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Abstract and Applied Analysis
Volume 2011, Article ID 361525, 16 pages
http://dx.doi.org/10.1155/2011/361525
Research Article

Nonsquareness in Musielak-Orlicz-Bochner Function Spaces

1Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
2Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China

Received 20 October 2010; Revised 3 January 2011; Accepted 14 February 2011

Academic Editor: Paul Eloe

Copyright © 2011 Shaoqiang Shang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The criteria for nonsquareness in the classical Orlicz function spaces have been given already. However, because of the complication of Musielak-Orlicz-Bochner function spaces, at present the criteria for nonsquareness have not been discussed yet. In the paper, the criteria for nonsquareness of Musielak-Orlicz-Bochner function spaces are given. As a corollary, the criteria for nonsquareness of Musielak-Orlicz function spaces are given.

1. Introduction

A lot of nonsquareness concepts in Banach spaces are known. Nonsquareness are the important notion in geometry of Banach space. One of reasons is that the property is strongly related to the fixed point property (see [1]). The criteria for nonsquareness in the classical Orlicz function spaces have been given in [2] already. However, because of the complication of Musielak-Orlicz-Bochner function spaces, at present the criteria for nonsquareness have not been discussed yet. The aim of this paper is to give criteria nonsquareness of Musielak-Orlicz-Bochner function spaces. As a corollary, the criteria for nonsquareness of Musielak-Orlicz function spaces are given. The topic of this paper is related to the topic of [38].

Let (𝑋,) be a real Banach space. 𝑆(𝑋) and 𝐵(𝑋) denote the unit sphere and unit ball, respectively. By 𝑋, denote the dual space of 𝑋. Let 𝑁,𝑅, and 𝑅+ denote the set natural number, reals, and nonnegative reals, respectively. Let us recall some geometrical notions concerning nonsquareness. A Banach space 𝑋 is said to be nonsquare space if for any 𝑥,𝑦𝑆(𝑋) we have min{(1/2)(𝑥+𝑦),(1/2)(𝑥𝑦)}<1. A Banach space 𝑋 is said to be uniformly nonsquare space if for any 𝑥,𝑦𝑆(𝑋), there exists 𝛿>0 such that min{(1/2)(𝑥+𝑦),(1/2)(𝑥𝑦)}<1𝛿.

Let (𝑇,,𝜇) be nonatomic measure space. Suppose that a function 𝑀𝑇×[0,)[0,] satisfies the following conditions:(1)for 𝜇-a.e, 𝑡𝑇, 𝑀(𝑡,0)=0, lim𝑢𝑀(𝑡,𝑢)= and 𝑀(𝑡,𝑢)< for some 𝑢>0,(2)for  𝜇-a.e, 𝑡𝑇, 𝑀(𝑡,𝑢) is convex on [0,) with respect to 𝑢,(3)for each 𝑢[0,), 𝑀(𝑡,𝑢) is a 𝜇-measurable function of 𝑡 on 𝑇.

Let 𝑒(𝑡)=sup{𝑢>0𝑀(𝑡,𝑢)=0}. It is well known that 𝑒(𝑡) is 𝜇-measurable (see [2]).

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 𝑢𝑋𝑇, define the modular of 𝑢 by 𝜌𝑀(𝑢)=𝑇𝑀(𝑡,𝑢(𝑡))𝑑𝑡.(1.1) Put𝐿𝑀(𝑋)=𝑢𝑋𝑇𝜌𝑀(𝜆𝑢)<forsome𝜆>0.(1.2) It is well known that Musielak-Orlicz-Bochner function space 𝐿𝑀(𝑋) is Banach spaces equipped with the Luxemburg norm𝑢=inf𝜆>0𝜌𝑀𝑢𝜆1(1.3) or Orlicz’s norm𝑢0=inf𝑘>01𝑘1+𝜌𝑀.(𝑘𝑢)(1.4) In particular, 𝐿𝑀(𝑅) and 𝐿0𝑀(𝑅) are said to be Musielak-Orlicz function space. Set1supp𝑢={𝑡𝑇𝑢(𝑡)0},𝐾(𝑢)=𝑘>0𝑘1+𝜌𝑀=(𝑘𝑢)𝑢0.(1.5) In particular, the set 𝐾(𝑢) can be nonempty. To show that, we give a proposition.

Proposition 1.1 (see [9]). If lim𝑢(𝑀(𝑡,𝑢)/𝑢)=  μ-a.e.   𝑡𝑇, then 𝐾(𝑢)𝜙 for any 𝑢𝐿0𝑀(𝑋).

We define a function1𝜎(𝑡)=sup𝑢0𝑀𝑡,2𝑢=12.𝑀(𝑡,𝑢)(1.6) Function 𝜎(𝑡) will be used in the further part of the paper. Moreover, 𝜎(𝑡) is 𝜇-measurable. To show that, we give a proposition.

Proposition 1.2. Function 𝜎(𝑡) is 𝜇-measurable.

