Let $\langle \mathbf{V},+,\cdot,||.|| \rangle$ be a normed vector space over $\mathbb{R}$.
Let $f : \mathbf{V} \to \mathbf{V}$ be an isometry that satisfies $f(\mathbf{0}) = \mathbf{0}$ .
What conditions on the vector space would or would not force $f$ to be linear?

examples: finite dimensional, complete, norm induced by an inner product, strictly convex

  • 1
  • I feel ... somewhat silly, although I would not have been able to find that on my own. I'll accept that if you post it as an answer. – user5810 Apr 20 '11 at 7:12
  • 1
    Ricky, I wouldn't worry too much. One of the goals of MO, in my opinion, is to match up people with natural (and good!) questions to people who happen to know the answer. – Yemon Choi Apr 20 '11 at 7:16
up vote 11 down vote accepted

If you assume $f$ to be surjective then $f$ has to be linear without any assumptions on $V$ by the Mazur-Ulam theorem. Wikipedia doesn't offer much more information than a link to the beautiful recent proof by J. Väisälä.

  • 4
    There is a nice generalization of the Mazur-Ulam theorem due to Figiel, T. Figiel, On nonlinear isometric embedding of normed linear spaces, Bull. Acad. Polon. Sei. Ser. Sei. Math. Astronom. Phys. 16 (1968), 185-188. If $f$ is an isometric embedding of the Banach space $X$ into the Banach space $Y$ and $f(0)=0$, then $X$ embeds isometrically isomorphically as a norm one complemented subspace of the closed linear span of $f[X]$. – Bill Johnson Apr 20 '11 at 10:06
  • Thank you Bill, this is indeed very nice! I'll have a closer look at it later. – Theo Buehler Apr 20 '11 at 10:11

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.