On Bisectors for Convex Distance Functions

Transcription

1 E etracta mathematicae Vol. 28, Núm. 1, On Bisectors for Conve Distance Functions Chan He, Horst Martini, Senlin Wu College of Management, Harbin University of Science and Technology, Harbin, China, chan Faculty of Mathematics, Chemnitz University of Technology, Chemnitz, Germany, Department of Applied Mathematics, Harbin University of Science and Technology, Harbin, China, Received February 26, 2013 Abstract: It is well known that the construction of Voronoi diagrams is based on the notion of bisector of two given points. Already in normed linear spaces, bisectors have a complicated structure and can, for many classes of norms, only be described with the help of topological methods. Even more general, we present results on bisectors for conve distance functions gauges. Let C, with the origin o from its interior, be the compact, conve set inducing a conve distance function gauge in the plane, and let B, be the bisector of and, i.e., the set of points z whose distance measured with the conve distance function induced by C to equals that to. For eample, we prove the following characterization of the Euclidean norm within the family of all conve distance functions: if the set L of points in the boundary C of C that create B, as a straight line has non-empty interior with respect to C, then C is an ellipse centered at the origin. For the subcase of normed planes we give an easier approach, etending the result also to higher dimensions. Key words: Birkhoff orthogonality, bisector, characterization of ellipse, conve distance function, Euclidean norm, gauge, isosceles orthogonality, Roberts orthogonality, Voronoi diagram. AMS Subject Class. 2010: 46B20, 52A Introduction It is well known that for the construction of Voronoi diagrams the notion of bisector of two given points is fundamental. To construct bisectors in Euclidean spaces is an elementary task, but to investigate them in general normed spaces can be, from the geometric and topological viewpoint, very difficult, since bisectors can even be full-dimensional sets. For results on bisectors and Voronoi diagrams in normed linear spaces we refer to the survey [14]. Here we want to prove some new theorems on bisectors for conve distance functions, which are more general than respective statements for norms. More precisely, for a planar conve body C taken as unit ball of a 57

2 58 c. he, h. martini, s. wu y γ C y y γ C o C Figure 1: A planar conve body and the corresponding gauge. conve distance function gauge we show the following: If the set L of all points in the boundary of C that create their bisectors B, see the definition below as straight lines has nonempty interior with respect to that boundary of C, then C is an ellipse centered at the origin. An easier approach to that characterization of Euclidean geometry, which is also available for higher dimensions, is applied to the subcase when C is centrally symmetric with respect to the origin, i.e., when the gauge γ C see the definition below is a norm. Let C R 2 be a conve body i.e., a compact, conve set with non-empty interior intc satisfying o intc, and C be the boundary of C. The gauge or Minkowski functional γ C, defined by γ C : R 2 R inf{λ > 0 : λc}, has the following properties see Figure 1 and, for instance, [7, p ]: 1. γ C 0 R 2, 2. γ C = 0 if and only if is the origin o, 3. γ C λ = λγ C R 2, λ 0, 4. γ C + y γ C + γ C y. The conve distance function d C, induced by C is then defined via γ C by: d C y, := γ C y. Clearly, d C, satisfies the following properties.

3 on bisectors for conve distance functions d C y, = d C y + z, + z, y, z R 2, 2. d C αy, α = αd C y, α 0. Notice that d C, is not necessarily symmetric, i.e., in general d C p, q d C q, p. The distance function d C, is symmetric if and only if C is symmetric with respect to the origin o, in which case γ C is a norm. By [p, q] we denote the segment possibly degenerate between two points p, q X, by [p, q the ray with starting point p passing through q p q, and by p, q the line passing through p and q p q. For each point C, we denote by the point in which the ray [, o intersects C. Also, each point C is associated with a number µ > 0 such that = µ. Let u and v be two linearly independent points in C. Then we call the set arcu, v := {λu + µv : λ, µ 0} C the minor arc of C connecting u and v. Let L be the set of points in C such that B, is a straight line, where B, is the bisector of and, which is defined by B, := { z R 2 : d C, z = d C, z }. More generally, the bisector Bp, q of two distinct points p and q is defined by Bp, q := { z R 2 : d C p, z = d C q, z }. Due to various applications such as Voronoi diagrams, see [5], [12], and [13], bisectors are deeply studied in Computational Geometry. But also in Minkowski Geometry they play an increasing role; see [8], [9], [14], and [16]. Of course, since the conve distance function is not necessarily symmetric, there are other ways to define bisectors. For eample, we can put the set B, := { z R 2 : d C z, = d C z, }. to be the bisector of the points and. One can easily verify that B, = B,. The set L introduced above can be empty, even if C is symmetric with respect to the origin. Such an eample can be found in [10, Eample 2.1]. The aim of the paper is to show that, if the interior of L with respect to C is not empty, then C is an ellipse centered at the origin.

