Junis Sawlet,

(Faculty of Mechanics and Mathematics,L.N.Gumilyov Eurasian National University,Astana,010008,Kazakhstan)

Abstract: In this paper,we extend the direct sum decomposition theorem of A-invariant subspaces in Lp(M)to the noncommutative Orlicz spaces associated with a growth function case.

Key words:Growth function;noncommutative Orlicz spaces;A-invariant subspace.

0 Introduction

Let M be a finite von Neumann algebra with a faithful normal tracial state τ.In[1],Arveson introduced the notion of finite,maximal,subdiagonal algebras A of M,as non-commutative analogues of weak-*Dirichlet algebras.Subsequently several authors studied the(non-commutative)Hp-spaces associated with such algebras.Blecher and Labuschagne[2]studied A-invariant subspaces ofLp(M)(1≤p<∞),obtained an A-invariant subspace has a direct sum decomposition.In[3],the author proved the above direct sum decomposition holds for A-invariant subspaces of some noncommutative symmetric spaces.

Let Φ be a growth function.In this paper,we will consider extend some results on A-invariant subspaces ofLp(M)in[2]to the noncommutativeLΦ(M)space case,this can be considered as a complement to the work in[2].

This paper is organized as follows.Section 1 contains some preliminary definitions.In section 2,we extend some results of[2]to the noncommutativeLΦ(M)space case.

1 Preliminaries and some results

Definition 1We say a function Φ is a growth function,if Φ is a continuous and nondecreasing function from[0,∞)onto itself.

In this paper we always assume that for a growth function Φ,tΦ(t)is also a growth function.For a growth function Φ,seta=sup{t:Φ(t)=0}.Thena<∞ and Φ(t)=0 for allt∈[0,a].Hence we may assume that Φ(t)>0 for allt>0(otherwise replace Φ by Φ(a+ ·)).We define the following quantitative indices:

Defniition 2 1.A growth function Φ obeys the ?2-condition for allt>0,often written as Φ ∈ ?2,if there is a constantK>1 such that Φ(2t)≤KΦ(t).

2.A growth function Φ said to satisfy the?12-condition for allt>0,denoted symbolically as Φ ∈ ?12,if there is a constant 0

Defniition 3 Two growth functions Φ1and Φ2are said to be equivalent,denoted symbolically as Φ1～ Φ2,if there exist positive constantsC1,C2,C3,C4such that

For a growth function Φ,Φ ∈ ?2∩if and only if 00.Let Φ be a growth function,1 ≤n< ∞.We set Φ(n)(t)= Φ(tn).ThenpΦ(n)=npΦ,qΦ(n)=nqΦ.We have the following result:

1.Φ0∈?2∩?12.

2.IfqΦ0α ≤ 1,then Φ(α)is a concave growth function.

3.IfpΦ0α ≥ 1,then Φ(α)is a convex growth function.

For more details see[4,5].

Let M be a fniite von Neumann subalgebra equipped with a normal fniite faithful tracial state τ.LetL0(M)denote the topological*-algebra of measurable operators with respect to(M,τ).The topology ofL0(M)is determined by convergence in measure.

Defniition 4 Let Φ be a growth function.We defnie the corresponding noncommutative Orlicz space on(M,τ)by

Forx∈LΦ(M),define

If Φ∈?2∩?12,thenLΦ(M)is a quasi-Banach space.

Given a von Neumann subalgebra N of M,an expectation E:M→N is defnied to be a positive linear map which preserves the identity and satisfeis E(xy)=xE(y)for allx∈N andy∈M.Since E is positive it is hermitian,i.e.E(x)?=E(x?)for allx∈M.Hence E(yx)=E(y)xfor allx∈N andy∈M.For a complete study of E,we refer to[1,6].

