Extreme points of unit ball of $l_1(\mathbb{N})$ Let $K$ be the closed unit ball of $l_1(\mathbb{N})$ over real numbers. Show that
$$
\operatorname{Ext}(K)= \{\pm e_n: e_n=(0,\ldots,1,0,\ldots)\}.
$$
My attempt:
I could prove that $\{\pm e_n: e_n=(0,\ldots,1,0,\ldots)\} \subset \operatorname{Ext}(K)$. But how do I prove that every extreme point is of the above form? Can anyone please guide me in this.
Let $x_1,x_2 \in K$ and $x \in \operatorname{Ext}(K)$. We know that $\|x\|=1$ and whenever $0 < \alpha <1$ and $x=\alpha x_1 +(1-\alpha)x_2$, we have $x_1=x_2=x$. How do I conclude that unless $x$ is of the form $\pm e_n$, this property cannot hold?
 A: Suppose $x$ is an extreme point. It is straightforward to see that
we must have $\sum_k |x_k| = 1$. Suppose there are two indices $i,j$ such that
$x_i \neq 0, x_j \neq 0$. Then consider
$p(t) = \sum_{k \notin \{i,j\}} x_k e_k + (x_i-(\operatorname{sgn}x_i)t)e_i + (x_i+(\operatorname{sgn}x_j)t)e_j$. We see that $p(0) = x$, $\|p(t)\| = 1$ for $t$ in
some non trivial interval $[-\delta,\delta]$, and that the
map $p:[-\delta,\delta] \to l_1$ is injective. Since
$x = {1 \over 2 } (p(\delta)+ p(-\delta))$, we see that $x$ is not an extreme
point, which is a contradiction.
A: Let $x = (x_n)_n \in K$ different from all $e_n$.
If $\|x\|_1 < 1$ then pick $\varepsilon \in \langle 0, 1\rangle$ such that $(1+\varepsilon)\|x\|_1 < 1$.
We have $\|(1+\varepsilon)x\|_1 < 1$ and $ \|(1-\varepsilon)x\|_1 < 1$ and 
$$x = \frac12(1+\varepsilon)x + \frac12 (1-\varepsilon)x$$
If $\|x\|_1 = 1$ then pick $i \in \mathbb{N}$ such that $x_i \in (0,1)$. We have
$$x = |x_i|(\operatorname{sgn} x_i \cdot e_i) + (1-|x_i|) \sum_{j\ne i}\frac{x_j}{1-|x_j|}e_j$$
where $\|\operatorname{sgn} x_i \cdot e_i\| = 1$ and $$\left\|\sum_{j\ne i}\frac{x_j}{1-|x_j|}e_j\right\|_1 = \frac{\sum_{j\ne i}|x_j|}{1-|x_j|} = 1$$
Therefore $x$ is not an extreme point of $K$.
A: If $x=(x_n)_n\in l_1$ with $\|x\|=1$ and if no $x_n$ is $\pm 1$ then $|x_n|<1$ for all $n$ and there exist $i,j$ with $i\ne j, $ such that $x_i\ne 0\ne x_j, $ because $\sum_{n\in \Bbb N}|x_n|=1.$
Let $d=\frac {1}{2}  \min(|x_i|, |x_j|, 1-|x_1|, 1-|x_j|).$
Let $y_i=x_i-d\cdot sgn(x_i)$ and $y_j=x_j+d\cdot sgn (x_j).$ Let $z_i=x_i+d\cdot sgn(x_i)$ and $z_j=x_j-d\cdot sgn (x_j).$ 
Let $y_n=z_n=x_n$ when $i\ne n\ne j.$
Then $y=(y_n)_n$ and $z=(z_n)_n$ each have norm $1,$ and $x=\frac {1}{2}y+\frac {1}{2}z.$ 
