15
$\begingroup$

I have encountered this when I was thinking about differentiability in Banach spaces. There, for $x\in X$ we usually need functionals $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$. This is a simple consequence of Hahn-Banach theorem and enables one to convert the problem at hand into a problem in ordinary calculus. Now My question is:

Suppose we have a Banach space $X$ whose dual $X^*$ separates points, i.e. for every nonzero $x\in X$ there is $u\in X^*$ such that $u(x)\neq 0$. Can one prove in ZF that for all nonzero $x\in X$ there is $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$ ?

I know that Hahn-Banach theorem is strictly weaker than axiom of choice, but I'm looking for a proof without using any "choicy" argument.

$\endgroup$
  • 1
    $\begingroup$ This doesn't answer your question, but I'd just like to point out that you don't need any choice principle to prove the Hahn-Banach theorem for separable Banach spaces (and one can make a case that this is all we care about in practice). $\endgroup$ – Nik Weaver Aug 16 '13 at 17:31
  • 3
    $\begingroup$ A suggestion of a possible counter-example: take X to be $\ell^\infty$ (trivial that dual separates points, even in ZF, because we can just use the point evaluations) and take your $x$ to be any sequence that does not attain its bound. Then it seems that to get a functional which attains its norm on $x$ one is going to need some choice or ultrafilter argument $\endgroup$ – Yemon Choi Aug 16 '13 at 17:31
  • $\begingroup$ @Yemon My own guess was also that it's not provable in ZF, but my knowledge of set theory is nowhere near proving an independence result. $\endgroup$ – Mohammad Safdari Aug 16 '13 at 18:02
  • $\begingroup$ @Nik That's interesting, can you give me a reference please? $\endgroup$ – Mohammad Safdari Aug 16 '13 at 18:04
  • $\begingroup$ @Mohammad: it's easy, just run through the usual proof of H-B, extending the functional one dimension at a time. At each step choose the maximum allowed value of the functional on the new vector --- so, no choice. If the space was separable to start with, that means we have a basis $(x_n)$ indexed by $\mathbb{N}$, so if we're extending from the subspace $E_0$, pass through the sequence of subspaces ${\rm span}(E_0, x_1, \ldots, x_n)$. No choice there either. $\endgroup$ – Nik Weaver Aug 16 '13 at 20:20
12
$\begingroup$

Yemon Choi suggested a counterexample in the comments which can easily be made into an actual counterexample.

It is consistent with ZF (see Asaf Karagila's answer here for a few references) that the dual space of $\ell_\infty$ is $\ell_1$ with the usual norm. The duality pairing is $$ \langle x,y \rangle = \sum x_n y_n. $$ Since the coordinate functionals separate points in $\ell_\infty$, the hypothesis in the question is satisfied.

On the other hand, for the sequence $x_n = 1-1/n$ in $\ell_\infty$ there obviously is no sequence $y \in \ell_1$ of norm one such that $\sum x_n y_n = 1 = \lVert x_n \rVert_\infty$.

$\endgroup$
  • 1
    $\begingroup$ See also mathoverflow.net/q/5351/45 on this site for the relationship between choice and the dual space of $\ell_\infty$. $\endgroup$ – Loop Space Aug 16 '13 at 19:27
  • $\begingroup$ Thanks. I'll accept this later to see if I can get any more information in the meantime. $\endgroup$ – Mohammad Safdari Aug 16 '13 at 19:52
  • 2
    $\begingroup$ @MohammadSafdari: You're welcome. There's surely a lot more that can be said, so feel free to wait :-) For further reading I recommend Eric Schechter's Handbook of Analysis and its Foundations for a thorough discussion of numerous weak forms of the axiom of choice and their uses in functional analysis. It is readable with only minimal background in set theory. The case of $(\ell_\infty)^\ast$ is discussed in the section on Pincus's Pathology. $\endgroup$ – Martin Aug 16 '13 at 21:45

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.