# Range-Kernel Orthogonality and Elementary Operators: The Nonsmoothness Case

Authors: A. Bachir; A. Segres; Nawal Sayyaf
Journal: Complexity
 (2020)

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## Abstract

The characterization of the points inC1ℋ, the trace class operators, that are orthogonal to the range of elementary operators has been carried out for certain kinds of elementary operators by many authors in the smooth case. In this note, we study that the characterization is a problem in nonsmoothness case for general elementary operators, and we give a counter example to S. Mecheri and M. Bounkhel results.


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## Body

## 1. Introduction

LetBℋ be the algebra of all bounded linear operators acting on a complex separable Hilbert space ℋ and A∈Bℋ be compact operator, and let s1A≥s2A≥⋯≥0 denote the eigenvalues of A=A∗A1/2 arranged in their decreasing order. A∈C1ℋ, the class trace, if(1)A1=∑i=1∞siA=trA<∞,where tr denotes the trace functional [1].We recall the definition of Birkhoff–James’s orthogonality in Banach spaces [2, 3].Definition 1.
IfA is a complex Banach space, then for any elements a,b∈A, we say that a is orthogonal to b, noted by a⊥b, iff for all λ,β∈ℂ there holds(2)λb+βa≥βa.
IfM and N are linear subspaces in A, we say that M is orthogonal to N, noted by M⊥N, if a+b≥a for all a∈M and all b∈N. If M=spana, we simply write a⊥N.(i)
The orthogonality in this sense is not symmetric(ii)
IfA is a Hilbert space with its inner product ⋅, then it follows from (2) that a,b=0 which means that Birkhoff–James’s orthogonality generalizes the usual sense in Hilbert space
We also recall the definition of the range-kernel orthogonality for a pair of operatorsE,T on Banach spaces introduced by R. Harte [4].Definition 2.
IfE:A⟶Y and T:ℬ⟶C are bounded linear operators between Banach spaces. T is called range-kernel orthogonal to E provided kerT is orthogonal to ranE in the sense of the Definition 1, i.e.,(3)s∈kerT⟹s+Ex≥s,for allx∈A.
IfT=E, we shall call that E is orthogonal.
The elementary operator is an operatorE defined on Banach A,ℬ-bimodule ℳ with its representation Ex=∑i=1naixbi, where a=aii∈An and b=bii∈ℬn are n-tuples of algebra elements. The length of E is defined to be the smallest number of multiplication terms required for any representation ∑jajxbj for E. The case of the elementary operator is restricted to the operator δA,B∈BBℋ, which is well known, as generalized derivation induced by fixed operators A and B in Bℋ and defined by δA,BX=AX−XB.
Anderson [5] proved that if A and B are normal operators then(4)for allX,S∈Bℋ:S∈kerδA,B⟹δA,BX+S≥S,where kerδA,B denotes the kernel of δA,B.
This means that the kernel ofδA,B is orthogonal to its range.
This result has been generalized in different directions, to nonnormal operators (see [6, 7]), to Cpℋ1≤p<∞, and to some elementary operators (see [8, 9]).
The main purpose of this paper is to characterize the elements that are orthogonal to the range of arbitrary elementary operators defined onC1ℋ in nonsmoothness case and give a counter example to the results of S. Mecheri and M. Bounkhel [10].
In addition to the notations and the definitions already introduced, we set, ifX is a normed linear space over a field K=ℝ or ℂ, we denote by BX the space of all linear bounded operator on X, the closure of the range of an operator T∈BX will be denoted by ran¯T, the restriction of T to an invariant subspace M will be denoted by TM, and the commutator AB−BA of the operators A,B will be denoted by A,B.


