uniform nonsquareness and locally uniform nonsquareness in orlicz–bochner function spaces and...

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 Chinab Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, PR China

a r t i c l e i n f o a b s t r a c t

Article history:Received 23 March 2013Accepted 30 July 2014Available online 11 August 2014Communicated by B. Chow

MSC:46E3046B20

Keywords:Uniform nonsquare spaceLocally uniform nonsquare spaceOrlicz–Bochner function spacesFixed point

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.

✩ 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.2014.07.0320022-1236/© 2014 Elsevier Inc. All rights reserved.

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S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076 2057

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 clas-sical 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 non-squareness. 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 proper-ties:

(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 Tto X, and for each u ∈ XT , we define the modular of u by

2058 S. Shang, Y. Cui / Journal of Functional Analysis 267 (2014) 2056–2076

ρM (u) =∫T

M(∥∥u(t)

∥∥)dt.Put

LM (X) ={u(t) ∈ XT :

∫T

M(∥∥λu(t)

∥∥)dt < ∞ for some λ > 0}.

It is well known that Orlicz–Bochner spaces LM (X) are Banach spaces when they are equipped with the Luxemburg norm

‖u‖ = inf{λ > 0 : ρM

(u

λ

)≤ 1

}

or with the Orlicz norm

‖u‖0 = infk>0

1k

[1 + ρM (ku)

].

In particular, LM (R) and L0M (R) are said to be Orlicz function spaces. It is well known

that ‖u‖ ≤ ‖u‖0 ≤ 2‖u‖. Set

K(u) ={k > 0 : 1

k

(1 + ρM (ku)

)= ‖u‖0

}.

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) = supu≥0

{u ≥ 0 : p(u) ≤ v

}

Then we call N(v) =∫ |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.


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 | vu | ≤ 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

min{‖x1 + x2‖, ‖x1 − x2‖

}≤

(‖x1‖ + ‖x2‖

)(1 − δmin{‖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

[1k1

M(‖k1x1‖

)+ 1

k2M

(‖k2x2‖

)]

≥ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)+ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

)

holds true for all x1, x2 ∈ X, k1, k2 ∈ R+ satisfying max{‖k1x1‖, ‖k2x2‖} ≥ w, m ≤min{‖k1x1‖, ‖k2x2‖}, max{k1, k2} ≤ l.


Proof. For any l ≥ m > 0 and w > 0, take x1, x2 ∈ X\{0} and k1, k2 ∈ R+ such that max{‖k1x1‖, ‖k2x2‖} ≥ w, m ≤ min{‖k1x1‖, ‖k2x2‖}, max{k1, k2} ≤ l. Moreover, we have

0 <k1

k1 + k2= 1

1 + k2k2

≤ 11 + m

l

= l

l + m< 1.

Similarly, we have 1 < k2/(k1 + k2) ≤ l/(l+m) < 1. Therefore, by Lemma 1, there exist numbers a = a(w) ∈ (0, 1) and γ = γ(a(w), l

l+m ) = γ(w, m, l) ∈ (0, 1) such that

M

(k2

k1 + k2u + k1

k1 + k2v

)≤ k2

k1 + k2(1 − γ)M(u) + k1

k1 + k2M(v), (1)

for all u ≥ w and v satisfying | vu | ≤ a. Moreover, we may assume without loss of generality that ‖k1x1‖ = max{‖k1x1‖, ‖k2x2‖}. For the clarity, we will divide the proof into two parts.

I. Suppose that ‖k2x2‖/‖k1x1‖ < a. Noticing that max{‖k1x1‖, ‖k2x2‖} ≥ w, m ≤min{‖k1x1‖, ‖k2x2‖}, max{k1, k2} ≤ l, by inequality (1), we have

k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

)

≤ k1 + k2

k1k2M

(k2

k1 + k2‖k1x1‖ + k1

k1 + k2‖k2x2‖

)

≤ k1 + k2

k1k2

[k2

k1 + k2(1 − γ)M

(‖k1x1‖

)+ k1

k1 + k2M

(‖k2x2‖

)]

≤ 1k1

(1 − γ)M(‖k1x1‖

)+ 1

k2M

(‖k2x2‖

)= 1

k1M

(‖k1x1‖

)+ 1

k2M

(‖k2x2‖

)− 1

k1γM

(‖k1x1‖

). (2)

