# Math Help - Hahn-Banach theorem, geometric version (question about proofstep)

1. ## Hahn-Banach theorem, geometric version (question about proofstep)

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 $x' \in X'$, $\alpha\in\mathbb{R}$, $\varepsilon>0$such that
$Re x'(y)+\varepsilon \le \alpha \le Re x'(x)-\varepsilon$ for all $y\in A,$ $x\in B$.

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

I don't understand the red part of the proof. How can one conclude that?
(I know that $||x'||=||Re x'||$, but I have no idea of how to get the "-"sign infront of the r.)

Regards,
engmaths

2. ## Re: Hahn-Banach theorem, geometric version (question about proofstep)

Define what you mean by $\text{Re}$. Surly, you cannot mean the real part of an $x\in X$, that would not make sense as $X$ is normed space not the complex numbers.

3. ## Re: Hahn-Banach theorem, geometric version (question about proofstep)

I do not follow your notation but I think I know what you want to prove.

We are given a normed vector space $(X, ~ | \cdot | )$. Let $f\in X^*$, in other words, $f$ is a linear functional in the dual space. A subset of $X$ is called a hyperplane if it has the form $\{ x\in X | f(x) = a \text{ for some } f\in X^*, f\not = 0\}$, where $a$ is a real (or complex) number. We use the simplified notation $(f=a)$ for the hyperplane described above. Given two subsets of $X$, call them $A$ and $B$, with $A\cap B = \emptyset$, we say that $A$ and $B$ can be "seperated by a hyperplane" if there exists a hyperplane $(f=a)$, for some $f\in X^*$ and $a\in \mathbb{R}$, such that $f(x)\leq a$ for $x\in A$ and $f(x)\geq b$ for $x\in B$, or the other way around, $f(x)\geq a$ for $x\in A$ and $f(x)\leq a$ for $x\in B$ (in other words each set lies in each side of the hyperplane). There is another similar definition for "strict seperatation". Given $A,B$ as before we say $A,B$ can be "strictly seperated by a hyperplane", if there is a hyperplane $(f=a)$ and $\varepsilon > 0$ such that $f(x)\leq a - \varepsilon$ for $x\in A$ and $f(x)\geq a + \varepsilon$ for all $x\in B$, or as before, the same property with $A$, $B$ reserved.

We prove the following theorem.

Theorem: Let $C,K$ be subsets of $X$. Both are convex, disjoint, $C$ is closed, and $K$ is compact. There exists a hyperplane that strictly separates $C$ and $K$.

Lemma: If $c\leq f(w)$ for all $|w|<1$ where $f$ is functional then $c\leq -|f|$.

Proof: This is just follows from definition of $|f|$:
$$|f| = \sup\{ |f(w)| : |w|\leq 1\} \geq \sup\{ |f(w)|: |w|<1\}$$
And the fact that $f(-w) = -f(w)$.
(Here $f$ is bounded linear functional).

The proof of this theorem is based on the following result. This is a variation on the theorem above but a bit different. We will assume this result in the proof of the theorem above.

Theorem: Let $A,\mathcal{O}$ be subsets of $X$. Both are convex, disjoint, $\mathcal{O}$ is open. There exists a hyperplane that separates $A$ and $\mathcal{O}$.

Proof: Define $D = C - K = \{ c - k ~ | ~ c\in C, k\in K\}$. As you said, $D$ is a closed convex set with $0\not \in D$ as $C,K$ are disjoint. Therefore, there is a radius $r>0$ such that $B_r(0)\cap D = \emptyset$. Thus, by assumed theorem, there is a hyperplane $(f=a)$ that seperates $D$ and the open ball. We have,
$$f(x-y) \leq a \leq f(z) \text{ for }x\in C,y\in K,z\in B_r(0)$$
Now we can rewrite $z=rw$ where $w\in B_1(0)$ and so,
$$f(x-y)\leq f(ru) = rf(w) \text{ for }x\in C,y\in K,w\in B_1(0)$$
Thus, by Lemma we have,
$$f(x-y)\leq -r||f|| \implies f(x) - f(y) \leq -r||f|| \implies f(x) \leq f(y) - r||f||$$
Now add $\tfrac{1}{2}r||f||>0$ to both sides to get,
$$f(x) + \tfrac{1}{2}r||f|| \leq f(y) - \tfrac{1}{2}r||f||$$
Thus, we see that,
$$\sup_{x\in C} f(x) + \tfrac{1}{2}r||f|| \leq \inf_{y\in K}f(y) - \tfrac{1}{2}r||f||$$
So we can choose a number $b$ in that interval $[ \sup ~ , \inf ]$ to get, that $f(x)+\tfrac{1}{2}r||f|| \leq b \leq f(x) -\tfrac{1}{2}r||f||$. This shows that $(f=b)$ is a hyperplane that strictly seperates the two sets.

4. ## Re: Hahn-Banach theorem, geometric version (question about proofstep)

Thanks for your answer, ThePerfectHacker. I didn't want to confuse anyone with my notation, I just thought that it was quite standard.

I understand everything you wrote, except the proof of the lemma. How do you conclude, from what you wrote, that c<=- |f| ?

5. ## Re: Hahn-Banach theorem, geometric version (question about proofstep)

Ok, I understood it now. It follows easily from inf M = - sup -M.

6. ## Re: Hahn-Banach theorem, geometric version (question about proofstep)

Hello,

could you explain me why C=A-B is closed. I will be grateful!

regards,
bill

7. ## Re: Hahn-Banach theorem, geometric version (question about proofstep)

Hello,

could you explain me why C=A-B is closed. I will be grateful!