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?
Best 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).