Proof. Pick a dense set {𝑟𝑖}𝑖=1 in [0,) and set 𝐵𝑘=1𝑡𝑇𝑀𝑡,2𝑟𝑘=12𝑀𝑡,𝑟𝑘,𝑞𝑘(𝑡)=𝑟𝑘𝜒𝐵𝑘(𝑡)(𝑘𝑁).(1.7) It is easy to see that for all 𝑘𝑁, 𝜎(𝑡)𝑞𝑘(𝑡)  μ-a.e on 𝑇. Hence, sup𝑘1𝑞𝑘(𝑡)𝜎(𝑡). For 𝜇-a.e 𝑡𝑇, arbitrarily choose 𝜀(0,𝜎(𝑡)). Then, there exists 𝑟𝑘(𝜎(𝑡)𝜀,𝜎(𝑡)) such that 𝑀(𝑡,(1/2)𝑟𝑘)=(1/2)𝑀(𝑡,𝑟𝑘), that is, 𝑞𝑘(𝑡)𝑟𝑘>𝜎(𝑡)𝜀. Since 𝜀 is arbitrary, we find sup𝑘1𝑞𝑘(𝑡)𝜎(𝑡). Thus, sup𝑘1𝑞𝑘(𝑡)=𝜎(𝑡).

It is easy to prove the following proposition.

Proposition 1.3. For any 𝛼(0,1), if 𝑢(𝑡)𝜎(𝑡), then 𝑀(𝑡,𝛼𝑢(𝑡))=𝛼𝑀(𝑡,𝑢(𝑡)).

Proposition 1.4. For any 𝛼(0,1), if 𝑢(𝑡)<𝜎(𝑡)<𝑣(𝑡), then 𝑀(𝑡,𝛼𝑢(𝑡)+(1𝛼)𝑣(𝑡))<𝛼𝑀(𝑡,𝑢(𝑡))+(1𝛼)𝑀(𝑡,𝑣(𝑡)).

Proposition 1.5. For any 𝛼(0,1), if 𝜎(𝑡)<𝑣(𝑡), then 𝑀(𝑡,𝛼𝑣(𝑡))<𝛼𝑀(𝑡,𝑣(𝑡)).

Definition 1.6 (see [2]). We say that 𝑀(𝑡,𝑢) satisfies condition Δ(𝑀Δ) if there exist 𝐾1 and a measureable nonnegative function 𝛿(𝑡) on 𝑇 such that 𝑇𝑀(𝑡,𝛿(𝑡))𝑑𝑡< and 𝑀(𝑡,2𝑢)𝐾𝑀(𝑡,𝑢) for almost all 𝑡𝑇 and all 𝑢𝛿(𝑡).

First, we give some results that will used in the further part of the paper.

Lemma 1.7 (see [2]). Suppose 𝑀Δ. Then 𝜌𝑀(𝑢)=1𝑢=1.

Lemma 1.8 (see [9]). Let 𝐿0𝑀(𝑋) be Musielak-Orlicz-Bochner function spaces, then, if 𝐾(𝑢)=𝜙, one has 𝑢0=𝑇𝐴(𝑡)𝑢(𝑡)𝑑𝑡, where 𝐴(𝑡)=lim𝑢(𝑀(𝑡,𝑢)/𝑢).

2. Main Results

Theorem 2.1. 𝐿𝑀(𝑋) is nonsquare space if and only if (a)𝑀Δ,(b)for any 𝑢,𝑣𝑆(𝐿𝑀(𝑋)), one has 𝜇{𝑡supp𝑢supp𝑣𝑢(𝑡)+𝑣(𝑡)>2𝑒(𝑡)}>0 or 𝜇({𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}{𝑡𝑇𝑣(𝑡)>𝜎(𝑡)})>0,(c)𝑋 is nonsquare space.

In order to prove the theorem, we give a lemma.

Lemma 2.2. Let 𝑋 be nonsquare space, then for any 𝑥,𝑦0, one has 𝑥+𝑦min{𝑥+𝑦,𝑥𝑦}>0.(2.1)

Proof. For any 𝑥,𝑦0, without loss of generality, we may assume 𝑥𝑦. Since 𝑋 is nonsquare space, we have 𝑥+𝑥=𝑥+𝑥𝑦𝑦>𝑚𝑖𝑛𝑥+𝑥𝑦,𝑦𝑥𝑥𝑦𝑦.(2.2) Therefore, by (2.2), we obtain 𝑥+𝑦𝑥+𝑥+𝑦𝑦1𝑥𝑦𝑦<𝑥+𝑥+𝑦𝑥=𝑥+𝑦(2.3) or 𝑥𝑦𝑥𝑥+𝑦𝑦1𝑥𝑦𝑦<𝑥+𝑥+𝑦𝑥=𝑥+𝑦.(2.4) This implies 𝑥+𝑦min{𝑥+𝑦,𝑥𝑦}>0. This completes the proof.

