Example of a non-linear isometry?

The note by Jussi Väisälä linked to by the Wikipedia article about the Mazur–Ulam theorem contains the following example:

An isometry need not be affine. To see this, let $E$ be the real line $\mathbf{R}$, let $F$ be the plane with the norm $\lVert x \rVert = \max(|x_1|, |x_2|)$, and let $\phi: R \to R$ be any function such that $|\phi(s)-\phi(t)| \le |s-t|$ for all $s, t \in\mathbf{R}$, for example, $\phi(t) = |t|$ or $\phi(t) = \sin t$. Setting $f(s) = (s, \phi(s))$ we get an isometry $f : E \to F$, which is usually not affine.

(But of course this is not a bijection.)

Yes.


Let $\mathbb{C}$ be a vector space over itself with absolute value as its norm.

Define $ \; \; f : \mathbb{C} \to \mathbb{C} \; \; $ by $ \; \; f(z) = \overline{z} \; \; $ .

$f$ is a non-linear (bijective) isometry that satisfies $\; f(0) = 0 \;$ .


See the Mazur–Ulam theorem.