On the positive commutator in the radical
aa r X i v : . [ m a t h . F A ] M a y ON THE POSITIVE COMMUTATOR IN THE RADICAL
MARKO KANDI ´C, KLEMEN ˇSIVIC
Abstract.
In this paper we prove that a positive commutator between a positive com-pact operator A and a positive operator B is in the radical of the Banach algebra gen-erated by A and B . Furthermore, on every at least three-dimensional Banach lattice weconstruct finite rank operators A and B satisfying AB ≥ BA ≥ AB − BA is not contained in the radical of the Banach algebra generated by A and B . These two results now completely answer to two open questions published in [4].We also obtain relevant results in the case of the Volterra and the Donoghue operator. Math. Subj. Classification (2010) : 47A10; 47A46; 47B07; 47B47.
Key words : Positive commutators, Positive compact operators, Invariant subspaces, Rad-ical. 1.
Introduction
Positive commutators of positive operators on Banach lattices were first considered in[4]. The authors proved [4, Theorem 2.2] that a positive commutator AB − BA of positivecompact operators A and B is necessarily quasi-nilpotent and furthermore, it is containedin the radical of the Banach algebra generated by A and B . They posed a question ifthe same is still true if we assume that only one of the operators A and B is compact.The first part of the problem was independently solved by Drnovˇsek [5] and Gao [10].Drnovˇsek proved even more. Theorem 1.1. [5, Theorem 3.1]
Let A and B be bounded operators on a Banach lattice E such that AB ≥ BA ≥ and AB is power-compact. Then AB − BA is an ideal-triangularizable quasi-nilpotent operator. Date : November 12, 2017.
This result is an improvement of [4, Theorem 2.4] where the authors considered opera-tors from appropriate Schatten ideals. In this paper we prove that a positive commutator AB − BA of positive operators A and B is in the radical of the Banach algebra gen-erated by A and B whenever one of the operators A and B is compact. We actuallyprove that AB − BA is contained in the radical of the possibly bigger algebra that con-tains the Banach algebra generated by A and B . This result now completely solves theopen problem [4, Open questions 3.7 (1)]. We also prove that on every at least three-dimensional Banach lattice E there exist finite rank operators A and B on E satisfying AB ≥ BA ≥
0, however the commutator AB − BA is not contained in the radical of theBanach algebra generated by A and B . Although this counterexample answers negativelyto [4, Open questions 3.7(2)] in the case of at least three-dimensional Banach lattices,there is a positive answer on two-dimensional Banach lattices.2. Preliminaries
Let E be a Riesz space, and let E + denote the set of all positive vectors in E . A linearsubspace J of E is said to be an ideal whenever 0 ≤ | x | ≤ | y | and y ∈ J imply x ∈ J . Anorder closed ideal is said to be a band . The set A d := { x ∈ E : | x | ∧ | a | = 0 for all a ∈ A } is called the disjoint complement of a set A in E . The disjoint complement of a nonemptyset in E is always a band in E . A band B of E is said to be a projection band if E = B ⊕ B d . A Riesz space E is said to be a normed Riesz space whenever E is equippedwith a lattice norm. If a normed Riesz space is a complete metric space with the metricinduced by the norm, then it is called a Banach lattice . It is well-known that every normedRiesz space is Archimedean.A nonzero vector a ∈ E + is an atom in a normed Riesz space E if 0 ≤ x, y ≤ a and x ∧ y = 0 imply either x = 0 or y = 0, or equivalently, if 0 ≤ x ≤ a implies x = λa for some λ ≥
0, i.e., the principal ideal B a generated by a is one-dimensional. It turnsout that B a is a projection band [13]. If every set of mutually disjoint nonzero vectorsof an Archimedean Riesz space E is finite, then E is a finite-dimensional atomic latticethat is order isomorphic to R n (where n = dim E ) with componentwise ordering. If an N THE POSITIVE COMMUTATOR IN THE RADICAL 3
Archimedean Riesz space E contains only finitely many pairwise disjoint atoms, then E isan order direct sum of the finite-dimensional atomic part of the lattice E and its disjointcomplement that is an atomless part of the lattice E .An operator T on a Banach lattice E is said to be positive whenever T maps the positivecone of E into itself. It is well-known that every positive operator on a Banach lattice isa bounded operator. By L ( X ) and L + ( E ) we denote the Banach algebra of all boundedoperators on a Banach space X and the set of all positive operators on a Banach lattice E , respectively. A family F of operators on a Banach space X is reducible if there existsa nontrivial closed subspace of X that is invariant under every operator from F . If thereexists a maximal chain C of closed subspaces of X such that every subspace from thechain C is invariant under every operator from F , then F is said to be triangularizable ,and C is called a triangularizing chain for F . A family F of operators on a Banach lattice E is said to be ideal-reducible if there exists a nontrivial closed ideal of E that is invariantunder every operator from F . Otherwise, we say that F is ideal-irreducible . A family F ofoperators on a Banach lattice E is said to be ideal-triangularizable whenever there existsa chain of closed ideals of E that is maximal as a chain in the lattice of all closed ideals of E and every ideal from the chain is invariant under F . Every ideal-triangularizable familyof operators is also triangularizable [6]. If C is a complete chain of closed subspaces of aBanach space X , the predecessor of M in C is denoted by M − . If every subspace M froma complete chain C of closed subspaces of X is invariant under the operator A ∈ L ( X ) and M 6 = M − , the induced operator A M on M / M − is defined by A M ( x + M − ) = Ax + M − for all x ∈ M . In the case of a finite-dimensional lattice R n we will use more common termsdecomposable, indecomposable and completely decomposable instead of ideal-reducible,ideal-irreducible and ideal-triangularizable, respectively. For a more detailed treatmenton triangularizability we refer the reader to [15].Positive operators A and B on a Banach lattice semi-commute whenever AB ≥ BA or AB ≤ BA.