Proof of Theorem 2.1. Necessity. (a) If 𝐿𝑀(𝑋) is nonsquare space, then 𝐿𝑀(𝑅) is nonsquare space, because 𝐿𝑀(𝑅) is isometrically embedded into 𝐿𝑀(𝑋). Since 𝐿𝑀(𝑅) is nonsquare space, then 𝑀Δ which follows from the theorem proved in more general case (see [10, 11]). Namely, if 𝑀Δ, then 𝐿𝑀(𝑅) contains isometric copy of 𝑙.
If (b) is not true, then there exist 𝑢,𝑣𝑆(𝐿𝑀(𝑋)) such that 𝜇{𝑡supp𝑢supp𝑣𝑢(𝑡)+𝑣(𝑡)>2𝑒(𝑡)}=0 and 𝜇({𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}{𝑡𝑇𝑣(𝑡))>𝜎(𝑡)})=0. Let 𝐺=supp𝑢supp𝑣. We have 12𝜌𝑀1(𝑢)+2𝜌𝑀=1(𝑣)2𝑇1𝑀(𝑡,𝑢(𝑡))𝑑𝑡+2𝑇=𝑀(𝑡,𝑣(𝑡))𝑑𝑡𝑇121𝑀(𝑡,𝑢(𝑡))+2𝑀(𝑡,𝑣(𝑡))𝑑𝑡𝐺𝑀1𝑡,2(1𝑢𝑡)+2(𝑣𝑡)𝑑𝑡+𝑇𝐺𝑀1𝑡,2(1𝑢𝑡)+2(𝑣𝑡)𝑑𝑡𝐺𝑀1𝑡,2𝑢(𝑡)+𝑣(𝑡)𝑑𝑡+𝑇𝐺𝑀1𝑡,2𝑢(𝑡)+𝑣(𝑡)𝑑𝑡=𝜌𝑀𝑢+𝑣2.(2.5) By 𝜇{𝑡supp𝑢supp𝑣𝑢(𝑡)+𝑣(𝑡)>2𝑒(𝑡)}=0 and 𝜇({𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}{𝑡𝑇𝑣(𝑡))>𝜎(𝑡)})=0, we obtain that two inequalities of (2.5) are equations. This implies (1/2)𝜌𝑀(𝑢)+(1/2)𝜌𝑀(𝑣)=𝜌𝑀((𝑢+𝑣)/2). By Lemma 1.7, we have 𝜌𝑀((𝑢+𝑣)/2)=1. Thus, (1/2)(𝑢+𝑣)=1. Similarly, we have (1/2)(𝑢𝑣)=1, a contradiction!
(c) Pick (𝑡)𝑆(𝐿𝑀(𝑋)), then there exists 𝑑>0 such that 𝜇𝐸>0, where 𝐸={𝑡𝑇(𝑡)𝑑}. Put 1(𝑡)=𝑑𝑥0𝜒𝐸(𝑡), where 𝑥0𝑆(𝑋). It is easy to see that 1(𝑡)𝐿𝑀(𝑋){0}. Hence, there exists 𝑘>0 such that 𝑘1(𝑡)𝑆(𝐿𝑀(𝑋)). By Lemma 1.7, we have 1=𝑇𝑀𝑡,𝑘1(𝑡)𝑑𝑡=𝐸𝑀𝑡,𝑘𝑑𝑥0𝑑𝑡.(2.6) Let 𝛼=𝑘𝑑. Then, 𝐸𝑀(𝑡,𝛼)𝑑𝑡=1. The necessity of (c) follows from the fact that 𝑋 is isometrically embedded into 𝐿𝑀(𝑋). Namely, defining the operator 𝐼𝑋𝐿𝑀(𝑋) by 𝐼(𝑥)=𝛼𝑥𝜒𝐸(𝑡),𝑥𝑋.(2.7) Hence, for any 𝑥𝑋{0}, we have 𝜌𝑀𝐼(𝑥)=𝑥𝐸𝑀𝑡,𝐼(𝑥)𝑥𝑑𝑡=𝐸𝑀𝑡,𝛼𝑥𝑥𝑑𝑡=𝐸𝑀(𝑡,𝛼)𝑑𝑡=1.(2.8) By Lemma 1.7, we have 𝐼(𝑥)/𝑥𝐿𝑀(𝑋)=1, hence 𝐼(𝑥)𝐿𝑀(𝑋)=𝑥.
Sufficiency. The proof requires the consideration of two cases separately.
