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Uniform nonsquareness and locally uniform nonsquareness in Orlicz–Bochner function spaces and applications
Journal of Functional Analysis 267 (2014) 2056–2076
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Journal of Functional Analysis www.elsevier.com/locate/jfa
Uniform nonsquareness and locally uniform nonsquareness in Orlicz–Bochner function spaces and applications ✩ Shaoqiang Shang a,∗ , Yunan Cui b a
Department of Mathematics, Northeast Forestry University, Harbin 150040, PR China b Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, PR China
a r t i c l e
i n f o
Article history: Received 23 March 2013 Accepted 30 July 2014 Available online 11 August 2014 Communicated by B. Chow MSC: 46E30 46B20
a b s t r a c t In this paper, criteria for uniform nonsquareness and locally uniform nonsquareness of Orlicz–Bochner function spaces equipped with the Orlicz norm are given. Although, criteria for uniform nonsquareness and locally uniform nonsquareness in Orlicz function spaces were known, we can easily deduce them from our main results. Moreover, we give a sufficient condition for an Orlicz–Bochner function space to have the fixed point property. © 2014 Elsevier Inc. All rights reserved.
Keywords: Uniform nonsquare space Locally uniform nonsquare space Orlicz–Bochner function spaces Fixed point
✩
Supported by “the Fundamental Research Funds for the Central Universities”, DL12BB36.
* Corresponding author. E-mail addresses: [email protected] (S. Shang), [email protected] (Y. Cui). http://dx.doi.org/10.1016/j.jfa.2014.07.032 0022-1236/© 2014 Elsevier Inc. All rights reserved.
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1. Introduction Uniform nonsquareness of Banach spaces has been defined by R.C. James as the geometric property which implies super-reflexivity (see [3,11]). So, proving this property of a Banach space we know, without any characterization of the dual space, that it is super-reflexive, so reflexive as well. Recently J. Garcia-Falset, E. Llorens-Fuster and E.M. Mazcuñan-Navarro have shown that uniformly nonsquare Banach spaces have the fixed point property (see [1]). Therefore, it was natural and interesting to look for criteria of non-squareness properties in various well-known classes of Banach spaces. In 2013, P. Foralewski, H. Hudzik and P. Kolwicz have given criteria for uniform nonsquareness and locally uniformly nonsquareness of Orlicz–Lorentz sequence spaces (see [2]). The topic of this paper is related to the topic of [5–10] and [17–29]. The criteria for uniform nonsquareness and locally uniform nonsquareness in the classical Orlicz function spaces have been given in [4,23,24] already. However, because of the complicated structure of Orlicz–Bochner function spaces equipped with Orlicz norm, up to now the criteria for uniform nonsquareness and locally uniform nonsquareness have not been discussed yet. The aim of this paper is to give criteria for uniform nonsquareness and locally uniform nonsquareness of Orlicz–Bochner function spaces equipped with Orlicz norm. Although, criteria for uniform nonsquareness and locally uniform nonsquareness in Orlicz function spaces were known, we can easily deduce them from our main results. Moreover, we give a sufficient condition for an Orlicz–Bochner function space to have the fixed point property. Let (X, · ) be a real Banach space. S(X) and B(X) denote the unit sphere and the unit ball of X, respectively. Let us recall some geometrical notions concerning nonsquareness. A Banach space X is said to be a nonsquare space if for any x, y ∈ S(X) we have min{x + y, x − y} < 2. A Banach space X is said to be a uniformly nonsquare space if for any x, y ∈ S(X), there exists δ > 0 such that min{x + y, x − y} < 2 − δ. A Banach space X is said to be a locally uniformly nonsquare space if for any x ∈ S(X), there exists δx > 0 such that min{x + y, x − y} < 2 − δx for any y ∈ S(X). 2. Preliminaries Definition 1. A function M : R → R+ is called an N -function if it has following properties: (1) if M is even, continuous convex and M (0) = 0; (2) M (u) > 0 for any u = 0; (3) limu→0 M (u)/u = 0 and limu→∞ M (u)/u = ∞. Let (T, Σ, μ) be a nonatomic finite measurable space. Moreover, for a given Banach space (X, · ), we denote by XT the set of all strongly μ-measurable function from T to X, and for each u ∈ XT , we define the modular of u by
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M u(t) dt.
ρM (u) = T
Put LM (X) =
u(t) ∈ XT :
M λu(t) dt < ∞ for some λ > 0 .
T
It is well known that Orlicz–Bochner spaces LM (X) are Banach spaces when they are equipped with the Luxemburg norm u ≤1 u = inf λ > 0 : ρM λ
or with the Orlicz norm u0 = inf
k>0
1
1 + ρM (ku) . k
In particular, LM (R) and L0M (R) are said to be Orlicz function spaces. It is well known that u ≤ u0 ≤ 2u. Set K(u) =
k>0:
1 1 + ρM (ku) = u0 . k
Since limu→∞ M (u)/u = ∞, the set K(u) are nonempty (see [4]). Let p(u) denote the right derivative of M (u) at u ∈ R+ , q(v) be the generalized inverse function of p(u) defined on R+ by
q(v) = sup u ≥ 0 : p(u) ≤ v u≥0
|v| Then we call N (v) = 0 q(s)ds the complementary function of M . It is well known that there holds the Young inequality uv ≤ M (u) + N (v) and uv = M (u) + N (v) ⇔ u = |q(v)| sign v or v = |p(u)| sign u. Moreover, it is well known that M and N are complementary to in the sense of Young each other. Definition 2. (See [4].) We say that an Orlicz function M satisfies condition Δ2 (M ∈ Δ2 ) if there exist K > 2 and u0 ≥ 0 such that M (2u) ≤ KM (u) (u ≥ u0 ) In this case, we write M ∈ Δ2 or N ∈ ∇2 .
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We know (see [4,21] and [23]) that if M ∈ ∇2 , then for any u0 > 0, α ∈ (0, 1) there exists δ > 0 such that M (αu) ≤ (α − δ)M (u) (u ≥ u0 ). First let us recall some results that will be used in the further part of the paper. Lemma 1. (See [4] and [21].) If N ∈ Δ2 , λ0 ∈ (0, 1), then for every w > 0, λ ∈ (0, λ0 ], there exist numbers a = a(w) ∈ (0, 1) and γ = γ(a(w), λ0 ) ∈ (0, 1) such that M λu + (1 − λ)v ≤ λ(1 − γ)M (u) + (1 − λ)M (v) holds true for all u ≥ w and v satisfying | uv | ≤ a. Lemma 2. (See [4].) (a) If N ∈ Δ2 , then set {K(u) : u ∈ S(L0M (X))} is bounded. (b) If M ∈ Δ2 and N ∈ Δ2 , then there exists δ ∈ (0, 1) such that inf K(u) : u ∈ S L0M (X) > 1 + δ. 3. Main results Theorem 1. The Orlicz–Bochner function space L0M (X) is a uniformly nonsquare space if and only if (a) M ∈ Δ2 and N ∈ Δ2 ; (b) X is a uniformly nonsquare space. In order to prove the theorem, we give some lemmas. Lemma 3. (See [27].) A Banach space X is uniformly nonsquare space if and only if there exists δ1 > 0 such that for any elements x1 , x2 ∈ X\{0}, we have
δ min{x1 , x2 } . 1− x1 + x2
min x1 + x2 , x1 − x2 ≤ x1 + x2
Lemma 4. Let X be a uniformly nonsquare space and M ∈ Δ2 , N ∈ Δ2 . Then for any l ≥ m > 0 and w > 0, there exists r = r(w, m, l) ∈ (0, 1) such that the inequality r
1 1 M k1 x1 + M k2 x2 k1 k2 k1 k2 k1 + k2 k1 k2 k1 + k2 M x1 + x2 + M x1 − x2 ≥ k1 k2 k1 + k2 k1 k2 k1 + k2
holds true for all x1 , x2 ∈ X, k1 , k2 ∈ R+ satisfying max{k1 x1 , k2 x2 } ≥ w, m ≤ min{k1 x1 , k2 x2 }, max{k1 , k2 } ≤ l.
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Proof. For any l ≥ m > 0 and w > 0, take x1 , x2 ∈ X\{0} and k1 , k2 ∈ R+ such that max{k1 x1 , k2 x2 } ≥ w, m ≤ min{k1 x1 , k2 x2 }, max{k1 , k2 } ≤ l. Moreover, we have 0<
k1 1 1 l < 1. = ≤ m = k 2 k1 + k2 1+ l l+m 1 + k2
Similarly, we have 1 < k2 /(k1 + k2 ) ≤ l/(l + m) < 1. Therefore, by Lemma 1, there exist l numbers a = a(w) ∈ (0, 1) and γ = γ(a(w), l+m ) = γ(w, m, l) ∈ (0, 1) such that
k1 k2 M u+ v k1 + k2 k1 + k2
≤
k2 k1 (1 − γ)M (u) + M (v), k1 + k2 k1 + k2
(1)
for all u ≥ w and v satisfying | uv | ≤ a. Moreover, we may assume without loss of generality that k1 x1 = max{k1 x1 , k2 x2 }. For the clarity, we will divide the proof into two parts. I. Suppose that k2 x2 /k1 x1 < a. Noticing that max{k1 x1 , k2 x2 } ≥ w, m ≤ min{k1 x1 , k2 x2 }, max{k1 , k2 } ≤ l, by inequality (1), we have k1 k2 k1 + k2 M x1 − x2 k1 k2 k1 + k2 k1 + k2 k1 k2 ≤ M k1 x1 + k2 x2 k1 k2 k1 + k2 k1 + k2 k2 k1 k1 + k2 (1 − γ)M k1 x1 + M k2 x2 ≤ k1 k2 k1 + k2 k1 + k2 1 1 ≤ (1 − γ)M k1 x1 + M k2 x2 k1 k2 1 1 1 = M k1 x1 + M k2 x2 − γM k1 x1 . k1 k2 k1
(2)
Similarly, we have k1 + k2 k1 k2 M x1 + x2 k1 k2 k1 + k2 1 1 1 ≤ M k1 x1 + M k2 x2 − γM k1 x1 . k1 k2 k1
(3)
Noticing that max{k1 x1 , k2 x2 } ≥ w, max{k1 , k2 } ≤ l and m ≤ min{k1 x1 , k2 x2 }, we have 1 m < , lk1 k1
1 m 1 m 1 = · < < . lk2 l k2 l k1
Therefore, by inequality (2)–(4), we obtain
(4)
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k1 + k2 k1 k2 k1 + k2 k1 k2 M x1 − x2 + M x1 + x2 k1 k2 k1 + k2 k1 k2 k1 + k2 2 1 1 2 ≤ M k1 x1 + M k2 x2 − γM k1 x1 − γM k1 x1 k1 k2 k1 k1 2 m 1 2 γM k1 x1 − γM k2 x2 ≤ M k1 x1 + M k2 x2 − k1 k2 lk1 k1 2 m m 2 γM k1 x1 − γM k2 x2 ≤ M k1 x1 + M k2 x2 − k1 k2 lk1 lk2 1 m 1 M k1 x1 + M k2 x2 = 2− γ l k1 k2 m 2 2 = 1− γ M k1 x1 + M k2 x2 . 2l k1 k2 Noticing that k1 + k2 k1 + k2 k1 k2 k1 k2 M x1 − x2 + M x1 + x2 > 0, k1 k2 k1 + k2 k1 k2 k1 + k2 we get 1 − (m/2l)γ > 0. Let r1 = 1 − 1 − (m/2l)γ. It is easy to see that r1 ∈ (0, 1). Therefore, r1
1 1 M k1 x1 + M k2 x2 k1 k2 k1 + k2 k1 k2 k1 + k2 k1 k2 ≥ M x1 + x2 + M x1 − x2 . k1 k2 k1 + k2 k1 k2 k1 + k2
II. Suppose that k2 x2 /k1 x1 ≥ a. Then k1 x1 /k2 x2 < 1/a. Moreover, we have k1 x1 1 mx1 ≤ < lx2 k2 x2 a
⇒
x1 l < x2 ma
⇒
max{x1 , x2 } l < . min{x1 , x2 } ma
Hence min{x1 , x2 } = x1 + x2 1+
1 max{x1 ,x2 } min{x1 ,x2 }
≥
1 am . = l l + am 1 + am
By Lemma 3, we may assume without loss of generality that δ min{x1 , x2 } . x1 − x2 ≤ x1 + x2 1 − x1 + x2 Therefore, using Lemma 3 and inequality (5), we get
(5)
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k1 k2 k1 k2 δ min{x1 , x2 } x1 − x2 ≤ x1 + x2 1 − k1 + k2 k1 + k2 x1 + x2 k1 k2 δam x1 + x2 ≤ 1− l + am k1 + k2 k2 k1 δam k1 x1 + k2 x2 . ≤ 1− l + am k1 + k2 k1 + k2 Let η = (δam)/(l + am). By the convexity of M , we have
k1 k2 M x1 − x2 k1 + k2
k1 k2 ≤ M (1 − η) k1 x1 + k2 x2 k1 + k2 k1 + k2 k1 k2 k1 x1 + k2 x2 ≤ (1 − η)M k1 + k2 k1 + k2 k1 k2 M k1 x1 + M k2 x2 . (6) ≤ (1 − η) k1 + k2 k1 + k2
Therefore, by inequality (7), we obtain k1 + k2 k1 k2 M (k1 x1 ) M (k2 x2 ) . M x1 − x2 ≤ (1 − η) + k1 k2 k1 + k2 k1 k2
(7)
Since M ∈ Δ2 , then there exists kl = kl (a, w) = kl (w) > 1 such that M ((1/a)u) ≤ kl M (u), u ≥ aw. Let βa = 1/kl , v = u/a. Then M (av) ≥ βa M (v), v ≥ w. By max{k1 x1 , k2 x2 } ≥ w and k1 x1 = max{k1 x1 , k2 x2 }, we have k1 x1 ≥ w. Hence M ak1 x1 ≥ βa M k1 x1 .
