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.

  • $\begingroup$ What is Adh(P)? $\endgroup$ – Teodorism Jul 15 at 16:43
  • 1
    $\begingroup$ I mean the closure, sorry for the frenchism. Will edit. $\endgroup$ – 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 $f^Tx=0$)


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