(2) 2. Abstract and Applied Analysis. In this paper, following the ideas of Vogt, we introduce a generalization of the concept of fuzzy isometry as follows.. which yields 𝜆𝑡1 ≥ 𝜆𝑡2 ≥ 𝜆𝑡1 . That is, 𝜆𝑡1 = 𝜆𝑡2 , which leads us to the following equalities:. 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,. 𝜆𝑡1 = 𝜆𝑡2 = ⋅ ⋅ ⋅ = 𝜆𝑡𝑛 = ⋅ ⋅ ⋅ .. 𝑁 (𝑥 − 𝑦, 𝑡) = 𝑁 (𝑥 − 𝑦 , 𝑡) ⇒ 𝑁 (𝑓 (𝑥) − 𝑓 (𝑦) , 𝑡) = 𝑁 (𝑓 (𝑥 ) − 𝑓 (𝑦 ) , 𝑡) .. (2). On the other hand, given 𝑔 ∈ 𝐼, we have 𝑡 𝑡 𝑁 (𝑐, ) ≤ 𝜆𝑡𝑛𝑡 ≤ 𝑁 (𝑔0 (𝑐) − 𝑐, ) ≤ 𝑁 (𝑔0 (𝑐) − 𝑐, 𝑡0 ) 2 𝑛𝑡 = 𝑘 < 1, (7). 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: let 𝑋, 𝑌 be fuzzy normed spaces and let 𝑓 : 𝑋 → 𝑌 be a fuzzy normpreserving 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. Let 𝐿(𝑐, 𝑁) := {𝑥 ∈ 𝑋 : 𝑁(𝑥, 𝑡) = 𝑁(2𝑐 − 𝑥, 𝑡) ≥ 𝑁(𝑐, 𝑡/2) for all 𝑡 > 0}. It is clear that 𝑐 ∈ 𝐿(𝑐, 𝑁) ≠ 0 and that 𝜙𝑐 (𝐿(𝑐, 𝑁)) ⊆ 𝐿(𝑐, 𝑁). Lemma 4. Every fuzzy isometry from 𝐿(𝑐, 𝑁) onto 𝐿(𝑐, 𝑁) fixes 𝑐. Proof. Let 𝐼 := {𝑔 : 𝐿(𝑐, 𝑁) → 𝐿(𝑐, 𝑁) : 𝑔 is an onto fuzzy isometry}. 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, . . ., 𝑡 𝜆𝑡𝑖 := inf {𝑁 (𝑔 (𝑐) − 𝑐, ) : 𝑔 ∈ 𝐼} . 𝑖. (3). If 𝑔 ∈ 𝐼, then we define 𝑔 := 𝑔−1 ∘ 𝜙𝑐 ∘ 𝑔, which is in 𝐼. Then 𝑁 (𝑔 (𝑐) − 𝑐, 𝑡) = 𝑁 ((𝑔−1 ∘ 𝜙𝑐 ∘ 𝑔) (𝑐) − 𝑐, 𝑡) (4). 𝑡 = 𝑁 (𝑔 (𝑐) − 𝑐, ) . 2. 𝑁 (𝑔0 (𝑐) − 𝑐, 𝑡0 ) = 𝑘 < 1.. (8). 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 𝑡 𝑡 𝑁 (𝑐, ) ≤ 𝜆𝑡𝑛𝑡 ≤ 𝑁 (𝑔0 (𝑐) − 𝑐, ) ≤ 𝑁 (𝑔0 (𝑐) − 𝑐, 𝑡0 ) 2 𝑛𝑡 = 𝑘 < 1, (9) but 𝑡 1 = lim 𝑁 (𝑐, ) ≤ 𝑘 < 1, 𝑡→∞ 2. (10). a contradiction which completes the proof. Remark 5. It can be checked that a fuzzy norm-preserving mapping is associated, for each 𝑡 > 0, with a function 𝜌𝑡 : 𝐷𝑡 ⊆ [0, 1] → [0, 1] such that 𝑁 (𝑓 (𝑥) − 𝑓 (𝑦) , 𝑡) = 𝜌𝑡 (𝑁 (𝑥 − 𝑦, 𝑡)) .. (11). It is then apparent that fuzzy isometries are fuzzy normpreserving mappings taking 𝜌𝑡 = 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. Fix 𝑥0 ∈ 𝑋 and let. := {𝑦 ∈ 𝑌 : 𝑁 (𝑦, 𝑡) = 𝑁 (2𝑓 (𝑥0 ) − 𝑦, 𝑡). (12). 𝑡 ≥ 𝑁 (𝑓 (𝑥0 ) , ) , 𝑡 > 0} . 2. Consequently, 𝑁(𝑔(𝑐) − 𝑐, 𝑡/2) ≥ 𝜆𝑡1 . Moreover, for all 𝑔 ∈ 𝐼, we have 𝑡 𝑁 (𝑔 (𝑐) − 𝑐, 𝑡) ≥ 𝑁 (𝑔 (𝑐) − 𝑐, ) ≥ 𝜆𝑡1 , 2. since 𝑔(𝑐) ∈ 𝐿(𝑐, 𝑁). Let us suppose that there exists 𝑔0 ∈ 𝐼 such that 𝑔0 (𝑐) ≠ 𝑐. Then, there exists 𝑡0 > 0 such that. 𝐿 (𝑓 (𝑥0 ) , 𝑁 ). = 𝑁 (𝜙𝑐 (𝑔 (𝑐)) − 𝑔 (𝑐) , 𝑡) = 𝑁 (2 (𝑔 (𝑐) − 𝑐) , 𝑡). (6). (5). We know that 𝑓(𝑥0 ) ∈ 𝐿(𝑓(𝑥0 ), 𝑁 ) ≠ 0. We now define map ℎ : 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)..

