# Does Bishop-Phelps Theorem hold for extreme points (slightly different version)?

Bishop-Phelps Theorem: Let $B\subseteq E$ be a bounded, closed, convex subset of a real Banach space $E.$ Then the set $$\{e^*\in E^*: e^* \text{ attains its supremum on } B \}$$ is norm-dense in the dual $E^*.$

Does the theorem hold for extreme points (slightly different version)? More precisely,

Question: Let $B\subseteq E$ be a bounded, closed and convex subset of a real Banach space $E.$ Is it true that the set $$\{e^*\in\text{ext} \left( B_{E^*} \right): e^*(e) = \|e\|\text{ for some }e\in B\}$$ norm-dense in $\text{ext}\left( B_{E^*} \right)?$

If there is an affirmative answer to my question, may I have reference?

Notation: $E^*$ is the conjugate space. $B_{E^*}$ is the closed unit ball of $E^*$. We say that $e^*$ is an extreme point of $B_{E^*}$ if it cannot be expressed as midpoint of two elements from $B_{E^*}.$ Denote $\text{ext}\left( B_{E^*}\right)$ to be the set of all extreme points of $B_{E^*}.$

• @FrancoisZiegler $B_{E^*}$ is the closed unit ball of $E^*,$ that is, $B_{E^*} = \{ e^*\in E^*: \|e^*\|\leq 1 \}$ where $\|e^*\|=\sup_{\|e\|\leq 1}|e^*(e)|.$ Jun 12, 2018 at 12:13

If $B$ is a bounded convex set that doesn't contain $0$, the Hahn-Banach separation theorem says there are $e^* \in E^*$ with $\| e^*\| = 1$ and $\epsilon > 0$ with $e^*(e) < -\epsilon$ for all $e \in B$; moreover, this is true (maybe with a smaller $\epsilon$) in a neighbourhood of $e^*$. If $E^*$ is strictly convex, so every point of its unit sphere is an extreme point, your set is not dense in $\text{ext}(B_{E^*})$.