The super right-commutant [ A i and the super left-commutant h A ] of a positiveoperator A on a Banach lattice E are defined by[ A i = { B ∈ L + ( E ) : BA ≥ AB } and h A ] = { B ∈ L + ( E ) : AB ≥ BA } , MARKO KANDI ´C, KLEMEN ˇSIVIC respectively. For the terminology and details not explained about Banach lattices andoperators on them we refer the reader to [2] and [16].The radical of a Banach algebra A is the intersection of all maximal modular left idealsof A which coincides with the intersection of all maximal modular right ideals of A . If A is a unital Banach algebra, then the radical of A is equal to the set { a ∈ A : r ( ba ) = 0 for all b ∈ A} = { a ∈ A : r ( ab ) = 0 for all b ∈ A} where r ( x ) denotes the spectral radius of an element x ∈ A . Therefore, the radical ofa unital Banach algebra is the largest among all ideals that consist of quasi-nilpotentelements. If a Banach algebra A is not unital, then we consider the spectrum of a ∈ A as the spectrum of the element a in the standard unitization of the algebra A . For theterminology not explained throughout the text regarding Banach algebras we refer thereader to [3]. 3. Counterexamples
In [4] the authors proved [4, Theorem 2.3] that the commutator AB − BA of realmatrices A and B is a nilpotent matrix whenever AB ≥ BA ≥
0. The authors alsoextended this result to the case when A ∈ C p and B ∈ C q are bounded operators on l with AB ≥ BA ≥ ≤ p, q ≤ ∞ and 1 /p + 1 /q = 1. In this case, the commutatoris a quasi-nilpotent trace class operator on l . See [4, Theorem 2.4]. They asked ([4, Openquestions 3.7(2)] if the commutator AB − BA is contained in the radical of the Banachalgebra generated by A and B . In this section we provide a counterexample on every atleast three-dimensional Banach lattice.For the construction of our counterexample to [4, Open questions 3.7(2)] we need somepreparation. The proof of the following result can be found in [13, Theorem 26.10]. Weprovide our proof for the sake of completness. Proposition 3.1.
Every infinite-dimensional Banach lattice has at least countably infin-itely many nonzero pairwise disjoint positive vectors.Proof.
Let us choose an aribtrary infinite-dimensional Banach lattice E . Suppose firstthat E does not contain atoms. We claim that there exists an increasing sequence of sets {A n } n ∈ N of positive nonzero pairwise disjoint vectors with |A n | = n and a sequence of N THE POSITIVE COMMUTATOR IN THE RADICAL 5 positive nonzero vectors { y n } n ∈ N such that y n is disjoint with A n for every positive integer n. Let us validate the statement above when n = 1. If y is an arbitrary nonzero positivevector in E , then the fact that E is without atoms implies that there exist nonzero positivevectors 0 ≤ x , y ≤ y satisfying x ∧ y = 0 . We choose A = { x } and y is disjoint with A . Suppose that there exist sets A ⊆ · · · ⊆ A n of positive nonzero pairwise disjointvectors with |A j | = j for 1 ≤ j ≤ n , and positive nonzero vectors y , . . . , y n such that y j ∈ A dj for 1 ≤ j ≤ n . Since y n is not atom, there exist positive nonzero vectors x n +1 and y n +1 satisfying 0 ≤ x n +1 , y n +1 ≤ y n and x n +1 ∧ y n +1 = 0. It is obvious that the vectors x n +1 and y n +1 are disjoint with A n . The set A n +1 := A n ∪ { x n +1 } obviously satisfies A n ⊆ A n +1 , |A n +1 | = n + 1 and y n +1 ∈ A dn +1 . Let A = S ∞ n =1 A n . The set A is countably infinite, and arbitrary two vectors from A are disjoint.Suppose now that E contains at least one atom. By Zorn’s lemma there exists amaximal set A of pairwise disjoint atoms of norm one. If A is infinite, then we are done.Suppose otherwise that A is a finite set, so that we have E = A dd ⊕ A d where A dd is afinite-dimensional atomic lattice and A d is an infinite-dimensional atomless lattice. Byalready proved, A d contains at least countably infinitely many nonzero positive pairwisedisjoint vectors. (cid:3) Lemma 3.2.
