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Biseparating maps between Lipschitz function spaces. (English) Zbl 1169.47024

Let $X,Y$ be bounded complete metric spaces and let $E,F$ be (real or complex) normed spaces. We write $\text{Lip}\left(X,E\right)=\left\{$all bounded $E$-valued Lipschitz functions}; $\text{Lip}\left(X\right)=\left\{$all bounded Lipschitz functionals}; ${L}^{\text{'}}\left(E,F\right)=\left\{$all linear bijections from $E$ to $F\right\}$. A map $T:\text{Lip}\left(X,E\right)\to \text{Lip}\left(Y,F\right)$ is said to be separating if $T$ is linear and $\parallel Tf\left(y\right)\parallel \phantom{\rule{0.166667em}{0ex}}\parallel Tg\left(y\right)\parallel =0$ for all $y\in Y$, whenever $f,g\in \text{Lip}\left(X,E\right)$ satisfy $\parallel fx<\parallel \parallel g\left(x\right)\parallel =0$ for all $x\in X$. $T$ is said to be biseparating if $T$ is bijective and both $T$ and ${T}^{-1}$ are separating. The authors establish the following results.

Proposition 1. Let $T:\text{Lip}\left(X,E\right)\to \text{Lip}\left(Y,F\right)$ be a biseparating map. Then there exists a bi-Lipschitz homeomorphism $h:Y\to X$ and a map $J:Y\to {L}^{\text{'}}\left(E,F\right)$ such that $Tf\left(y\right)=\left(Jy\right)\left(f\left(h\left(y\right)\right)\right)$ for all $f\in \text{Lip}\left(X,E\right)$ and $y\in Y$.

Proposition 2. Let $T:\text{Lip}\left(X\right)\to \text{Lip}\left(Y\right)$ be a bijective separating map. If $Y$ is compact, then $T$ is biseparating and continuous.

##### MSC:
 47B38 Operators on function spaces (general) 46E10 Topological linear spaces of continuous, differentiable or analytic functions 54C35 Function spaces (general topology)