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Pietro Majer
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Since $\|D\kappa(x) \|\le1$ and $\|D\kappa^{-1}(x) \|\le1$, both $\kappa$ and $\kappa^{-1}$ are $1$-Lipschitz, hence isometries. All isometries

By (a particular case of) the Mazur-Ulam theorem, any isometry on $\mathbb{R}^n$ areis affine, therefore so $\kappa(x)=v+Ux$ with $U\in O(n)$, $v\in\mathbb{R}^n$.

Since $\|D\kappa(x) \|\le1$ and $\|D\kappa^{-1}(x) \|\le1$, both $\kappa$ and $\kappa^{-1}$ are $1$-Lipschitz, hence isometries. All isometries on $\mathbb{R}^n$ are affine, therefore $\kappa(x)=v+Ux$ with $U\in O(n)$.

Since $\|D\kappa(x) \|\le1$ and $\|D\kappa^{-1}(x) \|\le1$, both $\kappa$ and $\kappa^{-1}$ are $1$-Lipschitz, hence isometries.

By (a particular case of) the Mazur-Ulam theorem, any isometry on $\mathbb{R}^n$ is affine, so $\kappa(x)=v+Ux$ with $U\in O(n)$, $v\in\mathbb{R}^n$.

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Pietro Majer
  • 60.6k
  • 4
  • 122
  • 269

Since $\|D\kappa(x) \|\le1$ and $\|D\kappa^{-1}(x) \|\le1$, both $\kappa$ and $\kappa^{-1}$ are $1$-Lipschitz, hence isometries. All isometries on $\mathbb{R}^n$ are affine, therefore $\kappa(x)=v+Ux$ with $U\in O(n)$.