## 2. Main Results

LetX be a normed linear space over the field K and X† its topological dual. Let D be the (multivalued) mapping defined from X to X† as(5)∀x∈X:Dx=φ∈X†:φx=x2;φ=x,called the normalized duality mapping. Hahn–Banach’s theorem ensures that there always exists at least one support functional (a support functional ψ at x∈X is a norm-one linear functional in X† such that ψx=x) at each vector x∈X and therefore Dx is nonempty for every x∈X. Moreover, it is well known that Dx is convex and weak∗-compact subset of X†. D is not linear in general but it is homogeneous, i.e., ∀α∈ℝ:Dαx=αDx.Theorem 1.
LetK be a closed linear subset of X and x∉K; then,(6)x⊥K⟺∃φ˜∈Dx:K⊆kerφ˜.Proof.
Letφ∈Dx such that φ,y=0, for all y∈K. Then,(7)φx+y=φx=x2,x2≤φx+y≤xx+y,that is, x≤x+y, for all y∈K. Hence, x⊥K.
Conversely, letx∉K such that x⊥K. Then, for all y∈K, x and y are linearly independent vectors. Let L be the closed subspace spaned by K and x, L=K,x, and define the function φ on L by φαx+βy=αx2 for all y∈K and all α,β∈ℂ. Clearly φ is linear (by the assumption that K is a linear subset of X). To prove the continuity of φ, let z∈L, then z=αx+βy and φz=αx2. By the definition of ⊥ and the assumption that x⊥K, we derive that z≥αx. If α≠0, we have(8)φz=αx2⋅zz≤αx2αxz=xz.
Ifα=0, then φz=φβy=0. Hence,(9)φz≤xz,for allz∈L.
Therefore,φ is continuous on L and φ=x. By Hahn–Banach theorem, there is a continuous linear functional φ˜ on X such that φ˜L=φ and φ˜=φ, where φ˜L is the restriction of φ˜ on L. It follows by the definition of φ and φ˜L=φ, that K⊆kerφ˜ and φ˜∈Dx.Notation 1.
LetK be a nonempty subset of X and T∈BX, and we denote the duality adjoint of T by T† and set K⊥r=x∈X:x⊥y;∀y∈K.Remark 1.
It is clear that ifxnn is a sequence in a subset K converging to y and x⊥xn, for all n, then x⊥y. Hence, x⊥K⟹x⊥K¯.Corollary 1.
(i)
∀x,y∈X,x⊥y⟺∃φ∈Dx:φy=0(ii)
∀x∈X,∀φ∈Dx:x⊥kerφ(iii)
s⊥ranT⟺∃φ∈Ds:φ∈kerT†Proof.
(i) and (ii) follow immediately from the previous Theorem1.
(iii) By Theorem1 and Remark 1, s⊥ranT is equivalent to the existence of φ∈Ds such that ranT⊆kerφ, that is, φTx=T†φx=0,∀x∈X. Hence,(10)s⊥ranT⟺∃φ∈Ds:φ∈kerT†.
It is well known thatC1ℋ not reflexive, even not smooth and its dual C1ℋ† is isometrically isomorphic to Bℋ. This isomorphism is given by(11)ϕ:Bℋ⟶C1ℋ†,R↦ϕR:ϕRX=trXR.
Then,(12)φ∈C1ℋ†⟺∃R∈Bℋ:ϕR=φ,φ=RandφX=trRX,∀X∈C1ℋ.
Letξ be the elementary operator defined by ξX=∑i=1,…,nAiXBi on C1ℋ, where A=A1,A2,…,AnandB=B1,B2,…,Bn are n-tuples in Bℋn. By isomorphism (11), we can assert that the duality of ξ on C1ℋ has the form ξ†X=∑i=1,…,nBiXAi. Indeed, let X∈C1ℋ and R∈Bℋ; then,(13)ϕRξX=tr∑i=1,2,…,nAiXBiR=trX∑i=1,2,…,nBiRAi=trXξ†R=ϕξ†RX.