Let E be a Banach lattice of dimension at least n and let x , . . . , x n benonzero positive pairwise disjoint vectors. Then there exist nonzero positive functionals ϕ , . . . , ϕ n on E such that ϕ i ( x j ) = δ ij for i, j = 1 , . . . , n. Proof.
Let us choose an arbitrary 1 ≤ i ≤ n and let J i be the ideal generated by the vector x i . By [18, Theorem 39.3.], there exists a positive functional ψ i on E with ψ i ( x i ) = 1 . Let ψ i | J i be the restriction of the functional ψ i on the ideal J i . By [18, Theorem 20.5.] andits following remark there exists a positive functional ϕ i on E which extends ψ i | J i and ϕ i is zero on J di . This implies ϕ i ( x j ) = δ ij for all j = 1 , . . . , n . (cid:3) MARKO KANDI ´C, KLEMEN ˇSIVIC
Although Lemma 3.2 is cruicial for Example 3.6 and Example 3.7, it is also of its owninterest. Before we indicate its importance let us recall first the well-known Jacobson’sdensity theorem for Banach algebras [3].
Theorem 3.3.
Let A be a Banach algebra and let π : A → L ( X ) be a continuousirreducible representation on a Banach space X . If x , . . . , x n are linearly independent in X and y , . . . , y n are arbitrary in X , then there exists a ∈ A such that π ( a ) x i = y i for all i = 1 , . . . , n. In the special case when π is the identity representation on L ( X ) Jacobson’s densitytheorem states that a finite set of linearly independent vectors can be mapped onto arbi-trary vectors by bounded operators. This fact can be also proved with an application ofthe Hahn-Banach theorem without applying the Jacobson’s density theorem. Now it isnatural to ask whether arbitrary linearly independent positive vectors in Banach latticescan be mapped to arbitrary positive vectors by positive operators. The following exampleshows that in general it cannot be done. Example 3.4.
The space R with a canonical ordering and equipped with the norm k · k is a Banach lattice. The only matrix mapping (cid:20) (cid:21) to (cid:20) (cid:21) and (cid:20) (cid:21) to (cid:20) (cid:21) is (cid:20) −
10 1 (cid:21) which is clearly not positive.
However, for a finite set of nonzero pairwise disjoint positive vectors we have a positiveresult.
Theorem 3.5.
Let E be a Banach lattice and let x , . . . , x n be arbitrary nonzero positivepairwise disjoint vectors in E . Then for arbitrary positive vectors y , . . . , y n in E thereexists a positive operator T on E such that T x j = y j for all j = 1 , . . . , n. Proof.
By Lemma 3.2 there exist positive functionals ϕ , . . . , ϕ n on E such that ϕ i ( x j ) = δ ij for all i, j = 1 , . . . , n. The positive operator T = n X i =1 y i ⊗ ϕ i N THE POSITIVE COMMUTATOR IN THE RADICAL 7 satisfies
T x j = n X i =1 ϕ i ( x j ) y i = n X i =1 δ ij y j = y j for all j = 1 , . . . , n. (cid:3) Throughout the rest of this section E is assumed to be at least three-dimensional Banachlattice. Let the vectors x , x and x and the functionals ϕ , ϕ and ϕ be as in Lemma3.2. The vectors x , x and x , in fact, exist since either E is finite-dimensional and so itis order isomorphic to R n with the componentwise order, or E is infinite-dimensional andwe may apply Proposition 3.1.The following example shows that the commutator AB − BA of compact operators A and B with AB ≥ BA ≥ A and B if we do not assume A ≥ Example 3.6.
An easy calculation shows that operators A = ( − x + x ) ⊗ ϕ + ( x + x ) ⊗ ϕ and B = x ⊗ ( ϕ + ϕ ) satisfy AB = ( x + x ) ⊗ ( ϕ + ϕ ) and BA = x ⊗ ϕ , so that AB and BA are positive operators on E . Since AB − BA = x ⊗ ( ϕ + ϕ ) + x ⊗ ϕ ≥ , we also have AB ≥ BA ≥ . The commutator AB − BA is not contained in the radicalof the Banach algebra A generated by the operators A and B since the operator A ( AB − BA ) = ( x − x ) ⊗ ϕ + ( x + x ) ⊗ ϕ has as its eigenvalue with a corresponding eigenvector x + x . The following example shows that the commutator AB − BA of compact operators A and B with AB ≥ BA ≥ A and B if we do not assume B ≥ Example 3.7.
