Can anyone prove that the spaces $ (\mathbb{R}^3, \|.\|_1) $ and $ (\mathbb{R}^3, \|.\|_\infty) $ cannot be isometric?
Thanks.
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Are you sure of this? Any two norms on a finite-dimensional space are equivalent, i.e., they generate the same topology. – user99680 Nov 23 '13 at 23:32
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1@user99680 This doesn't exclude the fact that an isometry cannot exist. – egreg Nov 23 '13 at 23:35
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@user99680 Yeah, I am. This is stronger than equivalency between norms on finite dimensional spaces. – Axiom Nov 23 '13 at 23:35
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Did you try to compare their unit spheres/balls? – Julien Nov 23 '13 at 23:36
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@egreg, right, I read my own problem: show they're not homeomorphic. I guess the homeomorphism between $(-1,1))$ and $\mathbb R$ is aan example. – user99680 Nov 23 '13 at 23:37
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@julien, Compare in what sense exactly? – Axiom Nov 23 '13 at 23:37
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@Shl As Platonic bodies. (Note that $(\mathbb{R}^2,\lVert\cdot\rVert_1)$ and $(\mathbb{R}^2,\lVert\cdot\rVert_\infty)$ are isometric.) – Daniel Fischer Nov 23 '13 at 23:37
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I'm not following, guys. Sorry. Could you be a little bit more clear? Thanks. – Axiom Nov 23 '13 at 23:38
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Try to draw them and count the extreme points. – Julien Nov 23 '13 at 23:46
2 Answers
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Let $B$ be a until ball in one of these spaces. (That is, pick a point and look at all points with distance less than or equal to one from that point.) Look at those points $x$ in $B$ such that there exists another unit ball $B'$ such that $B \cap B' = \{x\}$. In one of those spaces there are 8 of these points, in the other only 6.
Stephen Montgomery-Smith
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Here is how I would approach this problem.
Let $f:V\to W$ is an isometry between two normed-spaces. If $v_n$ converges to $v$ in $V$ then $|v_n - v| \to 0$ and since $f$ is an isometry it follows that $|f(v_n) - f(v)| = |f(v_n-v)| = |v_n-v| \to 0$. Thus, $f(v_n)$ converges to $f(v)$ in $W$.
Thus, if you can find a sequence in $\mathbb{R}^3$ which converges with respect to one norm but not the other it means the two normed spaces are not isometric.
Nicolas Bourbaki
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2All the norms on $\mathbb{R}^3$ are equivalent (that is, they induce the same topology). Therefore it's impossible to find a sequence that converges with respect to one norm and not the other. – egreg Nov 23 '13 at 23:29
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Yeah, exactly. I'm sorry, but that's not the way to go about this problem. – Axiom Nov 23 '13 at 23:31