Ann. Funct. Anal. 2 (2011), no. 2, A nnals of F unctional A nalysis ISSN: (electronic) URL:

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1 Ann. Funct. Anal. (0), no., 33 A nnals of F unctional A nalysis ISSN: (electronic) URL: SOME GEOMETRIC CONSTANTS OF ABSOLUTE NORMALIZED NORMS ON R HIROYASU MIZUGUCHI AND KICHI-SUKE SAITO Communicated by J. I. Fujii Abstract. We consider the Banach space X = (R, ) with a normalized, absolute norm. Our aim in this paper is to calculate the modified Neumann- Jordan constant C NJ (X) and the Zbăganu constant C Z(X).. Introduction and preliminaries Let X be a Banach space with the unit ball B X = {x X : x } and the unit sphere S X = {x X : x = }. Many geometric constants for a Banach space X have been investigated. In this paper we shall consider the following constants; { } x + y + x y C NJ (X) = sup ( x + y ) (x, y) (0, 0), { } 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). The constant C NJ (X), called the von Neumann-Jordan constant (hereafter referred to as NJ constant) have been considered in many papers ([3, 8, 0, ] and so on). The constant C NJ (X), called the modified von Neumann-Jordan constant (shortly, modified NJ constant) was introduced by Gao in [5] and does not necessarily coincide with C NJ (X) (cf. [, 4, 7]). The constant C Z (X) was introduced by Zbăganu ([5]) and was conjectured that C Z (X) coincides with Date: Received: 3 March 0; Accepted: 4 June 0. Corresponding author. 00 Mathematics Subject Classification. Primary 46B0; Secondary 46B5. Key words and phrases. Zbăganu constant, absolute norm, von Neumann-Jordan constant.

2 GEOMETRIC CONSTANTS OF R 3 the von Neumann-Jordan constant C NJ (X), but Alonso and Martin [] gave an example that C NJ (X) C Z (X) (cf.[6, 9]). A norm on R is said to be absolute if (a, b) = ( a, b ) for any (a, b) R, and normalized if (, 0) = (0, ) =. Let AN denote the family of all absolute normalized norm on R, and Ψ denote the family of all continuous convex function ψ on [0, ] such that ψ(0) = ψ() = and max{ t, t} ψ for all 0 t. As in [], it is well known that AN and Ψ are in a one-toone correspondence under the equation ψ = ( t, t) (0 t ). Denote ψ be an absolute normalized norm associated with a convex function ψ Ψ. For ψ, ϕ Ψ, we denote ψ ϕ if ψ ϕ for any t in [0, ]. Let ψ M = max 0 t ψ and M ψ = max 0 t ψ, where ψ = ( t, t) = ( t) + t corresponds to the l -norm. In [], Saito, Kato and Takahashi proved that, if ψ ψ (resp. ψ ψ ), then C NJ (C, ψ ) = M (resp. M ). We put X = (R, ψ ) for ψ Ψ. Our aim in this paper is to consider the conditions of ψ that C NJ (X) = C Z (X) or C NJ (X) = C NJ (X). In, we consider the modified von Neumann-Jordan 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 that C NJ (R, ψ ) = C NJ (R, ψ ) = M. Further, we consider the conditions that C NJ (R, ψ ) = C NJ (R, ψ ) = M M. In 3, we study the Zbăganu constant. First, we show that, if ψ ψ, then C Z (R, ψ ) = C NJ (R, ψ ) = M. If ψ ψ, then we give the necessarily and sufficient condition for that C Z (R, ψ ) = C NJ (R, ψ ) = M. Further we study the conditions that C Z (R, ψ ) = C NJ (R, ψ ) = M M. In 4, we calculate the modified NJ-constant C NJ (X) and the Zbăganu constant C Z (X) for some normed liner spaces.. The modified NJ constant of R In this section, we consider the Banach space X = (R, ψ ). 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. Proof. For any x, y S X, by [, 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 at t = t 0 (0 t 0 ), and put x = ψ(t 0 ) ( t 0, t 0 ), y = ψ(t 0 ) ( t 0, t 0 ).

