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volume 7, issue 5, article 168, 2006.

Received 05 December, 2005;

accepted 25 April, 2006.

Communicated by:K. Nikodem

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Journal of Inequalities in Pure and Applied Mathematics

ISOMETRIES ON LINEAR n-NORMED SPACES

CHUN-GIL PARK AND THEMISTOCLES M. RASSIAS

Department of Mathematics Hanyang University Seoul 133-791 Republic of Korea

EMail:[email protected] Department of Mathematics

National Technical University of Athens Zografou Campus

15780 Athens, Greece EMail:[email protected]

c

2000Victoria University ISSN (electronic): 1443-5756 385-05

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Isometries on Linearn-Normed Spaces

Chun-Gil Park and Themistocles M. Rassias

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J. Ineq. Pure and Appl. Math. 7(5) Art. 168, 2006

http://jipam.vu.edu.au

Abstract

The aim of this article is to generalize the Aleksandrov problem to the case of linearn-normed spaces.

2000 Mathematics Subject Classification:Primary 46B04, 46B20, 51K05.

Key words: Linearn-normed space,n-isometry,n-Lipschitz mapping.

1. Introduction

LetX andY be metric spaces. A mappingf :X →Y is called an isometry if f satisfies

d_Y f(x), f(y)

=d_X(x, y)

for allx, y ∈ X, whered_X(·,·)andd_Y(·,·)denote the metrics in the spacesX and Y, respectively. For some fixed number r > 0, suppose that f preserves distancer; i.e., for allx, y inXwithd_X(x, y) =r, we haved_Y f(x), f(y)

= r. Then r is called a conservative (or preserved) distance for the mapping f. Aleksandrov [1] posed the following problem:

Remark 1. Examine whether the existence of a single conservative distance for some mappingT implies thatT is an isometry.

The Aleksandrov problem has been investigated in several papers (see [3] – [10]). Th.M. Rassias and P. Šemrl [9] proved the following theorem for mappings satisfying the strong distance one preserving property (SDOPP), i.e., for every x, y ∈ X with kx −yk = 1 it follows that kf(x) −f(y)k = 1 and conversely.

Theorem 1.1 ([9]). Let X and Y be real normed linear spaces with dimen- sion greater than one. Suppose that f : X → Y is a Lipschitz mapping with Lipschitz constant κ = 1. Assume that f is a surjective mapping satisfying (SDOPP). Thenf is an isometry.

Definition 1.1 ([2]). LetXbe a real linear space withdimX ≥nandk·, . . . ,·k: Xⁿ →Ra function. Then(X,k·, . . . ,·k)is called a linearn-normed space if (nN1) kx1, . . . , xnk= 0⇐⇒x1, . . . , xnare linearly dependent

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(nN₂) kx₁, . . . , x_nk=kx_j₁, . . . , x_j_nk

for every permutation (j₁, . . . j_n) of (1, . . . , n)

(nN₃) kαx₁, . . . , x_nk=|α|kx₁, . . . , x_nk

(nN₄) kx+y, x₂, . . . , x_nk ≤ kx, x₂, . . . , x_nk+ky, x₂, . . . , x_nk

for allα ∈Rand allx, y, x₁, . . . , x_n∈ X. The functionk·, . . . ,·kis called the n-norm onX.

In [3], Chu et al. defined the notion of weak n-isometry and proved the Rassias and Šemrl’s theorem in linearn-normed spaces.

Definition 1.2 ([3]). We callf : X → Y a weakn-Lipschitz mapping if there is aκ≥0such that

kf(x₁)−f(x₀), . . . , f(x_n)−f(x₀)k ≤κkx₁−x₀, . . . , x_n−x₀k for all x₀, x₁, . . . , x_n ∈ X. The smallest suchκis called the weakn-Lipschitz constant.

Definition 1.3 ([3]). LetX andY be linearn-normed spaces andf :X →Y a mapping. We callf a weakn-isometry if

kx₁−x₀, . . . , x_n−x₀k=kf(x₁)−f(x₀), . . . , f(x_n)−f(x₀)k for allx0, x1, . . . , xn∈X.