LetK be a closed linear subspace of a hilbert space ℋ, and Pk be the orthogonal projection on K. For a bounded operator T on ℋ, the compression of T to K is the restriction K to the operator PkT. We denote by σpT for the point spectrum (set of eigenvalues of T) and by σpr for the reduced point spectrum of T, i.e.,(14)σprT=λ:Tx=λxandT∗x=λ¯x;x∈ℋ\0.Proposition 1.
LetT∈Bℋ and J be a two-sided ideal of Bℋ.
IfkerδT∩J⊆kerδ˜∩J, then(15)kerδT∩J≠0⟺σpT∩σpT∗∗=σprTandσprT≠∅,where δT=δT,T.Proof.
⟹(1)
Suppose that0≠S∈kerδT∩J, then TS=ST and thus T∗S=ST∗ or TS∗=S∗T. Hence, TS=ST, where S is a normal compact operator. Since σS≠0, then there is 0≠λ∈σS such that the Riesz projector Pλ is of finite rank and Pλ∈T0 (the bicommutant of T). It follows from S∈kerδT⊆kerδ˜T that Pλ∈kerδT⊆kerδ˜T and therefore the subspace ranPλ reduces T and mranPλ<∞. Let(16)Jλ=PλSranPλ:S∈J,wherePλSranPλ is the compression of S on ranPλ so that(17)kerδTranPλ∩Jλ⊆kerδ˜TranPλ∩Jλ.Indeed, let us consider the decompositionℋ=ranPλ⊕ranPλ⊥, then T can be written T=T1⊕T2, where T1=TranPλ.LetSλ∈kerδTranPλ∩Jλ, then T1Sλ=SλT1. Choosing S=Sλ⊕0 implies S∈kerδT∩J⊆kerδ˜T∩J. Thus, TS∗=S∗T and therefore T1Sλ∗=Sλ∗T1.It follows fromkerδTranPλ∩Jλ⊆kerδ˜TranPλ∩Jλ that TranPλ is a normal operator and therefore(18)σprTranPλ=σpTranPλ=σTranPλ,which implies thatσprT⊆σpT∩σpT∗∗ and σprT is not an emptyset.(2)
Letλ∈σpT∩σpT∗∗, then there exist x,y∈ℋ\0 such that Tx=λx and T∗y=λ¯y. Hence, there is S such that(19)S=x⊗y∈kerδT∩J⊆kerT˜∩J,Applying Fuglede–Putnam property [6], we get that ran¯S reduces T and Tran¯S is a normal operator. Since x∈kerTran¯S−λ, we derive that T∗x=λ¯x and therefore λ∈σprT. The inclusion is proved.Let us prove the reverse implication⇐.Letλ∈σpT∩σpT∗∗, then there exist x,y∈ℋ\0 such that Tx=λx and T∗y=λ¯y. Setting S=x⊗y, then S∈kerδT∩J.Corollary 2.
LetA be a normal operator in Bℋ with σprA=∅ and S∉kerδA be a positive operator in C1ℋ, then(20)δAX+S1≥S1,for allX∈C1ℋ.Proof.
SinceS is positive, then S=S, hence trS1S=trS1S=S2 so that there is T=S1I∈kerδA, and therefore by Corollary 1, isomorphisms (11) and (13), we get S is orthogonal to ranδA i.e.,(21)δAX+S1≥S1,for allX∈C1ℋ.
However,δA is a normal derivation satisfying kerδA⊆kerδ˜A and by the previous proposition, kerδ∩C1ℋ=0, and this means that S∉kerδA.
We recall that the characterization of the points inC1ℋ that are orthogonal to the range of the elementary operators has been carried out by certain authors for smooth case and for certain kind of elementary operators [9, 11]. In nonsmoothness case, S. Mecheri and M. Bounkhel [10] have obtained the following result.Proposition 2 (see [10])..
LetA,B be normal operators in Bℋ and let S=US∈C1ℋ. Then, S∈kerδA,B if and only if(22)dA,BX+S1≥S1,∀X∈C1ℋ,where dA,B=ΔA,B or dA,B=δA,B for ΔA,BX=AXB−X and δA,BX=AX−XB.Remark 2.
