aa r X i v : . [ m a t h . F A ] F e b LOCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES
JANKO BRA ˇCI ˇC
Abstract.
For an operator A on a complex Banach space X and a closed subspace M ⊆ X , the local commutant of A at M is the set C ( A ; M ) of all operators T on X suchthat T Ax = AT x for every x ∈ M . It is clear that C ( A ; M ) is a closed linear space ofoperators, however it is not an algebra, in general. For a given A , we show that C ( A ; M )is an algebra if and only if the largest subspace M A such that C ( A ; M ) = C ( A ; M A ) isinvariant for every operator in C ( A ; M ). We say that these are ultrainvariant subspacesof A . For several types of operators we prove that there exist non-trivial ultrainvariantsubspaces. For a normal operator on a Hilbert space, every hyperinvariant subspace isultrainvariant. On the other hand, the lattice of all ultrainvariant subspaces of a non-zeronilpotent operator can be strictly smaller than the lattice of all hyperinvariant subspaces. Introduction
Let X be a complex Banach space and let B ( X ) be the Banach algebra of all boundedlinear operators on X . The dual space of X will be denoted by X ∗ . For A ∈ B ( X ) and aclosed subspace M ⊆ X , the local commutant of A at M is C ( A ; M ) = { T ∈ B ( X ); T Ax = AT x for all x ∈ M } . It is clear that ( A ) ′ , the commutant of A , is a subset of C ( A ; M ).Of course, C ( A ; X ) = ( A ) ′ and C ( A ; { } ) = B ( X ).Local commutants were introduced by Larson [13, 6] more than twenty years ago,however there are only a few results related to them (see [4], for instance) although thatthere are some interesting open problems related to them. One among these problems isthe question when is C ( A ; M ) an algebra. Of course, for an arbitrary A and M which isequal either to { } or X the local commutant is an algebra. However, for A which is nota scalar multiple of I X , the identity operator on X , there exists a closed subspace M ⊆ X such that C ( A ; M ) is not a subalgebra of B ( X ) but just a linear subspace. Mathematics Subject Classification.
Primary 47A15 47L05 .
Key words and phrases.
Commutant, local commutant, hyperinvariant subspace, ultrainvariantsubspace.
Recall that a closed subspace M ⊆ X is said to be invariant for A ∈ B ( X ) if A M ⊆ M .It is well-known that the family of all closed subspaces which are invariant for A form alattice; the lattice operations are the intersection and the closed linear span. As usual,we will denote this lattice by Lat( A ). A sublattice of Lat( A ) is Lat h ( A ), the lattice ofall hyperinvariant subspaces of A : a subspace M ∈ Lat( A ) belongs to Lat h ( A ) if andonly if T M ⊆ M for every T ∈ ( A ) ′ . It is obvious that the trivial lattice {{ } , X } is asublattice of Lat h ( A ) and therefore of Lat( A ). In 1970s, Enflo showed that there existsa Banach space X and a bounded operator A on it such that Lat( A ) is trivial. Theresult was published much later in [9] and meanwhile Read published his example of anoperator without a non-trivial invariant subspace, see [17]. Nowadays several examples ofBanach spaces (including l ) with operators whose lattice of invariant subspaces is trivialare known. On the other hand, Lomonosov [15] proved that an operator has a non-trivialinvariant subspace if it commutes with an operator which is not a scalar multiple of I X andcommutes with a non-zero compact operator. It follows that for some infinite dimensionalBanach spaces X (see example given by Argyros and Haydon [2]) every operator in B ( X )has a non-trivial invariant subspaces.However for many Banach spaces (for instance, for the infinite dimensional separableHilbert space) it is not known if Lat( A ) is non-trivial for every operator A . Similarly, formore or less the same class of Banach spaces, it is not known if Lat h ( A ) can be trivialfor an operator A which is not a scalar multiple of I X . Work on these problems is stillvery active, see [20] and references therein. The reader is referred to [5] and [16] for moredetails about the problem of invariant subspaces.It turns out that our problem about the algebraic structure of C ( A ; M ) is connectedwith hyperinvariant subspaces of A . Our main observation is the following. Let A ∈ B ( X )and let M ⊆ X be a closed subspace. Denote by M A ⊆ X the set of all vectors x ∈ X suchthat T Ax = AT x for every T ∈ C ( A ; M ). Then M A is a closed subspace of X such that M ⊆ M A and C ( A ; M ) = C ( A ; M A ). The local commutant C ( A ; M ) is an algebra if andonly if M A is invariant for every operator in C ( A ; M ); we say that M A is an ultrainvariantsubspace of A . Since ( A ) ′ ⊆ C ( A ; M ) an ultrainvariant subspace is a hyperinvariantsubspace, as well. The opposite is not true, in general.Ultrainvariant subspaces are the main theme of the second part of this paper. Denoteby Lat u ( A ) the family of all ultrainvariant subspaces of A . It is a sublattice of Lat h ( A ). OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 3
For several types of operators A we show that Lat u ( A ) is not trivial. There are classesof operators for which we are able to characterize the lattice of ultrainvariant subspacescompletely. For instance, for algebraic operators, in particular, for operators on a finitedimensional Banach space, and for normal operators on a Hilbert space. If A is a unicel-lular operator on a Hilbert space, then Lat u ( A ) = Lat h ( A ) = Lat u ( A ). This holds, forinstance, for the Volterra operator and Donoghue operators.Several questions about ultrainvariant subspaces remain open. For instance, we do notknow if every compact operator has a non-trivial ultrainvariant subspace. Actually webelieve that it is hard to find an operator which has a non-trivial hyperinvariant subspacebut does not have a non-trivial ultrainvariant subspace.In the end of this section, let us say a few words about the terminology. Sz.-Nagy andFoia¸s initiated the study of hyperinvariant subspaces in their book Analyse harmoniquedes op´erateurs de l’espace Hilbert however they called them ultrainvariant subspaces. Inthis paper we use their terminology (see Definition 4.1) for a special type of hyperinvariantsubspaces. Another new term which we use is girder . The girder of C ( A ; M ) is the largestsubspace M A ⊆ X such that C ( A ; M ) = C ( A ; M A ). In our opinion the term girder issuitable as M A “carries” the whole space C ( A ; M ) in the sense that M A is the largestsubspace whose vectors give the information whether T belongs to C ( A ; M ) or not.2. Local commutants
Although that we are mainly interested in local commutants we begin our study byconsidering a slightly more general objects. Let X and Y be complex Banach spaces andlet B ( X , Y ) be the space of all bounded linear operators from X to Y . For A ∈ B ( X ) and B ∈ B ( Y ), let I ( A, B ) be the set of all operators S ∈ B ( X , Y ) which intertwine A and B , that is, I ( A, B ) = { S ∈ B ( X , Y ); SA = BS } . Of course, if X = Y and A = B , then I ( A, A ) = ( A ) ′ . Let M ⊆ X be a closed subspace. The space of operators which intertwine A and B locally at M is I ( A, B ; M ) = { S ∈ B ( X , Y ); SAx = BSx for every x ∈ M } . Inparticular, I ( A, A ; M ) = C ( A ; M ) is the local commutant of A at M .It is easily seen that I ( A, B ; M ) is closed in the strong operator topology and that I ( A, B ; M ) = I ( A − λI X , B − λI Y ; M ), for every λ ∈ C . For arbitrary invertible operators U ∈ B ( X ) and V ∈ B ( Y ), we have(1) I ( A, B ; M ) = V − I ( U AU − , V BV − ; U M ) U. J. BRA ˇCI ˇC If M , M are closed subspaces of X such that M ⊆ M , then I ( A, B ; M ) ⊆ I ( A, B ; M ).In particular, I ( A, B ) = I ( A, B ; X ) ⊆ I ( A, B ; M ) ⊆ I ( A, B ; { } ) = B ( X , Y ).Let { M j ; j ∈ J } be a non-empty family of closed subspaces of X . It is not hard to seethat _ j ∈ J { I ( A, B ; M j ); j ∈ J } ⊆ I ( A, B ; \ j ∈ J M j ) . On the other hand,(2) I ( A, B ; _ j ∈ J M j ) = \ j ∈ J I ( A, B ; M j ) . Here we denoted by W j ∈ J the closed linear span of the involved sets.To verify (2), note that I ( A, B ; W j ∈ J M j ) ⊆ T j ∈ J I ( A, B ; M j ) holds since M i ⊆ W j ∈ J M j andtherefore I ( A, B ; W j ∈ J M j ) ⊆ I ( A, B ; M i ), for every i ∈ J . For the opposite inclusion,observe that BSx j = SAx j for every x j ∈ M j and every j ∈ J if S ∈ T j ∈ J I ( A, B ; M j ).Hence S ∈ I ( A, B ; W j ∈ J M j ). (Here we used a simple fact that BSx = SAx holds for every x ∈ W U if it holds for every x ∈ U ⊆ X .)Let A ∈ B ( X ), B ∈ B ( Y ) and let M ⊆ X be a complemented closed subspace, that is,there exists a closed subspace N ⊆ X such that X = M ⊕ N . Let (cid:2) A A A A (cid:3) be the operatormatrix of A with respect to this decomposition. For arbitrary operators S ∈ B ( X , Y ) and T ∈ B ( X ), let [ S S ], respectively (cid:2) T T T T (cid:3) , be their operator matrices with respect tothe decomposition X = M ⊕ N . These assumptions and notation will be used throughoutthis section.A straightforward computation shows that S ∈ I ( A, B ) if and only if(3) BS = S A + S A and(4) BS = S A + S A . Similarly, T ∈ ( A ) ′ if and only if(5) T A + T A = A T + A T , T A + T A = A T + A T and(6) T A + T A = A T + A T , T A + T A = A T + A T . OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 5
Lemma 2.1.
