On the Aleksandrov-Rassias Problems on Linear 𝑛-Normed Spaces

Abstract

This paper generalizes T. M. Rassias' results in 1993 to 𝑛-normed spaces. If 𝑋 and 𝑌 are two real 𝑛-normed spaces and 𝑌 is 𝑛-strictly convex, a surjective mapping 𝑓∶𝑋→𝑌 preserving unit distance in both directions and preserving any integer distance is an 𝑛-isometry.

1. Introduction

Let 𝑋 and 𝑌 be two metric spaces. A mapping 𝑓∶𝑋→𝑌 is called an isometry if 𝑓 satisfies 𝑑_𝑌(𝑓(𝑥),𝑓(𝑦))=𝑑_𝑋(𝑥,𝑦) for all 𝑥,𝑦∈𝑋, where 𝑑_𝑋(⋅,⋅) and 𝑑_𝑌(⋅,⋅) denote the metrics in the spaces 𝑋 and 𝑌, respectively. For some fixed number 𝑟>0, suppose that 𝑓 preserves distance 𝑟, that is, for all 𝑥,𝑦∈𝑋 with 𝑑_𝑋(𝑥,𝑦)=𝑟, we have 𝑑_𝑌(𝑓(𝑥),𝑓(𝑦))=𝑟, then 𝑟 is called a conservative (or preserved) distance for the mapping 𝑓. In particular, we denote DOPP as 𝑓 preserving the one distance property and SDOPP as 𝑓 preserving the strong one distance property and also for 𝑓⁻¹.

In 1970 [1], Aleksandrov posed the following problem. Examine whether the existence of a single conservative distance for some mapping 𝑇 implies that 𝑇 is an isometry. This question is of great significance for the Mazur-Ulam Theorem [2].

In 1993, T. M. Rassias and P. Šemrl proved the following.

Theorem 1 (see [3]). Let 𝑋 and 𝑌 be two real normed linear spaces such that one of them has a dimension greater than one. Assume also that one of them is strictly convex. Suppose that 𝑓∶𝑋→𝑌 is a surjective mapping that satisfies SDOPP. Then, 𝑓 is an affine isometry (a linear isometry up to translation).

Theorem 2 (see [3]). Let 𝑋 and 𝑌 be two real normed linear spaces such that one of them has a dimension greater than one. Suppose that 𝑓∶𝑋→𝑌 is a Lipschitz mapping. Assume also that 𝑓 is a surjective mapping satisfying (SDOPP). Then, 𝑓 is an isometry.

Since 2004, the Aleksandrov problem in 𝑛-normed spaces (𝑛≥2) has been discussed, and some results are obtained [4–8].

Definition 3 (see [7]). Let 𝑋 be a real linear space with dim𝑋≥𝑛 and ‖⋅,…,⋅‖∶𝑋^𝑛→𝑅, a function, then (𝑋,‖⋅,…,⋅‖) is called a linear 𝑛-normed space if for any 𝛼∈𝑅 and all 𝑥,𝑦,𝑥₁,…,𝑥_𝑛∈𝑋 𝑛𝑁₁: ‖𝑥₁,…,𝑥_𝑛‖=0⇔𝑥₁,…,𝑥_𝑛 are linearly dependent, 𝑛𝑁₂: ‖𝑥₁,…,𝑥_𝑛‖=‖𝑥_𝑗1,…,𝑥_𝑗𝑛‖ for every permutation (𝑗₁,…,𝑗_𝑛) of (1,…,𝑛), 𝑛𝑁₃: ‖𝛼𝑥₁,…,𝑥_𝑛‖=|𝛼|‖𝑥₁,…,𝑥_𝑛‖, 𝑛𝑁₄: ‖𝑥+𝑦,𝑥₂,…,𝑥_𝑛‖≤‖𝑥,𝑥₂,…,𝑥_𝑛‖+‖𝑦,𝑥₂,…,𝑥_𝑛‖. The function ‖⋅,…,⋅‖ is called the 𝑛-norm on 𝑋.

