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.

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.

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.

Thank you very much! Also, I'm working very hard on perfecting my english which isn't my first language. If you find any written mistakes please correct them!

  • $\begingroup$ $C_{c}^{\infty}(\mathbb{R})$ is separable with respect to the supremum norm. For each $N \in \mathbb{N}$, pick countable dense subsets of $C_{c}^{\infty}([-N,N])$, then take the union. $\endgroup$ – fourierwho Dec 27 '17 at 0:08
  • $\begingroup$ What are the "problems we may encounter when using the axiom of choice" that you have in mind? Is it just the fact that it is not constructive? $\endgroup$ – JH vd Walt Dec 27 '17 at 4:56
  • $\begingroup$ I don't actually understand the question. What are you actually looking for? Examples of Hahn–Banach which are "not esoteric" somehow? $\endgroup$ – Asaf Karagila Dec 27 '17 at 11:11
  • $\begingroup$ @fourierwho I was speaking about $L^\infty$ which isn't separable. I will clarify my post. Thank you for your comment. $\endgroup$ – Alexis Leroux-Lapierre Dec 30 '17 at 19:06
  • 2
    $\begingroup$ The dual of $L^\infty$ is not the space of Radon measures. $\endgroup$ – Project Book Dec 30 '17 at 19:23

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