# A Generalized Mazur-Ulam Theorem for Fuzzy Normed Spaces

Authors: J. J. Font; J. Galindo; S. Macario; M. Sanchis
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We introducefuzzy norm-preserving maps, which generalize the concept of fuzzy isometry. Based on the ideas from Vogt, 1973, and Väisälä, 2003, we provide the following generalized version of the Mazur-Ulam theorem in the fuzzy context: let X, Y be fuzzy normed spaces and let f:X→Y be a fuzzy norm-preserving surjection satisfying f(0)=0. Then f is additive.


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## Body

## 1. Introduction

Studies on fuzzy normed spaces are relatively recent in the field of fuzzy functional analysis. It was Katsaras who in 1984 [1], while studying topological vector spaces, was the first to introduce the idea of fuzzy norm on a linear space. Eight years later, Felbin [2] offered an alternative definition. With this definition, the induced fuzzy metric is of Kaleva and Seikkala type [3]. In 1994, Cheng and Mordeson [4] defined another type of fuzzy norm on a linear space whose associated fuzzy metric is of Kramosil and Michalek type [5]. Finally, in [6] (see also [7]), Bag and Samanta redefined the concept of fuzzy norm given in [4] as follows.Definition 1.
LetX be a real linear space. A function N:X×R→[0,1] is said to be a fuzzy norm on X if, for all x,y∈X and all t,s,k∈R, it satisfies the following:(N1)
N(x,t)=0 for t≤0;(N2)
x=0 if and only if N(x,t)=1 for all t>0;(N3)
N(kx,t)=N(x,t/|k|) if k≠0;(N4)
N(x+y,t+s)≥min⁡{N(x,t),N(y,s)};(N5)
lim⁡t→∞N(x,t)=1.
The pair(X,N) is called a fuzzy normed space.We point out that classical normed spaces are strictly included in the class of fuzzy normed spaces (see [6]) and that (N2) and (N4) imply that, for a fixed x∈X, the function N(x,·) is nondecreasing. It is a well-known fact that every fuzzy norm on a real linear space X induces a topology on X defined as follows: a subbase for the neighborhood system at a point x∈X consists of the sets
(1)BN(x,ε,t)={y∈X:N(x-y,t)>1-ε}
for all 0<ε<1 and t>0. It is straightforward to verify that the filter of neighborhoods of the origin generated by the family {BN(0,ε,t):0<ε<1,t>0} satisfies the properties which make (X,N) a Hausdorff topological vector space.The theory of isometric mappings on classical normed spaces has its roots in the seminal paper by Mazur and Ulam ([8]; see also [9]), who proved that every bijective isometry between two real normed spaces is affine. It is known that the surjective assumption is essential in this result and that it is not true for complex normed spaces. The Mazur-Ulam theorem has been extended in many directions. For example, Baker [10] proved that the result remains true if we consider an (not necessarily onto) isometry between a real normed space and astrictly convex real normed space. Another direction was provided by Vogt [11], who replaced isometries by the more general notion ofequality of distance preserving maps (see also [12]).In this paper, following the ideas of Vogt, we introduce a generalization of the concept of fuzzy isometry as follows.Definition 2.
Let(X,N) and (Y,N′) be two fuzzy normed spaces. One says that f:X→Y is a fuzzy norm-preserving mapping if given x,y,x′,y′∈X, then, for all t>0,
(2)N(x-y,t)=N(x′-y′,t)⟹N′(f(x)-f(y),t)=N′(f(x′)-f(y′),t).
Let us recall here the definition of a fuzzy isometry.Definition 3.
Let(X,N) and (Y,N′) be two fuzzy normed spaces. It is said that f:X→Y is a fuzzy isometry if N′(f(x)-f(y),t)=N(x-y,t), for all x,y∈X and t>0.
We provide the following generalized version of the Mazur-Ulam theorem in the fuzzy context: letX, Y be fuzzy normed spaces and let f:X→Y be a fuzzy norm-preserving surjection satisfying f(0)=0. Then, f is additive. As a corollary, we deduce that if such an f is a fuzzy isometry, then f is affine.


