Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 22F (Linear Analysis)

LetThe map $R$ is reflection about $m$ and is an isometry. It interchanges $v,w$, so it preserves $S_1^{vw}$. Inductively, if it preserves $S_{n-1}^{vw}$, then it preserves its diameter and all distances in the defining condition for $S_n^{vw}$; hence it preserves every $S_n^{vw}$.

The midpoint belongs to $S_1^{vw}$ becauseSuppose $m\in S_{n-1}^{vw}$. For every $z\in S_{n-1}^{vw}$, reflection invariance gives $R(z)\in S_{n-1}^{vw}$ andThus $m\in S_n^{vw}$, so all the sets are nonempty and contain $m$.

Put $D_n=\mathrm{diam}(S_n^{vw})$. If $x,y\in S_n^{vw}$ with $n\ge 2$, then $y\in S_{n-1}^{vw}$ and the defining property of $x$ givesThereforeIf $z$ lies in every $S_n^{vw}$, then both $z$ and $m$ lie in $S_n^{vw}$, soHence $z=m$. We have proved the metric extraction of a midpoint by shrinking diameters:

Here “isometry” is used in the standard Mazur-Ulam theorem sense of a surjective distance-preserving map. Surjectivity is needed; a merely distance-preserving embedding need not preserve midpoints, as shown by a nonsurjective isometry need not preserve midpoints.

Distance preservation and surjectivity giveAssume inductively thatThen the two sets have the same diameter, and the universal distance condition defining the next set transfers through the bijection $u$. Hencefor every $n$.

Because $u$ is injective, it also preserves the intersection of this nested family. Part (a) therefore givesThus

Assume $u(0)=0$. Applying midpoint preservation to $2v$ and $0$ givesso $u(2v)=2u(v)$. Consequently,Thus $u$ is additive. It follows successively that

An isometry is continuous. For any $\lambda \in \mathbb{R}$, choose rationals $q_j\to \lambda$. ThenTogether with additivity, this proves that $u$ is real-linear. Hence the origin-fixing case of the Mazur-Ulam theorem givesfor all $a,b\in \mathbb{R}$ and $v,w\in V$.