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I am trying to prove that $\mathcal{c}$ and $\mathcal{c_0}$ are isomorphic but not isometrically isomorphic.

I've read on this forum that isometric isomorphism preserves extreme points, but I don't see why. I know that I should use the fact that isometries preserve all metric properties and being an extreme point is a metric property but I can't come up with any proof.

If $K$ is a convex set and $a \in K$ is its extreme point, then whenever $a = \frac{x+y}{2}, \ x,y \in K, $ we have that $x=y=a$.

What should I do now?

Could you help me with that?

Thanks.

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    $\begingroup$ Let $f\colon X \to Y$ be the isometric isomorphism. Let $C = f(K)$. You want to show that $f(a)$ is an extreme point of $C$. So suppose $f(a) = \frac{u+v}{2}$. Then - what corresponds to $u$ and $v$ in $X$? $\endgroup$ – Daniel Fischer Oct 4 '14 at 13:23
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    $\begingroup$ Yes, that's correct. And that shows that the image of an extreme point is an extreme point. Actually, we didn't need that $f$ be an isometric isomorphism to deduce that. $\endgroup$ – Daniel Fischer Oct 4 '14 at 13:47
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    $\begingroup$ It's right. But yes, you need more: The unit ball of $c$ has extreme points with norm $1$, the unit ball of $c_0$ doesn't. $\endgroup$ – David Mitra Oct 4 '14 at 14:02
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    $\begingroup$ An isometry would map the unit ball of $c$ onto the unit ball of $c_0$. $\endgroup$ – David Mitra Oct 4 '14 at 14:19
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    $\begingroup$ The second has a small typo. (I would argue by observing the only way $1=(a+b)/2$ with both $|a|\le1$ and $|b|\le1$ is if $a=b=1$; but this is essentially what you did.) $\endgroup$ – David Mitra Oct 4 '14 at 14:30

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