Thông tin tài liệu
Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2011, Article ID 875649, 15 pages doi:10.1155/2011/875649 Research Article Nonsquareness and Locally Uniform Nonsquareness in Orlicz-Bochner Function Spaces Endowed with Luxemburg Norm Shaoqiang Shang, 1 Yunan Cui, 2 and Yongqiang Fu 1 1 Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China 2 Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China Correspondence should be addressed to Shaoqiang Shang, sqshang@163.com Received 5 July 2010; Accepted 12 February 2011 Academic Editor: Nikolaos Papageorgiou Copyright q 2011 Shaoqiang Shang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Criteria for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces equipped with Luxemburg norm are given. We also prove that, in Orlicz-Bochner function spaces generated by locally uniform nonsquare Banach space, nonsquareness and locally uniform nonsquareness are equivalent. 1. Introduction A lot of nonsquareness concepts in Banach spaces are known see 1. Nonsquareness are important notions in geometry of Banach space. One of reasons is that these properties are strongly related to the fixed point property see 2. The criteria for nonsquareness and locally uniform nonsquareness in the classical Orlicz function spaces have been given in 3, 4 already. However, because of the complicated structure of Orlicz-Bochner function spaces equipped with the Luxemburg norm, the criteria for nonsquareness and locally uniform nonsquareness of them have not been found yet. The aim of this paper is to give criteria for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces equipped with Luxemburg norm. Let X, · be a real Banach space. SX and BX denote the unit sphere and unit ball, respectively. Let us recall some geometrical notions concerning nonsquareness. A Banach space X is said to be nonsquare if for any x, y ∈ SX we have min{xy/2, x−y/2} < 1. A Banach space X is said to be uniformly nonsquare if there exists δ>0 such that for any x, y ∈ SX,min{x y/2, x − y/2 } < 1 − δ. A Banach space X is said to be locally uniformly nonsquare if for any x ∈ SX, there exists δ x > 0 such that min{x y/2, x − y/2} < 1 − δ x , where y ∈ SX. 2 Journal of Inequalities and Applications Let R be set of real numbers. A function M : R → R is called an N-function if M is convex, even, M00, Mu > 0 u / 0 and lim u → 0 Mu/u0, and lim u → 0 Mu/u ∞. Let T, Σ,μ be a nonatomic measurable space. p denotes right derivative of M. Moreover, for a given Banach space X, ·, we denote by X T the set of all strongly μ- measurable function from T to X, and for each u ∈ X T , we define the modular of u by ρ M u G M u t dt. 1.1 Put L M u t ∈ X T : G M λu t dt < ∞ for some λ>0 . 1.2 The linear set L M endowed with the Luxemburg norm u inf λ>0:ρ M u λ ≤ 1 1.3 is a Banach space. We say that an Orlicz function M satisfies condition Δ 2 M ∈ Δ 2 if there exist K>2andu 0 ≥ 0 such that M 2u ≤ KM u u ≥ u 0 . 1.4 First let us recall a known result that will be used in the further part of the paper. Lemma 1.1 see 3. Suppose M ∈ Δ 2 .Then ρ M u n −→ 0 ⇐⇒ u n −→ 0,ρ M u n −→ 1 ⇐⇒ u n −→ 1 n −→ ∞ . 