Definition 4 (see [8]). Let 𝑋 and 𝑌 be two real linear 𝑛-normed spaces. (i)A mapping 𝑓∶𝑋→𝑌 is defined to be an 𝑛-isometry if for all 𝑥₁,…,𝑥_𝑛,𝑦₁,…,𝑦_𝑛∈𝑋, 𝑓𝑥₁−𝑓𝑦₁,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛. (1)(ii)A mapping 𝑓∶𝑋→𝑌 is called the 𝑛-distance one preserving property (𝑛-DOPP) if for 𝑥₁,…,𝑥_𝑛,𝑦₁,…,𝑦_𝑛∈𝑋, ‖𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛‖=1, it follows that ‖𝑓(𝑥₁)−𝑓(𝑦₁),…,𝑓(𝑥_𝑛)−𝑓(𝑦_𝑛)‖=1.(iii)A mapping 𝑓∶𝑋→𝑌 is called the 𝑛-strong distance one preserving property (𝑛-SDOPP) if for 𝑥₁,…,𝑥_𝑛,𝑦₁,…,𝑦_𝑛∈𝑋, ‖𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛‖=1, it follows that ‖𝑓(𝑥₁)−𝑓(𝑦₁),…,𝑓(𝑥_𝑛)−𝑓(𝑦_𝑛)‖=1 and conversely.(iv)A mapping 𝑓∶𝑋→𝑌 is called an 𝑛-Lipschitz if for all 𝑥₁,…,𝑥_𝑛,𝑦₁,…,𝑦_𝑛∈𝑋, 𝑓𝑥₁−𝑓𝑦₁,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛≤𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛. (2)

Definition 5 (see [7]). The points 𝑥₀,𝑥₁,…,𝑥_𝑛 of 𝑋 are called 𝑛-collinear if for every 𝑖, {𝑥_𝑗−𝑥_𝑖∶0≤𝑗≠𝑖≤𝑛} is linearly dependent.

Definition 6. 𝑋 is said to be 𝑛-strictly convex normed spaces if for any 𝑥₀,𝑥₁,𝑥₂,…,𝑥_𝑛∈𝑋,𝑥₂,…,𝑥_𝑛∉span{𝑥₀,𝑥₁}, and ‖𝑥₀+𝑥₁,𝑥₂,…,𝑥_𝑛‖=‖𝑥₀𝑥₂,…,𝑥_𝑛‖+‖𝑥₁𝑥₂,…,𝑥_𝑛‖ imply that 𝑥₀ and 𝑥₁ are linearly dependent.

C. Park and T. M. Rassias obtained the following.

Theorem 7 (see [8]). Let 𝑋 and 𝑌 be real linear 𝑛-normed spaces. If a mapping 𝑓∶𝑋→𝑌 satisfies the following conditions: (i)𝑓 has the 𝑛-DOPP,(ii)𝑓 is 𝑛-Lipschitz,(iii)𝑓 preserves the 2-collinearity,(iv)𝑓 preserves the 𝑛-collinearity, then 𝑓 is an 𝑛-isometry.

In 2009, Gao [6] researched another 𝑛-isometry and gave the 2-strictly convex concept [6].

In this paper, we generalize T. M. Rassias Theorems 1 and 7 on 𝑛-strictly convex normed spaces (𝑛>1).

2. Main Results

The proof of the following lemma was presented in [9], to be published; the proof is given again for the convenience of readers.

