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On the Mazur–Ulam problem in linear 2-normed spaces
J. Math. Anal. Appl. 327 (2007) 1041–1045 www.elsevier.com/locate/jmaa
On the Mazur–Ulam problem in linear 2-normed spaces Hahng-Yun Chu 1 Department of Mathematical Science, Seoul National University, Seoul 151-747, South Korea Received 21 January 2005 Available online 5 June 2006 Submitted by Steven G. Krantz
Abstract We study the notion of 2-isometry which is suitable to represent the concept of area preserving mapping in linear 2-normed spaces, and then prove the Mazur–Ulam problem in linear 2-normed spaces. © 2006 Elsevier Inc. All rights reserved. Keywords: Mazur–Ulam theorem; Linear 2-normed space; 2-Isometry
1. Introduction The main theme of this article is the relation between isometry and linearity, that is, the Mazur–Ulam theorem and its generalizations. The theory of isometry had its beginning in the important paper by Mazur and Ulam [6] in 1932. Let X and Y be 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; i.e., for all x, y in 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 . The basic problem of conservative distances is whether the existence of a single conservative distance for some f implies that f is an isometry of X into Y . It is called the Aleksandrov problem. E-mail address: [email protected]. 1 Present address: School of Mathematics, Korea Institute for Advanced Study, Seoul 130-722, Republic of Korea.
0022-247X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2006.04.053
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The Aleksandrov problem has been extensively studied by many authors [1,4,5,7–9]. Recently, Chu et al. [4] defined the concept of 2-isometry which is suitable to represent the notion of area preserving mappings in linear 2-normed spaces and proved the Rassias and Šemrl’s theorem [9] in linear 2-normed space. In [6], Mazur and Ulam proved the following theorem. Mazur–Ulam theorem. Every isometry of a real normed linear space onto a real normed linear space is a linear mapping up to translation. Also, Baker [2] investigated the Mazur–Ulam problem and obtained the following result. Theorem 1.1. [2] Let X and Y be real normed linear spaces and suppose that Y is strictly convex. If f : X → Y is an isometry, then f is affine. Recently, Väisälä [10] proved that every bijective isometry f : X → Y between normed linear spaces is affine. In this paper, we study the notion of 2-isometry in linear 2-normed spaces, and then prove that the Mazur–Ulam theorem holds when X and Y are linear 2-normed spaces, that is, the 2-isometry mapped to a linear 2-normed space is affine. From now on, unless otherwise specifications, let X and Y be linear 2-normed spaces, f : X → Y a mapping, and X and Y have dimensions greater than 1. 2. Definitions and lemmas Definition 2.1. [3] Let X be a real linear space with dim X > 1 and · , · : X 2 → R a function. Then (X, · , · ) is called a linear 2-normed space if (2N1 ) (2N2 ) (2N3 ) (2N4 )
x, y = 0 ⇔ x, y are linearly dependent; x, y = y, x; αx, y = |α|x, y; x, y + z x, y + x, z for all α ∈ R and x, y, z ∈ X.
Lemma 2.2. [3] For all γ ∈ R and all a, b ∈ X, a, b = a, b + γ a. As an immediate consequence of Lemma 2.2, we have the following: Remark 2.3. For all x, y, z ∈ X, x − z, x − y = x − z, y − z. For x, y, z ∈ X, we call x − z, y − z area of (the convex hull generated by) x, y and z. For f , consider the following condition which is called the area one preserving property (AOPP): (AOPP)
Let x, y, z ∈ X with x − z, y − z = 1, then f (x) − f (z), f (y) − f (z) = 1.
We call f a 2-isometry if x − z, y − z = f (x) − f (z), f (y) − f (z) for all x, y, z ∈ X. Definition 2.4. Let X be a real linear space and x, y, z mutually disjoint elements of X. Then x, y and z are said to be collinear if y − z = t (x − z) for some real number t.
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Definition 2.5. [4] We call f a 2-Lipschitz mapping if there is a κ 0 such that f (x) − f (z), f (y) − f (z) κx − z, y − z for all x, y, z ∈ X. In this case, the constant κ is called the 2-Lipschitz constant. Theorem 2.6. [4] Let f be a 2-Lipschitz mapping with the 2-Lipschitz constant κ 1. Assume that if x, y and z are collinear, then f (x), f (y) and f (z) are collinear, and that f satisfies (AOPP). Then f is a 2-isometry. Lemma 2.7. Let a, b be elements of X. Then u =
a+b 2
is the unique element of X satisfying
1 a − c, a − u = b − u, b − c = a − c, b − c 2 for some c ∈ X with a − c, b − c = 0 and u, a, b are collinear. Proof. Let u =
a+b 2 .
