# Nonsquareness in Musielak-Orlicz-Bochner Function Spaces

Authors: Shaoqiang Shang; Yunan Cui; Yongqiang Fu
Journal: Abstract and Applied Analysis
 (2011)

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## 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.


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## Body

## 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,∥·∥) 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,y∈S(X) we have min{∥(1/2)(x+y)∥,∥(1/2)(x-y)∥}<1. A Banach space X is said to be uniformly nonsquare space if for any x,y∈S(X), there exists δ>0 such that min{∥(1/2)(x+y)∥,∥(1/2)(x-y)∥}<1-δ.Let(T,∑,μ) be nonatomic measure space. Suppose that a function M:T×[0,∞)→[0,∞] satisfies the following conditions:(1)
forμ-a.e, t∈T, M(t,0)=0, limu→∞M(t,u)=∞ and M(t,u′)<∞ for some u′>0,(2)
forμ-a.e, t∈T, M(t,u) is convex on [0,∞) with respect to u,(3)
for eachu∈[0,∞), M(t,u) is a μ-measurable function of t on T.Lete(t)=sup{u>0:M(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 u∈XT, define the modular of u by (1.1)ρM(u)=∫TM(t,‖u(t)‖)dt.
Put(1.2)LM(X)={u∈XT:ρM(λu)<∞forsomeλ>0}.
It is well known that Musielak-Orlicz-Bochner function space LM(X) is Banach spaces equipped with the Luxemburg norm(1.3)‖u‖=inf⁡{λ>0:ρM(uλ)≤1}
or Orlicz’s norm(1.4)‖u‖0=inf⁡k>01k[1+ρM(ku)].
In particular, LM(R) and LM0(R) are said to be Musielak-Orlicz function space. Set(1.5)supp⁡u={t∈T:‖u(t)‖≠0},K(u)={k>0:1k(1+ρM(ku))=‖u‖0}.
In particular, the set K(u) can be nonempty. To show that, we give a proposition.Proposition 1.1 (see [9]).
Iflimu→∞(M(t,u)/u)=∞  μ-a.e.   t∈T, then  K(u)≠ϕ for any  u∈LM0(X).We define a function(1.6)σ(t)=sup⁡{u≥0:M(t,12u)=12M(t,u)}.
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.
Functionσ(t) is  μ-measurable.Proof.
Pick a dense set{ri}i=1∞ in [0,∞) and set
(1.7)Bk={t∈T:M(t,12rk)=12M(t,rk)},qk(t)=rkχBk(t)(k∈N).
It is easy to see that for all k∈N, σ(t)≥qk(t)  μ-a.e on T. Hence, supk≥1qk(t)≤σ(t). For μ-a.e t∈T, arbitrarily choose ɛ∈(0,σ(t)). Then, there exists rk∈(σ(t)-ɛ,σ(t)) such that M(t,(1/2)rk)=(1/2)M(t,rk), that is, qk(t)≥rk>σ(t)-ɛ. Since ɛ is arbitrary, we find supk≥1qk(t)≥σ(t). Thus, supk≥1qk(t)=σ(t).It is easy to prove the following proposition.Proposition 1.3.
For anyα∈(0,1), if  u(t)≤σ(t), then  M(t,αu(t))=αM(t,u(t)).Proposition 1.4.
For anyα∈(0,1), if  u(t)<σ(t)<v(t), then  M(t,αu(t)+(1-α)v(t))<αM(t,u(t))+(1-α)M(t,v(t)).Proposition 1.5.
For anyα∈(0,1), if  σ(t)<v(t), then  M(t,αv(t))<αM(t,v(t)).Definition 1.6 (see [2]).
We say thatM(t,u) satisfies condition Δ(M∈Δ) if there exist K≥1 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 t∈T and all u≥δ(t).First, we give some results that will used in the further part of the paper.Lemma 1.7 (see [2]).
SupposeM∈Δ. Then ρM(u)=1⇔∥u∥=1.Lemma 1.8 (see [9]).
