# On the Aleksandrov-Rassias Problems on Linearn-Normed Spaces

Authors: Yumei Ma
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

This paper generalizes T. M. Rassias' results in 1993 ton-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.


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## Body

## 1. Introduction

LetX and Y be two metric spaces. A mapping f:X→Y is called an isometry if f satisfies dY(f(x),f(y))=dX(x,y) for all x,y∈X, where dX(·,·) and dY(·,·) 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 dX(x,y)=r, we have dY(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 mappingTimplies thatTis 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]).
LetX 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]).
LetX 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 inn-normed spaces (n≥2) has been discussed, and some results are obtained [4–8].Definition 3 (see [7]).
LetX be a real linear space with dimX≥n and ∥·,…,·∥:Xn→R, a function, then (X,∥·,…,·∥) is called a linear n-normed space if for any α∈R and all x,y,x1,…,xn∈XnN1: ∥x1,…,xn∥=0⇔x1,…,xn are linearly dependent,nN2: ∥x1,…,xn∥=∥xj1,…,xjn∥ for every permutation (j1,…,jn) of (1,…,n),nN3: ∥αx1,…,xn∥=|α|∥x1,…,xn∥,nN4: ∥x+y,x2,…,xn∥≤∥x,x2,…,xn∥+∥y,x2,…,xn∥. The function ∥·,…,·∥ is called the n-norm on X.Definition 4 (see [8]).
LetX and Y be two real linear n-normed spaces. (i)
A mappingf:X→Y is defined to be an n-isometry if for all x1,…,xn,y1,…,yn∈X,
(1)∥f(x1)-f(y1),…,f(xn)-f(yn)∥=∥x1-y1,…,xn-yn∥.(ii)
A mappingf:X→Y is called the n-distance one preserving property (n-DOPP) if for x1,…,xn,y1,…,yn∈X, ∥x1-y1,…,xn-yn∥=1, it follows that ∥f(x1)-f(y1),…,f(xn)-f(yn)∥=1.(iii)
A mappingf:X→Y is called the n-strong distance one preserving property (n-SDOPP) if for x1,…,xn,y1,…,yn∈X, ∥x1-y1,…,xn-yn∥=1, it follows that ∥f(x1)-f(y1),…,f(xn)-f(yn)∥=1 and conversely.(iv)
A mappingf:X→Y is called an n-Lipschitz if for all x1,…,xn,y1,…,yn∈X,
(2)∥f(x1)-f(y1),…,f(xn)-f(yn)∥≤∥x1-y1,…,xn-yn∥.Definition 5 (see [7]).
The pointsx0,x1,…,xn of X are called n-collinear if for every i, {xj-xi: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,…,xn∈X,x2,…,xn∉span{x0,x1}, and ∥x0+x1,x2,…,xn∥=∥x0x2,…,xn∥+∥x1x2,…,xn∥ imply that x0 and x1 are linearly dependent.C. Park and T. M. Rassias obtained the following.Theorem 7 (see [8]).
LetX 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 Theorems1 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.
LetX be an n-normed space such that X has dimension greater than n and r>0. Suppose that 0<∥x1-y1,x2-y2,…,xn-yn∥≤2r for x1,…,xn,y1,…,yn∈X. Then, there exists ω∈X such that
(3)∥x1-ω,x2-y2,…,xn-yn∥=r,∥ω-y1,x2-y2,…,xn-yn∥=r.Proof.
Sincex1-y1,x2-y2,…,xn-yn are linearly independent and dimX>n, then there exists z0∈X∖span{x1-y1,…,xn-yn} with ∥z0,x2-y2,…,xn-yn∥=r.
Sety0=y1-x1. For any α∈R, we have
(4)∥z0+αy0,x2-y2,…,xn-yn∥≠0.
Let us define h(α) by
(5)h(α)=r(z0+αy0)∥z0+αy0,x2-y2,…,xn-yn∥,
then, we obtain
(6)∥h(α),x2-y2,…,xn-yn∥=r.
Set
(7)z1=-r(y1-x1)∥x1-y1,x2-y2,…,xn-yn∥,z2=r(y1-x1)∥x1-y1,x2-y2,…,xn-yn∥.
Clearly, z0≠z1,z2. And we have
(8)∥z1,x2-y2,…,xn-yn∥=r,∥z2,x2-y2,…,xn-yn∥=r.
On the other hand,
(9)limα→-∞h(α)=z1,limα→∞h(α)=z2.
Thus,
(10)h(-∞)=z1,h(+∞)=z2.
Define g:h(R)→R by
(11)g(z)=∥z-y0,x2-y2,…,xn-yn∥.