(8)
Moreover, by the convexity of M , we have M (k1 x1 ) M (k2 x2 ) k1 + k2 k1 k2 M x1 + x2 ≤ + . k1 k2 k1 + k2 k1 k2 Therefore, by inequalities (7)–(9) and k2 x2 /k1 x1 ≥ a, we obtain k1 k2 k1 + k2 k1 k2 k1 + k2 M x1 − x2 + M x1 + x2 k1 k2 k1 + k2 k1 k2 k1 + k2 2 2 M (k1 x1 ) M (k2 x2 ) + ≤ M k1 x1 + M k2 x2 − η k1 k2 k1 k2 1 2 2 1 M k1 x1 + M k2 x2 ≤ M k1 x1 + M k2 x2 − η k1 k2 l l 2 1 2 1 M ak1 x1 + M ak1 x1 ≤ M k1 x1 + M k2 x2 − η k1 k2 l l
(9)
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≤
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1 1 2 2 M k1 x1 + M k1 x1 M k1 x1 + M k2 x2 − ηβα k1 k2 l l
ηβα m 2 ηβα m 2 2 2 M k1 x1 + M k2 x2 − M k1 x1 − M k2 x2 k1 k2 2l k1 2l k2 2 2 ηβα m M k1 x1 + M k2 x2 . = 1− 2l k1 k2 ≤
Noticing that k1 k2 k1 + k2 k1 k2 k1 + k2 M x1 − x2 + M x1 + x2 > 0, k1 k2 k1 + k2 k1 k2 k1 + k2 we obtain that 1 − (ηβα m)/(2l) ∈ (0, 1). Finally, combining the considerations from Parts I and II and denoting ηβα m , r = max r1 , 1 − 2l
we get the inequality 1 1 r M k1 x1 + M k2 x2 k1 k2 k1 + k2 k1 + k2 k1 k2 k1 k2 ≥ M x1 + x2 + M x1 − x2 . k1 k2 k1 + k2 k1 k2 k1 + k2
This completes the proof. 2 Proof of Theorem 1. Necessity. Let us assume that L0M (X) is uniformly nonsquare. Since L0M (R) is a subspace of L0M (X) isomorphically, we obtain that L0M (R) is a uniformly nonsquare space. Hence we have that M ∈ Δ2 and N ∈ Δ2 . Suppose that (b) is not ∞ true. Then there exist {xn }∞ n=1 ⊂ S(X) and {yn }n=1 ⊂ S(X) such that xn + yn → 2 and xn − yn → 2 as n → ∞. Pick d > 0 and set u(t) = d · x · χT (t),
un (t) = d · xn · χT (t),
vn (t) = d · yn · χT (t),
n = 1, 2, ...
0 ∞ 0 0 It is easy to see that {un }∞ n=1 ⊂ LM (X), {vn }n=1 ⊂ LM (X) and u ∈ LM (X). Moreover, 0 0 0 we have u = un = vn , n = 1, 2, .... We may assume without loss of generality 0 ∞ 0 0 that {un }∞ n=1 ⊂ S(LM (X)), {vn }n=1 ⊂ S(LM (X)) and u ∈ S(LM (X)). Pick kn ∈ K(un ) ∞ and ln ∈ K(vn ). Since N ∈ Δ2 , by Lemma 2, we obtain that {kn }∞ n=1 and {ln }n=1 are bounded sequences. We may assume without loss of generality that kn → k0 and ln → l0 as n → ∞. Let hn = (kn ln )/(kn + ln ) and h0 = (k0 l0 )/(k0 + l0 ). By the Fatou Lemma, we have
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lim inf n→∞
1
1 1+ 1 + ρM kn (un + vn ) ≥ lim M hn un (t) + vn (t) dt n→∞ hn h0 T
1 = 1+ h0
M (h0 · 2d)dt
T
1 1 + M h0 2u(t) dt = h0 T
≥ 2u = 2. 0
Moreover, by the convexity of M , we have kn ln 1
1
kn + ln 1 + ρM (un + vn ) ≤ 1 + ρM (kn un ) + 1 + ρM (ln vn ) kn ln kn + ln kn ln = un 0 + vn 0 = 2. This means that lim sup n→∞
1
1 + ρM hn (un + vn ) ≤ 2. hn
Hence we have 1
1 + ρM hn (un + vn ) = lim un + vn 0 = 2. n→∞ hn n→∞ lim
Similarly, we have un − vn 0 → 2 as n → ∞, a contradiction. Hence we obtain that X is a uniformly nonsquare space. Sufficiency. Let l = sup{K(u) : u ∈ S(L0M (X))}, 1 + δ = inf{K(u) : u ∈ S(L0M (X))}. Since M ∈ Δ2 and N ∈ Δ2 , by Lemma 2, we have l < ∞ and δ > 0. Take u1 , u2 ∈ 1 ), E1 = {t ∈ T : k1 u1 (t) < S(L0M (X)), k1 ∈ K(u1 ), k2 ∈ K(u2 ). Let w = 1l M −1 ( 4·μT w} and E2 = {t ∈ T : k2 u2 (t) < w}. Then
δ 1 δ 1 1 1 dt ≤ dt ρM (ki ui )χEi ≤ M ki M −1 M M −1 ki ki l 4 · μT ki 4 · μT Ei
≤
1 ki
Ei
Ei
1 δ δ δ dt ≤ · μT ≤ · 4 · μT ki 4 · μT 4(1 + δ)
(10)
whenever i ∈ {1, 2}. We know that 1 + δ ≤ min{k1 , k2 } and max{k1 , k2 } ≤ l. By Lemma 4, there exists r = r(w, 1 + δ, l) ∈ (0, 1) such that the inequality 1 1 r M k1 x1 + M k2 x2 k1 k2 k1 + k2 k1 + k2 k1 k2 k1 k2 ≥ M x1 + x2 + M x1 − x2 (11) k1 k2 k1 + k2 k1 k2 k1 + k2
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holds true for all x1 , x2 ∈ X satisfying max{k1 x1 , k2 x2 } > w. Let G = E1 ∩ E2 . By inequality (10), it is easy to see that
1 1 ρM (ki ui )χG = ki ki
M ki ki ui (t) dt <
δ , 4(1 + δ)
(12)
G
whenever i ∈ {1, 2}. Noticing that E1 = {t ∈ T : k1 u1 (t) < w}, E2 = {t ∈ T : k2 u2 (t) < w} and G = E1 ∩ E2 , we have max{k1 u1 (t), k2 u2 (t)} ≥ w for all t ∈ T \ G. Therefore, by inequality (11), we have k1 k2 k1 k2 k1 + k2 u1 (t) − u2 (t) + k1 + k2 M u1 (t) + u2 (t) M k1 k2 k1 + k2 k1 k2 k1 + k2 1 1 ≤r M k1 u1 (t) + M k2 u2 (t) , k1 k12 where t ∈ T \ G. This implies that k1 + k2 k1 k2
T \G
= T \G
r
T \G
≤ 2r
T \G
k1 k2 k1 + k2 k1 k2 +M dt u1 (t) − u2 (t) u1 (t) + u2 (t) M k1 k2 k1 + k2 k1 + k2
≤
k1 k2 k1 k2 u1 (t) − u2 (t) dt + u1 (t) + u2 (t) dt M M k1 + k2 k1 + k2
1 k1
2 2 M k1 u1 (t) + M k2 u2 (t) dt k1 k2
T \G
1 M k1 u1 (t) dt + k2
M k2 u2 (t) dt .