Operator S ∈ B ( X , Y ) is in I ( A, B ; M ) if and only if (3) holds and T ∈ B ( X ) is in C ( A ; M ) if and only if (5) holds.Proof. Let x ∈ X be arbitrary and let x = x ⊕ x , where x ∈ M and x ∈ N . Then SAx = ( S A + S A ) x + ( S A + S A ) x and BSx = BS x + BS x . Hence, if S ∈ I ( A, B ; M ), that is, BS ( x ⊕
0) = SA ( x ⊕
0) for every x ∈ M , then BS x = ( S A + S A ) x for every x ∈ M which gives (3). It is clear that the opposite implication holds,as well. The second part of the assertion follows by the first one if we put B = (cid:2) A A A A (cid:3) , S = (cid:2) T T (cid:3) and S = (cid:2) T T (cid:3) in (3). (cid:3) It is clear that I ( A, B ; { } ) = B ( X , Y ) for arbitrary operators A and B . Let M = { } .The following corollary characterizes those pairs of operators A ∈ B ( X ) and B ∈ B ( Y )for which I ( A, B ; M ) = B ( X , Y ). Proposition 2.2.
Let M = { } . Then I ( A, B ; M ) = B ( X , Y ) if and only if A = λI M , B = λI Y , for some λ ∈ C , and A = 0 . In particular, C ( A ; M ) = B ( X ) if and only if A is a scalar multiple of I X .Proof. Assume that I ( A, B ; M ) = B ( X , Y ). Let S = [ S S ∈ B ( M , Y ) isarbitrary. By Lemma 2.1, BS = S A . Let y ∈ Y and 0 = ξ ∈ M ∗ be arbitrary. Then y ⊗ ξ ∈ B ( M , Y ) and therefore B ( y ⊗ ξ ) = ( y ⊗ ξ ) A . If x ∈ M is such that h x, ξ i = 1,then By = B ( y ⊗ ξ ) x = ( y ⊗ ξ ) A x = h A x, ξ i y . Hence, B ∈ B ( Y ) is an operator suchthat every 0 = y ∈ Y is its eigenvector. It follows that there exists λ ∈ C such that B = λI Y . Now we have λS = S A for every S ∈ B ( M , Y ) which gives A = λI M .Since every S = [ S S ] ∈ B ( X , Y ) satisfies (3) we have S A = 0 for every S ∈ B ( N , Y ).Hence, A = 0. The opposite implication is simple: if A = λI M , B = λI Y and A = 0,then (3) holds for every S = [ S S ].It follows from the first part of this corollary that in the case when X = Y and A = B we have C ( A ; M ) = B ( X ) if and only if A = λI X for some λ ∈ C . (cid:3) Equality I ( A, B ; M ) = I ( A, B ) holds if and only if every pair of operators S ∈ B ( M , Y )and S ∈ B ( N , Y ) which satisfies (3) satisfies (4), as well. In the following propositionwe determine those pairs A ∈ B ( X ) and B ∈ B ( Y ) for which I ( A, B ; M ) = I ( A, B ) holdsfor a given subspace M ∈ Lat( A ), M = X . We will use the following notation. For S ∈ B ( X , Y ), the kernel of S is denoted by ker( S ) and the range (image) of S by im( S ). J. BRA ˇCI ˇC If Z is a non-empty set of vectors in X , then by Z ⊥ we denote its annihilator in X ∗ , thatis, Z ⊥ = { ξ ∈ X ∗ ; h x, ξ i = 0 for all x ∈ Z } . Proposition 2.3.
Let M = X . Then I ( A, B ; M ) = I ( A, B ) and M ∈ Lat( A ) if and onlyif there exists λ ∈ C such that A = h A A λI N i , B = λI Y and im( A ) ⊆ im( λI M − A ) .In this case, I ( A, B ; M ) = { [ S S ] ∈ B ( X , Y ); im( λI M − A ) ⊆ ker( S ) , S ∈ B ( N , Y ) } .Proof. Assume that I ( A, B ; M ) = I ( A, B ), where X = M ∈ Lat( A ). Then the operatormatrix of A with respect to the decomposition X = M ⊕ N is A = (cid:2) A A A (cid:3) . Hence, byLemma 2.1, an operator S = [ S S ] ∈ B ( X , Y ) intertwines A and B at M if and only if itsatisfies (3) which in our case means(7) BS = S A . It follows that [0 S ] ∈ I ( A, B ; M ), and therefore [0 S ] ∈ I ( A, B ), for every S ∈ B ( N , Y ).Thus, [0 S ] satisfies (4), that is, BS = S A holds for every S ∈ B ( N , Y ). By Proposi-tion 2.2, there exists λ ∈ C such that A = λI N and B = λI Y . Now (7) reads as(8) S ( λI M − A ) = 0 . Hence, if [ S S ] is an operator in I ( A, λI Y ; M ), then im( λI M − A ) ⊆ ker( S ) and S isarbitrary. Note that [ S S ] is in I ( A, λI Y ) if and only if it satisfies (4) which reads as(9) S A = 0 . in our case. Since I ( A, λI Y ; M ) = I ( A, λI Y ) every S which satisfies (8) has to satisfy (9),as well. Suppose towards a contradiction that there exists x = A u ∈ im( A ) such that x im( λI M − A ). Let ξ ∈ M ∗ be such that ξ ∈ im( λI M − A ) ⊥ and h x, ξ i = 1. For any0 = y ∈ Y , the rank one operator y ⊗ ξ ∈ B ( M , Y ) satisfies ( y ⊗ ξ )( λI M − A ) = 0, whichmeans y ⊗ ξ ∈ I ( A, λI Y ; M ), however ( y ⊗ ξ ) A = 0 since ( y ⊗ ξ ) A u = h x, ξ i y = 0. Thisproves that im( A ) ⊆ im( λI M − A ).The opposite implication is easily checked and one gets that I ( A, B ; M ) = { [ S S ] ∈ B ( X , Y ); im( λI M − A ) ⊆ ker( S ) , S ∈ B ( N , Y ) } . (cid:3) Corollary 2.4. If M ∈ Lat( A ) , M = X , and A is not a scalar multiple of I X , then ( A ) ′ ( C ( A ; M ) . OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 7
Theorem 2.5.
Let A ∈ B ( X ) and assume that A is not a scalar multiple of I X . Let M ∈ Lat( A ) , { } 6 = M = X . The local commutant C ( A ; M ) is an algebra if and only if C ( A ; M ) = { T ∈ B ( X ); T ∈ ( A ) ′ and T = 0 } .Proof. We begin the proof with the following simple observation.
Claim.
Let X , X , X and X be Banach spaces. If T ∈ B ( X , X ) and R ∈ B ( X , X ) aresuch that R ( u ⊗ ξ ) T = 0 for all u ∈ X and ξ ∈ X ∗ , then either R = 0 or T = 0.Indeed! Assume that T = 0. Then there exists x ∈ X such that T x = 0. Let ξ ∈ X ∗ besuch that h T x, ξ i = 1. For an arbitrary u ∈ X we have, by the assumption, R ( u ⊗ ξ ) T = 0and therefore R ( u ⊗ ξ ) T x = Ru = 0. This shows that R = 0.Assume that C ( A ; M ) is an algebra. Since M ∈ Lat( A ) the operator matrix of A is ofthe form (cid:2) A A A (cid:3) . Hence, by Lemma 2.1, an operator T = (cid:2) T T T T (cid:3) is in C ( A ; M ) if andonly if(10) T A − A T = A T , T A − A T = 0and T ∈ B ( N , M ), T ∈ B ( N ) are arbitrary. In particular, the projection P onto M along N is in C ( A ; M ). Since C ( A ; M ) is assumed to be an algebra we see that operators P T P and ( I X − P ) T P are in C ( A ; M ) for every T ∈ C ( A ; M ). Note that the operatormatrices of P T P and ( I X − P ) T P are (cid:2) T
00 0 (cid:3) and (cid:2) T (cid:3) , respectively. Hence, if T ∈ C ( A ; M ), then the entries of its operator matrix have to satisfy(11) T A − A T = 0 , A T = 0 , T A − A T = 0 . Suppose that A = 0. Let u ∈ N and ξ ∈ N ∗ be arbitrary. Then U whose operatormatrix is (cid:2) u ⊗ ξ (cid:3) satisfies (10) and therefore U ∈ C ( A ; M ) which gives U ( I X − P ) T P ∈ C ( A ; M ). The operator matrix of U ( I X − P ) T P is (cid:2) u ⊗ ξ ) T (cid:3) and the entries of thismatrix satisfy (11); in particular, A ( u ⊗ ξ ) T = 0. Since u ⊗ ξ ∈ B ( N ) is an arbitraryoperator of rank at most 1 and A = 0 we conclude, by Claim, that T = 0.Now we may assume that A = 0. If A = λI M and A = µI N , then λ = µ as A is not a scalar multiple of I X . It follows, by the third equality in (11), that T = 0.Thus, it remains to consider the case when either A is not a scalar multiple of I M or A is not a scalar multiple of I N . Suppose that this holds. Let Q ∈ B ( N , M ) bearbitrary and let V ∈ B ( X ) be the operator whose operator matrix is (cid:2) Q (cid:3) . Then V ∈ C ( A ; M ) and therefore V ( I X − P ( T P ∈ C ( A ; M ) for every T ∈ C ( A ; M ). The J. BRA ˇCI ˇC operator matrix of V ( I X − P ( T P ∈ C ( A ; M ) is (cid:2) QT
00 0 (cid:3) . Hence, QT ∈ ( A ) ′ . Since T satisfies the third equality in (11) we have A QT = QT A = QA T which gives( A Q − QA ) T = 0. We have supposed that either A is not a scalar multiple of I M or A is not a scalar multiple of I N . Hence, by Proposition 2.2, there exists Q ∈ B ( N , M )such that R = A Q − Q A = 0. Thus, RT = 0 for every T ∈ C ( A ; M ). Let T befixed now. As in the previous paragraph, let u ∈ N and ξ ∈ N ∗ be arbitrary and let U ∈ C ( A ; M ) be the operator whose operator matrix is (cid:2) u ⊗ ξ (cid:3) . Then U ( I X − P ) T P ∈ C ( A ; M ) and therefore R ( u ⊗ ξ ) T = 0. Since u ⊗ ξ ∈ B ( N ) is an arbitrary operatorof rank at most 1 and R = 0 we conclude, by Claim, that T = 0. This proves that C ( A ; M ) = { T ∈ B ( X ); T ∈ ( A ) ′ and T = 0 } if C ( A ; M ) is assumed to be an algebra.The opposite implication is obvious. (cid:3) In the proof of Theorem 2.5 we have seen that the projection P which maps onto M along N is in C ( A ; M ) if M ∈ Lat( A ). It is not hard to see that the opposite implicationholds as well. Corollary 2.6.
