Ann. Funct. Anal. 2 (2011), no. 2, 22–33 A nnals of F unctional A nalysis ISSN: 2008-8752 (electronic) URL: www.emis.de/journals/AFA/ SOME GEOMETRIC CONSTANTS OF ABSOLUTE NORMALIZED NORMS ON R2 HIROYASU MIZUGUCHI AND KICHI-SUKE SAITO∗ Communicated by J. I. Fujii Abstract. We consider the Banach space X = (R2 , k · k) with a normalized, absolute norm. Our aim in this paper is to calculate the modified Neumann0 Jordan constant CN J (X) and the Zbăganu constant CZ (X). 1. Introduction and preliminaries Let X be a Banach space with the unit ball BX = {x ∈ X : kxk ≤ 1} and the unit sphere SX = {x ∈ X : kxk = 1}. Many geometric constants for a Banach space X have been investigated. In this paper we shall consider the following constants; kx + yk2 + kx − yk2 CN J (X) = sup (x, y) 6= (0, 0) , 2(kxk2 + kyk2 ) kx + yk2 + kx − yk2 0 CN J (X) = sup x, y ∈ SX , 4 kx + ykkx − yk CZ (X) = sup x, y ∈ X, (x, y) 6= (0, 0) . kxk2 + kyk2 The constant CN J (X), called the von Neumann-Jordan constant (hereafter referred to as NJ constant) have been considered in many papers ([3, 8, 10, 12] and so on). The constant CN0 J (X), called the modified von Neumann-Jordan constant (shortly, modified NJ constant) was introduced by Gao in [5] and does not necessarily coincide with CN J (X) (cf. [1, 4, 7]). The constant CZ (X) was introduced by Zbăganu ([15]) and was conjectured that CZ (X) coincides with Date: Received: 31 March 2011; Accepted: 14 June 2011. ∗ Corresponding author. 2010 Mathematics Subject Classification. Primary 46B20; Secondary 46B25. Key words and phrases. Zbăganu constant, absolute norm, von Neumann-Jordan constant. 22 GEOMETRIC CONSTANTS OF R2 23 the von Neumann-Jordan constant CN J (X), but Alonso and Martin [2] gave an example that CN J (X) 6= CZ (X) (cf.[6, 9]). A norm k · k on R2 is said to be absolute if k(a, b)k = k(|a|, |b|)k for any (a, b) ∈ R2 , and normalized if k(1, 0)k = k(0, 1)k = 1. Let AN2 denote the family of all absolute normalized norm on R2 , and Ψ2 denote the family of all continuous convex function ψ on [0, 1] such that ψ(0) = ψ(1) = 1 and max{1 − t, t} ≤ ψ(t) ≤ 1 for all 0 ≤ t ≤ 1. As in [11], it is well known that AN2 and Ψ2 are in a one-toone correspondence under the equation ψ(t) = k(1 − t, t)k (0 ≤ t ≤ 1). Denote k · kψ be an absolute normalized norm associated with a convex function ψ ∈ Ψ2 . For ψ, ϕ ∈ Ψ2 , we denote ψ ≤ ϕ if ψ(t) ≤ ϕ(t) for any t in [0, 1]. Let ψ(t) ψ2 (t) and M2 = max , 0≤t≤1 ψ2 (t) ψ(t) p where ψ2 (t) = k(1 − t, t)k2 = (1 − t)2 + t2 corresponds to the l2 -norm. In [11], Saito, Kato and