(13)
T \G
By inequality (13), we have k1 + k2 k1 + k2 k1 + k2 k1 + k2 ρM (u1 − u2 )χT \G + ρM (u1 + u2 )χT \G k1 k2 k1 k2 k1 k2 k1 k2
1 1 ≤ 2r ρM (k1 u1 )χT \G + ρM (k2 u2 )χT \G . k1 k2
(14)
Therefore, by inequalities (12)–(14) and 1 + δ = inf{K(u) : u ∈ S(L0M (X))}, we obtain k1 + k2 k1 + k2 1 + ρM (u1 − u2 )χT \G k1 k2 k1 k2 k1 + k2 k1 + k2 + 1 + ρM (u1 + u2 )χT \G k1 k2 k1 k2
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k1 + k2 k1 + k2 k1 + k2 + ρM (u1 − u2 )χT \G k1 k2 k1 k2 k1 k2 k1 + k2 k1 + k2 + ρM (u1 + u2 )χT \G k1 k2 k1 k2
k1 + k2 1 1 ≤2 + 2r ρM (k1 u1 )χT \G + ρM (k2 u2 )χT \G k1 k2 k1 k2
1 1 1 1 + 2r + ρM (k1 u1 )χT \G + ρM (k2 u2 )χT \G =2 k1 k2 k1 k2
1 1 = 2r 1 + ρM (k1 u1 )χT \G + 1 + ρM (k2 u2 )χT \G k1 k2 1 1 + + 2(1 − r) k1 k2
1 1 ≤ 2r 1 + ρM (k1 u1 )χT \G + 1 + ρM (k2 u2 )χT \G k1 k2 2 . + 2(1 − r) 1+δ
=2
(15)
Moreover, by the convexity of M and inequality (12), we have k1 + k2 k1 + k2 k1 + k2 k1 + k2 ρM (u1 − u2 )χG + ρM (u1 + u2 )χG k1 k2 k1 k2 k1 k2 k1 k2
2 2 ≤ ρM (k1 u1 )χG + ρM (k2 u2 )χG k1 k2 2δ ≤2· . 4(1 + δ)
(16)
Therefore, by inequalities (15) and (16), we have u1 − u2 0 + u1 + u2 0 k1 + k2 k1 + k2 k1 + k2 k1 + k2 1 + ρM 1 + ρM ≤ (u1 − u2 ) + (u1 + u2 ) k1 k2 k1 k2 k1 k2 k1 k2 k1 + k2 k1 + k2 1 + ρM = (u1 − u2 )χT \G k1 k2 k1 k2 k1 + k2 k1 + k2 + 1 + ρM (u1 + u2 )χT \G k1 k2 k1 k2 k1 + k2 k1 + k2 1 + ρM + (u1 − u2 )χG k1 k2 k1 k2 k1 + k2 k1 + k2 + 1 + ρM (u1 + u2 )χG k1 k2 k1 k2 1 2 1 ≤ 2r 1 + ρM (k1 u1 χT \G ) + 1 + ρM (k2 u2 χT \G ) + 2(1 − r) k1 k2 1+δ
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2 2 ρM (k1 u1 χG ) + ρM (k2 u2 χG ) k1 k2 1 2 1 1 + ρM (k1 u1 χT \G ) + 1 + ρM (k2 u2 χT \G ) + 2(1 − r) = 2r k1 k2 1+δ +
2r 2(1 − r) 2r ρM (k1 u1 χG ) + ρM (k2 u2 χG ) + ρM (k1 u1 χG ) k1 k2 k1 2(1 − r) + ρM (k2 u2 χG ) k2 1 2 1 1 + ρM (k1 u1 ) + 1 + ρM (k2 u2 ) + 2(1 − r) = 2r k1 k2 1+δ +
2(1 − r) 2(1 − r) ρM (k1 u1 χG ) + ρM (k2 u2 χG ) k1 k2 2 δ + 2(1 − r) ≤ 2r · 2 + 2(1 − r) 1+δ 4(1 + δ) 7δ = 2r · 2 + 2(1 − r) · 2 1 − 8(1 + δ) +
= 2r · 2 + 2(1 − r) · 2 − 4(1 − r) = 4 − 4(1 − r)
7δ 8(1 + δ)
7δ 8(1 + δ)
= 2u1 0 + 2u2 0 − 4(1 − r)
7δ . 8(1 + δ)
(17)
Denoting δ1 = (1 − r)
7δ 8(1 + δ)
it is easy to see that δ1 ∈ (0, 1). Therefore, by inequality (17), we have
min u1 − u2 0 , u1 + u2 0 ≤ 2 − δ1 . Hence we obtain that L0M (X) is a uniformly nonsquare space. This completes the proof. 2 Corollary 1. The Orlicz function space L0M (R) is a uniformly nonsquare space if and only if M ∈ Δ2 and N ∈ Δ2 . Theorem 2. The Orlicz–Bochner function space L0M (X) is a locally uniform nonsquare space if and only if (a) N ∈ Δ2 ; (b) X is a locally uniform nonsquare space.
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In order to prove the theorem, we give a lemma. Lemma 5. (See [5].) Let X be a locally uniform nonsquare space. Then: (a) For any x = 0, r1 ≥ r2 > 0, we have
inf x + y − min x + y, x − y : x ∈ X, r2 ≤ y ≤ r1 > 0;
y=0
(b) If xn → x, then limn→∞ δ(xn ) = δ(x), where
δ(x) = inf x + y − min x + y, x − y : x ∈ X, r2 ≤ y ≤ r1 . y=0
Proof of Theorem 2. Sufficiency. Suppose that the assumptions about M and X are 0 satisfied and that there exist u ∈ S(L0M (X)) and {vn }∞ n=1 ⊂ S(LM (X)) such that u + vn 0 → 2 and u − vn 0 → 2 as n → ∞. Let k ∈ K(u), ln ∈ K(vn ). Since N ∈ Δ2 , by Lemma 2, we obtain that {ln }∞ n=1 is a bounded sequence. We may assume without loss of generality that ln → l as n → ∞. We will derive a contradiction for each of the following two cases. Case 1. There exist ε0 > 0, σ0 > 0 such that μGn > ε0 , where Gn = {t ∈ T : ln vn (t) ≥ σ0 }. Since ln − 1 = ρM (ln vn ), we may assume without loss of generality that 2l − 1 ≥ ρM (ln vn ) whenever n ∈ N . Put Hn =
t ∈ T : σ0 ≤ ln vn (t) ≤ M −1
4 ε0 (2l − 1)
.
We have 2l − 1 ≥
M ln vn (t) dt ≥
M ln vn (t) dt ≥
Gn \Hn
T
4 μ(Gn \Hn ). ε0 (2l − 1)
This implies that μ(Gn \ Hn ) ≤ 14 ε0 . Hence, μHn > 12 ε0 . We define the function
η(t) = inf u(t) + y − min u(t) + y , u(t) − y : y=0
σ0 ≤ y ≤ M −1
4 ε0 (2l − 1)
on T0 , where T0 = {t ∈ T : u(t) = 0}. By Lemma 5, we have η(t) > 0 μ-a.e. on T0 . Let hn (t) → u(t) μ-a.e. on T0 , where hn (t) is a simple function. Hence ηn (t) = inf hn (t) + y − min hn (t) + y , hn (t) − y : y=0
σ0 ≤ y ≤ M −1
4 ε0 (2l − 1)
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is μ-measurable. By Lemma 5, we have ηn (t) → η(t) μ-a.e. on T0 . Then η(t) is μ-measurable. Using the inclusion T ⊃
∞ i=1
1 1 t ∈ T0 : < η(t) ≤ , i+1 i
we get that there exists η0 > 0 such that μH < 18 ε0 , where
H = t ∈ T : η(t) ≤ 2η0 . Let En = Hn \H, En1 = (Hn ∩ {t ∈ T : u(t) = 0})\H and En2 = (Hn ∩ {t ∈ T : u(t) = 0})\H. It is easy to see that En = En1 ∪ En2 , En1 ∩ En2 = ∅ and μEn ≥ 38 ε0 . By Lemma 3, we have u(t) + vn (t) − min u(t) + vn (t), u(t) − vn (t) ≥ η(t) ≥ 2η0 whenever t ∈ En1 . We may assume without loss of generality that there exists Fn1 ⊂ En1 such that μFn1 ≥
u(t) + vn (t) − u(t) + vn (t) ≥ η0
1 μE 1 , 2 n
t ∈ Fn1 .
Moreover, we know that for any u ≥ v > 0, λn ∈ (0, 1), we have
λn M (u) − M (λn u) − λn M (v) − M (λn v) v λn u u p(t)dt −
= λn 0
p(t)dt − λn 0
u 0
u = λn
0
v p(t)dt − λn
= λn
λn v p(t)dt − p(t)dt 0
λn u
λn v p(t)dt − p(t)dt
p(t)dt − 0
0
0
λn u p(t)dt − p(t)dt
v
λn v
λn u+(1−λ n )v
λn u p(t)dt − p(t)dt
≥ v
λn v
≥ 0. Noticing that t ∈ En2 , we have ln vn (t) ≥ σ0 > 0. Hence, k k k ln vn (t) ≥ k M (σ0 ) − M M ln vn (t) − M σ0 ≥ 0 k + ln k + ln k + ln k + ln
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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whenever n ∈ N . Let Fn = Fn1 ∪ En2 . Then 8μFn ≥ ε0 . Therefore, u0 + vn 0 − u + vn 0
k + ln 1
1
kln ≥ 1 + ρM (ku) + 1 + ρM (u + vn ) 1 + ρM (ln vn ) − k ln kln k + ln k + ln k ln kln = ρM (ku) + ρM (ln vn ) − ρM (u + vn ) kln k + ln k + ln k + ln ln k k + ln ≥ M ku(t) dt + M ln vn (t) dt kln k + ln k + ln Fn1
−
M
Fn1
+
k + ln kln
−
M
2 En
k + ln ≥ kln
M k + ln kln M
2 En
k + ln ≥ kln −
≥
2 En
u(t) + vn (t) dt
kln kln M u(t) + vn (t) dt k + ln k + ln kln u(t) + vn (t) dt k + ln ln M ku(t) dt + k + ln
2 En
kln k + ln
k M ln vn (t) dt k + ln
2 En
u(t) + vn (t) dt
kln kln M u(t) + vn (t) + η0 dt k + ln k + ln
Fn1
kln M u(t) + vn (t) dt k + ln
Fn1
+
k M ln vn (t) dt k + ln
−
kln k + ln
Fn1
Fn1
+
ln M ku(t) dt + k + ln
2 En
−
kln k + ln
Fn1
u(t) + vn (t) dt
k + ln kln
k + ln kln
2 En
M Fn1
kln M vn (t) dt − k + ln
kln η0 dt + k + ln
2 En
M
kln vn (t) dt k + ln
2 En
kln M vn (t) dt k + ln
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
−
M
2 En
=
k + ln kln
2071
kln vn (t) dt k + ln
M Fn1
k kln η0 dt + M ln vn (t) k + ln k + ln
ln vn (t) dt
2 En
k k + ln k kln k k + ln M η0 dt + M (σ0 ) − M σ0 dt. ≥ kln k + ln k + ln k + ln −M
Fn1
(18)
2 En
In consequence, by inequality (18) and Fatou Lemma, we have 0 = lim inf u0 + vn 0 − u + vn 0 n→∞ k + ln k kln k M η0 dt + M (σ0 ) − M σ0 dt ≥ lim inf n→∞ kln k + ln k + ln k + ln k+l ≥ kl
≥
k+l kl
kl η0 dt + M k+l
Fn1
≥
Fn1
min M 2 Fn1 ∪En
2 En
k k M (σ0 ) − M σ0 dt k+l k+l
2 En
k kl k η0 , M (σ0 ) − M σ0 dt k+l k+l k+l
k 1 k+l kl k · ε0 > 0, min M η0 , M (σ0 ) − M σ0 kl k+l k+l k+l 8
a contradiction. Case 2. For any ε > 0, σ > 0, there exists a natural number N such that μ{t ∈ T : ln vn (t) ≥ σ} < ε whenever n > N . By the Risez theorem, there exists a subsequence {n} of {n} such that ln vn (t) → 0 μ-a.e. on T . Using the inclusion ∞
t∈T : T0 = t ∈ T : ku(t) = 0 ⊃
n=1
1 1 < ku(t) ≤ , n+1 n
we get that there exists d > 0 such that 8μT1 < μT0 , where
T1 = t ∈ T0 : 0 < ku(t) < d ,
T0 = t ∈ T : ku(t) = 0 .