(3) Abstract and Applied Analysis. 3. Claim 1. 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 that ℎ does not depend on the choice of 𝑥 ∈ 𝑓−1 (𝑦). To this end, suppose that {𝑥 , 𝑥 } ⊆ 𝑓−1 (𝑦). Then,. Furthermore, 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 (𝑓 (𝑥1 − 𝑥 ) − 𝑓 (𝑥1 − 𝑥 ) , 𝑡). = 𝑁 (𝑦 , 𝑡). = 𝜌𝑡 (𝑁 (𝑥 − 𝑥 , 𝑡)) = 𝑁 (𝑓 (𝑥 ) − 𝑓 (𝑥 ) , 𝑡) (13). = 𝑁 (ℎ (𝑦 ) , 𝑡) .. . = 𝑁 (0, 𝑡) = 1,. (17). which is to say that 𝑓(𝑥1 − 𝑥 ) = 𝑓(𝑥1 − 𝑥 ). Let us next prove that ℎ is a fuzzy isometry. Suppose that 𝑓(𝑥 ) = 𝑦 and 𝑓(𝑥 ) = 𝑦 , with 𝑥 , 𝑥 ∈ 𝑋 and 𝑦 , 𝑦 ∈ 𝑌. Then,. 𝑁 (ℎ (𝑦 ) − ℎ (𝑦 ) , 𝑡) = 𝑁 (𝑓 (𝑥1 − 𝑥 ) − 𝑓 (𝑥1 − 𝑥 ) , 𝑡) . . = 𝜌𝑡 (𝑁 (𝑥 − 𝑥 , 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 Lemma 4 and conclude that ℎ fixes 𝑓(𝑥0 ). Hence, 𝑓 (𝑥0 ) = ℎ (𝑓 (𝑥0 )) = 𝑓 (𝑥1 − 𝑥0 ). (18). and, then, 1 = 𝑁 (𝑓 (𝑥0 ) − 𝑓 (𝑥1 − 𝑥0 ) , 𝑡). = 𝑁 (𝑓 (𝑥 ) − 𝑓 (𝑥 ) , 𝑡). = 𝜌𝑡 (𝑁 (2𝑥0 − 𝑥1 , 𝑡)). = 𝑁 (𝑦 − 𝑦 , 𝑡) .. = 𝑁 (𝑓 (2𝑥0 ) − 𝑓 (𝑥1 ) , 𝑡) (14). (19). = 𝑁 (𝑓 (2𝑥0 ) − 2𝑓 (𝑥0 ) , 𝑡) for all 𝑡 > 0, which is to say that. Finally, let us check that ℎ 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. 𝑓 (2𝑥0 ) = 2𝑓 (𝑥0 ) .. Next, let us define, for a fixed 𝑧 ∈ 𝑋, the following map: 𝑓𝑧 (𝑥) := 𝑓 (𝑥 + 𝑧) − 𝑓 (𝑧) ,. 𝑡 𝑁 (𝑦 , 𝑡) = 𝑁 (2𝑓 (𝑥0 ) − 𝑦 , 𝑡) ≥ 𝑁 (𝑓 (𝑥0 ) , ) . (15) 2. Hence,. (20). (21). for all 𝑥 ∈ 𝑋. It is clear that 𝑓𝑧 (0) = 0 and 𝑓𝑧 is surjective since 𝑓 is assumed to be also surjective. Furthermore, for all 𝑥, 𝑦 ∈ 𝑋 and 𝑡 > 0, 𝑁 (𝑓𝑧 (𝑥) − 𝑓𝑧 (𝑦) , 𝑡). . . . = 𝑁 (𝑓 (𝑥 + 𝑧) − 𝑓 (𝑦 + 𝑧) , 𝑡) = 𝜌𝑡 (𝑁 (𝑥 − 𝑦, 𝑡)) , (22). . 𝑁 (ℎ (𝑦 ) , 𝑡) = 𝑁 (𝑓 (𝑥1 − 𝑥 ) − 𝑓 (0) , 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. = 𝜌𝑡 (𝑁 (𝑥1 − 𝑥 − 0, 𝑡)) = 𝑁 (𝑓 (𝑥1 ) − 𝑓 (𝑥 ) , 𝑡) = 𝑁 (2𝑓 (𝑥0 ) − 𝑦 , 𝑡). (16). 𝑓 ((𝑥 − 𝑦) + 𝑦) − 𝑓 (𝑦). = 𝑁 (𝑦 , 𝑡). = 𝑓𝑦 (𝑥 − 𝑦) = 2𝑓𝑦 (. 𝑡 ≥ 𝑁 (𝑓 (𝑥0 ) , ) . 2. = 2 (𝑓 (. 𝑥−𝑦 ) 2. 𝑥−𝑦 + 𝑦) − 𝑓 (𝑦)) , 2. (23).

