Ann. Funct. Anal. (0), no., 33 A nnals of F unctional A nalysis ISSN: 008-875 (electronic) URL: www.emis.de/journals/afa/ 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.
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 ).
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).
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
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
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).
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.
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 =.
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:
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 ) = ( ).
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 ).
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