Geometrical Constants and. Norm Inequalities in Banach Spaces

Transcription

1 Geometrical Constants and Norm Inequalities in Banach Spaces Hiroyasu Mizuguchi Doctoral Program in Fundamental Sciences Graduate School of Science and Technology Niigata University March 013

2 Contents Introduction Acknowledgements Some geometrical constants of absolute normalized norms on R Introduction and preliminaries The modified NJ constant of absolute normalized norms on R The Zbăganu constant of absolute normalized norms on R Examples The characterization of the Dunkl-Williams constant.1 Introduction A characterization by Birkoff orthogonality A characterization by the frame of the unit ball Examples On some norm inequalities in Banach spaces Introduction The generalized Clarkson s inequality in the complex case The generalized Clarkson s inequality in the real case Another aspect of triangle inequality References 57 1

3 Introduction There are many important norm inequalities in mathematics such as triangle inequality, Clarkson s inequalities [1], Dunkl-Williams inequality [16] and so on. These inequalities are very useful to study geometrical structure of Banach spaces. For example, the notion of strict convexity is based on the occurrence of equality in the triangle inequality. In 1936, Clarkson [1] showed that L p -spaces with 1 < p < are uniformly convex by using Clarkson s inequalities. We have several geometrical constants of Banach spaces, for example, von Neumann-Jordan constant [13], James constant [19], Dunkl-Williams constant [5] and so on. Some of them are induced by norm inequalities. The von Neumann- Jordan constant and Dunkl-Williams constant measure how much the space is close (or far) to be a Hilbert space. The James constant represents non-squareness of the unit ball. These constants play an important role in the description of geometrical properties of Banach spaces, and have been investigated in many papers. In particular, many authors have calculated and estimated constants for Banach spaces. Our aim in this thesis is to present recent results on geometrical constants and norm inequalities. In Chapter 1, we study modified von Neuman-Jordan constant and Zbăganu constant of R with absolute normalized norms. In Chapter, we investigate how to calculate the Dunkl-Williams constant of Banach spaces. In Chapter 3, we describe some remarks on Clarkson s inequalities and triangle inequality. In Chapter 1, we consider some geometrical constants of R with absolute normalized norms. The notion of the von Neumann-Jordan constant (hereafter referred to as NJ constant) of Banach spaces was introduced by Clarkson in [13] and it has been studied by several authors ([6, 8, 37, 63, 66] and so on). The NJ constant C NJ (X) of a Banach space X is defined as { } x + y + x y C NJ (X) = sup ( x + y ) x, y X, (x, y) (0, 0). Some similar constants have been defined and studied: { } x + y C NJ(X) + x y = sup 4 x, y S X, { } x + y x y C Z (X) = sup x + y x, y X, (x, y) (0, 0),

4 where S X is the unit sphere of X. The constant C NJ (X), called the modified von Neumann-Jordan constant (shortly, modified NJ constant) was introduced by Gao in [18]. The constant C Z (X) was introduced by Zbăganu [70]. It has been shown that C NJ (X) and C Z(X) do not necessarily coincide with C NJ (X) (cf. [3, 4, 0, 1, 41]). In 000, Saito, Kato and Takahashi [58] calculated and estimated the NJ constant of C with absolute normalized norms. The results obtained in [58] also hold on R. We investigate the conditions of absolute normalized norms on R that the modified NJ constant coincides with NJ constant or that the Zbăganu constant coincides with NJ constant. Further we calculate modified NJ constant and Zbăganu constant for some Banach spaces. In Chapter, we study how to calculate the Dunkl-Williams constant. In 1964, Dunkl and Williams [16] showed that for any nonzero elements x, y in a Banach space X, x x y y 4 x y x + y. This inequality is called the Dunkl-Williams inequality and have been studied in many papers ([6, 31, 4, 43, 54, 60, 61] and so on). In the paper [16], the authors proved that the constant 4 can be replaced by if X is a Hilbert space. A bit later, Kirk and Smiley [34] completed this result by showing that the inequality with in place of 4 in fact characterizes Hilbert spaces. In [5], Jiménez-Melado et al. introduced the notion of the Dunkl-Williams constant of Banach spaces. The Dunkl-Williams constant DW (X) of a Banach space X is the smallest constant which can replace the 4 in the Dunkl-Williams inequality, that is, DW (X) = sup { x + y x y x x y } y x, y X, x, y 0, x y. With this notion, our previous comments can be written as: DW (X) 4 for any Banach space X, and X is a Hilbert space if and only if DW (X) =. On the other hand, from Baronti and Papini [7] we have that X is uniformly non-square if and only if DW (X) < 4. However, it is very difficult to calculate the Dunkl-Williams constant and so, except for Hilbert spaces, there exists no uniformly non-square Banach space in which the constant has been calculated. We introduce some notations related to Birkhoff orthogonality. Then we present a characterization of the Dunkl-Williams constant. We define the frame of the unit 3

5 ball of a Banach space which is related to the norming functionals for elements of the unit sphere. Then we improve the characterization. Thereafter, as an application, we calculate the Dunkl-Williams constant of the Day-James space l -l. In Chapter 3, we consider the generalized Clarkson s inequalities in the real and complex cases. In 1935, Jordan and von Neumann [6] characterized Hilbert spaces as Banach spaces satisfying the parallelogram law. In the next year 1936, as a generalization of the parallelogram law, Clarkson [1] proved famous norm inequalities for L p so called Clarkson s inequalities. To study the generalizations of Clarkson s inequalities, for two numbers p, q with 0 < p, q, the generalized Clarkson s inequality in the complex case have been considered: ( z + w q + z w q ) 1 q C( z p + w p ) 1 p for all z, w C. The best constant in this inequality is denoted by C p,q (C). Clarkson [1] proved that C p,p (C) = 1/p for 1 p. Later on the best constant C p,q (C) for the remaining pairs of p and q such that 0 < p, q were found by Koskela [40], Maligranda and Persson [44]. In 1997, Kuriyama et al. [35] obtained an elementary proof of the generalized Clarkson s inequality in the complex case. Moreover, we can consider the generalized Clarkson s inequality in the real case: ( a + b q + a b q ) 1 q C( a p + b p ) 1 p for all a, b R. As in the complex case, the best constant is denoted by C p,q (R). In 007, Maligranda and Sabourova [45] computed the best constant C p,q (R) for all 0 < p, q. We present an element proof of the generalized Clarkson s inequality in the real case. We also pay attention to the triangle inequality. The triangle inequality is one of the most fundamental inequalities in analysis. Several authors have been treating its generalizations, improvements and reverse inequalities ([16, 31, 4, 43, 49] and so on). We consider another aspect of the triangle inequality. For a Hilbert space H, the parallelogram law implies that the parallelogram inequality x + y ( x + y ) holds for all x, y H. Saitoh in [6] noted that the above inequality may be more suitable than the classical triangle inequality. Motivated by this, Belbachir et al. 4

6 [9] introduced the notion of q-norm (1 q < ). We introduce the notion of ψ- norm by considering the fact that an absolute normalized norm on R corresponds to a function ψ on the interval [0, 1] with some conditions (cf. [11, 58]). This is a generalization of the notion of q-norm. We show that a ψ-norm is a norm in the usual sense. Acknowledgements I would like to express the deepest gratitude to my supervisor Professor Kichi-Suke Saito at Niigata University for his great guidance and useful advice. I have studied under him during an undergraduate, a master s course and a doctoral course. He provided much support and instructions. I also wish to thank all my colleagues in Professor Saito laboratory. 5

