# 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.

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.

• 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