It follows that
(12)g(z1)=∥z1-y0,x2-y2,…,xn-yn∥=(1+r∥x1-y1,x2-y2,…,xn-yn∥)×∥x1-y1,x2-y2,…,xn-yn∥≥r,g(z2)={(1-r∥x1-y1,…,xn-yn∥)∥x1-y1,…,xn-yn∥,r∥x1-fcffy1,…,∥-if∥x1-y1,…,xn-yn∥>r(r∥x1-y1,…,xn-yn∥-1)∥x1-y1,…,xn-yn∥,(r∥x1-fy1,…,∥-)if∥x1-y1,…,xn-yn∥<r.
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.
Setω0=h(α0), ω=ω0+x1, we have
(13)∥ω0-y0,x2-y2,…,xn-yn∥=r.
And from ∥h(α),x2-y2,…,xn-yn∥=r, we have(14)∥ω-x1,x2-y2,…,xn-yn∥=r,∥ω-y1,x2-y2,…,xn-yn∥=∥ω0+x1-y1,x2-y2,…,xn-yn∥=∥ω0-y0,x2-y2,…,xn-yn∥=r.Lemma 9.
LetX 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,…,xn∈X such that x1-x0,…,xn-x0 are linearly independent. Then, ∥x1-x0,…,xn-x0∥≠0.
Set(15)z2∶=x0+x2-x0∥x1-x0,…,xn-x0∥.
Clearly,
(16)∥x1-x0,z2-x0,x3-x0,…,xn-x0∥=1.
Then
(17)∥f(x1)-f(x0),f(z2)-f(x0),f(x3)-f(x0),…,f(xn)-f(x0)∥=1.
This implies that f(x0)≠f(x1), which is a contradiction. Therefore, f is a bijective mapping.
Letx1,…,xn,y1,…,yn∈X and (k∈N∖{1}) satisfying
(18)∥x1-y1,x2-y2,…,xn-yn∥=1k.
By Lemma8, we can find w1∈X with
(19)∥x1-w1,x2-y2,…,xn-yn∥=1,∥w1-y1,x2-y2,…,xn-yn∥=1.
Set
(20)u1=w1+k(y1-w1),v1=w1+k(x1-w1).
Clearly, we have
(21)∥x1-v1,x2-y2,…,xn-yn∥=∥(k-1)(x1-w1),x2-y2,…,xn-yn∥=k-1∥w1-v1,x2-y2,…,xn-yn∥=k.
It follows from the hypothesis of f preserving any integer k; then,
(22)∥f(x1)-f(w1),f(x2-y2),…,f(xn)-f(yn)∥=1,∥f(x1)-f(v1),f(x2)-f(y2),…,f(xn)-f(yn)∥=k-1,∥f(w1)-f(v1),f(x2)-f(y2),…,f(xn)-f(yn)∥=k.
Clearly, we have
(23)∥f(w1)-f(v1),f(x2)-f(y2),…,f(xn)-f(yn)∥=∥f(x1)-f(v1),f(x2)-f(y2),…,f(xn)-f(yn)∥+∥f(x1)-f(w1),f(x2-y2),…,f(xn)-f(yn)∥.
We conclude that
(24)f(x2)-f(y2),…,f(xn)-f(yn)∉span{f(x1)-f(v1),f(x1)-f(w1)}.
Otherwise, if for some f(xi)-f(yi), we have μi,λi∈R with μi≠0 or λi≠0 such that
(25)f(xi)-f(yi)=μi(f(x1)-f(v1))+λi(f(x1)-f(w1)).
Suppose that λi≠0. Then,
(26)k-1=∥f(x1)-f(v1),…,f(xi)-f(yi),…,f(xn)-f(yn)∥=|λi|∥(x1)-f(v1),…,f(x1)-f(w1),…,f(xn)-f(yn)∥.
Assume that
(27)∥f(x1)-f(v1),…,f(x1)-f(w1),…,f(xn)-f(yn)∥≠0.
Set
(28)sj=f(xj)+(f(xj)-f(yj))×(∥f(x1)-f(v1),…,f(x1)-f(w1),…,f(xn)-f(yn)∥)-1,(j≥2).
Then, for j≠i,
(29)∥f(x1)-f(v1),…,sj-f(xj),…,f(x1)-f(w1),…,f(xn)-f(yn)∥=1.
Since f is bijective and preserves n-SDOPP on both directions. Then, there exists tj∈X with f(tj)=sj which satisfies that
(30)∥x1-v1,tj-xj,…,x1-w1,…,xn-yn∥=1.
However, by (20), x1-v1=(1-k)(x1-w1), and thus x1-v1, x1-w1 are linear dependent. Then,
(31)∥x1-v1,tj-xj,…,x1-w1,…,xn-yn∥=0.
