Mappings of preserving n -distance one in n -normed spaces
aa r X i v : . [ m a t h . F A ] S e p MAPPINGS OF PRESERVING n -DISTANCE ONE IN n -NORMEDSPACES XUJIAN HUANG AND DONGNI TAN ∗ Abstract.
We give a positive answer to the Aleksandrov problem in n -normed spacesunder the surjectivity assumption. Namely, we show that every surjective mapping pre-serving n -distance one is affine, and thus is an n -isometry. This is the first time to solvethe Aleksandrov problem in n -normed spaces with only surjective assumption even in theusual case n = 2. Finally, when the target space is n -strictly convex, we prove that everymapping preserving two n -distances with an integer ratio is an affine n -isometry. Introduction and Preliminaries
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 metric in the space X and Y , respectively.For some r >
0, suppose that f preserves distance r , i.e., 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 distance for the mapping f .In 1970, Aleksandrov [1] posed the following problem: Problem 1.1.
Under what conditions is a mapping of a metric space X into itself preserv-ing distance one an isometry? It is called the Aleksandrov problem. It has been extensively investigated by manyauthors (see [6, 7, 10, 16–20] and the references therein). This problem still remains openeven in the case where X = R n and Y = R m with 2 < n < m (see [19]).The study of n -normed spaces began early in the second half of the twentieth century(see [8, 9, 14, 15]), and it is also an widely-studied and interesting area even today (see e.g.[4–7]). Chu et al. [7] first generalized the Aleksandrov problem to n -normed spaces. Theirmain result [7, Theorem 2.10] proves that the weak n -distance one preserving mapping isan n -isometries under additional conditions (e.g. n -1-Lipschitz, preserving 2-collinearity).A natural question can be raised as a modified version of the Aleksandrov problem: Whathappens if two (or more) distances are preserved by a mapping between normed spaces? W.Benz [2] (see also [3]) investigated the case when the mapping preserves distances ρ and nρ for some ρ > n >
1. If the target space is strictly convex, they showed
Mathematics Subject Classification.
Primary 46A03; Secondary 51K05 .
Key words and phrases.
Aleksandrov problem, n -strictly convex, n -isometry, n -normed space. ∗ Corresponding author.The authors are supported by the Natural Science Foundation of China (Grant Nos. 11201337, 11201338,11371201, 11301384). The first author was supported by the Tianjin Science & Technology Fund 20111001. in [2] that this mapping is an affine isometry. If the mapping f preserves two distanceswith a non-integer ratio, it is an open problem whether or not f must be an isometry. Formore information we refer to [16–19]. Motivated by these results and also as an applicationof our main results we shall show that the result of W. Benz remains valid in n -normedspaces if the target space is n -strictly convex.In this paper, we show that every mapping between two n -normed spaces preserving afixed nonzero weak n -distance and 2-collinearity for the midpoint of a segment is affine, andthus is an n -isometry. By this we show that every surjective mapping preserving n -distanceone is an affine n -isometry. Finally, if the target space is n -strictly convex, we show thatevery mapping preserves two n -distances with an integer ratio is an affine n -isometry.Throughout this paper, all linear spaces will be assumed real. Let n ≥ X and Y betwo linear n -normed spaces whose dimensions greater than n − n -normed spacesand cite an example of n -normed spaces for the easy understanding of this kind of spaces.An n -norm on a real vector space X (of dimension at least n ) is a mapping k· , · · · , ·k : X n → R which satisfies the following four conditions:(a) k x , · · · , x n k = 0 if and only if x , · · · , x n are linearly dependent;(b) k x , · · · , x n k is invariant under permutation;(c) k αx , · · · , x n k = | α |k x , · · · , x n k for α ∈ R ;(d) k x + x , x , · · · , x n k ≤ k x , x , · · · , x n k + k x , x , · · · , x n k .The pair ( X, k· , · · · , ·k ) is called an n -normed space . Note that in this space, we have k x , x , · · · , x n k = k x + y, x , · · · , x n k for any linear combination y of x , · · · , x n ∈ X . Example 1.2. If X is a normed space with dual X ′ , then as formulated by ¨Gahler (see[9]) we may define an n -norm on X by k x , x , · · · , x n k := sup f j ∈ X ′ , k f j k≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( x ) · · · f ( x n ) ... . . . ... f n ( x ) · · · f n ( x n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = sup f j ∈ X ′ , k f j k≤ det[ f j ( x i )] . Meanwhile, if X is equipped with an inner product h· , ·i , we can define the standard n -normon X by k x , x , · · · , x n k := q det [ h x i , x j i ] , which can be interpreted as the volume of the n -dimensional parallelepiped spanned by x , x , · · · , x n ∈ X (see [11]).Recall some definitions in n -normed spaces. Definition 1.3.
