Recall the Bishop-Phelps Theorem.

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 $B_{E^*}.$

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

1 Answer 1


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 $\text{ext}(B_{E^*})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.