The Hahn-Banach theorem and the axiom of choice

Question

Let us recall the Hahn-Banach theorem about extensions of linear functionals:

Theorem: Let $E$ be a real vector space and $F$ a subspace. If $p:E\to \mathbb{R}$ is a sublinear function, and $g:F\to \mathbb{R}$ is a linear functional on $F$ which is dominated by $p$ on $F$ i.e. $$g(x)\leq p(x)\qquad \forall x\in F$$ then there exists a linear extension $f:E\to\mathbb{R}$ of $g$ to the whole space $E$, i.e., there exists a linear functional $f$ such that \begin{gather*} f(x)=g(x)\qquad \forall x\in F,\\ f(x)\leq p(x)\qquad \forall x\in E. \end{gather*}

During the proof of this theorem, we use Zorn's lemma. It has been shown that Hahn-Banach's theorem is not equivalent to the axiom of choice. There is also some work which has been done showing that for a separable Banach space, a more direct proof can be made. An article by Douglas K. Brown and Stephen G. Simpson studies these kinds of questions of equivalence of axioms in a logic oriented framework.

For my part, I am looking for a more practical example. Since the proof uses Zorn's lemma, we could (in theory) give a well enough built example of a linear functional which extension is "not so obvious". I will elaborate on what I mean by this.

As of most of functional analysis is, extending a linear functional is pretty straightforward in the finite dimensional case. For this reason, students do not have much trouble accepting a proof requiring this kind of machinery. Although, seeing how non-trivial linear functionals behave in the infinite dimensional case gave me the idea to look around for a "simple, yet troubling" example.

Edit: I think I should clarify what I want. The axiom of choice does not help us with the construction of an extension. I want an example where the extended linear functional is (almost) impossible to figure out/write down on paper. This would show students a problematic with non-constructive mathematics, i.e. they are unpractical in a computational framework.

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I've been doing my part of work, and I've looked up a bit. Here's what I came up with:

Let $g:C_c^{\infty}(\mathbb{R})\to \mathbb{R}$ be defined by $$g(\phi)\mapsto \phi(0)$$ where $C_c^\infty(\mathbb{R})$ is the set of compactly supported smooth functions from $\mathbb{R}$ to $\mathbb{R}$. We may see this real vector space as a subspace of $L^\infty(\mathbb{R})$. We know that $L^\infty$ is not separable and that it's dual is the space of Radon measures. We could extended $g$ to all of $L^\infty(\mathbb{R})$ using Hahn-Banach.

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I do not feel completely satisfied by this example because it may be just me who didn't think far enough to see an adequate extension since I did not spend enough time studying Radon measures. So, as a conclusion of this somewhat long post, I am asking for the following:

1. A well built example which clearly shows the problems we may encounter when using the axiom of choice
2. A computation of the extension of the $g$ I've given or a proof that the extension isn't constructible if the computation isn't possible.
3. Any other ideas which could be added as comments on the axiom of choice and Hahn-Banach's theorem, whatever viewpoint it may come from.

Answer 1

Let $\ell^\infty$ be the space of all bounded sequences with the supremum norm and $c$ the subspace of all convergent sequences. Clearly $\lim\colon c\to \mathbb K$ is a continuous linear functional. By Hahn-Banach, it has a continuous extension $f$ to $\ell^\infty$.

I claim there exists no absolutely summable sequence $(b_n)$ such that $f((a_n))=\sum_n a_n b_n$ for all $(a_n)\in \ell^\infty$. Indeed, if such a sequence existed, we could take $a_n=\delta_{n,k}$ to conclude $b_k=f((a_n))=\lim_n a_n=0$ for any $k$, hence $b=0$. But this would imply $f((a_n))=0$ for all $(a_n)\in \ell^\infty$, which is evidently not the case.

However, it is consistent with ZF that every continuous linear functional on $\ell^\infty$ is of the form $f((a_n))=\sum_n a_n b_n$ for some absolutely summable sequence $(b_n)$ (see here: $\ell^1$ vs. continuous dual of $\ell^{\infty}$ in ZF+AD). Thus we really need some form of the axiom of choice to get a continuous extension of $\lim$ to $\ell^\infty$.