Projection P is in C ( A ; M ) if and only if M is invariant for A . Let σ ( T ) denote the spectrum of T ∈ B ( X ). It follows from a well-known theoremproved by Rosenblum that I ( A, B ) = { } if σ ( A ) ∩ σ ( B ) = ∅ , see [18, Corollary 3.3]and [10, Theorem I.4.1]. Note that the condition σ ( A ) ∩ σ ( B ) = ∅ is not necessary for I ( A, B ) = { } . For instance, let B = 0 and let A ∈ B ( X ) be such that im( A ) = X andker( A ) = { } . Then 0 ∈ σ ( A ) ∩ σ ( B ), however I ( A,
0) = { } because SA = 0 holdsfor S ∈ B ( X , Y ) if and only if S = 0 as im( A ) = X . On the other hand, note that I (0 , A ) = { } . Indeed, if S ∈ B ( Y , X ) is such that im( S ) ⊆ ker( A ), then AS = 0. Corollary 2.7. If M ∈ Lat( A ) and σ ( A ) ∩ σ ( A ) = ∅ , then C ( A ; M ) is an algebra.Proof. By Lemma 2.1, T ∈ C ( A ; M ) if and only if (5) holds. Since A = 0 the equalitiesin (5) simplify to (10). Because of σ ( A ) ∩ σ ( A ) = ∅ , only T = 0 satisfies the secondequation in (10). Thus, C ( A ; M ) = { T ∈ B ( X ); T ∈ ( A ) ′ and T = 0 } and thereforeit is an algebra, by Theorem 2.5. (cid:3) In the following example we will see that C ( Q ; L ) can be an algebra for Q ∈ B ( X ) anda closed subspace L ⊆ X which is not invariant for Q . OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 9
Example 2.8.
Let X , X , X be closed subspaces such that X , X are non-trivial. Let X = X ⊕ X ⊕ X and suppose that there exists a surjective operator in B ( X , X ). Let Q ∈ B ( X ) be the projection onto X along X ⊕ X . Choose and fix a surjective operator U ∈ B ( X , X ). Let L ⊆ X be the subspace of all vectors x ∈ X which are of the form x = x ⊕ U x ⊕ x , where x ∈ X and x ∈ X are arbitrary. Denote by P j : X → X j ( j = 1 , ,
3) the operators given by P j ( x ⊕ x ⊕ x ) = x j . It is clear that P L = X and P L = X . Since U is surjective we have P L = X , as well. Note that L Lat( Q ).Namely, if x = x ⊕ U x ⊕ x ∈ L is such that U x = 0, then Qx = x ⊕ ⊕ L .Assume that T ∈ C ( Q ; L ). Let [ T ij ] i,j =1 be its operator matrix with respect to thedecomposition X = X ⊕ X ⊕ X . For an arbitrary x = x ⊕ U x ⊕ x ∈ L , we have T Qx = T x ⊕ T x ⊕ T x and QT x = ( T x + T U x + T x ) ⊕ ⊕
0. Since x ∈ X and x ∈ X are arbitrary and U is surjective it follows from T Qx = QT x that T = 0, T = 0, T = 0 and T = 0. Hence T ∈ ( Q ) ′ , that is C ( Q ; L ) = ( Q ) ′ ; in particular, C ( Q ; L ) is an algebra.Note that this example shows that C ( A ; M ) and ( A ) ′ can be equal for an operator whichis not a scalar multiple of I X and M = X which is not invariant for A (cf. Corollary 2.4). (cid:3) Although that the condition M ∈ Lat( A ) does not need to be fulfilled for C ( A ; M ) beingan algebra the invariance is not far away as we shall see in the following section.3. Multiplicative structure of the space of local intertwiners
Let A ∈ B ( X ) and let M ⊆ X be a closed subspace. We know that C ( A ; M ) is not analgebra in general. However it has some non-trivial multiplicative structure. For instance,let T ∈ ( A ) ′ be such that M ∈ Lat( T ). Then AST x = SAT x = ST Ax for all S ∈ C ( A ; M )and x ∈ M . Hence, C ( A ; M ) T ⊆ C ( A ; M ). The following proposition is a generalizationof this simple observation. Proposition 3.1.
Let A ∈ B ( X ) , B ∈ B ( Y ) and let M ⊆ X be a closed subspace.Inclusion I ( A, B ; T M ) T ⊆ I ( A, B ; M ) holds for every T ∈ C ( A ; M ) . If T is invertible,then I ( A, B ; T M ) T = I ( A, B ; M ) .Proof. Let T ∈ C ( A ; M ). For arbitrary S ∈ I ( A, B ; T M ) and x ∈ M , we have ST Ax = SAT x = BST x and therefore ST ∈ I ( A, B ; M ). Assume now that T ∈ C ( A ; M ) is invertible. Let S ∈ I ( A, B ; M ) be arbitrary. Thenfor every x ∈ M we have ST − AT x = SAx = BSx = BST − T x which implies that ST − ∈ I ( A, B ; T M ), that is, S ∈ I ( A, B ; T M ) T . (cid:3) For a non-empty set of operators
S ⊆ B ( X , Y ) and a closed subspace M ⊆ X , let S M denote the closed linear span in Y of all subspaces S M , where S ∈ S , that is, S M = W S ∈S S M . Theorem 3.2.
Let A ∈ B ( X ) , B ∈ B ( Y ) and let M ⊆ X be a closed subspace. Then C (cid:0) B ; I ( A, B ; M ) M (cid:1) is a subalgebra of B ( Y ) and it is the largest subalgebra of B ( Y ) suchthat I ( A, B ; M ) is a left module over it.Proof. Let us show that C (cid:0) B ; I ( A, B ; M ) M (cid:1) · I ( A, B ; M ) ⊆ I ( A, B ; M ). Assume that T ∈ C (cid:0) B ; I ( A, B ; M ) M (cid:1) and S ∈ I ( A, B ; M ). For every x ∈ M , we have BSx = SAx and therefore
BT Sx = T BSx = T SAx because Sx ∈ I ( A, B ; M ) M . We conclude that T S ∈ I ( A, B ; M ).Let T , T ∈ C (cid:0) B ; I ( A, B ; M ) M (cid:1) . If S ∈ I ( A, B ; M ), then T S ∈ I ( A, B ; M ), by theinclusion from the previous paragraph, and, by the same reason, T T S ∈ I ( A, B ; M ).Hence, for every S ∈ I ( A, B ; M ) and every x ∈ M , we have BT T Sx = T T SAx = T T BSx . Because of linearity and countinuity we may conclude that T T commuteswith B on the subspace I ( A, B ; M ) M of Y , that is, T T ∈ C (cid:0) B ; I ( A, B ; M ) M (cid:1) . Thus, C (cid:0) B ; I ( A, B ; M ) M (cid:1) is a subalgebra of B ( Y ) and I ( A, B ; M ) is a left module over it.Assume that T ∈ B ( Y ) is such that T I ( A, B ; M ) ⊆ I ( A, B ; M ). Let S ∈ I ( A, B ; M ) and x ∈ M be arbitrary. Then BT Sx = T SAx = T BSx which implies T ∈ C (cid:0) B ; I ( A, B ; M ) M (cid:1) .This shows that C (cid:0) B ; I ( A, B ; M ) M (cid:1) is the largest subalgebra of B ( Y ) such that I ( A, B ; M )is a left module over it. (cid:3) Corollary 3.3.