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For a mappingf :X →Y, consider the following condition which is called the weak n-distance one preserving property: For x0, x1, . . . , xn ∈ X with kx₁−y₁, . . . , x_n−y_nk= 1,kf(x₁)−f(x₀), . . . , f(x_n)−f(x₀)k= 1.

Theorem 1.2 ([3]). Letf : X → Y be a weakn-Lipschitz mapping with weak n-Lipschitz constantκ ≤1. Assume that ifx0, x1, . . . , xmare m-colinear then f(x₀), f(x₁), . . ., f(x_m)are m-colinear, m = 2, n, and that f satisfies the weakn−distance one preserving property. Thenf is a weakn-isometry.

In this paper, we introduce the concept ofn-isometry which is suitable for representing the notion of n-distance preserving mappings in linearn-normed spaces. We prove also that the Rassias and Šemrl theorem holds under some conditions whenXandY are linearn-normed spaces.

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2. The Aleksandrov Problem in Linear n-normed Spaces

In this section, letX andY be linearn-normed spaces with dimension greater thann−1.

Definition 2.1. Let X and Y be linear n-normed spaces andf : X → Y a mapping. We callf ann-isometry if

kx₁−y₁, . . . , x_n−y_nk=kf(x₁)−f(y₁), . . . , f(x_n)−f(y_n)k for allx₁, . . . , x_n, y₁, . . . , y_n ∈X.

For a mappingf :X →Y, consider the following condition which is called the n-distance one preserving property : For x1, . . . , xn, y1, . . . , yn ∈ X with kx₁−y₁, . . . , x_n−y_nk= 1,kf(x₁)−f(y₁), . . . , f(x_n)−f(y_n)k= 1.

Lemma 2.1 ([3, Lemma 2.3]). Let x₁, x₂, . . . , x_n be elements of a linear n- normed spaceXandγa real number. Then

kx₁, . . . , x_i, . . . , x_j, . . . , x_nk=kx₁, . . . , x_i, . . . , x_j+γx_i, . . . , x_nk.

for all1≤i6=j ≤n.

Definition 2.2 ([3]). The pointsx0, x1, . . . , xnofX are said to ben-colinear if for everyi,{x_j−x_i |0≤j 6=i≤n}is linearly dependent.

Remark 2. The pointsx₀, x₁ andx₂ are 2-colinear if and only if x₂−x₀ = t(x₁−x₀)for some real numbert.

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Theorem 2.2. Let f : X → Y be a weak n-Lipschitz mapping with weak n- Lipschitz constant κ ≤ 1. Assume that if x0, x1, . . . , xm are m-colinear then f(x₀), f(x₁), . . ., f(x_m)are m-colinear, m = 2, n, and that f satisfies the weakn−distance one preserving property. Thenf satisfies

kx₁−y₁, . . . , x_n−y_nk=kf(x₁)−f(y₁), . . . , f(x_n)−f(y_n)k for allx1, . . . , xn, y1, . . . yn∈Xwithx1, y1, yj 2-colinear forj = 2,3, . . . , n.

Proof. By Theorem1.2,f is a weakn-isometry. Hence

(2.1) kx₁−y, . . . , x_n−yk=kf(x₁)−f(y), . . . , f(x_n)−f(y)k for allx₁, . . . , x_n, y ∈X.

Ifx1, y1, y2 ∈X are 2-colinear then there exists at∈Rsuch thaty1−y2 = t(y₁−x₁). By Lemma2.1,

kx1−y1, x2−y2, . . . , xn−ynk

=kx₁ −y₁,(x₂−y₁) + (y₁−y₂), . . . , x_n−y_nk

=kx₁ −y₁,(x₂−y₁) + (−t)(x₁−y₁), . . . , x_n−y_nk

=kx₁ −y₁, x₂−y₁, . . . , x_n−y_nk

for allx₁, . . . , x_n, y₁, . . . , y_n ∈Xwithx₁, y₁, y₂2-colinear. By the same method as above, one can obtain that ifx₁, y₁, y_j are 2-colinear forj = 3, . . . , nthen

kx₁−y₁, x₂−y₂,x₃−y₃, . . . , x_n−y_nk (2.2)

=kx1−y1, x2−y1, x3−y3, . . . , xn−ynk

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=kx₁−y₁, x₂−y₁, x₃−y₁, . . . , x_n−y_nk

=· · ·

=kx₁−y₁, x₂−y₁, x₃−y₁, . . . , x_n−y₁k for allx₁, . . . , x_n, y₁, . . . y_n∈Xwithx₁, y₁, y_j 2-colinear forj = 2,3, . . . , n.

