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 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
 A: 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 $K_1,K_2$.
