# Hahn-Banach theorem: 2 versions

I have a question regarding the Hahn-Banach Theorem. Let the analytical version be defined as: Let $E$ be a vector space, $p: E \rightarrow \mathbb{R}$ be a sublinear function and $F$ be a subspace of E. Let $f: F\rightarrow \mathbb{R}$ be a linear function dominated by $p$ (by which I mean $\forall x \in F: f(x) \leq p(x)$). Then $f$ has a linear extension $g$ to $E$ with $g$ dominated by $p$.

Let the geometric version of Hahn-Banach Theorem be defined as: Let $E$ be a topological vector space, $\emptyset \neq A \subset E$ be open and convex. Let $M = V + x$ with $V$ a subspace of $E$ and $x \in E$. Suppose that $A \cap M = \emptyset$. Then there exists a closed hyperplane $H$ such that $M \subset H$ and $H \cap A = \emptyset$.

Now, I know that the analytical version is proved using Zorn's Lemma and that the geometric version can be derived from the analytical version. My question is: can the analytical version be derived from the geometric version ? I don't have a clue how to begin to prove this. (These versions hold for arbitrary vector spaces, not just finite dimensional ones).

Any help will be appreciated.

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Yes, the geometric and the analytic versions of the Hahn–Banach theorem follow from each other. Here's a proof of the direction you ask about:

Consider the space $X = E \times \mathbb{R}$ equipped with the product topology of the topology induced by $p$ on $E$ and the ordinary topology on $\mathbb{R}$. Note that (continuous) linear functionals $g$ on $E$ bijectively correspond to (closed) linear hyperplanes of $X$ not containing $0 \times \mathbb{R}$ via the identification of $g$ with its graph $\{(x,g(x))\,:\,x \in E\}$.

Let $\Gamma = \{(x,f(x))\,:x \in F\}$ be the graph of $f$. Then $\Gamma$ is disjoint from the convex cone $$C = \{(x,t)\,:\,p(x) \lt t\} \subset X$$ since $f$ is dominated by $p$: if $(x,t) \in \Gamma$ then $t = f(x) \leq p(x)$, so $(x,t) \notin C$.

Since the cone $C$ is open in $X$, the geometric version of the Hahn–Banach theorem implies that there exists a hyperplane $H \supset \Gamma$ of $X$, disjoint from $C$. In particular we must have for all $(x,t) \in H$ that $t \leq p(x)$. Note that $(0,0) \in \Gamma \subset H$, so $H$ is a linear subspace. Since $(0,1) \in C$ we have that $0 \times \mathbb{R}$ is not contained in $H$, so the hyperplane $H$ is the graph of a linear functional $g$ on $E$. Since $H \supset \Gamma$, we have found the graph of an extension $g$ of $f$ satisfying $g(x) \leq p(x)$ for all $x \in X$.

Note: It is possible to prove the geometric form of the Hahn–Banach theorem by a direct application of Zorn's lemma, see e.g. Schaefer's book on topological vector spaces, Chapter II, Theorem 3.1, page 46.

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It's alive! ALIVE!!!!!! –  Asaf Karagila Jan 31 '12 at 13:43
yes, it is - more or less :) but I don't have the time to hang around here a lot, especially not in chat... –  t.b. Jan 31 '12 at 13:48
So I heard. I'm just glad to see your name on my screen again... :) –  Asaf Karagila Jan 31 '12 at 13:50
@t.b.: Catch ya! –  Tim Jan 31 '12 at 13:50