• Then $K^* = \mathrm{clcone}(K)^*$ does not mean the conjugate cone of $K$. I was confused by the same notation. Indeed, I cannot understand how to use that relation to remove convexity assumption of $K_i$. Would you give me a hint or a reference about it? – flyingwith Aug 29 '16 at 14:00
• I do not get your first sentence. I said that the conjugate cone of $K$ is the same as the conjugate cone of $\operatorname{clcone}(K)$. If you have proven the theorem for convex cones, you can apply that theorem to $\operatorname{clcone}(K_i)$ (this is always convex). By the above relation, you can remove "$\operatorname{clcone}$" and get the theorem for $K_i$ (which might be non-convex). – gerw Aug 29 '16 at 19:10
• Now I understand your answer clearly. Sorry for my misunderstanding. I found that theorem from one of the Graduate Studies of Mathematics which I am self-studying. After this discovery, I became a little bit curious about how much I should trust the book. Anyway, thanks again. – flyingwith Aug 30 '16 at 12:42