Let A ∈ B ( X ) and let M ⊆ X be a closed subspace. Then C ( A ; C ( A ; M ) M ) is the largest algebra contained in C ( A ; M ) . If M ∈ Lat (cid:0) C ( A ; M ) (cid:1) , then C ( A ; M ) is analgebra.Proof. By Theorem 3.2, C ( A ; C ( A ; M ) M ) is the largest algebra contained in C ( A ; M ). If M ∈ Lat (cid:0) C ( A ; M ) (cid:1) , then C ( A ; M ) M = M and therefore C ( A ; M ) is an algebra. (cid:3) Now we consider the right module structure of I ( A, B ; M ). Let A ∈ B ( X ), B ∈ B ( Y )and let S ⊆ B ( X , Y ) be an arbitrary set of operators. It is clear that every S ∈ S OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 11 intertwines A and B at vector 0, that is, S ⊆ I ( A, B ; { } ). Hence, the family F of allclosed subspaces M ⊆ X such that S ⊆ I ( A, B ; M ) is not empty. Let M S = W M ∈ F M . By(2), S ⊆ T M ∈ F I ( A, B ; M ) = I ( A, B ; M S ), that is, M S ∈ F . Of course, M S is the largestsubspace in F . In particular, there exists the largest subspace M I ( A,B ; M ) of X on whichall operators from I ( A, B ; M ) intertwine A and B . We will simplify the notation to M A,B instead of M I ( A,B ; M ) (and M A instead of M A,A if X = Y and A = B ) and we will call thissubspace a girder , more precisely, M A,B is the girder of I ( A, B ; M ) (and M A is the girderof C ( A ; M )) induced by M . Of course, X is the girder of I ( A, B ). On the other hand, { } is not necessary a girder — think about I X . However, by Proposition 2.2, { } is thegirder of C ( A ; { } ) = B ( X ) if A is not a scalar multiple of I X .It is clear that T S ∈ I ( A,B ; M ) ker( BS − SA ) is a closed subspace of X such that BSx = SAx for every x in this subspace. Hence T S ∈ I ( A,B ; M ) ker( BS − SA ) ⊆ M A,B . On the other hand,if x ∈ M A,B , then
BSx = SAx for every S ∈ I ( A, B ; M ), by the definition of M A,B . Hence,(12) M A,B = \ S ∈ I ( A,B ; M ) ker( SA − BS ) . It follows that(13) I ( A, B ; M A,B ) = I ( A, B ; M ) and ( M A,B ) A,B = M A,B . Indeed, since M ⊆ M A,B we have I ( A, B ; M A,B ) ⊆ I ( A, B ; M ). On the other hand,the opposite inclusion follows by (12). Thus, I ( A, B ; M A,B ) = I ( A, B ; M ) and therefore( M A,B ) A,B = M A,B , by (12).
Proposition 3.4.
Let A ∈ B ( X ) , B ∈ B ( Y ) and let K , M ⊆ X be closed subspaces. If M ⊆ K , then M A,B ⊆ K A,B . In particular, if M ⊆ K ⊆ M A,B , then M A,B = K A,B andtherefore I ( A, B ; M A,B ) = I ( A, B ; K ) .Proof. Assume that M ⊆ K . Then I ( A, B ; M ) ⊇ I ( A, B ; K ). Hence, if x ∈ M A,B , then
SAx = BSx for every S ∈ I ( A, B ; K ) and therefore x ∈ K A,B .If M ⊆ K ⊆ M A,B , then I ( A, B ; M ) ⊇ I ( A, B ; K ) ⊇ I ( A, B ; M A,B ) = I ( A, B ; M ),that is, I ( A, B ; M A,B ) = I ( A, B ; K ). It follows from this and the definition of K A,B that K A,B = M A,B . (cid:3) For a closed subspace M ⊆ X , we denote by Alg( M ) the algebra of all operators T ∈ B ( X ) such that M ∈ Lat( T ). Theorem 3.5.
Let A ∈ B ( X ) , B ∈ B ( Y ) and let M ⊆ X be a closed subspace. Then C ( A ; M A,B ) ∩ Alg( M A,B ) is a subalgebra of B ( X ) and I ( A, B ; M ) is a right module over it.If X = Y and A = B , then C ( A ; M A ) ∩ Alg( M A ) is the largest subalgebra of B ( X ) suchthat C ( A ; M ) is a right module over it.Proof. Recall that M ⊆ M A,B , by the definition of M A,B , and I ( A, B ; M ) = I ( A, B ; M A,B ),by (13). Let S ∈ I ( A, B ; M ) and T ∈ C ( A ; M A,B ) ∩ Alg( M A,B ) be arbitrary. If x ∈ M ,then ST Ax = SAT x = BST x . The first equality holds because of T ∈ C ( A ; M A,B ) andthe second because of
T x ∈ M A,B and S ∈ I ( A, B ; M A,B ). We have proved that I ( A, B ; M ) · (cid:0) C ( A ; M A,B ) ∩ Alg( M A,B ) (cid:1) ⊆ I ( A, B ; M ) . If T , T ∈ C ( A ; M A,B ) ∩ Alg( M A,B ), then, of course, T T ∈ Alg( M A,B ). Let x ∈ M A,B be arbitrary. Since T ∈ C ( A ; M A,B ) we have T T Ax = T AT x . Because of T x ∈ M A,B we have AT T x = T AT x , as well. Hence, T T Ax = AT T x for every x ∈ M A,B , thatis, T T ∈ C ( A ; M A,B ). We have proved that C ( A ; M A,B ) ∩ Alg( M A,B ) is a subalgebra of B ( X ) and I ( A, B ; M ) is a right module over it.Assume now that X = Y and A = B . By the first part of the proof we already knowthat C ( A ; M A ) ∩ Alg( M A ) is a subalgebra of B ( X ) and C ( A ; M A ) is a right module over it.Let T ∈ B ( X ) be an arbitrary operator such that C ( A ; M A ) T ⊆ C ( A ; M A ). Since ( A ) ′ ⊆ C ( A ; M A ) and ( A ) ′ contains the identity operator I we see that T ∈ C ( A ; M A ). Let x ∈ M A and S ∈ C ( A ; M A ) be arbitrary. Then SAT x = ST Ax . Because of ST ∈ C ( A ; M A ) wealso have AST x = ST Ax . Hence ( SA − AS ) T x = 0 for all S ∈ C ( A ; M A ) which means,see (12), that T x ∈ M A . This proves that T ∈ Alg( M A ). Thus, C ( A ; M A ) ∩ Alg( M A ) isthe largest subalgebra of B ( X ) such that C ( A ; M ) is a right module over it. (cid:3) In the following theorem we list equivalent conditions for C ( A ; M ) being an algebra. Theorem 3.6.
Let A ∈ B ( X ) , A = λI X for every λ ∈ C , and let M ⊆ X be a closedsubspace. The following assertions are equivalent:(i) C ( A ; M ) is an algebra;(ii) C ( A ; M ) M ⊆ M A ;(iii) C ( A ; M ) M = M A ;(iv) M A ∈ Lat (cid:0) C ( A ; M ) (cid:1) . OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 13
Proof. ( i ) ⇒ ( ii ). Let S, T ∈ C ( A ; M ) be arbitrary. Since ST ∈ C ( A ; M ) we have( SA − AS ) T x = ST Ax − ST Ax = 0 for every x ∈ M . Hence, T x ∈ ker( SA − AS ). Since S ∈ C ( A ; M ) is arbitrary we have T x ∈ M A which means that T M ⊆ M A . Now we mayconclude that C ( A ; M ) M ⊆ M A .( ii ) ⇒ ( iii ). If C ( A ; M ) M = X , then C ( A ; M ) M = M A , of course. Assume thereforethat C ( A ; M ) M = X . Then there exist z ∈ X \ C ( A ; M ) M and ξ ∈ ( C ( A ; M ) M ) ⊥ suchthat h z, ξ i = 1. Since A is not a scalar multiple of I X there exists 0 = y ∈ X whichis not an eigenvector of A . Since Ax ∈ C ( A ; M ) M whenever x ∈ C ( A ; M ) M we have A ( y ⊗ ξ ) x = 0 = ( y ⊗ ξ ) Ax for every x ∈ C ( A ; M ) M . Hence, y ⊗ ξ ∈ C ( A ; M ). It followsfrom A ( y ⊗ ξ ) z = Ay and ( y ⊗ ξ ) Az = h Az, ξ i y and the assumption that y is not aneigenvector of A that A ( y ⊗ ξ ) z = ( y ⊗ ξ ) Az which implies z M A .( iii ) ⇒ ( i ). Let S, T ∈ C ( A ; M ) be arbitrary. Then, for every x ∈ M , we have( ST A − AST ) x = SAT x − SAT x = 0 since it follows by the assumption that
T x ∈ M A and we already know, see (13), that C ( A ; M ) = C ( A ; M A ). Hence, ST ∈ C ( A ; M ).( i ) ⇒ ( iv ). Since C ( A ; M ) = C ( A ; M A ) we have C ( A ; M A ) M A ⊆ (cid:0) M A (cid:1) A = M A , by theequivalence of assertions (i) and (ii). Hence, M A ∈ Lat (cid:0) C ( A ; M ) (cid:1) .( iv ) ⇒ ( i ). If M A ∈ Lat (cid:0) C ( A ; M ) (cid:1) , then C ( A ; M ) M ⊆ C ( A ; M ) M A ⊆ M A and therefore C ( A ; M ) is an algebra, by equivalence of assertions (i) and (ii). (cid:3) Ultrainvariant subspaces
We begin this section with a definition.
Definition 4.1.
A closed subspace M ⊆ X is ultrainvariant for A ∈ B ( X ) if it is invariantfor every operator in C ( A ; M ).Hence, every ultrainvariant subspace of A is a hyperinvariant subspace, as well. Wewill see that the opposite implication does not hold in general. For instance, if A = 0is a nilpotent operator, then im( A k ) is a hyperinvariant subspace of A for every k ∈ N ,however it is not necessary an ultrainvariant subspace (see Theorem 5.11). Proposition 4.2.