Since M is an N -function, we can choose 0 < h < d such that k k l l M (d) + M (c) − M d+ c >0 k+l k+l k+l k+l
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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whenever c ∈ (0, h]. Since vn (t) → 0 μ-a.e. on T , by the Egorov theorem, there exists a natural number N1 , if n > N1 , then ln vn (t) < h on F , where 8μ(T \F ) < μT1 . Moreover, we know that if u1 ≥ u2 ≥ v2 ≥ v1 > 0, then λn M (u1 ) + (1 − λn )M (v1 ) − M λn u1 + (1 − λn )v1 ≥ λn M (u2 ) + (1 − λn )M (v2 ) − M λn u3 + (1 − λn )v3 , where λn ∈ (0, 1). In fact, we have λn M (u1 ) + (1 − λn )M (v1 ) − M λn u1 + (1 − λn )v1
− λn M (u1 ) + (1 − λn )M (v2 ) − M λn u1 + (1 − λn )v2 = (1 − λn )M (v1 ) − M λn u1 + (1 − λn )v1
− (1 − λn )M (v2 ) − M λn u1 + (1 − λn )v2 v1 = (1 − λn )
λn u1 +(1−λ n )v1
p(t)dt −
p(t)dt
0
0
v2 − (1 − λn )
λn u1 +(1−λ n )v2
p(t)dt +
p(t)dt
0
0
v2 = −(1 − λn )
λn u1 +(1−λ n )v2
p(t)dt + v1
p(t)dt
λn u1 +(1−λn )v1
λn u1 +(1−λ n )v2
≥
v2
p(t)dt − λn u1 +(1−λn )v1
p(t)dt ≥ 0.
(19)
(1−λn )v1 +λn v2
Moreover, we have
λn M (u1 ) − M λn u1 + (1 − λn )v2 − λn M (u2 ) − M λn u2 + (1 − λn )v2 u1
λn u1 +(1−λ n )v2
p(t)dt −
= λn
p(t)dt − λn
0
0
u1
λn u2 +(1−λ n )v2
p(t)dt + 0
p(t)dt 0
λn u1 +(1−λ n )v2
p(t)dt −
= λn
u2
u2
p(t)dt
λn u2 +(1−λn )v2
λn u1 +(1−λ n )u2
λn u1 +(1−λ n )v2
p(t)dt −
≥ u2
λn u2 +(1−λn )v2
p(t)dt ≥ 0.
(20)
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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Therefore, by inequality (19) and (20), we have λn M (u1 ) + (1 − λn )M (v1 ) − M λn u1 + (1 − λn )v1 ≥ λn M (u2 ) + (1 − λn )M (v2 ) − M λn u3 + (1 − λn )v3 , where λn ∈ (0, 1). This means that if t ∈ T2 = (T0 \T1 )\(T \F ), then λn M ku(t) + (1 − λn )M ln vn (t) − M λn ku(t) + (1 − λn )ln vn (t) ≥ λn M (d) + (1 − λn )M (h) − M λn d + (1 − λn )h . It is easy to see that 4μT2 ≥ μT0 . Therefore, u0 + vn 0 − u + v0
k + ln kln 1
1
1 + ρM (ku) + 1 + ρM 1 + ρM (ln vn ) − (u + vn ) k ln kln k + ln k + ln k ln kln = ρM (ku) + ρM (ln vn ) − ρM (u + vn ) kln k + ln k + ln k + ln k ln k + ln M ku(t) + M ln vn (t) = kln k + ln k + ln ≥
T
− M λn ku(t) + (1 − λn )ln vn (t) dt ≥
k + ln kln
ln k M ku(t) + M ln vn (t) k + ln k + ln
T2
− M λn ku(t) + (1 − λn )ln vn (t) dt ≥
k + ln kln
ln k ln k M (d) + M (h) − M d+ h dt . k + ln k + ln k + ln k + ln
(21)
T
Therefore, by inequality (21) and Fatou Lemma, we have 0 = lim inf u0 + vn 0 − u + vn 0 n→∞
k + ln ≥ lim inf n→∞ kln k+l ≥ kl
T
T
k k ln ln M (d) + M (h) − M d+ h dt k + ln k + ln k + ln k + ln
l k l k M (d) + M (h) − M d+ h dt k+l k+l k+l k+l
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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≥
l k l k 1 k+l M (d) M (h) − M d+ h · μT0 kl k + l k+l k+l k+l 4
> 0, a contradiction. Hence we obtain that L0M (X) is a locally uniformly nonsquare spaces. Necessity. Since L0M (X) is a locally uniform nonsquare space and L0M (R) is embedded into L0M (X) isomorphically, we obtain that L0M (R) is a locally uniformly nonsquare space. This implies that N ∈ Δ2 . Suppose that X is not a locally uniformly nonsquare space. Then there exist x ∈ S(X) and {yn }∞ n=1 ⊂ S(X) such that x + yn → 2 and x − yn → 2 as n → ∞. Put u(t) = x · χT (t),
vn (t) = yn · χT (t),
n = 1, 2, ...
It is easy to see that u, vn ∈ L0M (X). We may assume without loss of generality that u, vn ∈ S(L0M (X)). Then vn (t) ∈ S(L0M (X)). Let ln ∈ K(vn ) and k0 ∈ K(u). Since N ∈ Δ2 , by Lemma 2, we obtain that {ln }∞ n=1 is a bounded sequence. Hence we may assume without loss of generality that ln → l0 as n → ∞. Let hn = (k0 ln )/(k0 + ln ) and h0 = (k0 l0 )/(k0 + l0 ). By the Fatou Lemma, we have 1 1
1+ lim inf 1 + ρM kn (un + vn ) ≥ lim M hn un (t) + vn (t) dt n→∞ hn n→∞ h0 T
1 = 1 + M (h0 · 2d)dt h0 T
=
1 1+ h0
M h0 2u(t) dt
T
≥ 2u = 2. 0
Moreover, by the convexity of M , we have 1
1
k0 + ln k0 ln 1 + ρM (u + vn ) ≤ 1 + ρM (k0 u) + 1 + ρM (ln vn ) k0 ln k0 + ln k0 ln = u0 + vn 0 = 2. This means that lim sup n→∞
1
1 + ρM hn (u + vn ) ≤ 2. hn
Hence we have lim
n→∞
1
1 + ρM hn (u + vn ) = lim u + vn 0 = 2. n→∞ hn
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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Similarly, we have u − vn 0 → 2 as n → ∞, a contradiction. Hence we obtain that X is a locally uniformly nonsquare space. This completes the proof. 2 Corollary 2. The Orlicz function space L0M (R) is a locally uniformly nonsquare space if and only if N ∈ Δ2 . 4. Applications to fixed point property of Orlicz–Bochner function spaces One of the classical problems of metric fixed point theory concerns existence of fixed points of nonexpansive mappings from nonempty bounded closed and convex sets into themselves. Let C be a nonempty bounded closed and convex subset of a Banach space X. A mapping T : C → C is nonexpansive if T x − T y ≤ x − y for all x, y ∈ C. A Banach space X is said to have the fixed point property (FPP) if every such mapping has a fixed point. Adding the assumption that C is weakly compact in this condition, we obtain the definition of the weak fixed point property (WFPP). In 1965, F. Browder [12] proved that Hilbert spaces have FPP. In the same year, Browder [13] and D. Gohde [14] showed independently that uniformly convex spaces have FPP, and W.A. Kirk [15] proved a more general result stating that all Banach spaces with weak normal structure have WFPP. Major progress in fixed point problems for nonexpansive mappings has been made recently. In 2003 (published in 2006), Jesús García-Falset and Enrique Llorens-Fuster, Eva M. Mazcuñan-Navarro (see [1]) solved a long-standing problem in the fixed point theory by proving FPP for all uniformly nonsquare Banach spaces. In 2004, Dowling, Lennard and Turett [16] proved that a nonempty closed bounded convex subset of c0 has FPP if and only if it is weakly compact. Using Theorem 1, we give a sufficient condition for an Orlicz–Bochner function space L0M (X) to have the fixed point property. Theorem 3. If (a) M ∈ Δ2 and N ∈ Δ2 ; (b) X is uniformly nonsquare space, then the Orlicz–Bochner function space L0M (X) have the fixed point property. Proof. By [1], we know that uniformly nonsquare Banach spaces have the fixed point property for nonexpansive mappings. Using Theorem 1, it is easy to see that L0M (X) have the fixed point property. 2 References [1] J. García-Falset, E. Llorens-Fuster, E.M. Mazcuñan-Navarro, Uniformly nonsquare Banach spaces have the fixed point property for nonexpansive mappings, J. Funct. Anal. 233 (2006) 494–514. [2] P. Foralewski, H. Hudzik, P. Kolwicz, Non-squareness properties of Orlicz–Lorentz sequence spaces, J. Funct. Anal. 264 (2013) 605–629. [3] R.C. James, Uniformly non-square Banach spaces, Ann. Math. 80 (1964) 542–550.
