Abstract and Applied Analysis

Abstract and Applied Analysis / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 361525 | https://doi.org/10.1155/2011/361525

Shaoqiang Shang, Yunan Cui, Yongqiang Fu, "Nonsquareness in Musielak-Orlicz-Bochner Function Spaces", Abstract and Applied Analysis, vol. 2011, Article ID 361525, 16 pages, 2011. https://doi.org/10.1155/2011/361525

Nonsquareness in Musielak-Orlicz-Bochner Function Spaces

Academic Editor: Paul Eloe
Received20 Oct 2010
Revised03 Jan 2011
Accepted14 Feb 2011
Published06 Apr 2011

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 [3–8].

Let (𝑋,β€–β‹…β€–) be a real Banach space. 𝑆(𝑋) and 𝐡(𝑋) denote the unit sphere and unit ball, respectively. By π‘‹βˆ—, 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 π‘₯,π‘¦βˆˆπ‘†(𝑋) we have min{β€–(1/2)(π‘₯+𝑦)β€–,β€–(1/2)(π‘₯βˆ’π‘¦)β€–}<1. A Banach space 𝑋 is said to be uniformly nonsquare space if for any π‘₯,π‘¦βˆˆπ‘†(𝑋), there exists 𝛿>0 such that min{β€–(1/2)(π‘₯+𝑦)β€–,β€–(1/2)(π‘₯βˆ’π‘¦)β€–}<1βˆ’π›Ώ.

Let βˆ‘(𝑇,,πœ‡) be nonatomic measure space. Suppose that a function π‘€βˆΆπ‘‡Γ—[0,∞)β†’[0,∞] satisfies the following conditions:(1)for πœ‡-a.e, π‘‘βˆˆπ‘‡, 𝑀(𝑑,0)=0, limπ‘’β†’βˆžπ‘€(𝑑,𝑒)=∞ and 𝑀(𝑑,π‘’ξ…ž)<∞ for some π‘’ξ…ž>0,(2)forβ€‰β€‰πœ‡-a.e, π‘‘βˆˆπ‘‡, 𝑀(𝑑,𝑒) is convex on [0,∞) with respect to 𝑒,(3)for each π‘’βˆˆ[0,∞), 𝑀(𝑑,𝑒) is a πœ‡-measurable function of 𝑑 on 𝑇.

Let 𝑒(𝑑)=sup{𝑒>0βˆΆπ‘€(𝑑,𝑒)=0}. It is well known that 𝑒(𝑑) is πœ‡-measurable (see [2]).

Moreover, for a given Banach space (𝑋,β€–β‹…β€–), we denote by 𝑋𝑇, the set of all strongly πœ‡-measurable function from 𝑇 to 𝑋, and for each π‘’βˆˆπ‘‹π‘‡, define the modular of 𝑒 by πœŒπ‘€(ξ€œπ‘’)=𝑇𝑀(𝑑,‖𝑒(𝑑)β€–)𝑑𝑑.(1.1) Put𝐿𝑀(𝑋)=π‘’βˆˆπ‘‹π‘‡βˆΆπœŒπ‘€ξ€Ύ(πœ†π‘’)<∞forsomeπœ†>0.(1.2) It is well known that Musielak-Orlicz-Bochner function space 𝐿𝑀(𝑋) is Banach spaces equipped with the Luxemburg norm‖𝑒‖=infπœ†>0βˆΆπœŒπ‘€ξ‚€π‘’πœ†ξ‚ξ‚‡β‰€1(1.3) or Orlicz’s norm‖𝑒‖0=infπ‘˜>01π‘˜ξ€Ί1+πœŒπ‘€ξ€».(π‘˜π‘’)(1.4) In particular, 𝐿𝑀(𝑅) and 𝐿0𝑀(𝑅) are said to be Musielak-Orlicz function space. Set1supp𝑒={π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–β‰ 0},𝐾(𝑒)=π‘˜>0βˆΆπ‘˜ξ€·1+πœŒπ‘€ξ€Έ=(π‘˜π‘’)‖𝑒‖0.(1.5) In particular, the set 𝐾(𝑒) can be nonempty. To show that, we give a proposition.

Proposition 1.1 (see [9]). If limπ‘’β†’βˆž(𝑀(𝑑,𝑒)/𝑒)=βˆžβ€‰β€‰ΞΌ-a.e. β€‰β€‰π‘‘βˆˆπ‘‡, then 𝐾(𝑒)β‰ πœ™ for any π‘’βˆˆπΏ0𝑀(𝑋).

We define a function1𝜎(𝑑)=sup𝑒β‰₯0βˆΆπ‘€π‘‘,2𝑒=12.𝑀(𝑑,𝑒)(1.6) Function 𝜎(𝑑) will be used in the further part of the paper. Moreover, 𝜎(𝑑) is πœ‡-measurable. To show that, we give a proposition.

Proposition 1.2. Function 𝜎(𝑑) is πœ‡-measurable.

Proof. Pick a dense set {π‘Ÿπ‘–}βˆžπ‘–=1 in [0,∞) and set π΅π‘˜=1π‘‘βˆˆπ‘‡βˆΆπ‘€π‘‘,2π‘Ÿπ‘˜ξ‚=12𝑀𝑑,π‘Ÿπ‘˜ξ€Έξ‚‡,π‘žπ‘˜(𝑑)=π‘Ÿπ‘˜πœ’π΅π‘˜(𝑑)(π‘˜βˆˆπ‘).(1.7) It is easy to see that for all π‘˜βˆˆπ‘, 𝜎(𝑑)β‰₯π‘žπ‘˜(𝑑)  μ-a.e on 𝑇. Hence, supπ‘˜β‰₯1π‘žπ‘˜(𝑑)β‰€πœŽ(𝑑). For πœ‡-a.e π‘‘βˆˆπ‘‡, arbitrarily choose πœ€βˆˆ(0,𝜎(𝑑)). Then, there exists π‘Ÿπ‘˜βˆˆ(𝜎(𝑑)βˆ’πœ€,𝜎(𝑑)) such that 𝑀(𝑑,(1/2)π‘Ÿπ‘˜)=(1/2)𝑀(𝑑,π‘Ÿπ‘˜), that is, π‘žπ‘˜(𝑑)β‰₯π‘Ÿπ‘˜>𝜎(𝑑)βˆ’πœ€. Since πœ€ is arbitrary, we find supπ‘˜β‰₯1π‘žπ‘˜(𝑑)β‰₯𝜎(𝑑). Thus, supπ‘˜β‰₯1π‘žπ‘˜(𝑑)=𝜎(𝑑).

It is easy to prove the following proposition.

Proposition 1.3. For any π›Όβˆˆ(0,1), if 𝑒(𝑑)β‰€πœŽ(𝑑), then 𝑀(𝑑,𝛼𝑒(𝑑))=𝛼𝑀(𝑑,𝑒(𝑑)).

