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 $x^\* = y\_1^\*=y\_2^\*.$




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> 
> **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?



Is it true that every Banach space has at least one extreme point that is normed by some point?
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 $B\_{X^\*}$.