Takahashi proved that, if ψ ≥ ψ2 (resp. ψ ≤ ψ2 ), then CN J (C2 , k · kψ ) = M12 (resp. M22 ). We put X = (R2 , k · kψ ) for ψ ∈ Ψ2 . Our aim in this paper is to consider the conditions of ψ that CN J (X) = CZ (X) or CN J (X) = CN0 J (X). In §2, we consider the modified von Neumann-Jordan constant. We prove that if ψ ≤ ψ2 , then CN0 J (X) = CN J (X) = M22 . If ψ ≥ ψ2 , then we present the necessarily and sufficient condition that CN0 J (R2 , k · kψ ) = CN J (R2 , k · kψ ) = M12 . Further, we consider the conditions that CN0 J (R2 , k · kψ ) = CN J (R2 , k · kψ ) = M12 M22 . In §3, we study the Zbăganu constant. First, we show that, if ψ ≥ ψ2 , then CZ (R2 , k · kψ ) = CN J (R2 , k · kψ ) = M12 . If ψ ≤ ψ2 , then we give the necessarily and sufficient condition for that CZ (R2 , k · kψ ) = CN J (R2 , k · kψ ) = M22 . Further we study the conditions that CZ (R2 , k · kψ ) = CN J (R2 , k · kψ ) = M12 M22 . In §4, we calculate the modified NJ-constant CN0 J (X) and the Zbăganu constant CZ (X) for some normed liner spaces. M1 = max 0≤t≤1 2. The modified NJ constant of R2 In this section, we consider the Banach space X = (R2 , k · kψ ). From the definition of the modified NJ constant, it is clear that CN0 J (X) ≤ CN J (X). In this section, we consider the condition that CN0 J (X) = CN J (X). Proposition 2.1. Let ψ ∈ Ψ2 . If ψ ≤ ψ2 , then CN0 J (X) = CN J (X) = M22 . Proof. For any x, y ∈ SX , by [11, Lemma 3], kx + yk2ψ + kx − yk2ψ ≤ kx + yk22 + kx − yk22 = 2 kxk22 + kyk22 ≤ 2M22 kxk2ψ + kyk2ψ = 4M22 . Now let ψ2 /ψ attain the maximum at t = t0 (0 ≤ t0 ≤ 1), and put x= 1 1 (1 − t0 , t0 ), y = (1 − t0 , −t0 ). ψ(t0 ) ψ(t0 ) 24 H. MIZUGUCHI, K.-S. SAITO Then x, y ∈ SX and 4(1 − t0 )2 + 4t20 ψ(t0 )2 ψ2 (t0 )2 =4 = 4M22 , 2 ψ(t0 ) kx + yk2ψ + kx − yk2ψ = which implies that CN0 J (X) = M22 . By [11, Theorem 1], we have this proposition. If ψ ≥ ψ2 , by [11, Theorem 1], then CN J (X) = M12 . We now give the necessarily and sufficient condition of CN0 J (X) = M12 . Theorem 2.2. Let ψ ∈ Ψ2 such that ψ ≥ ψ2 . Then CN0 J (X) = M12 if and only if there exist s, t ∈ [0, 1] (s < t) satisfying one of the following conditions: (1) ψ(s) = ψ2 (s), ψ(t) = ψ2 (t) and, if we put r = ψ(s)t+ψ(t)s , then ψψ(r) = ψ(s)+ψ(t) 2 (r) ψ(1−r) ψ2 (1−r) = M1 . (2) ψ(s) = ψ2 (s), ψ(t) = ψ2 (t) and, if we put r = ψ(1−r) ψ2 (1−r) ψ(t)s+ψ(s)t , ψ(t)+ψ(s)(2t−1) then ψ(r) ψ2 (r) = = M1 . Proof. (=⇒) Suppose that CN0 J (X) = M12 . First, for any x, y ∈ SX , by [11, Lemma 3], we have kx + yk2ψ + kx − yk2ψ ≤ M12 (kx + yk22 + kx − yk22 ) = 2M12 kxk22 + kyk22 ≤ 2M12 kxk2ψ + kyk2ψ = 4M12 . Since X = (R2 , k · kψ ) is finite dimensional, kx + yk2ψ + kx − yk2ψ 0 CN J (X) = max 4 x, y ∈ SX . Therefore, CN0 J (X) = M12 if and only if there exist x, y ∈ SX (x 6= y) such that kx + yk2ψ + kx − yk2ψ = 4M12 . From the above inequality, the elements x, y ∈ SX (x 6= y) satisfy kxkψ = kxk2 = 1, kykψ = kyk2 = 1 and kx + ykψ kx − ykψ = = M1 . kx + yk2 kx − yk2 Since k · kψ is absolute and x, y ∈ SX (x 6= y) satisfy kxk2 = kyk2 = 1, it is sufficient to consider the following three cases: (i) There exist s, t ∈ [0, 1] (s 6= t) satisfying x = ψ21(s) (1 − s, s) and y = 1 (1 − t, t). ψ2 (t) (ii) There exist s, t ∈ [0, 1] (s < t) satisfying x = ψ21(s) (1 − s, s) and y = 1 (−1 + t, t). ψ2 (t) GEOMETRIC CONSTANTS OF R2 25 (iii) There exist s, t ∈ [0, 1] (s > t) satisfying x = ψ21(s) (1 − s, s) and y = 1 (−1 + t, t). ψ2 (t) Case (i). We may suppose that s < t. Then there exist α, β ∈ [0, π2 ] (α < β) such that 1 1 x= (1 − s, s) = (cos α, sin α), y = (1 − t, t) = (cos β, sin β). ψ2 (s) ψ2 (t) Since kxk2 = kyk2 = 1, we have x+y =( s t α+β α+β 1−s 1−t + , + ) = ||x + y||2 (cos , sin ). ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t) 2 2 By [13, Propositions 2a and 2b], we remark that 1−s 1−t s t ≥ , ≤ . ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t) Since x − y is orthogonal to x + y in the Euclidean space (R2 , k · k2 ), we have 1−s 1−t s t x−y =( − , − ) ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t) α+β−π α+β−π = ||x − y||2 (cos , sin ) 2 2 α+β α+β = ||x − y||2 (sin , − cos ). 2 2 Thus we have α+β α+β , sin )||ψ ||x + y||ψ = ||x + y||2 ||(cos 2 2 sin α+β α+β α+β 2 = ||x + y||2 (cos + sin )ψ( ). 2 2 cos α+β + sin α+β 2 2 Since ||x + y||ψ = M1 ||x + y||2 , we have sin α+β α+β α+β 2 M1 = (cos + sin )ψ( ). α+β 2 2 cos 2 + sin α+β 2 Putting r = We also have cos sin α+β 2 α+β +sin α+β 2 2 , then it is clear that r = ||x − y||ψ = ||x − y||2 (sin ψ(s)t+ψ(t)s ψ(s)+ψ(t) and M1 = ψ(r) . ψ2 (r) cos α+β α+β α+β 2 + cos )ψ( ). α+β 2 2 cos 2 + sin α+β 2 Since ||x − y||ψ = M1 ||x − y||2 , we similarly have M1 = (sin cos α+β α+β α+β ψ(1 − r) 2 + cos )ψ( )= . α+β α+β 2 2 ψ2 (1 − r) sin 2 + cos 2 Case (ii). Then there exist α ∈ [0, π2 ] and β ∈ [ π2 , π] such that x= 1 1 (1 − s, s) = (cos α, sin α), y = (−1 + t, t) = (cos β, sin β). ψ2 (s) ψ2 (t) 26 H. MIZUGUCHI, K.-S. SAITO Since kxk2 = kyk2 = 1, we have x+y =( 1−s 1−t s t α+β α+β − , + ) = ||x + y||2 (cos , sin ). ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t) 2 2 By [13, Propositions 2a and 2b], we remark that 1−t s t 1−s ≥ , ≤ . ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t) Since x − y is orthogonal to x + y in the Euclidean space (R2 , k · k2 ), we have 1−s 1−t s t x−y =( + , − ) ψ2 (s) ψ2 (t) ψ2 (s) ψ2 (t) α+β−π α+β−π = ||x − y||2 (cos , sin ) 2 2 α+β α+β = ||x − y||2 (sin , − cos ). 2 2 Since cos α+β ≥ 0 and sin α+β ≥ 0, we have 2 2 α+β α+β , sin )||ψ 2 2 sin α+β α+β α+β 2 = ||x + y||2 (cos + sin )ψ( ). 2 2 cos α+β + sin α+β 2 2 ||x + y||ψ = ||x + y||2 ||(cos Since ||x + y||ψ = M1 ||x + y||2 , we have sin α+β α+β α+β 2 + sin )ψ( M1 = (cos ). α+β 2 2 cos 2 + sin α+β 2 Putting r = cos We also have sin α+β 2 α+β α+β +sin 2 2 , then it is clear that r = ||x − y||ψ = ||x − y||2 (sin ψ(t)s+ψ(s)t ψ(t)+ψ(s)(2t−1) and M1 = ψ(r) . ψ2 (r) cos α+β α+β α+β 2 ). + cos )ψ( α+β 2 2 cos 2 + sin α+β 2 Since ||x − y||ψ = M1 ||x − y||2 , we similarly have M1 = (sin cos α+β α+β α+β ψ(1 − r) 2 + cos )ψ( )= . α+β α+β 2 2 ψ2 (1 − r) sin 2 + cos 2 Case (iii). There exist s, t ∈ [0, 1] (s > t) satisfying x = ψ21(s) (1 − s, s) and y = ψ21(t) (−1 + t, t). Then, we put s0 = t and t0 = s. We define x0 , y0 in SX by x0 = 1 1 (1 − s0 , s0 ), y0 = (−1 + t0 , t0 ). ψ(s0 ) ψ(t0 ) Then we can reduce Case (ii). (⇐=). If we suppose (1) (resp. (2)), then we put x = ψ21(s) (1 − s, s) (resp. x = ψ21(s) (1 − s, s)) and y = ψ21(t) (1 − t, t) (resp. y = ψ21(t) (−1 + t, t)). Then we have ||x||ψ = ||x||2 = 1, ||y||ψ = ||y||2 = 1, ||x + y||ψ = M1 ||x + y||2 and GEOMETRIC CONSTANTS OF R2 27 ||x − y||ψ = M1 ||x − y||2 . Hence it is clear to prove that CN0 J (X) = M12 . This completes the proof. We next study the modified NJ constant in the general case. If ψ ∈ Ψ, then by [11, Therem 3], we have max{M12 , M22 } ≤ CN J (X) ≤ M12 M22 . However, by Theorem 2.2, there exist many ψ ∈ Ψ satisfying ψ ≥ ψ2 such that CN0 J (X) < max{M12 , M22 } = CN J (X). From [11, Theorem 3], CN J (X) = M12 M22 if either ψ/ψ2 or ψ2 /ψ attains a maximum at t = 1/2. Then, we have the following Proposition 2.3. Let ψ ∈ Ψ2 and let ψ(t) = ψ(1 − t) for all t ∈ [0, 1]. If ψ/ψ2 attains a maximum at t = 1/2, then CN0 J (X) = CN J (X) = M12 M22 . Proof. Suppose first M1 = ψ(1/2)/ψ2 (1/2). Take an arbitrary t ∈ [0, 1] and put x= 1 1 (t, 1 − t) , y = (1 − t, t). ψ(t) ψ(t) Then x, y ∈ SX and kx + ykψ = 2 1 2|2t − 1| 1 ψ( ) , kx − ykψ = ψ( ). ψ(t) 2 ψ(t) 2 Therefore we have ψ(1/2)2 kx + yk2ψ + kx − yk2ψ = (2t − 1)2 + 1 4 ψ(t)2 2 2 ψ(1/2) = 2ψ2 (t) ψ(t)2 2 ψ2 (t)2 ψ(1/2)2 2 ψ2 (t) = = M . 