7 1 Some geometrical constants of absolute normalized norms on R 1.1 Introduction and preliminaries Let X be a Banach space with dim X. We denote the unit sphere and the unit ball of a Banach space X by S X and B X, respectively. Many geometrical constants of a Banach space X have been investigated. In this chapter we shall consider the following constants: { } x + y + x y C NJ (X) = sup ( x + y ) x, y X, (x, y) (0, 0), { x + y C NJ(X) + x y = sup 4 { x + y x y C Z (X) = sup x + y x, y S X }, } x, y X, (x, y) (0, 0). The constant C NJ (X), called the von Neumann-Jordan constant (hereafter referred to as NJ constant) have been considered in many papers ([13, 6, 8, 37, 63, 66, 69] and so on). The constant C NJ (X), called the modified von Neumann-Jordan constant (shortly, modified NJ constant) was introduced by Gao in [18] and does not necessarily coincide with C NJ (X) (cf. [4, 0]). The constant C Z (X) was introduced by Zbăganu [70] and was conjectured that C Z (X) always coincides with the NJ constant C NJ (X), but Alonso and Martin [3] gave an example that C NJ (X) C Z (X) (cf.[1, 41]). A norm on R is said to be absolute if (x, y) = ( x, y ) for any (x, y) R, and normalized if (1, 0) = (0, 1) = 1. The l p -norms p are such examples: (x, y) p = { ( x p + y p ) 1 p (1 p < ), max{ x, y } (p = ). Let AN denote the family of all absolute normalized norms on R, and Ψ denote the family of all continuous convex functions 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 AN and Ψ are in a one-to-one correspondence under the equation ψ(t) = (1 t, t) 6

8 (0 t 1). Let ψ be an absolute normalized norm associated with a convex function ψ Ψ. For ψ, φ Ψ, we denote ψ φ if ψ(t) φ(t) for any 0 t 1. Let ψ(t) M 1 = max 0 t 1 ψ (t) and ψ (t) M = max 0 t 1 ψ(t), where ψ (t) = (1 t, t) = (1 t) + t corresponds to the l -norm. In 000, Saito, Kato and Takahashi [58] proved that, if ψ ψ (resp. ψ ψ ), then C NJ (C, ψ ) = M1 (resp. M ). The results obtained in [58] also hold on R. We put X = (R, ψ ) for ψ Ψ. Our aim in this chapter is to study the conditions of ψ that C NJ (X) = C NJ(X) or C Z (X) = C NJ (X). In Section 1., we consider the modified NJ constant. We prove that if ψ ψ, then C NJ (X) = C NJ (X) = M. If ψ ψ, then we present the necessarily and sufficient condition for that C NJ (X) = C NJ(X) = M1. Further, we consider the conditions that C NJ (X) = C NJ(X) = M1 M. In Section 1.3, we study the Zbăganu constant. First, we show that, if ψ ψ, then C Z (X) = C NJ (X) = M1. If ψ ψ, then we give the necessarily and sufficient condition for that C Z (X) = C NJ (X) = M. Further we study the conditions that C Z (X) = C NJ (X) = M1 M. In Section 1.4, we calculate modified NJ constant C NJ (X) and Zbăganu constant C Z(X) for some Banach spaces. 1. The modified NJ constant of absolute normalized norms on R In this section, we consider the Banach space X = (R, ψ ). The modified NJ constant C NJ (X) of a Banach space X is defined by { } x + y C NJ(X) + x y = sup 4 x, y S X. From the definition of the modified NJ constant, it is clear that C NJ (X) C NJ(X). In this section, we consider the condition that C NJ (X) = C NJ(X). Proposition Let ψ Ψ. If ψ ψ, then C NJ (X) = C NJ(X) = M. 7

9 Proof. For any x, y S X, by [58, Lemma 3], x + y ψ + x y ψ x + y + x y = ( ) x + y ( ) M x ψ + y ψ = 4M. Now let ψ /ψ attain the maximum M at t = t 0 (0 t 0 1), and put Then x, y S X and x = 1 ψ(t 0 ) (1 t 0, t 0 ), y = 1 ψ(t 0 ) (1 t 0, t 0 ). x + y ψ + x y ψ = 4(1 t 0) + 4t 0 ψ(t 0 ) = 4 ψ (t 0 ) ψ(t 0 ) = 4M, which implies that C NJ (X) = M. By [58, Theorem 1], we have this proposition. If ψ ψ, by [58, Theorem 1], then C NJ (X) = M 1. We now give the necessarily and sufficient condition for that C NJ (X) = M 1. Theorem 1... Let ψ Ψ such that ψ ψ. Then C NJ (X) = M 1 if and only if there exist s, t [0, 1] (s < t) satisfying one of the following conditions: (1) ψ(s) = ψ (s), ψ(t) = ψ (t) and, if we put r = ψ(s)t + ψ(t)s ψ(s) + ψ(t), then ψ(r) ψ (r) () ψ(s) = ψ (s), ψ(t) = ψ (t) and, if we put r = ψ(t)s + ψ(s)t ψ(t) + ψ(s)(t 1), then ψ(r) ψ (r) = ψ(1 r) ψ (1 r) = M 1. = ψ(1 r) ψ (1 r) = M 1. Proof. ( ) Suppose that C NJ (X) = M 1. First, for any x, y S X, by [58, Lemma 3], we have x + y ψ + x y ψ M 1 ( x + y + x y ) = M 1 M 1 ( x + y ) ( ) x ψ + y ψ = 4M 1. 8

10 Since X = (R, ψ ) is a finite dimensional Banach space, we have { x + y C NJ(X) ψ + x y } ψ = max 4 x, y S X. Therefore, C NJ (X) = M 1 if and only if there exist x, y S X (x y) such that x + y ψ + x y ψ = 4M 1. From the above inequality, the elements x, y S X satisfy x ψ = x = 1, y ψ = y = 1 and x + y ψ x + y = x y ψ x y = M 1. Since ψ is absolute and x, y S X satisfy x = y = 1, it is sufficient to consider the following three cases: (i) There exist s, t [0, 1] (s t) satisfying x = 1 1 (1 s, s) and y = (1 t, t). ψ (s) ψ (t) (ii) There exist s, t [0, 1] (s < t) satisfying x = 1 1 (1 s, s) and y = ( 1 + t, t). ψ (s) ψ (t) (iii) There exist s, t [0, 1] (s > t) satisfying x = 1 1 (1 s, s) and y = ( 1 + t, t). ψ (s) ψ (t) Case (i). We may suppose that s < t. Then there exist α, β [0, π ] (α < β) such that x = 1 1 (1 s, s) = (cos α, sin α), y = (1 t, t) = (cos β, sin β). ψ (s) ψ (t) Since x = y = 1, we have ( 1 s x + y = ψ (s) + 1 t ψ (t), s ψ (s) + t ) ( = x + y ψ (t) By [67, Propositions a and b], we remark that cos α + β, sin α + β ). 1 s ψ (s) 1 t ψ (t) and 9 s ψ (s) t ψ (t).

11 Since x y is orthogonal to x + y in the Euclidean space (R, ), we have ( 1 s x y = ψ (s) 1 t ψ (t), s ψ (s) t ) ψ (t) ( = x y cos α + β π, sin α + β π ) ( = x y sin α + β, cos α + β Thus we have ( x + y ψ = x + y cos α + β, sin α + β ) ψ = x + y ( cos α + β Since x + y ψ = M 1 x + y, we have Putting M 1 = then it is clear that We also have ( cos α + β r = + sin α + β ) ψ + sin α + β ) ψ r = ψ(s)t + ψ(t)s ψ(s) + ψ(t) x y ψ = x y ( sin α + β ( sin α+β, cos α+β + sin α+β and ( ). sin α+β cos α+β + sin α+β sin α+β cos α+β + sin α+β M 1 = ψ(r) ψ (r). ) + cos α + β ) ( cos α+β ψ cos α+β + sin α+β Since x y ψ = M 1 x y, we similarly have ( M 1 = sin α + β + cos α + β ) ( ) cos α+β ψ = sin α+β + cos α+β Case (ii). There exist α [0, π] and β [ π, π] such that. ) ) ψ(1 r) ψ (1 r). x = 1 1 (1 s, s) = (cos α, sin α), y = ( 1 + t, t) = (cos β, sin β). ψ (s) ψ (t) 10..

