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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?

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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 $p$.

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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).

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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
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