If $A, B \subset \mathbb R^n$ are closed, convex, and disjoint, is there a vector $a \in \mathbb R^n$ such that $a^t x < a^t y$ for all $x \in A$ and $y \in B$? I found many theorems requiring one of the sets to be compact or open or only implying that $a^t x \le c \le a^t y$ for some constant $c$. Furthermore, I found counterexamples for statements like $a^t x < c < a^t y$ which is however stronger than the condition in my question or for the infinite dimensional case. Does anyone know a source or a counterexample?
Thanks in advance
If one of the two sets is bounded, the statement is true. One even gets the stronger result you mentioned.
For unbounded sets it is false. I do not know if there are really simple counterexamples (e.g. twodimensional?), but I would try something like:
With $n=3$, set $A = \{x \colon x_2 = x_3 = 0\}$ and $$B= epi f = \{x \colon x_3 \geq f(x_1,x_2) \}.$$
If $f$ is convex, then both sets are convex and closed. Furthermore, as long as $f(x_1,0)>0$, one also has $A \cap B = \emptyset$.
Now set $f(x_1,x_2) = 0$ for $x_2 \geq e^{-x_1}$ and $f= (e^{-x_1} - x_2)^2$ for $x_2 \leq e^{-x_1}$. This should satisfy the properties above and you can see that any plane containing $A$ also intersects $B$.