12 Since x = y = 1, we have ( 1 s x + y = ψ (s) 1 t ψ (t), s ψ (s) + t ) ( = x + y ψ (t) By [67, Propositions a and b], we remark that cos α + β, sin α + β ). 1 s ψ (s) 1 t ψ (t) and s ψ (s) t ψ (t). Since x y is orthogonal to x + y in the Euclidean space (R, ), we have ( 1 s x y = ψ (s) + 1 t ψ (t), s ψ (s) t ) ψ (t) ( = x y cos α + β π, sin α + β π ) ( = x y sin α + β, cos α + β Since cos α+β 0 and sin α+β 0, we have ( x + y ψ = x + y cos α + β, sin α + β ) = x + y ( cos α + β Since x + y ψ = M 1 x + y, we have Putting M 1 = then it is clear that We also have ( cos α + β r = ψ + sin α + β ) ψ + sin α + β ) ψ r = ψ(t)s + ψ(s)t ψ(t) + ψ(s)(t 1) x y ψ = x y ( sin α + β ( sin α+β, cos α+β + sin α+β and ( ). sin α+β cos α+β + sin α+β sin α+β cos α+β + sin α+β M 1 = ψ(r) ψ (r). ) + cos α + β ) ( cos α+β ψ cos α+β + sin α+β 11. ) )..

13 Since x y ψ = M 1 x y, we similarly have M 1 = ( sin α + β + cos α + β ) ( cos α+β ψ sin α+β + cos α+β Case (iii). There exist s, t [0, 1] (s > t) satisfying x = 1 1 (1 s, s) and y = ( 1 + t, t). ψ (s) ψ (t) Then, we put s 0 = t and t 0 = s. We define x 0, y 0 in S X by Then we can reduce Case (ii). and x 0 = 1 ψ(s 0 ) (1 s 0, s 0 ), y 0 = 1 ψ(t 0 ) ( 1 + t 0, t 0 ). ( ) If we suppose (1) (resp. ()), then we put x = 1 (1 s, s) ψ (s) y = 1 (1 t, t) ψ (t) ) = ( resp. x = 1 ) (1 s, s) ψ (s) ( resp. y = 1 ) ( 1 + t, t). ψ (t) ψ(1 r) ψ (1 r). Then we have x ψ = x = 1, y ψ = y = 1, x + y ψ = M 1 x + y and x y ψ = M 1 x y. Hence it is clear to prove that C NJ (X) = M 1. We next study the modified NJ constant in the general case. If ψ Ψ, then by [58, Theorem ] we have max{m 1, M } C NJ (X) M 1 M. However, by Theorem 1.., there exist many ψ Ψ satisfying ψ ψ such that C NJ(X) < max{m 1, M } = C NJ (X). From [58, Theorem 3], C NJ (X) = M 1 M if either ψ/ψ or ψ /ψ attains a maximum at t = 1/. At first, we have the following Proposition Let ψ Ψ and let ψ(t) = ψ(1 t) for all t [0, 1]. If ψ/ψ attains a maximum M 1 at t = 1/, then C NJ (X) = C NJ(X) = M 1 M. 1

14 Proof. Suppose first M 1 = ψ(1/)/ψ (1/). Take an arbitrary t [0, 1] and put Then we have x, y S X and Therefore we have x = 1 1 (t, 1 t) and y = (1 t, t). ψ(t) ψ(t) x + y ψ = 4 ψ(t) ψ(1 ), x y ψ = x + y ψ + x y ψ 4 4 t 1 ψ( 1 ψ(t) ). = { (t 1) + 1 } ψ(1/) = ψ (t) ψ(1/) ψ(t) ψ(t) = ψ (t) ψ(1/) ψ(t) ψ (1/) = M ψ (t) 1 ψ(t). Since t is arbitrary, we have C NJ (X) M 1 M which prove that C NJ (X) = M 1 M. In the case that M = ψ (1/)/ψ(1/), C NJ (X) does not necessarily coincide with M 1 M. However, we have the following Theorem Let ψ Ψ and let ψ(t) = ψ(1 t) for all t [0, 1]. Assume that M = ψ (1/)/ψ(1/) and M 1 > 1. Then C NJ (X) = M 1 M if and only if there exist s, t [0, 1] (s < t) satisfying one of the following conditions: (1) ψ (s) = M ψ(s), ψ (t) = M ψ(t) and, if we put r = ψ(s)t + ψ(t)s ψ(s) + ψ(t), then ψ(r) = M 1ψ (r). () ψ (s) = M ψ(s), ψ (t) = M ψ(t) and, if we put r = Proof. For all x, y S X, we have ψ(t)s + ψ(s)t ψ(t) + ψ(s)(t 1), then ψ(r) = M 1ψ (r). x + y ψ + x y ψ M 1 = M 1 M 1 M ( x + y + x y ) ( x + y ) ( ) x ψ + y ψ = 4M 1 M. 13

15 From this inequality, C NJ (X) = M 1 M if and only if there exist x, y S X (x y) such that Thus, the elements x, y S X satisfy x + y ψ + x y ψ = 4M 1 M. x = y = M, x + y ψ = M 1 x + y, x y ψ = M 1 x y. Since ψ is absolute, it is sufficient to consider the following three cases: (i) There exist s, t [0, 1] (s t) satisfying x = 1 1 (1 s, s) and y = (1 t, t). ψ (s) ψ (t) (ii) There exist s, t [0, 1] (s < t) satisfying x = 1 1 (1 s, s) and y = ( 1 + t, t). ψ (s) ψ (t) (iii) There exist s, t [0, 1] (s > t) satisfying x = 1 1 (1 s, s) and y = ( 1 + t, t). ψ (s) ψ (t) As in the proof of Theorem 1.., we can prove this theorem. 1.3 The Zbăganu constant of absolute normalized norms on R The Zbăganu constant C Z (X) in [70] is defined by { } x + y x y C Z (X) = sup x + y x, y X, (x, y) (0, 0). Then it is clear that C Z (X) C NJ (X) for any Banach space X. In this section, we consider the condition that C Z (X) = C NJ (X) for X = (R, ψ ). Then, we have the following Proposition Let ψ Ψ. If ψ ψ, then C Z (X) = C NJ (X) = M1. 14

16 Proof. First, by [58, Lemma 3], we have for any x, y X. x + y ψ x y ψ x + y ψ + x y ψ ( M1 x + y + x y ) ( ) ( ) = M1 x + y M 1 x ψ + y ψ. Now let ψ/ψ attain the maximum M 1 at t = t 0 (0 t 0 1). Put x = (1 t 0, 0) and y = (0, t 0 ), respectively. Then we have x + y ψ + x y ψ = ψ(t 0 ) = M 1 ψ (t 0 ) = M 1 Since x + y ψ = ψ(t 0 ) = x y ψ, we have ( ) x ψ + y ψ. x + y ψ x y ψ = x + y ψ + x + y ψ = M 1 ( ) x ψ + y ψ. Therefore we have which implies that C Z (X) = M 1. x + y ψ x y ψ x ψ + y ψ = M 1, We next consider the case that ψ ψ. We remark that the Zbăganu constant C Z (X) can be in the following form: { } 4 x y C Z (X) = sup x + y + x y x, y X, (x, y) (0, 0). Then we have the following Theorem Let ψ Ψ. Assume that ψ ψ. Then C Z (X) = M if and only if there exist s, t [0, 1] (s < t) satisfying one of the following conditions: (1) ψ(s) = ψ (s), ψ(t) = ψ (t) and, if we put r = ψ(s)t + ψ(t)s ψ(s) + ψ(t), then ψ (r) ψ(r) = ψ (1 r) ψ(1 r) = M. 15

17 () ψ(s) = ψ (s), ψ(t) = ψ (t) and, if we put r = ψ(t)s + ψ(s)t ψ(t) + ψ(s)(t 1), then ψ (r) ψ(r) = ψ (1 r) ψ(1 r) = M. Proof. First, by [58, Lemma 3], for any x, y X, 4 x ψ y ψ ( ) x ψ + y ψ ( ) x + y = x + y + x y M ( ) x + y ψ + x y ψ. Since X = (R, ψ ) is a finite dimensional Banach space, { } 4 x ψ y ψ C Z (X) = max x + y ψ + x x, y X, (x, y) (0, 0). y ψ Then C Z (X) = M if and only if there exist x, y S X (x y) such that 4 x ψ y ψ x + y ψ + x y ψ = M. From the above inequality, x = x ψ = y ψ = y and x + y x + y ψ = x y x y ψ = M. Hence we may assume that x = x ψ = y ψ = y = 1. As in the proof of Theorem 1.., it is sufficient to consider the following three cases: (i) There exist s, t [0, 1] (s t) satisfying x = 1 1 (1 s, s) and y = (1 t, t). ψ (s) ψ (t) (ii) There exist s, t [0, 1] (s < t) satisfying x = 1 1 (1 s, s) and y = ( 1 + t, t). ψ (s) ψ (t) (iii) There exist s, t [0, 1] (s > t) satisfying x = 1 1 (1 s, s) and y = ( 1 + t, t). ψ (s) ψ (t) 16

