# Equivalent norms and isometries

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)?

• 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 $\ell^\infty$. – Julien 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. – Julien 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. – Julien May 13 '13 at 21:35

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.