Lemma 8. Let 𝑋 be an 𝑛-normed space such that 𝑋 has dimension greater than 𝑛 and 𝑟>0. Suppose that 0<‖𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖≤2𝑟 for 𝑥₁,…,𝑥_𝑛,𝑦₁,…,𝑦_𝑛∈𝑋. Then, there exists 𝜔∈𝑋 such that 𝑥₁−𝜔,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑟,𝜔−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑟. (3)

Proof. Since 𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛 are linearly independent and dim𝑋>𝑛, then there exists 𝑧₀∈𝑋⧵span{𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛} with ‖𝑧₀,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖=𝑟.
Set 𝑦₀=𝑦₁−𝑥₁. For any 𝛼∈𝑅, we have 𝑧₀+𝛼𝑦₀,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛≠0. (4) Let us define ℎ(𝛼) by ℎ(𝛼)=𝑟𝑧₀+𝛼𝑦₀𝑧₀+𝛼𝑦₀,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛, (5) then, we obtain ℎ(𝛼),𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑟. (6) Set 𝑧₁=−𝑟𝑦₁−𝑥₁𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛,𝑧₂=𝑟𝑦₁−𝑥₁𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛. (7) Clearly, 𝑧₀≠𝑧₁,𝑧₂. And we have 𝑧₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛𝑧=𝑟,₂,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑟. (8) On the other hand, lim_{𝛼→−∞}ℎ(𝛼)=𝑧₁,lim_𝛼→∞ℎ(𝛼)=𝑧₂. (9) Thus, ℎ(−∞)=𝑧₁,ℎ(+∞)=𝑧₂. (10) Define 𝑔∶ℎ(𝑅)→𝑅 by 𝑔(𝑧)=𝑧−𝑦₀,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛. (11) It follows that 𝑔𝑧₁=𝑧₁−𝑦₀,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑟1+𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛×𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛𝑔𝑧≥𝑟,₂=𝑟1−𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛,if𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛𝑟>𝑟𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛−1𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛,if𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛<𝑟. (12)
Thus, 𝑔(𝑧₂)≤𝑟.
Obviously, 𝑔(ℎ(𝛼)) is continuous on 𝑅. Using the mean value theorem, there exists 𝛼₀∈𝑅 such that 𝑔(ℎ(𝛼₀))=𝑟.
Set 𝜔₀=ℎ(𝛼₀), 𝜔=𝜔₀+𝑥₁, we have 𝜔₀−𝑦₀,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑟. (13) And from ‖ℎ(𝛼),𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖=𝑟, we have𝜔−𝑥₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑟,𝜔−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝜔₀+𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝜔₀−𝑦₀,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑟. (14)

Lemma 9. Let 𝑋 and 𝑌 be two real linear 𝑛-normed spaces whose dimensions are greater than 𝑛, and let 𝑌 be 𝑛-strictly convex normed space. Suppose that 𝑓∶𝑋→𝑌 is a surjective mapping satisfying (𝑛-SDOPP) with preserving distance 𝑘 for any 𝑘∈𝑁. Then, 𝑓 preserves distance 1/𝑘 for any 𝑘∈𝑁.