4 60 c. he, h. martini, s. wu In Section 2 we deal with the general planar case, namely when C is not necessarily symmetric with respect to the origin. Etending our investigations also to higher dimensions, we study in Section 3 the special case when C is symmetric with respect to the origin o. Benefiting from recent results in Functional Analysis, our approach in Section 3 is much easier. It is also working for the case when C is the unit ball of an infinite dimensional real normed linear space, in which case C is a bounded closed but not compact conve set with non-empty interior, which is symmetric with respect to the origin. 2. The general case The results in this section hold for R 2. We note that if L, then = µ L, which can be easily seen from the following equality: B µ, µ = µ B,. The following lemma describes the relation between the structure of a bisector and a property of C. Lemma 2.1. cf. [13, Lemma , Corollary ] Let be a point distinct from o. Then B, is homeomorphic to a line if and only if there is no non-trivial segment contained in C and parallel to,. By Lemma 2.1, if B, is homeomorphic to a line, then the two supporting lines of C which are parallel to the line, intersect C, in each case, in precisely one point. Lemma 2.2. If L, then there eist precisely two points n and s in C such that the lines n +, and s +, are the two supporting lines of C which are parallel to,. Moreover, o [n, s ]. Proof. First, since B, is a straigth line, there eist precisely two points n and s in C such that the lines n +, and s +, are two supporting lines of C. For each point z B, we have the inequality αz := d C, z = d C o, z + d C o, z

5 on bisectors for conve distance functions 61 Also, we have d C o, z = d C o, 1 2 z z d C o, 1 2 z + + d C o, 1 2 z 2 From 1 and 2 it follows that = 1 2 d C, z d C, z = 1 2 αz + 1 αz = αz. 2 1 d C o, 1 αz z d Co, z d C o, z + 1. Thus, when d C o, z tends to infinity, d C o, 1 αz z tends to 1. In the meantime, since αz = d C, z = d C, z, 1 d C o, αz z + 1 = d C o, αz z = 1. 1 Thus αz z is the midpoint of the chord [ 1 αz z +, 1 αz z ] of C which is parallel to,. Denote by H + and H the two open halfplanes bounded by, and containing n and s, respectively. If {z n } n=1 is a sequence contained in H + B, such that d C o, z n tends to infinity, then zn αz tends to n n since [ z n+ αz, zn ] n αz n is a chord of C parallel to,, whose midpoint z n αz n has distance to the origin tending to 1. Similarly, if {z n } n=1 is a sequence contained in H B, such that d C o, z n tends to infinity, then zn αz n tends to s. Net we show that o [n, s ]. Suppose that this is not true. Then we can suitably choose a pair of lines l and l from the following four lines: the two supporting lines of C which are parallel to n, s and the two lines parallel to n, s and passing through and, respectively. Suitably here means that the other two lines lie between them. Without loss of generality, we may assume that l and o are separated by the line n, s. We denote by X + the halfplane bounded by l which does not contain C. From foregoing discussions, there eist two points z 1 and z 2 in B, such that z 1 lies in the halfplane bounded by, containing n, and z 2 lies in the other halfplane containing s. These two points z 1 and z 2 can be chosen so that z 1 d C o,z 1 and z 2 d C o,z 2 are sufficiently close to n and s, respectively. Here sufficiently close to means that the rays [o, z 1 and [o, z 2 both intersect l.