Let X and Y be two n -normed spaces, and let f : X → Y be a mapping. (a) f is said to be an n -isometry if it satisfies k f ( x ) − f ( y ) , · · · , f ( x n ) − f ( y n ) k = k x − y , · · · , x n − y n k for all x , · · · , x n , y , · · · , y n ∈ X . In particular, if y = · · · = y n , f is said to be a weak n -isometry .(b) f is said to have the n -distance one preserving property ( n -DOPP), if k x − y , · · · , x n − y n k = 1 ⇒ k f ( x ) − f ( y ) , · · · , f ( x n ) − f ( y n ) k = 1 . APPINGS OF PRESERVING n -DISTANCE ONE IN n -NORMED SPACES 3 for all x , · · · , x n , y , · · · , y n ∈ X . In particular, if y = · · · = y n , f is said to have the weakn-distance one preserving property ( w - n -DOPP).(c) f is said to preserve ρ - n -distance for some ρ >
0, if k x − y , · · · , x n − y n k = ρ implies k f ( x ) − f ( y ) , · · · , f ( x n ) − f ( y n ) k = ρ for all x , · · · , x n , y , · · · , y n ∈ X . In particular, if y = · · · = y n , f is said to preserve w - ρ - n -distance .(d) f is called an n -Lipschitz mapping if there is a K ≥ k f ( x ) − f ( y ) , · · · , f ( x n ) − f ( y n ) k ≤ K k x − y , · · · , x n − y n k for all x , · · · , x n , y , · · · , y n ∈ X . In this case, the constant K is called the n -Lipschitzconstant. In particular, if y = · · · = y n , f is said to be a weak n -Lipschitz mapping .2. Isometry in n -normed spaces In this section we consider the Aleksandrov problem in n -normed spaces. We first intro-duce a weak case of preserving 2-collinearity. Then, we prove that the Aleksandrov problemholds in n -normed spaces under weaker hypothesis.Note that the points x, y, z of X are said to be if y − z = t ( x − z ) for somereal number t . The points x , x , · · · , x n of X are said to be n -collinear if for some i , thepoints x j − x i , ≤ j = i ≤ n are linearly dependent. Definition 2.1.
Let X and Y be two n -normed spaces, and let f be a mapping from X into Y . (a) f is said to preserve if x, y, z ∈ X are collinear, then f ( x ) , f ( y ) , f ( z )are collinear. In particular, if z = ( x + y ) / f is said to preserve .(b) f is said to preserve n -collinearity if x , x , · · · , x n of X are n -collinear, then f ( x ) , f ( x ) , · · · , f ( x n ) are n -collinear. That means that f preserves w -0-distance, i.e., if k x − x , · · · , x n − x k = 0, then k f ( x ) − f ( x ) , · · · , f ( x n ) − f ( x ) k = 0for all x , x , · · · , x n ∈ X .In the first step, we prove the following lemma indicating that a mapping f from an n -normed space X to an n -normed space Y , which preserves a nonzero weak n -distanceand 2-collinearity for the midpoint of a segment, satisfies Jensen’s equation: f ( x + y f ( x ) + f ( y )2 , ∀ x, y ∈ X. Lemma 2.2.
