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Inequalities for distances between points and distance preserving mappings
Nonlinear Analysis 62 (2005) 675 – 681 www.elsevier.com/locate/na
Inequalities for distances between points and distance preserving mappings Soon-Mo Jung∗ Mathematics Section, College of Science and Technology, Hong-Ik University, 339-701 Chochiwon, Korea Received 19 August 2004; accepted 6 April 2005
Abstract In this paper, we generalize the short diagonals lemma by proving a new inequality for distances between six points. Moreover, we apply this inequality to a partial solution to the Aleksandrov–Rassias problem. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 51K05 Keywords: Inequality; Parallelogram law; Short diagonals lemma; Aleksandrov problem; Aleksandrov–Rassias problem
1. Introduction Throughout this paper, let H be a real (or complex) inner √ product space, ·, · the inner product on H, and · the norm on H defined by x = x, x for all x ∈ H . If dim H 2, then the Pythagorean theorem states that for any x, y, z ∈ H , the equality x − z2 = y − x2 + z − y2 is true if and only if the vectors y − x and z − y are orthogonal to each other. This is one of the most important theorems in mathematics. In connection with this subject, we now ∗ Tel.: +82 41 8602584; fax: +82 42 3693657.
E-mail address: [email protected]. 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.04.003
676
S.-M. Jung / Nonlinear Analysis 62 (2005) 675 – 681
consider the case of four vectors x, y, z, w ∈ H . The parallelogram law states that the equality y − x2 + z − y2 + w − z2 + x − w2 = z − x2 + w − y2 holds true if and only if y − x, z − y, w − z, x − w are the sides of a (possibly degenerate) parallelogram with diagonals z − x and w − y. In Section 2 of this paper, using the short diagonals lemma, we prove an inequality for distances between six points. We devote Section 3 to a study of the Aleksandrov–Rassias problem (see Section 3 for details about the Aleksandrov–Rassias problem). 2. Inequality for distances between six points The parallelogram law can be generalized to an inequality which may be applied for any four points (vectors) in an inner product space. This inequality is known as the short diagonals lemma and proved in Lemma 15.4.2 of [5] for the case of Euclidean spaces. In this section, we will first prove the short diagonals lemma for the case of inner product spaces. Lemma 1. If H is a real (or complex) inner product space, then y − x2 + z − y2 + w − z2 + x − w2 z − x2 + w − y2 for all x, y, z, w ∈ H . Proof. We will prove the lemma when H is a complex inner product space. Let x, y, z, w be arbitrary vectors of H. Then, we get y − x2 + z − y2 + w − z2 + x − w2 − z − x2 − w − y2 = y − x, y − x + z − y, z − y + w − z, w − z + x − w, x − w − z − x, z − x − w − y, w − y = y − x, y − x + z, z − y, z − y, z + y, y + w − z, w − z + x, x − w, x − w, x + w, w − z, z + x, z + x, z − x, x − w, w + y, w + y, w − y, y = y − x, y − x + w − z, w − z + y, w − z + y, w − z + z − w, x + z − w, x = y − x, y − x + w − z, w − z + y − x, w − z + y − x, w − z = y − x + w − z, y − x + w − z 0, as required.
By applying Lemma 1 as well as the parallelogram law, we prove an inequality for distances between every two vectors among the given six vectors. The following theorem may be regarded as a generalization of the short diagonals lemma.