Let A ∈ B ( X ) , A = λI X for every λ ∈ C . A closed subspace M ⊆ X isultrainvariant for A if and only if C ( A ; M ) is an algebra and M is its girder.Proof. Assume that M is an ultrainvariant subspace of A . Let T , T ∈ C ( A ; M ) bearbitrary. Then T T Ax = T AT x = AT T x for every x ∈ M because T x ∈ M . Thus, C ( A ; M ) is an algebra. By Theorem 3.6, C ( A ; M ) M = M A . However, C ( A ; M ) M = M as M is invariant for every operator in C ( A ; M ), that is, M = M A . The opposite implicationfollows by Theorem 3.6. (cid:3) Subspaces { } and X are trivial ultrainvariant subspaces of every A ∈ B ( X ). If A is ascalar multiple of I X , then it is obvious that A has only trivial ultrainvariant subspacesas it has only trivial hyperinvariant subspaces.In this section we will prove a few general results about ultrainvariant subspaces. Ex-amples of ultrainvariant subspaces for some classes of operators will be given in sectionsthat follow. Proposition 4.3.
Let A ∈ B ( X ) and let U ∈ B ( X ) be an invertible operator. A closedsubspace M ⊆ X is an ultrainvariant subspace of A if and only if U M is an ultrainvariantsubspace of U AU − .Proof. Assume that M is ultrainvariant. By (1), every operator in C ( U AU − ; U M ) is of theform U T U − for some T ∈ C ( A ; M ). Hence, for every x ∈ M , we have ( U T U − )( U x ) =
U T x ∈ U M , that is, U M is invariant for every operator from C ( U AU − ; U M ) and istherefore ultrainvariant. It is clear that the opposite implication holds, as well. (cid:3) Recall that a closed subspace X ⊆ X is reducing for A ∈ B ( X ) if X ∈ Lat( A ) andthere exists X ∈ Lat( A ) such that X = X ⊕ X . In this case we will say that ( X , X ) is areducing pair for A . Note that A = A ⊕ A with respect to the decomposition X = X ⊕ X .Let P j : X → X j ( j = 1 ,
2) be given by P ( x ⊕ x ) = x and P ( x ⊕ x ) = x . Proposition 4.4.
Let A ∈ B ( X ) and let ( X , X ) be a reducing pair for A . If M isan ultrainvariant subspace of A , then M j = P j M is an ultrainvariant subspace of A j ( j = 1 , .Proof. Assume that M = M ⊕ M is ultrainvariant. Let T ∈ C ( A ; M ) be arbitrary andlet T ∈ B ( X ) be such that T = T ⊕ X = X ⊕ X .For an arbitrary x = x ⊕ x ∈ M , we have T Ax = T A x ⊕ A T x ⊕ AT x whichmeans that T ∈ C ( A ; M ). Since M is ultrainvariant it is invariant for T . In particular,for every x ∈ M we have T ( x ⊕
0) = T x ⊕ ∈ M ⊕ M , that is, M is invariant for T ∈ C ( A ; M ). We have seen that M is an ultrainvariant subspace of A and a similarproof shows that M is an ultrainvariant subspace of A . (cid:3) OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 15
Let n ≥ j ∈ { , . . . , n } , let X j be a Banach space and A j ∈ B ( X j ). Denote X = X ⊕ · · · ⊕ X n and A = A ⊕ · · · ⊕ A n . Let M j ⊆ X j be acomplemented closed subspace and let N j ⊆ X j be a closed subspace such that X j = M j ⊕ N j . Let (cid:20) A ( j )11 A ( j )12 A ( j )21 A ( j )22 (cid:21) be the operator matrix of A j with respect to the decomposition X j = M j ⊕ N j . Proposition 4.5.
If, for every j ∈ { , . . . , n } , M j is an ultrainvariant subspace of A j and σ ( A ( i )11 ) ∩ σ ( A j ) = ∅ whenever i = j , then M := M ⊕ · · · ⊕ M n is an ultrainvariantsubspace of A .Proof. It is enough to consider the case n = 2. We will identify X with M ⊕ N ⊕ M ⊕ N .Then M = M ⊕{ }⊕ M ⊕{ } . With respect to the decomposition X = M ⊕ N ⊕ M ⊕ N ,operator A has operator matrix A (1)11 A (1)12 A (1)22 A (2)11 A (2)12 A (2)22 . Let T ∈ C ( A ; M ) be arbitrary and let [ T ij ] i,j =1 be its operator matrix with respect to thedecomposition X = M ⊕ N ⊕ M ⊕ N . Hence, T Ax = AT x for every x = x ⊕ ⊕ x ⊕ ∈ M . If x = x ⊕ ⊕ ⊕
0, where x ∈ M is arbitrary, then it follows from T Ax = AT x that T A (1)11 = A (1)11 T + A (1)12 T T A (1)11 = A (1)22 T T A (1)11 = A (2)11 T + A (2)12 T T A (1)11 = A (2)22 T . (14)Let W ∈ B ( X ) be the operator whose operator matrix with respect to the decomposition X = M ⊕ N is (cid:2) T T T T (cid:3) . The first two equalities in (14) give W ∈ C ( A ; M ). Since M is an ultrainvariant subspace of A it is invariant for W and therefore T = 0. Hence, bythe first equality in (14), T ∈ (cid:0) A (1)11 (cid:1) ′ . Denote by W the operator in B ( X , X ) whoseoperator matrix with respect to the decompositions X = M ⊕ N and X = M ⊕ N is (cid:2) T T T T (cid:3) . It follows from the last two equalities in (14) that column (cid:2) T T (cid:3) intertwines A (1)11 and A . Since, by the assumption, σ ( A (1)11 ) ∩ σ ( A ) = ∅ we conclude that T = 0 and T = 0. Let now x ∈ M be arbitrary and x = 0 ⊕ ⊕ x ⊕
0. A similar reasoning as in theprevious paragraph shows that
T Ax = AT x if and only if T = 0, T = 0, T = 0and T ∈ (cid:0) A (2)11 (cid:1) ′ . Hence, the operator matrix of T with respect to the decomposition X = M ⊕ N ⊕ M ⊕ N is (cid:20) T T T T T T T T T T (cid:21) . It is not hard to check now that M = M ⊕ { } ⊕ M ⊕ { } is invariant for T . Since T ∈ C ( A ; M ) was arbitrary we conclude that M is an ultrainvariant subspace of A . (cid:3) Proposition 4.6.
Let A ∈ B ( X ) and let M ⊆ X be a closed subspace. Then C (cid:0) A ; C ( A ; M ) M (cid:1) C ( A ; M ) M is the smallest ultrainvariant subspace of A which contains M .Proof. By Corollary 3.3, C (cid:0) A ; C ( A ; M ) M (cid:1) is the largest algebra which is contained in C ( A ; M ). It follows, by Theorem 3.6, that (cid:0) C ( A ; M ) M (cid:1) A = C (cid:0) A ; C ( A ; M ) M (cid:1) C ( A ; M ) M and (cid:0) C ( A ; M ) M (cid:1) A is an ultrainvariant subspace of A . It is clear that M ⊆ (cid:0) C ( A ; M ) M (cid:1) A .Suppose that K ⊆ X is an ultrainvariant subspace of A such that M ⊆ K . Then C ( A ; K )is an algebra and C ( A ; M ) ⊇ C ( A ; K ). Since C (cid:0) A ; (cid:0) C ( A ; M ) M (cid:1) A (cid:1) = C (cid:0) A ; C ( A ; M ) M (cid:1) isthe largest algebra contained in C ( A ; M ) we have C ( A ; K ) ⊆ C (cid:0) A ; C ( A ; M ) M (cid:1) . Hence, if T ∈ C ( A ; K ), then it commutes with A at every vector from C ( A ; M ) M which means that C ( A ; M ) M ⊆ K A = K and therefore (cid:0) C ( A ; M ) M (cid:1) A ⊆ K . (cid:3) For A ∈ B ( X ), let Lat u ( A ) denote the family of all ultrainvariant subspaces of A . Proposition 4.7.
Let A ∈ B ( X ) and let { M j ; j ∈ J } be an arbitrary family of ultrain-variant subspaces of A . Then W j ∈ J M j and T j ∈ J M j are ultrainvariant subspaces of A . Hence, Lat u ( A ) is a sublattice of Lat h ( A ) .Proof. Denote K = W j ∈ J M j . By (2), we have C ( A ; K ) = T j ∈ J C ( A ; M j ). Hence, M j ( j ∈ J ) isinvariant for an arbitrary T ∈ C ( A ; K ). Assume that x ∈ K is such that x = x + · · · + x k ,where x i ∈ M j i ( i = 1 , . . . , k ). Then T x = T x + · · · + T x k ∈ M j + · · · + M j k ⊆ M . Itfollows that T K ⊆ K . Since T ∈ C ( A ; K ) is arbitrary we conclude that K is ultrainvariant.Denote L = T j ∈ J M j and let L ′ ⊆ X be the smallest ultrainvariant subspace of A suchthat L ⊆ L ′ (see Proposition 4.6). Since, for every j ∈ J , L ⊆ M j and M j is ultrainvariant OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 17 we have L ′ ⊆ M j . Hence L ′ ⊆ T j ∈ J M j = L , that is, L = L ′ , i.e., L is an ultrainvariantsubspace of A . (cid:3) The following example shows that even when M ∈ Lat( A ) is the girder of C ( A ; M )it is not necessary an ultrainvariant subspace of A , that is, C ( A ; M ) is not necessary analgebra. Example 4.8.
