Let A be an invertible bounded linear operator on a complex Banach space, {A}′ the commutant of A and Bᴀ the set of all operators T such that 〖sup〗_(ₙ≥₀)∥AⁿTA⁻ⁿ ∥ < +∞. Equality {A}′ = Bᴀ was studied by many authors for differents classes of operators. In this paper we investigate a local version of this equality and the case where A is a C₀–contraction.peerReviewe

Zarrabi, Mohamed

Extracta Mathematicae

English

Dehesa. Repositorio Institucional de la Universidad de Extremadura

E extracta mathematicae Vol. 27, Núm. 2, 175 – 185 (2012)The Commutant of an Operator with BoundedConjugation Orbits and C0–ContractionsMohamed Zarrabi ∗Université de Bordeaux, UMR 5251, 351 Cours de la Libération,F-33405 Talence Cedex, FranceMohamed.Zarrabi@math.u-bordeaux1.frPresented by Mostafa Mbekhta Received March 30, 2012Abstract : Let A be an invertible bounded linear operator on a complex Banach space, {A}′the commutant of A and BA the set of all operators T such that supn≥0 ∥AnTA−n∥ < +∞.Equality {A}′ = BA was studied by many authors for differents classes of operators. In thispaper we investigate a local version of this equality and the case where A is a C0–contraction.Key words: Operators, commutant, bounded conjugation orbit, C0–contraction.AMS Subject Class. (2010): 47B10, 47A10, 47A45.1. IntroductionDenote by L(X) the algebra of all bounded linear operators on the complexBanach space X. Let A ∈ L(X). We denote by σ(A) the spectrum of A andby {A}′ the commutant of A. Throughout this paper we assume that A isinvertible and we setBA ={T ∈ L(X) : supn≥0∥AnTA−n∥ < +∞}.Clearly we have {A}′ ⊂ BA. When X is of finite dimension, Deddens showedin [1] that equality {A}′ = BA holds if and only if σ(A) is a singleton. Heconjectured that the above result remains true when X is any Hilbert space.P.G. Roth gave a negative answer to this conjecture. Let V be the integralVolterra operator on H = L2([0, 1]) and let A be the operator acting on H⊕Hdefined byA =(1 + V 00 1).∗The author was partially supported by the ANR project ANR-09-BLAN-0058-01.175176 m. zarrabiIt is shown in [8, Example 2.7] that {A}′ ̸= BA. Moreover we have∥(A− 1)n∥ = ∥V n∥ = 1(n− 1)!,which implies that∥(A− 1)n∥1/n = O(1/n), n → +∞. (1)On the other hand side one can easily obtain from Theorem 2.3 and Theo-rem 2.5 of [8] the following result (see also Corollary 2.4).Theorem 1.1. Assume that for some complex number λ,∥(A− λ)n∥1/n = o(1/n), n → +∞. (2)Then BA = {A}′.The example given above shows that if we replace ”o” by ”O” in thecondition (2) then the conclusion of the theorem is false in general. Noticethat Williams showed in [11] that if σ(A) is a singleton then BA∩BA−1 = {A}′.This was improved by Drissi and Mbekhta in [2], replacing