3
$\begingroup$

Definition: Let $X$ be a Banach space and $X^*$ be its continuous dual of $X,$ that is, $X^*$ contains all bounded linear functionals on $X.$ Denote $$B_{X^*} = \{x^*\in X^*: \|x^*\|_{X^*}\leq 1\}.$$ We say that $x^*\in B_{X^*}$ is an extreme point of $B_{X^*}$ if whenever $$x^* = \frac{1}{2}(y_1^*+y_2^*)$$ for some $y_1^*,y_2^*$ with $\|y_i^*\|_{X^*} \leq 1$ for all $i=1,2,$ we have $x^* = y_1^*=y_2^*.$


Question: Given a Banach space $X.$ Is it true that there exists an extreme point $x^*$ of $B_{X^*}$ such that $$x^*(x) = 1$$ for some $x\in X$ with $\|x\|\leq 1?$

In other words, is it true that every Banach space has at least one extreme point that is normed by some point?

$\endgroup$
3
  • $\begingroup$ I do not like your wording. I would say you defined extreme point not of $B_X$ but of $B_{X^*}$. And then you ask a question about extreme functional not extreme point. I think you should re-write the question. $\endgroup$ Oct 24, 2018 at 13:35
  • $\begingroup$ @GeraldEdgar Thanks for pointing out my mistake. I have modified my question. Do you think it is okay now? $\endgroup$
    – Idonknow
    Oct 24, 2018 at 14:39
  • $\begingroup$ Yes, it looks good now. And Robert Israel answered already. $\endgroup$ Oct 24, 2018 at 14:43

1 Answer 1

7
$\begingroup$

Take any $x$ with $\|x\|=1$. $S(x) = \{x^* \in X^*: x^*(x) = \|x^*\| = 1 \}$ is a nonempty (by Hahn-Banach) weak-* compact convex set, so by Krein-Milman it has extreme points. Any extreme point of $S(x)$ is an extreme point of $B_{X^*}$.

$\endgroup$
1
  • $\begingroup$ Seriously, how can I forgot Krein-Milman theorem.... $\endgroup$
    – Idonknow
    Oct 24, 2018 at 14:45

Your Answer

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

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