Suppose $A,B$ are closed cvx cones. $A^*,B^*$ are their dual cones respectively.
How to show $$ (A^*+B^*)^*\subset A \cap B$$
My idea is:
$$A^* = \{x_1|x_1^Tx \geq 0, \forall x \in A\}$$
$$B^* = \{y_1|y_1^Ty \geq 0, \forall y \in B\}$$
$$(A^*+B^*)^* = \{z|z^T(x_1+y_1) \geq 0, \forall (x_1+y_1) \in (A^*+B^*)\}$$
So, to show "$\subset A \cap B$", we have to show $z \in A \cap B$.
My question is:
I am confused about $A \cap B$, since we cannot say $z \in A \cap B$ by proving $x_1^Tz \geq0$ and $y_1^Tz \geq 0$. It is because such $z$ could be ou of $A \cap B$.
The other way is by contradiction. However, in this way we have to consider three case: since $z \notin A \cap B$, it could be in $A$ but not in $B$, or $B$ not in $A$ or neither $A$ nor $B$. A little bit cumbersome
How to use the property of $A \cap B$?
Let $x\in(A^*+B^*)^*$. So $\forall{y}\in A^*+B^*, (x,y)\ge0$. It means that $\forall{y}\in A^*, (x,y)\ge0$ so $x\in A^{**}$. Since $A$ is closed $A^{**}=A$ and $x\in A$. Similarly we can show that $x\in B$. So $x\in A\cap B$