By the assumption, ifx₁, y₁, y₂ ∈Xare 2-colinear thenf(x₁), f(y₁), f(y₂)∈ Y are 2-colinear. So there exists at ∈ Rsuch thatf(y₁)−f(y₂) = t(f(y₁)− f(x₁)). By Lemma2.1,

kf(x₁)−f(y₁), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k

=kf(x1)−f(y1),(f(x2)−f(y1)) + (f(y1)−f(y2)), . . . , f(xn)−f(yn)k

=kf(x₁)−f(y₁),(f(x₂)−f(y₁)) + (−t)(f(x₁)−f(y₁)), . . . , f(x_n)−f(y_n)k

=kf(x₁)−f(y₁), f(x₂)−f(y₁), . . . , f(x_n)−f(y_n)k

for all x₁, . . . , x_n, y₁, . . . , y_n ∈ X with x₁, y₁, y₂ 2-colinear. If x₁, y₁, y_j are 2-colinear for j = 3, . . . , n then f(x1), f(y1), f(yj) are 2-colinear for j = 3, . . . , n. By the same method as above, one can obtain that iff(x₁), f(y₁), f(y_j) are 2-colinear forj = 3, . . . , n, then

kf(x₁)−f(y₁), f(x₂)−f(y₂), f(x₃)−f(y₃), . . . , f(x_n)−f(y_n)k (2.3)

=kf(x₁)−f(y₁), f(x₂)−f(y₁), f(x₃)−f(y₃), . . . , f(x_n)−f(y_n)k

=kf(x₁)−f(y₁), f(x₂)−f(y₁), f(x₃)−f(y₁), . . . , f(x_n)−f(y_n)k

=. . .

=kf(x1)−f(y1), f(x2)−f(y1), f(x3)−f(y1), . . . , f(xn)−f(y1)k for allx1, . . . , xn, y1, . . . yn∈Xwithx1, y1, yj 2-colinear forj = 2,3, . . . , n.

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By (2.1), (2.2) and (2.3),

kx₁−y₁, . . . , x_n−y_nk=kf(x₁)−f(y₁), . . . , f(x_n)−f(y_n)k

for all x₁, . . . , x_n, y₁, . . . y_n ∈ X withx₁, y₁, y_j 2-colinear forj = 2,3, . . . , n.

Now we introduce the concept of n-Lipschitz mapping and prove that the n-Lipschitz mapping satisfying then-distance one preserving property is ann- isometry under some conditions.

Definition 2.3. We callf :X → Y ann-Lipschitz mapping if there is aκ≥ 0 such that

kf(x₁)−f(y₁), . . . , f(x_n)−f(y_n)k ≤κkx₁−y₁, . . . , x_n−y_nk for allx₁, . . . , x_n, y₁, . . . , y_n∈X. The smallest suchκis called then-Lipschitz constant.

Lemma 2.3 ([3, Lemma 2.4]). Forx₁, x⁰₁ ∈ X, if x₁ and x⁰₁ are linearly de- pendent with the same direction, that is,x⁰₁ =αx₁ for someα >0, then

x₁+x⁰₁, x₂, . . . , x_n

=kx₁, x₂, . . . , x_nk+kx⁰₁, x₂, . . . , x_nk for allx₂, . . . , x_n∈X.

Lemma 2.4. Assume that ifx0, x1andx2are 2-colinear thenf(x0), f(x1)and f(x₂)are 2-colinear, and thatfsatisfies then-distance one preserving property.

Thenf preserves then-distancek for eachk ∈N.