BA−1 by the set{T ∈ L(X) : ∀ϵ > 0, ∥A−nTAn∥ = o(eϵ√n), n → +∞}. Notice that equalityBA∩BA−1 = {A}′ was extensively studied by Drissi and Mbekhta for differentclasses of operators and elements in Banach algebras (see [2], [3] and [4]).In this note, we are interested in a local version of the above results. Todescribe the results obtained here let us introduce some notations. For acomplex number λ, we setV(A,λ) ={x ∈ X : ∥(A− λ)nx∥1/n = o(1/n), n → +∞}.Notice that V(A,λ) is a linear subspace of X, invariant by A and is the zeroset if λ is not in the spectrum of A. We denote by CA the set of all operatorsT ∈ L(X) such that, for all λ ∈ σ(A) and x ∈ V(A,λ),∥∥(A− λ)iT (A−1 − λ−1)jx∥∥ 1i+j = o( 1i+ j), i+ j → +∞.Finally if (Vi)i∈I is a family of subsets of X we denote by Span(Vi, i ∈ I) thelinear subspace of X generated by ∪i∈IVi.We prove (Theorem 2.3) that if Span(V(A,λ), λ ∈ σ(A))is dense in X thenBA ∩ CA = {A}′. To prove this we introduce a local version of the sets {A}′the commutant of an operator 177and BA and we study the relationship between these sets, using a theorem of[5] about local properties of powers of an operator.In Section 3 we study the above equalities when A is a C0–contraction.Let H be a separable Hilbert space and let H∞ be the classical Hardy space ofall bounded and holomorphic functions on D. A contraction A on H is calleda C0–contraction (or in class C0) if it is completely nonunitary and thereexists a nonzero function θ ∈ H∞ such that θ(A) = 0. Notice that for a C0–contraction A on H, there exists a minimal inner function ΘA that annihilatesA, i.e., is such that ΘA(A) = 0 (see [7] and [6]). Let A be an invertible C0–contraction. We prove (Proposition 3.2) that if ΘA is a Blaschke productthen we have BA ∩ CA = {A}′. We obtain also (Proposition 3.3) that if ΘAis a Blaschke product then the equality BA = {A}′ holds if and only if ΘAis a Blaschke product with a unique root. If the singular part of ΘA is notconstant, we do not know when the above equalities holds.2. Bounded conjugation orbitLet A ∈ L(X) be invertible, x ∈ X and λ ∈ C\{0}. We denote by C(A,λ,x)the set of all operators T ∈ L(X) such that∥∥(A− λ)iT (A−1 − λ−1)jx∥∥ 1i+j = o( 1i+ j), i+ j → +∞.and C(A,λ) = ∩x∈V(A,λ)C(A,λ,x). Then we have CA = ∩λ∈σ(A)C(A,λ).We setB(A,x) ={T ∈ L(X) : supn≥0∥AnTA−nx∥ < +∞}and{(A, x)}′ ={T ∈ L(X) : ATx = TAx}.Clearly we have ∩n≥0{(A−n, x)}′ ⊂ B(A,x). We shall need the following ele-mentary result.Lemma 2.1. Let n, i and j be integers such that 0 ≤ i ≤ n and 0 ≤ j ≤ n.We setFn(i, j) =∑max{i,j}≤k≤n(−1)n−k(nk)(ki)(kj).178 m. zarrabiThen for every integer n ≥ 0, we haveFn(i, j) ={0 if i+ j < n,(n2n−i−j)(2n−i−jn−j)if i+ j ≥ n,and ∑0≤i,j≤nFn(i, j) = 3n.Proof. We set Q(x, y) = (x+ y + xy)n. Using the binomial formula twice,we haveQ(x, y) =n∑k=0(nk)(x+ y)kxn−kyn−k=n∑k=0k∑l=0(nk)(kl)xn+l−kyn−l=∑0≤i,j≤nn≤i+j(n2n− i− j)(2n− i− jn− j)xiyj .On the other hand side, writing Q(x, y) =((1 + x)(1 + y) − 1)nand usingagain the binomial formula, we getQ(x, y) =n∑k=0(−1)n−k(nk)(1 + x)k(1 + y)k=n∑k=0(−1)n−k(nk) ∑0≤i≤k0≤j≤k(ki)(kj)xiyj=∑0≤i,j≤nFn(i, j)xiyj .