1.5 2. Main Results Theorem 2.1. L M is nonsquare if and only if a M ∈ Δ 2 ; b X is nonsquare. In order to prove the theorem, we give a lemma. Lemma 2.2. If X is nonsquare, then for any x, y / 0, we have x y − min x y , x − y > 0. 2.1 Journal of Inequalities and Applications 3 Proof. Case 1. If x < y, then x y ≤ x x y · y 1 − x y · y < x x y − x x y 2.2 or x − y < x y . 2.3 Case 2. If x≥y, then x y ≤ y x · x y x y − 1 · y < y y x − y x y 2.4 or x − y < x y . 2.5 This implies x y−min{x y, x − y} > 0. This completes the proof. Proof of Theorem 2.1. a Necessity. Suppose that M/∈ Δ 2 , then there exist u ∈ SL M and δ>0 such that ρ M u1 − δ<1. Pick c>0 such that E {t ∈ T : ut≤c} is not a null set. Since M/∈ Δ 2 , there exist sequence {r n } ∞ n1 and disjont subsets {E n } ∞ n1 of E such that r n > 2nc, M 1 1 n r n > 2 n M 1 1 2n r n , 2 n M 1 1 2n r n μE n 2 −n δ. 2.6 Therefore, if we define v ∞ n1 r n χ E n , then for any l>1, we have ρ M lv ∞ n1 ρ M lr n χ E n ≥ ∞ nm ρ M 1 1 n r n χ E n > ∞ nm 2 n ρ M 1 1 2n r n χ E n ≥ ∞ nm 2 n M 1 1 2n r n μE n ∞ nm 2 n · 2 −n δ ∞, ρ M v ∞ n1 M r n μE n < ∞ n1 M r n c μE n < ∞ n1 M 1 1 2n r n μE n δ. 2.7 4 Journal of Inequalities and Applications This yields v 1,ρ M u ± v ≤ ρ M u ∞ n1 M r n c μE n 1 − δ δ 1. 2.8 Hence, u ± v≤1. But u v u − v≤2u 2, and we deduce that u v u − v 1. Moreover, we have 1/2u v1/2u − v 1and1/2u v − 1/2u − v} 1, a contradiction with nonsquareness of L M . If b is not true, then there exist x, y ∈ SX such that x y 1/2x y 1/2x − y.Pickα>0 such that T Mαdt 1. Put u t α · x · χ T t ,v t α · y · χ T t . 2.9 Then we have ρ M u T M αx dt T M α dt 1, ρ M v T M αy dt T M α dt 1. 2.10 It is easy to see u, v ∈ SL M . We know that u t v t 2 α · x y 2 · χ T t , u t − v t 2 α · x − y 2 · χ T t . 2.11 Hence, we have ρ M u v 2 T M α · x y 2 dt T M α dt 1, ρ M u − v 2 T M α · x − y 2 dt T M α dt 1. 2.12 It is easy to see 1/2u v, 1/2u − v ∈ SL M , a contradiction! Sufficiency. Suppose that there exists u, v ∈ SL M such that u v 1 2 u v 1 2 u − v 1. 2.13 We will derive a contradiction for each of the following two cases. Journal of Inequalities and Applications 5 Case 1. μ{t ∈ T : ut / 0}∩{t ∈ T : vt / 0}0. Let G {t ∈ T : ut / 0}. Hence, we have 1 2 ρ M u 1 2 ρ M v 1 2 G M u t dt 1 2 T\G M v t dt 1 2 G M u t v t dt 1 2 T\G M u t v t dt > G M 1 2 u t v t dt T\G M 1 2 u t v t dt T M 1 2 u t v t dt ρ M 1 2 u v . 2.14 Since M ∈ Δ 2 , we have ρ M uρ M v1. Hence, ρ M u v/2 < 1. This implies u v/2 < 1, a contradiction! Case 2. μ{t ∈ T : ut / 0}∩{t ∈ T : vt / 0} > 0. By Lemma 2.2, without loss of generality, we may assume that there exists T 1 ⊂{t ∈ T : ut / 0}∩{t ∈ T : vt / 0} such that ut vt > utvt, t ∈ T 1 and μT 1 > 0. Therefore, 1 2 ρ M u 1 2 ρ M v 1 2 T M u t dt 1 2 T M v t dt T 1 2 M u t 1 2 M v t dt ≥ T 1 M 1 2 u t 1 2 v t dt T\T 1 M 1 2 u t 1 2 v t dt > T 1 M 1 2 u t v t dt T\T 1 M 1 2 u t v t dt ρ M u v 2 . 2.15 Since M ∈ Δ 2 , we have ρ M uρ M v1. Hence, ρ M u v/2 < 1. This implies u v/2 < 1, a contradiction! Theorem 2.3. L M is locally uniformly nonsquare if and only if a M ∈ Δ 2 ; b X is locally uniformly nonsquare. 