# Extreme points of the unit ball of a Banach space

Let $$X$$ be a Banach space and let $$B_X$$ denote the closed unit ball of $$X$$. Let $$\mathrm{Ext}(B_X)$$ denote the set of all extreme points of the unit ball $$B_X$$.

Whenever $$X$$ is finite-dimensional, it is known that $$\mathrm{Ext}(B_X)\neq \emptyset.$$ Are there instances, where $$\mathrm{Ext}(B_X)=\emptyset$$?

Does completeness plays any role here? Is it true that we will always have $$\mathrm{Ext}(B_X)\neq \emptyset$$ for $$X$$ is Banach and $$\mathrm{Ext}(B_X)$$ may be empty in case of $$X$$ is not complete?

• Hint: consider $L^1[0,1]$ Commented Aug 7, 2021 at 14:27
• Or $c_0$.${}{}{}$ Commented Aug 7, 2021 at 14:28
• $c_0$ space is incomplete where as $L^1[0,1]$ is complete. So completeness has actually no role here. Am I correct?@DavidMitra
– pmun
Commented Aug 7, 2021 at 14:41
• $c_0$ is complete. One result is that the closed unit ball of every dual space has extreme points. Commented Aug 7, 2021 at 14:45
• You may want to look into Krein-Milman theorem. en.wikipedia.org/wiki/Krein%E2%80%93Milman_theorem
– daw
Commented Aug 8, 2021 at 17:11

As noted in the comments, for the two Banach spaces $$c_0$$ and $$L^1[0,1]$$, the unit ball has no extreme points.
Here is $$L^1[0,1]$$.
Write $$\|\cdot\|$$ for the $$L^1$$ norm. Write $$B := \{f \in L_1 : \|f\| \le 1\}$$ for the unit ball. Let $$h \in B$$. We claim $$h \not\in \operatorname{Ext}(B)$$.
Case 1. $$h = 0$$. Then $$h = \frac12(\mathbf1 + (-\mathbf1))$$ where $$\mathbf1$$ is the constant function with value $$1$$. Since $$\mathbf1 \in B$$ and $$-\mathbf1 \in B$$ and $$\mathbf1 \ne -\mathbf1$$, we conclude that $$h$$ is not an extreme point of $$B$$.
Case 2. $$0 < \|h\| \le 1$$. Define the function $$\phi : [0,1] \to \mathbb R$$ by $$\phi(t) = \int_0^t |h|$$ Then $$\phi$$ is continuous, $$\phi(0) = 0$$ and $$\phi(1) = \|h\| > 0$$. so there is $$t_0 \in (0,1)$$ so that $$\phi(t_0) = \frac12 \|h\|$$. Now $$h = \frac12(h_1+h_2)$$ where $$h_1 = 2 \mathbf1_{[0,t_0]} h, \qquad h_2 = 2 \mathbf1_{(t_0,1]} h$$ Then $$\|h_1\| = \|h_2\| = \|h\| \le 1$$ so $$h_1, h_2 \in B$$. Also, $$h_1 \ne h_2$$. So we again conclude that $$h$$ is not an extreme point of $$B$$.
Diestel, J.; Uhl, J. J., Vector measures, Mathematical Surveys. No. 15. Providence, R.I.: American Mathematical Society (AMS). XIII, 322 p. $35.60 (1977). ZBL0369.46039. • Thank you sir, for this detailed answer. Can we avoid the first case, as an extreme point must have norm$1$. @GEdgar – pmun Commented Aug 7, 2021 at 15:32 • +1 Very nice! @pmun as a slight generalisation, you may be interested in noting that$L^1(\mu )$admits extreme points if and only if$\mu\$ admits an atom of finite positive measure. Commented Aug 7, 2021 at 15:44