Is a non-euclidean-norm preserving map necessarily linear?

Question

Let $V$ and $W$ be two normed vector spaces and let $f:V \rightarrow W$ be a norm preserving map. I know that if both norms correspond to some inner product then $f$ is necessarily linear, but I can't find the answer for the more general case of normed vector spaces.

I suspect the answer is no, so I tried to come up with a counter-example involving "pseudo" rotations along non-euclideanly-spherical paths centered at the origin of $\mathbb R^2$, unsuccessfully.

I'd most importantly like an answer that does not assume $f$ to be surjective. However, any additional information about that particular case would be appreciated as well.

Answer 1

Take the map $\mathbb{R} \to \mathbb{R}, r \mapsto |r|$. This certainly preserves the Euclidean norm, but is not even additive, so not linear.

Another kind of dumb example is the trivial norm taking $0$ to $0$ and anything else to $1$: then any map taking $0_V$ to $0_W$ and $V \setminus\{0\}$ to $W \setminus\{0\}$ is norm preserving.

Or, you could take $V$ to have the trivial norm, and $W$ to be $\mathbb{C}$ with the Euclidean norm. Then any map $V \to W$ which maps $0$ to $0$ and maps $V \setminus \{0\}$ onto the unit circle is norm preserving. This certainly isn't surjective.

Answer 2

This is the example from J. A. Baker's paper for easy access.

Define $f:\mathbb{R}^2\to\mathbb{R}$ by $$f(x,y)=\begin{cases}y,&\text{ if }0\leq y\leq x\text{ or }x\leq y\leq 0\\x,&\text{ if }0\leq x\leq y\text{ or }y\leq x\leq 0\\0,&\text{ otherwise}\end{cases}$$

Then $f$ satisfies

1. $f(tx,ty)=tf(x,y)$

2. $|f(x,y)-f(u,v)|\leq\sqrt{(x-u)^2+(y-v)^2}$

3. $f$ is not linear

Put in $\mathbb{R}^2$ the usual norm, and $\mathbb{R}^3$ with the norm $$\|(x,y,z)\|=\max\left(\sqrt{x^2+y^2},|z|\right)$$

Define $F:\mathbb{R}^2\to\mathbb{R}^3$ by $F(x,y)=(x,y,f(x,y))$.

Then $F$ is a non-affine isometry.

Answer 3

Consider the map

$$f:\Bbb R\to\Bbb R^2,\; x\mapsto(x,\sin x)$$

with the standard metric on $\Bbb R$, but $\Bbb R^2$ equipped with the $\ell^\infty$-metric, that is, $\|(x,y)\|_\infty:=\max\{|x|,|y|\}$. Obviously, $f$ is non-lines (and non-affine). But it holds

\begin{align} \|f(x)-f(y)\|_\infty &= \|(x-y,\sin x-\sin y)\|_\infty \\ &=\max\{|x-y|,|\sin x-\sin y|\} \\ &= |x-y|. \end{align}

The last equality is because $|\sin' x|=|\cos x|<1\implies |\sin x-\sin y|\le |x-y|$.