18 As in the proof of Theorem 1.., we can similarly prove this theorem. We next study the Zbăganu constant C Z (X) in general case. If ψ Ψ, by [58, Theorem ], then we have max{m1, M } C NJ (X) M1 M. However, by Theorem 1.3., there exist many ψ Ψ satisfying ψ ψ such that C Z (X) < C NJ (X) max{m1, M }. From [58, Theorem 3], C NJ (X) = M1 M if either ψ/ψ or ψ /ψ attains a maximum at t = 1/. At first, we have the following Proposition Let ψ Ψ and let ψ(t) = ψ(1 t) for all t [0, 1]. If ψ /ψ attains a maximum M at t = 1/, then C Z (X) = C NJ (X) = M1 M. Proof. First, we know C Z (X) C NJ (X) = M1 M. Take an arbitrary t [0, 1] and put x = (t, 1 t) and y = (1 t, t). Then x ψ = y ψ = ψ(t), x + y ψ = (1, 1) ψ = ψ(1/) and x y ψ = (t 1, 1 t) ψ = t 1 ψ(1/). Hence we have 4 x ψ y ψ x + y ψ + x y ψ = ( x ψ + y ψ x + y ψ + x y ψ ) = = ψ(t) (1 + (t 1) ) ψ(1/) ψ(t) ψ (t) ψ(1/) = ψ(t) ψ (1/) ψ (t) ψ(1/) = M ψ(t) ψ (t). Since t is arbitrary, we have C Z (X) M 1 M. Therefore we have C Z (X) = M 1 M. In case that M 1 = ψ(1/)/ψ (1/), we have the following theorem as in the proof of Theorem 1.. and so omit the proof. 17

19 Theorem Let ψ Ψ and let ψ(t) = ψ(1 t) for all t [0, 1]. If M 1 = ψ(1/)/ψ (1/) and M > 1, then C Z (X) = M1 M if and only if there exist s, t [0, 1] (s < t) satisfying one of the following conditions: (1) ψ (s) = M ψ(s), ψ (t) = M ψ(t) and, if we put r = ψ(s)t + ψ(t)s ψ(s) + ψ(t), then ψ(r) = M 1ψ (r). () ψ (s) = M ψ(s), ψ (t) = M ψ(t) and, if we put 1.4 Examples r = ψ(t)s + ψ(s)t ψ(t) + ψ(s)(t 1), then ψ(r) = M 1ψ (r). In this section, we shall calculate C NJ (X) and C Z(X) for some Banach spaces X = (R, ψ ), where ψ Ψ. First, we consider the case that ψ = ψ p. Example Let 1 p and 1/p + 1/p = 1. We put t = min{p, p }. Then C NJ((R, p )) = C Z ((R, p )) = C NJ ((R, p )) = /t 1. Proof. Suppose that 1 p. Since ψ p ψ, we have C Z ((R, p )) = /p 1 by Proposition On the other hand, as in Theorem 1.., we take s = 0 and t = 1. Since we have r = ψ(0) 1 + ψ(1) 0 ψ(0) + ψ(1) = 1 and M 1 = ψ p(1/) ψ (1/) = 1/p 1/, C NJ((R, p )) = M 1 = /p 1. If p, then we similarly have, by Proposition 1..1 and Theorem 1.3., C NJ((R, p )) = C Z ((R, p )) = C NJ ((R, p )) = /p 1. In [68, Example], Yang and Li calculated the modified NJ constant of the following Banach space. From our theorems, we have Example Let λ > 0 and let X λ be the space R endowed with norm (x, y) λ = ( (x, y) p + λ (x, y) q) 1. 18

20 (i) If p q, then (ii) If 1 p q, then C NJ (X λ ) = C NJ(X λ ) = C Z (X λ ) = (λ + 1) /p + λ /q. C NJ (X λ ) = C NJ(X λ ) = C Z (X λ ) = /p + λ /q. (λ + 1) Proof. First, we remark that (p, q) is not necessarily a Hölder pair. We define a normalized norm 0 λ by (x, y) 0 λ = (x, y) λ λ + 1. Then 0 λ AN and so put the corresponding function ψ λ (t) = (1 t, t) 0 λ. (i) Suppose that p q. Since ψ λ ψ, by Proposition 1..1, we have C NJ (X λ ) = C NJ(X λ ) = M = (λ + 1) /p + λ /q. On the other hand, as in Theorem 1.3., we take s = 0 and t = 1. Then we have r = 1/ and ψ (1/)/ψ λ (1/) = M. Thus we have C Z (X λ ) = M = (λ + 1) /p + λ /q. (ii) Suppose that 1 p q. Since ψ λ ψ, by Theorem 1.. and Proposition 1.3.1, we similarly have (ii). Example Put ψ (t) (0 t 1/), ψ(t) = ( ) t + 1 (1/ t 1). Then C NJ((R, ψ )) < C Z ((R, ψ )) = C NJ ((R, ψ )) = ( 1). Proof. In fact, ψ Ψ and the norm ψ is given by x + y ( x y ), (x, y) ψ = ( ) 1 x + y ( x y ). 19

21 Since ψ ψ, by Proposition 1.3.1, we have ( ) C Z (R, ψ ) = M 1 = ( ) 1. We assume that C NJ ((R, ψ )) = M 1. r [0, 1] such that ψ(r) ψ (r) = ψ(1 r) ψ (1 r) = M 1. By Theorem 1.., we can choose This is impossible by the definition of ψ. Therefore we have C NJ ((R, ψ )) < M 1. Example Let 1/ β 1. We define a convex function ψ β Ψ by ψ β (t) = max{1 t, t, β}. (i) If 1/ β < 1/, then C Z ((R, ψβ )) < C NJ ((R, ψβ )) = C NJ((R, ψβ )) = β + (1 β) β. (ii) If 1/ β 1, then C Z ((R, ψβ )) = C NJ ((R, ψβ )) = C NJ((R, ψβ )) = {β + (1 β) }. Proof. By [58, Example 4], we have β + (1 β) ( ) 1/ β 1/, C NJ ((R, ψβ )) = β ( ) {β + (1 β) } 1/ β 1. Indeed, and have M 1 = ψ β (1/) ψ (1/) = 1 (1/ β 1/ ), β 1/ = β (1/ β 1). M = ψ (β) ψ β (β) = 1 β {(1 β) + β } 1/. Suppose that 1/ β 1/. Then ψ β ψ and so, by Proposition 1..1, we C NJ((R, ψβ )) = M = β + (1 β) β. 0

22 If 1/ β < 1/, then by Theorem 1.3., we have C Z ((R, ψβ )) < M. If β = 1/, then we have ψ β (1/) = 1/ = ψ (1/). As in Theorem 1.3., we take s = 0 and t = 1/. Since r = ψ β(0) 1/ + ψ β (1/) 0 ψ β (0) + ψ β (1/) = 1 1/ and we have ψ (1 1/ ) ψ β (1 1/ ) = ψ (1/ ) ψ β (1/ ) = M, C Z ((R, ψβ )) = M = β + (1 β) = ( 1) = {β + (1 β) }. β Assume that 1/ < β 1. Since M 1 = ψ β (1/)/ψ (1/), we have, by Proposition 1..3, C NJ((R, ψβ )) = M1 M = {β + (1 β) }. On the other hand, we take s = β and t = 1 β in Theorem Then we have r = ψ(β)(1 β) + ψ(1 β)β ψ(β) + ψ(1 β) = 1. Thus we have C Z ((R, ψβ )) = M 1 M = {β + (1 β) }. Example We consider ψ β in Example in the case of β = 1/. Using this ψ β, we define a convex function φ Ψ by ψ β (t) (0 t 1/), φ(t) = ψ (t) (1/ t 1). Then, as in Example 1.4.3, C Z ((R, φ )) < C NJ((R, φ )) = C NJ ((R, φ )) = M = ( ) 1. 1

