Chapter 4 Perpendicular Bisectors in Normed Spaces The center of the circumscribed circumference of a triangle is the intersec tion point of the three perpendicular bisectors of its three sides This i[.]
Chapter Perpendicular Bisectors in Normed Spaces The center of the circumscribed circumference of a triangle is the intersection point of the three perpendicular bisectors of its three sides This is a classical situation in an i.p.s What happens in a general normed linear space where various notions of orthogonality lead to alternative expressions for perpendicular bisectors? We explore such questions in this chapter 4.1 Definitions and basic properties In a real i.p.s (X, h·, ·i) with dimension greater than or equal to 2, given two linearly independent vectors x, y ∈ X, we define the perpendicular bisector of the linear segment [x, y] = {αx + (1 − α)y : ≤ α ≤ 1} as the set x+y + λu : λ∈R , where u 6= is a vector in the subspace generated by x and y (denoted by span(x, y)) and orthogonal to x − y If u := αx + βy, then by the requirement hx − y, ui = 0, we have and we may take α kxk2 − hy, xi = β(kyk2 − hx, yi), u = (kyk2 − hx, yi)x + (kxk2 − hy, xi)y 103 104 Norm Derivatives and Characterizations of Inner Product Spaces x+y y + λu u x Figure 4.1.1 In a real normed linear space (X, k · k), we can replace the inner product hx, yi by its generalizations ρ′± (x, y) Namely, we consider the vectors u± (x, y) = (kyk2 − ρ′± (x, y))x + (kxk2 − ρ′± (y, x))y (4.1.1) for all x, y in X, and define M± (x, y) := x+y + λu± (x, y) : λ ∈ R (4.1.2) It follows from the above that if x and y are linearly dependent, e.g y = λx for some λ ∈ R, then M± (x, λx) = x + λx We call the lines (4.1.2) perpendicular bisectors of the linear segment [x, y], and later we justify their associated orthogonal relations In what follows, we restrict ourselves to considerations related to M+ Furthermore, in the real normed linear space (X, k·k), we can also define the perpendicular bisector of the linear segment [x, y] as the set of metric nature: M ∗ (x, y) = {z ∈ X : z ∈ span (x, y), kz − xk = kz − yk} We have M ∗ (x, y) = M ∗ (y, x) and M+ (x, y) = M+ (y, x), but in general M+ (x, y) 6= M ∗ (x, y); for example, in (R2 , k·k+ ) if x = (3, 1) and y = (1, 1), Perpendicular Bisectors in Normed Spaces 105 then M+ (x, y) = {(2, 1) + λ(−4, 4) : λ ∈ R}, but M ∗ (x, y) = {(2, 1) + µ(0, 1) : µ ∈ R} The equality of the sets M+ and M ∗ is a characteristic property of inner product spaces Theorem 4.1.1 Let (X, k · k) be a real normed linear space with dim X ≥ Then X is an inner product space if and only if for all x, y in X M+ (x, y) = M ∗ (x, y) For the proof of this theorem, we need the following lemma based on the James orthogonality Lemma 4.1.1 In a real normed linear space (X, k · k) with dim X ≥ 2, for all vectors x and y in X, we have x+y + w : w ∈ X, w ⊥J x − y {z ∈ X : kz − xk = kz − yk} = 2 , Proof If z ∈ X and kz − xk = kz − yk, and we take w = z − x+y + w and then z = x+y 2 kw − (x − y)k = 2kz − xk = 2kz − yk = kw + (x − y)k Therefore, w ⊥J x − y Conversely, if z = kx − zk = x+y + 21 w with w ⊥J x − y, then 1 kx − y − wk = kx − y + wk = ky − zk 2 Proof of Theorem 4.1.1 Assume M+ (x, y) = M ∗ (x, y) for all x, y in X Fix v ∈ X and take ς ∈ X such that v ⊥J ς We can consider linearly 106 Norm Derivatives and Characterizations of Inner Product Spaces independent vectors x, y ∈ span (v, ς) and such that ς = x − y Then by the previous lemma x+y x+y + v ∈ + w : w ∈ X, v ⊥J x − y ∩ span (v, ς) 2 2 = {z ∈ X : kz − xk = kz − yk} ∩ span (x, y) = M ∗ (x, y) = M+ (x, y) Therefore, the vectors u+ (x, y) and v are linearly dependent, and x+y + λv : λ ∈ R = M+ (x, y) Now, if λ ∈ R, using M+ (x, y) = M ∗ (x, y), we have x + y + λv − x kς − 2λvk = kx − y − 2λvk = ky − x + 2λvk = x + y = 2 + λv − y = kx − y + 2λvk = kς + 2λvk Therefore, 2λv ⊥J ς for all λ ∈ R, i.e., the James orthogonality is homogeneous, which is a characterization of inner product space [James (1945); Amir (1986)] The converse implication is obvious 4.2 A new orthogonality relation In an inner product space, the perpendicular bisectors of the linear segments determined by the sides of a rhombus are its diagonals Indeed, this property characterizes inner products spaces Theorem 4.2.1 Let (X, k · k) be a real normed linear space with dim X ≥ Then X is an inner product space if and only if for all x, y in X with kxk = kyk one has M+ (x, y) = span (x + y) Proof If we assume that M+ (x, y) defined by (4.1.2) is the subspace generated by x + y, for all x, y in SX , the vector u+ (x, y) ∈ span (x + y), Perpendicular Bisectors in Normed Spaces 107 i.e., kyk2 − ρ′+ (x, y) = kxk2 − ρ′+ (y, x), so ρ′+ (x, y) = ρ′+ (y, x), and by the properties of ρ′+ , X is an inner product space Since in an inner product space, two vectors x and y are orthogonal if and only if they are orthogonal in the sense of James, if we take the rhombus with sides x + y and x − y, then by the last theorem M+ (x + y, x − y) = span (x) Based on this property, we define a new orthogonality relation Definition 4.2.1 In a real normed linear space (X, k·k) with dim X ≥ 2, we consider the perpendicular bisector orthogonality x ⊥M y if M+ (x + y, x − y) = span (x), or, equivalently, after using the definition of M+ and the properties of ρ′+ x ⊥M y Example 4.2.1 if and only if ρ′+ (x + y, x) = ρ′+ (x − y, x) (4.2.1) In the space (R2 , k · k+ ) for all x 6= 0, we have (x, 0) ⊥M (y1 , y2 ) if and only if y1 = and xy2 < The next lemma contains some basic properties of this orthogonality relation Proposition 4.2.1 The perpendicular bisector orthogonality ⊥M satisfies the following properties for all x, y in X and for all a, t in R: (i) (ii) (iii) (iv) (v) (vi) (vii) x ⊥M 0, ⊥M x; x ⊥M x if and only if x = 0; x ⊥M tx if and only if tx = 0; If x ⊥M y, then ax ⊥M ay; If x ⊥M y, then x ⊥M −y and −x ⊥M y; If x ⊥M y and x, y 6= 0, then x and y are linearly independent; If the norm derives from an inner product h·, ·i, the relation x ⊥M y is equivalent to the usual orthogonality hx, yi = 108 Norm Derivatives and Characterizations of Inner Product Spaces Proposition 4.2.2 If (X, k · k) is a real normed linear space with dim X ≥ 2, then for all pairs x, y in X\{0} there exist two unique real numbers t1 > and t2 < such that x + ti y ⊥M x − ti y, i ∈ {1, 2} (i.e., the ⊥M orthogonality admits diagonals) Proof Fix x, y in X\{0} For any t, the condition x + ty ⊥M x − ty is equivalent to kyk2 t2 + (ρ′sgn(t) (y, x) − ρ′sgn(t) (x, y))t − kxk2 = (4.2.2) So taking t1 = ρ′+ (x, y) − ρ′+ (y, x) + p ′ [ρ+ (x, y) − ρ′+ (y, x)]2 + 4kxk2 kyk2 >0 2kyk2 ρ′− (x, y) − ρ′− (y, x) − p ′ [ρ− (x, y) − ρ′− (y, x)]2 + 4kxk2 kyk2 be such that x + ty ⊥M x − ty If we assume (4.2.3), then x + ty ⊥J x − ty, i.e., kxk = tkyk and by (4.2.2) we also have ρ′+ (x, y) = ρ′+ (y, x) for x = or y = 0, which on the account of Theorem 2.1.1 means that X is an inner product space Corollary 4.2.2 Let (X, k · k) be a real normed linear space with dim X ≥ such that ρ′+ = ρ′− Then X is an inner product space if and only if the ⊥M -orthogonality is symmetric Perpendicular Bisectors