1 ψ(t)2 ψ2 (1/2)2 ψ(t)2 Since t is arbitrary, we have CN0 J (X) ≥ M12 M22 which prove that CN0 J (X) = M12 M22 . In the case that M2 = ψ2 (1/2)/ψ(1/2), CN0 J (X) does not necessarily coincide with M12 M22 . However, we have the following Theorem 2.4. Let ψ ∈ Ψ2 and let ψ(t) = ψ(1 − t) for all t ∈ [0, 1]. Assume that M2 = ψ2 (1/2)/ψ(1/2) and M1 > 1. Then CN0 J (X) = M12 M22 if and only if there exist s, t ∈ [0, 1] (s < t) satisfying one of the following conditions: , then (1) ψ2 (s) = M2 ψ(s), ψ2 (t) = M2 ψ(t) and, if we put r = ψ(s)t+ψ(t)s ψ(s)+ψ(t) ψ(r) = M1 ψ2 (r). ψ(t)s+ψ(s)t , then (2) ψ2 (s) = M2 ψ(s), ψ2 (t) = M2 ψ(t) and, if we put r = ψ(t)+ψ(s)(2t−1) ψ(r) = M1 ψ2 (r). 28 H. MIZUGUCHI, K.-S. SAITO Proof. (=⇒). For all x, y ∈ SX , we have kx + yk2ψ + kx − yk2ψ ≤ M12 kx + yk22 + kx − yk22 = 2M12 kxk22 + kyk22 ≤ 2M12 M22 kxk2ψ + kyk2ψ = 4M12 M22 . From this inequality, CN0 J (X) = M12 M22 if and only if there exist x, y ∈ SX (x 6= y) such that kx + yk2ψ + kx − yk2ψ = 4M12 M22 . Suppose that CN0 J (X) = M12 M22 . Then, the elements x, y ∈ SX (x 6= y) satisfy kxk2 = kyk2 = M2 , kx + ykψ = M1 kx + yk2 , kx − ykψ = M1 kx − yk2 . Since k · kψ is absolute, it is sufficient to consider the following three cases: 1 (1 − s, s) and y = (i) There exist s, t ∈ [0, 1] (s 6= t) satisfying x = ψ(s) 1 (1 − t, t). ψ(t) 1 (ii) There exist s, t ∈ [0, 1] (s < t) satisfying x = ψ(s) (1 − s, s) and y = 1 (−1 + t, t). ψ(t) 1 (iii) There exist s, t ∈ [0, 1] (s > t) satisfying x = ψ(s) (1 − s, s) and y = 1 (−1 + t, t). ψ(t) As in the proof of Theorem 2.2, we can prove this theorem. This completes the proof. 3. The Zbăganu constant of R2 The Zbăganu constant CZ (X) in [15] is defined by kx + ykkx − yk CZ (X) = sup x, y ∈ X, (x, y) 6= (0, 0) . kxk2 + kyk2 Then it is clear that CZ (X) ≤ CN J (X) for any Banach space X. In this section, we consider the condition that CZ (X) = CN J (X) for X = (R2 , k · kψ ). Then, we have the following Proposition 3.1. Let ψ ∈ Ψ2 . If ψ ≥ ψ2 , then CZ (X) = CN J (X) = M12 . Proof. For any x, y ∈ X, 2kx + ykψ kx − ykψ ≤ kx + yk2ψ + kx − yk2ψ ≤ M12 kx + yk22 + kx − yk22 = 2M12 kxk22 + kyk22 ≤ 2M12 kxk2ψ + kyk2ψ . Since ψ/ψ2 attains the maximum at t = t0 (0 ≤ t0 ≤ 1), we put x = (1 − t0 , 0) and y = (0, t0 ), respectively. Then we have kx + yk2ψ + kx − yk2ψ = 2ψ(t0 )2 = 2M12 ψ2 (t0 )2 = 2M12 kxk2ψ + kyk2ψ . GEOMETRIC CONSTANTS OF R2 29 Since kx + ykψ = ψ(t0 ) = kx − ykψ , we have 2kx + ykψ kx − ykψ = kx + yk2ψ + kx − yk2ψ = 2M12 kxk2ψ + kyk2ψ . Therefore we have kx + ykψ kx − ykψ = M12 , kxk2ψ + kyk2ψ which implies that CZ (X) = M12 . We next consider the case that ψ ≤ ψ2 . We remark that the Zbăganu constant CZ (X) is in the following form; 4kxkkyk CZ (X) = sup x, y ∈ X, (x, y) 