Similarly, we have

k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)

≤ 1k1

M(‖k1x1‖

)+ 1

k2M

(‖k2x2‖

)− 1

k1γM

(‖k1x1‖

). (3)

Noticing that max{‖k1x1‖, ‖k2x2‖} ≥ w, max{k1, k2} ≤ l and m ≤ min{‖k1x1‖, ‖k2x2‖}, we have

m

lk1<

1k1

,m

lk2= 1

l· mk2

<1l<

1k1

. (4)

Therefore, by inequality (2)–(4), we obtain


k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

)+ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)

≤ 2k1

M(‖k1x1‖

)+ 2

k2M

(‖k2x2‖

)− 1

k1γM

(‖k1x1‖

)− 1

k1γM

(‖k1x1‖

)≤ 2

k1M

(‖k1x1‖

)+ 2

k2M

(‖k2x2‖

)− m

lk1γM

(‖k1x1‖

)− 1

k1γM

(‖k2x2‖

)≤ 2

k1M

(‖k1x1‖

)+ 2

k2M

(‖k2x2‖

)− m

lk1γM

(‖k1x1‖

)− m

lk2γM

(‖k2x2‖

)

=(

2 − m

lγ

)[1k1

M(‖k1x1‖

)+ 1

k2M

(‖k2x2‖

)]

=(

1 − m

2l γ)[

2k1

M(‖k1x1‖

)+ 2

k2M

(‖k2x2‖

)].

Noticing that

k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

)+ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)> 0,

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

[1k1

M(‖k1x1‖

)+ 1

k2M

(‖k2x2‖

)]

≥ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)+ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

).

II. Suppose that ‖k2x2‖/‖k1x1‖ ≥ a. Then ‖k1x1‖/‖k2x2‖ < 1/a. Moreover, we have

m‖x1‖l‖x2‖

≤ ‖k1x1‖‖k2x2‖

<1a

⇒ ‖x1‖‖x2‖

<l

ma⇒ max{‖x1‖, ‖x2‖}

min{‖x1‖, ‖x2‖}<

l

ma.

Hence

min{‖x1‖, ‖x2‖}‖x1‖ + ‖x2‖

= 11 + max{‖x1‖,‖x2‖}

min{‖x1‖,‖x2‖}≥ 1

1 + lam

= am

l + am. (5)

By Lemma 3, we may assume without loss of generality that

‖x1 − x2‖ ≤(‖x1‖ + ‖x2‖

)(1 − δmin{‖x1‖, ‖x2‖}

‖x1‖ + ‖x2‖

).

Therefore, using Lemma 3 and inequality (5), we get


k1k2

k1 + k2‖x1 − x2‖ ≤ k1k2

k1 + k2

(‖x1‖ + ‖x2‖

)(1 − δmin{‖x1‖, ‖x2‖}

‖x1‖ + ‖x2‖

)

≤(

1 − δam

l + am

)k1k2

k1 + k2

(‖x1‖ + ‖x2‖

)

≤(

1 − δam

l + am

)(k2

k1 + k2‖k1x1‖ + k1

k1 + k2‖k2x2‖

).

Let η = (δam)/(l + am). By the convexity of M , we have

M

(k1k2

k1 + k2‖x1 − x2‖

)≤ M

((1 − η)

(k2

k1 + k2‖k1x1‖ + k1

k1 + k2‖k2x2‖

))

≤ (1 − η)M(

k2

k1 + k2‖k1x1‖ + k1

k1 + k2‖k2x2‖

)

≤ (1 − η)[

k2

k1 + k2M

(‖k1x1‖

)+ k1

k1 + k2M

(‖k2x2‖

)]. (6)

Therefore, by inequality (7), we obtain

k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

)≤ (1 − η)

(M(‖k1x1‖)

k1+ M(‖k2x2‖)

k2

). (7)

Since M ∈ Δ2, then there exists kl = kl(a, w) = kl(w) > 1 such that M((1/a)u) ≤klM(u), u ≥ aw. Let βa = 1/kl, v = u/a. Then M(av) ≥ βaM(v), v ≥ w. By max{‖k1x1‖, ‖k2x2‖} ≥ w and ‖k1x1‖ = max{‖k1x1‖, ‖k2x2‖}, we have ‖k1x1‖ ≥ w. Hence

M(a‖k1x1‖

)≥ βaM

(‖k1x1‖

). (8)