Case 1. 𝜇({𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}{𝑡𝑇𝑣(𝑡)>𝜎(𝑡)})>0. Without loss of generality, we may assume 𝜇{𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}>0. Let 𝐹={𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}. Put 𝐹1𝐹={𝑡𝐹𝑢(𝑡)+𝑣(𝑡)>𝑢(𝑡)+𝑣(𝑡)},2=𝐹{𝑡𝐹𝑢(𝑡)+𝑣(𝑡)>𝑢(𝑡)𝑣(𝑡)},3={𝑡𝐹𝑣(𝑡)=0}.(2.9) Since 𝑋 is nonsquare space, we have 𝜇(𝐹1𝐹3)>0 or 𝜇(𝐹2𝐹3)>0 by Lemma 2.2. Without loss of generality, we may assume 𝜇(𝐹1𝐹3)>0. Moreover, we have 12𝜌𝑀1(𝑢)+2𝜌𝑀(𝑣)𝜌𝑀12=1(𝑢+𝑣)2𝑇1𝑀(𝑡,𝑢(𝑡))𝑑𝑡+2𝑇𝑀(𝑡,𝑣(𝑡))𝑑𝑡𝑇𝑀1𝑡,2=𝑢(𝑡)+𝑣(𝑡)𝑑𝑡𝑇121𝑀(𝑡,𝑢(𝑡))+21𝑀(𝑡,𝑣(𝑡))𝑀𝑡,2𝑢(𝑡)+𝑣(𝑡)𝑑𝑡𝐹1𝐹3121𝑀(𝑡,𝑢(𝑡))+21𝑀(𝑡,𝑣(𝑡))𝑀𝑡,2(𝑢𝑡)+𝑣(𝑡)𝑑𝑡𝐹1𝐹3𝑀1𝑡,21𝑢(𝑡)+21𝑣(𝑡)𝑀𝑡,2𝑢(𝑡)+𝑣(𝑡)𝑑𝑡0.(2.10) Let 𝐸={𝑡𝐹1𝐹3𝑣(𝑡)𝜎(𝑡)}, 𝐸1={𝑡𝐸𝑣(𝑡)=𝜎(𝑡)=0}. Then, 𝜇((𝐹1𝐹3)(𝐸𝐸1))>0 or 𝜇(𝐸𝐸1)>0. By 𝐹={𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}(𝐹1𝐹3)((𝐹1𝐹3)(𝐸𝐸1)), we obtain 𝐹1𝐹3121𝑀(𝑡,𝑢(𝑡))+21𝑀(𝑡,𝑣(𝑡))𝑀𝑡,2(>𝑢𝑡)+𝑣(𝑡)𝑑𝑡𝐹1𝐹3𝑀1𝑡,21𝑢(𝑡)+21𝑣(𝑡)𝑀𝑡,2𝑢(𝑡)+𝑣(𝑡)𝑑𝑡,(2.11) whenever 𝜇((𝐹1𝐹3)(𝐸𝐸1))>0. By 𝐹={𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}(𝐹1𝐹3)(𝐸𝐸1), we obtain 𝐹1𝐹3𝑀1𝑡,2(1𝑢𝑡)+2(1𝑣𝑡)𝑀𝑡,2(𝑢𝑡)+𝑣(𝑡)𝑑𝑡>0,(2.12) whenever 𝜇(𝐸𝐸1)>0. This means that one of three inequalities of (2.10) is strict inequality. By 𝜌𝑀(𝑢)=𝜌𝑀(𝑣)=1, we have 𝜌𝑀((1/2)(𝑢+𝑣))<1. By Lemma 1.7, we have (1/2)(𝑢+𝑣)<1.Case 2. 𝜇({𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}{𝑡𝑇𝑣(𝑡))>𝜎(𝑡)})=0. By (b), we have 𝜇{𝑡supp𝑢supp𝑣𝑢(𝑡)+𝑣(𝑡)>2𝑒(𝑡)}>0. Let 𝐺={𝑡supp𝑢supp𝑣𝑢(𝑡)+𝑣(𝑡)>2𝑒(𝑡)}. Put 𝐺1𝐺={𝑡𝐺𝑢(𝑡)+𝑣(𝑡)>𝑢(𝑡)+𝑣(𝑡)},2={𝑡𝐺𝑢(𝑡)+𝑣(𝑡)>𝑢(𝑡)𝑣(𝑡)}.(2.13) Since 𝑋 is nonsquare space, we have 𝜇𝐺1>0 or 𝜇𝐺2>0 by Lemma 2.2. Without loss of generality, we may assume 𝜇𝐺1>0. Hence, 𝐺1𝑀1𝑡,2(1𝑢𝑡)+2(𝑣𝑡)𝑑𝑡>𝐺1𝑀1𝑡,2(𝑢𝑡)+𝑣(𝑡)𝑑𝑡.(2.14) Therefore, by (2.14), we have 12𝜌𝑀1(𝑢)+2𝜌𝑀=(𝑣)𝑇121𝑀(𝑡,𝑢(𝑡))+2𝑀(𝑡,𝑣(𝑡))𝑑𝑡𝐺1𝑀1𝑡,21𝑢(𝑡)+2𝑣(𝑡)𝑑𝑡+𝑇𝐺1𝑀1𝑡,21𝑢(𝑡)+2>𝑣(𝑡)𝑑𝑡𝐺1𝑀1𝑡,2𝑢(𝑡)+𝑣(𝑡)𝑑𝑡+𝑇𝐺1𝑀1𝑡,2𝑢(𝑡)+𝑣(𝑡)𝑑𝑡=𝜌𝑀𝑢+𝑣2.(2.15) By 𝜌𝑀(𝑢)=𝜌𝑀(𝑣)=1, we have 𝜌𝑀((𝑢+𝑣)/2)<1. By Lemma 1.7, we have (1/2)(𝑢+𝑣)<1. This completes the proof.

