I am trying to justify how the Hahn-Banach theorem was applied in the proof below. It looks like the proof is using the case for locally convex space (because the inequalities are strict). that corresponds to the second bullet point in the following screenshot form Wikipedia:
while in my book it is pretty much the same:
According to Wikipedia, Hahn-Banach requires the sets $A,B$ that are to be separated to be non-empty, convex, disjoint. Additionally, one of the sets should be closed while the other should be compact.
In the proof that I am reading it seems like $B=\text {epi} f$ which is closed and convex and non-empty by the arguments given at the beginning of the proof. Now who is $A$? I tried to find it like this:
Since $B$ is closed, $B^C$ is open, then for the point $(x_0,t_0)\in B^C$ ,$\exists$ an open $U \subseteq B^C$ such that $(x_0,t_0)\in U$. Then since the space is locally convex (every open neighborhood of a point contains a convex neighborhood of the point ), then $\exists A\in U, A$ convex, $(x_0,t_0)\in A$ . Obviously $A$ and $ B$ are disjoint, but still I am not able to argue that $A$ is compact. How do I do that? In case it is not possible how do I find an open A to use the first bullet, but in that case I don't know either how to justify the strict inequalities at both ends.
This is the initial part of the proof that I am reading:
(book: A Course on Large Deviations with an
Introduction to Gibbs Measures
Firas Rassoul-Agha, pag 53)
The compact set you'll be using is the singleton $\{(x_0, t_0)\}$. The theorem gives you the inequalities in the wrong way but you can just take $\phi := -f$ to reverse them if $f$ is the linear functional from the Hahn-Banach theorem (same for the middle constant(s)).