Proof. Firstly, 𝑓 is injective. Suppose, on the contrary, that there are 𝑥₀,𝑥₁∈𝑋, 𝑥₀≠𝑥₁, such that 𝑓(𝑥₀)=𝑓(𝑥₁). As dim𝑋>𝑛, it follows that there exist vectors 𝑥₂,…,𝑥_𝑛∈𝑋 such that 𝑥₁−𝑥₀,…,𝑥_𝑛−𝑥₀ are linearly independent. Then, ‖𝑥₁−𝑥₀,…,𝑥_𝑛−𝑥₀‖≠0.
Set 𝑧₂∶=𝑥₀+𝑥₂−𝑥₀𝑥₁−𝑥₀,…,𝑥_𝑛−𝑥₀. (15) Clearly, 𝑥₁−𝑥₀,𝑧₂−𝑥₀,𝑥₃−𝑥₀,…,𝑥_𝑛−𝑥₀=1. (16) Then 𝑓𝑥₁−𝑓𝑥₀,𝑓𝑧₂−𝑓𝑥₀,𝑓𝑥₃−𝑓𝑥₀,…,𝑓𝑥_𝑛−𝑓𝑥₀=1. (17) This implies that 𝑓(𝑥₀)≠𝑓(𝑥₁), which is a contradiction. Therefore, 𝑓 is a bijective mapping.
Let 𝑥₁,…,𝑥_𝑛,𝑦₁,…,𝑦_𝑛∈𝑋 and (𝑘∈𝑁⧵{1}) satisfying 𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=1𝑘. (18)
By Lemma 8, we can find 𝑤₁∈𝑋 with 𝑥₁−𝑤₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛𝑤=1,₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=1. (19) Set 𝑢₁=𝑤₁+𝑘𝑦₁−𝑤₁,𝑣₁=𝑤₁+𝑘𝑥₁−𝑤₁. (20) Clearly, we have 𝑥₁−𝑣₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=(𝑘−1)𝑥₁−𝑤₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛𝑤=𝑘−1₁−𝑣₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑘. (21) It follows from the hypothesis of 𝑓 preserving any integer 𝑘; then, 𝑓𝑥₁−𝑓𝑤₁,𝑓𝑥₂−𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛𝑓𝑥=1,₁−𝑓𝑣₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛𝑓𝑤=𝑘−1,₁−𝑓𝑣₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=𝑘. (22) Clearly, we have 𝑓𝑤₁−𝑓𝑣₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=𝑓𝑥₁−𝑓𝑣₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛+𝑓𝑥₁−𝑓𝑤₁,𝑓𝑥₂−𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛. (23) We conclude that 𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛∉span𝑓𝑥₁−𝑓𝑣₁,𝑓𝑥₁−𝑓𝑤₁. (24) Otherwise, if for some 𝑓(𝑥_𝑖)−𝑓(𝑦_𝑖), we have 𝜇_𝑖,𝜆_𝑖∈𝑅 with 𝜇_𝑖≠0 or 𝜆_𝑖≠0 such that 𝑓𝑥_𝑖−𝑓𝑦_𝑖=𝜇_𝑖𝑓𝑥₁−𝑓𝑣₁+𝜆_𝑖𝑓𝑥₁−𝑓𝑤₁. (25) Suppose that 𝜆_𝑖≠0. Then, 𝑘−1=𝑓𝑥₁−𝑓𝑣₁,…,𝑓𝑥_𝑖−𝑓𝑦_𝑖,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=𝜆_𝑖𝑥₁−𝑓𝑣₁,…,𝑓𝑥₁−𝑓𝑤₁,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛. (26) Assume that 𝑓𝑥₁−𝑓𝑣₁,…,𝑓𝑥₁−𝑓𝑤₁,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛≠0. (27) Set 𝑠_𝑗=𝑓𝑥_𝑗+𝑓𝑥_𝑗−𝑓𝑦_𝑗×𝑓𝑥₁−𝑓𝑣₁,…,𝑓𝑥₁−𝑓𝑤₁𝑓𝑥,…,_𝑛−𝑓𝑦_𝑛⁻¹,𝑗≥2. (28) Then, for 𝑗≠𝑖, 𝑓𝑥₁−𝑓𝑣₁,…,𝑠_𝑗−𝑓𝑥_𝑗,…,𝑓𝑥₁−𝑓𝑤₁,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=1. (29) Since 𝑓 is bijective and preserves 𝑛-SDOPP on both directions. Then, there exists 𝑡_𝑗∈𝑋 with 𝑓(𝑡_𝑗)=𝑠_𝑗 which satisfies that 𝑥₁−𝑣₁,𝑡_𝑗−𝑥_𝑗,…,𝑥₁−𝑤₁,…,𝑥_𝑛−𝑦_𝑛=1. (30) However, by (20), 𝑥₁−𝑣₁=(1−𝑘)(𝑥₁−𝑤₁), and thus 𝑥₁−𝑣₁, 𝑥₁−𝑤₁ are linear dependent. Then, 𝑥₁−𝑣₁,𝑡_𝑗−𝑥_𝑗,…,𝑥₁−𝑤₁,…,𝑥_𝑛−𝑦_𝑛=0. (31) This contradiction implies that 𝑓𝑥₁−𝑓𝑣₁,…,𝑓𝑥₁−𝑓𝑤₁,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=0. (32) This also contradicts with (26). Since 𝑌 is 𝑛-strictly convex, then there exists 𝛼>0 such that 𝑓𝑥₁−𝑓𝑣₁=𝛼𝑓𝑥₁−𝑓𝑤₁. (33) Then, 𝑓𝑥₁=1𝑓𝑣1+𝛼¹+𝛼𝑓𝑤1+𝛼¹. (34) Since 𝑓𝑥₁−𝑓𝑣₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛𝑓𝑥=𝑘−1,₁−𝑓𝑤₁,𝑓𝑥₂−𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=1, (35) then 𝛼=𝑘−1. Thus, 𝑓𝑥₁=1𝑘𝑓𝑣₁+𝑘−1𝑘𝑓𝑤₁, (36) Similarly, 𝑓𝑦₁=1𝑘𝑓𝑢₁+𝑘−1𝑘𝑓𝑤₁. (37) Hence, 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=1𝑘. (38)

