If $\Vert Tx-Ty \Vert = \Vert x-y \Vert$ for all $x,y \in X$ and $T(0)=0$ then T is a linear aplication.

Question

Problem: Lets $X$ and $Y$ normed vector spaces $T:X\rightarrow Y$ a aplication such that $\Vert Tx-Ty \Vert = \Vert x-y \Vert$ for all $x,y \in X$ and $T(0)=0$ then T is a linear aplication.

My attempt: If I evaluate in $0$: $$\Vert Tx \Vert = \Vert x\Vert \quad \mbox{for all $x \in X$}$$ Then, $$\Vert T(x+y) \Vert = \Vert x+y\Vert \leq \Vert x\Vert + \Vert y\Vert = \Vert Tx\Vert+\Vert Ty\Vert$$ But, I do not know how to continue.

In the linked questions, the space is assumed to be Euclidean, i.e., provided with an inner product structure. Here, the spaces are just normed. — daw, Jun 13 '17 at 06:05
see https://mathoverflow.net/questions/62380/when-do-0-preserving-isometries-have-to-be-linear — daw, Jun 13 '17 at 06:07
A short concise paper related to what you're asking is given here. — Aweygan, Jun 13 '17 at 15:20

score 6 · Accepted Answer · answered Jun 13 '17 at 06:13

This is not true in general: Set $X=\mathbb R$ and $Y:=\mathbb R^2$ with maximum norm $\|\cdot\|_\infty$.

Define $T$ by $$ T(x) := (x, \ |x|). $$ Then $T(0)=0$ and $$ \|T(x)-T(y)\| = \max( |x-y|, \big| |x|-|y| \big|) = |x-y| $$ due to $ \big||x|-|y| \big|\le |x-y|$.

The claim is true if $T$ is assumed to be bijective, see the answer to this question: https://mathoverflow.net/questions/62380/when-do-0-preserving-isometries-have-to-be-linear

In an inner product setting the answer is affirmative, see Showing that an Isometry on the Euclidean Plane fixing the origin is Linear

If $\Vert Tx-Ty \Vert = \Vert x-y \Vert$ for all $x,y \in X$ and $T(0)=0$ then T is a linear aplication.

1 Answers1