Let X and Y be two n -normed spaces, and let f : X → Y preserve w - ρ - n -distance for some ρ > . Then f is injective. Moreover if f preserves 2-collinearity for themidpoint of a segment, then f ( x ) − f (0) is additive. Proof:
For x = y ∈ X , the assumption that dim X ≥ n allows the existence of x , x , · · · , x n ∈ X such that k y − x, x − x, · · · , x n − x k = ρ. Since the mapping f preserves w - ρ - n -distance, we have k f ( y ) − f ( x ) , f ( x ) − f ( x ) , · · · , f ( x n ) − f ( x ) k = ρ. X. HUANG AND D. TAN
This implies f ( x ) = f ( y ), and thus f is injective. To see our second conclusion, it sufficesto prove that for all x, y ∈ X , we have f ( x + y f ( x ) + f ( y )2 . (1)To prove (1), set z = ( x + y ) / x, y ∈ X . Choose y , y , · · · , y n ∈ X such that k y − z, y − z, · · · , y n − z k = k x − z, y − z, · · · , y n − z k = ρ. Then clearly k f ( y ) − f ( z ) , f ( y ) − f ( z ) , · · · , f ( y n ) − f ( z ) k = ρ (2) k f ( x ) − f ( z ) , f ( y ) − f ( z ) , · · · , f ( y n ) − f ( z ) k = ρ. (3)Since f preserves 2-collinearity for the midpoint of a segment, there exists a real number t such that f ( y ) − f ( z ) = t ( f ( x ) − f ( z )) . By(2) and (3), we obtain that t = −
1, and hence f ( x + y f ( z ) = f ( x ) + f ( y )2 . ✷ One may wonder how to check that a mapping f from an n -normed space into anotherpreserves 2-collinearity. What interests us is that it only requires f to preserve w - n -DOPP(not necessarily surjective) and be a weak n -Lipschitz mapping or preserve n -collinearity.This has been indicated in [6, Lemma 3.2] which states that every n -isometry f preserves2-collinearity in n -normed spaces. For the convenience of readers and since the conditionis weaker, we here include a proof. Lemma 2.3.
Let X and Y be two n -normed spaces. Suppose that the mapping f : X → Y preserves w - ρ - n -distance for some ρ > . Then the following properties are equivalent: (a) f preserves n -collinearity; (b) f preserves 2-collinearity; (c) f preserves 2-collinearity for the midpoint of a segment. Proof:
For the implication ( a ) ⇒ ( b ) assume that, on the contrary, there are x , x , x ∈ X which are collinear such that f ( x ) − f ( x ) , f ( x ) − f ( x ) are linearly independent. Notethat x = x and f preserves w - ρ - n -distance. We can choose y , · · · , y n ∈ X such that k f ( x ) − f ( x ) , f ( y ) − f ( x ) , · · · , f ( y n ) − f ( x ) k = k x − x , y − x , · · · , y n − x k = ρ. Then the set A := { f ( x ) − f ( x ) : x ∈ X } contains n linearly independent vectors. Hencethere exist x , · · · , x n ∈ X such that k f ( x ) − f ( x ) , f ( x ) − f ( x ) , f ( x ) − f ( x ) , · · · , f ( x n ) − f ( x ) k 6 = 0 . Assume that f preserves n -collinearity. Then k x − x , x − x , · · · , x n − x k = 0 impliesthat k f ( x ) − f ( x ) , f ( x ) − f ( x ) , · · · , f ( x n ) − f ( x ) k = 0 . which is a contradiction. Thus f preserves 2-collinearity. APPINGS OF PRESERVING n -DISTANCE ONE IN n -NORMED SPACES 5 The implication ( b ) ⇒ ( c ) is clear.For the implication ( c ) ⇒ ( a ) without loss of generality we can assume that ρ = 1. Then f satisfies w - n -DOPP. Let g ( x ) = f ( x ) − f (0) for every x ∈ X . We first prove that g preserves distance m/k for all m, k ∈ N . Let x , x , · · · , x n be in X and m, k be in N suchthat k x , x , · · · , x n k = m/k. We see from Lemma 2.2 that g is Q -linear, and since g (0) = 0 and satisfies w - n -DOPP, wehave k g ( x ) , g ( x ) , · · · , g ( x n ) k = mk k g ( km x ) , g ( x ) , · · · , g ( x n ) k = mk . To see that g preserves n -collinearity, we only need to check that for all x , x , · · · x n ∈ X which are not all zero with k x , x , · · · , x n k = 0, k g ( x ) , g ( x ) , · · · , g ( x n ) k = 0 . Since k x , x , · · · , x n k = 0, we know that x , x , · · · , x n ∈ X are linearly dependent. Tosimplify the notation, the maximal linearly independent members of x , x , · · · , x n are stilldenoted by x , · · · , x k where 1 ≤ k < n . Choose y k +1 , · · · , y n ∈ X such that k x , · · · , x k , y k +1 , · · · , y n k = 1 . Then for every positive integer m , k x , · · · , x k , x k +1 + 1 m y k +1 , · · · , x n + 1 m y n k = 1 m n − k and by the above, k g ( x ) , · · · , g ( x k ) , g ( x k +1 ) + 1 m g ( y k +1 ) , · · · , g ( x n ) + 1 m g ( y n ) k = 1 m n − k . Triangle inequality hence gives k g ( x ) , · · · , g ( x k ) , g ( x k +1 ) , · · · , g ( x n ) k≤ k g ( x ) , · · · , g ( x k ) , g ( x k +1 ) + 1 m g ( y k +1 ) , · · · , g ( x n ) + 1 m g ( y n ) k + 1 m A m = 1 m n − k + 1 m A m , where A m = n − k X i =1 k g ( x ) , · · · , g ( x k + i − ) , g ( y k + i ) , g ( x k + i +1 ) , · · · , g ( x n ) k +1 m n − − k X i =1 k g ( x ) , · · · , g ( x k + i − ) , g ( y k + i ) , g ( y k + i +1 ) , g ( x k + i +2 ) , · · · , g ( x n ) k +1 m n − k − k g ( x ) , · · · , g ( x k ) , g ( y k +1 ) , g ( y k +2 ) , · · · , g ( y n ) k . Letting m → + ∞ we get the desired equation k g ( x ) , g ( x ) , · · · , g ( x n ) k = 0 . ✷ X. HUANG AND D. TAN
Since it has been showed that if f preserves a fixed nonzero weak n -distance and 2-collinearity for the midpoint of a segment then f ( x ) − f (0) is additive, it is natural to thinkof such mappings not far from being affine. It is clearly easy to prove f to be an n -isometryif it is affine. However it may not be an immediate result since continuity is not implied bypreserving nonzero weak n -distance. Proposition 2.4.
Let X and Y be two n -normed spaces. If f : X → Y preserves w - ρ - n -distance for some ρ > and preserves 2-collinearity for the midpoint of a segment, then f is an affine n -isometry. Proof:
We first prove that f is affine. For this purpose, we only need to show that themapping g : X → Y defined by g ( x ) = f ( x ) − f (0) is linear. By Lemmas 2.2 and 2.3, themapping g is injective, additive and preserves 2-collinearity. Let x ∈ X with x = 0 and t ∈ R with t = 1. Since 0 , x, tx are collinear, there exists a unique real number s such that g ( tx ) = sg ( x ) . We can define φ : R → R by φ ( t ) = s i.e., g ( tx ) = φ ( t ) g ( x ) , ∀ t ∈ R . Then clearly, the mapping φ is injective, additive with φ (0) = 0 and φ (1) = 1. Moreover φ does not depend on the choice of x under the assumption of linear independence. Indeed,choose y ∈ X such that x and y are linearly independent and let φ : R → R be a mappingsuch that g ( ty ) = φ ( t ) g ( y ) , ∀ t ∈ R . Since 0 , x + y, t ( x + y ) are collinear,0 , g ( x ) + g ( y ) , φ ( t ) g ( x ) + φ ( t ) g ( y )are collinear. Note that if g ( x ) and g ( y ) are linearly independent, then φ ( t ) = φ ( t ), asdesired. In fact, if n >
2, there exist x , · · · , x n ∈ X such that k g ( x ) , g ( y ) , g ( x ) , · · · , g ( x n ) k = k x, y, x , · · · , x n k = ρ. Then g ( x ) and g ( y ) are linearly independent. If n = 2, choose a real number a such that k g ( x ) , g ( ay ) k = k x, ay k = ρ. Then g ( x ) and g ( ay ) are linearly independent, and thus so are g ( x ) and g ( y ). We will provethat φ is an endomorphism. For any t, s ∈ R , 0 , x + sy, tx + tsy are collinear, and then0 , g ( x ) + φ ( s ) g ( y ) , φ ( t ) g ( x ) + φ ( ts ) g ( y )are collinear. It follows that φ ( st ) = φ ( s ) φ ( t ) for any t, s ∈ R . It is well-known thatthe every nonzero endomorphism of R is the identity. Then for any x ∈ X and t ∈ R , g ( tx ) = tg ( x ). Thus g is linear. It is easy to see that g is an n -isometry, and hence so is f .The proof is complete. ✷ Remark 2.5.
