# On dual cones: $(K\cap L)^+\subseteq K^+ + L^+$

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:

1. either $$x\in K^+ \cap L^+$$ or
2. 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?

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.

• What is Adh(P)? – Teodorism Jul 15 at 16:43
• I mean the closure, sorry for the frenchism. Will edit. – FXV Jul 15 at 16:45

This is only a partial answer.

First, the claim is obviously true if $$K\subset L$$, which implies $$L^+\subset K^+$$, and $$(K\cap L)^+ = K^+ = K^++L^+$$.

Second, the claim is true if $$K\cap L=\{0\}$$ and $$K,L$$ are closed convex cones. Assume the claim is not true. Then there is $$f \not\in K^++L^+$$. We can separate $$f$$ from $$K^++L^+$$, i.e., there exists non-zero $$x$$ such that $$f^Tx \le k'^Tx + l'^Tx \quad \forall k'\in K^+, l'\in L^+.$$ Since $$K^+$$ and $$L^+$$ are cones, it follows $$k'^Tx\ge 0$$ and $$l'^Tx\ge0$$ for all $$k'\in K^+$$, $$l'\in L^+$$. Hence, $$x\in K^{++} \cap L^{++}=K\cap L = \{0\}$$, contradiction.

I am not sure how to 'interpolate' between those extreme cases. The separation argument is going nowhere if $$K\cap L \ne\{0\}$$. (It yields $$f^Tx=0$$)