A precision on the concept of strict convexity in non-Archimedean analysis
aa r X i v : . [ m a t h . F A ] F e b A precision on the concept of strict convexity innon-Archimedean analysis
Javier Cabello S´anchez, Jos´e Navarro Garmendia*
Abstract.
We prove that the only non-Archimedean strictly convex spaces are the zero spaceand the one-dimensional linear space over Z / Z , with any of its trivial norms.
1. Introduction
In order to find a non-Archimedean version of the Mazur-Ulam theorem, M. Moslehian andG. Sadeghi introduced in [ ] the class of non-Archimedean strictly convex spaces.Later on, A. Kubzdela observed that the only non-Archimedean strictly convex space over afield with a non-trivial valuation is the zero space ([ , Theorem 2]).In this note, we prove: Proposition 2.3
The only non-Archimedean strictly convex spaces are the zero space and theone-dimensional linear space over Z / Z , with any of its trivial norms.
2. Non-Archimedean strictly convex spaces
Firstly, recall that on a non-Archimedean normed space X , “any triangle is isosceles”; that isto say, for any x, y ∈ X , k y k < k x k ⇒ k x + y k = k x k . Definition ]) . A non-Archimedean normed space X over a field K is strictly convex if (SC1) | | = 1 . (SC2) for any pair of vectors x, y ∈ X , k x k = k y k = k x + y k implies x = y . Observe that (SC2) may also be rephrased by saying that there are no equilateral triangles;that is to say, that for any pair of distinct vectors x = y ∈ X , k y k = k x k ⇒ k x + y k < k x k . Keywords: Mazur–Ulam Theorem; non-Archimedean normed spaces; strict convexity.Mathematics Subject Classification: 46S10, 26E30.
Example . If X is a one-dimensional normed linear space over a finite field, then its normis trivial; i.e., there exists a ∈ (0 , ∞ ) such that k x k = a for every nonzero vector x ∈ X .The one-dimensional linear space over Z / Z , with any of its possible trivial norms, is a strictlyconvex space, in the sense of the Definition above. Proposition . If X is a non-zero strictly convex space, then it is linearly isometric to aone-dimensional normed space over Z / Z . Proof.
First of all, observe that a strictly convex space can only occur in characteristic 3:for any vector x ∈ X , k x k = k x k = k − x k ;as 2 x + ( − x ) = x , condition (SC2) implies that 2 x = − x ; that is to say, 3 x = 0 for any vector x ,and we conclude that 3 = 0 in K .Now suppose there are two non-zero vectors x, y ∈ X such that y = ± x and we will arrive toa contradiction.Condition (SC1) implies that the characteristic of K is not 2, and, hence, x + y = x − y .Without loss of generality we may also assume that k y k ≤ k x k and k x − y k ≤ k x + y k . If k y k < k x k , then x + y and x − y are distinct vectors with the same norm, k x + y k = k x k = k x − y k , and whose sum mantains the norm k ( x + y ) + ( x − y ) k = k x k = k x k = k x + y k , in contradiction with (SC2).If k y k = k x k , then (SC2) implies the absurd chains of inequalities k x + y k < k x k = k x k = k ( x + y ) + ( x − y ) k ≤ max {k x + y k , k x − y k} , k x − y k < k x k = k x k = k ( x + y ) + ( x − y ) k ≤ max {k x + y k , k x − y k} . (cid:3) Acknowledgments
Supported in part by DGICYT projects MTM2016-76958-C2-1-P and PID2019-103961GB-C21(Spain), ERDF funds and Junta de Extremadura programs GR-15152, IB-16056 and IB-18087.
References [1]
Kubzdela, A.
Isometries, Mazur-Ulam theorem and Alexsandrov problem for non-Archimedean normedspaces.
Nonlinear Anal. , 2012, 2060–2068. https://doi:10.1016/j.na.2011.10.006 [2] Moslehian, M. S. and
Sadeghi, G.
A Mazur-Ulam theorem in non-Archimedean normed spaces.
NonlinearAnal. , 2008, 3405–3408. *Corresponding author: Departamento de Matem´aticas, Universidad de Extremadura, Avenidade Elvas s/n, 06006; Badajoz. Spain
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