Verify that the unit closed ball in $c_0$ has not extremal points

Question

I want to understand this proof of the fact that $c$ and $c_0$ aren't isometrically isomorphic, but I have very little experience in working with extremal points.

So how can I verify that the unit ball in $c_0$ didn't have one of these points? I tried to manipulate a generic sequence to obtain two difference sequences but the fact that I have to use a convex combination gives me problems.

And how can i prove that the extremal points are preserved via isometry?

Thanks in advance

Answer 1

If $x=(x_j )_{j\in\mathbb{N}} $ is any point of unit ball $B$ in $c_0 $, then there exists $k\in\mathbb{N} $ such that $|x_k|<\frac{1}{2} $ and therefore $ x=\frac{1}{2} u+\frac{1}{2} v $ where $$u_j=\begin{cases} x_j \mbox{ for } j\neq k\\ x_k-\frac{1}{2} \mbox{ for } j=k\end{cases}$$ $$v_j=\begin{cases} x_j \mbox{ for } j\neq k\\ x_k +\frac{1}{2} \mbox{ for } j=k\end{cases}$$ It is easy to see that $u,v\in B $ and thus $B$ have no extremal points.

If $I$ is an isometry between the Banach spaces $X$ and $Y$ then by Mazur - Ulam Theorem it must be affine, but affine injections preserved extremal points.

Answer 2

Suppose that $W\subset X$ is convex set, $T:X\rightarrow Y$ injective affine transformation and $x\in W$ is extremal point of $W$. Then $T(x)$ is extremal point of $T(W) .$

Indeed, suppose that $T(x) =sT(u) +t T(v) $ where $s,t\geq 0, s+t =1 , u\neq v\in W ,$ then $x=T^{-1} (sT(u) +t T(v) )=sT^{-1} T(u ) +t T^{-1} T (v) =su +tv $ and therefore $x$ is not an extreme point of $W.$ Contradiction.