Then u, a, b are collinear. By Lemma 2.2, u satisfies
1 a − c, a − u = b − u, b − c = a − c, b − c 2 for all c ∈ X with a − c, b − c = 0. To show the uniqueness, assume that v is an element of X satisfying 1 a − c, a − v = b − v, b − c = a − c, b − c 2 for some c ∈ X with a − c, b − c = 0 and v, a, b are collinear. Since v, a, b are collinear, there exists a real number t such that v = ta + (1 − t)b. In view of Lemma 2.2, we obtain 1 a − c, b − c = a − c, a − v 2 = a − c, a − ta + (1 − t)b = |1 − t|a − c, a − b = |1 − t|a − c, b − c and 1 a − c, b − c = b − v, b − c 2 = b − ta + (1 − t)b , b − c = −ta + tb, b − c = |t|a − b, b − c = |t|a − c, b − c. Since a − c, b − c = 0, we have 1 = |1 − t| = |t|. 2 Thus we obtain t = 12 and hence u = v. This completes the proof.
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3. Mazur–Ulam problem in linear 2-normed spaces Theorem 3.1. Assume that f (x), f (y) and f (z) are collinear when x, y and z are collinear. If f is a 2-isometry, then f is affine. Proof. Let g(x) = f (x) − f (0). Then g is a 2-isometry and g(0) = 0. Also it is true that if x, y and z are collinear, then g(x), g(y) and g(z) are collinear. Thus we may assume that f (0) = 0. Hence it suffices to show that f is linear. Let x, y ∈ X with x = y. Since dim X > 1, there exists an element z of X such that x − z, y − z = 0. Since f is a 2-isometry, we have f (x) − f (z), f (x) − f x + y 2 x +y = x − z, x − 2 x −y = x − z, 2 1 = x − z, x − y 2 1 = x − z, y − z 2 1 = f (x) − f (z), f (y) − f (z) 2 by Remark 2.3. Similarly, we can obtain f (y) − f x + y , f (y) − f (z) = 1 f (x) − f (z), f (y) − f (z). 2 2 x+y Since x+y 2 , x and y are collinear, f ( 2 ), f (x) and f (y) are also collinear. From Lemma 2.7, we have x +y f (x) + f (y) f = 2 2
for all x, y ∈ X. Since f (0) = 0, we can easily show that f is additive. It follows that f is Q-linear. Let r ∈ R+ with r = 1 and x ∈ X. Since 0, x and rx are collinear, f (0), f (x) and f (rx) are also collinear. Since f (0) = 0, there exists a real number k such that f (rx) = kf (x). Since dim X > 1, there exists an element y of X such that x, y = 0. Thus we have rx, y = rx, y = rx − 0, y − 0 = f (rx) − f (0), f (y) − f (0) = f (rx), f (y) = kf (x), f (y) = |k|f (x), f (y) = k f (x) − f (0), f (y) − f (0) = |k|x − 0, y − 0 = |k|x, y.
