4
$\begingroup$

If $E$ is a non-reflexive Banach space then there exists a linear functional $\lambda \in E^*$ of norm one such that $\lambda v < 1$ for all $v \in E$ of norm one. However, in the only non-reflexive examples that I somewhat understand ($\ell_1$, $c_0$) this does not happen when $\lambda$ is an extreme point of the closed unit ball of $E^*$.

Question: does there exist a Banach space $E$ and $\lambda \in E^*$ such that $\lambda$ is an extreme point of the unit ball and yet does not norm any (non-zero) vector of $E$?

$\endgroup$
2
$\begingroup$

Would you like a solution that does not involve any example that is new for you?? By James' theorem, which you mentioned, it would be enough to have a space $E$ that is not reflexive but every linear functional of norm one is an extreme point of the unit ball of $E^*$. In other words, you want the dual norm on $E^*$ to be strictly convex. But M. M. Day, Strict convexity and smoothness of normed spaces, Trans. Amer. Math. Soc. 78 (1955) 516–528, proved that every separable space $E$ has an equivalent norm whose dual norm is strictly convex. So you only need to know one example--a separable non reflexive space.

$\endgroup$
3
  • $\begingroup$ As an aside, is there then an analogue of the Bishop-Phelps theorem restricted to extreme norm one linear functionals? All I can prove is weak$^*$ density (which is enough for my needs). $\endgroup$ – Itaï BEN YAACOV Oct 16 '14 at 18:41
  • 1
    $\begingroup$ Yes, if the dual of the space is separable or, more generally, if every separable subspace has separable dual ("Asplund space"). Then the dual ball is the norm closed convex hull of its extreme points. No, in general (consider $C[0,1]). $\endgroup$ – Bill Johnson Oct 16 '14 at 19:01
  • $\begingroup$ Thanks but that is not what I meant - given an extreme point of the dual ball, can it be expressed as a norm limit of extreme points which norm something? No convex hull taken. $\endgroup$ – Itaï BEN YAACOV Oct 17 '14 at 6:09

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.