6 62 c. he, h. martini, s. wu Also, the numbers d C o, z 1 and d C o, z 2 can be chosen large enough such that z 1 and z 2 are both in X +. Then the line z 1, z 2, which is precisely B,, intersects the line, but does not touch the relative interior of the segment [, ]. This is impossible and leads to a contradiction. By Lemma 2.2, we can associate each L with two points n and s such that 1. n +, and s +, are the two supporting lines of C which are parallel to, ; 2. n is contained in the part of C that connects with counterclockwise. Lemma 2.3. If L, then B, = 1 µ 1 + µ + n, s. Proof. First we show that B, is parallel to the line n, s. Suppose to the contrary that this is not true. Let {z n } B, be a sequence, which is contained in the halfplane bounded by, and containing n, such that lim n d Co, z n =. We claim that lim n z n d C o, z n n, which is a contradiction to the fact that Lemma 2.2. Otherwise, lim n z n p d C o, z n = n, z n αz n tends to n cf. the proof of where p is the point of intersection of the lines n, s and B,. By Lemma 2.2, o [n, s ]. It follows that the lines n, s and B, coincide, which is a contradiction. Net we show that 1 µ 1+µ B,, which follows directly from the equations d C, 1 µ 1 + µ = d C o, 2µ 1 + µ = 2 = 2 d C o, 1 + µ 1 + µ = d C o, µ = µ d C o, µ = d C, 1 µ. 1 + µ

7 on bisectors for conve distance functions 63 z ref z n o s ref z z Figure 2: The definition of ref z. Lemma 2.4. If L, then, for each number β R, d C o, 2µ 2 + βn = d C o, + βn µ 1 + µ Proof. From Lemma 2.3 and the relation o [n, s ] it follows that, for each number β R, 1 µ 1 + µ + βn B,. Thus d C o, 1 µ + βn = d C o, 1 µ + βn +, 1 + µ 1 + µ from which 3 follows. Let be a point in L. For any point z = α+βn = 1 µ α +βn R 2, set see Figure 2 ref z = { 1 µ α + βn, α < 0, α + βn, α 0. Lemma 2.5. Let be a point in L. Then, for each point z R 2, we have 1. ref ref z = z, 2. d C o, z = d C o, ref z.

8 64 c. he, h. martini, s. wu Proof. Assume that z = α + βn. 1 ref ref z { ref 1 µ = α + βn, α < 0, ref µ α + βn, α 0, = α + βn. 2 If α < 0, then, by Lemma 2.4, d C o, α + βn = α 1 + µ d C o, 2µ β 2µ 1 + µ α 2µ n 1 + µ If α > 0, then = α 1 + µ 2µ d C o, = d C o, α µ + βn = d C o, ref z. d C o, α + βn = α 1 + µ d C o, µ β α µ + β α = α 1 + µ d C o, 2µ + β µ α = d C o, µ α + βn = d C o, ref z. 2µ n 1 + µ 2 n 1 + µ 2 n 1 + µ If α = 0, then d C o, z = d C o, βn = d C o, ref z. Lemma 2.6. If and y are two linearly independent points in C and arc, y L, then there eists a number γ 0 such that arcn, n y = γ 0 arcs, s y. Proof. We note that, since C is a conve curve, for each smooth point w arcn, n y there eists a unique point z arc, y such that w = n z. Let f 1 θcosθ, sinθ θ [θ 1, θ 2 ] and f 2 θ + πcosθ + π, sinθ + π