6 Journal of Inequalities and Applications In order to prove the theorem, we give a lemma. Lemma 2.4. If X is locally uniformly nonsquare, then a For any x / 0, r 1 ≥ r 2 > 0, we have inf y / 0 x y − min x y , x − y : x ∈ X, r 2 ≤ y ≤ r 1 > 0 2.16 b If x n → x,thenlim n →∞ δx n δx,where δ x inf y / 0 x y − min x y , x − y : x ∈ X, r 2 ≤ y ≤ r 1 . 2.17 Proof. a Since X is locally uniformly nonsquare, we have η x > 0andη λx λη x , where λ>0 and η x inf y x y − min x y , x − y : x y > 0 . 2.18 In fact, since X is locally uniformly nonsquare, we have η x inf y x y − min x y , x − y : x y > 0 x · inf y 2 − min x x y y , x x − y y : x y > 0 > 0, η λx inf y λx y − min λx y , λx − y : λx y > 0 λ · inf y x 1 λ y − min x 1 λ y , x − 1 λ y : x 1 λ y > 0 λ · inf y x y − min x y , x − y : x y > 0 λ · η x . 2.19 Journal of Inequalities and Applications 7 Case 1. If x≥y, then x y ≤ 1 − y x x y y x x ≤ 1 − y x x y y x x ≤ x − y y y − η y/xx ≤ x y − η r 2 /xx 2.20 or x − y ≤ x y − η r 2 /xx . 2.21 Case 2. If x < y, then x y ≤ x y · y x 1 − x y · y ≤ x x − η x y − x x y − η x ≤ x y − η x 2.22 or x − y ≤ x y − η x . 2.23 Therefore, we get, the following inequality inf y x y − min x y , x − y : x ∈ X ≥ min η r 2 /xx ,η x > 0 2.24 holds. b1 Suppose that lim sup n →∞ δx n >δx, where x n → x n →∞. Then there exist a>0 and subsequence {n} of {n}, such that δx n − δx ≥ a. By definition of δx, there exist y 0 ∈ X such that x y 0 − min x y 0 , x − y 0 <δ x a 8 ,r 1 ≤ y 0 ≤ r 2 . 2.25 We will derive a contradiction for each of the following two cases. 8 Journal of Inequalities and Applications Case 1. x y 0 x − y 0 . Since x n → x n →∞, there exists n 0 such that x n 0 − x <a/8. Therefore, x n 0 y 0 − x n 0 y 0 ≤ x x n 0 − x y 0 − x n 0 y 0 ≤ x x n 0 − x y 0 − x y 0 − x n 0 − x x y 0 − x y 0 2 x n 0 − x ≤ δ x a 8 2 x n 0 − x <δ x a 8 2 · a 8 δ x 3 8 a. 2.26 This implies δx n 0 ≤ δx3/8a, a contradiction! Case 2. x − y 0 / x y 0 . Without loss of generality, we may assume x − y 0 > x y 0 r, where r>0. Since x n → x n →∞, there exists n 0 such that x n 0 −x < min{1/8a, 1/8r}. Therefore, we have x n 0 − y 0 x − y 0 x n 0 − x ≥ x − y 0 − x n 0 − x ≥ x − y 0 − 1 8 r, x n 0 y 0 x y 0 x n 0 − x ≤ x y 0 x n 0 − x ≤ x y 0 1 8 r. 2.27 This implies x n 0 − y 0 ≥ x − y 0 − 1 8 r ≥ x y 0 r − 1 8 r ≥ x y 0 1 8 r ≥ x n 0 y 0 . 2.28 Similarly, we have x n 0 y 0 − x n 0 y 0 ≤ δ x 3 8 a. 2.29 Therefore, we have x n 0 y 0 − min x n 0 y 0 , x n 0 − y 0 ≤ δ x 3 8 a. 2.30 This implies δx n 0 ≤ δx3/8a, a contradiction! Hence, lim sup n →∞ δx n ≤ δx. Journal of Inequalities and Applications 9 b2 Suppose that lim inf n →∞ δx n <δx, where x n → x n →∞. Then there exist b>0 and subsequence {n} of {n}, such that δx − δx n ≥ b. Since x n → x n →∞, then there exist n 0 ∈ N such that x n 0 − x n < 1/8b, whenever n ≥ n 0 . By definition of δx n 0 , there exist y 0 ∈ X such that x n 0 y 0 − min x n 0 y 0 , x n 0 − y 0 <δ x n 0 b 8 ,r 2 ≤ y 0 ≤ r 1 . 2.31 Therefore, we have x n y 0 − min x n y 0 , x n − y 0 x n 0 − x n 0 x n y 0 − min x n 0 − x n 0 x n y 0 , x n 0 − x n 0 x n − y 0 ≤ x n 0 1 8 b y 0 − min x n 0 y 0 , x n 0 − y 0 1 8 b x n 0 y 0 − min x n 0 y 0 , x n 0 − y 0 1 4 b <δ x n 0 1 8 b 1 4 b <δ x n 0 3 8 b 2.32 whenever n ≥ n 0 . Since x n → x n →∞, there exists n 1 >n 0 such that |ηx − ηx n 1 | < 1/8b, where η x x y 0 − min x y 0 , x − y 0 . 