6= (0, 0) . kx + yk2 + kx − yk2 Then we have the following Theorem 3.2. Let ψ ∈ Ψ2 . Assume that ψ ≤ ψ2 . Then CZ (X) = M22 if and only if there exist s, t ∈ [0, 1] (s < t) satisfying one of the following conditions: 2 (r) (1) ψ(s) = ψ2 (s), ψ(t) = ψ2 (t) and, if we put r = ψ(s)t+ψ(t)s , then ψψ(r) = ψ(s)+ψ(t) ψ(1−r) ψ2 (1−r) = M2 . (2) ψ(s) = ψ2 (s), ψ(t) = ψ2 (t) and, if we put r = ψ(1−r) ψ2 (1−r) ψ(t)s+ψ(s)t , ψ(t)+ψ(s)(2t−1) then = M2 . Proof. For any x, y ∈ X, 4kxkψ kykψ ≤ 2 kxk2ψ + kyk2ψ ≤ 2 kxk22 + kyk22 = kx + yk22 + kx − yk22 ≤ M22 kx + yk2ψ + kx − yk2ψ . Since X = (R2 , k · kψ ) is finite dimensional, ( 4kxkψ kykψ CZ (X) = max kx + yk2ψ + kx − yk2ψ ) x, y ∈ X, (x, y) 6= (0, 0) . Then CZ (X) = M22 if and only if there exist x, y ∈ SX (x 6= y) such that 4kxkψ kykψ = M22 . 2 2 kx + ykψ + kx − ykψ From the above inequality, kxk2 = kxkψ = kykψ = kyk2 and kx + yk2 kx − yk2 = = M22 . kx + ykψ kx − ykψ Hence we may assume that kxk2 = kxkψ = kykψ = kyk2 = 1. ψ2 (r) ψ(r) = 30 H. MIZUGUCHI, K.-S. SAITO As in the proof of Theorem 2.2, it is sufficient to consider the following three cases: (i) There exist s, t ∈ [0, 1] (s 6= t) satisfying x = ψ21(s) (1 − s, s) and y = 1 (1 − t, t). ψ2 (t) (ii) There exist s, t ∈ [0, 1] (s < t) satisfying x = ψ21(s) (1 − s, s) and y = 1 (−1 + t, t). ψ2 (t) (iii) There exist s, t ∈ [0, 1] (s > t) satisfying x = ψ21(s) (1 − s, s) and y = 1 (−1 + t, t). ψ2 (t) As in the proof of Theorem 2.2, we can similarly prove this theorem. We next study the Zbăganu constant CZ (X) in general case. If ψ ∈ Ψ, by [11, Theorem 3], then we have max{M12 , M22 } ≤ CZ (X) ≤ CN J (X) ≤ M12 M22 . However, by Theorem 3.2, there exist many ψ ∈ Ψ satisfying ψ ≥ ψ2 such that CZ (X) < CN J (X) ≤ max{M12 , M22 }. From [11, Theorem 3], CN J (X) = M12 M22 if either ψ/ψ2 or ψ2 /ψ attains a maximum at t = 1/2. Then, we have the following Proposition 3.3. Let ψ ∈ Ψ2 and let ψ(t) = ψ(1 − t) for all t ∈ [0, 1]. If 2 (1/2) M2 = ψψ(1/2) , then CZ (X) = CN J (X) = M12 M22 . Proof. From the definition, we have CZ (X) ≤ CN J (X) = M12 M22 . Take an arbitrary t ∈ [0, 1] and put x = (t, 1 − t) and y = (1 − t, t). Then kxkψ = kykψ = ψ(t) and kx + ykψ = k(1, 1)kψ = 2ψ(1/2), kx − ykψ = k(2t − 1, 1 − 2t)kψ = 2|2t − 1|ψ(1/2). Hence we have 2 kxk2ψ + kyk2ψ 4kxkψ kykψ = kx + yk2ψ + kx − yk2ψ kx + yk2ψ + kx − yk2ψ ψ(t)2 (1 + (2t − 1)2 ) ψ(1/2)2 ψ(t)2 = 2ψ2 (t)2 ψ(1/2)2 2 ψ(t)2 ψ2 (1/2)2 2 ψ(t) = = M 2 ψ2 (t)2 ψ(1/2)2 ψ2 (t)2 = Since t is arbitrary, we have CZ (X) ≥ M12 M22 . Therefore we have CZ (X) = M12 M22 . This completes the proof. In case that