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

1en.wikipedia.org/wiki/Mazur%2DUlam_theorem – Theo Buehler Apr 20 '11 at 7:08

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

1Ricky, 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
If you assume $f$ to be surjective then $f$ has to be linear without any assumptions on $V$ by the MazurUlam theorem. Wikipedia doesn't offer much more information than a link to the beautiful recent proof by J. Väisälä.

4There is a nice generalization of the MazurUlam 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), 185188. 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