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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?

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  • $\begingroup$ 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. $\endgroup$
    – Gerald Edgar
    Commented 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
    Commented Oct 24, 2018 at 14:39
  • $\begingroup$ Yes, it looks good now. And Robert Israel answered already. $\endgroup$
    – Gerald Edgar
    Commented Oct 24, 2018 at 14:43

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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 $B_{X^*}$.

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

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