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?
 A: 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 $B$.

You can find a lot more about this interesting topic in
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.