2.33 Hence, we have η x n 1 >η x − 1 8 b ≥ δ x − 1 8 b ≥ δ x n 0 b − 1 8 b δ x n 0 7 8 b. 2.34 This implies x n 1 y 0 − min x n 1 y 0 , x n 1 − y 0 ≥ δ x n 0 7 8 b, 2.35 which contradict 2.32. Hence, lim inf n →∞ δx n ≥ δx. Combing b1 with b2, we get lim n →∞ δx n δx. This completes the proof. 10 Journal of Inequalities and Applications Proof of Theorem 2.3. Necessity.ByTheorem 2.1, M ∈ Δ 2 .Ifb is not true, then there exist x ∈ SX, {y n } ∞ n1 ⊂ SX such that 1/2x y n →1and1/2x − y n →1asn →∞. Pick α>0 such that T Mαdt 1. Put u t α · x · χ T t ,v n t α · y n · χ T t . 2.36 Then we have ρ M u T M αx dt T M α dt 1, ρ M v n T M αy n dt T M α dt 1. 2.37 It is easy to see u, v n ∈ SL M . We know that u t v n t 2 α · x y n 2 · χ T t , u t − v n t 2 α · x − y n 2 · χ T t . 2.38 Moreover, we have Mα·xy n /2 ≤ Mα, Mα·x−y n /2 ≤ Mα. By the dominated convergence theorem, we have lim n →∞ T M α · x y n 2 dt T lim n →∞ M α · x y n 2 dt T M α dt 1, lim n →∞ T M α · x − y n 2 dt T lim n →∞ M α · x − y n 2 dt T M α dt 1. 2.39 It is easy to see ρ M 1/2u v n → 1, ρ M 1/2u− v n → 1asn →∞.ByLemma 1.1,we have 1/2u v n →1and1/2u − v n →1asn →∞, a contradiction with locally uniform nonsquareness of L M . Sufficiency. Suppose that there exist u ∈ SL M , {v n } ∞ n1 ⊂ SL M such that 1/2u v n →1, 1/2u − v n →1asn →∞. We will derive a contradiction for each of the following two cases. Case 1. There exist ε 0 > 0, σ 0 > 0 such that μG n >ε 0 , where G n {t ∈ T : v n t≥σ 0 }.Put H n t ∈ T : σ 0 ≤ v n t ≤ M −1 4 ε 0 . 2.40 We have 1 T M v n t dt ≥ G n \H n M v n t dt ≥ 4 ε 0 · μ G n \ H n . 2.41 [...]... completes the proof Journal of Inequalities and Applications 15 Corollary 2.5 The following statements are equivalent: a LM is locally uniformly nonsquare if and only if LM is nonsquare; b X is locally uniformly nonsquare Acknowledgments The authors would like to thank the anonymous referee for some suggestions to improve the manuscript This work was supported by China Natural Science Fund under Grant...Journal of Inequalities and Applications 11 This implies μ Gn \ Hn ≤ 1/4 ε0 Hence, μHn > 1/2 ε0 We define a function η t inf y/0 y − min u t : σ0 ≤ y ≤ M−1 y , u t −y u t 4 ε0 2.42 on T0 , where T0 {t ∈ T : u t / 0} By Lemma 2.4, we have η t > 0 μ-a.e on T0 Let hn t → u t μ-a.e on T0 , where hn is simple function Hence, ηn t inf y/0 y −min hn t hn t : σ0 ≤ y ≤ M−1 y , hn t... 11061022 References 1 R C James, Uniform nonsquare Banach space,” Annals of Mathematics, vol 80, no 3, pp 542–550, 1964 2 J Garc´a-Falset, E Llorens-Fuster, and E M Mazcunan-Navarro, “Uniformly nonsquare Banach spaces ı ˜ have the fixed point property for nonexpansive mappings,” Journal of Functional Analysis, vol 233, no 2, pp 494–514, 2006 3 S T Chen, “Geometry of Orlicz spaces, ” Dissertationes Math,... min M η0 , M σ0 − M 2 2 Fn vn t u t −M 1 M vn t 2 vn t u t vn t ut M T 1 vn t 2 1 M ut 2 M T 1 M vn t 2 1 M ut 2 dt − M vn t 1 M vn t 2 1 M u t 2 2 En 1 2 1 M vn t 2 1 M u t 2 2 En ≥ dt dt dt 1 σ0 min M η0 , M σ0 − M 2 2 · μFn σ0 1 ≥ min M η0 , M σ0 − M 2 2 1 · ε0 8 By Lemma 1.1, we have ρM u contradiction with 1/2 ρM u M σ0 /2 } · 1/8 ε0 ρM vn 1, ρM u 1/2 ρM vn − ρM u vn /2 → 1 as n → ∞ This is in. .. 