# 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. [duplicate]

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.

## marked as duplicate by Batman, Claude Leibovici, user91500, Davide Giraudo functional-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 13 '17 at 17:06

• Done here. – José Carlos Santos Jun 13 '17 at 6:01
• 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 6:05
• – daw Jun 13 '17 at 6:07
• A short concise paper related to what you're asking is given here. – Aweygan Jun 13 '17 at 15:20

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