summary:Every Orlicz space equipped with Orlicz norm has weak sum property, therefore, it has weakly normal structure and fixed point property. A criterion of sum property also of normal structure for such spaces is given as well, which shows that every Orlicz space has isonormal structure

Chen, Shutao

Duan, Yanzheng

English

Institute of Mathematics AS CR, v. v. i.

Commentationes Mathematicae Universitatis CarolinaeShutao Chen; Yanzheng DuanNormal structure and weakly normal structure of Orlicz spacesCommentationes Mathematicae Universitatis Carolinae, Vol. 32 (1991), No. 2, 219--225Persistent URL: http://dml.cz/dmlcz/116959Terms of use:© Charles University in Prague, Faculty of Mathematics and Physics, 1991Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document mustcontain these Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://project.dml.czComment.Math.Univ.Carolin. 32,2 (1991)219–225 219Normal structure and weakly normal structureof Orlicz spacesShutao Chen, Yanzheng DuanAbstract. Every Orlicz space equipped with Orlicz norm has weak sum property, therefore,it has weakly normal structure and fixed point property. A criterion of sum property alsoof normal structure for such spaces is given as well, which shows that every Orlicz spacehas isonormal structure.Keywords: Orlicz space, normal structureClassification: 46B20Introduction.T. Landes [4] shows that, under certain conditions, an Orlicz sequence spacewith Luxemburg norm has normal or weakly normal structure iff it is separable.For Orlicz spaces with Orlicz norm, we will discover that the results are muchdifferent.We begin with some notations. A sequence (xn) of a Banach space X is calledlimit affine, if the limit λ(x) := limn ‖xn − x‖ > 0 exists for every x ∈ conv(xn),and λ is an affine function on conv(xn). (xn) is called growing, if λ(xn) ≤ λ(xn+1)for all n ∈ N . X is said to have sum property, if it contains no growing limit affinesequence. X is said to have weak sum property, if it contains no growing weaklyconverging limit affine sequence. X is said to have normal structure, if it containsno limit affine sequence (xn) with λ(xn) = λ(xn+1) > 0 for all n ∈ N . X is saidto have weakly normal structure, if it contains no weakly converging limit affinesequence (xn) with λ(xn) = λ(xn+1) > 0 for all n ∈ N . X is said to have isonormalstructure, if it is isomorphic to a Banach space with normal structure. X is said tohave fixed point property if every nonexpansive selfmapping on a weakly compactconvex subset of X has a fixed point.It is well known that sum property ⇒ normal structure and that weak sumproperty ⇒ weakly normal structure ⇒ fixed point property.Throughout this paper, we always denote by (G,Σ, µ) a complete, nonatomic,finite measure space. We sayM : R → R+ to be an N -function, if it is a continuous,convex, even function satisfying M(u) = 0, iff u = 0 and M(u)/u → 0 (resp. ∞)as u → 0 (resp. ∞). If M(u) is an N -function, then we denote by p(u) its right-hand derivative and by N(v) the conjugate N -function of M(u), i.e., N(v) :=maxu{uv − M(u)}.