Let X = X ⊕ X ⊕ X , where X j ( j = 1 , ,
3) are non-trivial subspaces of X . Let A be the projection on X ⊕ X along X and let M = X ⊕ { } ⊕ X , that is, in M are vectors of the form x ⊕ ⊕ x , where x ∈ X and x ∈ X are arbitrary. Hence, M is a non-trivial subspace of X and M ∈ Lat( A ).Let T ∈ C ( A ; M ) be arbitrary and let [ T ij ] i,j =1 be its operator matrix with respectto the decomposition X = X ⊕ X ⊕ X . For every x = x ⊕ ⊕ x ∈ M , we have T Ax = T x ⊕ T x ⊕ T x and AT x = ( T x + T x ) ⊕ ( T x + T x ) ⊕
0. It followsfrom
T Ax = AT x that T x = T x + T x , T x = T x + T x and T x = 0 for all x ∈ X and x ∈ X . Thus, T = 0, T = 0 and T = 0. An operator T is in C ( A ; M )if and only if its operator matrix with respect to the decomposition X = X ⊕ X ⊕ X isof the form h T T T T T T i , where T ij ∈ B ( X j , X i ) are arbitrary. It follows that C ( A ; M ) isnot an algebra.To determine M A , let x = x ⊕ x ⊕ x ∈ X be such that T Ax = AT x for every T ∈ C ( A ; M ). Since T Ax = ( T x + T x ) ⊕ ( T x + T x ) ⊕ T x and AT x =( T x + T x ) ⊕ ( T x + T x ) ⊕ T x = 0 for every T ∈ B ( X , X ).As the involved subspaces are non-trivial we conclude that x = 0. This proves that M A = M . However, M is not an ultrainvariant subspace of A because C ( A ; M ) is not analgebra. Moreover, C ( A ; M ) M = X . Namely, let x = x ⊕ x ⊕ x be an arbitrary vectorfrom X . Let y = y ⊕ ⊕ y ∈ M be such that y = 0 and y = 0. Then there exist T ∈ B ( X ), T ∈ B ( X , X ) and T ∈ B ( X ) such that T y = x , T y = x and T y = x . It is clear that T whose operator matrix with respect to the decomposition X = X ⊕ X ⊕ X is h T T T i is in C ( A ; M ) and T y = x . (cid:3) Operators with non-trivial ultrainvariant subspaces
Let A ∈ B ( X ) and assume that M ∈ Lat( A ) is complemented, say X = M ⊕ N . Let (cid:2) A A A (cid:3) be the operator matrix of A with respect this decomposition. Next propositionis basically a reformulation of Corollary 2.7. Proposition 5.1. If σ ( A ) ∩ σ ( A ) = ∅ , then M is an ultrainvariant subspace of A .Proof. If M is trivial, then there is nothing to prove. Assume therefore that { } 6 = M = X .By Corollary 2.7, C ( A ; M ) is an algebra. Hence, if (cid:2) T T T T (cid:3) is the operator matrix of T ∈ C ( A ; M ) with respect to the decomposition X = M ⊕ N , then T ∈ ( A ) ′ and T = 0 by Theorem 2.5. It follows that M is invariant for every T ∈ C ( A ; M ). (cid:3) Recall that the ascent α ( A ) of A ∈ B ( X ) is the smallest positive integer k such thatker( A k ) = ker( A k +1 ). If there is no positive integer with this property, then α ( A ) = ∞ .The descent δ ( A ) of A is the smallest positive integer k such that im( A k ) = im( A k +1 )and δ ( A ) = ∞ if there is no k with this property. Assume that A has finite ascent anddescent. Then α ( A ) = n = δ ( A ) for a positive integer n and, by [1, Lemma 2.21], im( A n ) isa closed complement of ker( A n ), that is X = ker( A n ) ⊕ im( A n ). Hence, (ker( A n ) , im( A n ))is a reducing pair of A and therefore A = A ⊕ A , where A ∈ B (ker( A n )) and A ∈ B (im( A n )). By [1, Theorem 2.23], A is nilpotent and A is invertible which implies σ ( A ) ∩ σ ( A ) = ∅ . It is clear now that the following corollary is a simple consequence ofProposition 5.1 Corollary 5.2. If A ∈ B ( X ) has finite ascent and descent n , then ker( A n ) and im( A n ) are ultrainvariant subspaces of A . It is not hard to see that for every λ ∈ C the kernel ker( A − λI X ) is an ultrainvariantsubspace of A ∈ B ( X ). Namely, if T ∈ C ( A ; ker( A − λI X )), then AT x = T Ax = λT x forevery x ∈ ker( A − λI X ), that is, ker( A − λI X ) is invariant for every T ∈ C ( A ; ker( A − λI X )).Hence, if A is not a scalar multiple of I X and the point spectrum σ p ( A ) is not empty, then A has a non-trivial ultrainvariant subspace. More can be said. We need the followinglemma. Lemma 5.3.
Let A ∈ B ( X ) and let A ⊆ B ( X ) be the strongly closed subalgebra generatedby A and I X . If M ∈ Lat( A ) , then C ( A ; M ) ⊆ C ( B ; M ) for every B ∈ A .Proof. Assume first that B = A − λI X , where λ ∈ C is such that B is an invertibleoperator. Hence, B M = M . It is clear that C ( A ; M ) = C ( B ; M ). Let us check that C ( B ; M ) ⊆ C ( B k ; M ) for every integer k ≥
0. If k = 0 or k = 1, then the assertion istrivial. Since B is invertible and B M = M we have C ( B ; M ) = C ( B ; M ) B , by Proposition3.1. Hence, if T ∈ C ( B ; M ), then T B ∈ C ( B ; M ) and, by induction, T B j ∈ C ( B ; M ) for OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 19 every j ≥
2. Let k ≥ T B j x = B j T x holds for every 0 ≤ j ≤ k − x ∈ M . Then T B k x = B k − T Bx and
T Bx = BT x for every x ∈ M . It followsthat T B k x = B k T x for every x ∈ M . Since A = B + λI X we conclude that T A k x = T k X j =0 (cid:18) kj (cid:19) λ j B k − j x = k X j =0 (cid:18) kj (cid:19) λ j B k − j T x = A k T x for every x ∈ M . It follows that
T p ( A ) x = p ( A ) T x for every x ∈ M and every polynomial p ∈ C [ z ]. Let B ∈ A be arbitrary. Let T ∈ C ( A ; M ) and x ∈ M . For every ε >
0, there exists apolynomial p such that k Bx − p ( A ) x k < ε and k BT x − p ( A ) T x k < ε . Hence k T Bx − BT x k ≤ k T kk Bx − p ( A ) x k + k BT x − p ( A ) T x k < ε ( k T k +1), which gives T Bx = BT x . (cid:3) Proposition 5.4.
Let A ∈ B ( X ) and let A ⊆ B ( X ) be the strongly closed subalgebragenerated by A and I X . If B ∈ A and M ∈ Lat( A ) is an ultrainvariant subspace of B , then M is an ultrainvariant subspace of A . In particular, every kernel ker( B − λI X )( λ ∈ C ) is an ultrainvariant subspace of A .Proof. By Lemma 5.3, C ( A ; M ) M ⊆ C ( B ; M ) M = M . Thus, M is invariant for everyoperator from C ( A ; M ), that is, it is an ultrainvariant subspace of A . (cid:3) It follows from the following proposition that a reducing subspace does not need to beultrainvariant.
Proposition 5.5.
Let A ∈ B ( X ) . Assume that ( M , N ) is a pair of reducing subspaces for A . Let A be equal to A M ⊕ A N with respect to the decomposition X = M ⊕ N . Then M isan ultrainvariant subspace of A if and only if I ( A M , A N ) = { } .Proof. Suppose that M is ultrainvariant and let S ∈ I ( A M , A N ) be arbitrary. Then T ( x ⊕ y ) = 0 ⊕ Sx , where x ∈ M , y ∈ N are arbitrary, defines a bounded linear operator on X . Since AT ( x ⊕
0) = 0 ⊕ A N Sx = 0 ⊕ SA M x = T A ( x ⊕
0) for all x ∈ M we have T ∈ C ( A ; M ). It follows from T M ⊆ M that Sx = 0 for every x ∈ M , that is, S = 0.Assume now that I ( A M , A N ) = { } . Let T ∈ C ( A ; M ) be arbitrary and let (cid:2) T T T T (cid:3) be its operator matrix with respect to the decomposition X = M ⊕ N . It follows from T A ( x ⊕
0) = AT ( x ⊕ x ∈ M is arbitrary, that T ∈ I ( A M , A N ), that is, T = 0.Hence, M is an ultrainvariant subspace of A . (cid:3) Let A ∈ B ( X ). A subset ∅ ( σ ( σ ( A ) is an isolated part of σ ( A ) if both σ and σ ( A ) \ σ are closed sets. It is well-known, see [10, § I.2], that for an isolated part σ of σ ( A ) there exists an idempotent P σ ∈ B ( X ), the Riesz projection of A corresponding to σ , such that M = im( P σ ) and N = im( I − P σ ) are in Lat( A ) and σ ( A | M ) = σ , σ ( A | N ) = σ ( A ) \ σ .Hence, the following result is an immediate consequence of Proposition 5.5. Corollary 5.6.