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[4] S.T. Chen, Geometry of Orlicz spaces, Dissertationes Math. 356 (1996) 1–204. [5] S. Shang, Y. Cui, Y. Fu, Nonsquareness and locally uniform nonsquareness in Orlicz–Bochner function spaces endowed with Luxemburg norm, J. Inequal. Appl. (2011) 1–15. [6] S. Shang, Y. Cui, Y. Fu, P -convexity of Orlicz–Bochner function spaces endowed with the Orlicz norm, Nonlinear Anal. TMA 75 (2012) 371–379. [7] P. Kolwicz, R. Pluciennik, P -convexity of Orlicz–Bochner function spaces, Proc. Amer. Math. Soc. 126 (1998) 2315–2322. [8] H. Hudzik, B.X. Wang, Approximative compactness in Orlicz spaces, J. Approx. Theory 95 (1998) 82–89. [9] S.T. Chen, H. Hudzik, On some convexities of Orlicz and Orlicz–Bochner spaces, Comment. Math. Univ. Carolin. 29 (1) (1988) 13–29. [10] P. Foralewski, Rotundity structure of local nature for generalized Orlicz–Lorentz function spaces, Nonlinear Anal. 74 (2011) 3912–3925. [11] R.C. James, Super-reflexive spaces with bases, Pacific J. Math. 41 (1972) 409–419. [12] F.E. Browder, Fixed point theorems for noncompact mappings in Hilbert space, Proc. Natl. Acad. Sci. USA 43 (1965) 1272–1276. [13] F.E. Browder, Nonexpansive nonlinear operators in a Banach space, Proc. Natl. Acad. Sci. USA 54 (1965) 1041–1044. [14] D. Gohde, Zum Prinzip der kontraktiven Abbildung, Math. Nachr. 30 (1965) 251–258. [15] W.A. Kirk, A fixed point theorem for mappings which do not increase distances, Amer. Math. Monthly 72 (1965) 1004–1006. [16] T. Domnguez Benavides, Weak uniform normal structure in direct sum spaces, Studia Math. 103 (1992) 283–290. [17] S. Shang, Y. Cui, Locally uniform convexity in Musielak–Orlicz function spaces of Bochner type endowed with the Luxemburg norm, J. Math. Anal. Appl. 378 (2011) 432–441. [18] S. Shang, Y. Cui, Y. Fu, Midpoint locally uniform rotundity of Musielak–Orlicz–Bochner function spaces endowed with the Luxemburg norm, J. Convex Anal. 19 (2012) 213–223. [19] Y.A. Cui, L. Jie, R. Pluciennik, Local uniform nonsquareness in Cesaro sequence spaces, Comment. Math. 37 (1997) 47–58. [20] P. Foralewski, H. Hudzik, L. Szymaszkiewicz, Local rotundity structure of generalized Orlicz– Lorentz sequence spaces, Nonlinear Anal. 68 (2008) 2709–2718. (1) [21] H. Hudzik, Uniformly non-ln Orlicz spaces with Luxemburg norm, Studia Math. 81 (1985) 271–284. [22] H. Hudzik, A. Kaminska, M. Mastyto, On the dual of Orlicz–Lorentz space, Proc. Amer. Math. Soc. 130 (6) (2002) 1645–1654. (1) [23] M. Denker, H. Hudzik, Uniformly non-ln Musielak–Orlicz sequence spaces, Proc. Indian Acad. Sci. Math. Sci. 101 (1991) 71–86. (1) [24] G. Alherk, H. Hudzik, Uniformly non-ln Musielak–Orlicz spaces of Bochner type, Forum Math. 1 (1989) 403–410. [25] P. Foralewski, H. Hudzik, P. Kolwicz, Non-squareness properties of Orlicz–Lorentz function spaces, J. Inequal. Appl. 2013 (32) (2013) 25 pp. [26] Y. Cui, H. Hudzik, M. Wista, K. Wlazlak, Non-squareness properties of Orlicz spaces equipped with the p-Amemiya norm, Nonlinear Anal. TMA 75 (2012) 3973–3993. (1) [27] A. Kaminska, B. Turett, Uniformly non-ln Musielak–Orlicz spaces, Bull. Acad. Pol. Sci., Sér. Sci. Math. 35 (3–4) (1987) 211–218. (1) [28] H. Hudzik, A. Kaminska, W. Kurc, Uniformly non-ln Musielak–Orlicz spaces, Bull. Acad. Polon. Sci. Math. 35 (7–8) (1987) 441–448. [29] H. Hudzik, Some class of uniformly non-square Orlicz–Bochner spaces, Comment. Math. Univ. Carolin. 26 (2) (1985) 269–274.
Contents lists available at ScienceDirect
Journal of Functional Analysis www.elsevier.com/locate/jfa
Uniform nonsquareness and locally uniform nonsquareness in Orlicz–Bochner function spaces and applications ✩ Shaoqiang Shang a,∗ , Yunan Cui b a
Department of Mathematics, Northeast Forestry University, Harbin 150040, PR China b Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, PR China
a r t i c l e
i n f o
Article history: Received 23 March 2013 Accepted 30 July 2014 Available online 11 August 2014 Communicated by B. Chow MSC: 46E30 46B20
a b s t r a c t In this paper, criteria for uniform nonsquareness and locally uniform nonsquareness of Orlicz–Bochner function spaces equipped with the Orlicz norm are given. Although, criteria for uniform nonsquareness and locally uniform nonsquareness in Orlicz function spaces were known, we can easily deduce them from our main results. Moreover, we give a sufficient condition for an Orlicz–Bochner function space to have the fixed point property. © 2014 Elsevier Inc. All rights reserved.
Keywords: Uniform nonsquare space Locally uniform nonsquare space Orlicz–Bochner function spaces Fixed point
✩
Supported by “the Fundamental Research Funds for the Central Universities”, DL12BB36.
* Corresponding author. E-mail addresses: [email protected] (S. Shang), [email protected] (Y. Cui). http://dx.doi.org/10.1016/j.jfa.2014.07.032 0022-1236/© 2014 Elsevier Inc. All rights reserved.
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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1. Introduction Uniform nonsquareness of Banach spaces has been defined by R.C. James as the geometric property which implies super-reflexivity (see [3,11]). So, proving this property of a Banach space we know, without any characterization of the dual space, that it is super-reflexive, so reflexive as well. Recently J. Garcia-Falset, E. Llorens-Fuster and E.M. Mazcuñan-Navarro have shown that uniformly nonsquare Banach spaces have the fixed point property (see [1]). Therefore, it was natural and interesting to look for criteria of non-squareness properties in various well-known classes of Banach spaces. In 2013, P. Foralewski, H. Hudzik and P. Kolwicz have given criteria for uniform nonsquareness and locally uniformly nonsquareness of Orlicz–Lorentz sequence spaces (see [2]). The topic of this paper is related to the topic of [5–10] and [17–29]. The criteria for uniform nonsquareness and locally uniform nonsquareness in the classical Orlicz function spaces have been given in [4,23,24] already. However, because of the complicated structure of Orlicz–Bochner function spaces equipped with Orlicz norm, up to now the criteria for uniform nonsquareness and locally uniform nonsquareness have not been discussed yet. The aim of this paper is to give criteria for uniform nonsquareness and locally uniform nonsquareness of Orlicz–Bochner function spaces equipped with Orlicz norm. Although, criteria for uniform nonsquareness and locally uniform nonsquareness in Orlicz function spaces were known, we can easily deduce them from our main results. Moreover, we give a sufficient condition for an Orlicz–Bochner function space to have the fixed point property. Let (X, · ) be a real Banach space. S(X) and B(X) denote the unit sphere and the unit ball of X, respectively. Let us recall some geometrical notions concerning nonsquareness. A Banach space X is said to be a nonsquare space if for any x, y ∈ S(X) we have min{x + y, x − y} < 2. A Banach space X is said to be a uniformly nonsquare space if for any x, y ∈ S(X), there exists δ > 0 such that min{x + y, x − y} < 2 − δ. A Banach space X is said to be a locally uniformly nonsquare space if for any x ∈ S(X), there exists δx > 0 such that min{x + y, x − y} < 2 − δx for any y ∈ S(X). 2. Preliminaries Definition 1. A function M : R → R+ is called an N -function if it has following properties: (1) if M is even, continuous convex and M (0) = 0; (2) M (u) > 0 for any u = 0; (3) limu→0 M (u)/u = 0 and limu→∞ M (u)/u = ∞. Let (T, Σ, μ) be a nonatomic finite measurable space. Moreover, for a given Banach space (X, · ), we denote by XT the set of all strongly μ-measurable function from T to X, and for each u ∈ XT , we define the modular of u by
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M u(t) dt.
ρM (u) = T
Put LM (X) =
u(t) ∈ XT :
M λu(t) dt < ∞ for some λ > 0 .
T
It is well known that Orlicz–Bochner spaces LM (X) are Banach spaces when they are equipped with the Luxemburg norm u ≤1 u = inf λ > 0 : ρM λ
or with the Orlicz norm u0 = inf
k>0
1
1 + ρM (ku) . k
In particular, LM (R) and L0M (R) are said to be Orlicz function spaces. It is well known that u ≤ u0 ≤ 2u. Set K(u) =
k>0:
1 1 + ρM (ku) = u0 . k
Since limu→∞ M (u)/u = ∞, the set K(u) are nonempty (see [4]). Let p(u) denote the right derivative of M (u) at u ∈ R+ , q(v) be the generalized inverse function of p(u) defined on R+ by
q(v) = sup u ≥ 0 : p(u) ≤ v u≥0
|v| Then we call N (v) = 0 q(s)ds the complementary function of M . It is well known that there holds the Young inequality uv ≤ M (u) + N (v) and uv = M (u) + N (v) ⇔ u = |q(v)| sign v or v = |p(u)| sign u. Moreover, it is well known that M and N are complementary to in the sense of Young each other. Definition 2. (See [4].) We say that an Orlicz function M satisfies condition Δ2 (M ∈ Δ2 ) if there exist K > 2 and u0 ≥ 0 such that M (2u) ≤ KM (u) (u ≥ u0 ) In this case, we write M ∈ Δ2 or N ∈ ∇2 .
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We know (see [4,21] and [23]) that if M ∈ ∇2 , then for any u0 > 0, α ∈ (0, 1) there exists δ > 0 such that M (αu) ≤ (α − δ)M (u) (u ≥ u0 ). First let us recall some results that will be used in the further part of the paper. Lemma 1. (See [4] and [21].) If N ∈ Δ2 , λ0 ∈ (0, 1), then for every w > 0, λ ∈ (0, λ0 ], there exist numbers a = a(w) ∈ (0, 1) and γ = γ(a(w), λ0 ) ∈ (0, 1) such that M λu + (1 − λ)v ≤ λ(1 − γ)M (u) + (1 − λ)M (v) holds true for all u ≥ w and v satisfying | uv | ≤ a. Lemma 2. (See [4].) (a) If N ∈ Δ2 , then set {K(u) : u ∈ S(L0M (X))} is bounded. (b) If M ∈ Δ2 and N ∈ Δ2 , then there exists δ ∈ (0, 1) such that inf K(u) : u ∈ S L0M (X) > 1 + δ. 3. Main results Theorem 1. The Orlicz–Bochner function space L0M (X) is a uniformly nonsquare space if and only if (a) M ∈ Δ2 and N ∈ Δ2 ; (b) X is a uniformly nonsquare space. In order to prove the theorem, we give some lemmas. Lemma 3. (See [27].) A Banach space X is uniformly nonsquare space if and only if there exists δ1 > 0 such that for any elements x1 , x2 ∈ X\{0}, we have
δ min{x1 , x2 } . 1− x1 + x2
min x1 + x2 , x1 − x2 ≤ x1 + x2
Lemma 4. Let X be a uniformly nonsquare space and M ∈ Δ2 , N ∈ Δ2 . Then for any l ≥ m > 0 and w > 0, there exists r = r(w, m, l) ∈ (0, 1) such that the inequality r
1 1 M k1 x1 + M k2 x2 k1 k2 k1 k2 k1 + k2 k1 k2 k1 + k2 M x1 + x2 + M x1 − x2 ≥ k1 k2 k1 + k2 k1 k2 k1 + k2
holds true for all x1 , x2 ∈ X, k1 , k2 ∈ R+ satisfying max{k1 x1 , k2 x2 } ≥ w, m ≤ min{k1 x1 , k2 x2 }, max{k1 , k2 } ≤ l.