Lemma 10. Let 𝑋 and 𝑌 be real 𝑛-normed spaces such that dim𝑋≥𝑛. If a mapping 𝑓∶𝑋→𝑌 preserves the distance 1/𝑘 for each 𝑘∈ℕ, then 𝑓 preserves the distance zero.

Proof. Choose 𝑥₁,…,𝑥_𝑛,𝑦₁,…,𝑦_𝑛∈𝑋 such that ‖𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛‖=0; that is, 𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛 are linearly dependent. Assume that {𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛} is a maximum linearly independent group of {𝑥₁−𝑦₁,…,𝑥_𝑛−𝑦_𝑛} (𝑚<𝑛). As dim𝑋≥𝑛, we can find a finite sequence of vectors 𝜔₁,𝜔₂,…,𝜔_𝑚∈𝑋 such that 𝑥₁−𝜔₁,…,𝑥_𝑚−𝜔_𝑚,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛 are linearly independent. Hence, it holds that 𝑥₁−𝜔₁,…,𝑥_𝑚−𝜔_𝑚,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛≠0. (39)
We will prove that 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛≤1𝑘, (40) for every 𝑘∈ℕ. Let 𝑚=1. We can find a vector 𝜔₁∈𝑋 such that 𝑥₁−𝜔₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛 are linearly independent. Set 𝜈₁=𝑥₁+𝑥₁−𝜔₁2𝑘𝑥₁−𝜔₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛, (41) for arbitrarily fixed 𝑘∈ℕ. Then, 𝑥₁−𝜈₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=1,𝜈2𝑘₁−𝑥₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛−𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛≤𝜈₁−𝑥₁+𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛≤𝜈₁−𝑥₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛+𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛. (42) Since ‖𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖=0, we get 𝜈₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=1.2𝑘 (43) Since 𝑓 preserves the distance 1/(2𝑘), we see that 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛≤𝑓𝑥₁−𝑓𝜈₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛+𝑓𝜈₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=112𝑘⋅2=𝑘. (44)