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Theorem 2. If H is a real (or complex) inner product space, then the inequality 2(z − u2 + x − v2 + y − w2 ) v − u2 + z − v2 + x − z2 + u − x2 + w − v2 + x − w2 + y − x2 + v − y2 + w − u2 + z − w2 + y − z2 + u − y2
(1)
is true for any six vectors u, v, w, x, y, z ∈ H . Moreover, the equality sign is true if and only if each {v − u, z − v, x − z, u − x}, {w − u, z − w, y − z, u − y}, {w − v, x − w, y − x, v − y} comprises the sides of an appropriate (possibly degenerate) parallelogram such that z − u and x − v, resp., z − u and y − w, resp., x − v and y − w are the diagonals of the corresponding parallelogram. Proof. If we applyLemma 1 toeachorderedquadruplet, (u, v, z, x), (u, w, z, y), (v, w, x, y) consecutively, then we get z − u2 + x − v2 v − u2 + z − v2 + x − z2 + u − x2 , z − u2 + y − w2 w − u2 + z − w2 + y − z2 + u − y2 , x − v2 + y − w2 w − v2 + x − w2 + y − x2 + v − y2 ,
(2)
respectively. By summing up the three inequalities, we easily obtain the desired inequality (1). We now assume that the equality sign holds true in (1). Lemma 1 implies that each inequality in (2) is true. Hence, it is easy to check that strict inequality in at least one of three inequalities of (2) implies the strict inequality sign in (1), a contradiction. Hence, we conclude that if the equality sign holds in (1) then the equality sign has to be true in each inequality of (2). That is, in view of the parallelogram law, we conclude that each {v − u, z − v, x − z, u − x}, {w − u, z − w, y − z, u − y}, {w − v, x − w, y − x, v − y} comprises the sides of an appropriate (possibly degenerate) parallelogram such that z − u and x −v, resp., z−u and y −w, resp., x −v and y −w are the diagonals of the corresponding parallelogram. Conversely, let us assume that each {v−u, z−v, x −z, u−x}, {w−u, z−w, y −z, u−y}, {w − v, x − w, y − x, v − y} comprises the sides of an appropriate (possibly degenerate) parallelogram such that z − u and x − v, resp., z − u and y − w, resp., x − v and y − w are the diagonals of the corresponding parallelogram. According to the parallelogram law, the equality sign is true in each inequality of (2), and hence the equality sign holds in (1). 3. Aleksandrov–Rassias problem Let X andY be normed spaces.A distance > 0 is said to be contractive (or non-expanding) by f : X → Y if x − y = always implies f (x) − f (y) . Similarly, a distance is said to be extensive (or non-shrinking) by f if the inequality f (x) − f (y) is true for all x, y ∈ X with x − y = . We say that is conservative (or preserved) by f if is contractive and extensive by f simultaneously. If f is an isometry, every distance > 0 is conservative by f, and conversely. At this point, we may raise a question: Is a mapping preserving certain distances an isometry?
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In 1970, Aleksandrov [1] had raised a question of whether a mapping f : X → X preserving a distance > 0 is an isometry, which is now known to us as the Aleksandrov problem. Without loss of generality, we may assume = 1 because X is a normed space (see [8]). Indeed, Beckman and Quarles [2] solved the Aleksandrov problem for any n-dimensional real Euclidean space E n : If a mapping f : E n → E n (2 n < ∞) preserves a distance > 0, then f is a linear isometry up to translation. They also constructed examples for non-isometric mappings which preserve unit distance for one-dimensional or for infinite-dimensional real Euclidean spaces (cf. [6]). Rassias [7] posed the following question: What happens if two (or more) distances are preserved by a mapping between normed spaces? Such a problem is called the Aleksandrov– Rassias problem. For strictly convex vector spaces, an answer was given by Benz [3] (see also [4]): Let X be a real normed space with dim X 2 and let Y be a real normed space which is strictly convex. Suppose f : X → Y is a mapping and N 2 is a fixed integer. If a distance > 0 is contractive and N is extensive by f, then f is a linear isometry up to translation. First, using Lemma 1 and the parallelogram law, we investigate the Aleksandrov–Rassias problem for the case when two distances are contractive and another one is extensive by a mapping. We may compare this theorem with Theorem 2.3 of Xiang [9]. In the theorem of Xiang, two contractive distances a and b have to satisfy an additional condition, 0 < a 2b. However, we can discard such a condition in the following theorem. Theorem 3. Let E n be an n-dimensional√real Euclidean space (2 n < ∞). If the distances a, b > 0 are contractive and the distance a 2 + b2 is extensive by a mapping f : E n → E n , then f is a linear isometry up to translation. Proof. Let us denote by uv the line segment between points u and v of E n and we denote by |uv| the length of uv. Consider a rectangle with |op| = |qr| = a and |pq| = |ro| = b:
r under f, respectively. Since the distance √ Denote by o , p , q , r the image of o, p, q,√ a 2 + b2 is extensive and since |oq| = |pr| = a 2 + b2 , we have
|o q |2 + |p r |2 |oq|2 + |pr|2 .
