Extreme points of unit ball of Banach spaces $\ell_1$, $c_0$, $\ell_\infty$

Question

Find extreme points of the unit balls of each Banach space, $l^1 $, $c_0$, $ l^\infty$

Can you help me with this one?

For the first space, $l^1$, I thought there was no extreme point, but apparently, this is not the answer :(

And I don't know if the fact that $l^\infty$ contains $c_0$ matters in this problem.

Thanks.

Answer 1

$c_0$ has no extreme points.

Let $x \in c_0$ unit ball. Then, $x_n \to 0$. Choose $N$ s.t. $|x_n| < 1/2$ for $n\geq N$ and use two sequences which match $x_n$ for $n \leq N$ and are $x_n + 2^{-n}$ and $x_n - 2^{-n}$ for $n > N$. Both of these sequences will be in the unit ball and their average will be $x$.

You can try $\ell^\infty$ on your own (the result is not the same as $c_0$) -- look at points which have all their coordinates with magnitude $1$.

Answer 2

Indeed, $\ell_1$, has a lot of extreme points. For example, take $$ e^{(n)}=(\underbrace{0, \cdots, 0}_{n-1},1, 0, \cdots) $$ And let $B$ be the closed unit ball in $\ell_1$. Clearly, $e^{(n)}\in B$. Now suppose that for $t \in (0,1)$, you have $b=\{ b_j \}_j , d=\{ d_j \}_j$ with $b,d \in B$ such that $$ e^{(n)} = t b+ ( 1- t )d $$ Then, you must have $1=tb_n+(1-t)d_n$ and $0=tb_j+(1-t)d_j$ for $j\neq n$ ,which gives that $$ b_j=d_j=\left\{ \begin{array}{cc} 1 & \text{if } j=n \\ 0 & \text{if } j\neq n \end{array} \right. $$ Thus $b=d=e^{(n)}$, proving that $e^{(n)}$ is an extreme point of $B$ for all $n$.

Now you basically only need to answer if there are any more extreme points than the $ e^{(n)}$ to see all the extreme points of the space $\ell_1$.