Corollary 2.3. 𝐿𝑀(𝑅) is nonsquare space if and only if (a)𝑀Δ,(b)for any 𝑢,𝑣𝑆(𝐿𝑀(𝑅)), one has 𝜇{𝑡supp𝑢supp𝑣|𝑢(𝑡)|+|𝑣(𝑡)|>2𝑒(𝑡)}>0 or 𝜇({𝑡𝑇|𝑢(𝑡)|>𝜎(𝑡)}{𝑡𝑇|𝑣(𝑡)|>𝜎(𝑡)})>0.

Theorem 2.4. Let 𝑒(𝑡)=0  μ-a.e on 𝑇. Then, 𝐿𝑀(𝑋) is nonsquare space if and only if (a)𝑀Δ,(b)𝜌𝑀(𝜎)<2,(c)𝑋 is nonsquare space.

Proof. Necessity. By Theorem 2.1, (a) and (c) are obvious. Suppose that 𝜌𝑀(𝜎)2. Then, there exists 𝐸Σ such that 𝜇𝐸>0, 𝜌𝑀(𝜎𝜒𝑇𝐸)=1 and 𝜌𝑀(𝜎𝜒𝐷)=1, where 𝐷𝑇𝐸. Set 𝑢(𝑡)=𝑥𝜎(𝑡)𝜒𝑇𝐸(𝑡),𝑣(𝑡)=𝑥𝜎(𝑡)𝜒𝐷(𝑡),(2.16) where 𝑥𝑆(𝑋). It is easy to see that 𝑢=𝑣=1, 𝜇{𝑡supp𝑢supp𝑣𝑢(𝑡)+𝑣(𝑡)>2𝑒(𝑡)}=0 and 𝜇({𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}{𝑡𝑇𝑣(𝑡)>𝜎(𝑡)})=0. Contradicting Theorem 2.1.
Sufficiency. We only need to prove that for any 𝑢,𝑣𝑆(𝐿𝑀(𝑋)), if 𝜇(supp𝑢supp𝑣)=0, then 𝜇({𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}{𝑡𝑇𝑣(𝑡)>𝜎(𝑡)})>0. Suppose that there exist 𝑢,𝑣𝑆(𝐿𝑀(𝑋)) such that 𝜇(supp𝑢supp𝑣)=0, 𝜇({𝑡𝑇𝑢(𝑡)>𝜎(𝑡)}{𝑡𝑇𝑣(𝑡)>𝜎(𝑡)})=0. By 𝜇(supp𝑢supp𝑣)=0, we have 𝜌𝑀(𝑢)+𝜌𝑀(𝑣)=𝜌𝑀(𝑢+𝑣)𝜌𝑀(𝜎)<2. This implies 𝜌𝑀(𝑢)<1 or 𝜌𝑀(𝑣)<1. Hence, 𝑢<1 or 𝑣<1, a contradiction! This completes the proof.

Theorem 2.5. 𝐿0𝑀(𝑋) is nonsquare space if and only if (a)for any 𝑢𝐿0𝑀(𝑋){0}, one has 𝐾(𝑢)𝜙,(b)at least one of the conditions (b1)𝑘𝑙/(𝑘+𝑙)𝐾(𝑢+𝑣)𝐾(𝑢𝑣), (b2)𝜇({𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)>0})>0, (b3)𝜇{𝑡supp𝑢supp𝑣𝑘𝑙/(𝑘+𝑙)(𝑢(𝑡)+𝑣(𝑡))>𝑒(𝑡)}>0is true, where 𝑘𝐾(𝑢), 𝑙𝐾(𝑣) and 𝑢,𝑣𝑆(𝐿0𝑀(𝑋)),(c)𝑋 is nonsquare space.

