# Hahn-Banach theorem (second geometric form)

I found in Brezis' Analyse Fonctionelle the Hahn-Banach theorem ("second geometric form") and there is a passage I can't understand. In newest versions of this book the proof has been modified.

Theorem. Let $A \subset E$ and $B \subset E$ be two nonempty convex subsets such that $A \cap B = \emptyset$. Assume that $A$ is closed and $B$ is compact. Then there exists a closed hyperplane that strictly separates $A$ and $B$.

Proof. For $\epsilon > 0$ let $A_\epsilon = A+B(0,\epsilon)$ and $B_\epsilon = B+B(0,\epsilon)$; those sets are convex and nonempty. They are disjoint too (otherwise, we can find a sequence $\epsilon_n \to 0$ and $x_n \in A$, $y_n \in B$ such that $\|x_n - y_n \|< 2\epsilon_n$; so we can extract a subsequence $y_n \to y \in A \cap B$). $\quad$ [CUT]

My question is:

How can we extract such a subsequence when we don't know if $A_\epsilon \cap B_\epsilon$ has elements in common with $A \cap B$?

## 4 Answers

The passage is worded somewhat ambiguously, but I think what is claimed is that there exists some $\epsilon > 0$ such that $A_\epsilon$ and $B_\epsilon$ are disjoint. For suppose the contrary: that for every $\epsilon > 0$ there exists $z \in A_\epsilon \cap B_\epsilon$. Choose a sequence $\epsilon_n \to 0$ and $z_n$ with $z_n \in A_{\epsilon_n} \cap B_{\epsilon_n}$. By definition of $A_\epsilon, B_\epsilon$ there exists $x_n \in A$ and $y_n \in B$ such that $\lVert x_n - z_n \rVert < \epsilon_n$, $\lVert y_n - z_n \rVert < \epsilon_n$, so by the triangle inequality $\lVert x_n - y_n \rVert < 2 \epsilon_n$. Since $B$ is compact, we can pass to a subsequence and assume $y_n$ converges to some $y \in B$. Then it follows from the triangle inequality that $x_n \to y$ as well. $A$ is closed, so $y \in A$, and thus $A,B$ are not disjoint, a contradiction.

• Ambiguously is being kind; it’s flat-out worded badly. But I certainly agree with your interpretation. – Brian M. Scott Apr 11 '12 at 2:10
• It is not ambiguous it is ok! – checkmath Apr 11 '12 at 2:39

The key here is that $B$ is compact. Then the sequence ${y_n}$ admits a convergent subsequence to an element $y\in B$. But since $y$ is arbitrarily close to elements of $A$ (i.e. the $x_n$) and $A$ is closed, then $y\in A$.

Another approach would be to say $f(x) = \text{dist}(x,A)$ (i.e. $\inf_{a \in A} \|x - a\|$) is a continuous function that is $0$ if and only if $x \in A$ (because $A$ is closed).
So $f(x) > 0$ on $B$. Since $B$ is compact $f$ attains a minimum on $B$, and the minimum value, call it $\eta$, is strictly positive. Now if $0 < \epsilon < \eta/2$, $A_{\epsilon}$ and $B_\epsilon$ are disjoint by the triangle inequality.

Another approach would be to consider the set $A-B = \{a-b|a\in A, b \in B\}$. The same reasoning (extracting a convergent subsequence) as above can be used to show that $A-B$ is closed (and convex, of course). Since the sets are disjoint, $0 \notin A-B$, and since the set is closed, you have $B(0,\epsilon)$ and $A-B$ are disjoint, for some $\epsilon > 0$. Then a simple estimate shows $A_{\frac{\epsilon}{2}}$ and $B_{\frac{\epsilon}{2}}$ are disjoint.