Defniition 5 Let A be a w*-closed unital subalgebra of M,and let E be a faithful,normal expectation from M onto the diagonal von Neumann subalgebra D=A∩J(A).Then A is a fniite subdiagonal algebra of M with respect to E if:

1.A+J(A)is w*-dense in M;

2.E(xy)=E(x)E(y),?x,y∈A;

3. τ?E=τ.

It is proved by Exel[7]that a finite subdiagonal algebra A is automatically maximal in the sense that if B is another subdiagonal algebra with respect to E containing A,then B=A.This maximality yields the following useful characterization of A:

where A0=A∩kerE(see[1]).

For growth function Φ,we defineHΦ(A)to be the closure of A inLΦ(M).These are the so-called Hardy spaces associated with A.LetKbe a subset ofLΦ(M).We setJ(K)={x?:x∈K}and denote the closed linear span ofKinLΦ(M)by[K]Φ.We will keep this notations throughout the paper.

Let(A)=[A0]Φ.Then(A)={x∈HΦ(A):E(x)=0}.

Definition 6 Let Φ be a growth function andKbe a closed subspace ofLΦ(M).We shall say thatKis right(or left)A-invariant subspace ofLΦ(M)ifKA?K(resp.AK?K).

In this paper,we only consider‘right A-invariant subspace’.For brevity,we simple call them ‘A-invariant subspace’.

2 A-invariant subspace of LΦ(M)

Leta=(an)n≥0be a finite sequence inLΦ(M),define

This gives two quasi norms on the family of all finite sequences inLΦ(M).To see this,denoting by B(2)the algebra of all bounded operators on2with its usual tracetr,let us consider the von Neumann algebra tensor product M?B(2)with the product trace τ?tr.τ?tris a semi-finite normal faithful trace,the associated noncommutativeLΦspace is denoted byLΦ(M?B(2)).Now,any finite sequencea=(an)n≥0inLΦ(M)can be regarded as an element inLΦ(M?B(2))via the following map

that is,the matrix ofT(a)has all vanishing entries except those in the first column which are thean’s.Such a matrix is called a column matrix,and the closure inLΦ(M?B(2))of all column matrices is called the column subspace ofLΦ(M?B(2)).Then

Therefore .LΦ(M,l2C)defines a quasi norm on the family of all finite sequences ofLΦ(M).The corresponding completion is a quasi Banach space,denoted byLΦ(M,).It is clear that a sequencea=(an)n≥0inLΦ(M)belongs toLΦ(M,)if and only if

Let{Xi:i∈I}be a collection of closed subspaces ofLΦ(M).We defnie the external columnLΦ-sumXito be the closure of the restricted algebraic sum in the norm.IfXis a subspace ofLΦM),and if{Xi:i∈I}is a collection of subspaces ofX,which together densely spanX,with the property that={0}ifij,then we say thatXis the internal columnLΦ-sumXi.Note that ifJis a fniite subset ofI,and ifxi∈Xfor alli∈J,then we have

This shows thatXis isometrically isomorphic to the external columnLΦ-sumXi.Since the p rojection onto the summands are clearly contractive,it follows b y routine arguments that if(xi)∈Xi,then the net(i∈J xi),indexed by the fniite subsetsJofI,converges in norm toi∈I xi.

Lemma 1 Let Φ be a growth function.If Φ∈?2∩,then

Let N=Mn(M)be the algebra ofn×nmatrices with entries from M and B=Mn(A)be the algebra ofn×nmatrices with entries from A.Forx∈N with entriesxi,j,defnie F(x)to be the matrix with entries E(xi,j)andThen(N,ν)is a fniite von Neaumann algebra and B is a fniite subdiagonal algebra of(N,ν).Leta=((x1,x2,···,xn,0···)),by Theorem 3.1 in[4],

Using(3),we obtain the desired inequality.

Let(A)=[A0]Φ.Then(A)={x∈HΦ(A):E(x)=0}.