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Due to the fact that o, p, q, r comprise the vertices of a rectangle such that oq and pr are the diagonals, the parallelogram law implies that |oq|2 + |pr|2 = |op|2 + |pq|2 + |qr|2 + |ro|2 . Further, we have |op|2 + |pq|2 + |qr|2 + |ro|2 |o p |2 + |p q |2 + |q r |2 + |r o |2 because both a and b are assumed to be contractive. In view of Lemma 1, we get |o p |2 + |p q |2 + |q r |2 + |r o |2 |o q |2 + |p r |2 . Our hypotheses obviously imply that |oq| |o q |,
|pr| |p r |,
|op| |o p |,
|pq| |p q |,
|qr| |q r |,
|ro| |r o |.
From those above relations, we conclude that |o q | = |p r | = a 2 + b2 , |o p | = |q r | = a,
|p q | = |r o | = b.
For any given o, p ∈ E n with |op| = a, we can select two points q, r in E n such that o, p, q, r comprise the vertices of a rectangle as the above figure indicates. On account of the √ above argument, we may conclude that |o p | = a. For other distances such as b 2 2 and √a + b , we may apply a similar argument. Therefore, f preserves the distances a, b, a 2 + b2 . Finally, we apply a theorem of Beckman and Quarles (see above or [2]) to our case, and we conclude that f is a linear isometry up to translation. Now, let H1 be a real (or complex) inner product space whose dimension is not less than 3, and let a1 , a2 , b1 , b2 , c1 , c2 , , , be positive numbers such that there exist vectors u, v, w, x, y, z ∈ H1 such that each of {v−u, z−v, x −z, u−x}, {w−u, z−w, y −z, u−y} and {w−v, x−w, y−x, v−y} determines a corresponding parallelogram, respectively, with v − u = x − z = a1 , z − v = u − x = a2 , w − u = y − z = b1 , z − w = u − y = b2 , w − v = y − x = c1 , x − w = v − y = c2 , z − u = , y − w = , x − v = , as we see in the following figure:
(We may say that the vectors u, v, w, x, y, z determine the above octahedron.)
(3)
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In the proof of the following theorem, we substantially follow the ideas of the proof of Theorem 3. Nevertheless, it seems to be better to write the entire proof. Theorem 4. Let H1 and H2 be either real inner product spaces or complex inner product spaces with dim H1 3 and dim H2 3. Assume that the distances a1 , a2 , b1 , b2 , c1 , c2 are contractive and the distances , , are extensive by a mapping f : H1 → H2 , where a1 , a2 , b1 , b2 , c1 , c2 , , , and are assumed to satisfy all the equalities in (3) for an octahedron in the above figure. Then, f preserves the distances a1 , a2 , b1 , b2 , c1 , c2 , , , and . Proof. Assume that the vectors u, v, w, x, y, z determine the octahedron in H1 satisfying (3) (see the above figure). Let us denote by u , v , w , x , y , z the image of u, v, w, x, y, z under f, respectively. Since {v −u, z−v, x −z, u−x}, {w−u, z−w, y −z, u−y} and {w−v, x −w, y −x, v −y} comprise the consecutive sides of appropriate parallelograms, our hypotheses together with Theorem 2 imply that 2(z − u 2 + x − v 2 + y − w 2 )
2(z − u2 + x − v2 + y − w2 ) = v − u2 + z − v2 + x − z2 + u − x2 + w − v2 + x − w2 + y − x2 + v − y2 + w − u2 + z − w2 + y − z2 + u − y2
v − u 2 + z − v 2 + x − z 2 + u − x 2 + w − v 2 + x − w 2 + y − x 2 + v − y 2 + w − u 2 + z − w 2 + y − z 2 + u − y 2
2(z − u 2 + x − v 2 + y − w 2 ). Since our hypotheses imply z − u z − u , x − v x − v , v − u v − u , z − v z − v , u − x u − x , w − v w − v , y − x y − x , v − y v − y , z − w z − w , y − z y − z ,
y − w y − w , x − z x − z , x − w x − w , w − u w − u , u − y u − y ,
