# Is it true that every Banach space has at least one extreme point that is normed by some point?

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?

• 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. – Gerald Edgar Oct 24 '18 at 13:35
• @GeraldEdgar Thanks for pointing out my mistake. I have modified my question. Do you think it is okay now? – Idonknow Oct 24 '18 at 14:39
• Yes, it looks good now. And Robert Israel answered already. – Gerald Edgar Oct 24 '18 at 14:43

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 $$B_{X^*}$$.