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Zbl 1169.47024
Araujo, JesÃºs; Dubarbie, Luis
Biseparating maps between Lipschitz function spaces.
(English)
[J] J. Math. Anal. Appl. 357, No. 1, 191-200 (2009). ISSN 0022-247X

Let $X,Y$ be bounded complete metric spaces and let $E,F$ be (real or complex) normed spaces. We write $\text{Lip}(X,E)= \{$all bounded $E$-valued Lipschitz functions\}; $\text{Lip}(X)= \{$all bounded Lipschitz functionals\}; $L'(E,F)=\{$all linear bijections from $E$ to $F\}$. A map $T:\text{Lip}(X,E)\to \text{Lip}(Y,F)$ is said to be separating if $T$ is linear and $\|Tf(y)\|\,\|Tg (y)\|=0$ for all $y\in Y$, whenever $f,g\in \text{Lip}(X,E)$ satisfy $\|fx<\|\|g(x) \|=0$ 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 $T:\text{Lip}(X,E)\to \text{Lip}(Y,F)$ be a biseparating map. Then there exists a bi-Lipschitz homeomorphism $h:Y \to X$ and a map $J:Y\to L'(E,F)$ such that $Tf(y)=(Jy) (f(h(y)))$ for all $f\in \text{Lip}(X,E)$ and $y\in Y$. Proposition 2. Let $T:\text{Lip}(X)\to \text{Lip}(Y)$ be a bijective separating map. If $Y$ is compact, then $T$ is biseparating and continuous.
[K. Chandrasekhara Rao (Kumbakonam)]
MSC 2000:
*47B38 Operators on function spaces
46E10 Topological linear spaces of functions with smoothness properties
54C35 Function spaces (general topology)

Keywords: biseparating map; disjointness preserving map; automatic continuity; Lipschitz function

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