23 The characterization of the Dunkl-Williams constant.1 Introduction In 1964, Dunkl and Williams [16] showed that for any nonzero elements x, y in a Banach space X, x x y y 4 x y x + y. This inequality is called the Dunkl-Williams inequality and have been studied in many papers ([6, 31, 4, 43, 54, 60, 61] and so on). In [16], it was proved that the constant 4 can be replaced by if X is a Hilbert space, and also that the value 4 is the best possible choice in the space (R, 1 ). A bit later, Kirk and Smiley [34] showed that a Banach space X is a Hilbert space if the inequality x x y x y y x + y holds for any nonzero elements x, y in X. Thus, the smallest number which can replace the 4 in the Dunkl-Williams inequality measures how much the space is close (or far) to be a Hilbert one. In 008, Jiménez-Melado et al. [5] introduced the Dunkl-Williams constant DW (X) of a Banach space X: { x + y DW (X) = sup x x y x y } y x, y X, x, y 0, x y. We summarize some basic properties of the Dunkl-Williams constant: (i) For any Banach space X, DW (X) 4 ([16]). (ii) DW (X) = X is a Hilbert space ([34]). (iii) DW (X) < 4 if and only if X is uniformly non-square, that is, there exits δ > 0 such that x + y (1 δ) and x = y = 1 imply x y (1 δ) ([7, 5]). From (iii), we have that DW (X) = 4 if and only if X is not uniformly non-square. However, to calculate the Dunkl-Williams constant is very difficult and so, except

24 for Hilbert spaces, there exists no uniformly non-square Banach space in which the Dunkl-Williams constant has been calculated. For x, y X, x is said to be Birkhoff orthogonal to y and denoted by x B y if x x + λy for any λ R ([10, ]). Birkhoff orthogonality coincides with the usual orthogonality in Hilbert spaces. Birkhoff orthogonality is always homogeneous, that is, x B y implies αx B βy for every α, β R. However it is not symmetric, that is, x B y does not necessarily imply y B x. In a Banach space of three or more dimension, Birkhoff orthogonality is symmetric if and only if the norm is induced by an inner product. More results about this orthogonality can be found in [1,, 5, 10, 15,, 3]. The Day-James space l p -l q is defined for two numbers p, q with 1 p, q as the space R with the norm { (x, y) p if xy 0, (x, y) p,q = (x, y) q if xy 0. James [] considered the space l p -l q with 1/p + 1/q = 1 as an example of a twodimensional Banach space where Birkhoff orthogonality is symmetric. Day [15] considered even more general spaces. In this chapter, we shall characterize the Dunkl-Williams constant and calculate DW (l -l ). In Section., we first recall a characterization of the Dunkl-Williams constant and introduce some notations related to Birkhoff orthogonality. Then we present a characterization of the Dunkl-Williams constant. To calculate the Dunkl- Williams constant, we also obtain several results about convergent sequences. In Section.3, we define the frame of the unit ball of a Banach space which is related to the norming functionals for elements of the unit sphere. Then we improve the characterization in Section.. We also consider the case of dim X = and prove that the frame of the unit ball of a two-dimensional Banach space X coincides with the set of all extreme point of the unit ball. In Section.4, as an application, we calculate the Dunkl-Williams constant of the Day-James space l -l. We first see that DW ((R, )) = and DW ((R, )) = 4. Furthermore we show that DW (l -l ) =. 3

25 . A characterization by Birkoff orthogonality Let X be a real Banach space with dim X. We denote the unit sphere and the unit ball of a Banach space X by S X and B X, respectively. Then we have a characterization of the Dunkl-Williams constant: Proposition..1. Let X be a real Banach space with dim X. Then { } u + v DW (X) = sup (1 t)u + tv u, v S X, u + v 0, 0 t 1 { } u + v = sup min 0 t 1 (1 t)u + tv u, v S X, u + v 0. For x, y S X, let z(t) = (1 t)x + ty for all t R. Lemma... Suppose that X is a Banach space with dim X. Let x, y S X with x + y 0. For t 0 [0, 1], the following are equivalent: (i) z(t 0 ) B x y. (ii) z(t) z(t 0 ) for all t R. Proof. Let λ R. Then we have z(t 0 ) + λ(x y) = z(t 0 λ). Hence z(t 0 ) B x y if and only if z(t 0 λ) z(t 0 ) for any λ R. Now we shall introduce some notations. For each x S X, we define the subset V (x) of X by V (x) = {y X x B y}. For each x S X and each y V (x), we put { } λ + µ Γ(x, y) = λ 0 µ, x + λy = x + µy and m(x, y) = sup{ x + γy γ Γ(x, y)}. We define the positive number M(x) by M(x) = sup{m(x, y) y V (x)} for each x S X. Proposition..3. Let X be a real Banach space with dim X and let x S X. Then the following holds: (i) 0 V (x). 4

26 (ii) If y V (x), then αy V (x) for any α R. (iii) m(x, 0) = 1 m(x, y) for each y V (x). (iv) m(x, αy) = m(x, y) for each y V (x) and each α 0. Proof. We directly have (i), (ii) and (iii). (iv) For each y V (x) and each α 0, take an arbitrary element γ Γ(x, αy). Then there exist λ, µ with λ 0 µ such that γ = (λ + µ)/ and x + λ(αy) = x + µ(αy). From the fact that x + (αλ)y = x + (αµ)y, we have αγ Γ(x, y). Hence x + γ(αy) = x + (αγ)y m(x, y). Thus we have m(x, αy) m(x, y). By the preceding paragraph, we also have m(x, y) = m(x, α 1 αy) m(x, αy). Therefore we obtain m(x, αy) = m(x, y). Lemma..4. Suppose that X is a real Banach space with dim X. Let x S X and y V (x) \ {0}. There exist s 1, s with s 1 0 s such that {t R x + ty = 1} = [s 1, s ]. Moreover, t x + ty is strictly decreasing on (, s 1 ] and strictly increasing on [s, ). Proof. By the convexity and the continuity of the function t x+ty and the fact that min t R x+ty = 1, {t R x+ty = 1} is a bounded closed convex subset of R. Thus there exist s 1, s with s 1 0 s such that {t R x+ty = 1} = [s 1, s ]. Assume that s > t > s and x+sy = x+ty > 1. Then there exists α (0, 1) such that t = (1 α)s + αs. By the convexity of t x + ty, we have x + sy = x + ty (1 α) + α x + sy < x + sy. This is a contradiction. Therefore t x + ty is strictly increasing on [s, ). One can similarly show that t x + ty is strictly decreasing on (, s 1 ]. 5

27 Proposition..5. Let X, Y be real Banach spaces with dim X = dim Y and let T be an isometric isomorphism from X onto Y. Then (i) For any x S X and any y V (x), m(t x, T y) = m(x, y). (ii) For any x S X, M(T x) = M(x). Proof. (i) Since Γ(T x, T y) = Γ(x, y), we have m(t x, T y) = sup{ T x + γt y γ Γ(x, y)} = sup{ x + γy γ Γ(x, y)} = m(x, y). (ii) From (i) and the fact that V (T x) = T (V (x)), we have M(T x) = sup{m(t x, T y) y V (x)} = sup{m(x, y) y V (x)} = M(x). Now we obtain a characterization of the Dunkl-Williams constant. Theorem..6. Let X be a real Banach space with dim X. Then DW (X) = sup{m(x) x S X }. Proof. Take any x, y S X with x + y 0. Assume that min 0 t 1 z(t) = z(t 0 ) for some t 0 [0, 1]. Then z(t 0 ) B x y by Lemma... Since Birkhoff orthogonality is homogeneous, we also have that z(t 0 )/ z(t 0 ) B x y. Since x = z(t 0 ) + t 0 (x y) and y = z(t 0 ) + (t 0 1)(x y), we have that z(t 0 ) z(t 0 ) + t 0 (x y) z(t 0 ) = 1 z(t 0 ) = z(t 0 ) z(t 0 ) + t 0 1 (x y) z(t 0 ). Thus we have and ( 1 t0 1 z(t 0 ) + t ) ( ) 0 z(t0 ) Γ z(t 0 ) z(t 0 ), x y x + y z(t 0 ) = z(t 0 ) z(t 0 ) + 1 ( t0 1 z(t 0 ) + t ) 0 (x y) z(t 0 ) ( ) z(t0 ) m z(t 0 ), x y ( ) z(t0 ) M. z(t 0 ) 6