1/8 ε0 ρM vn 1, ρM u 1/2 ρM vn − ρM u vn /2 → 1 as n → ∞ This is in vn /2 ≥ min{M η0 , 1/2 M σ0 − Journal of Inequalities and Applications 13 Case 2 For any ε > 0, σ > 0, there exists N such that μ{t ∈ T : vn t ≥ σ} < ε whenever n > N By the Riesz theorem, without loss of generality, we may assume that vn t → 0 μ− a.e on T Using {t ∈ T : u t ∞ / 0} ⊃ t∈T : n 1 1 n 1 ≤ < u t 1 , n 2.51 we get that there... Functional Analysis, vol 233, no 2, pp 494–514, 2006 3 S T Chen, “Geometry of Orlicz spaces, ” Dissertationes Math, vol 356, pp 1–204, 1996 4 C X Wu, T F Wang, S T Chen, and Y W Wang, Geometry Theory of Orlicz Spaces, H.I.T Print House, Harbin, China, 1986 ... {t ∈ T : u t / 0} 2.52 Since M is N -function, we can choose 0 < h < d such that 1/2 M d 1/2 M h − M d h /2 > 0 Since vn t → 0μ− a.e on T , by the Egorov theorem, there exists N1 such that vn t < h, t ∈ F whenever n > N1 , where F ⊂ T , μ T \ F < 1/8 μT0 Next, we will prove that if u1 ≥ u2 ≥ v2 ≥ v1 > 0, then 1 M u1 2 1 u1 v1 M v1 − M 2 2 ≥ 1 M u2 2 1 u2 v2 M v2 − M 2 2 2.53 In fact, we have 1 M u1... μ-measurable Using T⊃ ∞ t ∈ T0 : i 1 1 i 1 0 such that μH < 1/8 ε0 , where t ∈ T0 : η t ≤ 2η0 H 2.45 1 2 Let En Hn \ H, En Hn ∩ {t ∈ T : u t / 0} \ H, En Hn ∩ {t ∈ T : u t 0} \ H It 1 2 1 2 1 is easy to see En En ∪ En , En ∩ En φ and μEn ≥ 3/8 ε0 If t ∈ En , by Lemma 2.4, we have vn t u t − min{ u t vn t , u t − vn t } ≥ η t ≥ 2η0 2.46 1 1 Without loss... p t dt − p t dt 0 2.54 u1 v1 /2 p t dt − p t dt 0 14 Journal of Inequalities and Applications Moreover, we have 1 1 u1 v2 u2 v2 M u1 − M − M u2 − M 2 2 2 2 u1 1 2 0 u1 1 2 p t dt − u2 u1 v2 /2 p t dt − u2 v2 /2 p t dt − p t dt 0 0 0 2.55 u1 v2 /2 p t dt − p t dt u2 v2 /2 u2 u1 u2 /2 ≥ 1 2 u1 v2 /2 p t dt − u2 p t dt ≥ 0 u2 v2 /2 By 2.54 and 2.55 , we have 1 u1 v1 M v1 − M 2 2 1 M u1 2 This shows that... 1/2 u v p t dt − 0 u 1 2 1/2 u p t dt − 1 2 v 1 2 v 0 1/2 u p t dt − 0 1/2 v p t dt 0 1/2 u p t dt p t dt − 1/2 u 1/2 v 1/2 v p t dt − 1/2 v 1/2 v p t dt 0 p t dt ≥ 0 p t dt 0 2.48 12 Journal of Inequalities and Applications 2 Hence, if t ∈ En , then 1 M vn t 2 Let Fn 1 M vn t 2 1 σ0 M σ0 − M 2 2 ≥ 2.49 > 0 1 2 Fn ∪ En Then μFn ≥ 1/8 ε0 Therefore, 1 ρM u 2 u vn 1 ρM vn − ρM 2 2 1 2 T ≥ 1 Fn M u t . for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces equipped with Luxemburg norm are given. We also prove that, in Orlicz-Bochner function spaces generated by locally. Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2011, Article ID 875649, 15 pages doi:10.1155/2011/875649 Research Article Nonsquareness and Locally Uniform Nonsquareness. Uniform Nonsquareness in Orlicz-Bochner Function Spaces Endowed with Luxemburg Norm Shaoqiang Shang, 1 Yunan Cui, 2 and Yongqiang Fu 1 1 Department of Mathematics, Harbin Institute of Technology, Harbin 150001,
Ngày đăng: 21/06/2014, 05:20
Xem thêm: Báo cáo hóa học: " Research Article Nonsquareness and Locally Uniform Nonsquareness in Orlicz-Bochner Function Spaces Endowed with Luxemburg Norm" pot, Báo cáo hóa học: " Research Article Nonsquareness and Locally Uniform Nonsquareness in Orlicz-Bochner Function Spaces Endowed with Luxemburg Norm" pot