Similarly as in Example 3.6 we can see that operators A = ( x + x ) ⊗ ϕ and B = x ⊗ ϕ + ( x − x ) ⊗ ϕ + x ⊗ ϕ MARKO KANDI ´C, KLEMEN ˇSIVIC satisfy AB ≥ BA ≥ . An easy calculation also shows that ( AB − BA ) B ( x + 2 x ) = x + 2 x , so that is in the spectrum of the operator ( AB − BA ) B. Results
Although the counterexamples in Section 3 answer negatively to [4, Open questions3.7(2)] in the case of at least three-dimensional lattices, we now provide a positive resultin the two-dimensional case.
Proposition 4.1.
Suppose that matrices A and B in M ( R ) satisfy AB ≥ BA ≥ . Theneither AB = BA or A and B are simultaneously completely decomposable. If two matrices commute, then it is well-known that A and B are simultaneously trian-gularizable. Therefore we can conclude that real matrices A, B ∈ M ( R ) are simultane-ously triangularizable whenever they satisfy the conditions of the preceding proposition.It should be clear that the same result does not hold for n ≥
3. If matrices A and B are simultaneously triangularizable, then A and B are upper-triangular with respect tosome basis of the space C n . Now it is not difficult to see that the matrix A ( AB − BA ) isnilpotent. However, Example 3.6 shows that there exist matrices A, B ∈ M ( R ) satisfying AB ≥ BA ≥
0, and the matrix A ( AB − BA ) is not nilpotent. Proof.
We may assume B = 0, as otherwise A and B commute. Suppose first that thematrix AB is indecomposable. Then [1, Lemma 8.14], [1, Corollary 8.23] and the factthat we have r ( AB ) = r ( BA ) imply AB = BA .Assume now that AB is decomposable. The standard subspace invariant under AB is invariant under BA as well, since AB ≥ BA ≥
0. The fact that the space is two-dimensional implies that AB and BA are simultaneously similar to upper-triangularmatrices and that this similarity can be obtained by a permutation matrix. We mayobviously assume that AB and BA are already upper-triangular. Since AB ≥ BA ≥ AB and BA have the same diagonal. Let us write A = (cid:20) a bc d (cid:21) and B = (cid:20) e fg h (cid:21) . N THE POSITIVE COMMUTATOR IN THE RADICAL 9
From AB = (cid:20) α γ β (cid:21) and BA = (cid:20) α δ β (cid:21) we obtain α = ae + bg = ae + cf , β = cf + dh = bg + dh and ce + dg = ag + ch = 0 . Let us first consider the case g = 0. Then ce = cf = ch = 0 . If c = 0, then A and B are upper-triangular. If c = 0, then e = f = h = 0 and B = 0 which is impossible due toour assumption.Assume now that g = 0. Then we have b = fcg , d = − ceg and a = − chg . Now, a directcalculation shows γ = af + bh = 0 and δ = eb + f d = 0. Therefore AB = BA . (cid:3) Corollary 4.2.
Suppose that matrices A and B in M ( R ) satisfy AB ≥ BA ≥ . Then AB − BA is in the radical of the Banach algebra generated by A and B .Proof. If AB = BA , then there is nothing to prove. Otherwise, Proposition 4.1 impliesthat A and B are simultaneously completely decomposable. By McCoy’s theorem [15,Theorem 1.3.4] the matrix p ( A, B )( AB − BA ) is nilpotent for every polynomial p in twonon-commuting variables. Since the spectral radius is continuous on the set of all n × n matrices, a matrix C ( AB − BA ) is nilpotent for every C in the Banach algebra generatedby A and B which finishes the proof. (cid:3) Although real matrices A and B satisfying AB ≥ BA ≥ A and B are assumed to be positive. Proposition 4.3.