Proposition 2.4 has been shown in [13, Lemma 3.4]. Unfortunately theproof given in [13, Lemma 3.4] contains a mistake. The statement “lim k →∞ k g ( rx ) − g ( r k x ) , g ( x k ) , g ( x k ) , · · · , g ( x kn ) k = 0 (pp 978, line 11 of [13])” could not be obtained from thediscussing proof in [13]. For a counterexample, consider g to be the identity, i.e., g ( x ) = x for every x ∈ X . We may assume that r is an irrational number since the rational case issettled. For each k , choose x k , · · · , x kn such that k x, x k , x k , · · · , x kn k = (2+ [ | r − r k | ]) / | r − r k | . APPINGS OF PRESERVING n -DISTANCE ONE IN n -NORMED SPACES 7 Then clearly k x, x k , x k , · · · , x kn k > | r − r k | · k x, x k , x k , · · · , x kn k = 2 + [ | r − r k | ] is arational number as required in [13]. However, k g ( rx ) − g ( r k x ) , g ( x k ) , g ( x k ) , · · · , g ( x kn ) k = | r − r k | · k x, x k , x k , · · · , x kn k > k . Therefore the limit cannot be 0 as k goes toinfinity. The remaining results Lemma 3.5, Theorem 3.6, Corollary 3.7 and Corollary 3.8in [13] following from the the main lemma 3.4 need a new proof. For this and our mainresult (Theorem 2.6), we hence include a different proof in this paper.We are now ready to prove our main result that gives a positive answer to the Aleksandrovproblem in n -normed spaces. For a real vector space X , we denote the line joining twodifferent points x, y ∈ X by xy and affine( M ) by the affine subspace generated by M ⊂ X ,respectively. Theorem 2.6.
Let X and Y be two n -normed spaces. If a surjective mapping f : X → Y has n -DOPP, then f is an affine n -isometry. Proof:
In the following proof, without loss of generality we can assume that f (0) = 0. Wefirst prove that f − preserves 2-collinearity. This is equivalent to showing that if x, y, z ∈ X are not collinear then f ( x ) , f ( y ) , f ( z ) are not collinear. Indeed, choose x , · · · , x n ∈ X suchthat r = k x − z, y − z, x − z, · · · , x n − z k 6 = 0. Set u := z + ( x − z ) + ( y − z ) r . It is easy to check that k x − z, u − z, x − z, · · · , x n − z k = k y − z, u − z, x − z, · · · , x n − z k = 1 . Since f has n -DOPP, k f ( x ) − f ( z ) , f ( u ) − f ( z ) , f ( x ) − f ( z ) , · · · , f ( x n ) − f ( z ) k = 1 (4) k f ( y ) − f ( z ) , f ( u ) − f ( z ) , f ( x ) − f ( z ) , · · · , f ( x n ) − f ( z ) k = 1 . (5)If there exists some t ∈ R such that f ( x ) − f ( z ) = t ( f ( y ) − f ( z )). By (4),(5) and since f is injective, we obtain that t = − f ( z ) = ( f ( x ) + f ( y )) /
2. Similarly, f ( x ) =( f ( z ) + f ( y )) /
2. It follows that f ( x ) = f ( y ) = f ( z ), which is impossible.To see our conclusion, we shall show that f preserves 2-collinearity for the midpoint ofa segment. If this does not hold, then there exist x = y ∈ X with z = ( x + y ) / f ( x ) , f ( y ) , f ( z ) are not collinear. Now let w ∈ X such that f ( w ) = f ( x ) + f ( y )2 . Since f − preserves 2-collinearity, there exists a scalar t such that y − w = t ( x − w ). Wecan choose x , · · · , x n ∈ X satisfying k