## When do 0-preserving isometries have to be linear?

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

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en.wikipedia.org/wiki/Mazur%2DUlam_theorem – Theo Buehler Apr 20 2011 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. – Ricky Demer Apr 20 2011 at 7:12
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 2011 at 7:16

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.
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 2011 at 10:06