the above relation implies that v − u = x − z = a1 , z − v = u − x = a2 , w − u = y − z = b1 , z − w = u − y = b2 , w − v = y − x = c1 , x − w = v − y = c2 , z − u = , y − w = , x − v = . For arbitrarily given u, v ∈ H1 with v − u = a1 , we may select four vectors w, x, y, z in H1 such that u, v, w, x, y, z determine an octahedron as the above figure indicates. In
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view of the above argument, we may conclude that v − u = a1 . For other distances such as a2 , b1 , b2 , c1 , c2 , , , , we can apply a similar argument. Therefore, f preserves the distances a1 , a2 , b1 , b2 , c1 , c2 , , , and . √ If we set a1 = a2 = b1 = b2 = c1 = c2 = √and = = = 2 in Theorem 4, then we know that f preserves the distances and 2 . Due to Theorem 2.8 of Xiang [9], the following statement is true. Corollary 5. Let H1 and H2 be real Hilbert spaces with dim H1 3 and√dim H2 3. For a given > 0, assume that the distance is contractive and the distance 2 is extensive by a mapping f : H1 → H2 . Then, f is a linear isometry up to translation. Remark 6. In Theorem 4 and Corollary 5, H1 and H2 are allowed to be infinite-dimensional inner product spaces (or Hilbert spaces), while we assume in Theorem 3 that E n is a finitedimensional real Euclidean space. Acknowledgements This work was supported by Korea Research Foundation Grant (KRF-2003-015-C00023). References [1] [2] [3] [4] [5] [6] [7] [8] [9]
A.D. Aleksandrov, Mapping of families of sets, Sov. Math. Dokl. 11 (1970) 116–120. F.S. Beckman, D.A. Quarles, On isometries of Euclidean spaces, Proc. Amer. Math. Soc. 4 (1953) 810–815. W. Benz, Isometrien in normierten Räumen, Aequationes Math. 29 (1985) 204–209. W. Benz, H. Berens, A contribution to a theorem of Ulam and Mazur, Aequationes Math. 34 (1987) 61–63. J. Matoušek, Lectures on Discrete Geometry, Graduate Texts in Mathematics, vol. 212, Springer, New York, 2002. Th.M. Rassias, Is a distance one preserving mapping between metric spaces always an isometry?, Amer. Math. Monthly 90 (1983) 200. Th.M. Rassias, Mappings that preserve unit distance, Indian J. Math. 32 (1990) 275–278. Th.M. Rassias, Properties of isometries and approximate isometries, in: G.V. Milovanovic (Ed.), Recent Progress in Inequalities, Kluwer, Dodrecht, 1998, pp. 341–379. S. Xiang, Mappings of conservative distances and the Mazur-Ulam theorem, J. Math. Anal. Appl. 254 (2001) 262–274.
Inequalities for distances between points and distance preserving mappings Soon-Mo Jung∗ Mathematics Section, College of Science and Technology, Hong-Ik University, 339-701 Chochiwon, Korea Received 19 August 2004; accepted 6 April 2005
Abstract In this paper, we generalize the short diagonals lemma by proving a new inequality for distances between six points. Moreover, we apply this inequality to a partial solution to the Aleksandrov–Rassias problem. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 51K05 Keywords: Inequality; Parallelogram law; Short diagonals lemma; Aleksandrov problem; Aleksandrov–Rassias problem
1. Introduction Throughout this paper, let H be a real (or complex) inner √ product space, ·, · the inner product on H, and · the norm on H defined by x = x, x for all x ∈ H . If dim H 2, then the Pythagorean theorem states that for any x, y, z ∈ H , the equality x − z2 = y − x2 + z − y2 is true if and only if the vectors y − x and z − y are orthogonal to each other. This is one of the most important theorems in mathematics. In connection with this subject, we now ∗ Tel.: +82 41 8602584; fax: +82 42 3693657.