Proposition 1.4. For any π›Όβˆˆ(0,1), if 𝑒(𝑑)<𝜎(𝑑)<𝑣(𝑑), then 𝑀(𝑑,𝛼𝑒(𝑑)+(1βˆ’π›Ό)𝑣(𝑑))<𝛼𝑀(𝑑,𝑒(𝑑))+(1βˆ’π›Ό)𝑀(𝑑,𝑣(𝑑)).

Proposition 1.5. For any π›Όβˆˆ(0,1), if 𝜎(𝑑)<𝑣(𝑑), then 𝑀(𝑑,𝛼𝑣(𝑑))<𝛼𝑀(𝑑,𝑣(𝑑)).

Definition 1.6 (see [2]). We say that 𝑀(𝑑,𝑒) satisfies condition Ξ”(π‘€βˆˆΞ”) if there exist 𝐾β‰₯1 and a measureable nonnegative function 𝛿(𝑑) on 𝑇 such that βˆ«π‘‡π‘€(𝑑,𝛿(𝑑))𝑑𝑑<∞ and 𝑀(𝑑,2𝑒)≀𝐾𝑀(𝑑,𝑒) for almost all π‘‘βˆˆπ‘‡ and all 𝑒β‰₯𝛿(𝑑).

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

Lemma 1.7 (see [2]). Suppose π‘€βˆˆΞ”. Then πœŒπ‘€(𝑒)=1⇔‖𝑒‖=1.

Lemma 1.8 (see [9]). Let 𝐿0𝑀(𝑋) be Musielak-Orlicz-Bochner function spaces, then, if 𝐾(𝑒)=πœ™, one has ‖𝑒‖0=βˆ«π‘‡π΄(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑, where 𝐴(𝑑)=limπ‘’β†’βˆž(𝑀(𝑑,𝑒)/𝑒).

2. Main Results

Theorem 2.1. 𝐿𝑀(𝑋) is nonsquare space if and only if (a)π‘€βˆˆΞ”,(b)for any 𝑒,π‘£βˆˆπ‘†(𝐿𝑀(𝑋)), one has πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>2𝑒(𝑑)}>0 or πœ‡({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–>𝜎(𝑑)})>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 π‘₯,𝑦≠0, one has β€–π‘₯β€–+β€–π‘¦β€–βˆ’min{β€–π‘₯+𝑦‖,β€–π‘₯βˆ’π‘¦β€–}>0.(2.1)