(3)Now, by identifying the coefficients of xiyj in the two polynomial expressionsof Q(x, y) obtained above, we get the first equality of the lemma. To provethe second one it suffices to take x = y = 1 in (3).Theorem 2.2. Let A ∈ L(X) be invertible, x ∈ X and λ ∈ C\{0}. Thenwe haveB(A,x) ∩ C(A,λ,x) =∩n≥0{(A−n, x)}′ if and only if x ∈ V(A,λ).the commutant of an operator 179Proof. Assume that the equality B(A,x)∩C(A,λ,x) = ∩n≥0{(A−n, x)}′ holds.Then the identity operator belongs to C(A,λ,x) which implies that x ∈ V(A,λ).To prove the converse, we denote byW the bounded linear operator definedon L(X) by W (T ) = ATA−1. We have(W − I)nTx =n∑k=0(−1)n−k(nk)AkTA−kx,where I is the identity operator defined on L(X). We write Ak = (A−λ+λ)k,A−k = (A−1 − λ−1 + λ−1)k and we use the binomial formula. Then we get(W − I)n Tx =∑0≤i,j≤nλj−iFn(i, j)(A− λ)iT (A−1 − λ−1)jx, (4)where Fn(i, j) is defined in Lemma 2.1.Suppose that T ∈ C(A,λ,x). For every ϵ > 0, there exists a positive constantCϵ such that for all non–negative integers i and j, we have∥∥(A− λ)iT (A−1 − λ−1)jx∥∥ ≤ Cϵ ϵi+j(i+ j)i+j.Since by Lemma 2.1 we have Fn(i, j) = 0 for i + j < n, it follows from thelast inequality and (4) that∥∥(W − I)nTx∥∥ ≤ Cϵmax (|λ|n, |λ|−n) ∑0≤i,j≤nn≤i+jFn(i, j)ϵi+j(i+ j)i+j. (5)The function t → ϵttt decreases on the set [ϵe ,+∞[. So for i+ j ≥ n ≥ ϵ/e, wehave ϵi+j(i+j)i+j≤ ϵnnn . Combining this observation with the second equality inLemma 2.1 and (5) we get∥∥ (W − I)n Tx∥∥ ≤ Cϵ (3max (|λ|, |λ|−1) ϵ)nnn.This shows that ∥∥(W − I)nTx∥∥1/n = o (1/n) , n → ∞. (6)Let ℓ be an element in the dual X⋆ of X and let L be the continuous linearfunctional on L(X) defined by L(R) = ℓ(Rx), R ∈ L(X). By (6) we have∣∣L((W − I)nT )∣∣1/n = o (1/n) , n → ∞.180 m. zarrabiSuppose in addition that T ∈ B(A,x). Then |L (WnT ) | = O(1), n → ∞. Itfollows from Theorem 1.1 of [5] that L((W − I)nT ) = 0, n ≥ 1. Since ℓ is anyelement of X⋆ we obtain by the Hahn-Banach theorem that (W − I)nTx = 0,n ≥ 1. Now by induction on n, one checks easily that T ∈ {(A−n, x)}′ forevery n ≥ 1. So the inclusion B(A,x) ∩ C(A,λ,x) ⊂ ∩n≥0{(A−n, x)}′ holds forevery x ∈ X.Assume now that x ∈ V(A,λ). Let T ∈ ∩n≥0{(A−n, x)}′. Obviously T ∈B(A,x). On the other hand side, for all non-negative integers n,m, we haveA−nTA−mx = A−(n+m)Tx = TA−(n+m)x.Therefore we obtain(A− λ)iT (A−1 − λ−1)jx = (−λ)iAi(A−1 − λ−1)iT (A−1 − λ−1)jx= (−λ)iAiT (A−1 − λ−1)i+jx= (−λ)−jAiTA−i−j(A− λ)i+jx.Thus∥∥(A− λ)iT (A−1 − λ−1)jx∥∥ 1i+j ≤ ∥∥(−λ)−jAiTA−i−j∥∥ 1i+j ∥∥(A− λ)i+jx∥∥ 1i+j= o(1i+ j), i+ j → ∞and so T ∈ C(A,λ,x). Therefore we have ∩n≥0{(A−n, x)}′ ⊂ B(A,x) ∩ C(A,λ,x),which concludes the proof.Theorem 2.3. Let A ∈ L(X) be invertible. If Span(V(A,λ), λ ∈ σ(A))isdense in X thenBA ∩ CA = {A}′.Proof. Let T ∈ {A}′. For every x ∈ X, we have T ∈ ∩n≥0{(A, x)}′.It follows from Theorem 2.2 that if λ ∈ σ(A) and if x ∈ V(A,λ) then T ∈C(A,λ,x). Thus T ∈ ∩λ∈σ(A)(∩x∈V(A,λ) C(A,λ,x))= CA. So {A}′ ⊂ BA ∩ CA.Suppose now that T ∈ BA ∩ CA. For every λ ∈ σ(A) and x ∈ V(A,λ), wehave T ∈ B(A,x) ∩ C(A,λ,x). It follows again from Theorem 2.2 that for everyx ∈ V(A,λ), A−1Tx = TA−1x. Clearly this equality holds for every x ∈ X,since Span(V(A,λ), λ ∈ σ(A))is dense in X. So A−1 and A commute with T .the commutant of an operator 181Corollary 2.4. Let A be an invertible operator on X such that∥∥(A− λ)n∥∥1/n = o(1/n), n → +∞.for some complex number λ. Then BA = {A}′.Proof. We have V(A,λ) = X. By Theorem 2.3 we only need to prove thatCA = L(X). Let T ∈ L(X) and let ϵ > 0. There exists Cϵ such that for everynonnegative integer i ∥∥(A− λ)i∥∥ ≤ Cϵ ϵiiiNotice that the hypothesis implies that σ(A) = {λ} and so λ ̸= 0. We have∥∥(A−1 − λ−1)i∥∥ ≤ ∥∥λ−iA−i(λ−A)i∥∥≤ Cϵ(∥A−1∥|λ|)iϵiii.Thus for every x ∈ X and every nonnegative integers i and j, we get∥∥(A− λ)iT (A−1 − λ−1)jx∥∥ ≤ ∥T∥∥x∥C2ϵ (∥A−1∥|λ|)iϵi+jiijj.Since (i+ j)i+j ≤ ei+jiijj , we obtain∥∥(A− λ)iT (A−1 − λ−1)jx∥∥ ≤ C ′ϵ (ecϵ)i+j(i+ j)i+j ,where c = max{∥A−1∥|λ| , 1}and C ′ϵ is a constant independent of i and j. HenceCA = L(X).Remark. For a family of subspaces (Vi)i∈I of X, we setAlg (Vi, i ∈ I) = {T ∈ L(X) : TVi ⊂ Vi, (i ∈ I)}.If λ ∈ σ(A) and if V(A,λ) is closed, it follows from the Banach–Steinhausstheorem that ∥∥∥(A|V(A,λ) − λ)n∥∥∥1/n = o (1/n) , n → +∞.Then it follows that C(A,λ) = {T ∈ L(X) : TV(A,λ) ⊂ V(A,λ)}. So, if for everyλ ∈ σ(A), V(A,λ) is closed, then we have CA = Alg(V(A,λ), λ ∈ σ(A)).182 m. zarrabiLet A∈L(X) be invertible and with finite spectrum, σ(A)={λ1, λ2, . . . ,λp}.It is well known that X = F1+ · · ·+Fp, where F1, . . . , Fp are closed subspaceswhich are invariant by A and such that σ(A|Fi) = {λi}, 1 ≤ i ≤ p.Assume that ∥∥Q(A)n∥∥1/n = o(1/n), n → +∞,where Q(x) =∏1≤i≤p(x − λi). It is easily seen that Fi = V(A,λi), 1 ≤ i ≤ p.We deduce from Corollary 2.4 thatBA ∩Alg(V(A,λi), 1 ≤ i ≤ p)= {A}′.3. The case of C0–contractionsIn this section we investigate the equalities BA = {A} and BA∩CA = {A}′when A is a C0-contraction. Notice that when the spectrum of A is reducedto a single point, say σ(A) = {λ}, then A − λ is nilpotent if |λ| < 1 and∥(A − λ)n∥1/n = O(1/n) (n → +∞), if |λ| = 1 (see the proof at the end ofthis section).Henceforth we denote by A a C0-contraction on a separable Hilbert spaceH and by ΘA it’s minimal inner function (see [6]). We haveσ(A) = Clos Θ−1A (0) ∪ suppµ,where µ is the singular measure associated to the singular part of ΘA andsuppµ is the closed support of µ (see [6, p. 63]). The set of eigenvalues of Ais given by the formulaσp(A) = σ(A) ∩ D = Θ−1A (0).We see that A is invertible if and only if ΘA(0) ̸= 0. For λ ∈ D we setbλ(z) =|λ|λλ−z1−λz , z ∈ D. If ΘA(λ) = 0 we set kλ to be the multiplicity of λ.A typical example of a C0-contraction is the so-called model operator SΘ,where Θ is an inner function. Let KΘ = H2 ⊖ΘH2, where H2 is the classicalHardy space. SΘ is the compression of the shift, defined on KΘ by: f →PΘ(zf), where PΘ : H2 → KΘ is the orthogonal projection. Notice that theminimal inner function of SΘ is Θ.Lemma 3.1. We haveV(A,λ) ={{0} if λ ∈ σ(A) ∩ T,ker (A− λ)kλ if λ ∈ σ(A) ∩ D.the commutant of an operator 183Proof. Let λ ∈ σ(A)∩T and x ∈ V(A,λ). Since A is a contraction, it followsfrom [5, Theorem 1.1] that Ax = λx. Since A has no eigenvalue on T, we havex = 0. So V(A,λ) = {0}.Now let λ ∈ σ(A)∩D = σp(A). Notice that λ is an isolated point in σ(A).Denote by Fλ the (maximal) spectral subspace of A such that σ(A|Fλ)={λ} (see [6, p. 84–86]). The space H decomposes into the direct sum of twosubspaces Hλ and H′λ, hyperinvariant for A and so that Hλ = ker (A− λ)kλand λ /∈ σ(A|H′λ)(see [7, Theorem 7.1, p. 135]). So Fλ = ker (A− λ)kλ .On the other hand Fλ is the set of all x ∈ H such that the map z →(z − A)−1x admits an analytic extension to C \ {λ}. We have (z − A)−1x =∑n≥0(z − λ)−n−1(A − λ)nx for |z| sufficiently large. So Fλ = {x ∈ H :limn→+∞ ∥(A− λ)nx∥1/n = 0}. Thus we get ker (A− λ)kλ = V(A,λ) = Fλ.Proposition 3.2. Suppose that ΘA is a Blaschke product with ΘA(0) ̸=0. ThenBA ∩ CA = {A}′.Proof. By Lemma 3.1 we haveSpan{V(A,λ), λ ∈ σ(A)}= Span{ker (A− λ)kλ , λ ∈ ΛΘ}.Now the result follows from the completeness theorem ([7, Proposition 7.2,p. 135]) and Theorem 2.3.For u and v in H we denote by v ⊗ u the operator defined on H byx →< x, v > u.Proposition 3.3. Suppose that ΘA is a Blaschke product with ΘA(0) ̸=0. Then BA = {A}′ if and only if ΘA is a Blaschke product with a uniqueroot.Proof. If ΘA has the form bkλλ , |λ| < 1, then H = ker(A − λ)kλ and theequality BA = {A}′ follows from Corollary 2.4.For the converse assume that ΘA vanishes at two distincts points λ1 andλ2 with |λ1| ≤ |λ2| < 1. Notice that A∗ is a C0–contraction and ΘA∗(z) =ΘA(z), z ∈ D. So λ1 is an eigenvalue of A and λ2 is an eingenvalue of A∗.Let u and v be a nonzero vectors in H such that Au = λ1u and A∗v = λ2v.We have An(v ⊗ u)A−n =(λ1λ2)nv ⊗ u and so v ⊗ u ∈ BA \ {A}′.184 m. zarrabiWhen Θ is not a Blaschke product, we do not know when the equalitiesBA∩CA = {A}′ and BA = {A}′ hold. It is interesting to study the particularcase when ΘA(z) = ez+1z−1 , z ∈ D. Notice that in this case σ(A) = {1} and Asatisfies condition (1). Indeed, let, for ζ ∈ C \ {1},∆ζ(z) =ΘA(z)−ΘA(ζ)z − ζ, |z| < 1.Clearly ∆ζ ∈ H∞ and since ΘA(A) = 0, we have ∆ζ(A) = −ΘA(ζ) (A− ζ)−1.Hence ∥∥(A− ζ)−1∥∥ ≤ ∥∆ζ∥∞ΘA(ζ),where ∥ · ∥∞ is the supremum norm on D.If |z − ζ| ≥ 12 |ζ − 1| then |∆ζ(z)| ≤4|ζ−1| . On the other hand if |z − ζ| ≤12 |ζ − 1|, then |∆ζ(z)| ≤ supw∈[z,ζ] |Θ′A(w)| ≤8|ζ−1|2 . So∥∆ζ∥∞ ≤8|ζ − 1|2for |ζ − 1| ≤ 1.Thus ∥∥(A− ζ)−1∥∥ ≤ 8|ζ − 1|2e2|ζ−1| for |ζ − 1| ≤ 1. (7)Since σ(A) = {1}, we obtain from the Dunford–Riesz functional calculus thatfor every r > 0,(A− I)n = 12iπ∫|ζ−1|=r(ζ − 1)n(ζ −A)−1dζ.It follows from this and inequality (7) that∥∥(A− I)n∥∥ ≤ 8rn−1e 2r , 0 < r ≤ 1.Taking r = 1n , we see that A satisfies condition (1).References[1] J.A. Deddens, Another description of nest algebras, in “Hilbert Spaces Op-erators”, Lecture Notes in Mathematics 693, Springer-Verlag, Berlin, 1978,77 – 86.[2] D. Drissi, M. Mbekhta, Operators with bounded conjugation orbits, Proc.Amer. Math. Soc. 128 (9) (2000), 2687 – 2691.the commutant of an operator 185[3] D. Drissi, M. Mbekhta, Elements with generalized bounded conjugationorbits, Proc. Amer. Math. Soc. 129 (7) (2001), 2011 – 2016.[4] D. Drissi, M. Mbekhta, On the commutant and orbits of conjugation,Proc. Amer. Math. Soc. 134 (4) (2006), 1099 – 1106.[5] J. Esterle, M. Zarrabi, Local properties of powers of operators, Arch.Math. (Basel) 65 (1) (1995), 53 – 60.[6] N. Nikolski, “Treatise on the Shift Operator”, Grundlehren der Mathema-tischen Wissenschaften 273, Springer-Verlag, Berlin-Heidelberg 1986.[7] B.Sz.-Nagy, C. Foias, “Analyse Harmonique des Opérateurs de l’Espacede Hilbert”, Masson et Cie, Paris, 1967.[8] P.G. Roth, Bounded orbits of conjugation, analytic theory, Indiana Univ.Math. J. 32 (4) (1983), 491 – 509.[9] J.G. Stampfli, On a question of Deddens, in “Hilbert Space Operators”,Lecture Notes in Mathematics 693, Springer-Verlag, Berlin, 1978, 169 – 173.[10] J.G. Stampfli, J.P. Williams, Growth conditions and the numerical rangein a Banach algebra, Tôhôku Math. J. (2) 20 (4) (1968), 417 – 424.[11] J.P. Williams, On a boundedness condition for operators with a singletonspectrum, Proc. Amer. Math. Soc. 78 (1) (1980), 30 – 32.

The commutant of an operator with bounded conjugation orbits and C₀‒contractions

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The commutant of an operator with bounded conjugation orbits and C₀‒contractions

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