Find extremal points in the closed unit ball of : $c$, $l^p$ $L^p[0,1]$ , $L^\infty[0,1]$ I seek to find all extremeal points in the following closed unit balls:
All spaces are over reals and all sequences can be assumed to have real entries. 
$c$ the set of all sequences that converge.
intuitively I feel like the constant sequences 1,1,.... and -1,-1,... are the only extremal points but I cannot show this precisely.
$l^p$ , $1<p<\infty$ The space of p-summable sequences.
very stuck here, cannot even think of a candidate
I have done (and am quite certain of) the spaces; $c_0$, $C[0,1]$, $L_1[0,1]$ so if there is any way to extend these results I can use those. Thanks in advance. 
EDIT: Ok so I can show that the extreme points, if any exist, in $L^\infty$[0,1] are those on the unit sphere. I suspect any function on the unit sphere to be extreme but I cannot show this yet.
 A: I will assume that you are considering real sequences. The modifications for the complex case are quite easy. 
Answer for c: If $(x_n)$ has norm $1$ and $|x_k| <1$ for some $k$ then replace $x_k$ by $x_k+r$ and $x_k-r$ with $0<r<1-|x_k|$ (and keep all other coordinates unchanged) to get two elements $y$ and $z$ such that $x=\frac  {y+z} 2$. Hence the only extreme points are sequences with $x_n=\pm 1$ for all $n$. [ Such sequences are indeed extreme points]. 
In $\ell^{p}$ with $1<p<\infty$ every point with norm $1$ is an extreme point. Hint: There exists $n$ such that $|x_n| <1$. Try to write $(x_n)$ as an average of two sequences of norm $1$. 
A: Recall that if $X$ is a Banach space, and $x\in \overline {B_X(0,1)}$ is extreme, then $\|x\|=1.$ This is so because if $x=0$ then if we pick any norm $1\ y\in \overline {B_X(0,1)}$ then $x$ lies interior to the line segment $[-y,y].$ And if $0<x<1$, then $x$ lies in the interior of the line segment $\left[0,\frac{x}{\|x\|}\right]$ (because $x=(1-\|x\|)\cdot 0+\|x\|\cdot \frac{x}{\|x\|}$).
So it suffices to check only the points $x$ on the unit sphere, and since  $\overline {B_{\ell^p}(0,1)}$  is convex, we may consider, towards a contradiction, $z\neq w\in \overline {B_{\ell^p}(0,1)}$ such that $x=\frac{1}{2}z+\frac{1}{2}w.$ Then, using Minkowski's inequality, $1=\left \|\frac{z+w}{2}\right \|_p \le \left \|\frac{z}{2}\right \|_p+\left \|\frac{w}{2}\right \|_p$, which implies that $z$ and $w$ are norm $1$ vectors and therefore in fact, that equality holds in Minkowski. And this can only happen (recall, $p>1$), if there are numbers $r$ and $s$ such that $rz=sw$ and since $z$ and $w$ are unit vectors, we have $r=s$ and therefore $z=w$, which is a contradiction.