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Proof. For any l ≥ m > 0 and w > 0, take x1 , x2 ∈ X\{0} and k1 , k2 ∈ R+ such that max{k1 x1 , k2 x2 } ≥ w, m ≤ min{k1 x1 , k2 x2 }, max{k1 , k2 } ≤ l. Moreover, we have 0<
k1 1 1 l < 1. = ≤ m = k 2 k1 + k2 1+ l l+m 1 + k2
Similarly, we have 1 < k2 /(k1 + k2 ) ≤ l/(l + m) < 1. Therefore, by Lemma 1, there exist l numbers a = a(w) ∈ (0, 1) and γ = γ(a(w), l+m ) = γ(w, m, l) ∈ (0, 1) such that
k1 k2 M u+ v k1 + k2 k1 + k2
≤
k2 k1 (1 − γ)M (u) + M (v), k1 + k2 k1 + k2
(1)
for all u ≥ w and v satisfying | uv | ≤ a. Moreover, we may assume without loss of generality that k1 x1 = max{k1 x1 , k2 x2 }. For the clarity, we will divide the proof into two parts. I. Suppose that k2 x2 /k1 x1 < a. Noticing that max{k1 x1 , k2 x2 } ≥ w, m ≤ min{k1 x1 , k2 x2 }, max{k1 , k2 } ≤ l, by inequality (1), we have k1 k2 k1 + k2 M x1 − x2 k1 k2 k1 + k2 k1 + k2 k1 k2 ≤ M k1 x1 + k2 x2 k1 k2 k1 + k2 k1 + k2 k2 k1 k1 + k2 (1 − γ)M k1 x1 + M k2 x2 ≤ k1 k2 k1 + k2 k1 + k2 1 1 ≤ (1 − γ)M k1 x1 + M k2 x2 k1 k2 1 1 1 = M k1 x1 + M k2 x2 − γM k1 x1 . k1 k2 k1
(2)
Similarly, we have k1 + k2 k1 k2 M x1 + x2 k1 k2 k1 + k2 1 1 1 ≤ M k1 x1 + M k2 x2 − γM k1 x1 . k1 k2 k1
(3)
Noticing that max{k1 x1 , k2 x2 } ≥ w, max{k1 , k2 } ≤ l and m ≤ min{k1 x1 , k2 x2 }, we have 1 m < , lk1 k1
1 m 1 m 1 = · < < . lk2 l k2 l k1
Therefore, by inequality (2)–(4), we obtain
(4)
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k1 + k2 k1 k2 k1 + k2 k1 k2 M x1 − x2 + M x1 + x2 k1 k2 k1 + k2 k1 k2 k1 + k2 2 1 1 2 ≤ M k1 x1 + M k2 x2 − γM k1 x1 − γM k1 x1 k1 k2 k1 k1 2 m 1 2 γM k1 x1 − γM k2 x2 ≤ M k1 x1 + M k2 x2 − k1 k2 lk1 k1 2 m m 2 γM k1 x1 − γM k2 x2 ≤ M k1 x1 + M k2 x2 − k1 k2 lk1 lk2 1 m 1 M k1 x1 + M k2 x2 = 2− γ l k1 k2 m 2 2 = 1− γ M k1 x1 + M k2 x2 . 2l k1 k2 Noticing that k1 + k2 k1 + k2 k1 k2 k1 k2 M x1 − x2 + M x1 + x2 > 0, k1 k2 k1 + k2 k1 k2 k1 + k2 we get 1 − (m/2l)γ > 0. Let r1 = 1 − 1 − (m/2l)γ. It is easy to see that r1 ∈ (0, 1). Therefore, r1
1 1 M k1 x1 + M k2 x2 k1 k2 k1 + k2 k1 k2 k1 + k2 k1 k2 ≥ M x1 + x2 + M x1 − x2 . k1 k2 k1 + k2 k1 k2 k1 + k2
II. Suppose that k2 x2 /k1 x1 ≥ a. Then k1 x1 /k2 x2 < 1/a. Moreover, we have k1 x1 1 mx1 ≤ < lx2 k2 x2 a
⇒
x1 l < x2 ma
⇒
max{x1 , x2 } l < . min{x1 , x2 } ma
Hence min{x1 , x2 } = x1 + x2 1+
1 max{x1 ,x2 } min{x1 ,x2 }
≥
1 am . = l l + am 1 + am
By Lemma 3, we may assume without loss of generality that δ min{x1 , x2 } . x1 − x2 ≤ x1 + x2 1 − x1 + x2 Therefore, using Lemma 3 and inequality (5), we get
(5)
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k1 k2 k1 k2 δ min{x1 , x2 } x1 − x2 ≤ x1 + x2 1 − k1 + k2 k1 + k2 x1 + x2 k1 k2 δam x1 + x2 ≤ 1− l + am k1 + k2 k2 k1 δam k1 x1 + k2 x2 . ≤ 1− l + am k1 + k2 k1 + k2 Let η = (δam)/(l + am). By the convexity of M , we have
k1 k2 M x1 − x2 k1 + k2
k1 k2 ≤ M (1 − η) k1 x1 + k2 x2 k1 + k2 k1 + k2 k1 k2 k1 x1 + k2 x2 ≤ (1 − η)M k1 + k2 k1 + k2 k1 k2 M k1 x1 + M k2 x2 . (6) ≤ (1 − η) k1 + k2 k1 + k2
Therefore, by inequality (7), we obtain k1 + k2 k1 k2 M (k1 x1 ) M (k2 x2 ) . M x1 − x2 ≤ (1 − η) + k1 k2 k1 + k2 k1 k2
(7)
Since M ∈ Δ2 , then there exists kl = kl (a, w) = kl (w) > 1 such that M ((1/a)u) ≤ kl M (u), u ≥ aw. Let βa = 1/kl , v = u/a. Then M (av) ≥ βa M (v), v ≥ w. By max{k1 x1 , k2 x2 } ≥ w and k1 x1 = max{k1 x1 , k2 x2 }, we have k1 x1 ≥ w. Hence M ak1 x1 ≥ βa M k1 x1 .
(8)
Moreover, by the convexity of M , we have M (k1 x1 ) M (k2 x2 ) k1 + k2 k1 k2 M x1 + x2 ≤ + . k1 k2 k1 + k2 k1 k2 Therefore, by inequalities (7)–(9) and k2 x2 /k1 x1 ≥ a, we obtain k1 k2 k1 + k2 k1 k2 k1 + k2 M x1 − x2 + M x1 + x2 k1 k2 k1 + k2 k1 k2 k1 + k2 2 2 M (k1 x1 ) M (k2 x2 ) + ≤ M k1 x1 + M k2 x2 − η k1 k2 k1 k2 1 2 2 1 M k1 x1 + M k2 x2 ≤ M k1 x1 + M k2 x2 − η k1 k2 l l 2 1 2 1 M ak1 x1 + M ak1 x1 ≤ M k1 x1 + M k2 x2 − η k1 k2 l l
(9)
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≤
2063
1 1 2 2 M k1 x1 + M k1 x1 M k1 x1 + M k2 x2 − ηβα k1 k2 l l
ηβα m 2 ηβα m 2 2 2 M k1 x1 + M k2 x2 − M k1 x1 − M k2 x2 k1 k2 2l k1 2l k2 2 2 ηβα m M k1 x1 + M k2 x2 . = 1− 2l k1 k2 ≤
Noticing that k1 k2 k1 + k2 k1 k2 k1 + k2 M x1 − x2 + M x1 + x2 > 0, k1 k2 k1 + k2 k1 k2 k1 + k2 we obtain that 1 − (ηβα m)/(2l) ∈ (0, 1). Finally, combining the considerations from Parts I and II and denoting ηβα m , r = max r1 , 1 − 2l
we get the inequality 1 1 r M k1 x1 + M k2 x2 k1 k2 k1 + k2 k1 + k2 k1 k2 k1 k2 ≥ M x1 + x2 + M x1 − x2 . k1 k2 k1 + k2 k1 k2 k1 + k2
This completes the proof. 2 Proof of Theorem 1. Necessity. Let us assume that L0M (X) is uniformly nonsquare. Since L0M (R) is a subspace of L0M (X) isomorphically, we obtain that L0M (R) is a uniformly nonsquare space. Hence we have that M ∈ Δ2 and N ∈ Δ2 . Suppose that (b) is not ∞ true. Then there exist {xn }∞ n=1 ⊂ S(X) and {yn }n=1 ⊂ S(X) such that xn + yn → 2 and xn − yn → 2 as n → ∞. Pick d > 0 and set u(t) = d · x · χT (t),
un (t) = d · xn · χT (t),
vn (t) = d · yn · χT (t),
n = 1, 2, ...
0 ∞ 0 0 It is easy to see that {un }∞ n=1 ⊂ LM (X), {vn }n=1 ⊂ LM (X) and u ∈ LM (X). Moreover, 0 0 0 we have u = un = vn , n = 1, 2, .... We may assume without loss of generality 0 ∞ 0 0 that {un }∞ n=1 ⊂ S(LM (X)), {vn }n=1 ⊂ S(LM (X)) and u ∈ S(LM (X)). Pick kn ∈ K(un ) ∞ and ln ∈ K(vn ). Since N ∈ Δ2 , by Lemma 2, we obtain that {kn }∞ n=1 and {ln }n=1 are bounded sequences. We may assume without loss of generality that kn → k0 and ln → l0 as n → ∞. Let hn = (kn ln )/(kn + ln ) and h0 = (k0 l0 )/(k0 + l0 ). By the Fatou Lemma, we have
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lim inf n→∞
1
1 1+ 1 + ρM kn (un + vn ) ≥ lim M hn un (t) + vn (t) dt n→∞ hn h0 T
1 = 1+ h0
M (h0 · 2d)dt
T
1 1 + M h0 2u(t) dt = h0 T
≥ 2u = 2. 0
Moreover, by the convexity of M , we have kn ln 1
1
kn + ln 1 + ρM (un + vn ) ≤ 1 + ρM (kn un ) + 1 + ρM (ln vn ) kn ln kn + ln kn ln = un 0 + vn 0 = 2. This means that lim sup n→∞
1
1 + ρM hn (un + vn ) ≤ 2. hn
Hence we have 1
1 + ρM hn (un + vn ) = lim un + vn 0 = 2. n→∞ hn n→∞ lim
Similarly, we have un − vn 0 → 2 as n → ∞, a contradiction. Hence we obtain that X is a uniformly nonsquare space. Sufficiency. Let l = sup{K(u) : u ∈ S(L0M (X))}, 1 + δ = inf{K(u) : u ∈ S(L0M (X))}. Since M ∈ Δ2 and N ∈ Δ2 , by Lemma 2, we have l < ∞ and δ > 0. Take u1 , u2 ∈ 1 ), E1 = {t ∈ T : k1 u1 (t) < S(L0M (X)), k1 ∈ K(u1 ), k2 ∈ K(u2 ). Let w = 1l M −1 ( 4·μT w} and E2 = {t ∈ T : k2 u2 (t) < w}. Then
δ 1 δ 1 1 1 dt ≤ dt ρM (ki ui )χEi ≤ M ki M −1 M M −1 ki ki l 4 · μT ki 4 · μT Ei
≤
1 ki
Ei
Ei
1 δ δ δ dt ≤ · μT ≤ · 4 · μT ki 4 · μT 4(1 + δ)
(10)
whenever i ∈ {1, 2}. We know that 1 + δ ≤ min{k1 , k2 } and max{k1 , k2 } ≤ l. By Lemma 4, there exists r = r(w, 1 + δ, l) ∈ (0, 1) such that the inequality 1 1 r M k1 x1 + M k2 x2 k1 k2 k1 + k2 k1 + k2 k1 k2 k1 k2 ≥ M x1 + x2 + M x1 − x2 (11) k1 k2 k1 + k2 k1 k2 k1 + k2
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holds true for all x1 , x2 ∈ X satisfying max{k1 x1 , k2 x2 } > w. Let G = E1 ∩ E2 . By inequality (10), it is easy to see that
1 1 ρM (ki ui )χG = ki ki
M ki ki ui (t) dt <
δ , 4(1 + δ)
(12)
G
whenever i ∈ {1, 2}. Noticing that E1 = {t ∈ T : k1 u1 (t) < w}, E2 = {t ∈ T : k2 u2 (t) < w} and G = E1 ∩ E2 , we have max{k1 u1 (t), k2 u2 (t)} ≥ w for all t ∈ T \ G. Therefore, by inequality (11), we have k1 k2 k1 k2 k1 + k2 u1 (t) − u2 (t) + k1 + k2 M u1 (t) + u2 (t) M k1 k2 k1 + k2 k1 k2 k1 + k2 1 1 ≤r M k1 u1 (t) + M k2 u2 (t) , k1 k12 where t ∈ T \ G. This implies that k1 + k2 k1 k2
T \G
= T \G
r
T \G
≤ 2r
T \G
k1 k2 k1 + k2 k1 k2 +M dt u1 (t) − u2 (t) u1 (t) + u2 (t) M k1 k2 k1 + k2 k1 + k2
≤
k1 k2 k1 k2 u1 (t) − u2 (t) dt + u1 (t) + u2 (t) dt M M k1 + k2 k1 + k2
1 k1
2 2 M k1 u1 (t) + M k2 u2 (t) dt k1 k2
T \G
1 M k1 u1 (t) dt + k2
M k2 u2 (t) dt .
