Hello,

my question is about a proofstep in the proof of the second geometric version of the Hahn-Banach theorem. The theorem is:

Let X be a normed, real or complex vectorspace. Let A,B non-empty, convex subsets of X such that the intersection of A and B is empty. Let A be closed and B compact.

Then there exists an $\displaystyle x' \in X'$, $\displaystyle \alpha\in\mathbb{R}$, $\displaystyle \varepsilon>0 $such that

$\displaystyle Re x'(y)+\varepsilon \le \alpha \le Re x'(x)-\varepsilon$ for all $\displaystyle y\in A,$ $\displaystyle x\in B$.

The proof starts as follows:

$\displaystyle C:=A-B$ is convex and closed (that C is closed is showed later in the proof and can be assumed for now) with $\displaystyle 0\notin C$. So there exists a $\displaystyle r>0$ such that $\displaystyle B_r(0)\cap C =\varnothing$. By the first geometric version of Hahn-Banach there exists an $\displaystyle x'\ne 0$ such that $\displaystyle Re x'(y-x)<Re x'(rz)$ for all $\displaystyle x\in B, y\in A, z\in B_1(0)$.

That means: $\displaystyle Re x'(y-x)\le -r||x'|| $ for all $\displaystyle y\in A$, $\displaystyle x\in B$. [...]. qed.

I don't understand the red part of the proof. How can one conclude that?

(I know that $\displaystyle ||x'||=||Re x'||$, but I have no idea of how to get the "-"sign infront of the r.)

Thanks in advance for your help.

Regards,

engmaths