4
$\begingroup$

The question is completely contained in the title :) I can only add, that it is not difficult to give a counterexample for normed spaces, and also Banach-Steinhaus theorem implies the sequential closeness. However, I don't think that this argument can be extended beyond sequences. Thank you.

$\endgroup$
5
$\begingroup$

By the bipolar theorem, this is the case precisely when the topological and algebraic duals coincide. Whether this is true or not for infinite dimensional Banach spaces depends on the set theory you are using.

Edit in response to the comments. Using results of Solovay and Schwartz, the belgian mathematician Garnir showed (in the 70's) that there are set theoretical axiom systems under which every linear functional on a Banach space is continuous.

$\endgroup$
  • 4
    $\begingroup$ More precisely, the topological dual is weak$^*$ dense in the algebraic dual. The axiom of choice thus implies that it is not closed. $\endgroup$ – Jochen Wengenroth Mar 10 '16 at 7:19
  • 1
    $\begingroup$ Interesting. Since the Hahn-Banach theorem is equivalent to the assertion that every Banach space has a nonzero continuous functional, this means that it is equivalent to stating just that there is a nonzero functional. $\endgroup$ – Asaf Karagila Mar 10 '16 at 7:27
  • 1
    $\begingroup$ The proof I know of the fact that the topological dual is dense in the algebraic dual use the Hahn-Banach theorem. (the restriction of a linear form on a space of finite dimension is continuous and then you can extend it into a continus linear form on the all space...). I think the same thing apply to Oeiras answer as far as I know (it only work assuming Hahn Banach, so the last line on the fact it depends on the set theory might not hold...) $\endgroup$ – Simon Henry Mar 10 '16 at 9:35
  • 1
    $\begingroup$ @Oeiras: about your edit: I know that. My point was that the first part of your answer (that the topological dual is closed only if it is equal to the algebraic dual) use the Hahn Banach theorem, which is of form of axiom of choice, hence is not going to be true in any model of set theory. The model you are referring to clearly does not satisfies the Hahn Banach theorem as this theorem imply the existence of discontinuous linear form on any infinite dimensional Banach space. $\endgroup$ – Simon Henry Mar 10 '16 at 14:00
  • 2
    $\begingroup$ @oeiras: More specifically, in the model of Solovay (and in all models of automatic continuity) the Hahn-Banach theorem fails fairly spectacularly. If the proof of weak-* density relies on the Hahn-Banach theorem, then the proof cannot possibly work in these models of set theory that you mention. Either there is another proof, or the theorem is just false. To show it is false it is suffices to show that there is a Banach space without nonzero continuous linear functionals, but with a discontinuous linear functional. That does not sound far fetched. $\endgroup$ – Asaf Karagila Mar 10 '16 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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