y − w, x , · · · , x n k = 1 and 0 x intersects xy only inone point denoted by x . We claim that the f -image f (0 x ) belongs to a line 0 f ( x ) in Y .Otherwise, there are u, v ⊂ x such that f ( u ) , f ( v ) , f ( x ) are not collinear. Set E := affine( f ( u ) , f ( x ) , f ( v )) and F := affine( f ( x ) , f ( x ) , f ( y ) , f ( z )) . Since f − preserves 2-collinearity, we have f − ( E ) ⊂ x and f − ( F ) ⊂ xy . Observe that f ( x ) ∈ E ∩ F . Then E ∩ F contains infinity points. However, f − ( E ∩ F ) ⊂ f − ( E ) ∩ f − ( F ) ⊂ x ∩ xy = { x } . X. HUANG AND D. TAN
A contradiction since f is injective. By the claim, there are scalars s , s such that f ( tx ) = s f ( x ) and f ( − tx ) = s f ( x ). Since f has n -DOPP, we have k f ( y ) − f ( w ) , f ( x ) , · · · , f ( x n ) k = 12 k f ( x ) − f ( y ) , f ( x ) , · · · , f ( x n ) k = 1 , k f ( x ) − f ( w ) , f ( tx ) , · · · , f ( x n ) k = 12 k f ( x ) − f ( y ) , s f ( x ) , · · · , f ( x n ) k = 1 , k f ( x ) − f ( w ) , f ( − tx ) , · · · , f ( x n ) k = 12 k f ( x ) − f ( y ) , s f ( x ) , · · · , f ( x n ) k = 1 . It follows that | s i | = 1. Since f is injective, the only possibility is that s = − s = 1.Thus t = −
1. Therefore w = ( x + y ) /
2. A contradiction guarantees that f preserves2-collinearity for the midpoint of a segment. Proposition 2.4 thus completes the proof. ✷ Next, we shall show that the result of W. Benz holds in n -strictly convex spaces. Definition 2.7. An n -normed space X is said to be n -strictly convex space if for any x , x , · · · , x n ∈ X , x , · · · , x n / ∈ span { x , x } and k x + x , x , · · · , x n k = k x , x , · · · , x n k + k x , x , · · · , x n k > imply x = tx for some t ≥ . Theorem 2.8.
Let X and Y be two n -normed spaces, and let Y be n -strictly convex. If f : X → Y preserves two n -distances ρ and N ρ for some ρ > and some integer N > ,then f is an affine n -isometry. Proof:
It follows from Proposition 2.4 that we need only prove that f preserves 2-collinearity for the midpoint of a segment.(a) We first prove that f preserves 2 ρ - n -distance. Assume that N > f preserves n -distances ρ and N ρ . Let x , x · · · , x n , y , y · · · , y n be in X such that k x − y , x − y , · · · , x n − y n k = 2 ρ, and set ω i = y + i ( x − y , ∀ i ∈ N ∪ { } . Then ω = y , ω = x and ω i − ω i − = x − y , ∀ i ∈ N . It follows that k ω i − ω i − , x − y , · · · , x n − y n k = ρ, ∀ i ∈ N . and k ω N − y , x − y , · · · , x n − y n k = N ρ . Since f preserves ρ - n -distance, by the triangleinequality, we have k f ( x ) − f ( y ) , · · · , f ( x n ) − f ( y n ) k≤ k f ( ω ) − f ( ω ) , · · · , f ( x n ) − f ( y n ) k + k f ( ω ) − f ( ω ) , · · · , f ( x n ) − f ( y n ) k = 2 ρ and similarly, k f ( ω N ) − f ( x ) , · · · , f ( x n ) − f ( y n ) k ≤ ( N − ρ. APPINGS OF PRESERVING n -DISTANCE ONE IN n -NORMED SPACES 9 Therefore,
N ρ = k f ( ω N ) − f ( y ) , · · · , f ( x n ) − f ( y n ) k≤ k f ( ω N ) − f ( x ) , · · · , f ( x n ) − f ( y n ) k + k f ( x ) − f ( y ) , · · · , f ( x n ) − f ( y n ) k≤ ( N − ρ + 2 ρ = N ρ.
