This question just popped on my mind.
Let $A, B$ two disjoint, nonempty convex sets in the vector space $X$, can they be separated via a nonzero linear function in $X' = \{ f : X \to R ~ | \quad \text{f is linear} \} ?$ i.e., does there exist $f \in X' \setminus \{ 0\}$ such that
$$ f(a) \leq f(b) \quad \forall a\in A, ~ \forall b \in B $$
If not under what minimal condition one can separate them.
My Thought : Since $A \cap B = \emptyset $ using Zorn Lemma we can find two disjoint maximal convex sets, say $U, ~ V$ such that $ A \subseteq U, ~ B \subseteq V $ and through maximality of $U, V$ we can deduce that $U \cup V = X$ in other words $U,~ V$ make a convex partition of the space. Now from this, can we say that $U, ~V$ are two sides of a hyperplane ? i.e., $$ U \subseteq \{ x \in X ~ | \quad f(x) \leq \alpha \} , ~ V \subseteq \{ x \in X ~ | \quad f(x) \geq \alpha \} $$
for some $f \in X'$ and $\alpha \in \Bbb R$
Question #2: What if we assume $A, B$ are pointed cones with $A \cap B = \{0\}$
EDIT: I realized the answer of question # 1 is No generally see below link
https://math.stackexchange.com/q/929690/219176
But Still any answer regarding minimal conditions that guarantees separation is my main interest, and an answer for question #2.
Thank for your help.