LetLM0(X) be Musielak-Orlicz-Bochner function spaces, then, if K(u)=ϕ, one has ∥u∥0=∫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)
M∈Δ,(b)
for anyu,v∈S(LM(X)), one has μ{t∈suppu∩suppv:∥u(t)∥+∥v(t)∥>2e(t)}>0 or μ({t∈T:∥u(t)∥>σ(t)}∪{t∈T:∥v(t)∥>σ(t)})>0,(c)
X is nonsquare space.In order to prove the theorem, we give a lemma.Lemma 2.2.
LetX be nonsquare space, then for any x,y≠0, one has
(2.1)‖x‖+‖y‖-min⁡{‖x+y‖,‖x-y‖}>0.Proof.
For anyx,y≠0, without loss of generality, we may assume ∥x∥≤∥y∥. Since X is nonsquare space, we have
(2.2)‖x‖+‖x‖=‖x‖+‖‖x‖‖y‖y‖>min{‖x+‖x‖‖y‖y‖,‖x-‖x‖‖y‖y‖}.
Therefore, by (2.2), we obtain
(2.3)‖x+y‖≤‖x+‖x‖‖y‖⋅y‖+(1-‖x‖‖y‖)⋅‖y‖<‖x‖+‖x‖+‖y‖-‖x‖=‖x‖+‖y‖
or
(2.4)‖x-y‖≤‖x-‖x‖‖y‖⋅y‖+(1-‖x‖‖y‖)⋅‖y‖<‖x‖+‖x‖+‖y‖-‖x‖=‖x‖+‖y‖.
This implies ∥x∥+∥y∥-min{∥x+y∥,∥x-y∥}>0. This completes the proof.Proof of Theorem2.1.
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 M∈Δ which follows from the theorem proved in more general case (see [10, 11]). Namely, if M∉Δ, then LM(R) contains isometric copy of l∞.
If (b) is not true, then there existu,v∈S(LM(X)) such that μ{t∈suppu∩suppv:∥u(t)∥+∥v(t)∥>2e(t)}=0 and μ({t∈T:∥u(t)∥>σ(t)}∪{t∈T:∥v(t)∥)>σ(t)})=0. Let G=suppu∩suppv. We have
(2.5)12ρM(u)+12ρM(v)=12∫TM(t,‖u(t)‖)dt+12∫TM(t,‖v(t)‖)dt=∫T12M(t,‖u(t)‖)+12M(t,‖v(t)‖)dt≥∫GM(t,12‖u(t)‖+12‖v(t)‖)dt+∫T∖GM(t,12‖u(t)‖+12‖v(t)‖)dt≥∫GM(t,12‖u(t)+v(t)‖)dt+∫T∖GM(t,12‖u(t)+v(t)‖)dt=ρM(u+v2).
By μ{t∈suppu∩suppv:∥u(t)∥+∥v(t)∥>2e(t)}=0 and μ({t∈T:∥u(t)∥>σ(t)}∪{t∈T:∥v(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)(u-v)∥=1, a contradiction!
(c) Pickh(t)∈S(LM(X)), then there exists d>0 such that μE>0, where E={t∈T:∥h(t)∥≥d}. Put h1(t)=d·x0·χE(t), where x0∈S(X). It is easy to see that h1(t)∈LM(X)∖{0}. Hence, there exists k>0 such that k·h1(t)∈S(LM(X)). By Lemma 1.7, we have
(2.6)1=∫TM(t,‖k⋅h1(t)‖)dt=∫EM(t,‖k⋅d⋅x0‖)dt.
Let α=k·d. 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 I:X→LM(X) by
(2.7)I(x)=α⋅x⋅χE(t),x∈X.
Hence, for any x∈X∖{0}, we have
(2.8)ρM(I(x)‖x‖)=∫EM(t,‖I(x)‖x‖‖)dt=∫EM(t,α‖x‖‖x‖)dt=∫EM(t,α)dt=1.
By Lemma 1.7, we have ∥I(x)/∥x∥∥LM(X)=1, hence ∥I(x)∥LM(X)=∥x∥.
Sufficiency. The proof requires the consideration of two cases separately.
Case 1.
