Theorem:Let $K_1,\dots, K_m$ be convex cones in $R^n$ and let $K = K_1 \cap K_2 \cap \dots K_m$. If $K_1 \cap int(K_2) \cap \dots \cap int(K_m) \neq \emptyset$(regularity assumption), then $K^\circ = K_1^\circ + \dots + K_m^\circ$.
My question is why do we call the assumption regular? If it doesn't hold, $K$ could still be a nonempty convex cone I think. Or put it another way, what happens if $K \neq \emptyset$ but $K_1 \cap int(K_2) \dots \cap int(K_m) = \emptyset$? Why is this case irregular? Is it trivial?
I don't if I am in the right direction, but I'll put my thought here. If regular assumption doesn't hold, then $int(K) = \emptyset$, then $K$ must be contained in a lower dimension linear manifold(or affine set if you will). But even regular assumption holds, $int(K)$ could still be empty, since we can make $int(K_1) = \emptyset$. So I am not sure why we distinguish these two cases. Just for the theorem holds?
Yes, you need the regularity, otherwise the theorem may fail. It is quite insightful if you try to construct counterexamples.
If all $K_i$ are also closed, you have $$K^\circ = \operatorname{cl}(K_1^\circ + \ldots +K_m^\circ)$$ without further assumptions. It also helps to construct examples where the closure on the right-hand side is needed.