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Let $V$ and $W$ be two normed vector spaces and let $f: V \rightarrow W$ be a map that preserves distance as well as the origin, that is : $ \forall x,y \in V \quad \|f(x)-f(y)\|_W=\|x-y\|_V \quad $ and $\quad f(0_V)=0_W $

Is $f$ necessarily linear ?

I know that if the norm satisfies the parallelogram identity then the answer is yes.

By the Mazur–Ulam theorem, I know that if $f$ is surjective then the answer is also yes. Here I'm asking about the consequences of requiring origin preservation instead of surjectivity.

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No. Take $V = \mathbb{R}$ with the standard norm and $W = \mathbb{R}^2$ with the maximum norm. Then let $f(x) = (|x|,x)$, clearly distance and origin-preserving, but not linear.

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No, it may well not be linear. Consider the map$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb{R}^2\\&x&\mapsto&(x,|x|).\end{array}$$On $\mathbb{R}^2$, consider the norm $\bigl\|(x,y)\bigr\|=\max\bigl\{|x|,|y|\bigr\}$ and on $\mathbb R$ the usual norm. Then $f$ preserves distances and the origin, but it is not linear.

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