μ({t∈T:∥u(t)∥>σ(t)}∪{t∈T:∥v(t)∥>σ(t)})>0. Without loss of generality, we may assume μ{t∈T:∥u(t)∥>σ(t)}>0. Let F={t∈T:∥u(t)∥>σ(t)}. Put
(2.9)F1={t∈F:‖u(t)‖+‖v(t)‖>‖u(t)+v(t)‖},F2={t∈F:‖u(t)‖+‖v(t)‖>‖u(t)-v(t)‖},F3={t∈F:‖v(t)‖=0}.
Since X is nonsquare space, we have μ(F1∪F3)>0 or μ(F2∪F3)>0 by Lemma 2.2. Without loss of generality, we may assume μ(F1∪F3)>0. Moreover, we have
(2.10)12ρM(u)+12ρM(v)-ρM(12(u+v))=12∫TM(t,‖u(t)‖)dt+12∫TM(t,‖v(t)‖)dt-∫TM(t,12‖u(t)+v(t)‖)dt=∫T[12M(t,‖u(t)‖)+12M(t,‖v(t)‖)-M(t,12‖u(t)+v(t)‖)]dt≥∫F1∪F3[12M(t,‖u(t)‖)+12M(t,‖v(t)‖)-M(t,12‖u(t)+v(t)‖)]dt≥∫F1∪F3M(t,12‖u(t)‖+12‖v(t)‖)-M(t,12‖u(t)+v(t)‖)dt≥0.
Let E={t∈F1∪F3:∥v(t)∥≥σ(t)}, E1={t∈E:∥v(t)∥=σ(t)=0}. Then, μ((F1∪F3)∖(E∖E1))>0 or μ(E∖E1)>0. By F={t∈T:∥u(t)∥>σ(t)}⊃(F1∪F3)⊃((F1∪F3)∖(E∖E1)), we obtain
(2.11)∫F1∪F3[12M(t,‖u(t)‖)+12M(t,‖v(t)‖)-M(t,12‖u(t)+v(t)‖)]dt>∫F1∪F3M(t,12‖u(t)‖+12‖v(t)‖)-M(t,12‖u(t)+v(t)‖)dt,
whenever μ((F1∪F3)∖(E∖E1))>0. By F={t∈T:∥u(t)∥>σ(t)}⊃(F1∪F3)⊃(E∖E1), we obtain
(2.12)∫F1∪F3M(t,12‖u(t)‖+12‖v(t)‖)-M(t,12‖u(t)+v(t)‖)dt>0,
whenever μ(E∖E1)>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({t∈T:∥u(t)∥>σ(t)}∪{t∈T:∥v(t)∥)>σ(t)})=0. By (b), we have μ{t∈suppu∩suppv:∥u(t)∥+∥v(t)∥>2e(t)}>0. Let G={t∈suppu∩suppv:∥u(t)∥+∥v(t)∥>2e(t)}. Put
(2.13)G1={t∈G:‖u(t)‖+‖v(t)‖>‖u(t)+v(t)‖},G2={t∈G:‖u(t)‖+‖v(t)‖>‖u(t)-v(t)‖}.
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,
(2.14)∫G1M(t,12‖u(t)‖+12‖v(t)‖)dt>∫G1M(t,12‖u(t)+v(t)‖)dt.
Therefore, by (2.14), we have
(2.15)12ρM(u)+12ρM(v)=∫T12M(t,‖u(t)‖)+12M(t,‖v(t)‖)dt≥∫G1M(t,12‖u(t)‖+12‖v(t)‖)dt+∫T∖G1M(t,12‖u(t)‖+12‖v(t)‖)dt>∫G1M(t,12‖u(t)+v(t)‖)dt+∫T∖G1M(t,12‖u(t)+v(t)‖)dt=ρM(u+v2).
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)
M∈Δ,(b)
for anyu,v∈S(LM(R)), one has μ{t∈suppu∩suppv:|u(t)|+|v(t)|>2e(t)}>0 or μ({t∈T:|u(t)|>σ(t)}∪{t∈T:|v(t)|>σ(t)})>0.Theorem 2.4.
