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Let $X$ be a vector space, $||\cdot||_1$ and $||\cdot||_2$ two equivalent norms on $X$. Under what further assumptions can we prove there is an isometry between $(X,||\cdot||_1)$ and $(X,||\cdot||_2)$?

In particular, are $(\mathbb{R}^2,||\cdot||_1)$ and $(\mathbb{R}^2,||\cdot||_2)$ isometric (where the two norms are the maximum norm and the usual euclidean norm respectively)?

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Something that indicates that two normed vector spaces are not isometric is the comparison of their closed unit balls. In particular, properties like strict convexity should be conserved. What do you think of $\ell^1$ and $\ell^2$ now? Just draw their closed unit balls. Note that $\ell^1$ does not come with the max norm. That's $\ell^\infty$. –  1015 May 13 '13 at 20:19
    
@julien Although it's true that isometric bijections between normed spaces are automatically linear, this fact is not entirely obvious (and the strict convexity argument relies on it). Of course, it's possible that "isometry" was understood to be linear in the OP, I can't tell. –  75064 May 13 '13 at 21:28
    
@75064 Anyway, in this context (which sounds very much like Banach spaces/functional analysis basics), I am ready to bet that the OP meant an isometric (linear) isomorphism. –  1015 May 13 '13 at 21:31
    
Another option, which is essentially the same, is to consider the points where the norm is differentiable. The $\ell^2$ norm is differentiable everywhere but at $0$. Not the $\ell^1$ norm. –  1015 May 13 '13 at 21:35

2 Answers 2

up vote 3 down vote accepted

Normally, one proves that two normed linear spaces are isometric in a constructive™ way, by exhibiting an isometry between them. One common exception: since it's widely known that any two separable Hilbert spaces are isometrically isomorphic to each other, an explicit isometry is unnecessary in this case.

One elementary way to show that two spaces $X,Y$ are not isometrically isomorphic is to find a subset of $X$ that cannot be isometrically embedded into $Y$. For example, consider the subset $$A=\{(\pm 1,0), (0,\pm 1)\}\subset (\mathbb R^2,\|\cdot \|_1)$$ Any two points of $A$ are at distance $2$ from each other. Such a configuration is impossible in $(\mathbb R^2,\|\cdot \|_2)$. Indeed, three points at distance $2$ from one another must lie at the vertices of an equilateral triangle, and there is nowhere to put the fourth one.

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Neat minimal argument, +1. –  1015 May 13 '13 at 21:39
    
+1 Nice approach. –  copper.hat May 14 '13 at 7:01

These two spaces are not isometric. Unit sphere of the first space contains straight segment while unit sphere of second space doesn't. You can draw them to see this. But property of unit sphere to contain segments preserved under linear isometries, so there is no isometry between these two normed spaces. You can also check similar question asked earlier.

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That's what I had in mind, so +1. –  1015 May 13 '13 at 21:42

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