Is there a nonlinear map that preserves the norm but not the inner product?

Question

Linear maps between inner product spaces preserve the inner product if they preserve the norm.

What about nonlinear maps?

I can’t come up with a counterexample, nor can I prove it.

Answer 1

There are non-linear maps the preserve norms. For example: $$T : \Bbb{R} \to \Bbb{R}^2 : x \mapsto \begin{cases} -x & \text{if } x \in \Bbb{Q} \\ x & \text{if } x \in \Bbb{R} \setminus \Bbb{Q}.\end{cases}$$ Clearly, $|Tx| = |x|$ for all $x$. Indeed, other examples are not difficult to find either; the map simply needs to stabilise each sphere centred at the origin, but is free to permute the points within the spheres freely.

If you are looking for an isometry, i.e. a map that preserves distance, i.e. one such that $\|Tx - Ty\| = \|x - y\|$ for all $x, y$, then Mazur-Ulam Theorem states that such a map must be affine. If we also impose that the map preserves norms (which is not guaranteed by the above condition, as it is satisfied by a non-trivial translation), then the map must be linear (and hence preserve inner products, by the polarisation identity).