Let A and B be positive power-compact operators on a complex Ba-nach lattice E with a positive commutator AB − BA . Then A and B are simultaneouslytriangularizable.Proof. Assume that A and B are power-compact. Since a closed subspace is invariantunder A and B if and only if it is invariant under A and A + B , by [8, Lemma 2.2] wemay assume that 0 ≤ A ≤ B as we can replace B with A + B . We claim that A and B are simultaneously triangularizable. Let C be a maximal chain of closed ideals that areinvariant under B . Since 0 ≤ A ≤ B , every ideal in C is also invariant under A . Forevery J ∈ C the power-compact operator B J is ideal-irreducible. [8, Theorem 1.3] and[8, Corollary 3.3] imply that A J and B J commute on J / J − . Let C ′ be a maximal chainof closed subspaces invariant under A and B which contains the chain C . We claim that C ′ is a triangularizing chain for operators A and B . Since the chain C ′ is maximal as a chain of subspaces invariant under A and B , (i) and (ii) in [15, Theorem7.1.9] are satisfied. Therefore we only need to prove that the dimension of M / M − is atmost one for an arbitrary subspace M in the chain C ′ .Choose an arbitrary subspace M ∈ C ′ . We claim that there exists a closed ideal J in C satisfying J − ⊆ M − ⊆ M ⊆ J where J − and M − are predecessors of J and M in chains C and C ′ , respectively. Let J be the intersection of all closed ideals in C thatcontain M . Let J − and M − be the predecessors of J and M in the chains C and C ′ ,respectively. Assume first that J − = J . If I is an arbitrary closed ideal in C that isproperly contained in J , then I is contained in M , so that J − is also contained in M .Since M ⊆ J and J − ⊆ M , we have J = J − = M = M − . Assume now that J − is aproper subset of J . Since C ′ is a chain, then J − is a subset of M , as J is the intersectionof all closed ideals of C that contain M , and J − is a closed linear span of all closed idealsof C that are contained in M . If J − = M , then J − is actually equal to J which isimpossible due to our assumption J − = J . Therefore J − is a proper subset of M whichimplies J − ⊆ M − so that we have J − ⊆ M − ⊆ M ⊆ J and the claim is proved.Assume now that the dimension of M / M − is at least 2. Since A J and B J commute on J / J − and J − ⊆ M − , the induced operators A M and B M of the operators A and B onthe quotient Banach space M / M − also commute. Suppose first that at least one of A M and B M is not a scalar multiple of the identity operator. Without any loss of generalityassume that A M is not a scalar multiple of the identity operator. If A M is nilpotent, thenits kernel is invariant under both A M and B M , so that the preimage of the kernel of theoperator A M is a closed subspace invariant under A and B that is properly containedbetween M − and M . This contradicts the maximality of C ′ . If A M is not nilpotent, thensome power ( A M ) n is a nonzero compact operator. By Lomonosov’s theorem [12] thereexists a nontrivial closed subspace of M / M − invariant under A M and B M . Similarly asabove this is again a contradiction with maximality of C ′ . If both A M and B M are scalarmultiples of the identity operator, then every subspace of M / M − is invariant under A M and B M which would (similarly as above) lead to a contradiction with the maximality ofthe chain C ′ . N THE POSITIVE COMMUTATOR IN THE RADICAL 11
Therefore the dimension of M / M − is less than or equal to one and the chain C ′ is atriangularizing chain for the operators A and B . (cid:3) In the proof of the preceding Proposition we applied the Lomonosov theorem to provethat the chain C ′ is a triangularizing chain for operators A and B . In the real case thiscannot be done always. If the dimension of M / M − is finite and at least three, then wecan proceed as in the proof of Proposition 4.3 also in the real case, since every real matrixon R n with n at least three has a nontrivial hyperinvariant subspace, which leads to acontradiction with the maximality of the chain C ′ . However, the problem occurs when thedimension of M / M − is precisely two. In the following example we construct a positiveindecomposable matrix A ∈ R with a two dimensional invariant subspace M such thatthe restriction of A to M is irreducible. Example 4.4.
Take the indecomposable matrix A = and M the vector spacespanned by − − and − . A straightforward calculations show that subspace M is invariant under A , and the restriction of A to M is irreducible, since the only (real)eigenspace of A is spanned by the vector which does not belong to M . It should be noted that every positive matrix on R is triangularizable since the spectralradius of a positive matrix is always in its spectrum.The following theorem is the main result of this paper. It provides a positive answer tothe remaining part of [4, Open questions 3.7(1)] and at the same time it tells us actuallythat the commutator AB − BA of a positive compact operator and a positive operatoris contained in the radical of a bigger Banach algebra which contains the Banach algebragenerated by A and B . Theorem 4.5.
Let A be a positive compact operator on a Banach lattice E and let A and A be the Banach algebras generated by h A ] and [ A i , respectively. (a) The operator AB − BA is in the radical of the Banach algebra A for every B ∈ A . (b) The operator AB − BA is in the radical of the Banach algebra A for every B ∈ A . Proof.
Let A be the algebra generated by h A ] and let us choose B and C from A .The operators B and C are finite linear combinations of words in letters from h A ].Let B , . . . , B n and C , . . . , C m be operators in h A ] which occur as letters in wordswhose appropriate linear combinations are equal to B and C , respectively. The oper-ator D := A + B + · · · + B n + C + · · · + C m satisfies 0 ≤ A ≤ D and the commutator AD − DA = n X i =1 ( AB i − B i A ) + m X j =1 ( AC j − C j A )is obviously a positive operator on E . Let C be a maximal chain of closed ideals invariantunder D . Maximality of the chain C implies that C is complete and that for every J ∈ C with dim( J / J − ) > D J is ideal-irreducible on J / J − . Also, every closedideal in C is invariant under A , B i and C j for all 1 ≤ i ≤ n and 1 ≤ j ≤ m , so that it isalso invariant under A , B and C . [10, Corollary 3.5] implies that A J and D J commuteon J / J − . This implies, in particular, that A J commutes with ( B i ) J for all 1 ≤ i ≤ n , sothat actually A J and B J commute on J / J − . By Ringrose’s theorem for complete chains[15, Theorem 7.2.7], it follows that σ ( C ( AB − BA )) = [ J ∈ CJ 6 = J − σ ( C J ( A J B J − B J A J )) ∪ { } = { } . If B and C in A are arbitrary, then there exist sequences of operators { B n } n ∈ N and { C n } n ∈ N in A converging to B and C , respectively, so that C n ( AB n − B n A ) convergesto C ( AB − BA ) . Since the norm limit of compact quasi-nilpotent operators is quasi-nilpotent, C ( AB − BA ) is quasi-nilpotent which finishes the proof of (a). The proof of(b) is similar so we omit it. (cid:3) Corollary 4.6.