E-mail address: [email protected]. 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.04.003
676
S.-M. Jung / Nonlinear Analysis 62 (2005) 675 – 681
consider the case of four vectors x, y, z, w ∈ H . The parallelogram law states that the equality y − x2 + z − y2 + w − z2 + x − w2 = z − x2 + w − y2 holds true if and only if y − x, z − y, w − z, x − w are the sides of a (possibly degenerate) parallelogram with diagonals z − x and w − y. In Section 2 of this paper, using the short diagonals lemma, we prove an inequality for distances between six points. We devote Section 3 to a study of the Aleksandrov–Rassias problem (see Section 3 for details about the Aleksandrov–Rassias problem). 2. Inequality for distances between six points The parallelogram law can be generalized to an inequality which may be applied for any four points (vectors) in an inner product space. This inequality is known as the short diagonals lemma and proved in Lemma 15.4.2 of [5] for the case of Euclidean spaces. In this section, we will first prove the short diagonals lemma for the case of inner product spaces. Lemma 1. If H is a real (or complex) inner product space, then y − x2 + z − y2 + w − z2 + x − w2 z − x2 + w − y2 for all x, y, z, w ∈ H . Proof. We will prove the lemma when H is a complex inner product space. Let x, y, z, w be arbitrary vectors of H. Then, we get y − x2 + z − y2 + w − z2 + x − w2 − z − x2 − w − y2 = y − x, y − x + z − y, z − y + w − z, w − z + x − w, x − w − z − x, z − x − w − y, w − y = y − x, y − x + z, z − y, z − y, z + y, y + w − z, w − z + x, x − w, x − w, x + w, w − z, z + x, z + x, z − x, x − w, w + y, w + y, w − y, y = y − x, y − x + w − z, w − z + y, w − z + y, w − z + z − w, x + z − w, x = y − x, y − x + w − z, w − z + y − x, w − z + y − x, w − z = y − x + w − z, y − x + w − z 0, as required.
By applying Lemma 1 as well as the parallelogram law, we prove an inequality for distances between every two vectors among the given six vectors. The following theorem may be regarded as a generalization of the short diagonals lemma.
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Theorem 2. If H is a real (or complex) inner product space, then the inequality 2(z − u2 + x − v2 + y − w2 ) v − u2 + z − v2 + x − z2 + u − x2 + w − v2 + x − w2 + y − x2 + v − y2 + w − u2 + z − w2 + y − z2 + u − y2
(1)
is true for any six vectors u, v, w, x, y, z ∈ H . Moreover, the equality sign is true if and only if each {v − u, z − v, x − z, u − x}, {w − u, z − w, y − z, u − y}, {w − v, x − w, y − x, v − y} comprises the sides of an appropriate (possibly degenerate) parallelogram such that z − u and x − v, resp., z − u and y − w, resp., x − v and y − w are the diagonals of the corresponding parallelogram. Proof. If we applyLemma 1 toeachorderedquadruplet, (u, v, z, x), (u, w, z, y), (v, w, x, y) consecutively, then we get z − u2 + x − v2 v − u2 + z − v2 + x − z2 + u − x2 , z − u2 + y − w2 w − u2 + z − w2 + y − z2 + u − y2 , x − v2 + y − w2 w − v2 + x − w2 + y − x2 + v − y2 ,
(2)
respectively. By summing up the three inequalities, we easily obtain the desired inequality (1). We now assume that the equality sign holds true in (1). Lemma 1 implies that each inequality in (2) is true. Hence, it is easy to check that strict inequality in at least one of three inequalities of (2) implies the strict inequality sign in (1), a contradiction. Hence, we conclude that if the equality sign holds in (1) then the equality sign has to be true in each inequality of (2). That is, in view of the parallelogram law, we conclude that each {v − u, z − v, x − z, u − x}, {w − u, z − w, y − z, u − y}, {w − v, x − w, y − x, v − y} comprises the sides of an appropriate (possibly degenerate) parallelogram such that z − u and x −v, resp., z−u and y −w, resp., x −v and y −w are the diagonals of the corresponding parallelogram. Conversely, let us assume that each {v−u, z−v, x −z, u−x}, {w−u, z−w, y −z, u−y}, {w − v, x − w, y − x, v − y} comprises the sides of an appropriate (possibly degenerate) parallelogram such that z − u and x − v, resp., z − u and y − w, resp., x − v and y − w are the diagonals of the corresponding parallelogram. According to the parallelogram law, the equality sign is true in each inequality of (2), and hence the equality sign holds in (1). 3. Aleksandrov–Rassias problem Let X andY be normed spaces.A distance > 0 is said to be contractive (or non-expanding) by f : X → Y if x − y = always implies f (x) − f (y) . Similarly, a distance is said to be extensive (or non-shrinking) by f if the inequality f (x) − f (y) is true for all x, y ∈ X with x − y = . We say that is conservative (or preserved) by f if is contractive and extensive by f simultaneously. If f is an isometry, every distance > 0 is conservative by f, and conversely. At this point, we may raise a question: Is a mapping preserving certain distances an isometry?
