I was wondering if we can find $n\in\mathbb{N}$ such that there exists an isometry $(\mathbb{R}^3,\left\|{\cdot}\right\|_{\infty})\to (\mathbb{R}^n,\left\|{\cdot}\right\|_2)$.

I found out a theorem that states if $f:V\to W$ is a surjective isometry between real normed spaces then $f$ carries extreme points to extreme points, because $f$ is affine. Then if the isometry I mention exists it can't be surjective (well I know if $n>3$ this was trivial).

So, could it be that there is an (not surjective) isometry $(\mathbb{R}^3,\left\|{\cdot}\right\|_{\infty})\to (\mathbb{R}^n,\left\|{\cdot}\right\|_2)$?

Thank you.


Every subspace of the Euclidean space is Euclidean hence, in particular, strictly convex. As the max-norm is clearly not strictly convex, there is no such embedding.


Isometries are linear and injective so $f(\mathbb{R}^3)$ must be a subspace of $\mathbb{R}^n$ with dimension $3$, which ---as any real vector space with dimension $3$--- will be isomorphic to $\mathbb{R}^3$. This allows us to reduce the question to the case $n=3$. Actually, I think it is easier to prove the stronger statement: there exists no isometry $f:(\mathbb{R}^2,||\cdot||_\infty)\to(\mathbb{R}^2,||\cdot||_2)$:

Suppose there is such an isometry $f$. Call $i=(1,0)$, $j=(0,1)$; $f_i=f(i)$, $f_j=f(j)$. As $$ ||i||_\infty=||i+j||_\infty=||i-j||_\infty=1, $$ we must have that $$||f_i||_2=||f_i+f_j||_2=||f_i-f_j||_2=1,$$ where I've used the linearity and norm-preservation of $f$. This tells us that $f_i, f_i+f_j$ and $f_i-f_j$ lie on the (usual $||\cdot||_2-$) circle of radius 1, which is a contradiction as they lie on a line (notice that $f_j\neq0$ by the injetivity of $f$).

  • $\begingroup$ @Talexius It is my guess that you've already proved this for the case $n=3$ with that theorem you speak about in the question. I just felt this was a fun and quick proof. :) $\endgroup$ – user378947 Nov 1 '16 at 1:28

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