3 4 H. MIZUGUCHI, K.-S. SAITO Then x, y S X and x + y ψ + x y ψ = 4( t 0) + 4t 0 ψ(t 0 ) = 4 ψ (t 0 ) ψ(t 0 ) = 4M, which implies that C NJ (X) = M. By [, Theorem ], we have this proposition. If ψ ψ, by [, Theorem ], then C NJ (X) = M. We now give the necessarily and sufficient condition of C NJ (X) = M. Theorem.. Let ψ Ψ such that ψ ψ. Then C NJ (X) = M if and only if there exist s, t [0, ] (s < t) satisfying one of the following conditions: () ψ = ψ, ψ = ψ and, if we put r = ψt+ψs ψ(r), then ψ+ψ ψ = (r) ψ( r) ψ ( r) = M. () ψ = ψ, ψ = ψ and, if we put r = ψs+ψt ψ( r) ψ ( r) = M. ψ+ψ(t ), then ψ(r) ψ (r) = Proof. (= ) Suppose that C NJ (X) = M. First, for any x, y S X, by [, Lemma 3], we have x + y ψ + x y ψ M ( x + y + x y ) ( = M x + y ) M x ψ + y ψ = 4M. Since X = (R, ψ ) is finite dimensional, { x + y C NJ(X) ψ + x y } ψ = max 4 x, y S X. Therefore, C NJ (X) = M if and only if there exist x, y S X (x y) such that x + y ψ + x y ψ = 4M. From the above inequality, the elements x, y S X (x y) satisfy x ψ = x =, y ψ = y = and x + y ψ x + y = x y ψ x y = M. Since ψ is absolute and x, y S X (x y) satisfy x = y =, it is sufficient to consider the following three cases: (i) There exist s, t [0, ] (s t) satisfying x = ψ ( s, s) and y = ψ ( t, t). (ii) There exist s, t [0, ] (s < t) satisfying x = ψ ( s, s) and y = ψ ( + t, t).

4 GEOMETRIC CONSTANTS OF R 5 (iii) There exist s, t [0, ] (s > t) satisfying x = ψ ( s, s) and y = ψ ( + t, t). Case (i). We may suppose that s < t. Then there exist α, β [0, π ] (α < β) such that x = ( s, s) = (cos α, sin α), y = ( t, t) = (cos β, sin β). ψ ψ Since x = y =, we have x + y = ( s ψ + t ψ, s ψ + By [3, Propositions a and b], we remark that s ψ t ψ, t ψ ) = x + y (cos α + β s ψ t ψ., sin α + β ). Since x y is orthogonal to x + y in the Euclidean space (R, ), we have Thus we have x y = ( s ψ t ψ, s ψ = x y (cos α + β π = x y (sin α + β x + y ψ = x + y (cos α + β = x + y (cos α + β Since x + y ψ = M x + y, we have Putting r = We also have M = (cos α + β cos α+β sin α+β +sin α+β + sin α + β x y ψ = x y (sin α + β t ψ ), sin α + β π ), cos α + β )., sin α + β ) ψ + sin α + β sin α+β cos α+β + sin α+β sin α+β cos α+β + sin α+β, then it is clear that r = ψt+ψs ψ+ψ + cos α + β Since x y ψ = M x y, we similarly have M = (sin α + β + cos α + β cos α+β sin α+β + cos α+β ). cos α+β cos α+β + sin α+β ) = ). and M = ψ(r) ψ (r). ). ψ( r) ψ ( r). Case (ii). Then there exist α [0, π] and β [ π, π] such that x = ( s, s) = (cos α, sin α), y = ( + t, t) = (cos β, sin β). ψ ψ

