# Hahn-Banach and hyperplane separation

Let's take a look at the following result, based on Hahn-Banach (extension of bounded linear (real-valued) functionals):

Let $X$ be a normed space, $U$ a subspace of $X$ and $u_0 \in X$ with $inf_{u \in U} \Vert u - u_0\Vert = d > 0$.

Then there exists an $f: X \to \mathbb{R}$, which is bounded and linear, and the following holds:

1. $f(u) = 0$ for all $u \in U$.
2. $f(u_0) = d$
3. $\Vert f \Vert = 1$

I've been told that this corollary is (or implies) a separation theorem.

To my knowledge, the separation theorem states, that for a nonempty closed convex $M \subset X$ and $x_0 \in X-M$ there is a bounded linear $f: X \to \mathbb{R}$ and an $\alpha \in \mathbb{R}$ such that $f(M) \leq \alpha$ and $f(x_0) > \alpha$.

So I have to check if $M$ satiesfies $inf_{u \in M} \Vert u - x_0\Vert = d > 0$ to get the separation theorem from the corollary (with $\alpha = 0$).

But doesn't this already follow from $M$ being closed? Where do I need that $M$ in convex? Or is my separation theorem wrong? If I picture it, I do need convexity, so $M$ can't be something like a sphere around $x_0$.

Any help is appreciated.

• $M$ closed does indeed imply that $d(x_0,M) > 0$. But note that the theorem also requires that $U$ be a subspace, which $M$ is not. – Paul Sinclair Nov 26 '15 at 13:51
• @PaulSinclair But that means my separation theorem is wrong, am I right? Any closed Ball not including $0$ could be chosen as $M$ and this is no subspace. :( – hank789 Nov 26 '15 at 14:30
• No. It just means it doesn't follow immediately from the stated theorem. – Paul Sinclair Nov 26 '15 at 14:34
• Sorry, but it isn't obvious to me how you would use that theorem to prove the result. That is why I replied in a comment rather than an answer. I'll think about it. – Paul Sinclair Nov 26 '15 at 14:56
• One needs the Hahn-Banach theorem with sublinear majorant $p$ which, in your case, is the Minkowski functional of an open convex set not containing $x_0$. – Jochen Nov 26 '15 at 15:09