Let A ∈ B ( X ) . If σ is an isolated part of σ ( A ) , then im( P σ ) is anultrainvariant subspace of A . Corollary 5.6 is a special case of the next result for which we need some notions fromthe local spectral theory. Let A ∈ B ( X ) and x ∈ X . The local resolvent set ρ A ( x ) of A at x is the union of all open subsets U ⊆ C for which there exists an analytic function f : U → X such that ( A − λI X ) f ( λ ) = x for all λ ∈ U . The local spectrum of A at x is then defined as σ A ( x ) = C \ ρ A ( x ). It is obvious that σ A ( x ) is a closed subset of thespectrum σ ( A ). For an arbitrary F ⊆ C , the local spectral subspace of A correspondingto F is X A ( F ) = { x ∈ X ; σ A ( x ) ⊆ F } . By [14, Proposition 1.2.16 (a)], X A ( F ) is ahyperinvariant subspace of A . Proposition 5.7. X A ( F ) is an ultrainvariant subspace of A .Proof. Let T ∈ C ( A ; X A ( F )) be arbitrary. Let x ∈ X A ( F ) and let λ ∈ ρ A ( x ). Thenthere is an open neighborhood U ⊆ ρ A ( x ) of λ and an analytic function f : U → X suchthat ( A − λI X ) f ( λ ) = x for all λ ∈ U . By [14, Lemma 1.2.14 ], σ A ( f ( λ )) = σ A ( x ) forevery λ ∈ U . Hence, f ( λ ) ∈ X A ( F ) for all λ ∈ U . It follows that ( A − λI X ) T f ( λ ) = T ( A − λI X ) f ( λ ) = T x for all λ ∈ U . Since λ T f ( λ ) is an analytic function on U weconclude that λ ∈ σ A ( T x ) and therefore σ A ( T x ) ⊆ σ A ( x ) ⊆ F . Thus, X A ( F ) is invariantfor T . (cid:3) Recall that A ∈ B ( X ) is a decomposable operator if for every open cover C = U ∪ V there exist spaces M , N ∈ Lat( A ) such that X = M + N and σ ( A | M ) ⊆ U , σ ( A | N ) ⊆ V (seeDefinition 1.1.1 in [14]). If A is decomposable, then the local spectral subspace X A ( F ) isclosed whenever F is a closed subset of C (see Definition 1.2.18 and Theorem 1.2.29 in[14]) and σ ( A ) = ∪{ σ A ( x ); x ∈ X } (see Proposition 1.3.2 in [14]). Corollary 5.8. If A ∈ B ( X ) is a decomposable operator such that the spectrum σ ( A ) contains at least two points, then A has a proper non-trivial ultrainvariant subspace.Proof. Let λ = λ be points in σ ( A ) and let C = U ∪ V be an open cover such that λ ∈ U and λ U . Then X A ( U ) and X A ( V ) are non-trivial and proper closed subspaces of X . OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 21
Indeed, by [14, Theorem 1.2.23], X = X A ( U ) + X A ( V ). Since σ A ( x ) ⊆ U ∩ σ ( A ) ( σ ( A )and σ ( A ) = ∪{ σ A ( x ); x ∈ X } there are vectors which are not in X A ( U ). Similarly, thereare vectors which are not in X A ( V ). Hence, X A ( U ) and X A ( V ) are proper and non-trivial.By Proposition 5.7, these subspaces are ultrainvariant. (cid:3) Assume that A ∈ B ( X ) is a nilpotent operator of order n ≥
2, that is, A n = 0 and A n − = 0. It is well-known that { } = ker( A ) ⊆ ker( A ) ⊆ ker( A ) ⊆ · · · ⊆ ker( A n − ) ⊆ ker( A n ) = X and { } = im( A n ) ⊆ im( A n − ) ⊆ im( A n − ) ⊆ · · · ⊆ im( A ) ⊆ im( A ) = X are two chains of hyperinvariant subspaces of A . By [3, Remarque p. 317], subspacesim( A n − ) and ker( N n − ) are the smallest, respectively the largest, non-trivial hyperin-variant subspaces of A . For every j = 0 , , . . . , n −
1, we have im( A n − j ) ( im( A n − j − )and ker( A j ) ( ker( A j +1 ). Of course, im( A n − j ) ⊆ ker( A j ). Later we will need the followingsimple facts. Lemma 5.9. (i) For every j = 1 , . . . , n − , there exists u ∈ im( A n − j ) such that u ker( A j − ) .(ii) Let e ∈ X be such that A k e = 0 for an integer k ∈ { , , . . . , n − } . Then A k e W { e, Ae, . . . , A k − e } , that is, e, Ae, . . . , A k e are linearly independent.Proof. (i) If im( A n − j ) ⊆ ker( A j − ), then we would have A j − ( N n − j x ) = 0 for every x ∈ X ,that is, it would be A n − = 0.(ii) Towards a contradiction assume that there exist numbers α , . . . , α k − such that A k e = α e + α Ae + · · · + α k − A k − e. Since A k e = 0 there exists the smallest index 0 ≤ j ≤ k − α j = 0. Hence, A k e = α j A j e + α j +1 A j +1 e + · · · + α k − A k − e. Let m be the integer such that A m e = 0 and A m +1 e = 0. Multiply the above equality by A m − j . Since j < k we have0 = A m − j + k e = α j A m e + α j +1 A m +1 + · · · + α k − A m − j + k − e = α j A m e. It follows that α j = 0 as A m e = 0. This is a contradiction. (cid:3) Lemma 5.10.
Let j ∈ { , . . . , n } . If e ∈ ker( A j ) and ξ ∈ X ∗ , then ( e ⊗ ξ ) A j − + A ( e ⊗ ξ ) A j − + · · · + A j − ( e ⊗ ξ ) ∈ C ( A ; im( A n − j )) . Proof.
It is not hard to check that ( e ⊗ ξ ) A j − + A ( e ⊗ ξ ) A j − + · · · + A j − ( e ⊗ ξ ) and A commute locally at every vector A n − j x ∈ im( A n − j ), where x ∈ X is arbitrary. (cid:3) Theorem 5.11.
Let A ∈ B ( X ) be a nilpotent operator of order n ≥ . A closed subspace M ⊆ X is ultrainvariant for A if and only if M = ker( A j ) for some j ∈ { , , . . . , n } .Proof. By Proposition 5.4, every kernel ker( A j ) ( j = 0 , , . . . , n ) is an ultrainvariant sub-space of A . For the opposite implication, suppose that M is an ultrainvariant subspacefor A . Then M is hyperinvariant for A and therefore, by [3, Lemma 5], there exists aunique j ∈ { , , . . . , n } such that im( A n − j ) ⊆ M ⊆ ker( A j ).We claim that C ( A, im( A n − j )) im( A n − j ) = ker( A j ), for every j ∈ { , , . . . , n } . If j = 0or j = n , then the assertion is trivial. Assume therefore that 1 ≤ j ≤ n −
1. Let T ∈ C ( A, im( A n − j )) and x ∈ im( A n − j ) be arbitrary. Let z ∈ X be such that x = A n − j z .By Lemma 5.3, A j T x = T A j x = T A n z = 0. This proves that C ( A, im( A n − j )) im( A n − j ) ⊆ ker( A j ) . For the opposite inclusion, assume that x ∈ ker( A j ). Let e ∈ im( A n − j ) besuch that A j − e = 0 (it exists by Lemma 5.9 (i)). Since, by Lemma 5.9 (ii), vectors e, Ae, . . . , A j − e are linearly independent there exists ξ ∈ X ∗ such that h A j − e, ξ i = 1and h A i e, ξ i = 0 for i = 0 , . . . , j −
2. Denote T = ( x ⊗ ξ ) A j − + A ( x ⊗ ξ ) A j − + · · · + A j − ( x ⊗ ξ ). By Lemma 5.10, T ∈ C ( A, im( A n − j )). Since T e = x the equality C ( A, im( A n − j )) im( A n − j ) = ker( A j ) is proved. By Proposition 4.6, C (cid:0) A ; C ( A, im( A n − j ))im( A n − j ) (cid:1) C ( A, im( A n − j ))im( A n − j ) = C (cid:0) A ; ker( A j ) (cid:1) ker( A j ) = ker( A j )is the smallest ultrainvariant subspace of A which contains im( A n − j ). Hence, M =ker( A j ). (cid:3) An operator A ∈ B ( X ) is algebraic if there exists a non-zero polynomial p such that p ( A ) = 0 . It is not hard to see that for an algebraic operator A there exists a uniquemonic polynomial q A of the minimal degree, called the minimal polynomial of A , suchthat q A ( A ) = 0. Every operator on a finite dimensional Banach space is algebraic. Let q A ( z ) = ( z − λ ) n · · · ( z − λ k ) n k with λ , . . . , λ k distinct. For every j = 1 , . . . , k , let X j = ker (cid:0) ( A − λ j I X ) n i (cid:1) . Then X = X ⊕ · · · ⊕ X k and with respect to this decomposition A = ( λ I X + A ) ⊕ · · · ⊕ ( λ k I X k + A k ), where A j ∈ B ( X j ) is a nilpotent operator of order OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 23 n j (if n j = 1, then A j = 0). The reader is referred to [19, § Corollary 5.12.