Moreover, by the convexity of M , we have

k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)≤ M(‖k1x1‖)

k1+ M(‖k2x2‖)

k2. (9)

Therefore, by inequalities (7)–(9) and ‖k2x2‖/‖k1x1‖ ≥ a, we obtain

k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

)+ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)

≤ 2k1

M(k1‖x1‖

)+ 2

k2M

(k2‖x2‖

)− η

(M(k1‖x1‖)

k1+ M(k2‖x2‖)

k2

)

≤ 2k1

M(k1‖x1‖

)+ 2

k2M

(k2‖x2‖

)− η

(1lM

(k1‖x1‖

)+ 1

lM

(k2‖x2‖

))

≤ 2M

(k1‖x1‖

)+ 2

M(k2‖x2‖

)− η

(1M

(ak1‖x1‖

)+ 1

M(ak1‖x1‖

))
k1 k2 l l


≤ 2k1

M(k1‖x1‖

)+ 2

k2M

(k2‖x2‖

)− ηβα

(1lM

(k1‖x1‖

)+ 1

lM

(k1‖x1‖

))

≤ 2k1

M(k1‖x1‖

)+ 2

k2M

(k2‖x2‖

)− ηβαm

2l2k1

M(k1‖x1‖

)− ηβαm

2l2k2

M(k2‖x2‖

)

=(

1 − ηβαm

2l

)(2k1

M(k1‖x1‖

)+ 2

k2M

(k2‖x2‖

)).

Noticing that

k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

)+ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)> 0,

we obtain that 1 − (ηβαm)/(2l) ∈ (0, 1). Finally, combining the considerations from Parts I and II and denoting

r = max{r1, 1 − ηβαm

2l

},

we get the inequality

r

[1k1

M(‖k1x1‖

)+ 1

k2M

(‖k2x2‖

)]

≥ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 + x2‖

)+ k1 + k2

k1k2M

(k1k2

k1 + k2‖x1 − x2‖

).

This completes the proof. �Proof of Theorem 1. Necessity. Let us assume that L0

M (X) is uniformly nonsquare. Since L0M (R) is a subspace of L0

M (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‖ → 2and ‖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, ...

It is easy to see that {un}∞n=1 ⊂ L0M (X), {vn}∞n=1 ⊂ L0

M (X) and u ∈ L0M (X). Moreover,

we have ‖u‖0 = ‖un‖0 = ‖vn‖0, n = 1, 2, .... We may assume without loss of generality that {un}∞n=1 ⊂ S(L0

M (X)), {vn}∞n=1 ⊂ S(L0M (X)) and u ∈ S(L0

M (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 → l0as n → ∞. Let hn = (knln)/(kn + ln) and h0 = (k0l0)/(k0 + l0). By the Fatou Lemma, we have


lim infn→∞

1hn

[1 + ρM

(kn(un + vn)

)]≥ 1

h0

[1 +

∫T

limn→∞

M(hn

∥∥un(t) + vn(t)∥∥)dt]

= 1h0

[1 +

∫T

M(h0 · 2d)dt]

= 1h0

[1 +

∫T

M(h0

∥∥2u(t)∥∥)dt]

≥ 2‖u‖0 = 2.


kn + lnknln

[1 + ρM

(knln

kn + ln(un + vn)

)]≤ 1

kn

[1 + ρM (knun)

]+ 1

ln

[1 + ρM (lnvn)

]= ‖un‖0 + ‖vn‖0 = 2.

This means that

lim supn→∞

1hn

[1 + ρM

(hn(un + vn)

)]≤ 2.

Hence we have

limn→∞

1hn

[1 + ρM

(hn(un + vn)

)]= lim

n→∞‖un + vn‖0 = 2.

Similarly, we have ‖un − vn‖0 → 2 as n → ∞, a contradiction. Hence we obtain that Xis a uniformly nonsquare space.

Sufficiency. Let l = sup{K(u) : u ∈ S(L0M (X))}, 1 + δ = inf{K(u) : u ∈ S(L0

M (X))}. Since M ∈ Δ2 and N ∈ Δ2, by Lemma 2, we have l < ∞ and δ > 0. Take u1, u2 ∈S(L0

M (X)), k1 ∈ K(u1), k2 ∈ K(u2). Let w = 1lM

−1( 14·μT ), E1 = {t ∈ T : ‖k1u1(t)‖ <

w} and E2 = {t ∈ T : ‖k2u2(t)‖ < w}. Then

1kiρM

[(kiui)χEi

]≤ 1

ki

∫Ei

M

(ki

1lM−1

(δ

4 · μT

))dt ≤ 1

ki

∫Ei

M

(M−1

(δ

4 · μT

))dt

≤ 1ki

∫Ei

δ

4 · μT dt ≤ 1ki

· δ

4 · μT · μT ≤ δ

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

r

[1k1

M(‖k1x1‖

)+ 1

k2M

(‖k2x2‖

)]

≥ k1 + k2M

(k1k2 ‖x1 + x2‖

)+ k1 + k2

M

(k1k2 ‖x1 − x2‖

)(11)

k1k2 k1 + k2 k1k2 k1 + k2


holds true for all x1, x2 ∈ X satisfying max{‖k1x1‖, ‖k2x2‖} > w. Let G = E1 ∩E2. By inequality (10), it is easy to see that

1kiρM

[(kiui)χG

]= 1

ki

∫G

M(ki∥∥kiui(t)

∥∥)dt < δ

4(1 + δ) , (12)

whenever i ∈ {1, 2}. Noticing that E1 = {t ∈ T : ‖k1u1(t)‖ < w}, E2 = {t ∈ T :‖k2u2(t)‖ < w} and G = E1 ∩ E2, we have max{‖k1u1(t)‖, ‖k2u2(t)‖} ≥ w for all t ∈ T \G. Therefore, by inequality (11), we have

k1 + k2

k1k2M

(k1k2

k1 + k2

∥∥u1(t) − u2(t)∥∥) + k1 + k2

k1k2M

(k1k2

k1 + k2

∥∥u1(t) + u2(t)∥∥)

≤ r

[1k1

M(∥∥k1u1(t)

∥∥) + 1k12

M(∥∥k2u2(t)

∥∥)],where t ∈ T \G. This implies that

k1 + k2

k1k2

[ ∫T\G

M

(k1k2

k1 + k2

∥∥u1(t) − u2(t)∥∥)dt +

∫T\G

M

(k1k2

k1 + k2

∥∥u1(t) + u2(t)∥∥)dt]

=∫

T\G

k1 + k2

k1k2

[M

(k1k2

k1 + k2

∥∥u1(t) − u2(t)∥∥) + M

(k1k2

k1 + k2

∥∥u1(t) + u2(t)∥∥)]dt

≤∫

T\G

r

[2k1

M(∥∥k1u1(t)

∥∥) + 2k2

M(∥∥k2u2(t)

∥∥)]dt

≤ 2r[

1k1

∫T\G

M(∥∥k1u1(t)

∥∥)dt + 1k2

∫T\G

M(∥∥k2u2(t)

∥∥)dt]. (13)

By inequality (13), we have

k1 + k2

k1k2ρM

(k1 + k2

k1k2(u1 − u2)χT\G

)+ k1 + k2

k1k2ρM

(k1 + k2

k1k2(u1 + u2)χT\G

)

≤ 2r(

1k1

ρM[(k1u1)χT\G

]+ 1

k2ρM

[(k2u2)χT\G

]). (14)

Therefore, by inequalities (12)–(14) and 1 + δ = inf{K(u) : u ∈ S(L0M (X))}, we obtain

k1 + k2

k1k2

[1 + ρM

(k1 + k2


)]

+ k1 + k2[1 + ρM

(k1 + k2 (u1 + u2)χT\G

)]
k1k2 k1k2


= 2k1 + k2

k1k2+ k1 + k2

k1k2ρM

(k1 + k2


)

+ k1 + k2

k1k2ρM

(k1 + k2

k1k2(u1 + u2)χT\G

)

≤ 2k1 + k2

k1k2+ 2r

(1k1

ρM[(k1u1)χT\G

]+ 1

k2ρM

[(k2u2)χT\G

])

= 2[

1k1

+ 1k2

]+ 2r

(1k1

ρM[(k1u1)χT\G

]+ 1

k2ρM

[(k2u2)χT\G

])

= 2r[

1k1

(1 + ρM

[(k1u1)χT\G

])+ 1

k2

(1 + ρM

[(k2u2)χT\G

])]

+ 2(1 − r)[

1k1

+ 1k2

]