Proof. For any π‘₯,𝑦≠0, without loss of generality, we may assume β€–π‘₯‖≀‖𝑦‖. Since 𝑋 is nonsquare space, we have β€–β€–β€–β€–π‘₯β€–+β€–π‘₯β€–=β€–π‘₯β€–+β€–π‘₯‖𝑦‖‖‖‖‖‖‖𝑦‖>π‘šπ‘–π‘›π‘₯+β€–π‘₯‖𝑦‖‖‖,‖‖‖‖𝑦‖π‘₯βˆ’β€–π‘₯‖𝑦‖‖‖‖𝑦‖.(2.2) Therefore, by (2.2), we obtain β€–β€–β€–β€–π‘₯+𝑦‖≀π‘₯+β€–π‘₯β€–β€–β€–β€–+‖𝑦‖⋅𝑦1βˆ’β€–π‘₯‖⋅‖𝑦‖‖𝑦‖<β€–π‘₯β€–+β€–π‘₯β€–+β€–π‘¦β€–βˆ’β€–π‘₯β€–=β€–π‘₯β€–+‖𝑦‖(2.3) or β€–β€–β€–β€–π‘₯βˆ’π‘¦β€–β‰€π‘₯βˆ’β€–π‘₯β€–β€–β€–β€–+‖𝑦‖⋅𝑦1βˆ’β€–π‘₯‖⋅‖𝑦‖‖𝑦‖<β€–π‘₯β€–+β€–π‘₯β€–+β€–π‘¦β€–βˆ’β€–π‘₯β€–=β€–π‘₯β€–+‖𝑦‖.(2.4) This implies β€–π‘₯β€–+β€–π‘¦β€–βˆ’min{β€–π‘₯+𝑦‖,β€–π‘₯βˆ’π‘¦β€–}>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 π‘€βˆˆΞ” which follows from the theorem proved in more general case (see [10, 11]). Namely, if π‘€βˆ‰Ξ”, then 𝐿𝑀(𝑅) contains isometric copy of π‘™βˆž.
If (b) is not true, then there exist 𝑒,π‘£βˆˆπ‘†(𝐿𝑀(𝑋)) such that πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>2𝑒(𝑑)}=0 and πœ‡({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–)>𝜎(𝑑)})=0. Let 𝐺=suppπ‘’βˆ©supp𝑣. We have 12πœŒπ‘€1(𝑒)+2πœŒπ‘€=1(𝑣)2ξ€œπ‘‡1𝑀(𝑑,‖𝑒(𝑑)β€–)𝑑𝑑+2ξ€œπ‘‡=ξ€œπ‘€(𝑑,‖𝑣(𝑑)β€–)𝑑𝑑𝑇121𝑀(𝑑,‖𝑒(𝑑)β€–)+2β‰₯ξ€œπ‘€(𝑑,‖𝑣(𝑑)β€–)𝑑𝑑𝐺𝑀1𝑑,2(1‖𝑒𝑑)β€–+2(ξ‚ξ€œβ€–π‘£π‘‘)‖𝑑𝑑+𝑇⧡𝐺𝑀1𝑑,2(1‖𝑒𝑑)β€–+2(β‰₯ξ€œβ€–π‘£π‘‘)‖𝑑𝑑𝐺𝑀1𝑑,2ξ‚ξ€œβ€–π‘’(𝑑)+𝑣(𝑑)‖𝑑𝑑+𝑇⧡𝐺𝑀1𝑑,2‖𝑒(𝑑)+𝑣(𝑑)‖𝑑𝑑=πœŒπ‘€ξ‚€π‘’+𝑣2.(2.5) By πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>2𝑒(𝑑)}=0 and πœ‡({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–)>𝜎(𝑑)})=0, we obtain that two inequalities of (2.5) are equations. This implies (1/2)πœŒπ‘€(𝑒)+(1/2)πœŒπ‘€(𝑣)=πœŒπ‘€((𝑒+𝑣)/2). By Lemma 1.7, we have πœŒπ‘€((𝑒+𝑣)/2)=1. Thus, β€–(1/2)(𝑒+𝑣)β€–=1. Similarly, we have β€–(1/2)(π‘’βˆ’π‘£)β€–=1, a contradiction!
(c) Pick β„Ž(𝑑)βˆˆπ‘†(𝐿𝑀(𝑋)), then there exists 𝑑>0 such that πœ‡πΈ>0, where 𝐸={π‘‘βˆˆπ‘‡βˆΆβ€–β„Ž(𝑑)β€–β‰₯𝑑}. Put β„Ž1(𝑑)=𝑑⋅π‘₯0β‹…πœ’πΈ(𝑑), where π‘₯0βˆˆπ‘†(𝑋). It is easy to see that β„Ž1(𝑑)βˆˆπΏπ‘€(𝑋)⧡{0}. Hence, there exists π‘˜>0 such that π‘˜β‹…β„Ž1(𝑑)βˆˆπ‘†(𝐿𝑀(𝑋)). By Lemma 1.7, we have ξ€œ1=𝑇𝑀‖‖𝑑,π‘˜β‹…β„Ž1(β€–β€–ξ€Έξ€œπ‘‘)𝑑𝑑=𝐸𝑀‖‖𝑑,π‘˜β‹…π‘‘β‹…π‘₯0‖‖𝑑𝑑.(2.6) Let 𝛼=π‘˜β‹…π‘‘. Then, βˆ«πΈπ‘€(𝑑,𝛼)𝑑𝑑=1. The necessity of (c) follows from the fact that 𝑋 is isometrically embedded into 𝐿𝑀(𝑋). Namely, defining the operator πΌβˆΆπ‘‹β†’πΏπ‘€(𝑋) by 𝐼(π‘₯)=𝛼⋅π‘₯β‹…πœ’πΈ(𝑑),π‘₯βˆˆπ‘‹.(2.7) Hence, for any π‘₯βˆˆπ‘‹β§΅{0}, we have πœŒπ‘€ξ‚΅πΌ(π‘₯)ξ‚Ά=ξ€œβ€–π‘₯‖𝐸𝑀‖‖‖𝑑,𝐼(π‘₯)β€–β€–β€–ξ‚Άξ€œβ€–π‘₯‖𝑑𝑑=𝐸𝑀𝑑,𝛼‖π‘₯β€–ξ‚Άξ€œβ€–π‘₯‖𝑑𝑑=𝐸𝑀(𝑑,𝛼)𝑑𝑑=1.(2.8) By Lemma 1.7, we have ‖𝐼(π‘₯)/β€–π‘₯‖‖𝐿𝑀(𝑋)=1, hence ‖𝐼(π‘₯)‖𝐿𝑀(𝑋)=β€–π‘₯β€–.
Sufficiency. The proof requires the consideration of two cases separately.
Case 1. πœ‡({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–>𝜎(𝑑)})>0. Without loss of generality, we may assume πœ‡{π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}>0. Let 𝐹={π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}. Put 𝐹1𝐹={π‘‘βˆˆπΉβˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>‖𝑒(𝑑)+𝑣(𝑑)β€–},2=‖𝐹{π‘‘βˆˆπΉβˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>‖𝑒(𝑑)βˆ’π‘£(𝑑)},3={π‘‘βˆˆπΉβˆΆβ€–π‘£(𝑑)β€–=0}.(2.9) Since 𝑋 is nonsquare space, we have πœ‡(𝐹1βˆͺ𝐹3)>0 or πœ‡(𝐹2βˆͺ𝐹3)>0 by Lemma 2.2. Without loss of generality, we may assume πœ‡(𝐹1βˆͺ𝐹3)>0. Moreover, we have 12πœŒπ‘€1(𝑒)+2πœŒπ‘€(𝑣)βˆ’πœŒπ‘€ξ‚€12=1(𝑒+𝑣)2ξ€œπ‘‡1𝑀(𝑑,‖𝑒(𝑑)β€–)𝑑𝑑+2ξ€œπ‘‡ξ€œπ‘€(𝑑,‖𝑣(𝑑)β€–)π‘‘π‘‘βˆ’π‘‡π‘€ξ‚€1𝑑,2=ξ€œβ€–π‘’(𝑑)+𝑣(𝑑)‖𝑑𝑑𝑇121𝑀(𝑑,‖𝑒(𝑑)β€–)+2ξ‚€1𝑀(𝑑,‖𝑣(𝑑)β€–)βˆ’π‘€π‘‘,2β‰₯ξ€œβ€–π‘’(𝑑)+𝑣(𝑑)‖𝑑𝑑𝐹1βˆͺ𝐹3121𝑀(𝑑,‖𝑒(𝑑)β€–)+2ξ‚€1𝑀(𝑑,‖𝑣(𝑑)β€–)βˆ’π‘€π‘‘,2(β‰₯ξ€œβ€–π‘’π‘‘)+𝑣(𝑑)‖𝑑𝑑𝐹1βˆͺ𝐹3𝑀1𝑑,2β€–1𝑒(𝑑)β€–+2‖1𝑣(𝑑)β€–βˆ’π‘€π‘‘,2‖𝑒(𝑑)+𝑣(𝑑)‖𝑑𝑑β‰₯0.(2.10) Let 𝐸={π‘‘βˆˆπΉ1βˆͺ𝐹3βˆΆβ€–π‘£(𝑑)β€–β‰₯𝜎(𝑑)}, 𝐸1={π‘‘βˆˆπΈβˆΆβ€–π‘£(𝑑)β€–=𝜎(𝑑)=0}. Then, πœ‡((𝐹1βˆͺ𝐹3)⧡(𝐸⧡𝐸1))>0 or πœ‡(𝐸⧡𝐸1)>0. By 𝐹={π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βŠƒ(𝐹1βˆͺ𝐹3)βŠƒ((𝐹1βˆͺ𝐹3)⧡(𝐸⧡𝐸1)), we obtain ξ€œπΉ1βˆͺ𝐹3121𝑀(𝑑,‖𝑒(𝑑)β€–)+2ξ‚€1𝑀(𝑑,‖𝑣(𝑑)β€–)βˆ’π‘€π‘‘,2(>ξ€œβ€–π‘’π‘‘)+𝑣(𝑑)‖𝑑𝑑𝐹1βˆͺ𝐹3𝑀1𝑑,2β€–1𝑒(𝑑)β€–+2‖1𝑣(𝑑)β€–βˆ’π‘€π‘‘,2‖𝑒(𝑑)+𝑣(𝑑)‖𝑑𝑑,(2.11) whenever πœ‡((𝐹1βˆͺ𝐹3)⧡(𝐸⧡𝐸1))>0. By 𝐹={π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βŠƒ(𝐹1βˆͺ𝐹3)βŠƒ(𝐸⧡𝐸1), we obtain ξ€œπΉ1βˆͺ𝐹3𝑀1𝑑,2(1‖𝑒𝑑)β€–+2(1‖𝑣𝑑)β€–βˆ’π‘€π‘‘,2(‖𝑒𝑑)+𝑣(𝑑)‖𝑑𝑑>0,(2.12) whenever πœ‡(𝐸⧡𝐸1)>0. This means that one of three inequalities of (2.10) is strict inequality. By πœŒπ‘€(𝑒)=πœŒπ‘€(𝑣)=1, we have πœŒπ‘€((1/2)(𝑒+𝑣))<1. By Lemma 1.7, we have β€–(1/2)(𝑒+𝑣)β€–<1.Case 2. πœ‡({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–)>𝜎(𝑑)})=0. By (b), we have πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>2𝑒(𝑑)}>0. Let 𝐺={π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>2𝑒(𝑑)}. Put 𝐺1𝐺={π‘‘βˆˆπΊβˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>‖𝑒(𝑑)+𝑣(𝑑)β€–},2=β€–{π‘‘βˆˆπΊβˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>‖𝑒(𝑑)βˆ’π‘£(𝑑)}.(2.13) Since 𝑋 is nonsquare space, we have πœ‡πΊ1>0 or πœ‡πΊ2>0 by Lemma 2.2. Without loss of generality, we may assume πœ‡πΊ1>0. Hence, ξ€œπΊ1𝑀1𝑑,2(1‖𝑒𝑑)β€–+2(ξ‚ξ€œβ€–π‘£π‘‘)‖𝑑𝑑>𝐺1𝑀1𝑑,2(‖𝑒𝑑)+𝑣(𝑑)‖𝑑𝑑.(2.14) Therefore, by (2.14), we have 12πœŒπ‘€1(𝑒)+2πœŒπ‘€=ξ€œ(𝑣)𝑇121𝑀(𝑑,‖𝑒(𝑑)β€–)+2β‰₯ξ€œπ‘€(𝑑,‖𝑣(𝑑)β€–)𝑑𝑑𝐺1𝑀1𝑑,21‖𝑒(𝑑)β€–+2ξ‚ξ€œβ€–π‘£(𝑑)‖𝑑𝑑+𝑇⧡𝐺1𝑀1𝑑,21‖𝑒(𝑑)β€–+2>ξ€œβ€–π‘£(𝑑)‖𝑑𝑑𝐺1𝑀1𝑑,2ξ‚ξ€œβ€–π‘’(𝑑)+𝑣(𝑑)‖𝑑𝑑+𝑇⧡𝐺1𝑀1𝑑,2‖𝑒(𝑑)+𝑣(𝑑)‖𝑑𝑑=πœŒπ‘€ξ‚€π‘’+𝑣2.(2.15) By πœŒπ‘€(𝑒)=πœŒπ‘€(𝑣)=1, we have πœŒπ‘€((𝑒+𝑣)/2)<1. By Lemma 1.7, we have β€–(1/2)(𝑒+𝑣)β€–<1. This completes the proof.