(13)
T \G
By inequality (13), we have k1 + k2 k1 + k2 k1 + k2 k1 + k2 ρM (u1 − u2 )χT \G + ρM (u1 + u2 )χT \G k1 k2 k1 k2 k1 k2 k1 k2
1 1 ≤ 2r ρM (k1 u1 )χT \G + ρM (k2 u2 )χT \G . k1 k2
(14)
Therefore, by inequalities (12)–(14) and 1 + δ = inf{K(u) : u ∈ S(L0M (X))}, we obtain k1 + k2 k1 + k2 1 + ρM (u1 − u2 )χT \G k1 k2 k1 k2 k1 + k2 k1 + k2 + 1 + ρM (u1 + u2 )χT \G k1 k2 k1 k2
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k1 + k2 k1 + k2 k1 + k2 + ρM (u1 − u2 )χT \G k1 k2 k1 k2 k1 k2 k1 + k2 k1 + k2 + ρM (u1 + u2 )χT \G k1 k2 k1 k2
k1 + k2 1 1 ≤2 + 2r ρM (k1 u1 )χT \G + ρM (k2 u2 )χT \G k1 k2 k1 k2
1 1 1 1 + 2r + ρM (k1 u1 )χT \G + ρM (k2 u2 )χT \G =2 k1 k2 k1 k2
1 1 = 2r 1 + ρM (k1 u1 )χT \G + 1 + ρM (k2 u2 )χT \G k1 k2 1 1 + + 2(1 − r) k1 k2
1 1 ≤ 2r 1 + ρM (k1 u1 )χT \G + 1 + ρM (k2 u2 )χT \G k1 k2 2 . + 2(1 − r) 1+δ
=2
(15)
Moreover, by the convexity of M and inequality (12), we have k1 + k2 k1 + k2 k1 + k2 k1 + k2 ρM (u1 − u2 )χG + ρM (u1 + u2 )χG k1 k2 k1 k2 k1 k2 k1 k2
2 2 ≤ ρM (k1 u1 )χG + ρM (k2 u2 )χG k1 k2 2δ ≤2· . 4(1 + δ)
(16)
Therefore, by inequalities (15) and (16), we have u1 − u2 0 + u1 + u2 0 k1 + k2 k1 + k2 k1 + k2 k1 + k2 1 + ρM 1 + ρM ≤ (u1 − u2 ) + (u1 + u2 ) k1 k2 k1 k2 k1 k2 k1 k2 k1 + k2 k1 + k2 1 + ρM = (u1 − u2 )χT \G k1 k2 k1 k2 k1 + k2 k1 + k2 + 1 + ρM (u1 + u2 )χT \G k1 k2 k1 k2 k1 + k2 k1 + k2 1 + ρM + (u1 − u2 )χG k1 k2 k1 k2 k1 + k2 k1 + k2 + 1 + ρM (u1 + u2 )χG k1 k2 k1 k2 1 2 1 ≤ 2r 1 + ρM (k1 u1 χT \G ) + 1 + ρM (k2 u2 χT \G ) + 2(1 − r) k1 k2 1+δ
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2 2 ρM (k1 u1 χG ) + ρM (k2 u2 χG ) k1 k2 1 2 1 1 + ρM (k1 u1 χT \G ) + 1 + ρM (k2 u2 χT \G ) + 2(1 − r) = 2r k1 k2 1+δ +
2r 2(1 − r) 2r ρM (k1 u1 χG ) + ρM (k2 u2 χG ) + ρM (k1 u1 χG ) k1 k2 k1 2(1 − r) + ρM (k2 u2 χG ) k2 1 2 1 1 + ρM (k1 u1 ) + 1 + ρM (k2 u2 ) + 2(1 − r) = 2r k1 k2 1+δ +
2(1 − r) 2(1 − r) ρM (k1 u1 χG ) + ρM (k2 u2 χG ) k1 k2 2 δ + 2(1 − r) ≤ 2r · 2 + 2(1 − r) 1+δ 4(1 + δ) 7δ = 2r · 2 + 2(1 − r) · 2 1 − 8(1 + δ) +
= 2r · 2 + 2(1 − r) · 2 − 4(1 − r) = 4 − 4(1 − r)
7δ 8(1 + δ)
7δ 8(1 + δ)
= 2u1 0 + 2u2 0 − 4(1 − r)
7δ . 8(1 + δ)
(17)
Denoting δ1 = (1 − r)
7δ 8(1 + δ)
it is easy to see that δ1 ∈ (0, 1). Therefore, by inequality (17), we have
min u1 − u2 0 , u1 + u2 0 ≤ 2 − δ1 . Hence we obtain that L0M (X) is a uniformly nonsquare space. This completes the proof. 2 Corollary 1. The Orlicz function space L0M (R) is a uniformly nonsquare space if and only if M ∈ Δ2 and N ∈ Δ2 . Theorem 2. The Orlicz–Bochner function space L0M (X) is a locally uniform nonsquare space if and only if (a) N ∈ Δ2 ; (b) X is a locally uniform nonsquare space.
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In order to prove the theorem, we give a lemma. Lemma 5. (See [5].) Let X be a locally uniform nonsquare space. Then: (a) For any x = 0, r1 ≥ r2 > 0, we have
inf x + y − min x + y, x − y : x ∈ X, r2 ≤ y ≤ r1 > 0;
y=0
(b) If xn → x, then limn→∞ δ(xn ) = δ(x), where
δ(x) = inf x + y − min x + y, x − y : x ∈ X, r2 ≤ y ≤ r1 . y=0
Proof of Theorem 2. Sufficiency. Suppose that the assumptions about M and X are 0 satisfied and that there exist u ∈ S(L0M (X)) and {vn }∞ n=1 ⊂ S(LM (X)) such that u + vn 0 → 2 and u − vn 0 → 2 as n → ∞. Let k ∈ K(u), ln ∈ K(vn ). Since N ∈ Δ2 , by Lemma 2, we obtain that {ln }∞ n=1 is a bounded sequence. We may assume without loss of generality that ln → l as n → ∞. We will derive a contradiction for each of the following two cases. Case 1. There exist ε0 > 0, σ0 > 0 such that μGn > ε0 , where Gn = {t ∈ T : ln vn (t) ≥ σ0 }. Since ln − 1 = ρM (ln vn ), we may assume without loss of generality that 2l − 1 ≥ ρM (ln vn ) whenever n ∈ N . Put Hn =
t ∈ T : σ0 ≤ ln vn (t) ≤ M −1
4 ε0 (2l − 1)
.
We have 2l − 1 ≥
M ln vn (t) dt ≥
M ln vn (t) dt ≥
Gn \Hn
T
4 μ(Gn \Hn ). ε0 (2l − 1)
This implies that μ(Gn \ Hn ) ≤ 14 ε0 . Hence, μHn > 12 ε0 . We define the function
η(t) = inf u(t) + y − min u(t) + y , u(t) − y : y=0
σ0 ≤ y ≤ M −1
4 ε0 (2l − 1)
on T0 , where T0 = {t ∈ T : u(t) = 0}. By Lemma 5, we have η(t) > 0 μ-a.e. on T0 . Let hn (t) → u(t) μ-a.e. on T0 , where hn (t) is a simple function. Hence ηn (t) = inf hn (t) + y − min hn (t) + y , hn (t) − y : y=0
σ0 ≤ y ≤ M −1
4 ε0 (2l − 1)
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is μ-measurable. By Lemma 5, we have ηn (t) → η(t) μ-a.e. on T0 . Then η(t) is μ-measurable. Using the inclusion T ⊃
∞ i=1
1 1 t ∈ T0 : < η(t) ≤ , i+1 i
we get that there exists η0 > 0 such that μH < 18 ε0 , where
H = t ∈ T : η(t) ≤ 2η0 . Let En = Hn \H, En1 = (Hn ∩ {t ∈ T : u(t) = 0})\H and En2 = (Hn ∩ {t ∈ T : u(t) = 0})\H. It is easy to see that En = En1 ∪ En2 , En1 ∩ En2 = ∅ and μEn ≥ 38 ε0 . By Lemma 3, we have u(t) + vn (t) − min u(t) + vn (t), u(t) − vn (t) ≥ η(t) ≥ 2η0 whenever t ∈ En1 . We may assume without loss of generality that there exists Fn1 ⊂ En1 such that μFn1 ≥
u(t) + vn (t) − u(t) + vn (t) ≥ η0
1 μE 1 , 2 n
t ∈ Fn1 .
