5
$\begingroup$

The algebraic dual of a normed vector space is the space of all linear functionals to the ground field (either $\mathbb{R}$ or $\mathbb{C}$ for this question). The continuous dual is the subspace of all continuous linear functionals. Under certain conditions, it's quite easy to find an element of the algebraic dual of a space that is not in the continuous dual:

  1. We can start with a particular vector space, say $\ell^0$, and put on it two inequivalent norms, say $\|-\|_1$ and $\|-\|_2$. As they are inequivalent, there will be some functional that is continuous with respect to the one but not the other, in this case the functional $(x_n) \to \sum x_n$ will do.

  2. We can start with a normed vector space which has a topological basis which we then extend to a Hamel basis. Assuming that this extension is non-trivial, we can define a functional which is zero on the original topological basis but non-zero on some element of the extension. Unfortunately, the existence of a Hamel basis depends on some version of choice (not sure if it's equivalent to AC or not, but that's not important).

  3. We can modify the previous construction to avoid Hamel bases by using the Hahn-Banach theorem.

What I would like is an example of a non-continuous functional on a Banach space that can be explicitly written down. So I don't want to use choice or HBT. If there were an example of a normed vector space with two inequivalent norms (may as well assume that $\|-\|_a$ is strictly weaker than $\|-\|_b$) so that the space is a Banach space with respect to the weaker norm then method (1) would work, but that would contradict the open mapping theorem (unless there's some subtlety that I can't see right now).

This builds on What’s an example of a space that needs the Hahn-Banach Theorem?. I'm not interested in throwing choice or HBT out of the window, but it's useful to know (from a pedagogical angle if nothing else) exactly where they are needed and how far one can go without them. One thing that surprised me a little in the answers to that question was that you need choice/HBT to see that $\ell^1$ is not the continuous dual to $\ell^\infty$. That made me wonder about the algebraic dual and whether the same is true for that, or not.

Improve this question
$\endgroup$
8
  • $\begingroup$ I thought that any unbounded operator on an infinite dimensional banach space is not continuous. So couldn't you just come up with an unbounded operator? I may have misunderstood the question. $\endgroup$
    – Harry Gindi
    Dec 10, 2009 at 11:09
  • 1
    $\begingroup$ @Harry Gindi, To come up with an unbounded, everywhere defined operator one typically uses a Hamel basis. @Whomever, I suspect there is no way to do this without a Hamel basis, and no Hamel basis of any Banach space without some form of AC, but I will be happy to be shown wrong. $\endgroup$
    – Jonas Meyer
    Dec 10, 2009 at 11:21
  • 1
    $\begingroup$ @Harry (1): Write one down and I'll accept it as the answer! @Jonas and Harry (2): that's my understanding too, but I don't have any sense of where the line should be drawn. As an example, we almost always prove HBT using Zorn's lemma but it's strictly weaker. Similarly, we often say "By HBT there is a functional ..." when in concrete applications one can often just write it down. Pedagogically, I'd rather say "In this case, here's the functional, in general we need to use HBT and here's an example where it's needed.". $\endgroup$
    – Andrew Stacey
    Dec 10, 2009 at 11:47
  • 1
    $\begingroup$ @Harry, What is the norm you're considering on C^1? I don't think there is a Banach space norm you can put on C^1 with respect to which differentiation is unbounded. $\endgroup$
    – Jonas Meyer
    Dec 10, 2009 at 12:18
  • 2
    $\begingroup$ @Harry: No. To make sense of that, I have to assume that you are using the C^0 norm on C^1 functions in which case it's not a Banach space. $\endgroup$
    – Andrew Stacey
    Dec 10, 2009 at 12:19

2 Answers 2

Reset to default
5
$\begingroup$

I believe that part of Greg Kuperberg's answer to this other question answers your question by showing that no such "explicit" functional exists. Namely, he cites an assertion in an article by N. Brunner that implies that continuity of all functionals on all Banach spaces is consistent with ZF (a more precise statement is found at the link).

Improve this answer
$\endgroup$
1
  • $\begingroup$ I suppose I ought to be mildly embarrassed that I didn't spot that answer! But I'm not. That is totally bizarre. Now, I wonder if there's a system wherein this holds as well as HBT. That really would be a strange axiomatic setup. $\endgroup$
    – Andrew Stacey
    Dec 10, 2009 at 12:25
3
$\begingroup$

Alternatively (or maybe it is the same...?) in Solovay's model (which has DC, so I don't have to tell you what I mean by "continuous") where every subset of R has the property of Baire, it follows that every linear functional on a Banach space is continuous.

Improve this answer
$\endgroup$
1
  • $\begingroup$ From another answer of Greg Kuperberg at mathoverflow.net/questions/5351/…, it appears Solovay's model also implies that HB does not hold, which is relevant to Andrew's comment on my answer. I don't know how this relates to the situation in Brunner's paper. $\endgroup$
    – Jonas Meyer
    Dec 10, 2009 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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