in Normed Spaces 109 Proof Assume that ⊥M -orthogonality is symmetric For each pair of vectors x and y in X, let t > be such that x + ty ⊥M x − ty, so (4.2.2) is satisfied, i.e., kyk2 t2 + (ρ′+ (y, x) − ρ′+ (x, y))t − kxk2 = Analogously, by the symmetry of ⊥M , x − ty ⊥M x + ty and kyk2 t2 − (ρ′− (y, x) − ρ′− (x, y))t − kxk2 = 0, then from last equations and the equality of ρ′+ and ρ′− , we deduce ρ′+ (x, y) = ρ′+ (y, x), and X is an inner product space The converse is immediate In general, the perpendicular bisector orthogonality is not symmetric; for example, in (R2 , k · k+ ), if x = (1, 2) and y = (−1, 1), then x ⊥M y, but y 6⊥M x Based on the properties of the ⊥M orthogonality and those of the perpendicular bisectors, we obtain a new geometrical characterizations of inner product spaces Theorem 4.2.2 Let (X, k · k) be a real normed linear space with dim X ≥ The following conditions are equivalent: (i) X is an inner product space; (ii) For all x, y in X\{0}, if x ⊥M y, then M (x, x + y) = y + span (x), M (y, x + y) = x + span (y); and (iii) For all u, v in SX , we have, u + v ⊥M u − v; (iv) For all x, y ∈ X\{0}, if for some α in R we have x + αy ⊥M x − αy, then kxk = kαyk (4.2.4) Note that (ii) gives a characterization of the perpendicular bisectors of the rectangle sides (see Figure 4.2.1), (iii) states that in a rhombus the diagonals are orthogonal, and (iv) establishes that if the diagonals of a parallelogram are orthogonal then it is a rhombus 110 Norm Derivatives and Characterizations of Inner Product Spaces M (x, x + y) x+ x y y Figure 4.2.1 Proof If we assume condition (ii), let x, y be two vectors in X\{0} such that x ⊥M y, then x y M (x, x + y) = + span (x), M (y, x + y) = + span (y), 2 i.e., kxk2 = ρ′+ (x + y, x), kyk2 = ρ′+ (x + y, y), then kyk2 = ρ′+ (x + y, y) = kx + yk2 − ρ′− (x + y, x) and therefore, kx + yk2 = kyk2 + ρ′− (x + y, x) ≤ kyk2 + ρ′+ (x + y, x) = kyk2 + kxk2 Furthermore, by (v) of Proposition 4.2.1, x ⊥M −y, and we also obtain kx − yk2 ≤ kyk2 + kxk2 Finally, from last two inequalities, we deduce kx + yk2 + kx − yk2 ≤ 2(kxk2 + kyk2 ), which together with the fact that the ⊥M -orthogonality admits diagonals and by Theorem 1.4.4, we infer that X is an inner product space [Amir (1986); Ben´ıtez and del Rio (1984)] Perpendicular Bisectors in Normed Spaces 111 If we assume condition (iii), then for all u, v in SX if u + v ⊥M u − v, by a straightforward computation ρ′+ (u, v) = ρ′+ (v, u) and X is an i.p.s For condition (iv), given x, y in X\{0}, if ρ′+ (x, y) ≥ ρ′+ (y, x), let p ρ′+ (x, y) − ρ′+ (y, x) + (ρ′+ (y, x) − ρ′+ (x, y))2 + 4kxk2 kyk2 α := 2kyk2 be the unique positive root of (4.2.2) such that x + αy ⊥ x − αy Then, by (4.2.4) α= kxk , kyk and consequently, 4kxkkyk(ρ′+(x, y) − ρ′+ (y, x)) = 0, so ρ′+ (x, y) = ρ′+ (y, x) and X is an i.p.s Corollary 4.2.3 Let (X, k · k) be a real normed linear space with dim X ≥ Then X is an inner product space if and only if for all u, v in SX , u + v ⊥M u − v 4.3 Relations between perpendicular bisectors and classical orthogonalities If (X, h·, ·i) is an i.p.s and x, y are two vectors in X, we can consider the x+y + λu(x, y) with the vector u(x, y) orthogonal perpendicular bisector to x − y Now we show how this property characterizes i.p.s., whenever we consider in a real normed space (X, k · k) the classical orthogonal relations and u := u+ , where u+ is defined by (4.1.1) Proposition 4.3.1 Let (X, k · k) be a strictly convex space with dim X ≥ Then the following conditions are equivalent: (i) (ii) (iii) (iv) (v) X is an inner