For 𝑚≥2, we set 𝜈₁=𝑥₁+𝑥₁−𝜔₁×2^𝑚𝑘𝑥₁−𝜔₁,…,𝑥_𝑚−𝜔_𝑚,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛⁻¹,𝜈_𝑖=2𝑥_𝑖−𝜔_𝑖, (45)(46) for any 𝑖∈{2,3,…,𝑚}. Then, we have 𝑥_𝑖−𝜈_𝑖=𝜔_𝑖−𝑥_𝑖,𝜈_𝑖−𝑦_𝑖=𝑥_𝑖−𝜔_𝑖+𝑥_𝑖−𝑦_𝑖, (47) for each 𝑖∈{2,3,…,𝑚}. Since 𝑥_𝑖−𝑦_𝑖,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛 are linearly dependent, we get …,𝑥_𝑖−𝑦_𝑖,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛=0, (48) and hence, …,𝑥_𝑖−𝜔_𝑖,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛−…,𝑥_𝑖−𝑦_𝑖,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛≤…,𝑥_𝑖−𝜔_𝑖+𝑥_𝑖−𝑦_𝑖,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛≤…,𝑥_𝑖−𝜔_𝑖,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛+…,𝑥_𝑖−𝑦_𝑖,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛, (49) which together with (48) implies that …,𝜈_𝑖−𝑦_𝑖,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛=…,𝑥_𝑖−𝜔_𝑖,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛, (50) for all 𝑖∈{2,3,…,𝑚}. By a similar argument, we further obtain that 𝜈₁−𝑦₁,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛=𝜈₁−𝑥₁,…,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛. (51)
In view of (45), (50), and (51), we conclude that 𝜈₁−𝑦₁,𝜇₂,…,𝜇_𝑚,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛=𝑥₁−𝜈₁,𝑥₂−𝜔₂,…,𝑥_𝑚−𝜔_𝑚,𝑥_𝑚+1−𝑦_𝑚+1,…,𝑥_𝑛−𝑦_𝑛=12^𝑚𝑘, (52) where 𝜇_𝑖 denotes either 𝜈_𝑖−𝑦_𝑖 or 𝑥_𝑖−𝜈_𝑖 for 𝑖∈{2,3,…,𝑚}.
Since 𝑓 preserves the distance 1/(2^𝑚𝑘) for any 𝑘∈ℕ, it follows from (52) that 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,𝑓𝑥₃−𝑓𝑦₃,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛≤𝑓𝑥₁−𝑓𝜈₁,𝑓𝑥₂−𝑓𝜈₂,…,𝑓𝑥_𝑚−1−𝑓𝜈_𝑚−1,𝑓𝑥_𝑚−𝑓𝜈_𝑚,𝑓𝑥_𝑚+1−𝑓𝑦_𝑚+1,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛+𝑓𝑥₁−𝑓𝜈₁,𝑓𝑥₂−𝑓𝜈₂,…,𝑓𝑥_𝑚−1−𝑓𝜈_𝑚−1,𝑓𝜈_𝑚−𝑓𝑦_𝑚,𝑓𝑥_𝑚+1−𝑓𝑦_𝑚+1,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛+𝑓𝑥₁−𝑓𝜈₁,𝑓𝑥₂−𝑓𝜈₂,…,𝑓𝜈_𝑚−1−𝑓𝑦_𝑚−1,𝑓𝑥_𝑚−𝑓𝜈_𝑚,𝑓𝑥_𝑚+1−𝑓𝑦_𝑚+1,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛+𝑓𝑥₁−𝑓𝜈₁,𝑓𝑥₂−𝑓𝜈₂,…,𝑓𝜈_𝑚−1−𝑓𝑦_𝑚−1,𝑓𝜈_𝑚−𝑓𝑦_𝑚,𝑓𝑥_𝑚+1−𝑓𝑦_𝑚+1,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛+𝑓𝜈+⋯+₁−𝑓𝑦₁,𝑓𝜈₂−𝑓𝑦₂,…,𝑓𝜈_𝑚−1−𝑓𝑦_𝑚−1,𝑓𝜈_𝑚−𝑓𝑦_𝑚,𝑓𝑥_𝑚+1−𝑓𝑦_𝑚+1,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=12^𝑚𝑘⋅2^𝑚=1𝑘, (53) where 𝑘 is an arbitrary positive integer. Hence, we conclude that 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=0, (54) which implies that 𝑓 preserves the distance zero.

Remark 11. In Lemma 2.2 ([9] to be published), we give the same method under the condition of 𝑓 preserving 2-colinear.