Suppose that A and B be positive operators on a Banach lattice E witha positive commutator AB − BA . If at least one of the operators A and B is compact,then AB − BA is contained in the radical of the Banach algebra generated by A and B .Proof. Suppose that A is a compact operator. Let us denote by A the Banach algebragenerated by A and B , and let us denote by A the Banach algebra generated by h A ]. Since AB − BA is in the radical of A by Theorem 4.5(a), it follows that r ( C ( AB − BA )) = 0 foran arbitrary operator C ∈ A . The fact that we have A ⊆ A implies r ( C ( AB − BA )) = 0 N THE POSITIVE COMMUTATOR IN THE RADICAL 13 for every C ∈ A which means that AB − BA is in the radical of A . For the proof when B is compact we repeat the proof above and apply Theorem 4.5(b). (cid:3) In the case of power-compact operators we can obtain the following result.
Theorem 4.7. If A and B are positive power-compact operators on a Banach lattice witha positive commutator AB − BA , then some power of AB − BA is in the radical of theBanach algebra generated by A and B .Proof. Let A be the Banach algebra generated by A and B . By [8, Lemma 2.3], theoperator AB − BA is a power-compact operator. If AB − BA is nilpotent, then it is clearthat the apropriate power of the operator AB − BA is in the radical of A . Thefore wemay assume that AB − BA is not nilpotent. There exists a positive integer n such that( AB − BA ) n is a compact operator.Suppose first that the lattice E is a complex Banach lattice. Proposition 4.3 impliesthat A and B are simultaneously triangularizable. Let C be one of the triangularizingchains for operators A and B . It is not hard to see that C is a triangularizing chain forevery operator C from A . Since every diagonal coefficient of the operator AB − BA iszero, every diagonal coefficient of the operator C ( AB − BA ) n is zero as well. Compactnessof the operator ( AB − BA ) n and [15, Theorem 7.2.3] implies that C ( AB − BA ) n is quasi-nilpotent, and since C ∈ A was arbitrary we obtain that ( AB − BA ) n is contained in theradical of A .Assume now that E is a real Banach lattice. Let E C be the complexification of theBanach lattice E and let A C and B C be the standard complexifications of operators A and B on E C , respectively. It is not hard to see that we have A C B C ≥ B C A C on E C , sothat by Proposition 4.3 operators A C and B C are simultaneously triangularizable. Sincewe have ( A C B C − B C A C ) n = (( AB − BA ) C ) n = (( AB − BA ) n ) C , the operator ( A C B C − B C A C ) n is a nonzero compact operator on E C . By the proof abovewe see that ( A C B C − B C A C ) n is contained in the radical of the algebra A C which impliesthat r ( C C ( A C B C − B C A C ) n ) = 0 for all C C ∈ A C . This implies that r ( C ( AB − BA ) n ) = 0for all C ∈ A finishing the proof. (cid:3) A result for the Volterra operator and the Donoghue operator
An operator on a Banach space is said to be unicellular if its lattice of closed invari-ant subspaces is totally ordered. An application of the Lomonosov theorem shows thatevery compact operator is triangularizable which implies that every unicellular compactoperator has a unique triangularizing chain. The Volterra operator acting on L [0 , V f )( x ) = R x f ( t ) dt is a well-known example of a unicellular operator [14]with its lattice of closed invariant subspaces consisting of subspaces of the form L [ t, L [0 ,
1] which are zero almost everywhere on the interval [0 , t ] . Volterra operator has a lot of interesting properties. For an example, if a normal operator N commutes with a Volterra operator, then N is a multiple of the identity operator [9].An operator T on a Banach space X is said to be a strong quasi-affinity whenever forarbitrary closed subspace M invariant under T we have that T M is dense in M .The Volterra operator has this remarkable property. It is not hard to see that therange of V is dense in L [0 , . The function x x − t − t induces a topological isomorphismΦ : L [0 , → L [ t,
1] defined by (Φ f )( x ) = f (cid:18) x − t − t (cid:19) . Let us denote by V t the restriction of the Voterra operator to its invariant subspace L [ t, . It is not hard to see that V t satisfies( V t Φ f )( s ) = (1 − t ) Z s − t − t f ( v ) dv = (1 − t )(Φ V f )( s )for all s ∈ [ t, . Since V has a dense range, V t also has a dense range.The other example of a unicellular operator is the Donoghue operator on a separableHilbert space H . Let { e n } n ∈ N be an orthonormal basis of H and let { w n } n ∈ N be asequence of nonzero complex numbers such that {| w n |} n ∈ N is monotone decreasing andis in the space l . The
Donoghue operator S with a weight sequence { w n } n ∈ N is definedby Se = 0 and Se n = w n e n − for n > . The operator S is a unicellular operator [14]with its lattice of closed invariant subspaces equal to {M n } n ∈ N where M n is the linearspan of the vectors e , . . . , e n . An easy calculation shows that S M n = M n − , so that theinduced operator by the operator S on the quotient Banach space M n / M m is nonzerowhenever the dimension of M n / M m is at least 2. The later fact will be considered in N THE POSITIVE COMMUTATOR IN THE RADICAL 15 general in Proposition 5.1. We will be only concerned with the positive (the weights arepositive) Donoghue operator acting on l ordered componentwise.It is well-known that the Volterra operator and the Donoghue operator are compactoperators. Proposition 5.1.
