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)?

  • $\begingroup$ Something is wrong with your condition: $x$ cannot be "parallel" to every extreme point of $E_{X^{*}}$. $\endgroup$ – erz Feb 19 '19 at 11:03
  • $\begingroup$ @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 $E_Y$. $\endgroup$ – user534666 Feb 19 '19 at 11:09
  • $\begingroup$ 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 $E_X$. $\endgroup$ – Dirk Werner Feb 19 '19 at 11:22
  • $\begingroup$ @user534666 Should $|f^*(x)|=1$ read as $|f(x)|=1$? $\endgroup$ – 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.

I seem to have avoided the closure issue noted by Dirk Werner in the comment section. Is this correct?

  • $\begingroup$ @DirkWerner: I don't think so. It's just the set of extreme points norming all the extreme functionals that is closed; usually it's only a small portion of the set of extreme points. $\endgroup$ – Mateusz Wasilewski Feb 19 '19 at 12:39
  • $\begingroup$ Good point! (I have deleted my erroneous comment.) $\endgroup$ – Dirk Werner Feb 19 '19 at 14:14

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