Lete(t)=0  μ-a.e on T. Then, LM(X) is nonsquare space if and only if (a)
M∈Δ,(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 E∈Σ such that μE>0, ρM(σ·χT∖E)=1 and ρM(σ·χD)=1, where D⊂T∖E. Set
(2.16)u(t)=x⋅σ(t)⋅χT∖E(t),v(t)=x⋅σ(t)⋅χD(t),
where x∈S(X). It is easy to see that ∥u∥=∥v∥=1, μ{t∈suppu∩suppv:∥u(t)∥+∥v(t)∥>2e(t)}=0 and μ({t∈T:∥u(t)∥>σ(t)}∪{t∈T:∥v(t)∥>σ(t)})=0. Contradicting Theorem 2.1.
Sufficiency. We only need to prove that for any u,v∈S(LM(X)), if μ(suppu∩suppv)=0, then μ({t∈T:∥u(t)∥>σ(t)}∪{t∈T:∥v(t)∥>σ(t)})>0. Suppose that there exist u,v∈S(LM(X)) such that μ(suppu∩suppv)=0, μ({t∈T:∥u(t)∥>σ(t)}∪{t∈T:∥v(t)∥>σ(t)})=0. By μ(suppu∩suppv)=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.
LM0(X) is nonsquare space if and only if (a)
for anyu∈LM0(X)∖{0}, one has K(u)≠ϕ,(b)
at least one of the conditions
(b1)
kl/(k+l)∉K(u+v)∩K(u-v),(b2)
μ({t∈T:k∥u(t)∥-σ(t)>0}∪{t∈T:l∥v(t)∥-σ(t)>0})>0,(b3)
μ{t∈suppu∩suppv:kl/(k+l)(∥u(t)∥+∥v(t)∥)>e(t)}>0is true, wherek∈K(u), l∈K(v) and u,v∈S(LM0(X)),(c)
X is nonsquare space.Proof.
Necessity. (a) Suppose that there exists u∈LM0(X)∖{0} such that K(u)=ϕ, then ∥u∥0=∫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
(2.17)u1(t)=2u(t)χT1,u2(t)=2u(t)χT2.
Obviously, u=(1/2)(u1+u2). Pick sequence {kn}n=1∞⊂R+ such that kn→∞ as n→∞. Let T0=suppu1. By Levi theorem, we have
(2.18)‖u1‖0≤lim⁡n→∞1kn[1+∫TM(t,‖knu1(t)‖)dt]=lim⁡n→∞∫T0M(t,‖knu1(t)‖)‖knu1(t)‖‖u1(t)‖dt=∫T0lim⁡n→∞M(t,‖knu1(t)‖)‖knu1(t)‖‖u1(t)‖dt=∫TA(t)‖u1(t)‖dt=∫T1A(t)‖2u(t)‖dt.
Therefore,
(2.19)‖u1‖0≤∫T1A(t)⋅‖2u(t)‖dt=∫T1A(t)⋅‖u(t)‖dt+∫T1A(t)⋅‖u(t)‖dt+∫T2A(t)⋅‖u(t)‖dt-∫T2A(t)⋅‖u(t)‖dt=∫T1A(t)⋅‖u(t)‖dt+∫T2A(t)⋅‖u(t)‖dt=∫TA(t)⋅‖u(t)‖dt=‖u‖0.
Similarly, we have ∥u2∥0≤∥u∥0. By u=(1/2)(u1+u2), we obtain ∥u∥0=∥u1∥0=∥u2∥0=∥(1/2)(u1+u2)∥0. By u1(t)-u2(t)=2u(t)χT1+(-2u(t))χT2, we have ∥u∥0=∥(1/2)(u1-u2)∥0. Therefore,
(2.20)‖u1‖0=‖u2‖0=‖12(u1+u2)‖0=‖12(u1-u2)‖0=‖u‖0.
This implies
(2.21)‖u1‖0+‖u2‖0=min⁡{‖u1+u2‖0,‖u1-u2‖0},
a contradiction!
If (b) is not true, then there existu,v∈S(LM0(X)) such that kl/(k+l)∈K(u+v)∩K(u-v), μ({t∈T:k∥u(t)∥-σ(t)>0}∪{t∈T:l∥v(t)∥-σ(t)>0})=0 and μ{t∈suppu∩suppv:kl/(k+l)(∥u(t)∥+∥v(t)∥)>e(t)}=0, where k∈K(u), l∈K(v). Let E=T∖(suppu∩suppv). It is easy to see that if t∈E, then ∥u(t)∥·∥v(t)∥=0 on E. This implies
(2.22)M(t,klk+l‖u(t)‖+klk+l‖v(t)‖)=M(t,klk+l‖u(t)+v(t)‖)t∈E.
