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The Mazur–Ulam theorem shows that a surjective isometry between normed spaces is affine. My problem is not with the proof (which I haven't yet understood), simply how an affine isometric function between normed spaces can be anything other than linear ? My reasoning (which is presumably wrong in view of the literature) is.....

Let $f: A \to B$ be affine, then by definition, $f(a) = g(a) + f(0)$ where $g$ is linear. Since $f$ is isometric, $||f(0)|| = ||0|| = 0$, so by the property of the norm, $||f(0)||= 0 \implies $f(0) = 0 $ and so $f(a) = g(a)$ a linear function.


Ah ... I think I found my error (thanks for comment).

If $f$ is an isometry (not necessarily linear), it doesn't follow that $||f(a)||= ||a ||$, only that $||f(a_1) - f(a_2)||=||a_1 - a_2 ||$ and so not necessarily $||f(0)||= 0 $.

( I see while I was typing this the comment expanded to say the same).

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    $\begingroup$ consider translations. your reasoning is false because $||f(a)-f(b)||=||a-b||$ does not imply $||f(a)||=||a||$(i.e., an isometry need not be norm-preserving). $\endgroup$ – HyJu Jan 6 '17 at 9:46
  • $\begingroup$ @HyJu. many thanks. $\endgroup$ – Tom Collinge Jan 6 '17 at 9:57

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