## 2. The Results

Let(X,N) be a fuzzy normed space. Fix c∈X and define a function ϕc:X→X as ϕc(x):=2c-x. It is apparent that ϕc is bijective; indeed, ϕc∘ϕc=Id, where Id stands for the identity map on X. Furthermore, ϕc is a fuzzy isometry since N(ϕc(x)-ϕc(y),t)=N(2c-x-2c+y,t)=N(x-y,t), for all x,y∈X and t>0.LetL(c,N):={x∈X:N(x,t)=N(2c-x,t)≥N(c,t/2)forallt>0}. It is clear that c∈L(c,N)≠∅ and that ϕc(L(c,N))⊆L(c,N).Lemma 4.
Every fuzzy isometry fromL(c,N) onto L(c,N) fixes c.Proof.
LetI:={g:L(c,N)→L(c,N):gisanontofuzzyisometry}. This is a nonempty set since the identity map belongs to it. Fix t>0 and let us define, for each i=1,2,3,…,
(3)λit:=inf⁡{N(g(c)-c,ti):g∈I}.
If g∈I, then we define g′:=g-1∘ϕc∘g, which is in I. Then
(4)N(g′(c)-c,t)=N((g-1∘ϕc∘g)(c)-c,t)=N(ϕc(g(c))-g(c),t)=N(2(g(c)-c),t)=N(g(c)-c,t2).
Consequently, N(g(c)-c,t/2)≥λ1t.
Moreover, for allg∈I, we have
(5)N(g(c)-c,t)≥N(g(c)-c,t2)≥λ1t,
which yields λ1t≥λ2t≥λ1t. That is, λ1t=λ2t, which leads us to the following equalities:
(6)λ1t=λ2t=⋯=λnt=⋯.
On the other hand, giveng∈I, we have
(7)N(g(c)-c,t)≥min⁡{N(g(c),t2),N(c,t2)}=N(c,t2),
since g(c)∈L(c,N).
Let us suppose that there existsg0∈I such that g0(c)≠c. Then, there exists t0>0 such that
(8)N(g0(c)-c,t0)=k<1.
In addition, for each t>0, there exists nt such that N(g0(c)-c,t/nt)≤N(g0(c)-c,t0). Hence, by (6) and (7), we have that
(9)N(c,t2)≤λntt≤N(g0(c)-c,tnt)≤N(g0(c)-c,t0)=k<1,
but
(10)1=lim⁡t→∞N(c,t2)≤k<1,
a contradiction which completes the proof.Remark 5.
It can be checked that a fuzzy norm-preserving mapping is associated, for eacht>0, with a function ρt:Dt⊆[0,1]→[0,1] such that
(11)N′(f(x)-f(y),t)=ρt(N(x-y,t)).
It is then apparent that fuzzy isometries are fuzzy norm-preserving mappings takingρt=Id, t>0.Theorem 6.
Let(X,N) and (Y,N′) be two fuzzy normed spaces and let f:X→Y be a fuzzy norm-preserving surjection satisfying f(0)=0. Then f is additive.Proof.
Fixx0∈X and let
(12)L(f(x0),N′):={(f(x0),t2)y∈Y:N′(y,t)=N′(2f(x0)-y,t)y∈Y:N′(y,t)≥N′(f(x0),t2),t>0}.
We know thatf(x0)∈L(f(x0),N′)≠∅.
We now define maph:L(f(x0),N′)→L(f(x0),N′) as h(y):=f(x1-x′) for a fixed x1∈f-1(2f(x0)) and x′∈f-1(y).
Claim1. h is a fuzzy isometry from L(f(x0),N′) to L(f(x0),N′) which does not depend on the choice of x′∈f-1(y).
Let us first check thath does not depend on the choice of x′∈f-1(y). To this end, suppose that {x′,x′′}⊆f-1(y). Then,
(13)N′(f(x1-x′)-f(x1-x′′),t)=ρt(N(x′′-x′,t))=N′(f(x′′)-f(x′),t)=N′(0,t)=1,
which is to say that f(x1-x′)=f(x1-x′′).
Let us next prove thath is a fuzzy isometry. Suppose that f(x′)=y′ and f(x′′)=y′′, with x′,x′′∈X and y′,y′′∈Y. Then,
(14)N′(h(y′)-h(y′′),t)=N′(f(x1-x′)-f(x1-x′′),t)=ρt(N(x′′-x′,t))=N′(f(x′′)-f(x′),t)=N′(y′′-y′,t).
Finally, let us check thath maps L(f(x0),N′) onto L(f(x0),N′). Fix y′∈L(f(x0),N′) and let x′∈X such that f(x′)=y′. From the definition of L(f(x0),N′), we know that
(15)N′(y′,t)=N′(2f(x0)-y′,t)≥N′(f(x0),t2).
Hence,
(16)N′(h(y′),t)=N′(f(x1-x′)-f(0),t)=ρt(N(x1-x′-0,t))=N′(f(x1)-f(x′),t)=N′(2f(x0)-y′,t)=N′(y′,t)≥N′(f(x0),t2).
Furthermore,
(17)N′(2f(x0)-h(y′),t)=N′(2f(x0)-f(x1-x′),t)=N′(f(x1)-f(x1-x′),t)=ρt(N(x1-x1+x′-0,t))=N′(f(x′)-f(0),t)=N′(y′,t)=N′(h(y′),t).
As a consequence, we deduce that h(L(f(x0),N′))⊆L(f(x0),N′). Since it is a routine matter to verify that h-1=h, we infer that h(L(f(x0),N′))=L(f(x0),N′), and the claim is proved.
Thanks to Claim 1, we can apply Lemma4 and conclude that h fixes f(x0). Hence,
(18)f(x0)=h(f(x0))=f(x1-x0)
and, then,
(19)1=N′(f(x0)-f(x1-x0),t)=ρt(N(2x0-x1,t))=N′(f(2x0)-f(x1),t)=N′(f(2x0)-2f(x0),t)
for all t>0, which is to say that
(20)f(2x0)=2f(x0).
Next, let us define, for a fixedz∈X, the following map:
(21)fz(x):=f(x+z)-f(z),
for all x∈X.
It is clear thatfz(0)=0 and fz is surjective since f is assumed to be also surjective. Furthermore, for all x,y∈X and t>0,
(22)N′(fz(x)-fz(y),t)=N′(f(x+z)-f(y+z),t)=ρt(N(x-y,t)),
which is to say that fz is also a fuzzy norm-preserving map. Hence, by (20), we infer that fz(2x)=2fz(x), for all x∈X. Then, for all x,y∈X, we have
(23)f((x-y)+y)-f(y)=fy(x-y)=2fy(x-y2)=2(f(x-y2+y)-f(y)),
which yields
(24)f(x)+f(y)=2f(x+y2)=f(x+y);
that is, f is additive.Let us recall here that additivity yieldsQ-linearity, which, in presence of continuity, implies linearity. Hence, as a straightforward corollary of Theorem 6, we obtain a fuzzy version of the Mazur-Ulam theorem.Corollary 7.
Let(X,N) and (Y,N′) be two fuzzy normed spaces and let f:X→Y be a surjective fuzzy isometry. Then, f is affine.Proof.
It suffices to apply Theorem6 to g(x):=f(x)-f(0), x∈X.



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*Source: 624920-2014-06-02.xml*