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Since x, y = 0, k = ±r. Assume that k = −r, that is, f (rx) = −rf (x). Since r is a positive real number, there are positive rational numbers q1 , q2 such that q1 < r < q2 . Since dim X > 1, there exists an element z of X such that rx − q2 x, z − q2 x = 0. Then we obtain that (r + q2 )f (x), f (z) − f (q2 x) = rf (x) + q2 f (x), f (z) − f (q2 x) = −f (rx) + f (q2 x), f (z) − f (q2 x) = f (rx) − f (q2 x), f (z) − f (q2 x) = rx − q2 x, z − q2 x = |r − q2 |x, z − q2 x = (q2 − r)x, z − q2 x (q2 − q1 )x, z − q2 x = q1 x − q2 x, z − q2 x = f (q1 x) − f (q2 x), f (z) − f (q2 x) = q1 f (x) − q2 f (x), f (z) − f (q2 x) = |q1 − q2 |f (x), f (z) − f (q2 x) = (q2 − q1 )f (x), f (z) − f (q2 x). Since rx −q2 x, z −q2 x = 0, f (rx)−f (q2 x), f (z)−f (q2 x) = 0. So (r +q2 )f (x), f (z)− f (q2 x) = 0 and then f (x), f (z) − f (q2 x) = 0. Thus we have r + q2 q2 − q1 , which is a contradiction. Therefore k = r, that is, f (rx) = rf (x) for all positive real number r. Hence for every real number r, f (rx) = rf (x). This completes the proof. 2 A direct application of Theorems 2.6 and 3.1 yields the following corollary. Corollary 3.2. Let f be a 2-Lipschitz mapping with the 2-Lipschitz constant κ 1. Assume that f (x), f (y) and f (z) are collinear when x, y and z are collinear. If f satisfies (AOPP), then f is an affine 2-isometry. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
A.D. Aleksandrov, Mappings of families of sets, Soviet Math. Dokl. 11 (1970) 116–120. J.A. Baker, Isometries in normed spaces, Amer. Math. Monthly (1971) 655–658. Y.J. Cho, P.C.S. Lin, S.S. Kim, A. Misiak, Theory of 2-Inner Product Spaces, Nova Science, New York, 2001. H.Y. Chu, C.G. Park, W.G. Park, The Aleksandrov problem in linear 2-normed spaces, J. Math. Anal. Appl. 289 (2004) 666–672. H.Y. Chu, K.H. Lee, C.G. Park, On the Aleksandrov problem in linear n-normed spaces, Nonlinear Anal. 59 (2004) 1001–1011. S. Mazur, S. Ulam, Sur les transformationes isométriques d’espaces vectoriels normés, C. R. Acad. Sci. Paris 194 (1932) 946–948. Th.M. Rassias, Properties of isometric mappings, J. Math. Anal. Appl. 235 (1999) 108–121. Th.M. Rassias, On the A.D. Aleksandrov problem of conservative distances and the Mazur–Ulam theorem, Nonlinear Anal. 47 (2001) 2597–2608. Th.M. Rassias, P. Šemrl, On the Mazur–Ulam problem and the Aleksandrov problem for unit distance preserving mappings, Proc. Amer. Math. Soc. 118 (1993) 919–925. J. Väisälä, A proof of the Mazur–Ulam theorem, Amer. Math. Monthly 110 (2003) 633–635.
On the Mazur–Ulam problem in linear 2-normed spaces Hahng-Yun Chu 1 Department of Mathematical Science, Seoul National University, Seoul 151-747, South Korea Received 21 January 2005 Available online 5 June 2006 Submitted by Steven G. Krantz
Abstract We study the notion of 2-isometry which is suitable to represent the concept of area preserving mapping in linear 2-normed spaces, and then prove the Mazur–Ulam problem in linear 2-normed spaces. © 2006 Elsevier Inc. All rights reserved. Keywords: Mazur–Ulam theorem; Linear 2-normed space; 2-Isometry
1. Introduction The main theme of this article is the relation between isometry and linearity, that is, the Mazur–Ulam theorem and its generalizations. The theory of isometry had its beginning in the important paper by Mazur and Ulam [6] in 1932. Let X and Y be 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; i.e., for all x, y in 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 . The basic problem of conservative distances is whether the existence of a single conservative distance for some f implies that f is an isometry of X into Y . It is called the Aleksandrov problem. E-mail address: [email protected]. 1 Present address: School of Mathematics, Korea Institute for Advanced Study, Seoul 130-722, Republic of Korea.
0022-247X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2006.04.053
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H.-Y. Chu / J. Math. Anal. Appl. 327 (2007) 1041–1045
The Aleksandrov problem has been extensively studied by many authors [1,4,5,7–9]. Recently, Chu et al. [4] defined the concept of 2-isometry which is suitable to represent the notion of area preserving mappings in linear 2-normed spaces and proved the Rassias and Šemrl’s theorem [9] in linear 2-normed space. In [6], Mazur and Ulam proved the following theorem. Mazur–Ulam theorem. Every isometry of a real normed linear space onto a real normed linear space is a linear mapping up to translation. Also, Baker [2] investigated the Mazur–Ulam problem and obtained the following result. Theorem 1.1. [2] Let X and Y be real normed linear spaces and suppose that Y is strictly convex. If f : X → Y is an isometry, then f is affine. Recently, Väisälä [10] proved that every bijective isometry f : X → Y between normed linear spaces is affine. In this paper, we study the notion of 2-isometry in linear 2-normed spaces, and then prove that the Mazur–Ulam theorem holds when X and Y are linear 2-normed spaces, that is, the 2-isometry mapped to a linear 2-normed space is affine. From now on, unless otherwise specifications, let X and Y be linear 2-normed spaces, f : X → Y a mapping, and X and Y have dimensions greater than 1. 2. Definitions and lemmas Definition 2.1. [3] Let X be a real linear space with dim X > 1 and · , · : X 2 → R a function. Then (X, · , · ) is called a linear 2-normed space if (2N1 ) (2N2 ) (2N3 ) (2N4 )
x, y = 0 ⇔ x, y are linearly dependent; x, y = y, x; αx, y = |α|x, y; x, y + z x, y + x, z for all α ∈ R and x, y, z ∈ X.