This contradiction implies that
(32)∥f(x1)-f(v1),…,f(x1)-f(w1),…,f(xn)-f(yn)∥=0.
This also contradicts with (26). Since Y is n-strictly convex, then there exists α>0 such that
(33)f(x1)-f(v1)=α(f(x1)-f(w1)).
Then,
(34)f(x1)=11+αf(v1)+α1+αf(w1).
Since
(35)∥f(x1)-f(v1),f(x2)-f(y2),…,f(xn)-f(yn)∥=k-1,∥f(x1)-f(w1),f(x2-y2),…,f(xn)-f(yn)∥=1,
then α=k-1. Thus,
(36)f(x1)=1kf(v1)+k-1kf(w1),
Similarly,
(37)f(y1)=1kf(u1)+k-1kf(w1).
Hence,
(38)∥f(x1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥=1k.Lemma 10.
LetX 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.
Choosex1,…,xn,y1,…,yn∈X such that ∥x1-y1,…,xn-yn∥=0; that is, x1-y1,…,xn-yn are linearly dependent. Assume that {xm+1-ym+1,…,xn-yn} is a maximum linearly independent group of {x1-y1,…,xn-yn} (m<n). As dimX≥n, we can find a finite sequence of vectors ω1,ω2,…,ωm∈X such that x1-ω1,…,xm-ωm,xm+1-ym+1,…,xn-yn are linearly independent. Hence, it holds that
(39)∥x1-ω1,…,xm-ωm,xm+1-ym+1,…,xn-yn∥≠0.
We will prove that(40)∥f(x1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥≤1k,
for every k∈ℕ. Let m=1. We can find a vector ω1∈X such that x1-ω1,x2-y2,…,xn-yn are linearly independent. Set
(41)ν1=x1+x1-ω12k∥x1-ω1,x2-y2,…,xn-yn∥,
for arbitrarily fixed k∈ℕ. Then,
(42)∥x1-ν1,x2-y2,…,xn-yn∥=12k,∥ν1-x1,x2-y2,…,xn-yn∥-∥x1-y1,x2-y2,…,xn-yn∥≤∥(ν1-x1)+(x1-y1),x2-y2,…,xn-yn∥≤∥ν1-x1,x2-y2,…,xn-yn∥+∥x1-y1,x2-y2,…,xn-yn∥.
Since ∥x1-y1,x2-y2,…,xn-yn∥=0, we get
(43)∥ν1-y1,x2-y2,…,xn-yn∥=12k.
Since f preserves the distance 1/(2k), we see that
(44)∥f(x1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥≤∥f(x1)-f(ν1),f(x2)-f(y2),…,f(xn)-f(yn)∥+∥f(ν1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥=12k·2=1k.
Form≥2, we set
(45)ν1=x1+(x1-ω1)×(2mk∥x1-ω1,…,xm-ωm,xm+1-ym+1,…,xn-yn∥)-1,(46)νi=2xi-ωi,
for any i∈{2,3,…,m}. Then, we have
(47)xi-νi=ωi-xi,νi-yi=(xi-ωi)+(xi-yi),
for each i∈{2,3,…,m}. Since xi-yi,xm+1-ym+1,…,xn-yn are linearly dependent, we get
(48)∥…,xi-yi,…,xm+1-ym+1,…,xn-yn∥=0,
and hence,
(49)∥…,xi-ωi,…,xm+1-ym+1,…,xn-yn∥-∥…,xi-yi,…,xm+1-ym+1,…,xn-yn∥≤∥…,(xi-ωi)+(xi-yi),…,xm+1-ym+1,…,xn-yn∥≤∥…,xi-ωi,…,xm+1-ym+1,…,xn-yn∥+∥…,xi-yi,…,xm+1-ym+1,…,xn-yn∥,
which together with (48) implies that
(50)∥…,νi-yi,…,xm+1-ym+1,…,xn-yn∥=∥…,xi-ωi,…,xm+1-ym+1,…,xn-yn∥,
for all i∈{2,3,…,m}. By a similar argument, we further obtain that
(51)∥ν1-y1,…,xm+1-ym+1,…,xn-yn∥=∥ν1-x1,…,xm+1-ym+1,…,xn-yn∥.
In view of (45), (50), and (51), we conclude that
(52)∥ν1-y1,μ2,…,μm,xm+1-ym+1,…,xn-yn∥=∥x1-ν1,x2-ω2,…,xm-ωm,xm+1=cc-ym+1,…,xn-yn∥=12mk,
where μi denotes either νi-yi or xi-νi for i∈{2,3,…,m}.