Proof. Necessity. (a) Suppose that there exists 𝑢𝐿0𝑀(𝑋){0} such that 𝐾(𝑢)=𝜙, then 𝑢0=𝑇𝐴(𝑡)𝑢(𝑡)𝑑𝑡 by Lemma 1.8. Decompose 𝑇 into disjoint sets 𝑇1 and 𝑇2 such that 𝑇1𝐴(𝑡)𝑢(𝑡)𝑑𝑡=𝑇2𝐴(𝑡)𝑢(𝑡)𝑑𝑡. Put 𝑢1(𝑡)=2𝑢(𝑡)𝜒𝑇1,𝑢2(𝑡)=2𝑢(𝑡)𝜒𝑇2.(2.17) Obviously, 𝑢=(1/2)(𝑢1+𝑢2). Pick sequence {𝑘𝑛}𝑛=1𝑅+ such that 𝑘𝑛 as 𝑛. Let 𝑇0=supp𝑢1. By Levi theorem, we have 𝑢10lim𝑛1𝑘𝑛1+𝑇𝑀𝑘𝑡,𝑛𝑢1(𝑡)𝑑𝑡=lim𝑛𝑇0𝑀𝑘𝑡,𝑛𝑢1(𝑡)𝑘𝑛𝑢1𝑢(𝑡)1=(𝑡)𝑑𝑡𝑇0lim𝑛𝑀𝑘𝑡,𝑛𝑢1(𝑡)𝑘𝑛𝑢1𝑢(𝑡)1=(𝑡)𝑑𝑡𝑇𝑢𝐴(𝑡)1=(𝑡)𝑑𝑡𝑇1𝐴(𝑡)2𝑢(𝑡)𝑑𝑡.(2.18) Therefore, 𝑢10𝑇1=𝐴(𝑡)2𝑢(𝑡)𝑑𝑡𝑇1𝐴(𝑡)𝑢(𝑡)𝑑t+𝑇1𝐴(𝑡)𝑢(𝑡)𝑑𝑡+𝑇2𝐴(𝑡)𝑢(𝑡)𝑑𝑡𝑇2=𝐴(𝑡)𝑢(𝑡)𝑑𝑡𝑇1𝐴(𝑡)𝑢(𝑡)𝑑𝑡+𝑇2=𝐴(𝑡)𝑢(𝑡)𝑑𝑡𝑇𝐴(𝑡)𝑢(𝑡)𝑑𝑡=𝑢0.(2.19) Similarly, we have 𝑢20𝑢0. By 𝑢=(1/2)(𝑢1+𝑢2), we obtain 𝑢0=𝑢10=𝑢20=(1/2)(𝑢1+𝑢2)0. By 𝑢1(𝑡)𝑢2(𝑡)=2𝑢(𝑡)𝜒𝑇1+(2𝑢(𝑡))𝜒𝑇2, we have 𝑢0=(1/2)(𝑢1𝑢2)0. Therefore, 𝑢10=𝑢20=12𝑢1+𝑢20=12𝑢1𝑢20=𝑢0.(2.20) This implies 𝑢10+u20𝑢=min1+𝑢20,𝑢1𝑢20,(2.21) a contradiction!
If (b) is not true, then there exist 𝑢,𝑣𝑆(𝐿0𝑀(𝑋)) such that 𝑘𝑙/(𝑘+𝑙)𝐾(𝑢+𝑣)𝐾(𝑢𝑣), 𝜇({𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)>0})=0 and 𝜇{𝑡supp𝑢supp𝑣𝑘𝑙/(𝑘+𝑙)(𝑢(𝑡)+𝑣(𝑡))>𝑒(𝑡)}=0, where 𝑘𝐾(𝑢), 𝑙𝐾(𝑣). Let 𝐸=𝑇(supp𝑢supp𝑣). It is easy to see that if 𝑡𝐸, then 𝑢(𝑡)𝑣(𝑡)=0 on 𝐸. This implies 𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)=𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑣(𝑡)𝑡𝐸.(2.22) Therefore, by (2.22), we have 𝐸𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢𝑡)+𝑘𝑙(𝑘+𝑙𝑣𝑡)𝑑𝑡=𝐸𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢𝑡)+𝑣(𝑡)𝑑𝑡.(2.23) By 𝜇{𝑡supp𝑢supp𝑣(𝑘𝑙/(𝑘+𝑙))(𝑢(𝑡)+𝑣(𝑡))>𝑒(𝑡)}=0, we have 0=𝑇𝐸𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢t)+𝑘𝑙(=𝑘+𝑙𝑣𝑡)𝑑𝑡𝑇𝐸𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑣(𝑡)𝑑𝑡.(2.24) By (2.23) and (2.24), we have 𝑇𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢𝑡)+𝑘𝑙(𝑘+𝑙𝑣𝑡)𝑑𝑡=𝑇𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢𝑡)+𝑣(𝑡)𝑑𝑡.(2.25) By 𝜇({𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)>0})=0, we have 𝑇𝑙𝑘𝑘+𝑙𝑀(𝑡,𝑘𝑢(𝑡))+=𝑘+𝑙𝑀(𝑡,𝑙𝑣(𝑡))𝑑𝑡𝑇𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)𝑑𝑡.(2.26) Therefore, by (2.25) and (2.26), we have 𝑢0+𝑣0=1𝑘1+𝜌𝑀+1(𝑘𝑢)𝑙1+𝜌𝑀=(𝑙𝑣)𝑘+𝑙𝑙𝑘𝑙1+𝜌𝑘+𝑙𝑀𝑘(𝑘𝑢)+𝜌𝑘+𝑙𝑀=(𝑙𝑣)𝑘+𝑙𝑙𝑘𝑙1+𝑘+𝑙𝑇𝑘𝑀(𝑡,𝑘𝑢(𝑡))𝑑𝑡+𝑘+𝑙𝑇=𝑀(𝑡,𝑙𝑣(𝑡))𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑙𝑘𝑘+𝑙𝑀(𝑡,𝑘𝑢(𝑡))+)=𝑘+𝑙𝑀(𝑡,𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙=𝑘+𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙=𝑘+𝑙𝑢(𝑡)+𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝜌𝑀𝑘𝑙𝑘+𝑙(𝑢+𝑣)=𝑢+𝑣0.(2.27) Similarly, we have 𝑢0+𝑣0=𝑢𝑣0. This implies 𝑢0+𝑣0=𝑢+𝑣0=𝑢𝑣0, a contradiction!
(c) Pick (𝑡)𝑆(𝐿0𝑀(𝑋)), then there exists 𝑑>0 such that 𝜇𝐸>0, where 𝐸={𝑡𝑇(𝑡)𝑑}. Put 1(𝑡)=𝑑𝑥0𝜒𝐸(𝑡), where 𝑥0𝑆(𝑋). It is easy to see that 1(𝑡)𝐿0𝑀(𝑋){0}. Hence, there exists 𝑙>0 such that 𝑙1(𝑡)𝑆(𝐿0𝑀(𝑋)). The necessity of (c) follows from the fact that 𝑋 is isometrically embedded into 𝐿0𝑀(𝑋). Namely, defining the operator 𝐼𝑋𝐿0𝑀(𝑋) by 𝐼(𝑥)=𝑙𝑑𝑥𝜒𝐸(𝑡),𝑥𝑋.(2.28) It is easy to see that 𝐼(𝑥0)𝑆(𝐿0𝑀(𝑋)). Hence, for any 𝑥𝑋{0}, we have 𝐼(𝑥)0=inf𝑘>01𝑘1+𝜌𝑀(𝑘𝐼(𝑥))=inf𝑘>01𝑘1+𝐸𝑀(𝑡,𝑘𝑙𝑑𝑥)𝑑𝑡=inf𝑘>01𝑘1+𝐸𝑀𝑥𝑡,𝑘𝑥𝑙𝑑0𝑑𝑡=inf𝑘>01𝑘1+𝜌𝑀𝑥𝑘𝑥𝐼0=𝑥𝑥𝐼00𝐼𝑥=𝑥00=𝑥.(2.29)