28 By Proposition..1, we obtain DW (X) sup{m(x) x S X }. For each x S X, take any y V (x). Let λ, µ with λ 0 µ such that x + λy = x + µy. If λ = 0 or µ = 0, then we have x + λy = x + µy = 1 and hence x + λ + µ y = DW (X). We assume that λ < 0 < µ. Let t 1 = λ/(µ λ) (0, 1). Put z = x + λy x + λy and w = x + µy x + µy. Then (1 t 1 )z + t 1 w = x x + λy and hence (1 t 1)z + t 1 w = 1 x + λy. Thus we have x + λ + µ y = x + λy x + λy x + λy + x + µy x + µy = z + w (1 t 1 )z + t 1 w DW (X). Therefore DW (X) sup{m(x) x S X }. Lemma..7. Let X be a real Banach space with dim X. Then Γ(x, y) is a bounded subset of R for any x S X and any y V (x) \ {0}. Proof. Let x S X and y V (x) \ {0}. By Theorem..6 and the fact that DW (X) 4, we have m(x, y). Hence we have x+γy for any γ Γ(x, y). Thus Γ(x, y) is bounded. Proposition..8. Let X be a real Banach space with dim X. For each x S X and each y V (x) \ {0}, m(x, y) = max{ x + αy, x + βy }, where α = inf Γ(x, y) and β = sup Γ(x, y). 7

29 Proof. Let x S X and y V (x) \ {0}. Take an arbitrary element γ Γ(x, y). Then there exists t [0, 1] such that γ = (1 t)α + tβ. Thus we have x + γy (1 t) x + αy + t x + βy max{ x + αy, x + βy }. On the other hand, there exists {α n } in Γ(x, y) such that α n α. Then we have We have x + βy m(x, y) similarly. x + αy = lim n x + α n y m(x, y). To calculate the Dunkl-Williams constant, we need the following result. Theorem..9. Let X be a real Banach space with dim X, x S X y V (x). Suppose that {x n } in S X converges to x. If there exists {y n } with y n V (x n ) for all n N which converges to y, then m(x, y) lim m(x n, y n ). n Proof. By Proposition..3 (iii), it is clear if m(x, y) = 1, and so we may assume that m(x, y) > 1. For any ε (0, (m(x, y) 1)), there exist real numbers λ ε, µ ε with λ ε < 0 < µ ε such that x + λ ε y = x + µ ε y and x + λ ε + µ ε y > m(x, y) ε > 1. For arbitrary n N, there exists a nonnegative number µ n such that x n + λ ε y n = x n + µ n y n. and Since xn + µ n y n x + µ ε y = xn + λ ε y n x + λ ε y 0, it follows that x n + µ n y n x + µ ε y. From the fact that x + µn y x + µ ε y x + µ n y x n + µ n y n + x n + µ n y n x + µ ε y x x n + µ n y n y + xn + µ n y n x + µ ε y 0, 8

30 we also have x+µ n y x+µ ε y. Thus, by Lemma..4 and the fact x+µ ε y > 1, we obtain µ n µ ε. It follows from x n + λ ε + µ n y n x + λ ε + µ ε y that there exists n 0 N such that x n + λ ε + µ n y n > x + λ ε + µ ε y ε > m(x, y) ε for any n n 0. Therefore we obtain lim m(x n, y n ) inf m(x n, y n ) m(x, y) ε. n n n 0 Corollary..10. Let X be a real Banach space with dim X and let x S X. Suppose that C is a subset of V (x) satisfying M(x) = sup{m(x, y) y C} and that D is a dense subset of C. Then M(x) = sup{m(x, y) y D}. Proof. For any y 0 C, there exists a sequence {y n } in D converging to y 0. Thus m(x, y 0 ) lim m(x, y n ) sup{m(x, y) y D} n by Theorem..9. Therefore we obtain M(x) sup{m(x, y) y D}. Corollary..11. Suppose that X is a real Banach space with dim X. Let x S X and let {x n } in S X converging to x. Assume that, for any y V (x), there exists {y n } with y n V (x n ) \ {0} for all n N, converging to y. Then M(x) lim M(x n ). n Proof. Take an arbitrary element y V (x). By the assumption, there exists {y n } such that y n V (x n ) \ {0} for all n N and y n y. By Theorem..9, we have Thus we obtain M(x) lim M(x n ). n m(x, y) lim m(x n, y n ) lim M(x n ). n n 9

31 .3 A characterization by the frame of the unit ball The dual space of X is denoted by X. In [3], James proved the following result about Birkhoff orthogonality. Lemma.3.1 ([3]). Let X be a real Banach space with dim X. For x, y X, x B y if and only if there exists f S X such that f(x) = x and f(y) = 0. In this section, we characterize the Dunkl-Williams constant with the norming functionals. For each x 0 S X, a functional f S X is said to be norming if f(x 0 ) = x 0 = 1. Let ν(x 0 ) = {f S X f(x 0 ) = 1}. We note that, from the Hahn-Banach theorem, ν(x) for any x S X. For each x 0 and each f ν(x 0 ), let F (x 0, f) = S X (x 0 + ker f) and let E(x 0, f) be the relative boundary of F (x 0, f) in x 0 + ker f. We define the frame fr(b X ) of B X by fr(b X ) = {E(x, f) x S X, f ν(x)}. Lemma.3.. Suppose that X is a real Banach space with dim X. Let x 0 S X and f ν(x 0 ). Then ker f V (x) for any x F (x 0, f). Proof. Let x F (x 0, f). Take an arbitrary element y ker f. Then we have x = 1 = f(x) = f(x + λy) x + λy for all λ R. Thus y V (x). This means that ker f V (x). Let x 0 S X and f ν(x 0 ). By Lemmas..4 and.3., for each x F (x 0, f) and each y ker f \ {0}, we have s 1, s with s 1 0 s such that {t R x + ty = 1} = [s 1, s ]. Lemma.3.3. Suppose that X is a real Banach space with dim X. Let x 0 S X and f ν(x 0 ). Assume that x F (x 0, f) and that there exists y ker f \ {0} such that s 1 = 0 or s = 0 for s 1, s with {t R x + ty = 1} = [s 1, s ]. Then x E(x 0, f). Proof. Suppose that s = 0. Then, for any t > 0, we have x + ty > 1 and hence x + ty (x 0 + ker f) \ F (x 0, f). Thus we obtain x E(x 0, f). If s 1 = 0, we similarly have x E(x 0, f). 30

32 Remark.3.4. A Banach space X is said to be strictly convex if x, y S X x y imply x + y <. In a strictly convex Banach space X, and F (x 0, f) = E(x 0, f) = {x 0 } holds for each x 0 S X and each f ν(x 0 ) (cf. [46]), and hence fr(b X ) = S X. An element x S X is called an extreme point of B X if y, z S X and x = (y+z)/ imply x = y = z. The set of all extreme points of B X is denoted by ext(b X ). Proposition.3.5. Let X be a real Banach space with dim X. Then ext(b X ) fr(b X ). Proof. Let x S X \ fr(b X ) and let f ν(x). Since x fr(b X ), we have x F (x, f) \ E(x, f). Let y ker f \ {0}. Then, by Lemmas..4 and.3.3, there exist s 1, s with s 1 < 0 < s such that {t R x + ty = 1} = [s 1, s ]. Putting s 0 = min{ s 1, s }, then we have x ± s 0 y S X and x = 1 {(x + s 0y) + (x s 0 y)}. Therefore x ext(b X ). Lemma.3.6. Suppose that X is a real Banach space with dim X. Let x 0 S X and f ν(x 0 ). If x 0 F (x 0, f) \ E(x 0, f), then V (x 0 ) = ker f. Proof. By Lemma.3. and the fact that x 0 F (x 0, f), we have ker f V (x 0 ). Conversely, take any y V (x 0 ). Letting z = y f(y)x 0, then z 0 and z ker f. Since x 0 F (x 0, f) \ E(x 0, f), by Lemmas..4 and.3.3, we have s 1, s with s 1 < 0 < s such that {t R x 0 + tz = 1} = [s 1, s ]. For any t [s 1, s ], we have (1 tf(y)) x 0 B y and hence 1 tf(y) (1 tf(y)) x 0 + ty = x 0 + tz = 1. Thus f(y) = 0. Therefore we obtain V (x 0 ) = ker f. Theorem.3.7. Let X be a real Banach space with dim X. Then DW (X) = sup{m(x) x fr(b X )}. 31