Moreover, we know that for any u ≥ v > 0, λn ∈ (0, 1), we have
λn M (u) − M (λn u) − λn M (v) − M (λn v) v λn u u p(t)dt −
= λn 0
p(t)dt − λn 0
u 0
u = λn
0
v p(t)dt − λn
= λn
λn v p(t)dt − p(t)dt 0
λn u
λn v p(t)dt − p(t)dt
p(t)dt − 0
0
0
λn u p(t)dt − p(t)dt
v
λn v
λn u+(1−λ n )v
λn u p(t)dt − p(t)dt
≥ v
λn v
≥ 0. Noticing that t ∈ En2 , we have ln vn (t) ≥ σ0 > 0. Hence, k k k ln vn (t) ≥ k M (σ0 ) − M M ln vn (t) − M σ0 ≥ 0 k + ln k + ln k + ln k + ln
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whenever n ∈ N . Let Fn = Fn1 ∪ En2 . Then 8μFn ≥ ε0 . Therefore, u0 + vn 0 − u + vn 0
k + ln 1
1
kln ≥ 1 + ρM (ku) + 1 + ρM (u + vn ) 1 + ρM (ln vn ) − k ln kln k + ln k + ln k ln kln = ρM (ku) + ρM (ln vn ) − ρM (u + vn ) kln k + ln k + ln k + ln ln k k + ln ≥ M ku(t) dt + M ln vn (t) dt kln k + ln k + ln Fn1
−
M
Fn1
+
k + ln kln
−
M
2 En
k + ln ≥ kln
M k + ln kln M
2 En
k + ln ≥ kln −
≥
2 En
u(t) + vn (t) dt
kln kln M u(t) + vn (t) dt k + ln k + ln kln u(t) + vn (t) dt k + ln ln M ku(t) dt + k + ln
2 En
kln k + ln
k M ln vn (t) dt k + ln
2 En
u(t) + vn (t) dt
kln kln M u(t) + vn (t) + η0 dt k + ln k + ln
Fn1
kln M u(t) + vn (t) dt k + ln
Fn1
+
k M ln vn (t) dt k + ln
−
kln k + ln
Fn1
Fn1
+
ln M ku(t) dt + k + ln
2 En
−
kln k + ln
Fn1
u(t) + vn (t) dt
k + ln kln
k + ln kln
2 En
M Fn1
kln M vn (t) dt − k + ln
kln η0 dt + k + ln
2 En
M
kln vn (t) dt k + ln
2 En
kln M vn (t) dt k + ln
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
−
M
2 En
=
k + ln kln
2071
kln vn (t) dt k + ln
M Fn1
k kln η0 dt + M ln vn (t) k + ln k + ln
ln vn (t) dt
2 En
k k + ln k kln k k + ln M η0 dt + M (σ0 ) − M σ0 dt. ≥ kln k + ln k + ln k + ln −M
Fn1
(18)
2 En
In consequence, by inequality (18) and Fatou Lemma, we have 0 = lim inf u0 + vn 0 − u + vn 0 n→∞ k + ln k kln k M η0 dt + M (σ0 ) − M σ0 dt ≥ lim inf n→∞ kln k + ln k + ln k + ln k+l ≥ kl
≥
k+l kl
kl η0 dt + M k+l
Fn1
≥
Fn1
min M 2 Fn1 ∪En
2 En
k k M (σ0 ) − M σ0 dt k+l k+l
2 En
k kl k η0 , M (σ0 ) − M σ0 dt k+l k+l k+l
k 1 k+l kl k · ε0 > 0, min M η0 , M (σ0 ) − M σ0 kl k+l k+l k+l 8
a contradiction. Case 2. For any ε > 0, σ > 0, there exists a natural number N such that μ{t ∈ T : ln vn (t) ≥ σ} < ε whenever n > N . By the Risez theorem, there exists a subsequence {n} of {n} such that ln vn (t) → 0 μ-a.e. on T . Using the inclusion ∞
t∈T : T0 = t ∈ T : ku(t) = 0 ⊃
n=1
1 1 < ku(t) ≤ , n+1 n
we get that there exists d > 0 such that 8μT1 < μT0 , where
T1 = t ∈ T0 : 0 < ku(t) < d ,
T0 = t ∈ T : ku(t) = 0 .
Since M is an N -function, we can choose 0 < h < d such that k k l l M (d) + M (c) − M d+ c >0 k+l k+l k+l k+l
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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whenever c ∈ (0, h]. Since vn (t) → 0 μ-a.e. on T , by the Egorov theorem, there exists a natural number N1 , if n > N1 , then ln vn (t) < h on F , where 8μ(T \F ) < μT1 . Moreover, we know that if u1 ≥ u2 ≥ v2 ≥ v1 > 0, then λn M (u1 ) + (1 − λn )M (v1 ) − M λn u1 + (1 − λn )v1 ≥ λn M (u2 ) + (1 − λn )M (v2 ) − M λn u3 + (1 − λn )v3 , where λn ∈ (0, 1). In fact, we have λn M (u1 ) + (1 − λn )M (v1 ) − M λn u1 + (1 − λn )v1
− λn M (u1 ) + (1 − λn )M (v2 ) − M λn u1 + (1 − λn )v2 = (1 − λn )M (v1 ) − M λn u1 + (1 − λn )v1
− (1 − λn )M (v2 ) − M λn u1 + (1 − λn )v2 v1 = (1 − λn )
λn u1 +(1−λ n )v1
p(t)dt −
p(t)dt
0
0
v2 − (1 − λn )
λn u1 +(1−λ n )v2
p(t)dt +
p(t)dt
0
0
v2 = −(1 − λn )
λn u1 +(1−λ n )v2
p(t)dt + v1
p(t)dt
λn u1 +(1−λn )v1
λn u1 +(1−λ n )v2
≥
v2
p(t)dt − λn u1 +(1−λn )v1
p(t)dt ≥ 0.
(19)
(1−λn )v1 +λn v2
Moreover, we have
λn M (u1 ) − M λn u1 + (1 − λn )v2 − λn M (u2 ) − M λn u2 + (1 − λn )v2 u1
λn u1 +(1−λ n )v2
p(t)dt −
= λn
p(t)dt − λn
0
0
u1
λn u2 +(1−λ n )v2
p(t)dt + 0
p(t)dt 0
λn u1 +(1−λ n )v2
p(t)dt −
= λn
u2
u2
p(t)dt
λn u2 +(1−λn )v2
λn u1 +(1−λ n )u2
λn u1 +(1−λ n )v2
p(t)dt −
≥ u2
λn u2 +(1−λn )v2
p(t)dt ≥ 0.
(20)
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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Therefore, by inequality (19) and (20), we have λn M (u1 ) + (1 − λn )M (v1 ) − M λn u1 + (1 − λn )v1 ≥ λn M (u2 ) + (1 − λn )M (v2 ) − M λn u3 + (1 − λn )v3 , where λn ∈ (0, 1). This means that if t ∈ T2 = (T0 \T1 )\(T \F ), then λn M ku(t) + (1 − λn )M ln vn (t) − M λn ku(t) + (1 − λn )ln vn (t) ≥ λn M (d) + (1 − λn )M (h) − M λn d + (1 − λn )h . It is easy to see that 4μT2 ≥ μT0 . Therefore, u0 + vn 0 − u + v0
k + ln kln 1
1
1 + ρM (ku) + 1 + ρM 1 + ρM (ln vn ) − (u + vn ) k ln kln k + ln k + ln k ln kln = ρM (ku) + ρM (ln vn ) − ρM (u + vn ) kln k + ln k + ln k + ln k ln k + ln M ku(t) + M ln vn (t) = kln k + ln k + ln ≥
T
− M λn ku(t) + (1 − λn )ln vn (t) dt ≥
k + ln kln
ln k M ku(t) + M ln vn (t) k + ln k + ln
T2
− M λn ku(t) + (1 − λn )ln vn (t) dt ≥
k + ln kln
ln k ln k M (d) + M (h) − M d+ h dt . k + ln k + ln k + ln k + ln
(21)
T
Therefore, by inequality (21) and Fatou Lemma, we have 0 = lim inf u0 + vn 0 − u + vn 0 n→∞
k + ln ≥ lim inf n→∞ kln k+l ≥ kl
T
T
k k ln ln M (d) + M (h) − M d+ h dt k + ln k + ln k + ln k + ln
l k l k M (d) + M (h) − M d+ h dt k+l k+l k+l k+l
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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≥
l k l k 1 k+l M (d) M (h) − M d+ h · μT0 kl k + l k+l k+l k+l 4
> 0, a contradiction. Hence we obtain that L0M (X) is a locally uniformly nonsquare spaces. Necessity. Since L0M (X) is a locally uniform nonsquare space and L0M (R) is embedded into L0M (X) isomorphically, we obtain that L0M (R) is a locally uniformly nonsquare space. This implies that N ∈ Δ2 . Suppose that X is not a locally uniformly nonsquare space. Then there exist x ∈ S(X) and {yn }∞ n=1 ⊂ S(X) such that x + yn → 2 and x − yn → 2 as n → ∞. Put u(t) = x · χT (t),
vn (t) = yn · χT (t),
n = 1, 2, ...
It is easy to see that u, vn ∈ L0M (X). We may assume without loss of generality that u, vn ∈ S(L0M (X)). Then vn (t) ∈ S(L0M (X)). Let ln ∈ K(vn ) and k0 ∈ K(u). Since N ∈ Δ2 , by Lemma 2, we obtain that {ln }∞ n=1 is a bounded sequence. Hence we may assume without loss of generality that ln → l0 as n → ∞. Let hn = (k0 ln )/(k0 + ln ) and h0 = (k0 l0 )/(k0 + l0 ). By the Fatou Lemma, we have 1 1
1+ lim inf 1 + ρM kn (un + vn ) ≥ lim M hn un (t) + vn (t) dt n→∞ hn n→∞ h0 T
1 = 1 + M (h0 · 2d)dt h0 T
=
1 1+ h0
M h0 2u(t) dt
T
≥ 2u = 2. 0
Moreover, by the convexity of M , we have 1
1
k0 + ln k0 ln 1 + ρM (u + vn ) ≤ 1 + ρM (k0 u) + 1 + ρM (ln vn ) k0 ln k0 + ln k0 ln = u0 + vn 0 = 2. This means that lim sup n→∞
1
1 + ρM hn (u + vn ) ≤ 2. hn
Hence we have lim
n→∞
1
1 + ρM hn (u + vn ) = lim u + vn 0 = 2. n→∞ hn
S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076
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Similarly, we have u − vn 0 → 2 as n → ∞, a contradiction. Hence we obtain that X is a locally uniformly nonsquare space. This completes the proof. 2 Corollary 2. The Orlicz function space L0M (R) is a locally uniformly nonsquare space if and only if N ∈ Δ2 . 4. Applications to fixed point property of Orlicz–Bochner function spaces One of the classical problems of metric fixed point theory concerns existence of fixed points of nonexpansive mappings from nonempty bounded closed and convex sets into themselves. Let C be a nonempty bounded closed and convex subset of a Banach space X. A mapping T : C → C is nonexpansive if T x − T y ≤ x − y for all x, y ∈ C. A Banach space X is said to have the fixed point property (FPP) if every such mapping has a fixed point. Adding the assumption that C is weakly compact in this condition, we obtain the definition of the weak fixed point property (WFPP). In 1965, F. Browder [12] proved that Hilbert spaces have FPP. In the same year, Browder [13] and D. Gohde [14] showed independently that uniformly convex spaces have FPP, and W.A. Kirk [15] proved a more general result stating that all Banach spaces with weak normal structure have WFPP. Major progress in fixed point problems for nonexpansive mappings has been made recently. In 2003 (published in 2006), Jesús García-Falset and Enrique Llorens-Fuster, Eva M. Mazcuñan-Navarro (see [1]) solved a long-standing problem in the fixed point theory by proving FPP for all uniformly nonsquare Banach spaces. In 2004, Dowling, Lennard and Turett [16] proved that a nonempty closed bounded convex subset of c0 has FPP if and only if it is weakly compact. Using Theorem 1, we give a sufficient condition for an Orlicz–Bochner function space L0M (X) to have the fixed point property. Theorem 3. If (a) M ∈ Δ2 and N ∈ Δ2 ; (b) X is uniformly nonsquare space, then the Orlicz–Bochner function space L0M (X) have the fixed point property. Proof. By [1], we know that uniformly nonsquare Banach spaces have the fixed point property for nonexpansive mappings. Using Theorem 1, it is easy to see that L0M (X) have the fixed point property. 2 References [1] J. García-Falset, E. Llorens-Fuster, E.M. Mazcuñan-Navarro, Uniformly nonsquare Banach spaces have the fixed point property for nonexpansive mappings, J. Funct. Anal. 233 (2006) 494–514. [2] P. Foralewski, H. Hudzik, P. Kolwicz, Non-squareness properties of Orlicz–Lorentz sequence spaces, J. Funct. Anal. 264 (2013) 605–629. [3] R.C. James, Uniformly non-square Banach spaces, Ann. Math. 80 (1964) 542–550.
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