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We know, Mazur–Ulam theorem states that if $V$ and $W$ are normed spaces over $\mathbb{R}$ and the mapping $f\colon V\to W$ is a surjective isometry, then $f$ is affine.

Can somebody say counterexample when $f\colon V\to W$ is no surjective then f is not affine.

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up vote 4 down vote accepted

Define $f \colon \mathbb R \to \mathbb R^2$ by $f(x) = (x, \left|x\right|)$. If $\mathbb R$ is endowed with the usual norm given by the absolute value, and $\mathbb R^2$ with the maximum norm $\left|x\right|_\infty = \max\{|x_1|, |x_2|\}$, then $f$ is an isometry.

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I think that $f\colon \mathbb R\to L^1(\mathbb R)$ defined by

$$f(x)=\begin{cases}1_{[0,x]}&\text{if}\ x\geq 0\\ -1_{[0,-x]}&\text{if}\ x<0\end{cases}$$

is also a counterexemple.

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An other example: the space of the bounded serie provided of the norm of the supremum. We can consider the map $\varphi$ who at the serie $u_0,u_1,u_2,...$ associate the serie $\sin(u_0),u_0,u_1,u_2,...$.

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