Therefore, by (2.22), we have
(2.23)∫EM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt=∫EM(t,klk+l‖u(t)+v(t)‖)dt.
By μ{t∈suppu∩suppv:(kl/(k+l))(∥u(t)∥+∥v(t)∥)>e(t)}=0, we have
(2.24)0=∫T∖EM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt=∫T∖EM(t,klk+l‖u(t)+v(t)‖)dt.
By (2.23) and (2.24), we have
(2.25)∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt=∫TM(t,klk+l‖u(t)+v(t)‖)dt.
By μ({t∈T:k∥u(t)∥-σ(t)>0}∪{t∈T:l∥v(t)∥-σ(t)>0})=0, we have
(2.26)∫T(lk+lM(t,k‖u(t)‖)+kk+lM(t,l‖v(t)‖))dt=∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt.
Therefore, by (2.25) and (2.26), we have
(2.27)‖u‖0+‖v‖0=1k[1+ρM(ku)]+1l[1+ρM(lv)]=k+lkl[1+lk+lρM(ku)+kk+lρM(lv)]=k+lkl[1+lk+l∫TM(t,k‖u(t)‖)dt+kk+l∫TM(t,l‖v(t)‖)dt]=k+lkl[1+∫T(lk+lM(t,k‖u(t)‖)+kk+lM(t,l‖v(t)‖))dt]=k+lkl[1+∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt]=k+lkl[1+∫TM(t,klk+l‖u(t)+v(t)‖)dt]=k+lkl[1+ρM(klk+l(u+v))]=‖u+v‖0.
Similarly, we have ∥u∥0+∥v∥0=∥u-v∥0. This implies ∥u∥0+∥v∥0=∥u+v∥0=∥u-v∥0, a contradiction!
(c) Pickh(t)∈S(LM0(X)), then there exists d>0 such that μE>0, where E={t∈T:∥h(t)∥≥d}. Put h1(t)=d·x0·χE(t), where x0∈S(X). It is easy to see that h1(t)∈LM0(X)∖{0}. Hence, there exists l>0 such that l·h1(t)∈S(LM0(X)). The necessity of (c) follows from the fact that X is isometrically embedded into LM0(X). Namely, defining the operator I:X→LM0(X) by
(2.28)I(x)=ld⋅x⋅χE(t),x∈X.
It is easy to see that I(x0)∈S(LM0(X)). Hence, for any x∈X∖{0}, we have
(2.29)‖I(x)‖0=inf⁡k>01k[1+ρM(k⋅I(x))]=inf⁡k>01k[1+∫EM(t,k⋅ld‖x‖)dt]=inf⁡k>01k[1+∫EM(t,k⋅‖x‖ld‖x0‖)dt]=inf⁡k>01k[1+ρM(k⋅‖x‖I(x0))]=‖‖x‖⋅I(x0)‖0=‖x‖⋅‖I(x0)‖0=‖x‖.
Sufficiency.Suppose that there exists u,v∈S(LM0(X)) such that ∥u∥0=∥v∥0=∥(1/2)(u+v)∥0=∥(1/2)(u-v)∥0=1. Let k∈K(u), l∈K(v). We will derive a contradiction for each of the following two cases.
Case 1.
μ[({t∈T:∥u(t)∥≠0}∪{t∈T:∥v(t)∥≠0})∖{t∈T:σ(t)>0}]=0. By Lemma 2.2, we have ∥u(t)∥+∥v(t)∥>min{∥u(t)+v(t)∥,∥u(t)-v(t)∥}t∈suppu∩suppv. Put
(2.30)T1={t∈supp⁡u∩supp⁡v:‖u(t)‖+‖v(t)‖>‖u(t)+v(t)‖},T2={t∈supp⁡u∩supp⁡v:‖u(t)‖+‖v(t)‖>‖u(t)-v(t)‖}.
