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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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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 '11 at 7:12
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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
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1 Answer 1

up vote 9 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.

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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
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