Determine the extreme points of the unit ball of $l^1(\mathbb{Z^+})$ and $L^1[0,1]$.
My attempt: I know the definition but I don't know how to find these extreme points.Please help me to solve this problem.Thanks in advance.
Extreme point:An element $f$ of the convex subset $K$ of $X$ is said to be an extreme point of
$K$ if for no distinct pair $f_1$ and $f_2$ in $K$ is $f=\frac{f_1+f_2}{2}$
Edit: The answer is the extreme points of the unit ball of $L^1[0,1]$ are the delta functions and the unit ball of $l^1(\mathbb{Z^+})$ has no extreme points .
2nd Edit: $l^1(\mathbb{Z^+})$ denote the collection of all complex functions $f$ on $\mathbb{Z^+}$ such that $\sum_{n=0}^{\infty}|f(n)|<\infty$.It is exercise number 9 of first chapter of Banach algebra techniques in operator theory by Douglas.
3rd Edit: My previous answer was wrong,the correct answer is the extreme points of the unit ball of that $l^1(\mathbb{Z^+})$ are the delta functions and the unit ball of $L^1[0,1]$ has no extreme points .Can now someone prove this?
Best Answer
Wrong. There is no such thing as a delta function in $L^1[0,1]$. There are no extreme points of the unit ball in $L^1$. Reason: given a unit-norm function $f\in L^1$, you can find $b\in [0,1]$ such that both $f\chi_{[0,b]}$ and $f\chi_{[b,1]}$ have norm $1/2$. (Use the intermediate value theorem).
Wrong. One extreme point is $(1,0,0,0,\dots)$ and there are other such.
The difference between the two cases is that the measure space in the second cases has atoms, whereas $[0,1]$ does not. This is why the support of an $L^1$ function can be divided in two subsets of positive measure, but the support of a sequence sometimes cannot be.