M1 = ψ(1/2)/ψ2 (1/2), we have the following theorem as in the proof of Theorem 2.2 and so omit the proof. Theorem 3.4. Let ψ ∈ Ψ2 and let ψ(t) = ψ(1−t) for all t ∈ [0, 1]. If M1 = ψψ(1/2) 2 (1/2) and M2 > 1, then CZ (X) = M12 M22 if and only if there exist s, t ∈ [0, 1] (s < t) satisfying one of the following conditions: GEOMETRIC CONSTANTS OF R2 31 (1) ψ2 (s) = M2 ψ(s), ψ2 (t) = M2 ψ(t) and, if we put r = ψ(s)t+ψ(t)s , then ψ(s)+ψ(t) ψ(r) = M1 ψ2 (r). ψ(t)s+ψ(s)t (2) ψ2 (s) = M2 ψ(s), ψ2 (t) = M2 ψ(t) and, if we put r = ψ(t)+ψ(s)(2t−1) , then ψ(r) = M1 ψ2 (r). 4. Examples In this section, we calculate CN0 J (X) and CZ (X) of some Banach spaces X = (R2 , k · kψ ), where ψ ∈ Ψ. First, we consider the case that ψ = ψp . Example 4.1. Let 1 ≤ p ≤ ∞ and 1/p + 1/q = 1. We put t = min(p, q). Then 2 CN0 J (R2 , k · kp ) = CZ (R2 , || · ||p ) = CN J (R2 , k · kp ) = 2 t −1 . 2 Suppose that 1 ≤ p ≤ 2. Since ψp ≥ ψ2 , we have CZ (R2 , || · ||p ) = 2 p −1 by Proposition 3.1. On the other hand, as in Theorem 2.2, we take s = 0 and t = 1. 1 − 12 1 p Since r = ψ(0)·1+ψ(1)·0 = and M = ψ (1/2)/ψ (1/2) = 2 , by Theorem 2.2, 1 p 2 ψ(0)+ψ(1) 2 2 we have CN0 J (R2 , k · kp ) = M12 = 2 p −1 . If 2 ≤ p ≤ ∞, then we similarly have, by Proposition 2.1 and Theorem 3.2, 2 0 CN J (R2 , k · kp ) = CZ (R2 , || · ||p ) = CN J (R2 , k · kp ) = 2 p −1 . In [14, Example], C. Yang and H. Li calculated the modified NJ constant of the following normed linear space. From our theorems, we have Example 4.2. Let λ > 0 and Xλ = R2 endowed with norm ||(x, y)||λ = (||(x, y)||2p + λ||(x, y)||2q )1/2 . (i) If 2 ≤ p ≤ q ≤ ∞, then CN J (Xλ ) = CN0 J (Xλ ) = CZ (Xλ ) = 2(λ+1) . 22/p +λ22/q 2/p 2/q 2 +λ2 . 2(λ+1) (ii) If 1 ≤ p ≤ q ≤ 2, then CN J (Xλ ) = CN0 J (Xλ ) = CZ (Xλ ) = To see this, first, we remark that (p, q) is not necessarily a Hölder pair. We define the normalized norm || · ||0λ by ||(x, y)||λ ||(x, y)||0λ = √ . 1+λ Then ||·||0λ is absolute and so put the corresponding function ψλ (t) = ||(1−t, t)||0λ . (i) Suppose that 2 ≤ p ≤ q ≤ ∞. Since ψλ ≤ ψ2 , by Proposition 2.1, we have 2(λ+1) CN J (Xλ ) = CN0 J (Xλ ) = M22 = 22/p . On the other hand, in Theorem 3.2, we +λ22/q take s = 0 and t = 1. Then we have r = 1/2 and M22 ψ2 (1/2) ψλ (1/2) = M2 . Thus we have 22/p +λ22/q . 2(λ+1) CZ (Xλ ) = = (ii) Suppose that 1 ≤ p ≤ q ≤ 2. Since ψλ ≥ ψ2 , by Theorem 2.2 and Proposition 3.1, we similarly have (ii). Example 4.3. Put ψ2 (t) (0 ≤ t ≤ 1/2), ψ(t) = 2 − √2 t + √2 − 1 (1/2 ≤ t ≤ 1). √ √ Then CN0 J (R2 , k · kψ ) < CZ (R2 , k · kψ ) = CN J (R2 , k · kψ ) = 2 2( 2 − 1). 