Let X be a Banach space and let T be a compact unicellular operatoron X . Then for an arbitrary subspace M from the triangularizing chain C of T only oneof the following statements hold. (a) M = T M . (b) M = M − and T M is of codimension in M . (c) T M = M − is of codimension in M .Moreover, if C is a continuous chain, then T is a strong quasi-affinity. It should be noted that Proposition 5.1 immediately implies that V is a strong quasi-affinity and that V is quasi-nilpotent by [15, Example 7.2.5]. Proof.
Assume that T M is a proper subspace of M . We claim that the dimension of thequotient space M /T M is at most one. Otherwise, one can find two proper incomparableclosed subspaces N and N of M that properly contain T M . Since N and N are alsoinvariant under T , this contradicts unicellularity of T . Unicelullarity of T also implies that T M ∈ C . Since the codimension of M − in M is also at most one, we have M − = T M or M − = M . If C is a continuous chain, then it is obvious that (a) holds. (cid:3) Proposition 5.2.
Let E be a Banach lattice and let A be a positive compact unicellularoperator on E such that the unique triangularizing chain for T consists of closed idealsof E . If a positive operator B semi-commutes with A , then A and B are simultaneouslytriangularizable.Proof. Let C be a maximal chain of closed subspaces invariant under both A and B . Bythe assumption, every subspace in C is an ideal of E . Due to the maximality, C is acomplete chain. Suppose now that the dimension of J / J − is at least two for some J ∈ C .Since the operator A J + B J is ideal-irreducible on J / J − and semi-commutes with the operator A J , [10, Corollary 3.5] implies( A J + B J ) A J = A J ( A J + B J ) , so that we have A J B J = B J A J . We claim that A J is nonzero operator on J / J − .Otherwise we have A J ⊆ J − , and since Proposition 5.1 implies that the codimension of A J is at most one, we have that the dimension of J / J − is at most one as well and wereached to a contradiction.The operator A J is not a scalar multiple of the identity operator on J / J − as otherwise A would have at least two incomparable closed invariant subspaces which contradicts itsunicellularity. By [17, Corollary 2.4] (if E is a real lattice) or [12] (if E is a complexlattice), the operator A J has a nontrivial hyperinvariant subspace. However, this is incontradiction with the maximality of the chain C . Therefore, C is a maximal chain ofclosed subspaces of E . (cid:3) Whenever one of the operators A and B in Proposition 4.3 is the Volterra operator ora Donoghue operator, then the other operator does not need to have any compactnessproperties. Corollary 5.3.
Let V be the Volterra operator on L [0 , and let S be the Donoghueoperator on l . Then the following statements hold. (a)
If a positive operator T on L [0 , semi-commutes with V then V and T aresimultaneously ideal-triangularizable. (b) If a positive operator T on l semi-commutes with S , then S and T are simulta-neously ideal-triangularizable. In [4, Example 3.3], a Banach algebra generated by the Volterra operator V and amultiplication operator M on L [0 ,
1] defined by (
M f )( x ) = xf ( x ) was considered. Sincethe relation M V − V M = V implies M V ≥ V M ≥
0, Corollary 4.6 implies that
M V − V M is contained in the radical of the Banach algebra generated by V and M .The authors [4] observed that this actually follows from the fact that V and M aresimultaneously ideal-triangularizable with the unique triangularizing chain { L [ t, } t ∈ [0 , . The following result states that the Volterra operator is actually in the radical of thepossibly much bigger Banach algebra generated by h V ] and [ V i . N THE POSITIVE COMMUTATOR IN THE RADICAL 17
Theorem 5.4.