≤ 2r[

1k1

(1 + ρM

[(k1u1)χT\G

])+ 1

k2

(1 + ρM

[(k2u2)χT\G

])]

+ 2(1 − r) 21 + δ

. (15)

Moreover, by the convexity of M and inequality (12), we have

k1 + k2

k1k2ρM

(k1 + k2

k1k2(u1 − u2)χG

)+ k1 + k2

k1k2ρM

(k1 + k2

k1k2(u1 + u2)χG

)

≤ 2k1

ρM[(k1u1)χG

]+ 2

k2ρM

[(k2u2)χG

]≤ 2 · 2δ

4(1 + δ) . (16)

Therefore, by inequalities (15) and (16), we have

‖u1 − u2‖0 + ‖u1 + u2‖0

≤ k1 + k2

k1k2

[1 + ρM

(k1 + k2

k1k2(u1 − u2)

)]+ k1 + k2

k1k2

[1 + ρM

(k1 + k2

k1k2(u1 + u2)

)]

= k1 + k2

k1k2

[1 + ρM

(k1 + k2


)]

+ k1 + k2

k1k2

[1 + ρM

(k1 + k2

k1k2(u1 + u2)χT\G

)]

+ k1 + k2

k1k2

[1 + ρM

(k1 + k2

k1k2(u1 − u2)χG

)]

+ k1 + k2

k1k2

[1 + ρM

(k1 + k2

k1k2(u1 + u2)χG

)]

≤ 2r[

1 (1 + ρM (k1u1χT\G)

)+ 1 (

1 + ρM (k2u2χT\G))]

+ 2(1 − r) 2
k1 k2 1 + δ


+ 2k1

ρM (k1u1χG) + 2k2

ρM (k2u2χG)

= 2r[

1k1

(1 + ρM (k1u1χT\G)

)+ 1

k2

(1 + ρM (k2u2χT\G)

)]+ 2(1 − r) 2

1 + δ

+ 2rk1

ρM (k1u1χG) + 2rk2

ρM (k2u2χG) + 2(1 − r)k1

ρM (k1u1χG)

+ 2(1 − r)k2

ρM (k2u2χG)

= 2r[

1k1

(1 + ρM (k1u1)

)+ 1

k2

(1 + ρM (k2u2)

)]+ 2(1 − r) 2

1 + δ

+ 2(1 − r)k1

ρM (k1u1χG) + 2(1 − r)k2

ρM (k2u2χG)

≤ 2r · 2 + 2(1 − r) 21 + δ

+ 2(1 − r)(

δ

4(1 + δ)

)

= 2r · 2 + 2(1 − r) · 2[1 − 7δ

8(1 + δ)

]

= 2r · 2 + 2(1 − r) · 2 − 4(1 − r) 7δ8(1 + δ)

= 4 − 4(1 − r) 7δ8(1 + δ)

= 2‖u1‖0 + 2‖u2‖0 − 4(1 − r) 7δ8(1 + δ) . (17)

Denoting

δ1 = (1 − r) 7δ8(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. �Corollary 1. The Orlicz function space L0

M (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.


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

infy �=0

{‖x‖ + ‖y‖ − min

{‖x + y‖, ‖x− y‖

}: x ∈ X, r2 ≤ ‖y‖ ≤ r1

}> 0;

(b) If xn → x, then limn→∞ δ(xn) = δ(x), where

δ(x) = infy �=0

{‖x‖ + ‖y‖ − min

{‖x + y‖, ‖x− y‖

}: x ∈ X, r2 ≤ ‖y‖ ≤ r1

}.

Proof of Theorem 2. Sufficiency. Suppose that the assumptions about M and X are satisfied and that there exist u ∈ S(L0

M (X)) and {vn}∞n=1 ⊂ S(L0M (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 :‖lnvn(t)‖ ≥ σ0}. Since ln−1 = ρM (lnvn), we may assume without loss of generality that 2l − 1 ≥ ρM (lnvn) whenever n ∈ N . Put

Hn ={t ∈ T : σ0 ≤

∥∥lnvn(t)∥∥ ≤ M−1

(4

ε0(2l − 1)

)}.

We have

2l − 1 ≥∫T

M(∥∥lnvn(t)

∥∥)dt ≥ ∫Gn\Hn

M(∥∥lnvn(t)

∥∥)dt ≥ 4ε0(2l − 1)μ(Gn\Hn).