9 on bisectors for conve distance functions 65 θ [θ 1, θ 2 ] be the polar equations of arcn, n y and arcs, s y, respectively. Then f 1 θ f 1 θ = f 2 θ + π f 2 θ + π holds for all θ [θ 1, θ 2 ] with a countable set of eceptions. By integration, 4 yields f 1 θ f 1 θ 1 = f 2θ + π f 2 θ 1 + π, θ [θ 1, θ 2 ]. 4 Thus f 1 θ f 2 θ + π = f 1θ 1 f 2 θ 1 + π, θ [θ 1, θ 2 ]. Hence there eists a number γ 0 such that arcn, n y = γ 0 arcs, s y. To continue our discussion we need the forthcoming Lemma 2.8, which proves a seemingly obvious fact. In its proof we use the following lemma from [13]. Lemma 2.7. cf. Lemma in [13] For three points a 1, a 2, a 3, we have Ba 1, a 2, a 3 := Ba 1, a 2 Ba 1, a 3 = if and only if either a 3 is contained in the interior of the set F G 12 or a 3 lies on one of the boundary line segments of F G 12 and the tangent to C, where this line segments stems from, does not contain a boundary line segment of C. Lemma 2.8. Let C R 2 be a conve body containing the origin o in its interior, and d C, be the conve distance function induced by C. Then, for any two pairs of distinct points a 1 and a 2, a 3 and a 4 such that the segment [a 1, a 2 ] is not parallel to [a 3, a 4 ], Ba 1, a 2 and Ba 3, a 4 cannot be two parallel straight lines. Proof. Suppose to the contrary that Ba 1, a 2 and Ba 3, a 4 are two parallel straight lines. First, observe that for any α > 0 and any vectors p, q, z, the following equalities hold

10 66 c. he, h. martini, s. wu U T 12 T t 12 t G 12 a 1 F 12 F 21 a 2 G 21 L C 1 C 2 D 12 d 12 d 21 D 21 Figure 3: F G 12 is the shaded region. Bαp + z, αq + z = {w : d C αp + z, w = d C αq + z, w} = {w : d C p + 1 α z, 1 α w = d C q + 1 α z, 1 } α w = {w : d C p, 1 α w z = d C q, 1 } α w z = {αv + z : d C p, v = d C q, v} = α{v : d C p, v = d C q, v} + z = αbp, q + z, Thus we may assume, without loss of generality, that a 4 = a 1. Net we take some terminology from [13], to apply Lemma 2.7. Without loss of generality we assume that the line a 1, a 2 is horizontal. Let t 12 and d 12 be the top and bottom point of C 1 := C + a 1, respectively; t 21 and d 21 be the top and bottom point of C 2 := C + a 2, respectively; and let U and L be the upper and lower common supporting lines of C 1 and C 2, respectively. Let T 21 be the steepest tangent to C 2 at t 21, and let T 12 be the least steep tangent to C 1 at t 12. Correspondingly, let D 12 be the steepest tangent to C 1 at d 12, and let D 21 be the least steep tangent to C 2 at d 21. We consider the four cones with ape a 1 defined by the lines through a 1 parallel to T 12 and D 12, respectively. Let F 12 denote the cone bounded by the line parallel to D 12 from above and by the line parallel to T 12 from below, and let G 12 be the opposite cone. Analogously we have F 21 and G 21 with ape a 2. For brevity let F G 12 denote the set G 12 F 12 F 21 G 21, see Figure 3.

11 on bisectors for conve distance functions 67 On the line a 1, a 3 there is precisely one point a such that the bisectors Ba 1, a 2 and Ba 1, a intersect. Since the lines a 1, a 2 and a 1, a do not coincide, there eists a point a 3 distinct from a on the line a 1, a \F G 12. Then Ba 1, a 2, a 3 := Ba 1, a 2 Ba 1, a 3 =, since Ba 1, a 3 is a straight line parallel to Ba 1, a 3. This is a contradiction to Lemma 2.7. Lemma 2.9. If lies in the closure of the interior of L with respect to C, then µ = 1. Proof. First suppose that is in the interior of L in C. Then there eist two linearly independent points u, v L such that arcu, v L. If n u = n, then, by Lemma 2.2, s u = s. By Lemma 2.3, B u, u is parallel to B,, which is in contradiction to Lemma 2.8. Thus n u n. Similarly, n v n. Without loss of generality, these two points u and v are chosen such that n v = ref n u and n u = α + βn, where α, β > 0. Then from the definition of ref it follows that Thus and n v = µ α + βn. Lemma 2.6 shows that there eists a number γ 0 > 0 such that arcs u, s v = γ 0 arcn u, n v. 5 s u = γ 0 n u = γ 0 α γ 0 βn s v = γ 0 n v = γ 0 µ α γ 0 βn. From 5 it follows that the line s u, s v is parallel to the line,. By Lemma 2.1, {s u, s v } = s u, s v C. Since s u, s v C, by the definition of ref and Lemma 2.5 we have γ 0 µ α γ 0 βn = s v = ref s u = ref γ 0 α γ 0 βn = 1 µ γ 0 α γ 0 βn, which implies that µ = 1 µ. Thus µ = 1. One can easily verify that µ is continuous with respect to. Thus, for each point y in the closure of the interior of L with respect to C, µ y = 1.