Corollary 2.3. 𝐿𝑀(𝑅) is nonsquare space if and only if (a)π‘€βˆˆΞ”,(b)for any 𝑒,π‘£βˆˆπ‘†(𝐿𝑀(𝑅)), one has πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆ|𝑒(𝑑)|+|𝑣(𝑑)|>2𝑒(𝑑)}>0 or πœ‡({π‘‘βˆˆπ‘‡βˆΆ|𝑒(𝑑)|>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆ|𝑣(𝑑)|>𝜎(𝑑)})>0.

Theorem 2.4. Let 𝑒(𝑑)=0  μ-a.e on 𝑇. Then, 𝐿𝑀(𝑋) is nonsquare space if and only if (a)π‘€βˆˆΞ”,(b)πœŒπ‘€(𝜎)<2,(c)𝑋 is nonsquare space.

Proof. Necessity. By Theorem 2.1, (a) and (c) are obvious. Suppose that πœŒπ‘€(𝜎)β‰₯2. Then, there exists 𝐸∈Σ such that πœ‡πΈ>0, πœŒπ‘€(πœŽβ‹…πœ’π‘‡β§΅πΈ)=1 and πœŒπ‘€(πœŽβ‹…πœ’π·)=1, where π·βŠ‚π‘‡β§΅πΈ. Set 𝑒(𝑑)=π‘₯β‹…πœŽ(𝑑)β‹…πœ’π‘‡β§΅πΈ(𝑑),𝑣(𝑑)=π‘₯β‹…πœŽ(𝑑)β‹…πœ’π·(𝑑),(2.16) where π‘₯βˆˆπ‘†(𝑋). It is easy to see that ‖𝑒‖=‖𝑣‖=1, πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>2𝑒(𝑑)}=0 and πœ‡({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–>𝜎(𝑑)})=0. Contradicting Theorem 2.1.
Sufficiency. We only need to prove that for any 𝑒,π‘£βˆˆπ‘†(𝐿𝑀(𝑋)), if πœ‡(suppπ‘’βˆ©supp𝑣)=0, then πœ‡({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–>𝜎(𝑑)})>0. Suppose that there exist 𝑒,π‘£βˆˆπ‘†(𝐿𝑀(𝑋)) such that πœ‡(suppπ‘’βˆ©supp𝑣)=0, πœ‡({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–>𝜎(𝑑)}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–>𝜎(𝑑)})=0. By πœ‡(suppπ‘’βˆ©supp𝑣)=0, we have πœŒπ‘€(𝑒)+πœŒπ‘€(𝑣)=πœŒπ‘€(𝑒+𝑣)β‰€πœŒπ‘€(𝜎)<2. This implies πœŒπ‘€(𝑒)<1 or πœŒπ‘€(𝑣)<1. Hence, ‖𝑒‖<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 π‘’βˆˆπΏ0𝑀(𝑋)⧡{0}, one has 𝐾(𝑒)β‰ πœ™,(b)at least one of the conditions (b1)π‘˜π‘™/(π‘˜+𝑙)βˆ‰πΎ(𝑒+𝑣)∩𝐾(π‘’βˆ’π‘£), (b2)πœ‡({π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)>0})>0, (b3)πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆπ‘˜π‘™/(π‘˜+𝑙)(‖𝑒(𝑑)β€–+‖𝑣(𝑑)β€–)>𝑒(𝑑)}>0is true, where π‘˜βˆˆπΎ(𝑒), π‘™βˆˆπΎ(𝑣) and 𝑒,π‘£βˆˆπ‘†(𝐿0𝑀(𝑋)),(c)𝑋 is nonsquare space.