Let V be the Volterra operator on L [0 , and let S be the Donoghueoperator on l . Then V is in the radical of the Banach algebra generated by h V ] and [ V i ,and S is in the radical of the Banach algebra generated by h S ] and [ S i . Proof.
By Corollary 5.3 the set h V ] ∪ [ V i is triangularizable, so that the closure of thealgebra A generated h V ] and [ V i is triangularizable as well. [15, Theorem 7.2.4] impliesthat the operator AV is quasi-nilpotent for all A ∈ A which finishes the proof.For the proof in the case of the Donoghue operator note first that it is quasi-nilpotentby [11, Problem 96] and its solution. [15, Theorem 7.2.3] implies that every diagonalcoefficient of the operator S is zero. Since h S ] ∪ [ S i is triangularizable by Corollary 5.3,the Banach algebra B generated by h S ] ∪ [ S i is triangularizable as well. For all B ∈ B everydiagonal coefficient of the operator BS is zero, so that the operator BS is quasi-nilpotentby [15, Theorem 7.2.3] and the proof is finished. (cid:3) Remark 5.5.
Let V be the Volterra operator and let A be the Banach algebra generated byall compact operators in h V ] ∪ [ V i . Then A is triangularizable with the unique triangular-izing chain { L [ t, } t ∈ [0 , . By [15, Corollary 7.2.4] every operator in A is quasi-nilpotent,so that A is a radical Banach algebra. Remark 5.6.
Suppose that A and B be positive operators on a Banach lattice E with apositive commutator AB − BA . If one of the operators is power compact, then AB − BA isquasi-nilpotent by [5, Corollary 3.2] . In particular, this implies that a positive commutator AB − BA of positive power compact operators A and B is quasi-nilpotent which also followsfrom Theorem 4.7. We finish this paper with two open questions.(a) How big is the Banach algebra generated by h V ] ∪ [ V i ?(b) Can we assume in Theorem 4.7 that only one of the operators A and B is powercompact? Acknowledgments.
This work was supported in part by the Slovenian Research Agency.
References [1] Y. A. Abramovich, C. D. Aliprantis,
An invitation to operator theory , American MathematicalSociety, Providence, 2002.[2] C.D. Aliprantis, O. Burkinshaw,
Positive operators , reprint of the 1985 original, Springer, Dor-drecht, 2006.[3] F. F. Bonsall, J. Duncan,
Complete normed algebras , Ergebnisse der Mathematik und ihrer Gren-zgebiete, Band 80. Springer-Verlag, New York-Heidelberg, 1973.[4] J. Braˇciˇc, R. Drnovˇsek, Y. B. Farforovskaya, E. L. Rabkin, J. Zem´anek,
On positive commutators ,Positivity Once more on positive commutators , Studia Math.
11 (2012), 241–245.[6] R. Drnovˇsek,
Triangularizing semigroups of positive operators on an atomic normed Riesz spaces ,Proc. Edin. Math. Soc. (2000), 43–55.[7] R. Drnovˇsek, M. Kandi´c, Ideal-triangularizability of semigroups of positive operators , IntegralEquations and Operator Theory (2009), 539–552.[8] R. Drnovˇsek, M. Kandi´c, More on positive commutators , J. Math. Anal. Appl. (2011), 580–584.[9] A. F. M. ter Elst, M. Sauter, J. Zem´anek,
Generation and commutation properties of the Volterraoperator , Arch. Math. (2012), no. 5, 467–479.[10] N. Gao, On commuting and semi-commuting positive operators , Proc. Amer. Math. Soc., in press.[11] P. Halmos,
A Hilbert space problem book , Second edition, Graduate Texts in Mathematics,Springer-Verlag, New York-Berlin, 1982[12] V. Lomonosov,
Invariant subspaces of the family of operators that commute with a completelycontinuous operator . (Russian) Funkcional. Anal. i Priloen. (1973), no. 3, 55–56.[13] W. A. J. Luxemburg, A. C. Zaanen, Riesz spaces I , North Holland, Amsterdam, 1971.[14] H. Radjavi, P. Rosenthal,
Invariant subspaces , Ergebnisse der Mathematik und ihrer Grenzgebiete,Band 77. Springer-Verlag, New York-Heidelberg, 1973.[15] H. Radjavi, P. Rosenthal,
Simultaneous Triangularization , Springer-Verlag, New York, 2000.[16] H. H. Schaefer,
Banach lattices , Springer-Verlag, Berlin-Heidelberg-New York, 1974.[17] G. Sirotkin,
A Version of The Lomonosov Subspace Theorem for Real Banach Spaces
IndianaUniv. Math. J. (2005), no. 1, 257–262.[18] A.C. Zaanen, Introduction to Operator Theory in Riesz Spaces , Springer-Verlag Berlin Heidelberg,1997., Springer-Verlag Berlin Heidelberg,1997.