# Regarding extreme point in a Banach space

Let $$X$$ be a Banach space. And let $$X^*$$ be the dual space of $$X$$. Let $$E_X$$ and $$E_{X^*}$$ denote the extreme points of the unit ball of $$X$$ and $$X^*$$. Let $$x\in X$$ and $$|f(x)|=1$$ for every $$f\in E_{X^*}.$$ Does that imply $$x\in E_X?$$

Can anyone suggest a text to study theory of extreme points of a convex set in a Banach space(for beginners)?

• Something is wrong with your condition: $x$ cannot be "parallel" to every extreme point of $E_{X^{*}}$. – erz Feb 19 '19 at 11:03
• @erz this question is in context to the proof of Proposition 2 inthis paper sciencedirect.com/science/article/pii/S0022247X03005961 I am not understanding how they make the statement. ‘In particular, $Tx$ is in $E_Y$. – user534666 Feb 19 '19 at 11:09
• The condition is reasonable; e.g., the constant-1 function in $C[0,1]$ is such an $x$ (and it is extreme). Note that the set of all vectors satisfying the condition in the question is closed; so the best one can hope for is that $x$ is in the (norm-) closure of $E_X$. – Dirk Werner Feb 19 '19 at 11:22
• @user534666 Should $|f^*(x)|=1$ read as $|f(x)|=1$? – Skeeve Feb 19 '19 at 15:38

Suppose that $$x$$ satisfies the condition and is not extreme in the unit ball of $$X$$. It means that we can write $$x = \frac{y+z}{2}$$ for some $$y,z \in B_{X}$$ different from $$x$$. Since $$|\frac{1}{2}f(y)+\frac{1}{2}f(z)|=1$$, we get that $$f(y)=f(z)$$, so $$f(y-z)=0$$ and $$y-z\neq0$$. It holds for any $$f$$ in $$E_{X^{\ast}}$$, so it also holds for arbitrary $$f\in B_{X^{\ast}}$$ by the Krein-Milman theorem. This is a contradiction, since the dual separates points.