Consider the Banach space $c$ of convergent sequences over $\mathbb{C}$ with the infinity norm defined by $\lvert\lvert(x_n)_{n\in\mathbb{N}}\rvert\rvert_\infty = \operatorname{sup}(\{x_n\in\mathbb{C}:n\in\mathbb{N}\})$. I am trying to characterize all extreme points of the closed unit ball. So far I have found that every extreme point must have norm $1$, i.e. it contains a $1$ or its limit is $1$.
I hope that these would be precisely the extreme points, but I am unable to show that either of these assumptions (containing a $1$ or converging to $1$) necessarily implies being an extreme point.
When we assume such a sequence is not extreme, then we can only find that certain $x_n$ are extreme points in the closed unit ball of $\mathbb{C}$, but not all of them. So we cannot conclude that $x$ is extreme. My question is which sequences are the extreme points or any hints that will help me along.
Best Answer
As noted by kimchi, if there is $i \in \mathbb{N}$ with $|x_i| < 1$ then $x$ is not an extreme point.
Now consider $x \in c$ with absolute value of all of its coordinates equal to $1$. Suppose $x = \frac{1}{2} y + \frac{1}{2} z$ for some $y,z \in c$. For arbitrary $n \in \mathbb{N}$ we have $|x_n| = 1$ hence $x_n$ in an extreme point of the unit ball $B_{\mathbb{C}}$ of $\mathbb{C}$. As $y_n,z_n \in B_{\mathbb{C}}$ and $x_n = \frac{1}{2}y_n + \frac{1}{2}z_n$, we get $x_n =y_n=z_n$. Now $n\in \mathbb{N}$ was arbitrary, hence, $x=y=z$ and $x$ is an extreme point.
To sum up the extreme points of $c$ are exactly the convergent sequences of complex units.