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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?

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

1 Answer 1

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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.

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    $\begingroup$ Thank you sir, for this detailed answer. Can we avoid the first case, as an extreme point must have norm $1$. @GEdgar $\endgroup$
    – pmun
    Commented Aug 7, 2021 at 15:32
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    $\begingroup$ +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. $\endgroup$
    – Jose Avilez
    Commented Aug 7, 2021 at 15:44

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