Lemma 2 Let Φ be a growth function.Assume Φ ∈ ?2∩.IfKis right A-invariant subspace ofLΦ(M),then[K∩M]Φ=K.

Proof Let Φ0=tΦ(t)for allt≥ 0.(1)Consider the casepΦ>1,pΦ0>1.It follows from Theorem 1.5 in[4]that Φ is convex function.Let Ψ be complementary function of Φ.By Theorem 3.7 in[9],LΦ(M)?=LΨ(M).If[K∩M]ΦK,then there exist an elementz∈Kandx∈LΨ(M)such that τ(zx) 0 and τ(yx)=0 for everyy∈[K∩M]Φ.Setw=(x?x+1)12.Thenw∈LΦ(M)andw?1∈M,by the Szeg? factorization in[9],there exist a unitary anda∈HΦ(n)(A)such thata?1∈ A andw=ua.Sincew?w≥x?x,there exists a contraction ν∈ M such thatx= νw.On the other hand,sinceKis right A-invariant subspace ofLΦ(M),we haveza?1b∈K∩M for everyb∈ A which implies τ(xza?1b)=0.Hence,xza?1∈(A)∩LΨ(M).By Lemma 4.1 in[9],xza?1∈(A),soxz∈(A)Therefore τ(zx)=0.This is a contradiction.

(2)By Theorem 1.5 in[4]we know that there exist an∈Nsuch that growth function Φ(n)is convex function andpΦ(n)>1.For everyx∈K,we setw=.Thenw∈LΦ(M)andw?1∈ M.Sincew?w≥x?x,there exists a contraction ν∈M such thatx=νw.

Write

wherew1=υandwk=for 2≤k≤n.Sincewk∈(M)and∈M,by the Szeg? factorization in[9],we have

withun∈M a unitary,andxn∈HΦ(n)(A)such that∈A.Repeating this argument,we get a similar factorization forwn?1un:

and then forwn?2un?1,and so on.In this way,we obtain a factorization:

whereu∈ M is a unitary,andxk∈HΦ(n)(A)such that∈ A.Settinga=x2...xn,then by Theorem 2.4 in[4],(A)anda?1∈A.Lety=νux1.Theny∈LΦ(n)(M)andx=ya.So we havey∈K∩LΦ(n)(M).On other hand,by virtue of(1),(A)? [K∩M]Φ,we getx∈[K∩M]Φ.

The following lemma is the special case of Lemma 5.3 in[3].

Lemma 3 Let Φ be a growth function.Assume Φ ∈ ?2∩.Ifeis a projection in M,thenv∈(M)satisfeisfor alld∈ M,if and only ifv=e.

IfKis a right A-invariant subspace ofLΦ(M),we say thatKis type 2 if the wandering quotientK/[KA0]Φis trivial.We will show that the wandering quotientK/[KA0]Φis isometric to a canonically defnied subspace ofKwhich can be called the right wandering subspace ofK.We say thatKis type 1 if this subspace generatesKas an A-module.

Defniition 7 Let Φ be a growth function satisfying Φ∈?2∩?12.LetKbe an A-invariant subspace ofLΦ(M).Letr=max{2,qΦ},we defnie the right wandering subspace ofKto be theLΦ-closure of the right wandering subspace ofK∩Lr(M).

Theorem 1 Let Φ be a growth function satisfying Φ ∈ ?2∩?12.Suppose thatKis an A-invariant subspace ofLΦ(M).

(i)Kmay be written as a columnLΦ-sumwhereZis a type 2 subspace ofLΦ(M),and whereuiare partial isometries in M ∩Kwithu?jui=0 ifij,and with∈ D.Moreover,for eachi,={0},left mul tiplication by theare contractive projection fromKonto the summandsuiHΦ(A),and left multiplication by 1?i uiis a contractive projection fromKontoZ.

(ii)The wandering quotientK/[KA0]Φis isometrically isomorphic to the right wandering subspace ofK;and the latter also equals toui[D]Φ,whereuiare from(i).

