# 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. – copper.hat May 27 '16 at 19:18
• Unless you are in finite dimensions the closed unit ball is not compact. – copper.hat May 27 '16 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 $\|\cdot \|_1$. – copper.hat May 27 '16 at 19:21
• Right... Then, this problem cannot apply Krein Milman Theorem... In infinite dimension, Does the proposition hold? – asfajaf May 27 '16 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. – copper.hat May 27 '16 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? – asfajaf May 27 '16 at 19:34
• The weak-* topology would be on $(l_1)^*$. – copper.hat May 27 '16 at 19:35
• No $\ell_1 =(c_0)^*$ therefore we can consider on $\ell_1$ a weak star topology. – MotylaNogaTomkaMazura May 27 '16 at 19:38
• I think that unit ball is compact in weak-* topology on $(l^1)^{*}$. – asfajaf May 27 '16 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? – asfajaf May 27 '16 at 19:50
• $e_k$ is the $k$th unit vector. Is that what you are asking? – copper.hat May 27 '16 at 19:51
• Why $x_n$ always express finite linear combination? – asfajaf May 27 '16 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 $\pm e_1,...,\pm e_n$. – copper.hat May 27 '16 at 20:01