Lemma 2.2. [3] For all γ ∈ R and all a, b ∈ X, a, b = a, b + γ a. As an immediate consequence of Lemma 2.2, we have the following: Remark 2.3. For all x, y, z ∈ X, x − z, x − y = x − z, y − z. For x, y, z ∈ X, we call x − z, y − z area of (the convex hull generated by) x, y and z. For f , consider the following condition which is called the area one preserving property (AOPP): (AOPP)
Let x, y, z ∈ X with x − z, y − z = 1, then f (x) − f (z), f (y) − f (z) = 1.
We call f a 2-isometry if x − z, y − z = f (x) − f (z), f (y) − f (z) for all x, y, z ∈ X. Definition 2.4. Let X be a real linear space and x, y, z mutually disjoint elements of X. Then x, y and z are said to be collinear if y − z = t (x − z) for some real number t.
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Definition 2.5. [4] We call f a 2-Lipschitz mapping if there is a κ 0 such that f (x) − f (z), f (y) − f (z) κx − z, y − z for all x, y, z ∈ X. In this case, the constant κ is called the 2-Lipschitz constant. Theorem 2.6. [4] Let f be a 2-Lipschitz mapping with the 2-Lipschitz constant κ 1. Assume that if x, y and z are collinear, then f (x), f (y) and f (z) are collinear, and that f satisfies (AOPP). Then f is a 2-isometry. Lemma 2.7. Let a, b be elements of X. Then u =
a+b 2
is the unique element of X satisfying
1 a − c, a − u = b − u, b − c = a − c, b − c 2 for some c ∈ X with a − c, b − c = 0 and u, a, b are collinear. Proof. Let u =
a+b 2 .
Then u, a, b are collinear. By Lemma 2.2, u satisfies
1 a − c, a − u = b − u, b − c = a − c, b − c 2 for all c ∈ X with a − c, b − c = 0. To show the uniqueness, assume that v is an element of X satisfying 1 a − c, a − v = b − v, b − c = a − c, b − c 2 for some c ∈ X with a − c, b − c = 0 and v, a, b are collinear. Since v, a, b are collinear, there exists a real number t such that v = ta + (1 − t)b. In view of Lemma 2.2, we obtain 1 a − c, b − c = a − c, a − v 2 = a − c, a − ta + (1 − t)b = |1 − t|a − c, a − b = |1 − t|a − c, b − c and 1 a − c, b − c = b − v, b − c 2 = b − ta + (1 − t)b , b − c = −ta + tb, b − c = |t|a − b, b − c = |t|a − c, b − c. Since a − c, b − c = 0, we have 1 = |1 − t| = |t|. 2 Thus we obtain t = 12 and hence u = v. This completes the proof.