This implies that μ(Gn \Hn) ≤ 14ε0. Hence, μHn > 1

2ε0. We define the function

η(t) = infy �=0

{∥∥u(t)∥∥ + ‖y‖ − min

{∥∥u(t) + y∥∥, ∥∥u(t) − y

∥∥} :

σ0 ≤ ‖y‖ ≤ M−1(

4ε0(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) = infy �=0

{∥∥hn(t)∥∥ + ‖y‖ − min

{∥∥hn(t) + y∥∥, ∥∥hn(t) − y

∥∥} :

σ0 ≤ ‖y‖ ≤ M−1(

4)}

ε0(2l − 1)


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

{t ∈ T0 : 1

i + 1 < η(t) ≤ 1i

},

we get that there exists η0 > 0 such that μH < 18ε0, where

H ={t ∈ T : η(t) ≤ 2η0

}.

Let En = Hn\H, E1n = (Hn∩{t ∈ T : ‖u(t)‖ �= 0})\H and E2

n = (Hn∩{t ∈ T : ‖u(t)‖ =0})\H. It is easy to see that En = E1

n ∪E2n, E1

n ∩E2n = ∅ and μEn ≥ 3

8ε0. By Lemma 3, we have

∥∥u(t)∥∥ +

∥∥vn(t)∥∥− min

{∥∥u(t) + vn(t)∥∥, ∥∥u(t) − vn(t)

∥∥} ≥ η(t) ≥ 2η0

whenever t ∈ E1n. We may assume without loss of generality that there exists F 1

n ⊂ E1n

such that

μF 1n ≥ 1

2μE1n,

∥∥u(t)∥∥ +

∥∥vn(t)∥∥−

∥∥u(t) + vn(t)∥∥ ≥ η0 t ∈ F 1

n .

Moreover, we know that for any u ≥ v > 0, λn ∈ (0, 1), we have

λnM(u) −M(λnu) −[λnM(v) −M(λnv)

]

= λn

u∫0

p(t)dt−λnu∫0

p(t)dt−[λn

v∫0

p(t)dt−λnv∫0

p(t)dt]

=[λn

u∫0

p(t)dt− λn

v∫0

p(t)dt]−

[ λnu∫0

p(t)dt−λnv∫0

p(t)dt]

= λn

u∫v

p(t)dt−λnu∫

λnv

p(t)dt

≥λnu+(1−λn)v∫

v

p(t)dt−λnu∫

λnv

p(t)dt

≥ 0.

Noticing that t ∈ E2n, we have ‖lnvn(t)‖ ≥ σ0 > 0. Hence,

kM

(∥∥lnvn(t)∥∥)−M

(k ∥∥lnvn(t)

∥∥) ≥ kM(σ0) −M

(k

σ0

)≥ 0

k + ln k + ln k + ln k + ln


whenever n ∈ N . Let Fn = F 1n ∪ E2

n. Then 8μFn ≥ ε0. Therefore,

‖u‖0 + ‖vn‖0 − ‖u + vn‖0

≥ 1k

[1 + ρM (ku)

]+ 1

ln

[1 + ρM (lnvn)

]− k + ln

kln

[1 + ρM

(kln

k + ln(u + vn)

)]

= k + lnkln

[ln

k + lnρM (ku) + k

k + lnρM (lnvn) − ρM

(kln

k + ln(u + vn)

)]

≥ k + lnkln

[∫F 1

n

lnk + ln

M(∥∥ku(t)

∥∥)dt +∫F 1

n

k

k + lnM

(∥∥lnvn(t)∥∥)dt

−∫F 1

n

M

(kln

k + ln

∥∥u(t) + vn(t)∥∥)dt]

+ k + lnkln

[∫E2

n

lnk + ln

M(∥∥ku(t)

∥∥)dt +∫E2

n

k

k + lnM


−∫E2

n

M

(kln

k + ln

∥∥u(t) + vn(t)∥∥)dt]

≥ k + lnkln

[∫F 1

n

M

(kln

k + ln

∥∥u(t)∥∥ + kln

k + ln

∥∥vn(t)∥∥)dt

−∫F 1

n

M

(kln

k + ln

∥∥u(t) + vn(t)∥∥)dt]

+ k + lnkln

[∫E2

n

lnk + ln

M(∥∥ku(t)