33 Proof. Let x S X \ fr(b X ) and let f ν(x). Since x fr(b X ), we have x F (x, f) \ E(x, f). Let y V (x) \ {0}. By Lemmas..4 and.3.3, we have s 1, s with s 1 < 0 < s such that {t R x + ty = 1} = [s 1, s ]. For this s, we have y V (x + s y) by Lemmas.3. and.3.6. It follows from {t R (x + s y) + ty = 1} = [s 1 s, 0] that x + s y E(x, f) fr(b X ). We prove that m(x, y) m(x + s y, y). Let λ, µ be real numbers with λ 0 µ such that x + λy = x + µy. Case 1. Suppose that 0 µ s. From the fact that x + λy = x + µy = 1, we have s 1 λ 0. Thus we obtain (λ + µ)/ [s 1, s ] and hence x + λ + µ y = 1 m(x + s y, y). Case. If s < µ, then we have λ s < 0 < µ s. From the fact that x + s y + (λ s )y = x + s y + (µ s )y, we have Thus 1 {(λ s ) + (µ s )} Γ(x + s y, y). x + λ + µ y = x + s y + 1 {(λ s ) + (µ s )}y m(x + s y, y). Hence we have m(x, y) m(x + s y, y) M(x + s y) sup{m(x) x fr(b X )}. Thus and hence sup{m(x) x S X } sup{m(x) x fr(b X )} sup{m(x) x S X } = sup{m(x) x fr(b X )}. Therefore, by Theorem..6, we obtain DW (X) = sup{m(x) x fr(b X )}. Now we consider the case of dim X =. 3

34 Lemma.3.8. Let X be a two-dimensional real Banach space. Then ext(b X ) = fr(b X ). Proof. Let x S X \ ext(b X ). Then there exist distinct elements y, z S X such that x = (y + z)/. For any f ν(x), it follows from 1 = f(x) = 1 (f(y) + f(z)) that f(y) = f(z) = 1 and hence y, z F (x, f). Thus we have that x+t(y z) = 1 for all t [ 1/, 1/]. On the other hand, from the fact dim ker f = 1, we have ker f = [y z] which is the closed linear span of {y z}. Hence, for any w ker f \ {0}, there exists a unique nonzero real number α such that w = α(y z). Thus we have [ 1 ] α, 1 {t R x + tw = 1} α and so we obtain x E(x, f). Therefore x fr(b X ). By Theorem.3.7 and Lemma.3.8, we have the following result. Theorem.3.9. Let X be a two-dimensional real Banach space. Then DW (X) = sup{m(x) x ext(b X )}..4 Examples In this section, as an application of the results in above sections, we calculate the Dunkl-Williams constant of the Day-James space l -l. We first see that DW ((R, )) = and DW ((R, )) = 4. Example.4.1. DW ((R, )) = ([16, 34]). Proof. From Remark.3.4, we have ext ( B (R, )) = S(R, ). Letting e 1 = (1, 0), by Proposition..5 (i) and the fact that the mapping (r cos α, r sin α) (r cos(α + θ), r sin(α + θ)) is isometric isomorphism from (R, ) onto itself for all θ [0, π], it is enough to consider M(e 1 ). From the fact that, for x, y (R, ), x B y if and only if x y in the usual sense, V (e 1 ) = {(0, t) t R}. 33

35 Putting e = (0, 1), we have M(e 1 ) = m(e 1, e ) by Proposition..3 (iv). It is easy to check that m(e 1, e ) = 1. Thus, by Theorem..6, we obtain DW ((R, )) =. Example.4.. DW ((R, )) = 4. ([7, 5]) Proof. Let x 0 = (1, 1). Then it is clear that x 0 ext ( B (R, )). For t (0, 1), letting y t = (1, t), then y t V (x 0 ). For λ 0 and µ 0, we have ( 1 λt x 0 + λy t = λ 0), 1 t ( ) (1 + λ) λ 1 t and x 0 + µy t = 1 + µ. We note that, for any µ 0, there exists a unique λ 0 such that x 0 + λy t = x 0 + µy t. For µ t/(1 t), put λ = µ/t. Then x 0 + λy t = x 0 + µy t. We have λ + µ = 1 t µ and t x 0 + λ + µ y t = t µ. For µ t/(1 t), put λ = ( + µ). Then x 0 + λy t = x 0 + µy t. We have λ + µ = 1 and x 0 + λ + µ y t = 1 + t. Thus m(x 0, y t ) = 1 + t for any t (0, 1) and hence, by Theorem.3.9, DW ((R, )) M(x 0 ) sup{1 + t 0 < t < 1} = 4. From the fact that DW ((R, )) 4, we have DW ((R, )) = 4. The Day-James space l -l is defined as the space R with the norm (x, y) (xy 0), (x, y), = (x, y) (xy 0). In the rest of this chapter, we shall consider the space l -l and calculate the Dunkl- Williams constant DW (l -l ). From [55, Theorem.5], we have the following: Lemma.4.3. There is an isometric isomorphism that identifies (l -l ) l -l 1 such that if f (l -l ) is identified with the element (u, v) l -l 1, then f(x, y) = ux + vy for all (x, y) l -l. 34 with

36 Lemma.4.4. For any a [0, 1], put b = 1 a. Then V ((a, b)) = {α(b, a) α R}. Proof. It is clear that (a, b) B (b, a) and hence by Proposition..3 (ii). {α(b, a) α R} V ((a, b)) Conversely, take any y = (y 1, y ) V ((a, b)). Then, by Lemma.3.1, there exists f S (l -l ) such that f((a, b)) = 1 and f((y 1, y )) = 0. For this f, by Lemma.4.3, there exists (u, v) l -l 1 such that (u, v),1 = f (l -l ) f((x, y)) = ux + vy for all (x, y) l -l. By the Cauchy-Schwarz inequality, we have 1 = f((a, b)) = ua + vb (u, v) (a, b) (u, v),1 (a, b), = 1 = 1 and and hence (u, v) = (a, b). From the fact that f(y 1, y ) = 0, we obtain ay 1 + by = 0. Thus y = (y 1, y ) {α(b, a) α R}. Lemma.4.5. V ((1, 1)) = {(a, b) ab 0}. Proof. Take any y = (y 1, y ) V ((1, 1)). Then, by Lemma.3.1, there exists f S (l -l ) such that f((1, 1)) = 1 and f((y 1, y )) = 0. For this f, by Lemma.4.3, there exists (u, v) l -l 1 such that (u, v),1 = f (l -l ) = 1 and f((x, y)) = ux+ vy for all (x, y) l -l. From the fact that f((1, 1)) = 1, we have v = (1 u). We note that 0 u 1, since (u, (1 u)),1 = 1. It follows from f((y 1, y )) = 0 that uy 1 (1 u)y = 0. If u = 0, then y = 0 and hence y 1 y = 0. If u 0, then we have Thus y 1 y = 1 u u y 0. y = (y 1, y ) {(a, b) ab 0} and hence V ((1, 1)) {(a, b) ab 0}. Conversely, take any (a, b) with ab 0. If ab = 0, then we have (a, b) V ((1, 1)), immediately. Hence we may assume that ab > 0. We define a linear functional f by f((s, t)) = b a + b s a a + b t. 35

