showing to be extreme subset (might use Hahn decomposition Theorem) I am studying Functional analysis by myself and stumbled this question and am completely at a loss.
We want to show that $\{ f \in L^1 [0,1 ] : ||f|| =1 \}$ is an extreme subset of $\{ \mu \in C[0,1]^{\ast} : ||\mu||=1$.
In the hint it says to use Hahn decomposition theorem but I am not sure how to go about this type of problem. Any help would be appreciated.
Thnx.  
 A: Suppose that $f\in L^1[0,1]$ is non-negative with $\|f\|_1=1$. If $f=t\mu_1+(1-t)\mu_2$ for measures with $\|\mu_j\|=1$, this means that for every $g\in C[0,1]$
$$\tag{1}
\int_0^1 gf\,dm=t\int_0^1 g\,d\mu_1+(1-t)\int_0^1 g\,d\mu_2.
$$
Taking $g=1$, we get $1=t\mu_1(1)+(1-t)\mu_2(1)$. As $|\mu_j(1)|\leq\|\mu_j\|=1$, we get $\mu_j(1)=1$, $j=1,2$. This forces $\mu_1,\mu_2$ to be positive (since a unital functional of norm one is positive). 
Using dominated convergence and the fact that continuous funtions are dense in $L^1$, we can push (1) to the case where $g\in L^\infty[0,1]$ (in particular, $g$ can be any bounded measurable function). 
Taking characteristic functions, we get
$$
\int_\Delta\,f\,dm=t\mu_1(\Delta)+(1-t)\mu_2(\Delta).
$$
This shows that $\mu_1,\mu_2$ are absolutely continuous with respect to $f\,dm$. So there exist Radon-Nikodym derivatives $h_1,h_2\geq0$ such that 
$$
\mu_j(\Delta)=\int_\Delta\,h_jf\,dm
$$
So now we have 
$$
\int gfdm=t\int gh_1fdm+(1-t)\int gh_2fdm=\int gf(th_1+(1-t)h_2)\,dm
$$
for every bounded measurable function $g$. So $f(th_1+(1-t)h_2)=f$ a.e., from where we conclude that $th_1+(1-t)h_2=1$ a.e. (their value on the set where $f=0$ is irrelevant and we can make it zero). As $h_1,h_2\geq0$ (from $\mu_1,\mu_2\geq0$), we deduce that $h_1,h_2\leq1$ a.e., i.e. they are in $L^\infty[0,1]$. Then $h_jf\in L^1[0,1]$ and $\mu_j=h_jf\,dm$.
Now consider a general $f$ (i.e. not necessarily positive) with $\|f\|_1=1$. Write its Hahn Decomposition as $f=f^+-f^-$. This means that there exists $E\subset[0,1]$ with $1_E\,f=f^+$, $-1_{E^c}f=f^-$. 
If $f\,dm=t\mu_1+(1-t)\mu_2$, then 
$$
f^+\,dm=t\mu_1(E\cap\cdot)+(1-t)\mu_2(E\cap\cdot),\ \ f^-\,dm=t(-\mu_1(E^c\cap\cdot))+(1-t)(-\mu_2(E^c\cap\cdot))
$$
Now we can apply the first part of the proof to normalized versions of each of $f^+,f^-$ (so that they have $\|\cdot\|_1=1$) and we get that $\mu_j=(h_j^+-h_j^-)f\,dm$, $j=1,2$. So the unit sphere is extreme. 