Moreover, we have
(2.31)‖u‖0+‖v‖0=1k[1+ρM(ku)]+1l[1+ρM(lv)]=k+lkl[1+lk+l∫TM(t,k‖u(t)‖)dt+kk+l∫TM(t,l‖v(t)‖)dt]=k+lkl[1+∫T(lk+lM(t,k‖u(t)‖)+kk+lM(t,l‖v(t)‖))dt]≥k+lkl[1+∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt]≥k+lkl[1+∫T1M(t,klk+l‖u(t)+v(t)‖)dt]+k+lkl[1+∫T∖T1M(t,klk+l‖u(t)+v(t)‖)dt]=k+lkl[1+ρM(klk+l(u+v))]≥‖u+v‖0.
By ∥u∥0+∥v∥0=∥u+v∥0, three inequalities of (2.31) are equation. This implies
(2.32)k+lkl[1+∫T(lk+lM(t,k‖u(t)‖)+kk+lM(t,l‖v(t)‖))dt]=k+lkl[1+∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt].
Next, we will prove μ({t∈T:k∥u(t)∥-σ(t)>0}∪{t∈T:l∥v(t)∥-σ(t)>0})=0. Suppose that μ({t∈T:k∥u(t)∥-σ(t)>0}∪{t∈T:l∥v(t)∥-σ(t)>0})>0. Without loss of generality, we may assume μ{t∈T:k∥u(t)∥-σ(t)>0}>0. Therefore, by (2.32), we have l∥v(t)∥-σ(t)≥0  μ-a.e. on {t∈T:k∥u(t)∥-σ(t)>0}. Hence, μ({t∈T:k∥u(t)∥-σ(t)>0}∩{t∈T:l∥v(t)∥-σ(t)≥0})>0. Since three inequalities of (2.31) are equation, we deduce
(2.33)∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt=∫TM(t,klk+l‖u(t)+v(t)‖)dt.
Moreover, it is easy to see
(2.34)klk+l‖u(t)‖+klk+l‖v(t)‖>σ(t)≥e(t)
on {t∈T:k∥u(t)∥-σ(t)>0}∩{t∈T:l∥v(t)∥-σ(t)≥0}. Therefore, by (2.33) and (2.34), we have
(2.35)klk+l‖u(t)‖+klk+l‖v(t)‖=klk+l‖u(t)+v(t)‖μ-a.e. on {t∈T:k∥u(t)∥-σ(t)>0}∩{t∈T:l∥v(t)∥-σ(t)≥0}. Since X is nonsquare space, we have
(2.36)klk+l‖u(t)‖+klk+l‖v(t)‖>min{klk+l‖u(t)+v(t)‖,klk+l‖u(t)-v(t)‖}
on {t∈T:k∥u(t)∥-σ(t)>0}∩{t∈T:l∥v(t)∥-σ(t)≥0}. Thus,
(2.37)klk+l‖u(t)‖+klk+l‖v(t)‖>klk+l‖u(t)-v(t)‖μ-a.e. on {t∈T:k∥u(t)∥-σ(t)>0}∩{t∈T:l∥v(t)∥-σ(t)≥0}. Therefore, by (2.34) and (2.37), we have
(2.38)‖u‖0+‖v‖0=1k[1+ρM(ku)]+1l[1+ρM(lv)]≥k+lkl[1+∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt]>k+lkl[1+∫TM(t,klk+l‖u(t)-v(t)‖)dt]=k+lkl[1+ρM(klk+l(u-v))]≥‖u-v‖0.
This implies ∥u∥0+∥v∥0>∥u-v∥0, a contradiction! Hence, μ({t∈T:k∥u(t)∥-σ(t)>0}∪{t∈T:l∥v(t)∥-σ(t)>0})=0. Since three inequalities of (2.31) are equation, we deduce kl/(k+l)∈K(u+v) and μ{t∈T1:(kl/(k+l))(∥u(t)∥+∥v(t)∥)>e(t)}=0. By (b), we have kl/(k+l)∉K(u-v) or μ{t∈T2:(kl/(k+l))(∥u(t)∥+∥v(t)∥)>e(t)}>0. If kl/(k+l)∉K(u-v), then
(2.39)‖u‖0+‖v‖0=1k[1+ρM(ku)]+1l[1+ρM(lv)]≥k+lkl[1+∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt]≥k+lkl[1+∫TM(t,klk+l‖u(t)-v(t)‖)dt]=k+lkl[1+ρM(klk+l(u-v))]>‖u-v‖0.
