# Tag Info

0

I think it's worth mentioning the well-known fact that, in a finite-dimensional euclidean space (of dimension larger than 2), just preserving one distance (say, $\|f(x)-f(y\|=1$ as soon as $\|x-y\|=1$) is sufficient for $f \colon X \to X$ to be an isometry.

0

Maybe It is time to sum up the work that has been down up to now : Let $X$ and $Y$ be normed spaces, and $f$ a function $X\to Y$. Denote by $\cal C$ the condition "$||x-y|| = n\in \mathbb N \Rightarrow ||f(x)-f(y)|| = n$". 1) If $\dim(X)=1$, no proper condition can be added to $\cal C$, and even to the stronger condition set by the asker, that would imply ...

6

Terry Tao's argument generalizes to show that $f$ must be an isometry whenever $X$ has dimension greater than 1 and $Y$ is strictly convex. We start by proving a series of lemmas: Lemma 1: Let $x,y\in X$, and let $a,b\geq0$ be such that $a+b\geq\|x-y\|$ and $|a-b|\leq\|x-y\|$. Then there exists $z\in X$ such that $\|x-z\|=a$ and $\|y-z\|=b$. Proof: Let ...

17

So it turns out my earlier intuition was incorrect, and one can leverage the order properties of ${\bf R}$ to show: Theorem. Let $X, Y$ be real Hilbert spaces, with the dimension of $X$ at least two, and let $f: X \to Y$ be a function such that $\|f(x)-f(y)\|=\|x-y\|$ whenever $\|x-y\|$ is a natural number. (We do not assume $f$ to be continuous.) Then ...

2

Partial answer, in the case where the norm space = $R$. Let us define condition $\cal C$ by $||x-y||= n$ ⇔ $||f(x)-f(y)|| = n$. Assume that the normed space is $R$, so that the norm is of the form $\lambda|*|$ with $\lambda > 0$. w.l.g, I assume that the norm is the usual absolute value. Let $h(t)$ be a continuous bijection $[0,1]\to [0,1]$, such that ...

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Here is one possible condition: If $f(x)$ is continuous and fulfills $||f(kx)-f(ky)|| = k||f(x)-f(y)||$, for every $k \in N$. This works because : 1) the equivalence you wrote above will be true not only for $n\in N$ but also for every positive rational number $n$ (easy to check) 2) it will be in fact true for all positive real number $n$ : recall that ...

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