Sufficiency. Suppose that there exists 𝑢,𝑣𝑆(𝐿0𝑀(𝑋)) such that 𝑢0=𝑣0=(1/2)(𝑢+𝑣)0=(1/2)(𝑢𝑣)0=1. Let 𝑘𝐾(𝑢), 𝑙𝐾(𝑣). We will derive a contradiction for each of the following two cases.
Case 1. 𝜇[({𝑡𝑇𝑢(𝑡)0}{𝑡𝑇𝑣(𝑡)0}){𝑡𝑇𝜎(𝑡)>0}]=0. By Lemma 2.2, we have 𝑢(𝑡)+𝑣(𝑡)>𝑚𝑖𝑛{𝑢(𝑡)+𝑣(𝑡),𝑢(𝑡)𝑣(𝑡)}𝑡supp𝑢supp𝑣. Put 𝑇1𝑇={𝑡supp𝑢supp𝑣𝑢(𝑡)+𝑣(𝑡)>𝑢(𝑡)+𝑣(𝑡)},2={𝑡supp𝑢supp𝑣𝑢(𝑡)+𝑣(𝑡)>𝑢(𝑡)𝑣(𝑡)}.(2.30) Moreover, we have 𝑢0+𝑣0=1𝑘1+𝜌𝑀+1(𝑘𝑢)𝑙1+𝜌𝑀=(𝑙𝑣)𝑘+𝑙𝑙𝑘𝑙1+𝑘+𝑙𝑇𝑘𝑀(𝑡,𝑘𝑢(𝑡))𝑑𝑡+𝑘+𝑙𝑇=𝑀(𝑡,𝑙𝑣(𝑡))𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑙𝑘𝑘+𝑙𝑀(𝑡,𝑘𝑢(𝑡))+𝑘+𝑙𝑀(𝑡,𝑙𝑣(𝑡))𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇1𝑀𝑡,𝑘𝑙+𝑘+𝑙𝑢(𝑡)+𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑇1𝑀𝑡,𝑘𝑙=𝑘+𝑙𝑢(𝑡)+𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝜌𝑀𝑘𝑙𝑘+𝑙(𝑢+𝑣)𝑢+𝑣0.(2.31) By 𝑢0+𝑣0=𝑢+𝑣0, three inequalities of (2.31) are equation. This implies 𝑘+𝑙𝑘𝑙1+𝑇𝑙𝑘𝑘+𝑙𝑀(𝑡,𝑘𝑢(𝑡))+)=𝑘+𝑙𝑀(𝑡,𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙.𝑘+𝑙𝑣(𝑡)𝑑𝑡(2.32) Next, we will prove 𝜇({𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)>0})=0. Suppose that 𝜇({𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)>0})>0. Without loss of generality, we may assume 𝜇{𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}>0. Therefore, by (2.32), we have 𝑙𝑣(𝑡)𝜎(𝑡)0  μ-a.e. on {𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}. Hence, 𝜇({𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)0})>0. Since three inequalities of (2.31) are equation, we deduce 𝑇𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢𝑡)+𝑘𝑙(𝑘+𝑙𝑣𝑡)𝑑𝑡=𝑇𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢𝑡)+𝑣(𝑡)𝑑𝑡.(2.33) Moreover, it is easy to see 𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)>𝜎(𝑡)𝑒(𝑡)(2.34) on {𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)0}. Therefore, by (2.33) and (2.34), we have 𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)=𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑣(𝑡)(2.35)𝜇-a.e. on {𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)0}. Since 𝑋 is nonsquare space, we have 𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)>𝑚𝑖𝑛𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑣(𝑡),𝑘𝑙𝑘+𝑙𝑢(𝑡)𝑣(𝑡)(2.36) on {𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)0}. Thus, 𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)>𝑘𝑙𝑘+𝑙𝑢(𝑡)𝑣(𝑡)(2.37)𝜇-a.e. on {𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)0}. Therefore, by (2.34) and (2.37), we have 𝑢0+𝑣0=1𝑘1+𝜌𝑀+1(𝑘𝑢)𝑙1+𝜌𝑀(𝑙𝑣)𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙>𝑘+𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙=𝑘+𝑙𝑢(𝑡)𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝜌𝑀𝑘𝑙𝑘+𝑙(𝑢𝑣)𝑢𝑣0.