(1)
The previous result shows thatrandA,B⊥∩C1cH=kerdA,B∩C1ℋ. In what follows, we show that this is not true, i.e., ranδA⊥∩C1ℋ⊈kerδA∩C1cH. Indeed, let ℋ=l2ℤ, A∈BH:Aen+1=en;n∈ℤ, and S=e1⊗e1, where en:n∈ℤ is an orthonormal basis for ℋ. Hence, A is a unitary operator (bilateral shift) with σprA=∅ and S∈C1ℋ. By simple calculus, we obtain(23)Ae1⊗e1−e1⊗e1A=e0⊗e1−e1⊗A∗e1=e0⊗e1−e1⊗e2≠0.Thus,S∉kerδA.On the contrary,ϕIS=tre1⊗e1=1 and ϕI=I=1, where I is the identity operator on ℋ. Thus, ϕI∈DS. Since I∈kerδA, we get by (13), ϕI∈kerδA† so that it follows from Corollary 1 that S⊥ranδA.(2)
Furthermore, Corollary2 provides a class of operators as counterexample to the orthogonal decomposition C1ℋ=ran¯δA⊕kerδA so that the problem of characterization of operators in C1ℋ that are orthogonal to the range of elementary operators is still open.
In this note, as an application of the previous section, we shall study this problem in different directions for arbitrary elementary operators.Theorem 2.
Letξ be the elementary operator defined on C1ℋ by
ξX=∑i=1,…,nAiXBi, where A=A1,…,AnandB=B1,…,Bn are n-tuples of operators in Bℋn and let S∈C1ℋ with the condition kerξ=kerξ˜. Let R=UR=S or S∗, then(1)
IfkerR is invariant under R∗,Ai and Bi∗ for all i=1,…,n, then there holds(24)∀X∈C1ℋ:ξX+R1≥R1⟹ξU=0.IfkerR reduces R,Ai and Bi∗ for all i=1,…,n, then there holds(25)ξU=0⟺∀X∈C1ℋ:ξX+R1≥R1.(2)
IfkerR is invariant under R∗,Ai and ranR,Bi=0 for all i=1,…,n, then there holds(26)∀X∈C1ℋ:ξX+R1≥R1⟹ξR=0.IfkerR reduces R,Ai and R,Bi=0 for all i=1,…,n, then there holds(27)ξR=0⟺∀X∈C1ℋ:ξX+R1≥R1.Proof.
(1)
We prove only the case whenR=S=US, the case R=S∗ is similar to the previous one.IfkerS is invariant under S∗,Ai and Bi∗ for all n=1,…,n, then with respect to the decomposition ℋ=kerS⊥⊕kerS, we get the following S=S1⊕0,U=U1⊕0 and(28)Ai=Ai1Ai20Ai3,Bi=Bi10Bi2Bi3,1≤i≤n.It is clear that(29)X=X1000∈C1ℋ⟹X1∈C1kerS⊥andξX=ξ1X1⊕0, where ξ1X1=∑i=1,…,nAi1X1Bi1. Hence,(30)∀X∈C1ℋ:ξX+S1≥S1⟹ξ1X1+S11≥S11,∀X1∈C1kerS⊥.SinceS1 is injective and S1=U1S1, then S1 is a smooth point in C1kerS⊥. So, it follows by Corollary 1 and the uniqueness of T that T=S11U1∗ and then by ismorphism (11), ∑i=1,…,nBi1U1∗Ai1=0=∑i=1,…,nAi∗U1Bi1.From the hypothesiskerξ=kerξ˜, we derive that kerξ1=kerξ˜1 and then ∑i=1,…,nAi1U1Bi1=0; therefore, ξU=∑i=1,…,nAiUBi=0 which means that U∈kerξ.For the reverse implication, ifξU=0, then ξ1U1=0, and since kerξ1=kerξ˜1, we get ξ˜1U1=0=ξ1†U1∗. We derive from Corollary 1 that(31)ξ1X1+S11≥S11;∀C1kerS⊥,and sincekerS reduces both Ai and Bi for all i=1,…,n, we obtain, for X=Xkk=1,…,4∈C1ℋ and Ai=Ai1⊕Ai2,Bi=Bi1⊕Bi2, that(32)ξX+S1=ξ1X1+S1∗∗∗1≥ξ1X1+S11≥S11=S1.(2)
IfkerS is invariant under S∗,Ai and S,Bi=0 for all i=1,…,n, then kerS reduces B, following the same steps as in the previous case, we get ∑i=1,…,nAiUBi=0. Hence, S,Bi=0 for all i=1,…,n implying ξS=0. Since S1 is injective, then ξU1=0. We argue similarly as in the previous case to finish the proof.Remark 3.