5 6 H. MIZUGUCHI, K.-S. SAITO Since x = y =, we have x + y = ( s ψ t ψ, s ψ + By [3, Propositions a and b], we remark that s ψ t ψ, t ψ ) = x + y (cos α + β s ψ t ψ., sin α + β ). Since x y is orthogonal to x + y in the Euclidean space (R, ), we have x y = ( s ψ + t ψ, s ψ = x y (cos α + β π = x y (sin α + β Since cos α+β 0 and sin α+β 0, we have x + y ψ = x + y (cos α + β = x + y (cos α + β Since x + y ψ = M x + y, we have Putting r = We also have M = (cos α + β cos α+β sin α+β +sin α+β + sin α + β x y ψ = x y (sin α + β t ψ ), sin α + β π ), cos α + β )., sin α + β ) ψ + sin α + β sin α+β cos α+β + sin α+β sin α+β cos α+β + sin α+β, then it is clear that r = ψs+ψt ψ+ψ(t ) and M = ψ(r) ψ (r). + cos α + β Since x y ψ = M x y, we similarly have M = (sin α + β + cos α + β cos α+β sin α+β + cos α+β ). cos α+β cos α+β + sin α+β ) = ). ). ψ( r) ψ ( r). Case (iii). There exist s, t [0, ] (s > t) satisfying x = ψ ( s, s) and y = ( + t, t). Then, we put s ψ 0 = t and t 0 = s. We define x 0, y 0 in S X by x 0 = ψ(s 0 ) ( s 0, s 0 ), y 0 = ψ(t 0 ) ( + t 0, t 0 ). Then we can reduce Case (ii). ( =). If we suppose () (resp. ()), then we put x = ψ ( s, s) (resp. x = ψ ( s, s)) and y = ψ ( t, t) (resp. y = ψ ( + t, t)). Then we have x ψ = x =, y ψ = y =, x + y ψ = M x + y and

6 GEOMETRIC CONSTANTS OF R 7 x y ψ = M x y. Hence it is clear to prove that C NJ (X) = M. This completes the proof. We next study the modified NJ constant in the general case. If ψ Ψ, then by [, Therem 3], we have max{m, M } C NJ (X) M M. However, by Theorem., there exist many ψ Ψ satisfying ψ ψ such that C NJ(X) < max{m, M } = C NJ (X). From [, Theorem 3], C NJ (X) = M M if either ψ/ψ or ψ /ψ attains a maximum at t = /. Then, we have the following Proposition.3. Let ψ Ψ and let ψ = ψ( t) for all t [0, ]. If ψ/ψ attains a maximum at t = /, then C NJ (X) = C NJ(X) = M M. Proof. Suppose first M = ψ(/)/ψ (/). Take an arbitrary t [0, ] and put Then x, y S X and Therefore we have x = (t, t), y = ( t, t). ψ ψ x + y ψ = ψ ψ( ), x y ψ = x + y ψ + x y ψ 4 t ψ( ψ ). = { (t ) + } ψ(/) ψ = ψ ψ(/) ψ = ψ ψ(/) ψ ψ (/) = M ψ ψ. Since t is arbitrary, we have C NJ (X) M M which prove that C NJ (X) = M M. In the case that M = ψ (/)/ψ(/), C NJ (X) does not necessarily coincide with M M. However, we have the following Theorem.4. Let ψ Ψ and let ψ = ψ( t) for all t [0, ]. Assume that M = ψ (/)/ψ(/) and M >. Then C NJ (X) = M M if and only if there exist s, t [0, ] (s < t) satisfying one of the following conditions: () ψ = M ψ, ψ = M ψ and, if we put r = ψt+ψs, then ψ+ψ ψ(r) = M ψ (r). () ψ = M ψ, ψ = M ψ and, if we put r = ψs+ψt, then ψ+ψ(t ) ψ(r) = M ψ (r).