product space; x − y ⊥ρ u(x, y) for all x, y in X; x − y ⊥ρ u(x, y) for all x, y in X; x − y ⊥P u(x, y) for all x, y in X; x − y ⊥J u(x, y) for all x, y in X; 112 Norm Derivatives and Characterizations of Inner Product Spaces (vi) x − y ⊥B u(x, y) for all x, y in X Proof To prove this theorem, it is only necessary to check that each of properties (ii), (iii), (iv), (v) and (vii) implies (i) If we assume condition (ii), i.e., for all x, y in X ρ′+ (x − y, u(x, y)) = 0, (4.3.1) replacing y by x − z and using Theorem 2.1.1, we obtain for all x, z in X (4.3.2) ρ′+ z, (ρ′+ (x − z, x) − kxk2 )z + F (z, x)x = 0, where F (z, x) := kx − zk2 + ρ′− (x, z) − ρ′+ (x − z, x) However, lim F (z, λx) = kzk2 > λ→0+ (4.3.3) So, for λ > in a neighborhood of zero, F (z, λx) > 0, and replacing x by λx and using Theorem 2.1.1 in (4.3.2), we have (ρ′+ (λx − z, x) − λkxk2 )kzk2 + F (z, λx)ρ′+ (z, x) = Taking limit when λ tends to zero in the last equality, by using Proposition 2.1.6 and (4.3.3), we obtain ρ′+ = ρ′− On the other hand, by the substitutions z := x − y, y := y in (4.3.1), by Theorem 2.1.1 and using ρ′+ = ρ′− , we have for all z, y in X, = (kyk2 − ρ′+ (z + y, y))kzk2 + (kz + yk2 − ρ′+ (z + y, y) − ρ′+ (y, z))ρ′+ (z, y) Now take linearly independent unit vectors w, v in X, and substitute z := w, y := v in the above equality: ρ′+ (w + v, v)(1 + ρ′+ (w, v)) = + kw + vk2 ρ′+ (w, v) − ρ′+ (v, w)ρ′+ (w, v) (4.3.4) By interchanging the roles of w and v in (4.3.1), using Theorem 2.1.1, and the equality ρ′+ (w+v, w) = ρ′+ (w+v, w+v−v) = kw+vk2 −ρ′− (w+v, v) as well as the fact that ρ′+ = ρ′− , we obtain ρ′+ (w + v, v)(1 + ρ′+ (v, w)) = −1 + kw + vk2 + ρ′+ (v, w)ρ′+ (w, v) (4.3.5) By comparing ρ′+ (w + v, v) in (4.3.4) and (4.3.5), removing common factor − ρ′+ (w, v)ρ′+ (v, w) (which is different from zero because X is strictly 113 Perpendicular Bisectors in Normed Spaces convex and w, v are linearly independent) and dividing by the common factor, we obtain kw + vk2 = + ρ′+ (v, w) + ρ′+ (w, v) Replacing v by −v and adding, we get kw + vk2 + kw − vk2 = 4, which on account of Theorem 1.4.3 gives a characterization of an i.p.s To prove that (iii) implies (i), we first use the same argument as that used for orthogonality ⊥ρ in order to prove that ρ′+ = ρ′− (considering F is not necessary) The rest follows from the previous implication If condition (iv) holds, then for all x, y in X ku(x, y)k2 + kx − yk2 = ku(x, y) + x − yk2 , and replacing x by tx and y by ty for all t > 0, using u(tx, ty) = t3 u(x, y), we have t4 ku(x, y)k2 + kx − yk2 = kt2 u(x, y) + x − yk2 Then for all γ > 0, γ ku(x, y)k2 = kγu(x, y)+x−yk2 −kx−yk2, dividing by 2γ and taking limit when γ tends to zero, finally using the definition of ρ′+ , we obtain condition (ii) and X is an i.p.s If we assume condition (v), i.e., for all x, y in X ku(x, y) − x + yk = ku(x, y) + x − yk, replacing x by tx and y by ty for all t > 0, kt2 u(x, y) − x + yk2 = kt2 u(x, y) + x − yk2 , and for all γ > kγu(x, y) − x + yk2 = kγu(x, y) + x − yk2 , and subtracting kx − yk2 from the both sides of the last equality, dividing by 2γ and taking limit when γ tends to zero, using the definition of ρ′± we obtain, ρ′+ (x − y, u(x, y)) + ρ′− (x − y, u(x, y)) = (4.3.6) By the substitutions, z := x − y, x := x in (4.3.6) and using (v) from Theorem 2.1.1 2(ρ′+ (x − z, x) − kxk2 )kzk2 + ρ′+ (z, F (z, x)x) + ρ′− (z, F (z, x)x) = 0, 114 Norm Derivatives and Characterizations of Inner Product Spaces where F (z, x) is defined in (4.3.2) Replacing x by λx, λ > 0, dividing by λ, taking limit when λ tends