IfS is a smooth point in C1ℋ, then kerS or kerS∗ are trivial, then relations (11) and (13) hold for any elementary ξ satisfying kerξ=kerξ˜, and relation (13) hold under the conditions S,Bi=0 for sll i=1,…,n for any operator ξX=∑i=1,…,nAiXBi with kerξ=kerξ˜.Lemma 1.
LetA=A1,…,An,B=B1,…,Bn with AiandBi be operators in Bℋ such that ∑iAiAi∗≤1,∑iAi∗Ai≤1,∑iBi∗Bi≤1, and ∑iBiBi∗≤1. Let ξ be the elementary operator defined on Cp:1≤p<∞ℋ by(33)ξX=∑i=1,…,nAiXBi−X.
Then,(i)
kerξ=kerξ˜⟹kerξ†=kerξ†˜(ii)
IfξS=0=ξ˜S for some compact operator S, then S,Bi=0 for all 1≤i≤nProof.
(i)
We haveξ†S=0⟺ξ˜S∗. Since kerξ=kerξ˜, it follows that(34)ξ˜S=0⟺ξS∗=0⟺ξ†˜S=0.(ii)
See [11].
As an application of Theorem2 and by using Lemma 1, we obtain the following result which generalizes that given by B. P. Duggal in the smooth case [9].Corollary 3.
LetξX=∑i=1,…,nAiXBi−X such that(35)∑iAiAi∗≤1,∑iAi∗Ai≤1,∑iBi∗Bi≤1,∑iBiBi∗≤1,with kerξ=kerξ˜ and kerξB=kerξ˜B. For all S∈C1ℋ, the following assertions hold:(i)
IfkerS reduces S, then(36)S⊥ranξ⟹S∈kerξ.(ii)
IfkerS reduces S and S∈kerξB, then(37)S⊥ranξ⟹S∈kerξ.In what follows, for an elementary operatorξ defined on C1ℋ by(38)ξX=∑i=1,…,nAiXBi,where A=A1,…,AnandB=B1,…,Bn are n-tuples of operators in Bℋn, we shall prove that there are operators, not necessary in the kernel, which are orthogonal to the range without the condition kerξ⊆kerξ˜.Recall that ifX is a Banach space, then the function ψx,y defined by(39)ψx,yt=x+ty−xtis increasing on 0,∞, convex for all x,y∈X, and limt⟶0+ψx,yt exists for all x,y∈X.So, from the properties ofψx,y and the fact that limt⟶0+ψx,yt=limt⟶0−ψx,−yt, it follows that(40)x⊥B−Jy⟺limt⟶0−ψx,yt≤0≤limt⟶0+ψx,yt,∀t>0.Indeed, ifx⊥B−Jy, then x+λy−x≥0, for all λ∈ℂ. Therefore, by setting λ=teiθ, where t=λ and θ∈ℝ and since ψx,y is increasing on 0,∞, then 0≤limt⟶0+ψx,yt. The converse holds from the convexity of ψx,y.Lemma 2.
