TITLE: On dual cones: $(K\cap L)^+\subseteq K^+ + L^+$
QUESTION [5 upvotes]: I am trying to derive the conditions under which $(K\cap L)^+\subseteq K^+ + L^+$  where $K^+$ (and $L^+$) denotes the dual cone of convex cone $K$ (and $L$), and $K+L$ denotes the Minkowski sum of $K$ and $L$. (I've read somewhere that the condition is $K\cap \textrm{int} L\neq \emptyset.$)
To start off, suppose $x\in(K\cap L)^+ $. Then, $\langle\ x, \phi\rangle \geq 0,\ \forall \phi\in K\cap L$. In order for $x$ to be in $K^+ + L^+$, it should be decomposable into $x=x_1+x_2$ such that $x_1\in K^+,\ x_2\in L^+$. One way I can think of to find such a decomposition is to project $x $ on to $K^+$ and  $L^+$ and (since dual cones are cones themselves) use appropriate scalars on the respective projections to get $x_1$ and $x_2$ such that their sum is equal to $x$. For these projections to add up nicely to $x$, I'm thinking that:

either $x\in K^+ \cap  L^+$ or
if $ x\notin K^+ \cap  L^+$, $x$  should make an acute angle with the closest boundaries/faces of the dual cones.

Where am I going wrong?

REPLY [2 votes]: It is straightforward that $P^+\cap Q^+ \subset (P+Q)^+$.
Conversely if $x \in (P+Q)^+$, assuming that $P$ and $Q$ are non-empty cones you have that $0\in Cl(P)$ and $0 \in Cl(Q)$ whence $x\in P^+$ and $x\in Q^+$.
So we have $P^+\cap Q^+=(P+Q)^+$ for any two non-empty cones $P, Q$.
Taking the dual cone we get that:
$$Cl(P+Q) = (P^+ \cap Q^+)^+$$
so if you let $P=K^+$ and $Q=L^+$ (which are two closed clones) and using the fact that the bidual is the closure you get that:
$$K^+ + L^+ = (Cl(K) \cap Cl(L))^+ $$
So the question you are left with is: can the dual cone of $Cl(K)\cap Cl(L)$ be smaller than that of $K\cap L$?
It can be the case for instance if $K$ is a half-line on the boundary of $L$ (open), in which case $K\cap L=\emptyset$ and $Cl(K)\cap Cl(L)=K$.
A sufficient condition is obviously that $K$ and $L$ are closed.
I don't think the condition $K\cap Int(L) \not= \emptyset$ is necessary, for instance you could have $K$ and $L$ two distinct half-lines - they are closed but their interior is empty.
A necessary condition is that $Cl(K)\cap Cl(L)=Cl(K\cap L)$, using again the fact that the bidual is the closure. Because the dual of $K\cap L$ and that of its closure are the same, this condition is also sufficient. Not sure we can make this condition more explicit.