to zero, and using (4.3.3) and Proposition 2.1.6, we get ρ′+ = ρ′− Then by (4.3.6) ρ′+ (x − y, u(x, y)) = 0, and therefore we have condition (ii) and X is an i.p.s Finally, if condition (vi) holds, then by Proposition 2.1.7 for all x, y in X ρ′− (x − y, u(x, y)) ≤ ≤ ρ′+ (x − y, u(x, y)) (4.3.7) If we first make the substitution z := x−y, x := x and then z := −ty, t > 0, using (iv) from Theorem 2.1.1, ρ′+ (x + ty, x) = kx + tyk2 − tρ′− (x + ty, y) and dividing by t2 , we obtain ρ′− (y, Ft (x, y)x) ≤ (kxk2 − ρ′+ (x + ty, x))kyk2 ≤ ρ′+ (y, Ft (x, y)x), (4.3.8) where Ft (x, y) = ρ′+ (x, y) − ρ′− (x + ty, y) Now we consider two cases: ρ′+ (x, y) = ρ′− (x, y) or ρ′+ (x, y) > ρ′− (x, y) In this last case, by using Proposition 2.1.6 and Corollary 2.1.1, we get lim Ft (x, y) > and for t > in a neighborhood of zero Ft (x, y) > 0, so t→0+ by (4.3.8) and Theorem 2.1.1, Ft (x, y)ρ′− (y, x) ≤ (kxk2 − ρ′+ (x + ty, x))kyk2 ≤ Ft (x, y)ρ′+ (y, x) Taking limit when t tends to zero, we obtain ρ′− (y, x) ≤ ≤ ρ′+ (y, x) Then for all x, y in X, ρ′+ (x, y) = ρ′− (x, y) or ρ′− (y, x) ≤ ≤ ρ′+ (y, x) Now, for all t > 0, if we consider x and x + ty using (iii) and (v) from Theorem 2.1.1, we have ρ′+ (x, y) = ρ′− (x, y) or ρ′− (x + ty, x) ≤ ≤ ρ′+ (x + ty, x) If ρ′+ (x, y) > ρ′− (x, y), for all t > 0, ρ′− (x + ty, x) ≤ ≤ ρ′+ (x + ty, x) and taking limit when t tends to zero, we obtain the contradiction kxk = Then ρ′+ = ρ′− and by (4.3.7) we have ρ′+ (x − y, u(x, y)) = 0, i.e., condition (ii) holds and X is an i.p.s Now we generalize another property of perpendicular bisectors Corollary 4.3.1 Let (X, k ·k) be a strictly convex space with dim X ≥ Then X is an i.p.s if and only if for all x, y in X and for all z in M+ (x, y), we have kx − zk = ky − zk Perpendicular Bisectors in Normed Spaces Proof 115 By hypothesis for all λ > x + y x + y = , + λu(x, + λu(x, y) − x y) − y i.e., x − y ⊥J 2λu(x, y), and if λ = 12 , by our last proposition X is an i.p.s 4.4 On the radius of the circumscribed circumference of a triangle If (X, h·, ·i) is an i.p.s and x, y are two linearly independent vectors in X, in the triangle of sides x, y and x − y, the radius R of the circumscribed circumference is given by the formula kxkkykkx − yk p , s(s − kxk)(s − kyk)(s − kx − yk) where s = (kxk + kyk + kx − yk) /2 is the semiperimeter of the triangle Moreover, another equivalent expression for R is kyk2 kxk2 − hx, yi x + kxk2 kyk2 − hy, xi y , 2kxk2 kyk2 − hx, yi2 − hy, xi2 which we get by finding R as kαx+βyk, and determining α and β by means of the condition that αx+βy− 12 x must be orthogonal to x and αx+βy− 12 y must be orthogonal to y Then, in a real normed space (X, k · k), given two linearly independent vectors x and y, we can define the radius of the circumscribed circumference in the triangle of sides x, y and x − y as: kyk2 kxk2 − ρ′− (x, y) x + kxk2 kyk2 − ρ′− (y, x) y R(x, y) := 2kxk2 kyk2 − ρ′− (x, y)2 − ρ′− (y, x)2 This definition is only possible if ρ′− (x, y)2 + ρ′− (y, x)2 < 2kxk2 kyk2 , i.e., ρ′− (x, y) < kxkkyk or ρ′− (y, x) < kxkkyk For this reason we assume that X is strictly convex Theorem 4.4.1 Let (X, k · k) be a strictly convex space with dim X ≥ Then X is an i.p.s if and only if for all linearly independent vectors x, y 116 Norm Derivatives and Characterizations of Inner Product Spaces in X, kxkkykkx − yk , R(x, y) = p s(s − kxk)(s − kyk)(s − kx − yk) where s = (kxk + kyk + kx − yk) /2 Proof If we let y := tz with t > and we take limit when t tends to zero, we obtain kxk2 kzk2x − ρ′− (z, x)z = lim R(x, tz) 2kxk2 kzk2 − ρ′− (x, z)2 − ρ′− (z, x)2 t→0+ 1/2 kxk2 t2 kzk2kx − tzk2 = lim t→0+ [kx − tzk2 − (tkzk − kxk)2 ] [(kxk + tkzk2 ) − kx − tzk2 ] 1/2 tkx − tzk = kxkkzk lim t→0+ kx − tzk2 − (tkzk − kxk)2 !1/2 tkx − tzk × lim+ t→0 (kxk + tkzk) − kx − tzk2 s s kxk kxk = kxkkzk −2ρ′− (x, z) + 2kxkkzk 2kxkkzk + 2ρ′− (x, z) kzkkxk2 , = p kxk2 kzk2 − ρ′− (x, z)2 and therefore, 2 2 − ρ′ x z z x , kxk − ρ′− kzk x − kxk , kzk x z z ′ r kxk − ρ− kzk , kxk kzk = 2 x z ′ − ρ− kxk , kzk If u := v in SX , x z , v := , then for all linearly independent vectors u and kxk kzk ′ (u, v)2 − ρ′− (v, u)2 u − ρ′− (v, u)v = − ρ− p − ρ′− (u, v)2 Now, if ρ′− (v, u) = 0, then q − ρ′− (u, v)2 = − ρ′− (u, v)2 Therefore, ρ′− (u, v) = 0, and consequently, (see (19.6) in [Amir (1986)]) X is an i.p.s Perpendicular Bisectors in Normed Spaces 117 Analogous definitions of R(x, y) can be given by replacing the role of ρ′− by ρ′+ or by changing the order of the arguments appearing in ρ′− For example, assume that the radius of the circumscribed circumference is given by kyk2 kxk2 − ρ′ (x, y) x + kxk2 kyk2 − ρ′ (y, x) y + + ˆ y) := R(x, kxk2 kyk2 − ρ′− (x, y)2 (which is equal to R(x, y) in an i.p.s.) Then a strictly convex space X with dim X ≥ is an i.p.s if and only if kxkkykkx − yk ˆ y) = p R(x, s(s − kxk)(s − kyk)(s − kx − yk) (4.4.1) The proof is immediate and uses the fact that the right-hand side ˆ y) are symmetric, which implies of (4.4.1) and the numerator of R(x, ρ′− (x, y)2 = ρ′− (y, x)2 for all x, y in X and X is an i.p.s 4.5 Circumcenters in a triangle Let (X, k·k) be a real normed linear space with dim X ≥ In the following, when speaking about points we will always refer to vectors with the initial point at the origin Then, consider the triangle ∆xyz with sides x − y, y − z, x − z, where x, y, z belong to X, x, y are linearly independent and z is in the plane determined by x and y (see Figure 4.5.1) → z → y △xyz → x Figure 4.5.1 In an i.p.s., the perpendicular bisectors of the three sides of a triangle all pass through the circumcenter, which is the center of the circumscribed circle (see [Coxeter (1969)], p 12-13 and [Puig-Adam (1986)], p 92) Then, in a natural way we have the following 118 Norm Derivatives and Characterizations of Inner Product Spaces Definition 4.5.1 The triangle ∆xyz has a circumcenter C if and only if M+ (x, y), M+ (x, z) and M+ (y, z) meet at the point C First we consider the class of triangles ∆xy(x + y), where we can show the following result: Theorem 4.5.1 Let (X, k · k) be a strictly convex space with dim X ≥ Then X is an i.p.s if and only if there exists the circumcenter of ∆xy(x+y) for all linearly independent vectors x, y in X Proof If we consider ∆xy(x + y), using the linear independence of x and y, it is a straightforward computation to prove that the three straight lines M+ (x, y), M+ (x, x + y) and M+ (y, x + y) meet at a point if and only if the following equality holds (kyk2 − ρ′+ (x, y))(kxk2 − ρ′+ (x + y, x))(ky + xk2 − kyk2 − ρ′+ (y, x)) = (kyk2 − ρ′+ (x + y, y)) (ky + xk2 − kxk2 − ρ′+ (x, y))(kxk2 − ρ′+ (y, x)) (4.5.1) + 2(kxk2 − ρ′+ (x + y, x))(kxk2 − ρ′+ (y, x) − kyk2 + ρ′+ (x, y) If we replace x by λx (λ > 0), then by using the properties of ρ′+ , simplifying λ; dividing by 2λ the term ky+λxk2 −kyk2 −λρ′+ (y, x) in the left part of the last equality and dividing also by 2λ the term ky||2 −ρ′+ (λx+y, y) in the right part of last equality; using the equality ρ′+ (λx + y, y) = kλx + yk2 − λρ′− (λx + y, x), and finally by taking limit when λ tends to zero, by Proposition 2.1.6, Corollary 2.1.1 and the definition