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S.-M. Jung / Nonlinear Analysis 62 (2005) 675 – 681
In 1970, Aleksandrov [1] had raised a question of whether a mapping f : X → X preserving a distance > 0 is an isometry, which is now known to us as the Aleksandrov problem. Without loss of generality, we may assume = 1 because X is a normed space (see [8]). Indeed, Beckman and Quarles [2] solved the Aleksandrov problem for any n-dimensional real Euclidean space E n : If a mapping f : E n → E n (2 n < ∞) preserves a distance > 0, then f is a linear isometry up to translation. They also constructed examples for non-isometric mappings which preserve unit distance for one-dimensional or for infinite-dimensional real Euclidean spaces (cf. [6]). Rassias [7] posed the following question: What happens if two (or more) distances are preserved by a mapping between normed spaces? Such a problem is called the Aleksandrov– Rassias problem. For strictly convex vector spaces, an answer was given by Benz [3] (see also [4]): Let X be a real normed space with dim X 2 and let Y be a real normed space which is strictly convex. Suppose f : X → Y is a mapping and N 2 is a fixed integer. If a distance > 0 is contractive and N is extensive by f, then f is a linear isometry up to translation. First, using Lemma 1 and the parallelogram law, we investigate the Aleksandrov–Rassias problem for the case when two distances are contractive and another one is extensive by a mapping. We may compare this theorem with Theorem 2.3 of Xiang [9]. In the theorem of Xiang, two contractive distances a and b have to satisfy an additional condition, 0 < a 2b. However, we can discard such a condition in the following theorem. Theorem 3. Let E n be an n-dimensional√real Euclidean space (2 n < ∞). If the distances a, b > 0 are contractive and the distance a 2 + b2 is extensive by a mapping f : E n → E n , then f is a linear isometry up to translation. Proof. Let us denote by uv the line segment between points u and v of E n and we denote by |uv| the length of uv. Consider a rectangle with |op| = |qr| = a and |pq| = |ro| = b:
r under f, respectively. Since the distance √ Denote by o , p , q , r the image of o, p, q,√ a 2 + b2 is extensive and since |oq| = |pr| = a 2 + b2 , we have
|o q |2 + |p r |2 |oq|2 + |pr|2 .
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Due to the fact that o, p, q, r comprise the vertices of a rectangle such that oq and pr are the diagonals, the parallelogram law implies that |oq|2 + |pr|2 = |op|2 + |pq|2 + |qr|2 + |ro|2 . Further, we have |op|2 + |pq|2 + |qr|2 + |ro|2 |o p |2 + |p q |2 + |q r |2 + |r o |2 because both a and b are assumed to be contractive. In view of Lemma 1, we get |o p |2 + |p q |2 + |q r |2 + |r o |2 |o q |2 + |p r |2 . Our hypotheses obviously imply that |oq| |o q |,
|pr| |p r |,
|op| |o p |,
|pq| |p q |,
|qr| |q r |,
|ro| |r o |.
From those above relations, we conclude that |o q | = |p r | = a 2 + b2 , |o p | = |q r | = a,
|p q | = |r o | = b.