∗Supported by the Special Scientific Research Fund of China for University Doctoral Instruction.220 Shutao Chen, Yanzheng DuanLet M be an N -function. For every µ-measurable function x : G → R we define̺M (x) =∫G M(x(t)) dµ, and(1)LM ={x : ̺M (βx) < ∞ for some β > 0},EM ={x : ̺M (βx) < ∞ for all β > 0},‖x‖ =‖x‖M = infk>0[1 + ̺M (kx)]/k, x ∈ LM .Then the Orlicz space (LM , ‖ · ‖) and its subspace (EM , ‖ · ‖) are Banach spaces.Main results.Lemma 1. Suppose xn ∈ LM , ‖xn‖ ≤ K, n ∈ N and xn(t) → x(t) in measure asn → ∞, then x ∈ LM .Proof: Since ‖xn/K‖ ≤ 1, by [1], ̺M (xn/K) ≤ ‖xn/K‖ ≤ 1, n ∈ N . Without lossof generality, we may assume xn(t) µ-a.e. on G (pass a subsequence, if necessary),then, by Fatou’s lemma, ̺M (x/K) ≤ lim infn̺M (xn/K) ≤ 1, i.e., x ∈ LM . Lemma 2. If xn → 0 weakly in LM and xn(t)→ y(t) in measure, then y = 0.Proof: Again, we may assume xn(t) → y(t) µ-a.e. on G. Let F = {t ∈ G; y(t)6= 0}. If µF > 0, then there exists E ∈ Σ with µE < µF and xn(t) → y(t) uni-formly on G \ E. Define v(t) = signy(t)χF\E(t), then v ∈ LM∗ and 〈v, xn〉 =∫F\E v(t)xn(t) dµ →∫F\E |y(t)| dµ > 0 contradicting the hypothesis xn → 0weakly. We say that an interval [a, b] is a structural affine interval of the N -function M ,if M is affine on [a, b] and it is neither affine on [a − ε, b] nor on [a, b + ε] for anyε > 0.Theorem 1. For any N -function M , LM has weak sum property, therefore, it hasweakly normal structure.Theorem 2. The following are equivalent,(i) LM has sum property,(ii) LM has normal structure,(iii) there exist a > 0, C > 1 such that for any structural affine interval [u, v] ofM with u ≥ a, we have v/u ≤ C.Proof of Theorems 1 and 2: For any limit affine sequence (xn) in LM withxi 6= xj whenever i 6= j, by [1], the “inf” in (1) is attainable for all x 6= 0. Therefore,for all i 6= j, we may find kij > 0 such that(2) ‖xi − xj‖ = [1 + ̺M (kij(xi − xj))]/kij .First we show that there exists a subsequence N1 of N such that for any j ∈N1, {kij}i∈N1 is bounded. Indeed, if {kij}i is bounded for all j ∈ N , then we letN1 = N . Otherwise, there exist some m ∈ N and a subsequence I of N such thatNormal structure and weakly normal structure of Orlicz spaces 221kim → ∞ as i(∈ I) → ∞. Hence, for any σ > 0, if we define Gi = {t ∈ G :|xi(t)− xm(t)| > σ}, then by (2), we have‖xi − xm‖ > ̺M (|xi − xm|kim)/kim ≥ [M(σkim)/kim]µGi .Since λ(xm) < ∞ and M(u)/u → ∞ as u → ∞, we must have µGi → 0 asi(∈ I)→ ∞. This shows that {xi}i∈I converges to xm in measure. We may assumethat I does not contain