12 68 c. he, h. martini, s. wu B, o Figure 4: A bisector B, which is a line not passing through the origin. Corollary If C lies in the interior of L in C, then 1. B, contains the origin, 2. ref z = α + βn holds for any point z = α + βn, and 3. ref is linear. Proof. 1 Since, by Lemma 2.3, 1 µ 1 + µ B,, it follows from Lemma 2.9 that o B,. 2 This is an easy consequence of the definition of ref and the fact that µ = 1. 3 Let z 1 = α 1 + β 1 n and z 2 = α 2 + β 2 n be two arbitrary points in R 2, and λ, γ be two arbitrary real numbers. Then ref λz 1 + γz 2 = ref λα1 + γα 2 + λβ 1 + γβ 2 n This implies that ref is linear. = λα 1 + γα 2 + λβ 1 + γβ 2 n = λref α 1 + β 1 n + γref α 2 + β 2 n = λref z 1 + γref z 2.

13 on bisectors for conve distance functions 69 Remark Without further assumptions, B, being a straight line does not imply that B, is a straight line containing the origin. Here is an eample: Let C be a conve body in R 2, the polar coordinate ρ = ρθ of whose boundary C satisfies the following conditions cf. Figure 4: 1. For each θ [0, π/2] and each θ [3π/2, 2π, ρθ = For each θ π/2, π, ρθ = sinarctan3 tanπ θ/ sinπ θ. And ρπ = For each θ π, 3π/2, ρθ = sinarctan3 tanθ π/ sinθ π. Let = 3, 0 and o = 1.5, 0. We claim that B, is the straight line o + 0, 1, 0, 1. In fact, for any z o + [o, 0, 1, the polar angles θ and θ of vectors z and z, respectively, satisfy θ [0, π/2 and tan θ = z o E o = z o E o E o E E o = 1 tan θ, E 3 where p q E stands for the Euclidean distance between two points p and q. Then θ = π + arctan 1 1 π ] 3 tan θ = π arctan 3 tan θ 2, π. For any point z o + [o, 0, 1 \{o }, the polar angle θ of z is in π/2, π. In this case we have the equation It follows that ρθ = sin θ sinπ θ. d C, z d C, z = z E ρθ ρθ z E z = E sin θ z E sinπ θ sinπ θ ρθ sin θ ρθ = z o E z o sinπ θ ρθ E sin θ ρθ = 1. For the case of o, the polar angle of o is π. Thus d C, o d C, o = o E ρ0 ρπ o = 1.5 E = 1,

14 70 c. he, h. martini, s. wu which implies that o B,. Hence In a similar way, the inclusion can be proved. Hence o + [o, 0, 1 B,. o + [o, 0, 1 B, o + 0, 1, 0, 1 B,. Notice that, by Lemma 2.1, B, is homeomorphic to a straight line. Thus B, = o + 0, 1, 0, 1. Also, it can be seen from Figure 4 that, even if B, is a straight line, the set of midpoints of chords of C, which are parallel to,, is not necessarily contained in a line. Lemma If C lies in the interior of L with respect to C and y is a point in L, then ref y L. Proof. From Lemma 2.5 and Corollary 2.10 it follows that B y, y = {z : d C y, z = d C y, z} = {z : d C o, z + y = d C o, z y} = { z : d C o, ref z + y = d C o, ref z y } = { z : d C o, ref z + ref y = d C o, ref z ref y } = { z : d C ref y, ref z = d C ref y, ref z }. Thus ref B y, y = B ref y, ref y. Since ref is linear, B ref y, ref y is also a straight line. ref y L. Therefore Theorem If the interior of L with respect to C is not empty, then C is an ellipse centered at the origin.