Proof. Necessity. (a) Suppose that there exists π‘’βˆˆπΏ0𝑀(𝑋)⧡{0} such that 𝐾(𝑒)=πœ™, then ‖𝑒‖0=βˆ«π‘‡π΄(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑 by Lemma 1.8. Decompose 𝑇 into disjoint sets 𝑇1 and 𝑇2 such that βˆ«π‘‡1𝐴(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑=βˆ«π‘‡2𝐴(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑. Put 𝑒1(𝑑)=2𝑒(𝑑)πœ’π‘‡1,𝑒2(𝑑)=2𝑒(𝑑)πœ’π‘‡2.(2.17) Obviously, 𝑒=(1/2)(𝑒1+𝑒2). Pick sequence {π‘˜π‘›}βˆžπ‘›=1βŠ‚π‘…+ such that π‘˜π‘›β†’βˆž as π‘›β†’βˆž. Let 𝑇0=supp𝑒1. By Levi theorem, we have ‖‖𝑒1β€–β€–0≀limπ‘›β†’βˆž1π‘˜π‘›ξ‚Έξ€œ1+π‘‡π‘€ξ€·β€–β€–π‘˜π‘‘,𝑛𝑒1β€–β€–ξ€Έξ‚Ή(𝑑)𝑑𝑑=limπ‘›β†’βˆžξ€œπ‘‡0π‘€ξ€·β€–β€–π‘˜π‘‘,𝑛𝑒1β€–β€–ξ€Έ(𝑑)β€–β€–π‘˜π‘›π‘’1‖‖‖‖𝑒(𝑑)1β€–β€–=ξ€œ(𝑑)𝑑𝑑𝑇0limπ‘›β†’βˆžπ‘€ξ€·β€–β€–π‘˜π‘‘,𝑛𝑒1(‖‖𝑑)β€–β€–π‘˜π‘›π‘’1‖‖‖‖𝑒(𝑑)1β€–β€–=ξ€œ(𝑑)𝑑𝑑𝑇‖‖𝑒𝐴(𝑑)1β€–β€–=ξ€œ(𝑑)𝑑𝑑𝑇1𝐴(𝑑)β€–2𝑒(𝑑)‖𝑑𝑑.(2.18) Therefore, ‖‖𝑒1β€–β€–0β‰€ξ€œπ‘‡1=ξ€œπ΄(𝑑)β‹…β€–2𝑒(𝑑)‖𝑑𝑑𝑇1ξ€œπ΄(𝑑)⋅‖𝑒(𝑑)‖𝑑t+𝑇1ξ€œπ΄(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑+𝑇2ξ€œπ΄(𝑑)⋅‖𝑒(𝑑)β€–π‘‘π‘‘βˆ’π‘‡2=ξ€œπ΄(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑𝑇1ξ€œπ΄(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑+𝑇2=ξ€œπ΄(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑𝑇𝐴(𝑑)⋅‖𝑒(𝑑)‖𝑑𝑑=‖𝑒‖0.(2.19) Similarly, we have ‖𝑒2β€–0≀‖𝑒‖0. By 𝑒=(1/2)(𝑒1+𝑒2), we obtain ‖𝑒‖0=‖𝑒1β€–0=‖𝑒2β€–0=β€–(1/2)(𝑒1+𝑒2)β€–0. By 𝑒1(𝑑)βˆ’π‘’2(𝑑)=2𝑒(𝑑)πœ’π‘‡1+(βˆ’2𝑒(𝑑))πœ’π‘‡2, we have ‖𝑒‖0=β€–(1/2)(𝑒1βˆ’π‘’2)β€–0. Therefore, ‖‖𝑒1β€–β€–0=‖‖𝑒2β€–β€–0=β€–β€–β€–12𝑒1+𝑒2ξ€Έβ€–β€–β€–0=β€–β€–β€–12𝑒1βˆ’π‘’2ξ€Έβ€–β€–β€–0=‖𝑒‖0.(2.20) This implies ‖‖𝑒1β€–β€–0+β€–β€–u2β€–β€–0‖‖𝑒=min1+𝑒2β€–β€–0,‖‖𝑒1βˆ’π‘’2β€–β€–0,(2.21) a contradiction!
If (b) is not true, then there exist 𝑒,π‘£βˆˆπ‘†(𝐿0𝑀(𝑋)) such that π‘˜π‘™/(π‘˜+𝑙)∈𝐾(𝑒+𝑣)∩𝐾(π‘’βˆ’π‘£), πœ‡({π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)>0})=0 and πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆπ‘˜π‘™/(π‘˜+𝑙)(‖𝑒(𝑑)β€–+‖𝑣(𝑑)β€–)>𝑒(𝑑)}=0, where π‘˜βˆˆπΎ(𝑒), π‘™βˆˆπΎ(𝑣). Let 𝐸=𝑇⧡(suppπ‘’βˆ©supp𝑣). It is easy to see that if π‘‘βˆˆπΈ, then ‖𝑒(𝑑)‖⋅‖𝑣(𝑑)β€–=0 on 𝐸. This implies 𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™β€–ξ‚ξ‚€π‘˜+𝑙‖𝑣(𝑑)=𝑀𝑑,π‘˜π‘™β€–ξ‚π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)π‘‘βˆˆπΈ.(2.22) Therefore, by (2.22), we have ξ€œπΈπ‘€ξ‚€π‘‘,π‘˜π‘™(π‘˜+𝑙‖𝑒𝑑)β€–+π‘˜π‘™(ξ‚ξ€œπ‘˜+𝑙‖𝑣𝑑)‖𝑑𝑑=𝐸𝑀𝑑,π‘˜π‘™(ξ‚π‘˜+𝑙‖𝑒𝑑)+𝑣(𝑑)‖𝑑𝑑.(2.23) By πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆ(π‘˜π‘™/(π‘˜+𝑙))(‖𝑒(𝑑)β€–+‖𝑣(𝑑)β€–)>𝑒(𝑑)}=0, we have ξ€œ0=𝑇⧡𝐸𝑀𝑑,π‘˜π‘™(π‘˜+𝑙‖𝑒t)β€–+π‘˜π‘™(=ξ€œπ‘˜+𝑙‖𝑣𝑑)‖𝑑𝑑𝑇⧡𝐸𝑀𝑑,π‘˜π‘™ξ‚π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)‖𝑑𝑑.(2.24) By (2.23) and (2.24), we have ξ€œπ‘‡π‘€ξ‚€π‘‘,π‘˜π‘™(π‘˜+𝑙‖𝑒𝑑)β€–+π‘˜π‘™(ξ‚ξ€œπ‘˜+𝑙‖𝑣𝑑)‖𝑑𝑑=𝑇𝑀𝑑,π‘˜π‘™(ξ‚π‘˜+𝑙‖𝑒𝑑)+𝑣(𝑑)‖𝑑𝑑.(2.25) By πœ‡({π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)>0})=0, we have ξ€œπ‘‡ξ‚€π‘™π‘˜π‘˜+𝑙𝑀(𝑑,π‘˜β€–π‘’(𝑑)β€–)+=ξ€œπ‘˜+𝑙𝑀(𝑑,𝑙‖𝑣(𝑑)β€–)𝑑𝑑𝑇𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™β€–ξ‚π‘˜+𝑙‖𝑣(𝑑)𝑑𝑑.