I am trying to prove the following exercise, which is a strong version of Hahn-Banach separation theorem in the case of finite dimensional spaces.
Let's suppose $X$ is a normed finite dimensional space, and $A$ and $B$ two disjoint convex sets. Sea $\{a_n\}$ a dense subset of $A$ and for each $n$ consider $A_n = conv\{a_1,…,a_n\}$ the convex hull of $\{a_1,…,a_n\}$. Prove that:
a) Each $A_n$ is compact, $A_n \subset A_{n+1} \ \forall n$, y $\cup_{n}A_n$ is a dense in $A$.
b)There exists $\varphi_n \in X^*_{\mathbb{R}}$ y $\alpha_n \in \mathbb{R}$ such that $||\varphi_n|| = 1 \ \forall n$ and $\varphi_n(a) < \alpha_n < \varphi_n(b) \ \forall a \in A_n, b \in B$
c)There exist $\varphi \in X^*_{\mathbb{R}}$ y $\alpha \in \mathbb{R}$ such that $||\varphi|| = 1 \ \forall n$ and $\varphi(a) \leq \alpha \leq \varphi(b) \ \forall a \in A, b \in B$
I manage to prove a): that each $A_n$ is compact follows from the fact that $A_n = f_n(S_n)$ being $f_n:\mathbb{R}^n \to X$ with $f_n(\lambda_1,…,\lambda_n) = \sum_{i=1}^n \lambda_ia_i$ and $S_n$ the compact set $\{(\lambda_1,…,\lambda_n): \lambda_i\geq 0 \ \forall i, \ \sum_{i=1}^n\lambda_i = 1\}$, since $f_n$ is continuos $A_n$ is compact. That the union is dense is in $A$ is obvious since it contains $\{a_n\}$ which is dense in A. Also is direct that $A_n \subset A_{n+1}$
When I started to prove $b)$ I found a problem with the exercise, I think that we cant obtain the strict inequality in b), for example take $X = \mathbb{R}^2$, $A = (1,0)$ and $B = B(0,1)$ the open ball of radio 1. They are disjoint and convex but we cant find a line separating them strictly. So here I assume that the correct version is with $\leq$ instead of $<$. Despite this, I couldn't prove it, I am not being able to construct $\varphi_n$. My first Idea was trying to use the Hahn-Banach separation theorem, but I only know that $A_n$ is compact, $B$ may be not closed. Even in the case of $n=1$ this is a problem. It would be great to put $A_1$ in a open ball disjoint from $B$ and then I can apply H-B, but it is impossible to do this because $a_1$ could be on the closure of $B$ and once again the problem is that $B$ is not closed. I am stuck on this, I would appreciate any idea to continue.
Best Answer
Well, a nice counterexample shows that b) can only be true if we put $$\varphi_n(a)\leq \alpha_n \leq \varphi_n(b) \ \forall a \in A_n, b \in B$$
Think of $\mathbb{R}^2$, $B = \{x=0,y>0\} \cup \{x<0\}$, and $A = \{x=0,y\leq0\}$ and $a_1 = (0,0)$, then you will see that the only way to separate the convex sets $B$ and $A_n$ is to take the line $x=0$, and here we got that equality holds for some points.
Respect to the new statement, I managed to prove it using Hahn Banach between $A_n$ and $B_m$, this provide me a funciontal $\varphi_{nm}$ separating this sets. Then I divide by the norm of $\varphi_{nm}$. Finally, for every fixed $n$ I have a sequence on $m$ of functionals living in the compact set $\overline{B}(0,1) \subset X^*_{\mathbb{C}}$ and this means that there is a subsequent that converges to a certaing $\varphi_n$. Obviously this is true because $\textrm{dim }X<\infty$, I think that the key is that the ball is compact. Working this way I prove the statement. Similarly for c). If someone is interested in more details I can give a better explanation