# Application of the Krein Milman Theorem about ball $l^1$

Show that ball $l^1$ is the norm closure of the convex hull of its extreme point

Define the extrem point. A point in $K$ is an extreme point of $K$ if there is no proper open line segment that contains point and lies entirely in $K$.

The Krein Milman Theorem: If $K$ is a nonempty compact convex subset of a locally convex subspace X, then $K=\bar{co}$(ext $K$). where ext($K$): the set of extreme points of K and co($K$): convex hall of K

Then, It suffices to prove that ball $l^1$ is nonempty compact convex set.

But, I don't know how to prove. Can I get some hints?

• You mean the closed ball presumably. Commented May 27, 2016 at 19:18
• Unless you are in finite dimensions the closed unit ball is not compact. Commented May 27, 2016 at 19:19
• Why don't you first figure out what the extreme points are. Use finite dimensions as a guide. Or, in fact, the definition of $\|\cdot \|_1$. Commented May 27, 2016 at 19:21
• Right... Then, this problem cannot apply Krein Milman Theorem... In infinite dimension, Does the proposition hold? Commented May 27, 2016 at 19:29
• The statement is true. You need to figure out what the extreme points are first. Not too hard to guess by looking at the planar case. Commented May 27, 2016 at 19:30

Consider the space $\ell_1$ with the weak star topology. It is a locally convex topological wector space, and by Banach - Alaoglu theorem the unit ball is compact in this space. Hence by Krein - Milman theorem it is a closed convex envelope of its extreme points.

• But, unit ball is compact in weak* topology... Can we apply the Krein-Milman theorem? Commented May 27, 2016 at 19:34
• The weak-* topology would be on $(l_1)^*$. Commented May 27, 2016 at 19:35
• No $\ell_1 =(c_0)^*$ therefore we can consider on $\ell_1$ a weak star topology.
– user235708
Commented May 27, 2016 at 19:38
• I think that unit ball is compact in weak-* topology on $(l^1)^{*}$. Commented May 27, 2016 at 19:45

Here is a simple solution (albeit I am partial to sledgehammer approaches):

It is not too hard to show that the extreme points of the closed unit ball are $\pm e_k$.

Since $\overline{B}(0,1)$ is convex and closed it follows that $\operatorname{\overline{co}} \{ \pm e_k \}_k \subset \overline{B}(0,1)$.

Suppose $x \in \overline{B}(0,1)$, and let $x_n =\sum_{k=1}^n x(k) e_k$. Note that $x_n \in \operatorname{co} \{ \pm e_1,...,\pm e_n \}$, and $x_n \to x$ (in norm). Since $\operatorname{\overline{co}}\{ \pm e_k \}_k$ is closed, it follows that $x \in \operatorname{\overline{co}} \{ \pm e_k \}_k$.

• Can you explain about $e_k$ in $l^1$?? And, Do you assume that ball $l^1$ in finite dimension? Commented May 27, 2016 at 19:50
• $e_k$ is the $k$th unit vector. Is that what you are asking? Commented May 27, 2016 at 19:51
• Why $x_n$ always express finite linear combination? Commented May 27, 2016 at 19:57
• $x_n$ is just the first $n$ components of $x$, and $\sum_{k=1}^n |x(k)| \le 1$. Hence you can write $x_n = \sum_{k=1}^n |x(k)| (\operatorname{sgn} x(k)) e_k + (1-\sum_{k=1}^n |x(k)|) {1 \over 2}(e_1+ (-e_1)$. It is not too hard to check that this is a convex combination of $\pm e_1,...,\pm e_n$. Commented May 27, 2016 at 20:01