(2.26) Therefore, by (2.25) and (2.26), we have ‖𝑒‖0+‖𝑣‖0=1π‘˜ξ€Ί1+πœŒπ‘€ξ€»+1(π‘˜π‘’)𝑙1+πœŒπ‘€ξ€»=(𝑙𝑣)π‘˜+π‘™ξ‚ƒπ‘™π‘˜π‘™1+πœŒπ‘˜+π‘™π‘€π‘˜(π‘˜π‘’)+πœŒπ‘˜+𝑙𝑀=(𝑙𝑣)π‘˜+π‘™ξ‚Έπ‘™π‘˜π‘™1+ξ€œπ‘˜+π‘™π‘‡π‘˜π‘€(𝑑,π‘˜β€–π‘’(𝑑)β€–)𝑑𝑑+ξ€œπ‘˜+𝑙𝑇=𝑀(𝑑,𝑙‖𝑣(𝑑)β€–)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+π‘‡ξ‚€π‘™β€–π‘˜π‘˜+𝑙𝑀(𝑑,π‘˜β€–π‘’(𝑑))+β€–)=π‘˜+𝑙𝑀(𝑑,𝑙‖𝑣(𝑑)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™ξ‚ξ‚Ή=π‘˜+𝑙‖𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™ξ‚ξ‚Ή=π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚ƒπ‘˜π‘™1+πœŒπ‘€ξ‚€π‘˜π‘™π‘˜+𝑙(𝑒+𝑣)=‖𝑒+𝑣‖0.(2.27) Similarly, we have ‖𝑒‖0+‖𝑣‖0=β€–π‘’βˆ’π‘£β€–0. This implies ‖𝑒‖0+‖𝑣‖0=‖𝑒+𝑣‖0=β€–π‘’βˆ’π‘£β€–0, a contradiction!
(c) Pick β„Ž(𝑑)βˆˆπ‘†(𝐿0𝑀(𝑋)), then there exists 𝑑>0 such that πœ‡πΈ>0, where 𝐸={π‘‘βˆˆπ‘‡βˆΆβ€–β„Ž(𝑑)β€–β‰₯𝑑}. Put β„Ž1(𝑑)=𝑑⋅π‘₯0β‹…πœ’πΈ(𝑑), where π‘₯0βˆˆπ‘†(𝑋). It is easy to see that β„Ž1(𝑑)∈𝐿0𝑀(𝑋)⧡{0}. Hence, there exists 𝑙>0 such that π‘™β‹…β„Ž1(𝑑)βˆˆπ‘†(𝐿0𝑀(𝑋)). The necessity of (c) follows from the fact that 𝑋 is isometrically embedded into 𝐿0𝑀(𝑋). Namely, defining the operator πΌβˆΆπ‘‹β†’πΏ0𝑀(𝑋) by 𝐼(π‘₯)=𝑙𝑑⋅π‘₯β‹…πœ’πΈ(𝑑),π‘₯βˆˆπ‘‹.(2.28) It is easy to see that 𝐼(π‘₯0)βˆˆπ‘†(𝐿0𝑀(𝑋)). Hence, for any π‘₯βˆˆπ‘‹β§΅{0}, we have ‖‖𝐼(π‘₯)0=infπ‘˜>01π‘˜ξ€Ί1+πœŒπ‘€ξ€»(π‘˜β‹…πΌ(π‘₯))=infπ‘˜>01π‘˜ξ‚Έξ€œ1+𝐸𝑀(𝑑,π‘˜β‹…π‘™π‘‘β€–π‘₯β€–)𝑑𝑑=infπ‘˜>01π‘˜ξ‚Έξ€œ1+𝐸𝑀‖‖π‘₯𝑑,π‘˜β‹…β€–π‘₯‖𝑙𝑑0‖‖𝑑𝑑=infπ‘˜>01π‘˜ξ€Ί1+πœŒπ‘€ξ€·ξ€·π‘₯π‘˜β‹…β€–π‘₯‖𝐼0=β€–β€–ξ€·π‘₯ξ€Έξ€Έξ€»β€–π‘₯‖⋅𝐼0ξ€Έβ€–β€–0‖‖𝐼π‘₯=β€–π‘₯β€–β‹…0ξ€Έβ€–β€–0=β€–π‘₯β€–.(2.29)
Sufficiency. Suppose that there exists 𝑒,π‘£βˆˆπ‘†(𝐿0𝑀(𝑋)) such that ‖𝑒‖0=‖𝑣‖0=β€–(1/2)(𝑒+𝑣)β€–0=β€–(1/2)(π‘’βˆ’π‘£)β€–0=1. Let π‘˜βˆˆπΎ(𝑒), π‘™βˆˆπΎ(𝑣). We will derive a contradiction for each of the following two cases.
Case 1. πœ‡[({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–β‰ 0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–β‰ 0})⧡{π‘‘βˆˆπ‘‡βˆΆπœŽ(𝑑)>0}]=0. By Lemma 2.2, we have ‖𝑒(𝑑)β€–+‖𝑣(𝑑)β€–>π‘šπ‘–π‘›{‖𝑒(𝑑)+𝑣(𝑑)β€–,‖𝑒(𝑑)βˆ’π‘£(𝑑)β€–}π‘‘βˆˆsuppπ‘’βˆ©supp𝑣. Put 𝑇1𝑇={π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>‖𝑒(𝑑)+𝑣(𝑑)β€–},2=β€–{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>‖𝑒(𝑑)βˆ’π‘£(𝑑)}.(2.30) Moreover, we have ‖𝑒‖0+‖𝑣‖0=1π‘˜ξ€Ί1+πœŒπ‘€ξ€»+1(π‘˜π‘’)𝑙1+πœŒπ‘€ξ€»=(𝑙𝑣)π‘˜+π‘™ξ‚Έπ‘™π‘˜π‘™1+ξ€œπ‘˜+π‘™π‘‡β€–π‘˜π‘€(𝑑,π‘˜β€–π‘’(𝑑))𝑑𝑑+ξ€œπ‘˜+𝑙𝑇‖=𝑀(𝑑,𝑙‖𝑣(𝑑))π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+π‘‡ξ‚€π‘™π‘˜π‘˜+𝑙𝑀(𝑑,π‘˜β€–π‘’(𝑑)β€–)+β‰₯π‘˜+𝑙𝑀(𝑑,𝑙‖𝑣(𝑑)β€–)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™ξ‚ξ‚Ήβ‰₯π‘˜+𝑙‖𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇1𝑀𝑑,π‘˜π‘™ξ‚ξ‚Ή+π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇⧡𝑇1𝑀𝑑,π‘˜π‘™ξ‚ξ‚Ή=π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚ƒπ‘˜π‘™1+πœŒπ‘€ξ‚€π‘˜π‘™β‰₯π‘˜+𝑙(𝑒+𝑣)‖𝑒+𝑣‖0.(2.31) By ‖𝑒‖0+‖𝑣‖0=‖𝑒+𝑣‖0, three inequalities of (2.31) are equation. This implies π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+π‘‡ξ‚€π‘™β€–π‘˜π‘˜+𝑙𝑀(𝑑,π‘˜β€–π‘’(𝑑))+β€–)=π‘˜+𝑙𝑀(𝑑,𝑙‖𝑣(𝑑)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™ξ‚ξ‚Ή.