(4) 4. Abstract and Applied Analysis. which yields 𝑓 (𝑥) + 𝑓 (𝑦) = 2𝑓 (. 𝑥+𝑦 ) = 𝑓 (𝑥 + 𝑦) ; 2. (24). that is, 𝑓 is additive. Let us recall here that additivity yields Q-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 Theorem 6 to 𝑔(𝑥) := 𝑓(𝑥) − 𝑓(0), 𝑥 ∈ 𝑋.. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.. Acknowledgments This research is supported by the Spanish Ministry of Education and Science (Grant no. MTM2011-23118) and by Bancaixa (Project P1-1B2011-30).. References [1] A. K. Katsaras, “Fuzzy topological vector spaces. II,” Fuzzy Sets and Systems, vol. 12, no. 2, pp. 143–154, 1984. [2] C. Felbin, “Finite-dimensional fuzzy normed linear space,” Fuzzy Sets and Systems, vol. 48, no. 2, pp. 239–248, 1992. [3] O. Kaleva and S. Seikkala, “On fuzzy metric spaces,” Fuzzy Sets and Systems, vol. 12, no. 3, pp. 215–229, 1984. [4] S. C. Cheng and J. N. Mordeson, “Fuzzy linear operators and fuzzy normed linear spaces,” Bulletin of the Calcutta Mathematical Society, vol. 86, no. 5, pp. 429–436, 1994. [5] I. Kramosil and J. Michálek, “Fuzzy metrics and statistical metric spaces,” Kybernetika, vol. 11, no. 5, pp. 326–334, 1975. [6] T. Bag and S. K. Samanta, “Finite dimensional fuzzy normed linear spaces,” Journal of Fuzzy Mathematics, vol. 11, no. 3, pp. 687–705, 2003. [7] T. Bag and S. K. Samanta, “Fuzzy bounded linear operators,” Fuzzy Sets and Systems, vol. 151, no. 3, pp. 513–547, 2005. [8] S. Mazur and S. Ulam, “Sur les transformations isometriques d’espaces vectoriels normes,” Comptes Rendus de l’Académie des Sciences, vol. 194, pp. 946–948, 1932. [9] J. Väisälä, “A proof of the Mazur-Ulam theorem,” The American Mathematical Monthly, vol. 110, no. 7, pp. 633–635, 2003. [10] J. A. Baker, “Isometries in normed spaces,” The American Mathematical Monthly, vol. 78, pp. 655–658, 1971. [11] A. Vogt, “Maps which preserve equality of distance,” Studia Mathematica, vol. 45, pp. 43–48, 1973. [12] R. Hu, “On the maps preserving the equality of distance,” Journal of Mathematical Analysis and Applications, vol. 343, no. 2, pp. 1161–1165, 2008..

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