This implies ∥u∥0+∥v∥0>∥u-v∥0, a contradiction! If μ{t∈T2:(kl/(k+l))(∥u(t)∥+∥v(t)∥)>e(t)}>0, then
(2.40)‖u‖0+‖v‖0=1k[1+ρM(ku)]+1l[1+ρM(lv)]≥k+lkl[1+∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt]>k+lkl[1+∫T2M(t,klk+l‖u(t)-v(t)‖)dt]+k+lkl[1+∫T∖T2M(t,klk+l‖u(t)-(t)‖)dt]=k+lkl[1+ρM(klk+l(u-v))]≥‖u-v‖0.
This implies ∥u∥0+∥v∥0>∥u-v∥0, a contradiction!Case 2.
μ[({t∈T:∥u(t)∥≠0}∪{t∈T:∥v(t)∥≠0})∖{t∈T:σ(t)>0}]>0. Let E=({t∈T:∥u(t)∥≠0}∪{t∈T:∥v(t)∥≠0})∖{t∈T:σ(t)>0}. Put
(2.41)E1={t∈E:‖u(t)‖⋅‖v(t)‖=0},E2={t∈E:‖u(t)‖⋅‖v(t)‖>0}.
Then, μE1>0 or μE2>0. If μE1>0, then
(2.42)∫E1(lk+lM(t,k‖u(t)‖)dt+kk+lM(t,l‖v(t)‖))dt>∫E1M(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt.
Therefore, by (2.42), we have
(2.43)‖u‖0+‖v‖0=1k[1+ρM(ku)]+1l[1+ρM(lv)]=k+lkl[1+∫T(lk+lM(t,k‖u(t)‖)+kk+lM(t,l‖v(t)‖))dt]>k+lkl[1+∫E1M(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt]+k+lkl[1+∫T∖E1M(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt]≥k+lkl[1+∫TM(t,klk+l‖u(t)+v(t)‖)dt]=k+lkl[1+ρM(klk+l(u+v))]≥‖u+v‖0.
This implies ∥u∥0+∥v∥0>∥u+v∥0, a contradiction! If μE2>0, then μE21>0 or μE22>0 by Lemma 2.2, where
(2.44)E21={t∈E2:‖u(t)‖+‖v(t)‖>‖u(t)+v(t)‖},E22={t∈E2:‖u(t)‖+‖v(t)‖>‖u(t)-v(t)‖}.
Without loss of generality, we may assume μE21>0. Hence,
(2.45)∫E21M(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt>∫E21M(t,klk+l‖u(t)+v(t)‖)dt.
Therefore, by (2.45), we have
(2.46)‖u‖0+‖v‖0=1k[1+ρM(ku)]+1l[1+ρM(lv)]≥k+lkl[1+∫TM(t,klk+l‖u(t)‖+klk+l‖v(t)‖)dt]>k+lkl[1+∫E21M(t,klk+l‖u(t)+v(t)‖)dt]+k+lkl[1+∫T∖E21M(t,klk+l‖u(t)+v(t)‖)dt]=k+lkl[1+ρM(klk+l(u+v))]≥‖u+v‖0.
This implies ∥u∥0+∥v∥0>∥u+v∥0, a contradiction! This completes the proof.Corollary 2.6.
LM0(R) is nonsquare space if and only if (a)
for anyu∈LM0(R), one has K(u)≠ϕ,(b)
at least one of the conditions
(b1)
kl/(k+l)∉K(u+v)∩K(u-v),(b2)
μ({t∈T:k|u(t)|-σ(t)>0}∪{t∈T:l|v(t)|-σ(t)>0})>0,(b3)
μ{t∈suppu∩suppv:(kl/(k+l))(|u(t)|+|v(t)|)>e(t)}>0is true, wherek∈K(u), l∈K(v) and u,v∈S(LM0(R)).



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*Source: 361525-2011-04-06.xml*