π‘˜+𝑙‖𝑣(𝑑)‖𝑑𝑑(2.32) Next, we will prove πœ‡({π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)>0})=0. Suppose that πœ‡({π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)>0})>0. Without loss of generality, we may assume πœ‡{π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}>0. Therefore, by (2.32), we have 𝑙‖𝑣(𝑑)β€–βˆ’πœŽ(𝑑)β‰₯0  μ-a.e. on {π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}. Hence, πœ‡({π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}∩{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)β‰₯0})>0. Since three inequalities of (2.31) are equation, we deduce ξ€œπ‘‡π‘€ξ‚€π‘‘,π‘˜π‘™(π‘˜+𝑙‖𝑒𝑑)β€–+π‘˜π‘™(ξ‚ξ€œπ‘˜+𝑙‖𝑣𝑑)‖𝑑𝑑=𝑇𝑀𝑑,π‘˜π‘™(ξ‚π‘˜+𝑙‖𝑒𝑑)+𝑣(𝑑)‖𝑑𝑑.(2.33) Moreover, it is easy to see π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™π‘˜+𝑙‖𝑣(𝑑)β€–>𝜎(𝑑)β‰₯𝑒(𝑑)(2.34) on {π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}∩{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)β‰₯0}. Therefore, by (2.33) and (2.34), we have π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™π‘˜+𝑙‖𝑣(𝑑)β€–=π‘˜π‘™β€–π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)(2.35)πœ‡-a.e. on {π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}∩{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)β‰₯0}. Since 𝑋 is nonsquare space, we have π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™ξ‚†π‘˜+𝑙‖𝑣(𝑑)β€–>π‘šπ‘–π‘›π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)β€–,π‘˜π‘™β€–ξ‚‡π‘˜+𝑙‖𝑒(𝑑)βˆ’π‘£(𝑑)(2.36) on {π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}∩{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)β‰₯0}. Thus, π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™π‘˜+𝑙‖𝑣(𝑑)β€–>π‘˜π‘™β€–π‘˜+𝑙‖𝑒(𝑑)βˆ’π‘£(𝑑)(2.37)πœ‡-a.e. on {π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}∩{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)β‰₯0}. Therefore, by (2.34) and (2.37), we have ‖𝑒‖0+‖𝑣‖0=1π‘˜ξ€Ί1+πœŒπ‘€ξ€»+1(π‘˜π‘’)𝑙1+πœŒπ‘€ξ€»β‰₯(𝑙𝑣)π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™β€–ξ‚ξ‚Ή>π‘˜+𝑙‖𝑣(𝑑)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™ξ‚ξ‚Ή=π‘˜+𝑙‖𝑒(𝑑)βˆ’π‘£(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚ƒπ‘˜π‘™1+πœŒπ‘€ξ‚€π‘˜π‘™π‘˜+𝑙(π‘’βˆ’π‘£)β‰₯β€–π‘’βˆ’π‘£β€–0.(2.38) This implies ‖𝑒‖0+‖𝑣‖0>β€–π‘’βˆ’π‘£β€–0, a contradiction! Hence, πœ‡({π‘‘βˆˆπ‘‡βˆΆπ‘˜β€–π‘’(𝑑)β€–βˆ’πœŽ(𝑑)>0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆπ‘™β€–π‘£(𝑑)β€–βˆ’πœŽ(𝑑)>0})=0. Since three inequalities of (2.31) are equation, we deduce π‘˜π‘™/(π‘˜+𝑙)∈𝐾(𝑒+𝑣) and πœ‡{π‘‘βˆˆπ‘‡1∢(π‘˜π‘™/(π‘˜+𝑙))(‖𝑒(𝑑)β€–+‖𝑣(𝑑)β€–)>𝑒(𝑑)}=0. By (b), we have π‘˜π‘™/(π‘˜+𝑙)βˆ‰πΎ(π‘’βˆ’π‘£) or πœ‡{π‘‘βˆˆπ‘‡2∢(π‘˜π‘™/(π‘˜+𝑙))(‖𝑒(𝑑)β€–+‖𝑣(𝑑)β€–)>𝑒(𝑑)}>0. If π‘˜π‘™/(π‘˜+𝑙)βˆ‰πΎ(π‘’βˆ’π‘£), then ‖𝑒‖0+‖𝑣‖0=1π‘˜ξ€Ί1+πœŒπ‘€ξ€»+1(π‘˜π‘’)𝑙1+πœŒπ‘€ξ€»β‰₯(𝑙𝑣)π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™β€–ξ‚ξ‚Ήβ‰₯π‘˜+𝑙‖𝑣(𝑑)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™ξ‚ξ‚Ή=π‘˜+𝑙‖𝑒(𝑑)βˆ’π‘£(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚ƒπ‘˜π‘™1+πœŒπ‘€ξ‚€π‘˜π‘™π‘˜+𝑙(π‘’βˆ’π‘£)>β€–π‘’βˆ’π‘£β€–0.