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Proof. Suppose that there exist x₀, x₁ ∈ X with x₀ 6= x₁ such that f(x₀) = f(x1). Since dimX ≥ n, there are x2, . . . , xn ∈ X such thatx1 −x0, x2 − x₀, . . . , x_n−x₀are linearly independent. Sincekx₁−x₀, x₂−x₀, . . . , x_n−x₀k 6=

0, we can set

z2 :=x0+ x₂−x₀

kx₁−x₀, x₂−x₀, . . . , x_n−x₀k. Then we have

kx₁−x₀, z₂−x₀, x₃−x₀, . . . , x_n−x₀k

=

x₁−x₀, x₂−x₀

kx₁−x₀, x₂−x₀, . . . , x_n−x₀k, x₃−x₀, . . . , x_n−x₀

= 1.

Sincef preserves then-distance 1,

kf(x₁)−f(x₀), f(z₂)−f(x₀), . . . , f(x_n)−f(x₀)k= 1.

But it follows fromf(x₀) =f(x₁)that

kf(x1)−f(x0), f(z2)−f(x0), . . . , f(xn)−f(x0)k= 0, which is a contradiction. Hencef is injective.

Letx₁, . . . , x_n, y₁, . . . , y_n ∈X,k∈Nand

kx₁−y₁, x₂−y₂, . . . , x_n−y_nk=k.

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We put

z_i =y₁+ i

k(x₁−y₁), i= 0,1, . . . , k.

Then

kz_i+1−z_i, x₂−y₂, . . . , x_n−y_nk

=

y₁+i+ 1

k (x₁ −y₁)−

y₁+ i

k(x₁−y₁)

, x₂−y₂, . . . , x_n−y_n

= 1

k(x₁−y₁), x₂−y₂, . . . , x_n−y_n

= 1

kkx₁−y₁, x₂−y₂, . . . , x_n−y_nk= k k = 1

for all i = 0,1, . . . , k − 1. Since f satisfies the n-distance one preserving property,

(2.4) kf(z_i+1)−f(z_i), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k= 1

for all i = 0,1, . . . , k −1. Sincez₀, z₁ andz₂ are 2-colinear,f(z₀), f(z₁) and f(z₂)are also 2-colinear. Thus there is a real numbert₀ such that

f(z₂)−f(z₁) = t₀(f(z₁)−f(z₀)).

By (2.4),

kf(z1)−f(z0),f(x2)−f(y2), . . . , f(xn)−f(yn)k

=kf(z₂)−f(z₁), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k

=kt₀(f(z₁)−f(z₀)), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k

=|t₀|kf(z₁)−f(z₀), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k.

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So we havet₀ =±1. Ift₀ =−1,f(z₂)−f(z₁) =−f(z₁) +f(z₀), that is, f(z2) =f(z0).

Sincef is injective,z₂ =z₀, which is a contradiction. Thust₀ = 1. Hence f(z₂)−f(z₁) =f(z₁)−f(z₀).

Similarly, one can obtain that

f(z_i+1)−f(z_i) = f(z_i)−f(zi−1) for alli= 2,3, . . . , k−1. Thus

f(z_i+1)−f(z_i) = f(z₁)−f(z₀) for alli= 1,2, . . . , k−1. Hence

f(x₁)−f(y₁) = f(z_k)−f(z₀)

=f(z_k)−f(zk−1) +f(zk−1)−f(zk−2) +· · ·+f(z₁)−f(z₀)

=k(f(z₁)−f(z₀)).

Therefore,

kf(x₁)−f(y₁),f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k

=kk(f(z1)−f(z0)), f(x2)−f(y2), . . . , f(xn)−f(yn)k

=kk(f(z₁)−f(z₀)), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k

=k, which completes the proof.

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Theorem 2.5. Letf :X →Y be ann-Lipschitz mapping withn-Lipschitz con- stant κ = 1. Assume that ifx0, x1, x2 are 2-colinear thenf(x0), f(x1), f(x2) are 2-colinear, and that if x₁ − y₁, . . . , x_n −y_n are linearly dependent then f(x₁)− f(y₁), . . . , f(x_n) −f(y_n) are linearly dependent. If f satisfies the n-distance one preserving property, thenf is ann-isometry.

