2
$\begingroup$

How can one show that a norm-preserving map $T: X \rightarrow X'$ where $X,X'$ are vector spaces and $T(0) = 0$ is linear? Thanks in advance.

  • 0
    Related threads: http://math.stackexchange.com/q/121046, http://math.stackexchange.com/q/81086, http://math.stackexchange.com/q/169652012-05-17

2 Answers 2

5

The claim is false. For example, take $X=X'=\mathbb{R}$, and $T(x)=|x|$ for all $x$.

  • 1
    But $n$ot conversely.2012-05-17
2

This claim is true if your map $T$ is surjective and an isometry. In this case simply apply Mazur-Ulam theorem.

If we require that map to be surjective, but only norm-preserving, then we can construct counterexample $ T:\mathbb{C}\to\mathbb{C}:z\mapsto z e^{i |z|} $ where $\mathbb{C}$ is consideres as vector spaces over $\mathbb{R}$.

  • 0
    The Mazur-Ulam theorem is a result that only holds for *real* vector spaces. On complex vector spaces you have many counterexamples, perhaps worth mentioning: $z \mapsto \bar{z}$.2012-05-17