37 Then we have f((1, 1)) = 1, f((a, b)) = 0 and ( b f (l -l ) = a + b, a a + b),1 Thus, by Lemma.3.1, (a, b) V ((1, 1)) and hence Lemma.4.6. = b a + b + {(a, b) ab 0} V ((1, 1)). a a + b = 1 DW (l -l ) = max{sup{m((a, b), (b, a)) a + b = 1, 0 < b < 1/ < a < 1}, sup{m((1, 1), (a, b)) a + b = 1, 0 < b < 1/ < a < 1}}. Proof. By the definition of the norm,, we clearly have ext(b l -l ) = {±(a, b) a + b = 1, 0 a, b 1} {±(1, 1)}. Since the mapping x x is an isometric isomorphism from l -l onto itself, by Proposition..5 (ii), we have M( x) = M(x) for all x ext(b l -l ). By Theorem.3.9, we have DW (l -l ) = sup{m(x) x ext(b l -l )} have = sup{m(x) x {(a, b) a + b = 1, 0 a, b 1} {(1, 1)}} = max{sup{m((a, b)) a + b = 1, 0 a, b 1}, M((1, 1))}. Take an arbitrary element (a, b) with a + b = 1 and 0 a, b 1. Then we V ((a, b)) = {α(b, a) α R} by Lemma.4.4. Hence we have M((a, b)) = m((a, b), (b, a)) by Proposition..3 (iv). Thus sup{m((a, b)) a + b = 1, 0 a, b 1} = sup{m((a, b), (b, a)) a + b = 1, 0 a, b 1}. Since the mapping (s, t) (t, s) is an isometric isomorphism from l -l onto itself, by Proposition..5 (i), we have m((b, a), (a, b)) = m((a, b), (b, a)) 36

38 for all (a, b) satisfying a + b = 1 and 0 a, b 1. Thus sup{m((a, b), (b, a)) a + b = 1, 0 a, b 1} = sup{m((a, b), (b, a)) a + b = 1, 0 b 1/ a 1}. By Corollary..11, we obtain Thus sup{m((a, b)) a + b = 1, 0 a, b 1} = sup{m((a, b), (b, a)) a + b = 1, 0 < b < 1/ < a < 1}. For all (a, b) (0, 0) with ab 0, by Proposition..3 (iv), we have ( ) (a, b) m((1, 1), (a, b)) = m (1, 1),. a + b M((1, 1)) = sup{m((1, 1), (a, b)) a + b = 1, 0 a, b 1}. Take an arbitrary element (a, b) with a + b = 1 and 0 a, b 1. Since the mapping (s, t) ( t, s) is an isometric isomorphism from l -l onto itself, we have by Proposition..5 (i). Hence we have by Proposition..3 (iv). Thus m((1, 1), (a, b)) = m((1, 1), ( b, a)) m((1, 1), (a, b)) = m((1, 1), (b, a)) sup{m((1, 1), (a, b)) a + b = 1, 0 a, b 1} = sup{m((1, 1), (a, b)) a + b = 1, 0 b 1/ a 1}. By Corollary..10, we obtain M(1, 1) = sup{m((1, 1), (a, b)) a + b = 1, 0 < b < 1/ < a < 1}. Henceforth, for a (1/, 1), we put b = 1 a and let x a = (a, b), y a = (b, a). We also put u = (1, 1). 37

39 Lemma.4.7. The following holds. (i) For each µ 0, there exists a unique λ 0 such that x a + λy a, = x a + µy a,. (ii) For each µ 0, there exists a unique λ 0 such that u + λx a, = u + µx a,. Proof. (i) We define a function f from R into R by f(t) = x a + ty a,. Then we have that min t R f(t) = 1 is attained at only t = 0. Hence, by Lemma..4, f(t) is strictly increasing on [0, ) and strictly decreasing on (, 0]. Thus we obtain (i). The proof of (ii) is similar and so we omit it. Lemma.4.8. Γ(x a, y a ) = [ [ 0, b ] a 0, a + b ab a b ] (a b), (a > b). Proof. Γ(x a, y a ) is defined by { } λ + µ Γ(x a, y a ) = λ 0 µ, x a + λy a, = x a + µy a,. Hence, by Lemma.4.7 (i), for each µ 0, it is enough to find λ 0 such that x a + λy a, = x a + µy a, and calculate (λ + µ)/. For λ 0 and µ 0, we have b λa (λ a x a + λy a, = ), b 1 + λ ( a λ 0), b Related to x a + µy a, = x a a b y a = 1, b and 1 + µ (0 µ b a ), a + µb ( b a µ a+b a b ), µa b ( a+b a b µ). x a + a + b a b y a = 1, a b, 38

40 we have to consider the following two cases. Case 1. a b. Then x a a b y a, = 1 b 1 a b = x a + a + b a b y a and we obtain that b/a (1 ab)/b (a + b)/(a b) and x a + 1 ab y a = 1, b = xa a b y a Thus we consider the following four subcases. b,, Subcase µ b/a. Put λ = µ. Then we have a/b λ 0,. x a + λy a, = 1 + λ = 1 + µ = x a + µy a, and (λ + µ)/ = 0. Subcase 1.. b/a µ (1 ab)/b. Put λ = abµ + (µ 1)b. Then we have a/b λ 0, x a + λy a, = 1 + abµ + (µ 1)b = a + abµ + µ b = a + µb = x a + µy a, and λ + µ = µ abµ + (µ 1)b. Subcase 1.3. (1 ab)/b µ (a + b)/(a b). Put λ = (a b + µb)/a. Then we have λ a/b, x a + λy a, = b ( a b + µb ) a a = b + (a b + µb) = a + µb = x + µy a, and Notice that λ + µ = µ 1 1 ab b 1 = (a b)(µ 1). a a(a b) b 0. 39

41 Subcase 1.4. (a + b)/(a b) µ. Put λ = (µa b)/a. Then we have λ a/b, ( ) µa b x a + λy a, = b a a and (λ + µ)/ = b/a. We define a function f on R + by = b + (µa b) = µa b = x + µy a, f(µ) = µ abµ + (µ 1)b. Then it is easy to check that f(µ) is increasing in the interval [b/a, (1 ab)/b ]. Hence we have Γ(x a, y a ) = [0, b/a]. Case. a > b. Then x a a b y a, = 1 b > 1 a b = x a + a + b a b and we obtain that (1 + b )/ab (a + b)/(a b) and x a b y a = ab, 1 b = xa a b y a Thus we consider the following four subcases.,, Subcase.1. 0 µ b/a. Similarly to Subcase 1.1, we have (λ + µ)/ = 0. Subcase.. b/a µ (a + b)/(a b). Similarly to Subcase 1., we have λ + µ = µ abµ + (µ 1)b. Subcase.3. (a + b)/(a b) µ (1 + b )/ab. Put λ = abµ + (µ 1)a. Then we have a/b λ 0, x a + λy a, = 1 + λ = 1 abµ + (µ 1)a. = b abµ + µ a = b µa = x a + µy a, and λ + µ = µ abµ + (µ 1)a. 40

42 Subcase.4. (1 + b )/ab µ. Similarly to Subcase 1.4, we have (λ + µ)/ = b/a. We define a function g on R + by g(µ) = µ abµ + (µ 1)a. Then it is clear that g(µ) is decreasing in the interval [(a + b)/(a b), (1 + b )/ab]. Hence we have [ Γ(x a, y a ) = 0, g ( )] [ a+b a b = 0, a + b ] ab. a b Lemma.4.9. Γ(u, x a ) = [ (a b), 0]. Proof. Γ(u, x a ) is defined by { } λ + µ Γ(u, x a ) = λ 0 µ, u + λx a, = u + µx a,. Hence, by Lemma.4.7 (ii), for each µ 0, it is enough to find λ 0 such that u + λx a, = u + µx a, and calculate (λ + µ)/. For λ 0 and µ 0, we have + λ + (a b)λ (λ 1/a), u + λx a, = 1 λb ( 1/a λ 0), 1 + µa (0 µ 1/b), u + µx a, = + µ + (a b)µ (1/b µ). For any 1/ < a < 1, u 1 a x a = 1 + b, a < 1 + a b = u + 1 b x a holds and so we obtain 0 b/a 1/b and u + b a x a = 1 + b, a = u 1 a x a., Thus we consider the following three cases., Case 1. 0 µ b/a. Put λ = aµ/b. Then we have 1/a λ 0, u + λx a, = 1 ( a ) b µ b = 1 + µa = u + µx a, 41