7 8 H. MIZUGUCHI, K.-S. SAITO Proof. (= ). For all x, y S X, we have ( x + y ψ + x y ψ M x + y + x y ) ( = M x + y ) M M x ψ + y ψ = 4M M. From this inequality, C NJ (X) = M M if and only if there exist x, y S X (x y) such that x + y ψ + x y ψ = 4M M. Suppose that C NJ (X) = M M. Then, the elements x, y S X (x y) satisfy x = y = M, x + y ψ = M x + y, x y ψ = M x y. Since ψ is absolute, it is sufficient to consider the following three cases: (i) There exist s, t [0, ] (s t) satisfying x = ( s, s) and y = ψ ( t, t). ψ (ii) There exist s, t [0, ] (s < t) satisfying x = ψ ( + t, t). ψ (iii) There exist s, t [0, ] (s > t) satisfying x = ψ ( + t, t). ψ ( s, s) and y = ( s, s) and y = As in the proof of Theorem., we can prove this theorem. This completes the proof. 3. The Zbăganu constant of R The Zbăganu constant C Z (X) in [5] 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 3.. Let ψ Ψ. If ψ ψ, then C Z (X) = C NJ (X) = M. Proof. For any x, y X, x + y ψ x y ψ x + y ψ + x y ψ ( M x + y + x y ) ( = M x + y ) M x ψ + y ψ. Since ψ/ψ attains the maximum at t = t 0 (0 t 0 ), we put x = ( t 0, 0) and y = (0, t 0 ), respectively. Then we have x + y ψ + x y ψ = ψ(t 0 ) = M ψ (t 0 ) = M x ψ + y ψ.

8 Since x + y ψ = ψ(t 0 ) = x y ψ, we have Therefore we have which implies that C Z (X) = M. GEOMETRIC CONSTANTS OF R 9 x + y ψ x y ψ = x + y ψ + x y ψ = M x ψ + y ψ. x + y ψ x y ψ x ψ + y ψ = M, We next consider the case that ψ ψ. We remark that the Zbăganu constant C Z (X) is 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 3.. Let ψ Ψ. Assume that ψ ψ. Then C Z (X) = M if and only if there exist s, t [0, ] (s < t) satisfying one of the following conditions: () ψ = ψ, ψ = ψ and, if we put r = ψt+ψs, then ψ (r) = ψ+ψ ψ(r) ψ( r) = M ψ ( r). () ψ = ψ, ψ = ψ and, if we put r = ψs+ψt, then ψ (r) = ψ+ψ(t ) ψ(r) ψ( r) = M ψ ( r). Proof. 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 finite dimensional, { } 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 Hence we may assume that x + y x + y ψ = x y x y ψ = M. x = x ψ = y ψ = y =.

9 30 H. MIZUGUCHI, K.-S. SAITO As in the proof of Theorem., it is sufficient to consider the following three cases: (i) There exist s, t [0, ] (s t) satisfying x = ψ ( s, s) and y = ψ ( t, t). (ii) There exist s, t [0, ] (s < t) satisfying x = ψ ( s, s) and y = ψ ( + t, t). (iii) There exist s, t [0, ] (s > t) satisfying x = ψ ( s, s) and y = ψ ( + t, t). As in the proof of Theorem., we can similarly prove this theorem. We next study the Zbăganu constant C Z (X) in general case. If ψ Ψ, by [, Theorem 3], then we have max{m, M } C Z (X) C NJ (X) M M. However, by Theorem 3., there exist many ψ Ψ satisfying ψ ψ such that C Z (X) < C NJ (X) max{m, M }. From [, Theorem 3], C NJ (X) = M M if either ψ/ψ or ψ /ψ attains a maximum at t = /. Then, we have the following Proposition 3.3. Let ψ Ψ and let ψ = ψ( t) for all t [0, ]. If M = ψ (/) ψ(/), then C Z(X) = C NJ (X) = M M. Proof. From the definition, we have C Z (X) C NJ (X) = M M. Take an arbitrary t [0, ] and put x = (t, t) and y = ( t, t). Then x ψ = y ψ = ψ and x + y ψ = (, ) ψ = ψ(/), x y ψ = (t, t) ψ = t ψ(/). Hence we have 4 x ψ y ψ x + y ψ + x = ( x ψ + ) y ψ y ψ x + y ψ + x y ψ ψ = ( + (t ) ) ψ(/) ψ = ψ ψ(/) = ψ ψ (/) ψ ψ(/) = M ψ ψ Since t is arbitrary, we have C Z (X) M M. Therefore we have C Z (X) = M M. This completes the proof. In case that M = ψ(/)/ψ (/), we have the following theorem as in the proof of Theorem. and so omit the proof. Theorem 3.4. Let ψ Ψ and let ψ = ψ( t) for all t [0, ]. If M = ψ(/) ψ (/) and M >, then C Z (X) = M M if and only if there exist s, t [0, ] (s < t) satisfying one of the following conditions:

10 GEOMETRIC CONSTANTS OF R 3 () ψ = M ψ, ψ = M ψ and, if we put r = ψt+ψs, then ψ+ψ ψ(r) = M ψ (r). () ψ = M ψ, ψ = M ψ and, if we put r = ψs+ψt, then ψ+ψ(t ) ψ(r) = M ψ (r). 4. Examples In this section, we calculate C NJ (X) and C Z(X) of some Banach spaces X = (R, ψ ), where ψ Ψ. First, we consider the case that ψ = ψ p. Example 4.. Let p and /p + /q =. We put t = min(p, q). Then C NJ (R, p ) = C Z (R, p ) = C NJ (R, p ) = t. Suppose that p. Since ψ p ψ, we have C Z (R, p ) = p by Proposition 3.. On the other hand, as in Theorem., we take s = 0 and t =. Since r = ψ(0) +ψ() 0 ψ(0)+ψ() = and M = ψ p (/)/ψ (/) = p, by Theorem., we have C NJ (R, p ) = M = p. If p, then we similarly have, by Proposition. and Theorem 3., C NJ (R, p ) = C Z (R, p ) = C NJ (R, p ) = p. In [4, Example], C. Yang and H. Li calculated the modified NJ constant of the following normed linear space. From our theorems, we have Example 4.. Let λ > 0 and X λ = R endowed with norm (x, y) λ = ( (x, y) p + λ (x, y) q) /. (i) If p q, then C NJ (X λ ) = C NJ (X λ) = C Z (X λ ) = (λ+) /p +λ /q. (ii) If p q, then C NJ (X λ ) = C NJ (X λ) = C Z (X λ ) = /p +λ /q (λ+). To see this, first, we remark that (p, q) is not necessarily a Hölder pair. We define the normalized norm 0 λ by (x, y) 0 λ = (x, y) λ + λ. Then 0 λ is absolute and so put the corresponding function ψ λ = ( t, t) 0 λ. (i) Suppose that p q. Since ψ λ ψ, by Proposition., we have C NJ (X λ ) = C NJ (X λ) = M = (λ+). On the other hand, in Theorem 3., we /p +λ /q take s = 0 and t =. Then we have r = / and ψ (/) = M ψ λ (/). Thus we have C Z (X λ ) = M = /p +λ /q. (λ+) (ii) Suppose that p q. Since ψ λ ψ, by Theorem. and Proposition 3., we similarly have (ii). Example 4.3. Put ψ (0 t /), ψ = ( ) t + (/ t ). Then C NJ (R, ψ ) < C Z (R, ψ ) = C NJ (R, ψ ) = ( ).

