1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ thank you! can you provide me some hints about how to prove that affine transformations preserve extremal points? $\endgroup$ – Riccardo Nov 29 '13 at 8:28
  • $\begingroup$ done, I've work out a proof! thanks! $\endgroup$ – Riccardo Nov 29 '13 at 8:45
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.