m. We claim that N1 = I satisfies our requirement. Infact, if kij → ∞ as i(∈ I1)→ ∞ for some j ∈ I and some subsequence I, of I, thenin the same way we can show that xi → xj in measure as i(∈ I1) → ∞. This isimpossible since xj 6= xm.By the diagonal method, we can find a subsequence N2 of N1 such that kij →kj < ∞ as i(∈ N2)→ ∞ for each j ∈ N1. We claim that kj → ∞ as j(∈ N2)→ ∞.In fact, if this is not true, then N2 contains a subsequence N3 such that kj → k < ∞as j(∈ N3)→ ∞. By (1) and (2), for all n, i, j ∈ N3, n 6= i, j,(3)‖xn − xi‖+ ‖xn − xj‖ − ‖2xn − xi − xj‖ ≥≥ [1 + ̺M (kni(xn − xi))]/kni + [1 + ̺M (knj(xn − xj))]/knj−− [1 + ̺M ((2xn − xi − xj)kniknj/(kni + knj))](kni + knj)/kniknj ==∫G[M((xn(t)− xi(t))kni)/kni +M((xn(t)− xj(t))knj)/knj−− M((2xn(t)− xi(t)− xj(t))kniknj/(kni + knj))(kni + knj)/kniknj ] dµ.Denote the last integrand in (3) by fnij(t), then by the convexity of M, fnij(t) ≥ 0for all t ∈ G. Since λ is affine on conv(xk), let n → ∞, by (3),∫G fnij(t) dµ→ 0, therefore, fnij(t) → 0 in measure. By the diagonal method, we can choosea subsequence N4 of N3 such that fnij(t)→ 0 µ-a.e. on G as n(∈ N4)→ ∞ for alli, j ∈ N3.For each t ∈ G, choose a subsequence {nτ = nτ (t)} of N4 such that(∗) |v(t)| = lim infn∈N4|xn(t)|, limτxnτ (t) = v(t),then by Fatou’s lemma, |v(t)| < ∞ µ-a.e. on G (one may prove this analogously asin Lemma 1). Let τ → ∞, by the continuity of M ,(4)0 = limτfnτij(t) ==M((v(t) − xi(t))ki)/ki +M((v(t) − xj(t))kj)/kj−− M((2v(t)− xi(t)− xj(t))kikj/(ki + kj))(ki + kj)/kikjµ-a.e. on G. Since for µ-a.e. t ∈ G, (4) holds for all i, j ∈ N3, by replacing j by nτin (4) and taking τ → ∞, for each t ∈ G, we have(5)M((v(t)− xi(t))ki)/ki ==M((v(t)− xi(t))kik/(ki + k))(ki + k)/kik222 Shutao Chen, Yanzheng Duanµ-a.e. on G. Since 0 < k/(ki + k) < 1, and u 6= 0, 0 < α < 1 implies M(αu) <αM(u), for all t ∈ G satisfying v(t) 6= xi(t), we haveM((v(t)− xi(t))kik/(ki + k))(ki + k)/kik < M((v(t)− xi(t))ki)/ki .Since this inequality contradicts (5), we have xi(t) = v(t) µ-a.e. on G for all i ∈ N3,which contradicts the assumption xi 6= xj whenever i 6= j.Now, we prove (iii)⇒ (i) in Theorem 2. If LM does not have sum property, thenthere exists a growing limit affine sequence. Without loss of generality, we mayassume that every two points in the sequence are different. By the above discussion,it contains a subsequence (xn) satisfying kij → kj < ∞ as i → ∞ and kj → ∞ asj → ∞, where kij satisfies (2), i, j ∈ N . Since M(u)/u → ∞ as u → ∞, for theconstant a > 0 in (iii), we can find b > a such that M(12 (a+ b)) <12 [M(a)+M(b)].Since M is convex, by (iii)(6) M(αu+ (1 − α)v) < αM(u) + (1− α)M(v)for all 0 < α < 1 and all u ≤ a, v ≥ b or u ≥ a, v ≥ Cu. If we define v(t) as in (∗),then by (4) and (6), for µ-a.e. t ∈ G, if ki|v(t)−xi(t)| ≤ a, then kj |v(t)−xj(t)| ≤ b;if ki|v(t)− xi(t)| > a, then kj |v(t)− xj(t)| ≤ Cki|v(t)− xi(t)|. Therefore, for µ-a.e.t ∈ G,(7) kj |v(t)− xj(t)| ≤ max{b, Cki|v(t)− xi(t)|} := ui(t).By (2) and Fatou’s lemma, we have(8) λ(xj) ≥ [1 + ̺M (kj(v − xj))]/kj ≥ ‖v − xj‖.Thus, v − xj ∈ LM , therefore, ui ∈ LM . Since λ > 0, we have lim inf ‖v − xj‖ :=τ > 0. It follows from (7) thatkj = ‖kj(v − xj)‖/‖v − xj‖ ≤ ‖ui‖/‖v − xj‖.Let j → ∞, we get a contradiction ∞ ≤ ‖ui‖/τ < ∞.Next, we turn to Theorem 1. If LM does not have weak sum property, thenby (3), there exists a weakly converging (to zero) limit affine sequence (xn) with‖xn‖ → 1 and λ(xn) → 1. By the first part of the proof, passing a subsequenceif necessary, we may assume kij → kj < ∞ as i → ∞ and kj → ∞ as j → ∞,where kij satisfies (2). It follows from (8) that xj → v in measure (similarly verifiedas in the first part of the proof). Therefore, by Lemma 2, v = 0. We may alsoassume xj → 0 µ-a.e. on G. We prove the theorem by showing limλ(xj) ≥ 4/3contradicting the assumption λ(xj)→ 1.For each j ∈ N , we choose a set Gj ∈ Σ such that xj is bounded on Gj and[1 + ̺M (kjxjχGj )]/kj > [1 + ̺M (kjxj)]/kj − 1/kj ,Normal structure and weakly normal structure of Orlicz spaces 223then by (1) and (8),λ(xj) = [1 + ̺M (kjxj)]/kj == [1 + ̺M (kjxjχGj )]/kj + [1 + ̺M (kjxjχG\Gj )]/kj − 1/kj >> [1 + ̺M (kjxj)]/kj − 1/kj + ‖xjχG\Gj‖ − 1/kj ≥≥ ‖xj‖+ ‖xjχG\Gj‖ − 2/kj ,i.e.‖xjχG\Gj‖ < λ(xj)− ‖xj‖+ 2/kj .(9)It follows that(10) ‖xjχGj‖ ≥ ‖xj‖ − ‖xjχG\Gj‖ > 2‖xj‖ − λ(xj)− 2/kj .Since xj is bounded on Gj , there exists δ = δ(j) > 0 such that(11) ‖xjχE‖ < 1/kj whenever E ⊂ Gj and µE < δ.Since xi → 0 µ-a.e. on G, there exists F ∈ Σ with µF < δ such that xi → 0uniformly on G \ F . Hence, there exists I = I(j) ∈ N such that for all i > I, wehave(12) ‖xiχG\F ‖ < 1/kj .It follows that(13) ‖xiχF ‖ ≥ ‖xi‖ − ‖xiχG\F ‖ > ‖xi‖ − 1/kj .Hence, by (1), (2), (9)–(13),‖xi − xj‖ = [1 + ̺M (kij(xi − xj)χG\(Gj\F ))]/kij++ [1 + ̺M (kij(xi − xj)χGj\F )]/kij − 1/kij ≥≥ ‖(xi − xj)χG\(Gj\F )‖+ ‖(xi − xj)χGj\F ‖ − 1/kij ≥≥ ‖xiχG\(Gj\F )‖ − ‖xjχG\(Gj\F )‖++ ‖xjχGj\F ‖ − ‖xiχGj\F ‖ − 1/kij == ‖xiχG\(Gj\F )‖ − ‖xjχG\Gj + xjχGj\F ‖++ ‖xjχGj − xjχGj\F ‖ − ‖xiχGj\F ‖ − 1/kij >> (‖xi‖ − 1/kj)− (λ(xj)− ‖xj‖+ 2/kj + 1/kj)++ (2‖xj‖ − λ(xj)− 2/kj − 1/kj)− 1/kj − 1/kij == ‖xi‖+ 3‖xj‖ − 2λ(xj)− 8/kj − 1/kij .224 Shutao Chen, Yanzheng DuanLet i → ∞, we haveλ(xj) ≥ 1 + 3‖xj‖ − 2λ(xj)− 9/kj .Let j → ∞, then limλ(xj) ≥ 4/3.Finally, we prove (ii)⇒ (iii) in Theorem 2. If (iii) does not hold, then there existthe sequences {uj}, {vj} such that M(u1)µG > 1, uj+1 > 2juj , vj > 2juj and p(u)is constant on [uj , vj ], j ∈ N . By the first two assumptions, we can choose disjointsets gj ∈ Σ such that µG \ Uj∈NGj > 0 and(14) 2−j = ujp(uj)µGj = [M(uj) +N(p(uj))]µGj(the last equality holds by the special case of Young’s inequality). Hence, we canfind u0 large enough so that there is G0 satisfyingΣj∈NN(p(uj))µGj +N(p(u0))µG0 = 1.