Extreme points of the unit ball of $l^1(\mathbb{Z^+})$ and $L^1[0,1]$ 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?
 A: I'm answering my question in the comments, show explicitly what @user147263 mentioned above.
$\,$
Pick any $f \in B_1(0,L^1([0,1]))$, the closed unit ball of $L^1([0,1])$.
The function $F : \mathbb R \rightarrow \mathbb R$, $t \mapsto \int_0^t |f(x)|\,dx$, is continuous, since for all $t_0, \, t_1 \in [0,1]$ there holds 
$$|F(t_0) - F(t_1) | = \left| \int_{t_1}^{t_0} |f(x)| \, dx \right| \leq ||f||_{L^1([0,1])} \, |t_0 - t_1|, $$ hence $||F||_{L(\mathbb R)} \leq ||f||_{L^1([0,1])} < \infty$.
Using the intermediate value theorem we can find $c \in [0,1]$, such that $F(c) = \frac{1}{2} ||f||_{L^1}$. 
Let now $g_1 := 2 \, f|_{\chi_{[0,c]}}$, $g_2 := 2\, f|_{\chi_{[c,1]}}$. Then $g_1, \, g_2 \in B_1(0,L^1([0,1]))$ and $g_1 \neq f \neq g_2$, on the other hand $$f = \frac{1}{2} g_1 + \frac{1}{2} g_2.$$
Thus $f$ is not extremal, therefore there are no extremal points in $B_1(0,L^1([0,1]))$.
A: 
The answer is the extreme points of the unit ball of $L^1[0,1]$ are the delta functions

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).

the unit ball of $l^1(\mathbb Z^+)$ has no extreme points 

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.