This implies that k f ( x ) − f ( y ) , · · · , f ( x n ) − f ( y n ) k = 2 ρ. (b) Let z = ( x + y ) / x, y ∈ X . Let g ( x ) = f ( x ) − f (0). Then g preservestwo n -distances ρ and 2 ρ . Thus, there is no loss of generality in assuming that f (0) = 0.We shall prove that there exist x , x , · · · , x n ∈ X such that k y − z, x , · · · , x n k = ρ and f ( x i ) span { f ( y ) − f ( z ) , f ( x ) − f ( z ) } for i = 2 , , · · · , n. Choose y , y , · · · , y n ∈ X such that k y − z, y , · · · , y n k = ρ . We define the set C to consistof all elements ν in X such that k y − z, ν, y , · · · , y n k = ρ , that is C := { ν ∈ X : k y − z, ν, y , · · · , y n k = ρ } . We can choose x ∈ C such that f ( x ) span { f ( y ) − f ( z ) , f ( x ) − f ( z ) } . Otherwise, assume that for every ν ∈ C there exist α, β ∈ R such that f ( ν ) = α ( f ( y ) − f ( z )) + β ( f ( x ) − f ( z )) . (6)Note that k y − z, ν, y · · · , y n k = ρ . It follows that k x − z, ν, y , · · · , y n k = ρ. Then k f ( y ) − f ( z ) , f ( ν ) , f ( y ) , · · · , f ( y n ) k = ρ, (7) k f ( x ) − f ( z ) , f ( ν ) , f ( y ) , · · · , f ( y n ) k = ρ, (8)It follows from (6), (7) and (8) that | β |k f ( y ) − f ( z ) , f ( x ) − f ( z ) , · · · , f ( y n ) k = ρ | α |k f ( x ) − f ( z ) , f ( y ) − f ( z ) , · · · , f ( y n ) k = ρ. This yields | α | = | β | . Moreover, | α | is a fixed positive real number. Therefore, there areat most four elements in f ( C ). This is impossible, because the set C contains “enough”elements. This follows from Lemma 2.2 that f is injective and for each r ∈ R , the element ν r := y + r ( y − z ) belongs to C . So there exists x ∈ C such that f ( x ) span { f ( y ) − f ( z ) , f ( x ) − f ( z ) } . Next, set C := { ν ∈ X : k y − z, x , ν, y , · · · , y n k = ρ } . By the same method as above, we can choose x ∈ C such that f ( x ) span { f ( y ) − f ( z ) , f ( x ) − f ( z ) } . This process can be repeated until we obtain the promised x , x , · · · , x n ∈ X such that k y − z, x , · · · , x n k = ρ and f ( x i ) span { f ( y ) − f ( z ) , f ( x ) − f ( z ) } for i = 2 , , · · · , n. (c) We are now ready to show the desired result that f preserves 2-collinearity for thethe midpoint of a segment. Let z = ( x + y ) / x, y ∈ X . Let x , x , · · · , x n bein X such that k y − z, x , · · · , x n k = ρ and f ( x i ) span { f ( y ) − f ( z ) , f ( x ) − f ( z ) } for i = 2 , , · · · , n. Then we deduce from the fact that f preserves n -distances ρ and 2 ρ that k f ( y ) − f ( x ) , f ( x ) , · · · , f ( x n ) k = k f ( y ) − f ( z ) , f ( x ) , · · · , f ( x n ) k + k f ( x ) − f ( z ) , f ( x ) , · · · , f ( x n ) k . Since Y is n -strictly convex, there exists a real number t > f ( y ) − f ( z ) = t ( f ( z ) − f ( x )) . This completes the proof. ✷ Remark 2.9. [12, Theorem 11] tried to generalize Benz’s Theorem on n -normed spaces.However, on the part (d) of the proof of [12, Theorem 11] the statement that f ( p ) − f ( p ) = t ( f ( p ) − f ( p )) for some t cannot follow just from k f ( p ) − f ( p ) , f ( y ) − f ( x ) , · · · , f ( y n ) − f ( x ) k = k f ( p ) − f ( p ) , f ( y ) − f ( x ) , · · · , f ( y n ) − f ( x ) k + k f ( p ) − f ( p ) , f ( y ) − f ( x ) , · · · , f ( y n ) − f ( x ) k = 2 ρ. It remains to check that f ( y i ) − f ( x ) span { f ( p ) − f ( p ) , f ( p ) − f ( p ) } for i = 2 , , · · · , n (It is the demand from the definition of n strictly convexity ([12, definition 3] or Definition2.7 of our paper)). It is a hard and key step which cannot be missed. Acknowledgements.
The authors wish to express their appreciation to Guanggui Dingfor many very helpful comments regarding isometric theory in Banach spaces.
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Department of Mathematics, Tianjin University of Technology, 300384 Tianjin, China
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