LetS,T∈Cp1≤p<∞ and K be a closed subspace of ℋ. If K reduces S, then(41)SK⊥B−JT11andSK⊥⊥B−JT22⟹S⊥B−JT,where T11 and T22 are the compression of T to the subsapces K and K⊥, respectively.Proof.
With respect to the decompositionℋ=K⊕K⊥, we can write S=SK⊕SK⊥ and T=Tiji,j=1,2, then(42)S+λT=ScK+λT11λT12λT21SK⊥+λT22,∀λ∈ℂ.
Since thep−norm od an operator matrix always dominates the p−norm of its diagonal parts [12], then(43)S+λTp≥SK+λT11p+SK⊥+λT22pSp=SKp+SK⊥p.
Settingλ=teiθ, where t=λ and θ∈ℝ, then(44)S+teiθTp−Sp≤SK+teiθT11p−SKp+SK⊥+teiθT22p−SK⊥p.
Thus,(45)limt⟶0+ψS,eiθTt≤limt⟶0+ψS∣K,eiθT11t+limt⟶0+ψSK⊥,eiθT22t.
So, the result follows from (40).Proposition 3.
Letξ be the elementary operator defined on C1ℋ by ξ=LARB, where A=A1,…,An,B=B1,…,Bn are n-tuples of operators in Bℋ. If S=US such that kerS is invariant under S∗ and Ai for all i=1,…,n (resp. kerS∗ is invariant under S and Ai for all i=1,…,n), then(46)LB11RA11U1∗=0⟹S⊥ranξresp.LB11RA11U1=0⟹S⊥ranξ,such that A11=A111,A112,…,A11n;Ak=Ai,ji,j=1,2,k=1,…,n (with the same notations for B), where A11kresp.B11k is the compression of Ak (resp.  Bk) to kerS⊥ and S1=U1S1=SkerS⊥.Proof.
Letℋ=kerS⊥⊕kerS; since kerS is invariant under S∗ and Ai for all i=1,…,n, we obtain(47)S=S1⊕0,Ak=A11k0A21kA22k,Bk=Bi,jki,j=1,2,for all k=1,…,n. Let X=Xi,ji,j=1,2∈C1ℋ, by simple calculation, we obtain that the compression of ξX to the subspace kerS⊥ is(48)ξ11X11=∑k=1nA11kX11B11k=LA11RB11X11,so that it follows from the previous lemma that if S1⊥ranξ11, then S⊥ranξ. Since S1 is a smooth point (injective operator) in C1kerS⊥ and U1S1 its polar decomposition, where U1 is an isometry in BkerS⊥. Applying Corollary 3 (ii), we get that if LB11RA11U1∗=0, then S⊥ranξ.Example 1.
Letℋ=l2ℤ and Aen+1=enn∈ℤ and(49)Sen+1=12n+1en,ifn≥0,0,ifn≤−1.
We can assert that(50)S∗en=12n+1en+1,ifn≥0,0ifn≤−1.
A is unitary with σprA=∅(see Remark 2), then kerδA∩C1ℋ=0. It is clear that kerS is spanned by the set ⋯,e−2,e−1,e0 and kerS∗ is spanned by the set ⋯,e−2,e−1, then kerS∗⊂kerS so that kerS∗ reduces S∗, also AkerS∗⊆kerS∗. Hence, S=SkerS∗⊥⊕0. If S=US and U1=UkerS∗⊥, then(51)Uen=en−1,ifn≥1,0ifn≤0,U1en=en−1,ifn≥1,0ifn=0.
IfA1 is the compression of A to kerS∗⊥, then A1en=en−1,ifn≥10ifn≤0.
We haveA1U1en=U1A1en=en−2 for all n≥2 and A1U1en=U1A1en=0 for n=0,1. Then, U1kerδA, and therefore by Proposition 3, we get S⊥ranδA. We have also S∗e0=1/2e1∉kerS which means that kerS is not invariant under S∗.



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