∥∥)dt +∫E2

n

k

k + lnM


−∫E2

n

M

(kln

k + ln

∥∥u(t) + vn(t)∥∥)dt]

≥ k + lnkln

[∫F 1

n

M

(kln

k + ln

∥∥u(t) + vn(t)∥∥ + kln

k + lnη0

)dt

−∫F 1

n

M

(kln

k + ln

∥∥u(t) + vn(t)∥∥)dt]

+ k + lnkln

[∫E2

n

klnk + ln

M(∥∥vn(t)

∥∥)dt− ∫E2

n

M

(kln

k + ln

∥∥vn(t)∥∥)dt]

≥ k + lnkln

[∫1

M

(kln

k + lnη0

)dt +

∫2

klnk + ln

M(∥∥vn(t)

∥∥)dt
Fn En


−∫E2

n

M

(kln

k + ln

∥∥vn(t)∥∥)dt]

= k + lnkln

∫F 1

n

M

(kln

k + lnη0

)dt +

∫E2

n

[k

k + lnM

(∥∥lnvn(t)∥∥)

−M

(k

k + ln

∥∥lnvn(t)∥∥)]dt

≥ k + lnkln

∫F 1

n

M

(kln

k + lnη0

)dt +

∫E2

n

[k

k + lnM(σ0) −M

(k

k + lnσ0

)]dt. (18)

In consequence, by inequality (18) and Fatou Lemma, we have

0 = lim infn→∞

(‖u‖0 + ‖vn‖0 − ‖u + vn‖0)

≥ lim infn→∞

[k + lnkln

∫F 1

n

M

(kln

k + lnη0

)dt +

∫E2

n

[k

k + lnM(σ0) −M

(k

k + lnσ0

)]dt

]

≥ k + l

kl

∫F 1

n

M

(kl

k + lη0

)dt +

∫E2

n

[k

k + lM(σ0) −M

(k

k + lσ0

)]dt

≥ k + l

kl

∫F 1

n∪E2n

min{M

(kl

k + lη0

),

k

k + lM(σ0) −M

(k

k + lσ0

)}dt

≥ k + l

klmin

{M

(kl

k + lη0

),

k

k + lM(σ0) −M

(k

k + lσ0

)}· 18ε0 > 0,

a contradiction.Case 2. For any ε > 0, σ > 0, there exists a natural number N such that μ{t ∈ T :

‖lnvn(t)‖ ≥ σ} < ε whenever n > N . By the Risez theorem, there exists a subsequence {n} of {n} such that lnvn(t) → 0 μ-a.e. on T . Using the inclusion

T0 ={t ∈ T :

∥∥ku(t)∥∥ �= 0

}⊃

∞⋃n=1

{t ∈ T : 1

n + 1 <∥∥ku(t)

∥∥ ≤ 1n

},

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

lM(d) + k

M(c) −M

(l

d + kc

)> 0

k + l k + l k + l k + l


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 ‖lnvn(t)‖ < h on F , where 8μ(T\F ) < μT1. Moreover, we know that if u1 ≥ u2 ≥ v2 ≥ v1 > 0, then

λnM(u1) + (1 − λn)M(v1) −M(λnu1 + (1 − λn)v1

)≥ λnM(u2) + (1 − λn)M(v2) −M

(λnu3 + (1 − λn)v3

),

where λn ∈ (0, 1). In fact, we have


)−

[λnM(u1) + (1 − λn)M(v2) −M

(λnu1 + (1 − λn)v2

)]= (1 − λn)M(v1) −M

(λnu1 + (1 − λn)v1

)−

[(1 − λn)M(v2) −M

(λnu1 + (1 − λn)v2

)]

= (1 − λn)v1∫0

p(t)dt−λnu1+(1−λn)v1∫

0

p(t)dt

− (1 − λn)v2∫0

p(t)dt +λnu1+(1−λn)v2∫

0

p(t)dt

= −(1 − λn)v2∫

v1


λnu1+(1−λn)v1

p(t)dt

≥λnu1+(1−λn)v2∫

λnu1+(1−λn)v1

p(t)dt−v2∫

(1−λn)v1+λnv2

p(t)dt ≥ 0. (19)

Moreover, we have

λnM(u1) −M(λnu1 + (1 − λn)v2

)−

[λnM(u2) −M

(λnu2 + (1 − λn)v2

)]