(15)Definev = Σj≥0p(uj)χGj ,xn = u0χG0 +Σj∈NvjχGj +Σj>nujχGj ,then by (15), ̺N (v) = 1, therefore, v ∈ L∗M and ‖v‖ = 1 (cf. [1]).First we show that xn ∈ EM for any n ∈ N . Given arbitrary K > 1, chooseJ > n such that 2j > K, then vj > 2juj > Kuj > uj for all j > J . ThereforeΣj>JM(Kuj)µGj = Σj>J [Kujp(Kuj)− N(p(Kuj))]µGj << Σj>JKujp(Kuj)µGj = Σj>JKujp(uj)µGj = KΣj>J2−j < ∞ .This implies ̺M (Kxn) < ∞. Since K > 1 is arbitrary, we have xn ∈ EM .Let kn = ‖xn‖ and yn = xn/kn, then yn ∈ EM and ‖yn‖ = 1. By (1) and (15),‖yn‖ ≥ 〈v, yn〉 = [u0p(u0)µG0 +Σj≤nvjp(uj)µGj +Σj>nujp(uj)µGj ]/kn == [̺N (v) + ̺M(knyn)]/kn ≥ ‖yn‖ = 1.Moreover, sincekn = ‖xn‖ ≥ 〈v, xn〉 > Σj≤nvjp(uj)µGj ≥ Σj≤n2jujp(uj)µGj = n,we have kn → ∞ as n → ∞.We complete the proof by showing λ = 2 on conv(yn). Indeed, for any y ∈conv(yn), there exist λi ≥ 0,Σi≤mλi = 1 such that y = Σi≤mλiyi. Since 〈v, yn〉 = 1,we have 〈v, y〉 = Σi≤mλi〈v, yn〉 = 1. For any ε > 0, since y ∈ EM , there existsI > m such that ‖yχF‖ < ε, where F = Ui>IGi. In view of xn(t) ≤ max{vI , u0}Normal structure and weakly normal structure of Orlicz spaces 225on G \ F and kn → ∞ as n → ∞, we can find n0 ∈ N such that ‖ynχG\F ‖ < ε forall n > n0. Define v0 = vχG\F − vχF , then ‖v0‖ = ‖v‖ = 1 and for all n > n0,2 ≥ ‖y‖+ ‖yn‖ ≥ ‖y − yn‖ ≥ 〈v0, y − yn〉 == 〈v0, yχG\F 〉+ 〈v0, yχF 〉 − 〈v0, ynχG\F 〉 − 〈v0, ynχF 〉 == 〈v, yχG\F 〉 − 〈v, yχF 〉 − 〈v, ynχG\F 〉+ 〈v, ynχF 〉 == 〈v, y〉 − 2〈v, yχF 〉 − 2〈v, ynχG\F 〉+ 〈v, yn〉 >> 1− 2‖yχF‖ − 2‖ynχG\F ‖+ 1 > 2− 4ε,which shows that λ(y) = 2. Theorem 3. LM has isonormal structure.Proof: If M is strictly convex, then the condition (iii) in Theorem 2 holds for allu 6= 0 and all C 6= 1. Therefore, LM has normal structure in this case. By [1], forany Orlicz function M and any ε > 0, we can construct a strictly convex Orliczfunction H such that‖x‖M ≤ ‖x‖H ≤ (1 + ε)‖x‖Mfor all x ∈ LM , which shows that LM is isomorphic to LH , i.e. LM has isonormalstructure. References[1] Wu Congxin, Wang Tingfu, Chen Shutao, Wang Yuwen, Geometry of Orlicz spaces, Harbin,1986.[2] Wu Congxin, Wang Tingfu, Orlicz spaces and their applications, Harbin, 1983.[3] Landes T., Permanence properties of normal structure, Pacific J. Math. 110, No. 1 (1984),125–143.Harbin Normal University, Department of Mathematics, Harbin, ChinaHarbin Institute of Technology, Department of Mathematics, Harbin, China(Received September 13, 1990, revised February 19, 1991)

Normal structure and weakly normal structure  of Orlicz spaces

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Normal structure and weakly normal structure of Orlicz spaces

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Institute of Mathematics AS CR, v. v. i.