Let A ∈ B ( X ) be algebraic and let A = ( λ I X + A ) ⊕· · ·⊕ ( λ k I X k + A k ) beits decomposition as described above. Then a closed subspace M ⊆ X is an ultrainvariantsubspace of A if and only if M = ker( A m ) ⊕ · · · ⊕ ker( A m k k ) for some ≤ m j ≤ n j ( j = 1 , . . . , k ) .Proof. Assume that M is an ultrainvariant subspace of A . Let M = M ⊕ · · · ⊕ M k , where M j ⊆ X j ( j = 1 , . . . , k ). Since every subspace X j is reducing subspace of A , it follows,by Proposition 4.4, that M j is an ultrainvariant subspace of λ I X j + A j . By Theorem5.11, there exists m j ∈ { , , . . . , n j } such that M j = ker( A m j j ). The opposite implicationfollows by Proposition 4.5 because σ ( λ i I X i + A i ) ∩ σ ( λ j I X j + A j ) = ∅ whenever i = j . (cid:3) Ultrainvariant subspaces of operators on a Hilbert space
In this section, H is the infinite dimensional separable complex Hilbert space. Theinner product of vectors x, y ∈ H is denoted by ( x, y ). For a closed subspace M ⊆ H , let M ⊥ be its orthogonal complement.Our first result in this section is the complete description of the lattice of ultrainvariantsubspaces of a normal operator. It turns out that Lat u ( A ) = Lat h ( A ) whenever A isnormal.Let A be a normal operator on H and let E be its spectral measure. For a Borel subset σ ⊆ σ ( A ), operator E ( σ ) ∈ B ( H ) is an orthogonal projection. Subspace M = im( E ( σ ))is reducing for A and M ⊥ = im( E ( σ ( A ) \ σ )) because E ( σ ( A ) \ σ ) = I − E ( σ ) (see SectionXXXI.7 in [11] for details). Theorem 6.1.
Let A ∈ B ( H ) be a normal operator and let E be its spectral measure. Asubspace M ⊆ H is an ultrainvariant subspace of A if and only if M is a hyperinvariantsubspace of A , that is, if and only if there exists a Borel subset σ ⊆ σ ( A ) such that M = im( E ( σ )) .Proof. By [7, Theorem 3], M ⊆ H is a hyperinvariant subspace of A if and only if thereexists a Borel subset σ ⊆ σ ( A ) such that M = im( E ( σ )). Hence, it remains to provethat for every Borel subset σ ⊆ σ ( A ) the subspace M = im( E ( σ )) is ultrainvariant. Wecan write A = A M ⊕ A M ⊥ with respect to the decomposition H = M ⊕ M ⊥ . We have to see that I ( A M , A M ⊥ ) = { } because then, by Proposition 5.5, M is an ultrainvariantsubspace of A . There are a few ways how to see that I ( A M , A M ⊥ ) = { } . We will showit as follows. Assume that S ∈ I ( A M , A M ⊥ ). Then operator T ∈ B ( H ) whose operatormatrix with respect to the decomposition H = M ⊕ M ⊥ is [ S ] commutes with A . Itfollows, by The Fuglede-Putnam Theorem, that T commutes with E ( σ ) (see TheoremsXXXI.7.4 and XXXI.7.9 in [11], for instance). It is easy to see now that T E ( σ ) = E ( σ ) T gives S = 0. (cid:3) Let 1 ≤ p < ∞ and let ( X, µ ) be a measure space such that L p ( X, µ ) is separable. For φ ∈ L ∞ ( X, µ ), let M φ be the multiplication by φ in L p ( X, µ ). By [12, Theorem 1], a closedsubspace M ⊆ L p ( X, µ ) is hyperinvariant for M φ if and only if M = M χ σ ◦ φ (cid:0) L p ( X, µ ) (cid:1) for some Borel set σ ⊆ X . Here χ σ denotes the characteristic function of σ and M χ σ ◦ φ is the multiplication operator induced by χ σ ◦ φ . Note that M χ σ ◦ φ is an idempotent.Hence M χ σ ◦ φ (cid:0) L p ( X, µ ) (cid:1) is complemented and its complement M χ X \ σ ◦ φ (cid:0) L p ( X, µ ) (cid:1) is ahyperinvariant subspace of M φ . A similar reasoning as in the proof of Theorem 6.1 givesthat every hyperinvariant subspace of M φ is an ultrainvariant subspace.For a closed subspace M ⊆ H , let P M be the orthogonal projection onto M . Let A ∈ B ( H ). A distance on Lat( A ) is the mapping θ which is given by θ ( M , N ) = k P M − P N k ( M , N ∈ Lat( A )). A subspace M ∈ Lat( A ) is said to be inaccessible if the only continuousmapping f : [0 , → Lat( A ) with f (0) = M is the constant map f ( t ) = M ( t ∈ [0 , Theorem 6.2.
Let A ∈ B ( H ) . If M ∈ Lat( A ) is inaccessible, then it is an ultrainvariantsubspace of A .Proof. Let 0 = T ∈ C ( A ; M ) be an arbitrary operator. Denote Λ = { λ ∈ R ; 0 ≤ λ < k T k − } . Then for each λ ∈ Λ, the operator I H − λT is invertible and therefore M λ = ( I H − λT ) M is a closed subspace of H . If y ∈ M λ , then there exists x ∈ M suchthat y = ( I H − λT ) x . Hence, Ay = A ( I H − λT ) x = ( I H − λT ) Ax ∈ M λ because Ax ∈ M .This shows that M λ ∈ Lat( A ) for every λ ∈ Λ. The map λ M λ is continuous (see theproof of Theorem 1 in [7]). Since M = M and M is inaccessible we have M = M λ foreach λ ∈ Λ. Thus, M = ( I H − λT ) M for a non-zero λ which implies that M is invariantfor T , that is, it is an ultrainvariant subspace of A . (cid:3) It is clear that M ∈ Lat( A ) is inaccessible if it is an isolated point. OCAL COMMUTANTS AND ULTRAINVARIANT SUBSPACES 25
Corollary 6.3.
Let A ∈ B ( H ) . If M ∈ Lat( A ) is such that P M commutes with every P N ( N ∈ Lat( A )) , then M is an ultrainvariant subspace of A .Proof. Assume that M ∈ Lat( A ) is such that P M P N = P N P M for every N ∈ Lat( A ).Then, for every N ∈ Lat( A ), P M P N is an orthogonal projection such that P M P N ≤ P M and P M P N ≤ P N . Suppose that N = M . Then either P M P N < P M or P M P N < P N . If the formerholds, then there exists x ∈ im( P M ), k x k = 1, such that P M P N x = 0. Since P M x = x itfollows that P N x = P N P M x = 0. Thus, k P M − P N k ≥ k P M x − P N x k = k x k = 1 which gives k P M − P N k = 1. We get the same conclusion if P M P N < P N holds. We conclude that M isan isolated point of Lat( A ). Hence, it is inaccessible and therefore it is an ultrainvariantsubspace of A , by Theorem 6.3. (cid:3) Recall that A ∈ B ( H ) is unicellular if Lat( A ) is totally ordered, that is, for any twosubspaces M , N ∈ Lat( A ), either M ⊆ N or M ⊇ N . Corollary 6.4. If A ∈ B ( H ) is unicellular, then every M ∈ Lat( A ) is an ultrainvariantsubspace of A .Proof. Assume that A ∈ B ( H ) is unicellular and let M ∈ Lat( A ) be arbitrary. If N ∈ Lat( A ), then either M ⊆ N or M ⊇ N which implies that P M and P N commute. ByCorollary 6.3, M is an ultrainvariant subspace of A . (cid:3) An example of a unicellular operator is the Volterra operator V on L [0 , V ) consists of subspaces M α = { f ∈ L [0 , f = 0 a.e. on [0 , α ] } , where α ∈ [0 , V is actually an ultrainvariant subspace.Let ( e n ) ∞ n =0 be an orthonormal basis of H and let w = ( w n ) ∞ n =1 be a bounded sequenceof complex numbers. Then w determines an operator W w ∈ H by W w e = 0 and W w e n = w n e n − for n ∈ N (hence, W w is the weighted backward shift determined by w ). For each n ≥
0, denote M n = W { e , . . . , e n } . Proposition 6.5.
Every M n ⊆ H ( n ≥ is an ultrainvariant subspace of W w .Proof. Since M = ker( W w ) and kernels are ultrainvariant (see the paragraph beforeLemma 5.3) we may assume that n ≥
1. Observe that W w x ∈ M k for some x ∈ H and k ≥ x ∈ M k +1 . Indeed, if x = ∞ P j =0 ( x, e j ) e j and ( x, e j ) = 0 for some j > k + 1, then W w x = P ∞ j =1 ( x, e j ) w j e j − M k . Let T ∈ C ( W w ; M n ) be arbitrary. Then W w T e = T W w e = 0 and therefore T e ∈ M .Suppose that for j < n we have T e j ∈ M j . Then W w T e j +1 = T W w e j +1 = w j T e j ∈ M j which gives T e j +1 ∈ M j +1 . We conclude that M n is invariant for every T ∈ C ( W w ; M n ),that is, M n is ultrainvariant. (cid:3) If w is a sequence of non-zero complex numbers such that | w | > | w | > · · · and ∞ P n =1 | w n | < ∞ , then W w is called the Donoghue operator with weight sequence w . By[16, Theorem 4.12] (a special case was considered by Donoghue [8, Example 1]), a closedsubspace M ⊆ H is a non-trivial invariant subspace of the Donoghue operator W w if andonly if M = M n for some n ≥
0. Hence the Donoghue operator is unicellular and thereforeevery M n is an ultrainvariant subspace of W w . Acknowledgments.
The author was supported by the Slovenian Research Agencythrough the research program P2-0268.
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Faculty of Natural Sciences and Engineering, University of Ljubljana, Aˇskerˇceva c.12, SI-1000 Ljubljana, Slovenia
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