banach-spacesfunctional-analysis
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.
I'm not sure what you are doing will lead anywhere.
Here's what you need to show:
If $f$ is in the unit ball of $C_0$, then there exists two functions $g$ and $h$ in the unit ball of $C_0$ such that $g\ne h$ and $f$ is a nontrivial convex combination of $g$ and $h$.
To show this, you could use the following
Hint: for $f$ in the unit ball of $C_0$, eventually $|f(x)|<1/2$. Using this, find two functions $g$ and $h$ in the unit ball of $C_0$ with $f={1\over2}(h+g)$.
Very informally, you can add a "bump" and subtract the same bump to $f$ over an interval on which the inequality first mentioned holds. $g$ and $h$ will have the forms $f+b$ and $f-b$, where $b$ is the "bump". Do this in such a way that the resulting functions are in the unit ball of $C_0$. Draw a picture here...
I hope that made sense...
You are quite close: don't look at $(1\pm r)f$, look at $f\pm r$ instead.
There is $\varepsilon \gt 0$ such that the set $U = \{x : \lvert f(x)\rvert \lt 1-\varepsilon\}$ is non-empty. Let $u \in U$ be arbitrary. Urysohn's lemma gives a function $g$ such that $g(u) = 1$ and $g|_{U^c} \equiv 0$. Take $r = \varepsilon g$. Then $f = \frac{1}{2}(f+r) + \frac{1}{2}(f-r)$ shows that $f$ is not extremal because $\lvert f(x) \pm r(x)\rvert \leq 1$ for all $x$.
Best Answer
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 $\ell_1$.