Proof. By Lemma 2.4, f preserves the n-distance k for each k ∈ N. For x₁, . . . , x_n, y₁, . . . , y_n∈X, there are two cases depending upon whetherkx₁− y₁, . . . , x_n −y_nk = 0 or not. In the case kx₁ −y₁, . . . , x_n−y_nk = 0, x₁ − y1, . . . , xn−ynare linearly dependent. By the assumption,f(x1)−f(y1), . . . , f(x_n)−f(y_n)are linearly dependent. Hence

kf(x1)−f(y1), . . . , f(xn)−f(yn)k= 0.

In the casekx₁−y₁, . . . , x_n−y_nk>0, there exists ann₀ ∈Nsuch that kx1−y1, . . . , xn−ynk< n0.

Assume that

kf(x₁)−f(y₁), . . . , f(x_n)−f(y_n)k<kx₁−y₁, . . . , x_n−y_nk.

Set

w=y₁+ n₀

kx1−y1, . . . , xn−ynk(x₁−y₁).

Then we obtain that

kw−y1, x2−y2, . . . , xn−ynk

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=

y₁+ n₀

kx₁−y₁, . . . , x_n−y_nk(x₁−y₁)−y₁, x₂−y₂, . . . , x_n−y_n

= n₀

kx₁−y₁, . . . , x_n−y_nkkx₁−y₁, . . . , x_n−y_nk=n₀. By Lemma2.4,

kf(w)−f(y₁), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k=n₀. By the definition ofw,

w−x₁ =

n₀

kx1−y1, . . . , xn−ynk −1

(x₁−y₁).

Since

n₀

kx1−y1, . . . , xn−ynk >1, w−x₁ andx₁−y₁have the same direction. By Lemma2.3,

kw−y₁, x₂−y₂, . . . , x_n−y_nk

=kw−x₁, x₂−y₂, . . . , x_n−y_nk+kx₁−y₁, x₂−y₂, . . . , x_n−y_nk.

So we have

kf(w)−f(x₁),f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k

≤ kw−x₁, x₂−y₂, . . . , x_n−y_nk

=n₀− kx₁−y₁, x₂−y₂, . . . , x_n−y_nk.

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By the assumption,

n₀ =kf(w)−f(y₁), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k

≤ kf(w)−f(x1), f(x2)−f(y2), . . . , f(xn)−f(yn)k

+kf(x₁)−f(y₁), f(x₂)−f(y₂), . . . , f(x_n)−f(y_n)k

< n₀− kx₁−y₁, x₂−y₂, . . . , x_n−y_nk+kx₁−y₁, x₂−y₂, . . . , x_n−y_nk

=n₀,

which is a contradiction. Hencef is ann-isometry.

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References

[1] A.D. ALEKSANDROV, Mappings of families of sets, Soviet Math. Dokl., 11 (1970), 116–120.

[2] Y.J. CHO, P.C.S. LIN, S.S. KIM AND A. MISIAK, Theory of 2-Inner Product Spaces, Nova Science Publ., New York, 2001.

[3] H. CHU, K. LEE AND C. PARK, On the Aleksandrov problem in linear n-normed spaces, Nonlinear Analysis – Theory, Methods & Applications, 59 (2004), 1001–1011.

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[6] B. MIELNIKANDTh.M. RASSIAS, On the Aleksandrov problem of con- servative distances, Proc. Amer. Math. Soc., 116 (1992), 1115–1118.

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[9] Th.M. RASSIAS ANDP. ŠEMRL, On the Mazur–Ulam problem and the Aleksandrov problem for unit distance preserving mappings, Proc. Amer.

Math. Soc., 118 (1993), 919–925.

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[10] Th.M. RASSIAS AND S. XIANG, On mappings with conservative dis- tances and the Mazur–Ulam theorem, Publications Faculty Electrical En- gineering, Univ. Belgrade, Series: Math., 11 (2000), 1–8.

[11] S. XIANG, Mappings of conservative distances and the Mazur–Ulam the- orem, J. Math. Anal. Appl., 254 (2001), 262–274.

[12] S. XIANG, On the Aleksandrov–Rassias problem for isometric mappings, in Functional Equations, Inequalities and Applications, Th.M. Rassias, Kluwer Academic Publ., Dordrecht, 2003, 191–221.