# Hahn-Banach sans Zorn

The Hahn-Banach Theorem in functional analysis is the theorem saying more or less that normed vector spaces have many bounded linear functionals. The theorem is usually proven via Zorn’s lemma, giving the impression that Hahn-Banach uses the full power of choice. I’ll here give a proof based on the ultrafilter lemma, that every filter can be extended to an ultrafilter. One of the benefits of this approach, besides relying on a (strictly) weaker assumption, is that we get a little more information about how the functionals are constructed. The idea is that the ultrafilters allow us to “glue” functionals together. A latex’ed version can be found here.

Let’s start off with reviewing what Hahn-Banach is actually saying. First of all, say that a seminorm on a vector space $X$ is a function $p:X\to\mathbb R$ such that

1. (Non-negative) $p(x)\geq 0$;
2. (Subadditivity) $p(x+y)\leq p(x)+p(y)$;
3. (Absolute homogeneity) $p(\lambda x)=|\lambda|p(x)$.

A functional is simply a linear function from a vector space to its underlying field.

Hahn-Banach Theorem. Let $X$ be a normed space, $Z\subseteq X$ a subspace, $f$ a functional on $Z$, $p$ a seminorm on $X$ such that $|f(z)|\leq p(z)$ for every $z\in Z$. Then there is a linear functional $\bar f$ on $X$ such that $f(z)=\bar f(z)$ for every $z\in Z$ and $|\bar f(x)|\leq p(x)$ for every $x\in X$.

To prove this we first show a weaker version, which can be proven in ZF.

Finite Extension Lemma (ZF). Let $X$, $Z$, $f$ and $p$ be as in the Hahn-Banach Theorem. Then for any $x_0\in X$ there is a linear functional $\bar f$ on $\text{span}(Z\cup\{x_0\})$ which extends $f$ and satisfies that $|\bar f(x)|\leq p(x)$ for every $x\in\text{dom }\bar f$.

The proof is roughly that in this finite case we get a closed interval of possible values of $\bar f(x_0)$, so we can simply pick one of those. For more details see e.g. my noteWe will prove Hahn-Banach in the case where $X$ is a real vector space, as the transition to the complex case is standard. Define the set $S$ as

$g\in S$ iff $g$ is a linear functional extending $f$, defined on a subspace $Y\subseteq X$ that contains $Z$, and which is absolutely bounded by $p$.

Then to any $x\in X$ set $A_x$ to be all those $g\in S$ with $x\in\text{dom }g$. The finite extension lemma implies that all the $A_x$‘s are non-empty and furthermore that finite intersections of them are non-empty as well. The $A_x$‘s then form a “subbasis” for a filter $\mathcal F$:

$\mathcal F:=\{A\subseteq S\mid\exists x_1\cdots\exists x_m:\bigcap_{n=1}^m A_{x_n}\subseteq A\}$.

Use the ultrafilter lemma to extend $\mathcal F$ to an ultrafilter $\mathcal U$ on $S$. It then holds that given any $x\in X$, $x\in\text{dom }g$ for $\mathcal U$-a.e. $g\in S$. Now form the (ill-founded) ultrapower $(\mathcal M,E):=(\text{Ult}(V,\mathcal U),\in_{\mathcal U})$ with associated ultrapower embedding

$j:(V,\in)\to(\mathcal M,E)$.

Then setting $h:=[\text{id}]_{\mathcal U}$ we see that given any $x\in X$, $x\ E\ \text{dom }h$, giving us a candidate for an extension! But note that $h$ is a function into the hyperreals $j(\mathbb R)$ (inside the ultrapower), so it’s not really a functional. But the hyperreals has the nifty feature that there is a homomorphism $\text{st}:(j(\mathbb R),E)\to(\mathbb R,\in)$ called the standard part. This is basically by the completeness of the reals, so that to any hyperreal we can associate a unique real such that their difference is infinitesimal. Then set $\bar f$ to be the functional on $X$ defined as

$\bar f:=\text{st}\circ h\circ j\upharpoonright X$.

As $g$ extends $f$ for $\mathcal U$-a.e. $g$, we get that $(h\text{ extends } j(f))^{\mathcal M}$. This means that

$\text{st}(h(j(z))=\text{st}(j(f)(j(z)))=\text{st}(j(f(z)))=f(z)$,

for every $z\in Z$, so $\bar f$ does in fact extend $f$. Furthermore, note that $|h(x)|\leq j(p)(x)$ for every $x\ E\ \text{dom }h$, so that

$|\bar f(x)|=|\text{st}(h(j(x)))|=\text{st}(|h(j(x))|)\leq\text{st}(j(p)(j(x)))=\text{st}(j(p(x)))=p(x)$,

showing that $\bar f$ is absolutely bounded by $p$ as well. QED

Note that the function $h$ in the proof was defined as $[\text{id}]_{\mathcal U}$, so that really it corresponds to all the functionals that we’re interested in, at the same time. The ultrafilters are thus used for this sort of “gluing”, that if we can just make sure that a given property holds for many partial functionals, there is a canonical functional with the property as well.

This note was inspired by Asaf Karagila’s note on Hahn-Banach and the Axiom of Choice.