(iii)Ifr=max{2,qΦ},thenKis type 1 if and only ifK∩Lr(M)is type 1,and if and only ifZ={0}in i);Kis type 2 if and only ifK∩Lr(M)is type 2,and if and only ifK=Z.

Proof (i)10Suppose thatqΦ≤2.By Lemma 2,Kis theLΦ-closure ofK∩L2(M).Theorem 2.1 in[2]gives a decompositionK∩L2(M)=Z⊕col[YA]2,whereZis type 2,andYis the right wandering subspace ofK∩L2(M).LetZ=[Z]Φ,thenZis type 2.Indeed,it is clear that[ZA0]Φ=[[Z]ΦA(chǔ)0]Φ=[ZA0]Φ.Thus

We have that

To see this,note thatZ?Z∩L2(M)andZ∩L2(M)=Z⊕([YA]2∩Z).Letz∈[YA]2∩Z,then there exist{zn}?Z,{yn}?YA such that

In the same way,we also obtain the second equality.

Sincez?y=0 forz∈Z,y∈YA,it follows that

From(4)and(5),it follows thatK=Z⊕col K1,whereK1=[YA]Φ.By Corollary 2.3 in[2],[YA]2=ui[A]2,foruias above.Thus theLΦ-closure ofuiA is all ofK1.On the other hand,(ui[A]Φ)?(uj[A]Φ)={0}ifij.SoK1=ui[A]Φ.

20Suppose that 2

30Since left multiplication byannihilatesZanduj[A]Φifij,left multiplication by theare contractive projections fromKontoui[A]Φ,and left multiplication by the 1?i uiis a contrac tive projection ontoZ.

(ii)Defnie a map π :K=Z⊕col(uiHΦ(A))→Kby π(w)=i uiE(xi)ifw=z+i uiE(xi)forz∈Zandxi∈HE(A).

Since∈D,by Lemma 1 we haveHence,π is well defnied contractive projection into its range,and hence induces an isometric map fromK/ker(π)ontouiLΦ(D). SincexiA0? [A0]Φ? ker(π)ifxi∈HΦ(A),it is clear that ker(π)?KA0and so ker(π)? [KA0]Φ.Conversely if π(z+i uixi)=0 thenE(xi)=0 for everyi.Thus∈ ker(E)∩HΦ(A)=[A0]Φ.Henceuixi∈ui[A0]Φ,and soz+i uixi∈[KA0]Φ.So ker(π)=[KA0]Φ.

(iii)IfK∩Lr(M)is type 1 then it is clear from the proof of(i)thatZ=Z={0}.Similarly,ifK∩Lr(M)is type 2 thenZ=K.It is obvious that ifK=ZthenKis type 2.Conversely,ifKis type 2 then the wandering quotient is{0}.Identifying this with the subspace ofKdescribed in(ii),all theuiare zero.ThusK=Z.AlsoK∩Lr(M)=Z∩Lr(M)=Zwhich is type 2.Since[(uiLΦ(D))A]Φ=uiHΦ(A),we have thatZ={0}if and only ifuiLΦ(D)generatesK.By(ii)this happens if and only ifKis type 1.

IfqΦ≤2,suppose thatK=uiHΦ(A)for partial isometriesuisatisfying the relations in(1),then[uiA]2=uiH2(A).LetX=K∩L2(M),then by Theorem 2.1 in[2],we have thatX=[XA0]2⊕col[YD]2=Z⊕col[YA]2, whereZis type 2,andYis th e right wandering subspace ofX.Ifx∈X,then there exist{xi}?HΦ(A)such thatx=i uixi.So for eachi,==∈HΦ(A)∩L2(M)=HpΦ(A)∩L2(M)=H2(A)and τxa)=0,?a∈A0.From this it follows thatui⊥[XA0]2,which forcesui∈[YD]2.HenceuiA∈[YD]2A?[YA]2.ThusuiH2(A)?[YA]2.SoK=uiHΦ(A)? [YA]Φ.On the other hand,by the proof of(i),K=[Z]Φ⊕[YA]Φ.HenceZ={0},which givesK∩L2(M)is type 1.