= λn

u1∫0


0

p(t)dt− λn

u2∫0


0

p(t)dt

= λn

u1∫u2


λnu2+(1−λn)v2

p(t)dt

≥λnu1+(1−λn)u2∫

u


p(t)dt ≥ 0. (20)
2 λnu2+(1−λn)v2


Therefore, by inequality (19) and (20), we have


)≥ λnM(u2) + (1 − λn)M(v2) −M

(λnu3 + (1 − λn)v3

),

where λn ∈ (0, 1). This means that if t ∈ T2 = (T0\T1)\(T\F ), then

λnM(∥∥ku(t)

∥∥) + (1 − λn)M(∥∥lnvn(t)

∥∥)−M(λn

∥∥ku(t)∥∥ + (1 − λn)

∥∥lnvn(t)∥∥)

≥ λnM(d) + (1 − λn)M(h) −M(λnd + (1 − λn)h

).

It is easy to see that 4μT2 ≥ μT0. Therefore,

‖u‖0 + ‖vn‖0 − ‖u + v‖0

≥ 1k

[1 + ρM (ku)

]+ 1

ln

[1 + ρM (lnvn)

]− k + ln

kln

[1 + ρM

(kln

k + ln(u + vn)

)]

= k + lnkln

[ln

k + lnρM (ku) + k

k + lnρM (lnvn) − ρM

(kln

k + ln(u + vn)

)]

= k + lnkln

[∫T

[ln

k + lnM

(∥∥ku(t)∥∥) + k

k + lnM

(∥∥lnvn(t)∥∥)

−M(λn

∥∥ku(t)∥∥ + (1 − λn)

∥∥lnvn(t)∥∥)]dt]

≥ k + lnkln

[∫T2

[ln

k + lnM

(∥∥ku(t)∥∥) + k

k + lnM

(∥∥lnvn(t)∥∥)

−M(λn

∥∥ku(t)∥∥ + (1 − λn)

∥∥lnvn(t)∥∥)]dt]

≥ k + lnkln

[∫T

[ln

k + lnM(d) + k

k + lnM(h) −M

(ln

k + lnd + k

k + lnh

)]dt

]. (21)

Therefore, by inequality (21) and Fatou Lemma, we have

0 = lim infn→∞

(‖u‖0 + ‖vn‖0 − ‖u + vn‖0)

≥ lim infn→∞

k + lnkln

[∫T

[ln

k + lnM(d) + k

k + lnM(h) −M

(ln

k + lnd + k

k + lnh

)]dt

]

≥ k + l

kl

[∫ [l

k + lM(d) + k

k + lM(h) −M

(l

k + ld + k

k + lh

)]dt

]
T


≥ k + l

kl

[l

k + lM(d) k

k + lM(h) −M

(l

k + ld + k

k + lh

)]· 14μT0

> 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 L0

M(R) is embedded into L0

M (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(L0

M (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 = (k0ln)/(k0 + ln) and h0 = (k0l0)/(k0 + l0). By the Fatou Lemma, we have

lim infn→∞

1hn

[1 + ρM

(kn(un + vn)

)]≥ 1

h0

[1 +

∫T

limn→∞

M(hn

∥∥un(t) + vn(t)∥∥)dt]

= 1h0

[1 +

∫T

M(h0 · 2d)dt]

= 1h0

[1 +

∫T

M(h0

∥∥2u(t)∥∥)dt]

≥ 2‖u‖0 = 2.


k0 + lnk0ln

[1 + ρM

(k0ln

k0 + ln(u + vn)

)]≤ 1

k0

[1 + ρM (k0u)

]+ 1

ln

[1 + ρM (lnvn)

]= ‖u‖0 + ‖vn‖0 = 2.

This means that

lim supn→∞

1hn

[1 + ρM

(hn(u + vn)

)]≤ 2.

Hence we have

lim 1 [1 + ρM

(hn(u + vn)

)]= lim ‖u + vn‖0 = 2.

n→∞ hn n→∞


Similarly, we have ‖u− vn‖0 → 2 as n → ∞, a contradiction. Hence we obtain that Xis a locally uniformly nonsquare space. This completes the proof. �Corollary 2. The Orlicz function space L0

M (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

‖Tx− Ty‖ ≤ ‖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 L0

M (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 L0

M (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 L0

M (X)have the fixed point property. �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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