11 3 H. MIZUGUCHI, K.-S. SAITO In fact, ψ Ψ and the norm of ψ is a + b ( a b ) (a, b) ψ = ( ) a + b ( a b ). Since ψ ψ, by Proposition 3., we have C Z (R, ψ ) = M = ( ). We assume that C NJ (R, ψ ) = M. By Theorem., we can choose r [0, ] such that ψ(r) = ψ( r) = M ψ (r) ψ ( r). This is impossible by the definition of ψ. Therefore we have C NJ (R, ψ ) < M. Example 4.4. Let / β. We define a convex function ψ β Ψ by ψ β = max{ t, t, β}. By [, Example 4], we have β + ( β) C NJ (R (β [, ψβ ) = β, ]) (β + ( β) ) (β (, ]). Indeed, and M = { (β [ ψ β (/) ψ (/) = β / = β (β ( M = ψ (β) ψ β (β) = β {( β) + β } /., ]), ]) If / β /, then ψ β ψ and so, by Proposition., we have C NJ(R, ψβ ) = M = β + ( β) β. By Theorem 3., we have C Z (R, ψβ ) < M. Assume that / < β. Since M = ψ β(/) ψ, we have, by Proposition.3, (/) C NJ(R, ψβ ) = M M = (β + ( β) ). On the other hand, we take s = β and t = β in Theorem 3.4. Then we have r = ψ(β)( β)+ψ( β)β = /. By Theorem 3.4, we have ψ(β)+ψ( β) C Z (R, ψβ ) = M M = (β + ( β) ). Example 4.5. We consider ψ β in Example 4.4 in case of β = /. Then we have C NJ(R, ψβ ) = C NJ (R, ψβ ) = M = ( ). On the other hand, we have C Z (R, ψβ ) = M = ( ). For this ψ β, define a convex function ϕ Ψ by { ψβ (0 t /), ϕ = ψ (/ t ).

12 As in Example 4., we similarly have GEOMETRIC CONSTANTS OF R 33 C Z (R, ϕ ) < C NJ(R, ϕ ) = C NJ (R, ϕ ) = M = ( ). Acknowledgement. The second author is supported in part by Grants-in- Aid for Scientific Research, Japan Society for the Promotion of Science (No ). The authors would also like to thank the referees for some helpful comments. References. J. Alonso, P. Martin and P.L. Papini, Wheeling around von Neumann-Jordan constant in Banach Spaces, Studia Math. 88 (008), no., J. Alonso and P. Martin, A counterexample for a conjecture of G. Zbâganu about the Neumann-Jordan constant, Rev. Roumaine Math. Pures Appl. 5 (006), J.A. Clarkson, The von Neumann-Jordan constant for the Lebesgue spaces, Ann. of Math. () 38 (937), no., J. Gao, A Pythagorean approach in Banach spaces, J. Inequal. Appl. (006), Art. ID 9498,. 5. J. Gao and K. Lau, On the geometry of spheres in normed linear spaces, J. Austral. Math. Soc., 48(990), J. Gao and S. Saejung, Normal structure and the generalized James and Zbăganu constant, Nonlinear Anal. 7 (009), no. 7-8, J. Gao and S. Saejung, Some geometric measures of spheres in Banach spaces, Appl. Math. Comput. 4 (009), no., P. Jordan and J. von Neumann, On inner products in linear metric spaces, Ann. of Math. () 36 (935), no. 3, E. Llorens-Fuster, E.M. Mazcuñán-Navarro and S. Reich, The Ptolemy and Zbăganu constants of normed spaces, Nonlinear Anal. 7 (00), no., M. Kato and Y. Takahashi, On sharp estimates concerning von Neumann-Jordan and James constants for a Banach space, Rend. Circ. Mat. Palermo, Serie II, Suppl. 8 (00), 7.. K.-S. Saito, M. Kato and Y. Takahashi, Von Neumann-Jordan constant of absolute normalized norms on C, J. Math. Anal. Appl. 44 (000), no., Y. Takahashi and M. Kato, A simple inequality for the von Neumann-Jordan and James constants of a Banach space, J. Math. Anal. Appl. 359 (009), no., Y. Takahashi, M. Kato and K. -S. Saito, Strict convexity of absolute norms on C and direct sums of Banach spaces, J. Inequal. Appl., 7(00), C. Yang and H. Li, An inequality between Jordan-con Neumann constant and James constant, Appl. Math. Lett. 3 (00), no. 3, G. Zbăganu, An inequality of M. Rǎdulescu and S. Rǎdulescu which characterizes inner product spaces, Rev. Roumaine Math. Puers Appl. 47 (00), Department of Mathematical Sciences, Graduate School of Science and Technology, Niigata University, Niigata, Japan. address: mizuguchi@m.sc.niigata-u.ac.jp address: saito@math.sc.niigata-u.ac.jp