IfqΦ>2,suppose thatK=uiHΦ(A)for partial isometriesuisatisfying the relations in(i),thenuiH2(A).LetX=[K∩LqΦ(M)]2,then by Theorem 2.1 in[2],we haveX=[XA0]2⊕col[YD]2=Z⊕[YA]2,whereZis type 2,andYis the right wandering subspace ofX.Using an argument in the proof of the caseqΦ≤2 we obtainK∩LqΦ(M)is type 1.

Corollary 1 Let Φ be a growth function satisfying Φ∈?2∩?12,and letr=max{2,qΦ}.SupposeKis a subspace ofLΦ(M)of the formK=K1⊕col K2whereK1is a type 1 andK2is type 2.ThenK1(respectivelyK2)theLΦ-closure of the type 1 part(respectively type 2)ofK∩Lr(M).

Proof It is clearly(K1∩Lr(M))+(K2∩Lr(M))?K∩Lr(M).On the other hand,ifx∈K∩Lr(M),then there existki∈K1such thatx=k1+k2holds.SinceK1is type 1,we can writeK1=uiHΦ(A)for someuias above.Sok1=i ui k1.Sinceui∈K1we haveu?i k2=0,and soki=i ui x.Thusk2=(1?i ui)x.From this,we have

Hence

SinceK1∩Lr(M)is type 1 andK2∩Lr(M)is type 2,by Corollary 4.6 in[2],we obtainK1∩Lr(M)(respectivelyK2∩Lr(M))is the type 1 part(respectively type 2)ofK∩Lr(M).By Lemma 2,we have thatK1=[K1∩Lr(M)]ΦandK2=[K2∩Lr(M)]Φ.

Corollary 2 Let Φ be a growth function satisfying Φ ∈ ?2∩ ?12,and letx∈LΦ(M).If[xA]Φis a type 1 A-invariant subspace ofLΦ(M),thenx=i uihi,whereuiare partial isometries in M∩[xA]Φwithu?jui=0 ifij,andu?i ui∈D,andhi∈LΦ(M)withu?i uihi=hiandu?i ui∈ [hiA]Φ.

The proof is similar to the proof of Corollary 4.9 in[2],therefore we omit the details.

Proposition 1 Let Φ be a growth function satisfying Φ ∈ ?2∩?12,and letr=max{2,qΦ}.Suppose thatKis an A-invariant subspace ofLΦ(M).The following are equivalent:

(i)K=uHΦ(A)for a unitaryu∈M.

(ii)The right wandering subspace ofKhas a cyclic and separating vector.

(iii)The right wandering subspace ofK∩Lr(M)has a cyclic and separating vector.

Proof (i)?(ii)and(iii).IfK=uHΦ(A),thenK∩Lr(M)=uHr(A)as in the proof of(iii)of Theorem 1.Hence the right wandering subspace ofK∩Lr(M)isuLr(D),and the right wandering subspace ofKisuLΦ(D).These both have cyclic and separating vectors.(iii)?(i).By Proposition 4.7 in[2],there exists a unitary u∈M such that K∩Lr(M)=uHr(A).Thus K=uHΦ(A).(ii)?(i).Let W be the right wandering subspace of K,then K=Z⊕col[WA]Φ.By an adaption of an argumentffff31 from p.13 in[10],there exists an isometric D-module isomorphism ψ :LΦ(D)→ W.We put ψ(1)=u,v=u?u ∈ (D),then we have

By Lemma 3,u?u=v=1,which impliesuis unitary andW=uLΦ(D).SinceZ?W=0,we getZ=0.ThusK=uHΦ(A).