I've been studying the proof of the Mazur-Ulam theorem in the pdf linked to at the end of this Wikipedia article. I'm struggling with some details in that pdf.(adsbygoogle = window.adsbygoogle || []).push({});

Theorem: Let E and F be arbitrary normed spaces. If ##f:E\to F## is a bijective isometry, then f is an affine map.

Some of the things I don't get:

1. Their definition of "affine map". How does ##f(tx+(1-t)y)=tf(x)+(1-t)f(y)## for all ##x,y\in X## and all ##t\in[0,1]## imply that f-f(0) is linear? I don't see how to deal with (f-f(0))(ax) where ##a\in\mathbb R## is arbitrary.

2. The claim that ##\psi## is an isometry. ##\psi:E\to E## is defined by ##\psi(x)=2z-x## for all ##x\in E##. This map sends an arbitrary point in E to the point that's "on the opposite side of z", i.e. the point y such that y=z+(z-x). They claim that ##\psi## is an isometry, but consider e.g. ##E=\mathbb R##, z=2, x=1. We have ##\psi(1)=2\cdot 2-1=3##, but ##\|\psi(1)\|=3\neq 1=\|1\|##.

3. If ##\psi## isn't an isometry, then I don't see a reason to think that the map the author denotes by g* should be an isometry either. It's defined by ##g^*=\psi\circ g^{-1}\circ\psi\circ g##, where g is a bijective isometry. The step ##\|g^*(z)-z\|\leq\lambda## relies on ##g^*## being an isometry.

4. All they're proving is that for all ##a,b\in E##, we have ##f\left(\frac{a+b}{2}\right)=\frac 1 2 f(a)+\frac 1 2 f(b)##. It's not obvious that this implies that f is affine.

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# Mazur-Ulam theorem (bijective isometries are affine maps)

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