(2.39) This implies ‖𝑒‖0+‖𝑣‖0>β€–π‘’βˆ’π‘£β€–0, a contradiction! If πœ‡{π‘‘βˆˆπ‘‡2∢(π‘˜π‘™/(π‘˜+𝑙))(‖𝑒(𝑑)β€–+‖𝑣(𝑑)β€–)>𝑒(𝑑)}>0, then ‖𝑒‖0+‖𝑣‖0=1π‘˜ξ€Ί1+πœŒπ‘€ξ€»+1(π‘˜π‘’)𝑙1+πœŒπ‘€ξ€»β‰₯(𝑙𝑣)π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™β€–ξ‚ξ‚Ή>π‘˜+𝑙‖𝑣(𝑑)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇2𝑀𝑑,π‘˜π‘™ξ‚ξ‚Ή+π‘˜+𝑙‖𝑒(𝑑)βˆ’π‘£(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇⧡𝑇2𝑀𝑑,π‘˜π‘™β€–ξ‚ξ‚Ή=π‘˜+𝑙‖𝑒(𝑑)βˆ’(𝑑)π‘‘π‘‘π‘˜+π‘™ξ‚ƒπ‘˜π‘™1+πœŒπ‘€ξ‚€π‘˜π‘™π‘˜+𝑙(π‘’βˆ’π‘£)β‰₯β€–π‘’βˆ’π‘£β€–0.(2.40) This implies ‖𝑒‖0+‖𝑣‖0>β€–π‘’βˆ’π‘£β€–0, a contradiction!Case 2. πœ‡[({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–β‰ 0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–β‰ 0})⧡{π‘‘βˆˆπ‘‡βˆΆπœŽ(𝑑)>0}]>0. Let 𝐸=({π‘‘βˆˆπ‘‡βˆΆβ€–π‘’(𝑑)β€–β‰ 0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆβ€–π‘£(𝑑)β€–β‰ 0})⧡{π‘‘βˆˆπ‘‡βˆΆπœŽ(𝑑)>0}. Put 𝐸1={π‘‘βˆˆπΈβˆΆβ€–π‘’(𝑑)‖⋅‖𝑣(𝑑)β€–=0},𝐸2={π‘‘βˆˆπΈβˆΆβ€–π‘’(𝑑)‖⋅‖𝑣(𝑑)β€–>0}.(2.41) Then, πœ‡πΈ1>0 or πœ‡πΈ2>0. If πœ‡πΈ1>0, then ξ€œπΈ1ξ‚€π‘™π‘˜π‘˜+𝑙𝑀(𝑑,π‘˜β€–π‘’(𝑑)β€–)𝑑𝑑+>ξ€œπ‘˜+𝑙𝑀(𝑑,𝑙‖𝑣(𝑑)β€–)𝑑𝑑𝐸1𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™ξ‚π‘˜+𝑙‖𝑣(𝑑)‖𝑑𝑑.(2.42) Therefore, by (2.42), we have ‖𝑒‖0+‖𝑣‖0=1π‘˜ξ€Ί1+πœŒπ‘€ξ€»+1(π‘˜π‘’)𝑙1+πœŒπ‘€ξ€»=(𝑙𝑣)π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+π‘‡ξ‚€π‘™β€–π‘˜π‘˜+𝑙𝑀(𝑑,π‘˜β€–π‘’(𝑑))+β€–)>π‘˜+𝑙𝑀(𝑑,𝑙‖𝑣(𝑑)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝐸1𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™ξ‚ξ‚Ή+π‘˜+𝑙‖𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇⧡𝐸1𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™β€–ξ‚ξ‚Ήβ‰₯π‘˜+𝑙‖𝑣(𝑑)π‘‘π‘‘π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™(=π‘˜+𝑙‖𝑒𝑑)+𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚ƒπ‘˜π‘™1+πœŒπ‘€ξ‚€π‘˜π‘™π‘˜+𝑙(𝑒+𝑣)β‰₯‖𝑒+𝑣‖0.(2.43) This implies ‖𝑒‖0+‖𝑣‖0>‖𝑒+𝑣‖0, a contradiction! If πœ‡πΈ2>0, then πœ‡πΈ12>0 or πœ‡E22>0 by Lemma 2.2, where 𝐸12=ξ€½π‘‘βˆˆπΈ2ξ€Ύ,πΈβˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>‖𝑒(𝑑)+𝑣(𝑑)β€–22=ξ€½π‘‘βˆˆπΈ2β€–ξ€Ύ.βˆΆβ€–π‘’(𝑑)β€–+‖𝑣(𝑑)β€–>‖𝑒(𝑑)βˆ’π‘£(𝑑)(2.44) Without loss of generality, we may assume πœ‡πΈ12>0. Hence, ξ€œπΈ12𝑀𝑑,π‘˜π‘™(π‘˜+𝑙‖𝑒𝑑)β€–+π‘˜π‘™(ξ‚ξ€œπ‘˜+𝑙‖𝑣𝑑)‖𝑑𝑑>𝐸12𝑀𝑑,π‘˜π‘™(ξ‚π‘˜+𝑙‖𝑒𝑑)+𝑣(𝑑)‖𝑑𝑑.(2.45) Therefore, by (2.45), we have ‖𝑒‖0+‖𝑣‖0=1π‘˜ξ€Ί1+πœŒπ‘€ξ€»+1(π‘˜π‘’)𝑙1+πœŒπ‘€ξ€»β‰₯(𝑙𝑣)π‘˜+π‘™ξ‚Έξ€œπ‘˜π‘™1+𝑇𝑀𝑑,π‘˜π‘™π‘˜+𝑙‖𝑒(𝑑)β€–+π‘˜π‘™β€–ξ‚ξ‚Ή>π‘˜+𝑙‖𝑣(𝑑)π‘‘π‘‘π‘˜+π‘™ξƒ¬ξ€œπ‘˜π‘™1+𝐸12𝑀𝑑,π‘˜π‘™ξ‚ξƒ­+π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξƒ¬ξ€œπ‘˜π‘™1+𝑇⧡𝐸12𝑀𝑑,π‘˜π‘™ξ‚ξƒ­=π‘˜+𝑙‖𝑒(𝑑)+𝑣(𝑑)β€–π‘‘π‘‘π‘˜+π‘™ξ‚ƒπ‘˜π‘™1+πœŒπ‘€ξ‚€π‘˜π‘™π‘˜+𝑙(𝑒+𝑣)β‰₯‖𝑒+𝑣‖0.(2.46) This implies ‖𝑒‖0+‖𝑣‖0>‖𝑒+𝑣‖0, a contradiction! This completes the proof.

Corollary 2.6. 𝐿0𝑀(𝑅) is nonsquare space if and only if (a)for any π‘’βˆˆπΏ0𝑀(𝑅), one has 𝐾(𝑒)β‰ πœ™,(b)at least one of the conditions (b1)π‘˜π‘™/(π‘˜+𝑙)βˆ‰πΎ(𝑒+𝑣)∩𝐾(π‘’βˆ’π‘£), (b2)πœ‡({π‘‘βˆˆπ‘‡βˆΆπ‘˜|𝑒(𝑑)|βˆ’πœŽ(𝑑)>0}βˆͺ{π‘‘βˆˆπ‘‡βˆΆπ‘™|𝑣(𝑑)|βˆ’πœŽ(𝑑)>0})>0,(b3)πœ‡{π‘‘βˆˆsuppπ‘’βˆ©suppπ‘£βˆΆ(π‘˜π‘™/(π‘˜+𝑙))(|𝑒(𝑑)|+|𝑣(𝑑)|)>𝑒(𝑑)}>0is true, where π‘˜βˆˆπΎ(𝑒), π‘™βˆˆπΎ(𝑣) and 𝑒,π‘£βˆˆπ‘†(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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Copyright Β© 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.


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