# Hahn-Banach separation theorem

Let $V_1,V_2$ be convex subsets of a normed space $X$ with $V_1^\circ\neq\emptyset$ and $V_1^\circ\cap V_2 =\emptyset$. Then there exists $x'\in X'\setminus\{0\}$ such that $$\operatorname{Re} x'(v_1)\leq \operatorname{Re}x'(v_2)\quad\forall v_1\in V_1, v_2\in V_2$$

I started like this:

Since $V_1$ is convex, $V_1^\circ$ is convex, too. Hence by Hahn-Banach's separation theorem there exists $x'\in X'$ such that $\operatorname{Re} x'(v_1)\leq \operatorname{Re}x'(v_2)$ for all $v_1\in V_1^\circ$ and $v_2\in V_2$.

My idea was now to use Hahn-Banachs extension theorem: Since $V_1^\circ\subset V_1$ there exists an extension $\tilde x'$ such that $\tilde x'|_{V_1^\circ}=x'$. But my problem is I don't know how to conclude from this that $\operatorname{Re}\tilde x'(v_1)\leq \operatorname{Re} \tilde x'(v_2)$ for all $v_1\in V_1$, $v_2\in V_2$ and why $\tilde x'\neq 0$.

How can I argue here?

• $x'$ is already defined on the entire space, there is nothing to extend. What you want is $$\sup_{v\in V_1} \operatorname{Re} x'(v) \leqslant \sup_{v\in V_1^\circ} \operatorname{Re} x'(v).$$ The continuity of $x'$ gives you that if $V_1 \subset \overline{V_1^\circ}$. Can you show the latter? – Daniel Fischer Jun 19 '14 at 16:17
• I'm not sure I understand the inequality. If $V_1\subset\overline{V_1^\circ}$ we have $\sup_{v\in V_1}\operatorname{Re} x'(v)\leq \sup_{v\in\overline{V_1^\circ}} \operatorname{Re} x'(v)$. Is the latter supremum the same as $\sup_{v\in V_1^\circ}$ since for every $v\in \overline{V_1^\circ}$ we can find a sequence $(v_n)\subset V_1^\circ$ such that $v_n\to v$? – dinosaur Jun 19 '14 at 16:35
• okay, after having done some research I found out that for convex sets we have $V^\circ = \overline{V^\circ}$. I think I get it now, thanks! – dinosaur Jun 19 '14 at 17:16
• Yes. Or, more generally, for a continuous $f$, one has $f(\overline{A}) \subset \overline{f(A)}$. – Daniel Fischer Jun 19 '14 at 17:16