(2.38) This implies 𝑢0+𝑣0>𝑢𝑣0, a contradiction! Hence, 𝜇({𝑡𝑇𝑘𝑢(𝑡)𝜎(𝑡)>0}{𝑡𝑇𝑙𝑣(𝑡)𝜎(𝑡)>0})=0. Since three inequalities of (2.31) are equation, we deduce 𝑘𝑙/(𝑘+𝑙)𝐾(𝑢+𝑣) and 𝜇{𝑡𝑇1(𝑘𝑙/(𝑘+𝑙))(𝑢(𝑡)+𝑣(𝑡))>𝑒(𝑡)}=0. By (b), we have 𝑘𝑙/(𝑘+𝑙)𝐾(𝑢𝑣) or 𝜇{𝑡𝑇2(𝑘𝑙/(𝑘+𝑙))(𝑢(𝑡)+𝑣(𝑡))>𝑒(𝑡)}>0. If 𝑘𝑙/(𝑘+𝑙)𝐾(𝑢𝑣), then 𝑢0+𝑣0=1𝑘1+𝜌𝑀+1(𝑘𝑢)𝑙1+𝜌𝑀(𝑙𝑣)𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙=𝑘+𝑙𝑢(𝑡)𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝜌𝑀𝑘𝑙𝑘+𝑙(𝑢𝑣)>𝑢𝑣0.(2.39) This implies 𝑢0+𝑣0>𝑢𝑣0, a contradiction! If 𝜇{𝑡𝑇2(𝑘𝑙/(𝑘+𝑙))(𝑢(𝑡)+𝑣(𝑡))>𝑒(𝑡)}>0, then 𝑢0+𝑣0=1𝑘1+𝜌𝑀+1(𝑘𝑢)𝑙1+𝜌𝑀(𝑙𝑣)𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙>𝑘+𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇2𝑀𝑡,𝑘𝑙+𝑘+𝑙𝑢(𝑡)𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑇2𝑀𝑡,𝑘𝑙=𝑘+𝑙𝑢(𝑡)(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝜌𝑀𝑘𝑙𝑘+𝑙(𝑢𝑣)𝑢𝑣0.(2.40) This implies 𝑢0+𝑣0>𝑢𝑣0, a contradiction!Case 2. 𝜇[({𝑡𝑇𝑢(𝑡)0}{𝑡𝑇𝑣(𝑡)0}){𝑡𝑇𝜎(𝑡)>0}]>0. Let 𝐸=({𝑡𝑇𝑢(𝑡)0}{𝑡𝑇𝑣(𝑡)0}){𝑡𝑇𝜎(𝑡)>0}. Put 𝐸1={𝑡𝐸𝑢(𝑡)𝑣(𝑡)=0},𝐸2={𝑡𝐸𝑢(𝑡)𝑣(𝑡)>0}.(2.41) Then, 𝜇𝐸1>0 or 𝜇𝐸2>0. If 𝜇𝐸1>0, then 𝐸1𝑙𝑘𝑘+𝑙𝑀(𝑡,𝑘𝑢(𝑡))𝑑𝑡+>𝑘+𝑙𝑀(𝑡,𝑙𝑣(𝑡))𝑑𝑡𝐸1𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)𝑑𝑡.(2.42) Therefore, by (2.42), we have 𝑢0+𝑣0=1𝑘1+𝜌𝑀+1(𝑘𝑢)𝑙1+𝜌𝑀=(𝑙𝑣)𝑘+𝑙𝑘𝑙1+𝑇𝑙𝑘𝑘+𝑙𝑀(𝑡,𝑘𝑢(𝑡))+)>𝑘+𝑙𝑀(𝑡,𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝐸1𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙+𝑘+𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝐸1𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙𝑘+𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙(=𝑘+𝑙𝑢𝑡)+𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝜌𝑀𝑘𝑙𝑘+𝑙(𝑢+𝑣)𝑢+𝑣0.(2.43) This implies 𝑢0+𝑣0>𝑢+𝑣0, a contradiction! If 𝜇𝐸2>0, then 𝜇𝐸12>0 or 𝜇E22>0 by Lemma 2.2, where 𝐸12=𝑡𝐸2,𝐸𝑢(𝑡)+𝑣(𝑡)>𝑢(𝑡)+𝑣(𝑡)22=𝑡𝐸2.𝑢(𝑡)+𝑣(𝑡)>𝑢(𝑡)𝑣(𝑡)(2.44) Without loss of generality, we may assume 𝜇𝐸12>0. Hence, 𝐸12𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢𝑡)+𝑘𝑙(𝑘+𝑙𝑣𝑡)𝑑𝑡>𝐸12𝑀𝑡,𝑘𝑙(𝑘+𝑙𝑢𝑡)+𝑣(𝑡)𝑑𝑡.(2.45) Therefore, by (2.45), we have 𝑢0+𝑣0=1𝑘1+𝜌𝑀+1(𝑘𝑢)𝑙1+𝜌𝑀(𝑙𝑣)𝑘+𝑙𝑘𝑙1+𝑇𝑀𝑡,𝑘𝑙𝑘+𝑙𝑢(𝑡)+𝑘𝑙>𝑘+𝑙𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝐸12𝑀𝑡,𝑘𝑙+𝑘+𝑙𝑢(𝑡)+𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝑇𝐸12𝑀𝑡,𝑘𝑙=𝑘+𝑙𝑢(𝑡)+𝑣(𝑡)𝑑𝑡𝑘+𝑙𝑘𝑙1+𝜌𝑀𝑘𝑙𝑘+𝑙(𝑢+𝑣)𝑢+𝑣0.(2.46) This implies 𝑢0+𝑣0>𝑢+𝑣0, a contradiction! This completes the proof.

Corollary 2.6. 𝐿0𝑀(𝑅) is nonsquare space if and only if (a)for any 𝑢𝐿0𝑀(𝑅), one has 𝐾(𝑢)𝜙,(b)at least one of the conditions (b1)𝑘𝑙/(𝑘+𝑙)𝐾(𝑢+𝑣)𝐾(𝑢𝑣), (b2)𝜇({𝑡𝑇𝑘|𝑢(𝑡)|𝜎(𝑡)>0}{𝑡𝑇𝑙|𝑣(𝑡)|𝜎(𝑡)>0})>0,(b3)𝜇{𝑡supp𝑢supp𝑣(𝑘𝑙/(𝑘+𝑙))(|𝑢(𝑡)|+|𝑣(𝑡)|)>𝑒(𝑡)}>0is true, where 𝑘𝐾(𝑢), 𝑙𝐾(𝑣) and 𝑢,𝑣𝑆(𝐿0𝑀(R)).

Acknowledgments

The authors would like to thank the anonymous referee for the suggestions to improve the manuscript. This work was supported by Academic Leaders Fund of Harbin University of Science and Technology and supported by China Natural Science Fund 11061022.

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