15 on bisectors for conve distance functions 71 Proof. If L = C then, by Corollary 2.10, o B, holds for every C. This implies that C is symmetric with respect to the origin o. This, together with the assumption that B, is a straight line for each point C, implies that C is an ellipse centered at the origin cf. [4, pp ]. Thus we only need to show that L = C. If this is not true, then there eists a point z C\{L}. It follows that z is also not in L. Let L be a relative interior point of L with respect to C, and L be the maimal connected component of L containing. Then L is contained in one of the open halfplanes bounded by the line z, z. Hence there eist two points u and v in C such that the closure of L is arcu, v. Now we know that the interior of arcu, v with respect to C is contained in L. a point in the relative interior of arc u, Let u be d C o,u+ with respect to C. By the definition of ref u, the line ref u, is parallel to the line u, u. Thus ref u arcu, v. Moreover, ref u arc u, = { ref u y : y arc u, } = arc ref u, u. From Lemma 2.12 it follows that arcu, ref u L. This contradicts the fact that L is the maimal connected component of L containing. 3. A characterization of inner product spaces In this section, C is the unit ball B X of a real normed linear space X with norm, whose unit sphere is the boundary of B X and denoted by S X. In this situation, C is a closed bounded conve body which is not necessarily compact with o as interior point and center of symmetry. A normed linear space is called a Banach space if it is complete. A Banach space X is a Hilbert space if the norm is compatible with an inner product on the linear space X. Although we study the general case when X is not necessarily finite dimensional, our method belongs to the geometry of finite dimensional Banach spaces or Minkowski spaces, cf. [15], [14], and the monograph [18]. Some notions of generalized orthogonality types in normed linear spaces are needed for the discussion in the sequel. Let and y be in X. We say that is isosceles orthogonal to y if the equality + y = y holds, and for this situation we write I y; is said to be Roberts orthogonal to y if the equality + αy = αy

16 72 c. he, h. martini, s. wu holds for any number α R, and we write R y for this case. The implication, y X, R y I y is trivial while its reverse, namely, y X, I y R y, forces X to be an inner product space cf. [10], [4], [2], or [3]. In other words, isosceles orthogonality is in general not homogeneous. We notice that B, = {z X : z = z + } = {z X : z I }. We also need the notion of Birkhoff orthogonality. is said to be Birkhoff orthogonal to y if the inequality + λy holds for any number λ R, and in this case we write B y. Geometrically, B y means that, when y = 0, there eists a line which contains, is parallel to the line passing through y and y, and supports the disc of radius centered at o in the two-dimensional subspace spanned by and y cf. Figure 5. One can also verify the implication, y X, R y B y. For more information about the structure of bisectors, also in view of relations to properties of generalized orthogonality types especially isosceles and Birkhoff orthogonality, we refer to [14], [16], [8], and [9]. Still we denote by L the set of points in S X such that B, is a hyperplane. Notice that, since B X is symmetric with respect to the origin, B, always contains the origin o. A subset R of a topological space T is said to be rare in T if the interior of the closure of R in T is empty. An operator on a real Banach space of the form s e,e : 2e e is called a reflection, where e X and e X the dual space of X satisfy e e = 1. For a point e S X there eists at most one e S X such that s e,e is an isometric reflection. When there eists such an e we say that e is a vector of isometric reflection and that e is the isometric reflection functional associated to e. We refer to [17], [6], and [1] for more about isometric reflections and isometric reflection vectors.