Sincef preserves the distance 1/(2mk) for any k∈ℕ, it follows from (52) that
(53)∥f(x1)-f(y1),f(x2)-f(y2),f(x3)-f(y3),…,f(xn)-f(yn)∥≤∥f(x1)-f(ν1),f(x2)-f(ν2),…,f(xm-1)-f(νm-1),f(xm)-f(νm),f(xm+1)-f(ym+1),…,f(xn)-f(yn)∥+∥f(x1)-f(ν1),f(x2)-f(ν2),…,f(xm-1)-f(νm-1),f(νm)-f(ym),f(xm+1)-f(ym+1),…,f(xn)-f(yn)∥+∥f(x1)-f(ν1),f(x2)-f(ν2),…,f(νm-1)-f(ym-1),f(xm)-f(νm),f(xm+1)-f(ym+1),…,f(xn)-f(yn)∥+∥f(x1)-f(ν1),f(x2)-f(ν2),…,f(νm-1)-f(ym-1),f(νm)-f(ym),f(xm+1)-f(ym+1),…,f(xn)-f(yn)∥+⋯++∥f(ν1)-f(y1),f(ν2)-f(y2),…,f(νm-1)-f(ym-1),f(νm)-f(ym),f(xm+1)-f(ym+1),…,f(xn)-f(yn)∥=12mk·2m=1k,
where k is an arbitrary positive integer. Hence, we conclude that
(54)∥f(x1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥=0,
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.
LetX 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,…,xn-yn∥>0 for x1,…,xn,y1,…,yn∈X.
Take positive integersk,m such that
(55)m-1k≤∥x1-y1,x2-y2,…,xn-yn∥≤mk.
Set
(56)pi=x1+ik·y1-x1∥x1-y1,x2-y2,…,xn-yn∥,
for i=0,1,…,m-2, and
(57)pm=y1.
Clearly, for i=1,…,m-2,
(58)∥pi-pi-1,x2-y2,…,xn-yn∥=1k,0<∥pm-pm-2,x2-y2,…,xn-yn∥=∥y1-x1-m-2k·y1-x1∥x1-y1,x2-y2,…,xn-yn∥,x2-y2,…,xn-ynm-2k·y1-x1∥x1-y1,x2-y2,…,xn-yn∥∥=(1-m-2k·1∥x1-y1,x2-y2,…,xn-yn∥)·∥x1-y1,x2-y2,…,xn-yn∥=∥x1-y1,x2-y2,…,xn-yn∥-m-2k≤mk-m-2k=2k.
According to Lemma 8, there exists pm-1∈X such that
(59)∥pm-1-pm-2,x2-y2,…,xn-yn∥=1k,∥pm-1-y1,x2-y2,…,xn-yn∥=1k.
It follows from Lemma 9 that we have
(60)∥f(pi)-f(pi-1),f(x2)-f(y2),…,f(xn)-f(yn)∥=1k,
for i=0,1,2,…,m.
On the other hand,(61)∥f(x1)-f(y1),f(x2)-f(y2)…,f(xn)-f(yn)∥≤∑i=1m‍∥f(pi)-f(pi-1),f(x2)-f(y2),…,f(xn)-f(yn)∥=mk.
Hence
(62)∥f(x1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥≤∥x1-y1,x2-y2,…,xn-yn∥.
Suppose that
(63)∥f(x1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥<∥x1-y1,x2-y2,…,xn-yn∥.
For any x1,y1,x2,y2,…,xn,yn∈X, with
(64)∥x1-y1,x2-y2,…,xn-yn∥≠0,
find a positive integer k0 satisfying ∥x1-y1,x2-y2,…,xn-yn∥<k0.
Setz1=x1+k0(y1-x1)/∥x1-y1,x2-y2,…,xn-yn∥. Clearly, ||z1-x1,x2-y2,…,xn-yn||=k0, and ∥z1-y1,x2-y2,…,xn-yn∥=k0-∥x1-y1,x2-y2,…,xn-yn∥.
It follows that∥f(z1)-f(x1),f(x2)-f(y2),…,f(xn)-f(yn)∥=k0 and
(65)k0=∥f(z1)-f(x1),f(x2)-f(y2),…,f(xn)-f(yn)∥≤∥f(z1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥+∥f(x1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥<k0-∥x1-y1,x2-y2,…,xn-yn∥+∥x1-y1,x2-y2,…,xn-yn∥=k0.
Then (63) is not valid. Hence,
(66)∥f(x1)-f(y1),f(x2)-f(y2),…,f(xn)-f(yn)∥=∥x1-y1,x2-y2,…,xn-yn∥.Corollary 13.
LetX 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.



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