Theorem 12. Let 𝑋 and 𝑌 be real 𝑛-normed spaces such that dim𝑋>𝑛 and 𝑌 is 𝑛-strictly convex. If a surjective mapping 𝑓∶𝑋→𝑌 has the 𝑛-SDOPP and preserves the distance 𝑘 for any 𝑘∈ℕ, then 𝑓 is an affine 𝑛-isometry.

Proof. Assume that ‖𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖>0 for 𝑥₁,…,𝑥_𝑛,𝑦₁,…,𝑦_𝑛∈𝑋.
Take positive integers 𝑘,𝑚 such that 𝑚−1𝑘≤𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛≤𝑚𝑘. (55) Set 𝑝_𝑖=𝑥₁+𝑖𝑘⋅𝑦₁−𝑥₁𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛, (56) for 𝑖=0,1,…,𝑚−2, and 𝑝_𝑚=𝑦₁. (57) Clearly, for 𝑖=1,…,𝑚−2, 𝑝_𝑖−𝑝_𝑖−1,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=1𝑘,0<𝑝_𝑚−𝑝_𝑚−2,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑦₁−𝑥₁−𝑚−2𝑘⋅𝑦₁−𝑥₁𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=1−𝑚−2𝑘⋅1𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛⋅𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛−𝑚−2𝑘≤𝑚𝑘−𝑚−2𝑘=2𝑘. (58) According to Lemma 8, there exists 𝑝_𝑚−1∈𝑋 such that 𝑝_𝑚−1−𝑝_𝑚−2,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=1𝑘,𝑝_𝑚−1−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=1𝑘. (59) It follows from Lemma 9 that we have 𝑓𝑝_𝑖−𝑓𝑝_𝑖−1,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=1𝑘, (60) for 𝑖=0,1,2,…,𝑚.
On the other hand, 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂…,𝑓𝑥_𝑛−𝑓𝑦_𝑛≤^𝑚_𝑖=1𝑓𝑝_𝑖−𝑓𝑝_𝑖−1,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=𝑚𝑘. (61) Hence 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛≤𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛. (62) Suppose that 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛<𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛. (63) For any 𝑥₁,𝑦₁,𝑥₂,𝑦₂,…,𝑥_𝑛,𝑦_𝑛∈𝑋, with 𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛≠0, (64) find a positive integer 𝑘₀ satisfying ‖𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖<𝑘₀.
Set 𝑧₁=𝑥₁+𝑘₀(𝑦₁−𝑥₁)/‖𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖. Clearly, ||𝑧₁−𝑥₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛||=𝑘₀, and ‖𝑧₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖=𝑘₀−‖𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛‖.
It follows that ‖𝑓(𝑧₁)−𝑓(𝑥₁),𝑓(𝑥₂)−𝑓(𝑦₂),…,𝑓(𝑥_𝑛)−𝑓(𝑦_𝑛)‖=𝑘₀ and 𝑘₀=𝑓𝑧₁−𝑓𝑥₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛≤𝑓𝑧₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛+𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛<𝑘₀−𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛+𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛=𝑘₀. (65) Then (63) is not valid. Hence, 𝑓𝑥₁−𝑓𝑦₁,𝑓𝑥₂−𝑓𝑦₂,…,𝑓𝑥_𝑛−𝑓𝑦_𝑛=𝑥₁−𝑦₁,𝑥₂−𝑦₂,…,𝑥_𝑛−𝑦_𝑛. (66)

Corollary 13. Let 𝑋 and 𝑌 be two real linear 𝑛-normed spaces. Suppose that mapping 𝑓∶𝑋→𝑌 preserves any positive integer 𝑘-distance and Lipschitz condition. Then, 𝑓 is an 𝑛-isometry.

Acknowledgments

This work is supported by the Fundamental Research Funds for the Central Universities in China and Education Department of Liaoning province in china.