The Equivalence Of TVS-cone Metric And Metric Andaleksandrov Problem

On the one hand, this article aims at TVS-cone metric space, then defines TVS-cone2-metric space and researches the equivalence relation between TVS-cone2-metric and2-metric based on the equivalence relation between TVS-cone metric and metric. On the other hand, this article aims at2-normed space, then researches the Aleksandrov problem and the Aleksandrov-Rassias problem, then we have the result that f is a2-isometry, three chapter have been made as follows:In the first chapter, we show the definition of TVS-cone2-metric. By a nonlinear scalarization functionζe (y)=inf{t∈R:y∈te-P} and the2-metric defined by dp=ζeod on TVS-cone2-metric, we have the property of converge and complete on TVS-cone2-metric space is also established on2-metric space. Our theroems extend some results in Du. We prove that the dp defined by dp=ζeod on TVS-cone2-metric space (X,d) is2-metric, the equivalence of cone2-metric and2-metric and the equivalence ofand dp=ξe(D(x,y,z)).In the second chapter, we show the Benz’s theorem, and we introduce the Aleksandrov problem without the condition" Y is strictly convex and dimX≥2" in normed space, then we instead normed space by2-normed space, the same result is established. That is X, Y are2-normed space and f:X×X→Y is a surjection, for all x,y,p,q∈X,satisfied: then f is an2-isometry. Especially,when f have AOPP or nDOPP and satisfy the above condition. Then f is an2-isometry.In the third chapter, we prove that the Aleksandrov-Rassias problem can extend in2-normed space. If X, Y is2-normed space, f have the AOPP and f is2-Lipschitz map-ping, then f is2-isometry.Moreover, we prove that when dim X≥2and Y is strictly convex, and f preserves the three distances1,a and1+a, f is an2-isometry.