Proof of $\left(K_{1} \cap K_{2}\right)^{*}=K_{1}^{*}+K_{2}^{*}:$ the dual of intersection of convex cones is the sum of their duals

Question

I'm trying to prove the following:

Let $K_{1}$ and $K_{2}$ be pointed closed convex cone in $E$ such that $K_{1 }∩ K_2 \neq \emptyset$.

Show that $\left(K_{1} \cap K_{2}\right)^{*}=K_{1}^{*}+K_{2}^{*}$

under an appropriate assumption on $K_{1}$ and $K_{2}$.

I have managed to show $K_{1}^{*}+K_{2}^{*} \subseteq \left(K_{1} \cap K_{2}\right)^{*} $ without making any assumptions, but I have a hard time showing $\left(K_{1} \cap K_{2}\right)^{*}\subseteq K_{1}^{*}+K_{2}^{*}$.

would appreciate some help

Answer 1

The proof only needs a basic property of dual cone: $$ \begin{align} K_1 \subseteq K_2 \implies K_1^* \supseteq K_2^*, \tag{1} \end{align} $$ which is easy to verify.

Apply (1) to $K_1 \cap K_2$, and $K_1,K_2$, we have \begin{align} \left.\begin{array}{l} (K_1 \cap K_2) \subseteq K_1 \implies (K_1 \cap K_2)^* \supseteq K_1^* \\ (K_1 \cap K_2) \subseteq K_2 \implies (K_1 \cap K_2)^* \supseteq K_2^* \\ \end{array}\right\} \implies (K_1 \cap K_2)^* \supseteq (K_1^* \cup K_2^*). \tag{2} \end{align} Take convex hull of both side of (2). Left side won't change, and right side becomes $K_1^*+K_2^*$: \begin{equation} (K_1 \cap K_2)^* \supseteq K_1^* + K_2^*. \tag{3} \end{equation}

Apply (1) to $K_1^* \cup K_2^*$, and $K_1^*,K_2^*$ similarly, we have \begin{align} \left.\begin{array}{l} K_1^* \subseteq (K_1^* \cup K_2^*) \implies K_1^{**} \supseteq (K_1^* \cup K_2^*)^* \\ K_2^* \subseteq (K_1^* \cup K_2^*) \implies K_2^{**} \supseteq (K_1^* \cup K_2^*)^* \end{array}\right\} \implies (K_1^{**} \cap K_2^{**}) \supseteq (K_1^* \cup K_2^*)^*. \tag{4} \end{align} Apply (1) to (4): \begin{align} (K_1^{**} \cap K_2^{**})^* & \subseteq (K_1^* \cup K_2^*)^{**} \\ \end{align} For any set $K$, dual cone of dual cone is the convex hull of origin set, i.e., $K^{**}=\operatorname{ConvexHull}(K)$, so \begin{align} \implies (K_1 \cap K_2)^* & \subseteq \operatorname{ConvexHull}(K_1^* \cup K_2^*) \\ \implies (K_1 \cap K_2)^* & \subseteq K_1^* + K_2^* \tag{5} \end{align}

(3)(5) implies $$(K_1 \cap K_2)^* = K_1^* + K_2^*$$

Actually, this holds not only for closed convex cone $K_1, K_2$, but also holds for arbitrary set $K_1,K_2$.