For any given o, p ∈ E n with |op| = a, we can select two points q, r in E n such that o, p, q, r comprise the vertices of a rectangle as the above figure indicates. On account of the √ above argument, we may conclude that |o p | = a. For other distances such as b 2 2 and √a + b , we may apply a similar argument. Therefore, f preserves the distances a, b, a 2 + b2 . Finally, we apply a theorem of Beckman and Quarles (see above or [2]) to our case, and we conclude that f is a linear isometry up to translation. Now, let H1 be a real (or complex) inner product space whose dimension is not less than 3, and let a1 , a2 , b1 , b2 , c1 , c2 , , , be positive numbers such that there exist vectors u, v, w, x, y, z ∈ H1 such that each of {v−u, z−v, x −z, u−x}, {w−u, z−w, y −z, u−y} and {w−v, x−w, y−x, v−y} determines a corresponding parallelogram, respectively, with v − u = x − z = a1 , z − v = u − x = a2 , w − u = y − z = b1 , z − w = u − y = b2 , w − v = y − x = c1 , x − w = v − y = c2 , z − u = , y − w = , x − v = , as we see in the following figure:
(We may say that the vectors u, v, w, x, y, z determine the above octahedron.)
(3)
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In the proof of the following theorem, we substantially follow the ideas of the proof of Theorem 3. Nevertheless, it seems to be better to write the entire proof. Theorem 4. Let H1 and H2 be either real inner product spaces or complex inner product spaces with dim H1 3 and dim H2 3. Assume that the distances a1 , a2 , b1 , b2 , c1 , c2 are contractive and the distances , , are extensive by a mapping f : H1 → H2 , where a1 , a2 , b1 , b2 , c1 , c2 , , , and are assumed to satisfy all the equalities in (3) for an octahedron in the above figure. Then, f preserves the distances a1 , a2 , b1 , b2 , c1 , c2 , , , and . Proof. Assume that the vectors u, v, w, x, y, z determine the octahedron in H1 satisfying (3) (see the above figure). Let us denote by u , v , w , x , y , z the image of u, v, w, x, y, z under f, respectively. Since {v −u, z−v, x −z, u−x}, {w−u, z−w, y −z, u−y} and {w−v, x −w, y −x, v −y} comprise the consecutive sides of appropriate parallelograms, our hypotheses together with Theorem 2 imply that 2(z − u 2 + x − v 2 + y − w 2 )
2(z − u2 + x − v2 + y − w2 ) = v − u2 + z − v2 + x − z2 + u − x2 + w − v2 + x − w2 + y − x2 + v − y2 + w − u2 + z − w2 + y − z2 + u − y2
v − u 2 + z − v 2 + x − z 2 + u − x 2 + w − v 2 + x − w 2 + y − x 2 + v − y 2 + w − u 2 + z − w 2 + y − z 2 + u − y 2
2(z − u 2 + x − v 2 + y − w 2 ). Since our hypotheses imply z − u z − u , x − v x − v , v − u v − u , z − v z − v , u − x u − x , w − v w − v , y − x y − x , v − y v − y , z − w z − w , y − z y − z ,
y − w y − w , x − z x − z , x − w x − w , w − u w − u , u − y u − y ,
the above relation implies that v − u = x − z = a1 , z − v = u − x = a2 , w − u = y − z = b1 , z − w = u − y = b2 , w − v = y − x = c1 , x − w = v − y = c2 , z − u = , y − w = , x − v = . For arbitrarily given u, v ∈ H1 with v − u = a1 , we may select four vectors w, x, y, z in H1 such that u, v, w, x, y, z determine an octahedron as the above figure indicates. In
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view of the above argument, we may conclude that v − u = a1 . For other distances such as a2 , b1 , b2 , c1 , c2 , , , , we can apply a similar argument. Therefore, f preserves the distances a1 , a2 , b1 , b2 , c1 , c2 , , , and . √ If we set a1 = a2 = b1 = b2 = c1 = c2 = √and = = = 2 in Theorem 4, then we know that f preserves the distances and 2 . Due to Theorem 2.8 of Xiang [9], the following statement is true. Corollary 5. Let H1 and H2 be real Hilbert spaces with dim H1 3 and√dim H2 3. For a given > 0, assume that the distance is contractive and the distance 2 is extensive by a mapping f : H1 → H2 . Then, f is a linear isometry up to translation. Remark 6. In Theorem 4 and Corollary 5, H1 and H2 are allowed to be infinite-dimensional inner product spaces (or Hilbert spaces), while we assume in Theorem 3 that E n is a finitedimensional real Euclidean space. Acknowledgements This work was supported by Korea Research Foundation Grant (KRF-2003-015-C00023). References [1] [2] [3] [4] [5] [6] [7] [8] [9]
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