Isometries, Mazur-Ulam theorem and Aleksandrov problem for non-Archimedean normed spaces
This paper deals with isometries in non-Archimedean normed spaces. Its main purpose is to study the behaviour in the non-Archimedean setting of the classical Mazur-Ulam theorem and the classical Aleksandrov problem, both extensively studied in the Archimedean literature.NEWLINENEWLINES. Mazur and S. Ulam proved in 1932 that if \(E,F\) are real normed spaces and \(T: E \rightarrow F\) is a surjective isometry then \(f\) is a linear map up to translation. Now, let us replace the field \(\mathbb{R}\) of real numbers by a complete non-Archimedean non-trivially valued field \(\mathbb{K}\). A non-Archimedean version of this theorem, for normed spaces over \(\mathbb{K}\), was given by \textit{M. S. Moslehian} and \textit{G. A. Sadeghi} [Nonlinear Anal., Theory Methods Appl. 69, No. 10, A, 3405--3408 (2008; Zbl 1160.46049)], showing also that the Marzur-Ulam theorem fails in the non-Archimedean case. In the present paper, the author proves the following stronger result: Let \(E,F\) be normed spaces over \(\mathbb{K}\). Assume that there exists a surjective isometry \(E \rightarrow F\). If every surjective isometry \(E \rightarrow F\) is an additive map up to translation, then \(E= F = \{ 0 \}\).NEWLINENEWLINEAlso, in 1970, A. D. Aleksandrov asked under what conditions a map of a metric space into itself satisfying the strong distance one preserving property (SDOPP) is an isometry (recall that if \(E,F\) are normed spaces, a map \(T: E \rightarrow F\) has the SDOPP if, for all \(x,y \in E\), one verifies \(\| x - y \| = 1\) if and only if \(\| T(x) - T(y )\| = 1\)). \textit{T. M. Rassias} and \textit{P. Šemrl} proved in [Proc. Am. Math. Soc. 118, No. 3, 919--925 (1993; Zbl 0780.51010)] that if \(T: E \rightarrow F\) is a map between real normed spaces \(E,F\) such that one of them has dimension \(>1\) andNEWLINENEWLINE(1) \(T\) is surjective,NEWLINENEWLINE(2) \(T\) is a \(1\)-Lipschitz map (i.e., \(\| T(x) - T(y) \| \leq \| x-y \|\) for all \(x,y \in E\)),NEWLINENEWLINE(3) \(T\) has the SDOPP,NEWLINENEWLINE\noindent then \(T\) is an isometry.NEWLINENEWLINEIn the present paper, the author shows that the situation in the non-Archimedean case differs substantially. In fact, he gets that the following are equivalent:NEWLINENEWLINE(i) For every finite-dimensional space \(E\) with norm-one elements, every map \(T: E \rightarrow E\) satisfying (1), (2), (3) and \(T(0)=0\) is an isometry. (ii) \(K\) is locally compact.NEWLINENEWLINETo finish the paper, the author proves that if \(E\) is a finite-dimensional space having norm-one elements, then every isometry \(E \rightarrow E\) is surjective if and only if \(\mathbb{K}\) is spherically complete and its residue class field is finite. This result extends the one given in 1984 by W. H. Schikhof, for \(E:= \mathbb{K}\).
scientific article; zbMATH DE number 6030594
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