17 on bisectors for conve distance functions 73 o y Figure 5: Birkhoff orthogonality. Lemma 3.1. Let X be a real Banach space, e S X, e S X, and s e,e be a reflection. Then s e,e is an isometric reflection if and only if e R H := { : X, e = 0}. Proof. First we recall that R y B y holds for any, y X. Now suppose that s e,e is an isometric reflection. Then for any point z H and any real number α we have that s e,e e + αz = αz e. Since s e,e is a linear isometry, we have e + αz = e αz, which implies that e R z. Thus e R H. Now suppose that e R H holds. Clearly, H is a hyperplane of X. Then for any point y X there eists a point z H such that y = e ye+z. Then s e,e y = e ye + z 2e e ye + ze = z e ye = e ye + z = y, which implies that s e,e is an isometry. Lemma 3.2. Let e be a point in S X. If B e, e is a hyperplane, then e is an isometric reflection vector.

18 74 c. he, h. martini, s. wu Proof. Let H be the hyperplane B e, e. First we show that e R H. For each point z H = B e, e, it is clear that e I z. Since H is a hyperplane passing through the origin o this is because o B e, e, we have {λz : λ R} H. Thus e I λz holds for each number λ R, which implies that e R z. Since z is arbitrary in H, e R H. Note that the relation e R H implies that e B H. Then there eists a point e in S X such that e e = 1 and cf. [11, Theorem 2.1] H = { : X, e = 0}. Then s e,e is a reflection and e R { : X, e = 0}. By Lemma 3.1, s e,e vector. is an isometric reflection and e is an isometric reflection Remark 3.3. The reverse of Lemma 3.2 is not true. Take, for eample, the normed or Minkowski plane X = R 2,. Let e = 1, 0. Then e is an isometric reflection vector. To see this, we put e = 1, 0. Then e S X, e e = 1, and e R {λ0, 1 : λ R} = { : X, e = 0}. However, B e, e is not a hyperplane, but even a set with nonempty interior cf. Figure 6. Lemma 3.4. cf. [6] A Banach space X is a Hilbert space if and only if the set of all isometric reflection vectors in X is not rare in S X. The following is the result that we announced. Theorem 3.5. A Banach space X is a Hilbert space or, B X is an ellipsoid if and only if the set L is not rare in S X. Proof. The necessity is obvious, so we only need to show the sufficiency. By Lemma 3.2, the set L is a subset of the set of isometric reflection vectors in X. Since L is not rare in S X, the set of isometric reflection vectors is also not rare in S X. Then it follows from Lemma 3.4 that X is a Hilbert space.

19 on bisectors for conve distance functions 75 e e Figure 6: B e, e is not a hyperplane. In the following we etend Theorem 2.13 to higher dimensional cases. Now we assume that C R n n 2 is a conve body containing the origin o in its interior. Let L be the set of points C such that B, is a hyperplane. Then we have the following theorem. Theorem 3.6. If the interior of the set L with respect to C is not empty, then C is an ellipsoid centered at the origin. Proof. By Theorem 2.13 we only need to deal with the subcase when n > 2, and by Theorem 3.5 it suffices to show that C is centered at the origin. Indeed, if this is true, then = γ C is a norm on R n and the interior of the set L with respect to S X = C is not empty. By Theorem 3.5, X = R n, is a Hilbert space and B X = C is an ellipsoid. Let be an interior point of L with respect to C, and z be an arbitrary point in C such that and z are linearly independent. We denote by X,z the two-dimensional subspace of R n spanned by and z, and by C 0 the intersection of C and X,z. Then it is clear that is an interior point of L C 0 with respect to C 0. By Theorem 2.13, C 0 is an ellipse centered at the origin. This implies that {, z} C 0 C. Since z is arbitrary, it follows that C is centered at the origin. Acknowledgements The first and third named authors are grateful for financial support from the following grants: National Nature Science Foundation of China NSFC grant number: , a foundation from the Ministry of Education of Heilongjiang Province grant number 1251H013, China Postdoctoral Science Foundation grant number 2012M520097, and the

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