Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question

3 Answers 3

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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 $\sin(u_0),u_0,u_1,u_2,...$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.