# Direct proof of the separation theorem of Hahn-Banach

The "extension" (or "analytic") form of the theorem of Hahn-Banach has a natural and yet elegant proof. In just any textbook I have ever seen, it is proved first; the "separation" (or "geometric") version of Hahn-Banach's theorem is proved as a kind of corollary of the former.

Question: Are the two theorems actually equivalent? If so, is any direct proof of the analytic version known that is instead based on the geometric one?

Yes, the two theorems are equivalent in the sense that one can easily be deduced from the other and both have direct proofs from scratch.

A standard textbook starting with a direct proof of the geometric version is Schaefer's Topological Vector Spaces, Chapter II, Section 3.

The statement Schaefer proves is:

Let $F$ be a subspace of a topological vector space $E$ and let $U$ be a nonempty open convex subset, disjoint from $F$. Then there is a closed hyperplane $H$ containing $F$ and disjoint from $U$.

The usual reductions via translation and taking the difference of the convex sets then yield the separation theorems of an open convex set from a point and of an open convex set from a compact convex set.

The proof starts by a simple geometric observation: Let $U$ be an open and convex subset of a Hausdorff topological vector space of dimension $\geq 2$. If $U$ does not contain $0$, then there is a one-dimensional subspace disjoint from $U$. This is easily reduced to the two-dimensional case, where it is rather clear.

To establish the above statement, a straightforward application of Zorn's lemma shows that there is a maximal (hence closed) subspace $M$ containing $F$ and disjoint from $U$. Since $U$ is non-empty, $E/M$ has dimension at least $1$. If the dimension of $E/M$ is $1$, then $M$ is a hyperplane and we're done. Suppose towards a contradiction that the dimension is at least $2$. The image of $U$ in $E/M$ does not contain $0$ and is open since the canonical projection $\pi \colon E \to E/M$ is open. Since $M$ is closed, $E/M$ is Hausdorff. Therefore there is a one-dimensional subspace $L$ of $E/M$ not meeting the image of $U$. The pre-image of $L$ contains $M$, is strictly larger and does not meet $U$, contradicting the maximality of $M$.

In order to get the analytic form, identify a linear functional on a subspace with its graph in $E \times \mathbb{R}$. Endow $E \times \mathbb{R}$ with the product topology induced by the sublinear functional $p$ and the usual topology on $\mathbb{R}$. The set $U = \lbrace (x,t) \mid p(x) \lt t\rbrace$ is an open convex cone in $E \times \mathbb{R}$, not containing $(0,0)$. A linear functional $f$ is dominated by $p$ iff $\operatorname{graph}(f)$ is disjoint from $U$. A maximal closed hyperplane $M$ containing $\operatorname{graph}(f)$ and disjoint from $U$ is then seen to be the graph of a linear functional $F$ which is obviously an extension of $f$ and dominated by $p$.

The separating version is ubiquitous in economics, and in my experience, most textbooks in mathematical economics prove it directly (though sometimes only in the finite dimensional case).

• Most important proofs in economics that use this, that I've seen, for the infinite dimensional case refer to Schaefer's Topological Vector Spaces (that's how I teach my students). This is because a famous paper by Bewely in 1972 cited Schaefer, and because Schaefer's is an amazing book. But I would be surprised if any recent economics PhD graduate from the US knows of a proof of the Hahn-Banach theorem. Europe, yes there are. But mathematical economics seems to be dead and buried in the US. – Rabee Tourky Jun 23 '13 at 0:31
• Rabee: I believe there are several recent economics PhD graduates from the (US) program in which I teach who can prove the Hahn-Banach theorem. – Steven Landsburg Jun 23 '13 at 13:20
• Steven, I'll send you a bottle of good New Zealand Shiraz if you can find a single current or recent US PhD student in Economics who can prove the second welfare theorem when the economy is in $L_1$ and prices are in its dual $L_\infty$, and preferences are strictly monotone and linear (It's a very nice Shiraz:-). The student must not have done a masters in western Europe. – Rabee Tourky Jun 23 '13 at 23:03
• Rabee: I can think of several (including one of your co-authors) who I believe are quite likely to meet your criteria, but I think this is probably the wrong place to list names and then argue about whose education is likely to be deficient. But I'll take a few days, make a couple of inquiries to make sure I'm right, and then send you an email. – Steven Landsburg Jun 24 '13 at 14:43