32 H. MIZUGUCHI, K.-S. SAITO In fact, ψ ∈ Ψ2 and the norm of k · kψ is p |a|2 + |b|2 (|a| ≥ |b|) k(a, b)kψ = √ 2 − 1 |a| + |b| (|a| ≤ |b|) . √ √ Since ψ ≥ ψ2 , by Proposition 3.1, we have CZ (R2 , k · kψ ) = M12 = 2 2 2 − 1 . We assume that CN0 J (R2 , k · kψ ) = M12 . By Theorem 2.2, we can choose r ∈ [0, 1] such that ψψ(r) = ψψ(1−r) = M1 . This is impossible by the definition of ψ. 2 (r) 2 (1−r) 0 2 Therefore we have CN J (R , k · kψ ) < M12 . Example 4.4. Let 1/2 ≤ β ≤ 1. We define a convex function ψβ ∈ Ψ2 by ψβ (t) = max{1 − t, t, β}. By [11, Example 4], we have 2 2 β + (1 − β) (β ∈ [ 12 , √12 ]) 2 2 β CN J (R , k · kψβ ) = 2 2(β + (1 − β)2 ) (β ∈ ( √12 , 1]). Indeed, M1 = ( 1 ψβ (1/2) ψ2 (1/2) = β√ 1/ 2 = √ (β ∈ [ 12 , √12 ]) 2β (β ∈ ( √12 , 1]) and ψ2 (β) 1 = {(1 − β)2 + β 2 }1/2 . ψβ (β) β √ If 1/2 ≤ β ≤ 1/ 2, then ψβ ≤ ψ2 and so, by Proposition 2.1, we have M2 = β 2 + (1 − β)2 . β2 By Theorem 3.2, we have CZ (R2 , k · kψβ ) < M22 . √ ψ (1/2) Assume that 1/ 2 < β ≤ 1. Since M1 = ψβ2 (1/2) , we have, by Proposition 2.3, CN0 J (R2 , k · kψβ ) = M22 = CN0 J (R2 , k · kψβ ) = M12 M22 = 2(β 2 + (1 − β)2 ). On the other hand, we take s = β and t = 1 − β in Theorem 3.4. Then we have = 1/2. By Theorem 3.4, we have r = ψ(β)(1−β)+ψ(1−β)β ψ(β)+ψ(1−β) CZ (R2 , k · kψβ ) = M12 M22 = 2(β 2 + (1 − β)2 ). √ Example 4.5. We consider ψβ in Example 4.4 in case of β = 1/ 2. Then we have √ √ CN0 J (R2 , k · kψβ ) = CN J (R2 , k · kψβ ) = M22 = 2 2( 2 − 1). √ √ On the other hand, we have CZ (R2 , k · kψβ ) = M22 = 2 2( 2 − 1). For this ψβ , define a convex function ϕ ∈ Ψ2 by ( ψβ (t) (0 ≤ t ≤ 1/2), ϕ(t) = ψ2 (t) (1/2 ≤ t ≤ 1). GEOMETRIC CONSTANTS OF R2 33 As in Example 4.2, we similarly have √ √ CZ (R2 , k · kϕ ) < CN0 J (R2 , k · kϕ ) = CN J (R2 , k · kϕ ) = M22 = 2 2 2−1 . Acknowledgement. The second author is supported in part by Grants-inAid for Scientific Research, Japan Society for the Promotion of Science (No. 23540189). The authors would also like to thank the referees for some helpful comments. References 1. J. Alonso, P. Martin and P.L. Papini, Wheeling around von Neumann-Jordan constant in Banach Spaces, Studia Math. 188 (2008), no. 2, 135–150. 2. J. Alonso and P. 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Rǎdulescu which characterizes inner product spaces, Rev. Roumaine Math. Puers Appl. 47 (2001), 253–257. Department of Mathematical Sciences, Graduate School of Science and Technology, Niigata University, Niigata, 950-2181 Japan. E-mail address: mizuguchi@m.sc.niigata-u.ac.jp E-mail address: saito@math.sc.niigata-u.ac.jp