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3. Mazur–Ulam problem in linear 2-normed spaces Theorem 3.1. Assume that f (x), f (y) and f (z) are collinear when x, y and z are collinear. If f is a 2-isometry, then f is affine. Proof. Let g(x) = f (x) − f (0). Then g is a 2-isometry and g(0) = 0. Also it is true that if x, y and z are collinear, then g(x), g(y) and g(z) are collinear. Thus we may assume that f (0) = 0. Hence it suffices to show that f is linear. Let x, y ∈ X with x = y. Since dim X > 1, there exists an element z of X such that x − z, y − z = 0. Since f is a 2-isometry, we have f (x) − f (z), f (x) − f x + y 2 x +y = x − z, x − 2 x −y = x − z, 2 1 = x − z, x − y 2 1 = x − z, y − z 2 1 = f (x) − f (z), f (y) − f (z) 2 by Remark 2.3. Similarly, we can obtain f (y) − f x + y , f (y) − f (z) = 1 f (x) − f (z), f (y) − f (z). 2 2 x+y Since x+y 2 , x and y are collinear, f ( 2 ), f (x) and f (y) are also collinear. From Lemma 2.7, we have x +y f (x) + f (y) f = 2 2
for all x, y ∈ X. Since f (0) = 0, we can easily show that f is additive. It follows that f is Q-linear. Let r ∈ R+ with r = 1 and x ∈ X. Since 0, x and rx are collinear, f (0), f (x) and f (rx) are also collinear. Since f (0) = 0, there exists a real number k such that f (rx) = kf (x). Since dim X > 1, there exists an element y of X such that x, y = 0. Thus we have rx, y = rx, y = rx − 0, y − 0 = f (rx) − f (0), f (y) − f (0) = f (rx), f (y) = kf (x), f (y) = |k|f (x), f (y) = k f (x) − f (0), f (y) − f (0) = |k|x − 0, y − 0 = |k|x, y.
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Since x, y = 0, k = ±r. Assume that k = −r, that is, f (rx) = −rf (x). Since r is a positive real number, there are positive rational numbers q1 , q2 such that q1 < r < q2 . Since dim X > 1, there exists an element z of X such that rx − q2 x, z − q2 x = 0. Then we obtain that (r + q2 )f (x), f (z) − f (q2 x) = rf (x) + q2 f (x), f (z) − f (q2 x) = −f (rx) + f (q2 x), f (z) − f (q2 x) = f (rx) − f (q2 x), f (z) − f (q2 x) = rx − q2 x, z − q2 x = |r − q2 |x, z − q2 x = (q2 − r)x, z − q2 x (q2 − q1 )x, z − q2 x = q1 x − q2 x, z − q2 x = f (q1 x) − f (q2 x), f (z) − f (q2 x) = q1 f (x) − q2 f (x), f (z) − f (q2 x) = |q1 − q2 |f (x), f (z) − f (q2 x) = (q2 − q1 )f (x), f (z) − f (q2 x). Since rx −q2 x, z −q2 x = 0, f (rx)−f (q2 x), f (z)−f (q2 x) = 0. So (r +q2 )f (x), f (z)− f (q2 x) = 0 and then f (x), f (z) − f (q2 x) = 0. Thus we have r + q2 q2 − q1 , which is a contradiction. Therefore k = r, that is, f (rx) = rf (x) for all positive real number r. Hence for every real number r, f (rx) = rf (x). This completes the proof. 2 A direct application of Theorems 2.6 and 3.1 yields the following corollary. Corollary 3.2. Let f be a 2-Lipschitz mapping with the 2-Lipschitz constant κ 1. Assume that f (x), f (y) and f (z) are collinear when x, y and z are collinear. If f satisfies (AOPP), then f is an affine 2-isometry. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
A.D. Aleksandrov, Mappings of families of sets, Soviet Math. Dokl. 11 (1970) 116–120. J.A. Baker, Isometries in normed spaces, Amer. Math. Monthly (1971) 655–658. Y.J. Cho, P.C.S. Lin, S.S. Kim, A. Misiak, Theory of 2-Inner Product Spaces, Nova Science, New York, 2001. H.Y. Chu, C.G. Park, W.G. Park, The Aleksandrov problem in linear 2-normed spaces, J. Math. Anal. Appl. 289 (2004) 666–672. H.Y. Chu, K.H. Lee, C.G. Park, On the Aleksandrov problem in linear n-normed spaces, Nonlinear Anal. 59 (2004) 1001–1011. S. Mazur, S. Ulam, Sur les transformationes isométriques d’espaces vectoriels normés, C. R. Acad. Sci. Paris 194 (1932) 946–948. Th.M. Rassias, Properties of isometric mappings, J. Math. Anal. Appl. 235 (1999) 108–121. Th.M. Rassias, On the A.D. Aleksandrov problem of conservative distances and the Mazur–Ulam theorem, Nonlinear Anal. 47 (2001) 2597–2608. Th.M. Rassias, P. Šemrl, On the Mazur–Ulam problem and the Aleksandrov problem for unit distance preserving mappings, Proc. Amer. Math. Soc. 118 (1993) 919–925. J. Väisälä, A proof of the Mazur–Ulam theorem, Amer. Math. Monthly 110 (2003) 633–635.