of ρ′+ , we obtain ρ′+ (y, x) = ρ′− (y, x) or ρ′+ (y, x) = Replacing x by −x, we have ρ′+ (y, x) = ρ′− (y, x) or ρ′− (y, x) = 0, and combining the four possible cases, we obtain ρ′+ = ρ′− Moreover, by substituting x := u, y := v−u in (4.5.1), using the equality ρ′+ = ρ′− and the properties of ρ′+ , operating and grouping in a suitable way, we get −kuk2 − kvk2 + ku − vk2 + 2ρ′+ (v, u) = ρ′+ (v − u, u)(kuk2 + kvk2 − ρ′+ (u, v) − ρ′+ (v, u)) = ku − vk2 (−kuk2 + ρ′+ (v, u)) + kuk2 kvk2 − ρ′+ (u, v)ρ′+ (v, u) By symmetry and equalities ρ′+ = ρ′− and ρ′+ (u − v, v) = −kv − uk2 − ρ′+ (v − u, u), (4.5.2) Perpendicular Bisectors in Normed Spaces 119 we have −kuk2 − kvk2 + ku − vk2 + 2ρ′+ (u, v) = 0, or ρ′+ (v − u, u)(kuk2 + kvk2 − ρ′+ (u, v) − ρ′+ (v, u)) = ku − vk2 (−kuk2 + ρ′+ (v, u)) − kuk2 kvk2 + ρ′+ (v, u)ρ′+ (u, v) (4.5.3) If (4.5.2) and (4.5.3) hold, then ρ′+ (u, v)ρ′+ (v, u) = kuk2 kvk2 , but by hypothesis, X is strictly convex and we obtain a contradiction (cf Theorem 2.1.1 (vii)) Then, for all u, v in X kuk2 + kvk2 − ku − vk2 ′ ∈ ρ+ (u, v), ρ′+ (v, u) , and changing v for −v and using ρ′+ = ρ′− , ku + vk2 − kuk2 − kvk2 ∈ {ρ′+ (u, v), ρ′+ (v, u)} Then for all t > 0, the function defined by f (t) := ku + tvk2 − kuk2 − t2 kvk2 2t is continuous in (0, +∞) and f (t) ∈ {ρ′+ (u, v), ρ′+ (v, u)} for all t > 0, lim f (t) = ρ′+ (u, v) Then ρ′+ (u, v) = ρ′+ (v, u) for all u, v in X, and t→0+ by Theorem 2.1.1 (x), X is an inner product space Note that Theorem 4.5.1 covers a general case because ∆xy(x + y) is equivalent to the triangle determined by x and y (i.e., with sides x, y and x − y) (see Figure 4.5.2) → y → x + → y △xy(x + y) → x Figure 4.5.2 120 Norm Derivatives and Characterizations of Inner Product Spaces Then, with a similar proof to that of the last theorem, we have the following Corollary 4.5.1 Let (X, k · k) be a strictly convex space with dim X ≥ such that there exists the circumcenter of ∆xy(αx + y) for all linearly independent vectors x, y in X and α in R fixed Then (X, k · k) is smooth In a real normed linear space (X, k · k) we consider the triangle ∆xyz with sides x − y, y − z and x − z, where x and y are linearly independent and z = z1 x + z2 y (z1 , z2 in R) is in the plane determined by x and y Then by a straightforward computation, we can prove that the intersections of the three perpendicular bisectors M+ (x, y), M+ (x, z) and M+ (y, z) give us three points A(z2 − 1) − Bz1 x+z + (Cx + Dz) C1 = 2(CB + DBz1 − ADz2 ) E(1 − z1 ) + F z2 x+y + C2 = (Ax + By) 2(BF z1 − AE − AF z2 ) D(z1 + z2 ) + C y+z C3 = + (Ey + F z), 2(−CE − CF z2 − DEz1 ) (4.5.4) where A = kyk2 − ρ′+ (x, y), B = kxk2 − ρ′+ (y, x), C = kzk2 − ρ′+ (x, z), D = kxk2 − ρ′+ (z, x), E = kzk2 − ρ′+ (y, z), F = kyk2 − ρ′+ (z, y), and by Theorem 4.5.1, in a real normed space, the three points are, in general, different Definition 4.5.2 The points C1 , C2 , C3 defined in (4.5.4) are called the circumcenter points of ∆xyz We consider two classical properties concerning the circumcenter of a triangle in the Euclidean plane, and we translate these properties into a real normed linear space considering the circumcenter point C1 to obtain new characterizations of inner product spaces Theorem 4.5.2 Let (X, k · k) be a strictly convex space with dim X ≥ and let x, y, z be vectors in X with kxk = kyk = kzk Then X is an inner product space if and only if the circumcenter point C1 of ∆xyz is the origin Perpendicular Bisectors in Normed Spaces 121 Proof The direct part is a well-known result Reciprocally, if for all x, y, z in X with kxk = kyk = kzk the circumcenter point C1 of ∆xyz is the origin, then in particular it is true for unit vectors such that x and y are linearly independent and z = z1 x + z2 y with z1 , z2 in R, z2 6= In such a case, using (4.5.4) and the linear independence of x and y, we infer that the system + z1 + z2 + A(z2 − 1) − Bz1 (C + Dz1 ) = 0, CB + DBz1 − ADz2 A(z2 − 1) − Bz1 − Dz2 = 0, CB + DBz1 − ADz2 where A, B, C and D are defined after (4.5.4) gives C = D, and therefore ρ′+ (x, z) = ρ′+ (z, x) for all x, z in SX As a consequence, by Theorem 2.1.1, (x), is an inner product space Theorem 4.5.3 Let (X, k · k) be a strictly convex space with dim X ≥ Then X is an inner product space if and only if the circumcenter point C1 of ∆xy(λx + y) is (x + y) kxk2 λ + ρ′+ (x, y) + ρ′+ (y, x) 2kxk2 + ρ′+ (x, y) + ρ′+ (y, x) whenever x, y are linearly independent vectors in X with kxk = kyk and λ belongs to R Proof The direct part of the statement is just a verification Reciprocally, consider ∆xy(λx + y), where x and y are unit vectors and assume C1 = kxk2 λ + ρ′+ (x, y) + ρ′+ (y, x) Then, using expression (4.5.4) of C1 (x + y) 2kxk2 + ρ′+ (x, y) + ρ′+ (y, x) (where z1 = λ and z2 = 1) and the linear independence of x and y, we have λ + ρ′+ (x, y) + ρ′+ (y, x) Bλ 1+λ − (C + Dλ) = 2(CB + DBλ − AD) + ρ′+ (x, y) + ρ′+ (y, x) Bλ = − D 2(CB + DBλ − AD) By the last equalities, with a long and tedious computation we can prove that ρ′+ (x, y) = ρ′+ (y, x) (ρ′+ (x, y) + ρ′+ (y, x) = 2(1 − λ) and or = ρ′+ (λx + y, x)) 122 Norm Derivatives and Characterizations of Inner Product Spaces Now, we claim that |ρ′+ (x, y)| = |ρ′+ (y, x)| for all x, y in SX Let x and y be two unit linearly independent vectors in X If λ = 1, the result is evident Assume that λ 6= If ρ′+ (x, y) + ρ′+ (y, x) = 2(1 − λ) we consider the vectors x and and = ρ′+ (λx + y, x), (4.5.5) λx + y , then we have two cases: kλx + yk ρ′+ (λx + y, x) = ρ′+ (x, λx + y) or ρ′+ (x, λx + y) + ρ′+ (λx + y, x) = 2(1 − λ)kλx + yk and = ρ′+ λx + y ,x λx + kλx + yk In the first case by Theorem 2.1.1 and (4.5.5), = λ + ρ′+ (x, y) and = − λ = ρ′+ (y, x) ρ′+ (x, y) λx + In the second case, we take the vectors x and λx + the process λx+y kλx+yk and repeat λx+y kλx+yk Summarizing, if we consider the sequence (bn ) defined by b1 := λx + bn−1 for all n ≥ 2, we have to bear in mind two y and bn := λx + kbn−1 k possibilities Possibility There exists n in N such that ρ′+ (x, y) + ρ′+ (y, x) = 2(1 − λ), = ρ′+ (b1 , x), ρ′+ (x, bk ) + = 2(1 − λ)kbk k, = ρ′+ (bk+1 , x) for ≤ k ≤ n − and ρ′+ (bn , x) = ρ′+ (x, bn ) We will prove, by induction, that we can infer ρ′+ (x, y) = ρ′+ (y, x) = − λ For n = we have proved that it is true If it is true for n − 1, we wish to prove that it is true for n = Using Theorem 2.1.1, we have = ρ′+ (x, bn ) = ρ′+ x, λx + kbbn−1 n−1 k ρ′+ (x,bn−1 ) ρ′ (x,b ) and kbn−1 k = + 1−λn−1 kbn−1 k So, ρ′+ (x, bn−1 ) + = 2ρ′+ (x, bn−1 ) and ρ′+ (x, bn−1 ) = and by induction hypothesis, ρ′+ (x, y) = ρ′+ (y, x) = − λ λ+ = ρ′+ (bn−1 , x), ... kxk2 = (4 .2. 2) So taking t1 = ρ′+ (x, y) − ρ′+ (y, x) + p ′ [ρ+ (x, y) − ρ′+ (y, x) ]2 + 4kxk2 kyk2 >0 2kyk2 ρ′− (x, y) − ρ′− (y, x) − p ′ [ρ− (x, y) − ρ′− (y, x) ]2 + 4kxk2 kyk2 and we take limit when t tends to zero, we obtain kxk2 kzk2x − ρ′− (z, x)z = lim R(x, tz) 2kxk2 kzk2 − ρ′− (x, z )2 − ρ′− (z, x )2 t→0+ 1 /2 kxk2 t2 kzk2kx... Theorem 2. 1.1 2( ρ′+ (x − z, x) − kxk2 )kzk2 + ρ′+ (z, F (z, x)x) + ρ′− (z, F (z, x)x) = 0, 114 Norm Derivatives and Characterizations of Inner Product Spaces where F (z, x) is defined in (4.3 .2) Replacing