We  study some local spectral properties for bounded   operators R, S, RS and SR in the case that R and  S  satisfy  the  operator  equation  RSR = R2. Among other results, we prove that S, R, SR and RS share Dunford?s property (C) when RSR = R2  and SRS = SSupported in part by MICINN (Spain), Grant MTM2013-45643

Aiena, Pietro

González Ortiz, Manuel

Extracta Mathematicae

We study some local spectral properties for bounded operators R, S, RS and SR in the case that R and S satisfy the operator equation RSR =R^2. Among other results, we prove that S, R, SR and RS share Dunford's property (C) when RSR = R^2and SRS =S^2peerReviewe

González, Manuel

Dehesa. Repositorio Institucional de la Universidad de Extremadura

E extracta mathematicae Vol. 31, Núm. 1, 37 – 46 (2016)Local Spectral Theory for Operators R and SSatisfying RSR = R2Pietro Aiena, Manuel GonzálezDipartimento di Metodi e Modelli Matematici, Facoltà di Ingegneria, Università diPalermo (Italia), paiena@unipa.itDepartamento de Matemáticas, Facultad de Ciencias, Universidad de Cantabria,E-39071 Santander (Spain), manuel.gonzalez@unican.esReceived February 5, 2016Abstract : We study some local spectral properties for bounded operators R, S, RS and SRin the case that R and S satisfy the operator equation RSR = R2. Among other results, weprove that S, R, SR and RS share Dunford’s property (C) when RSR = R2 and SRS = S2.Key words: Local spectral subspace, Dunford’s property (C), operator equation.AMS Subject Class. (2010): 47A10, 47A11, 47A53, 47A55.1. Introduction and preliminariesThe equivalence of Dunford’s property (C) for products RS and SR ofoperators R ∈ L(Y,X) and S ∈ L(X,Y ), X and Y Banach spaces, has beenstudied in [2]. As noted in [13] the proof of Theorem 2.5 in [2] contains a gapwhich was filled up in [13, Theorem 2.7]. In [2] it was also studied property(C) for operators R,S ∈ L(X) which satisfy the operator equationsRSR = R2 and SRS = S2. (1)A similar gap exists in the proof of Theorem 3.3 in [2], which states theequivalence of property (C) for R, S, RS and SR, when R, S satisfy (1).In this paper we give a correct proof of this result and we prove furtherresults concerning the local spectral theory of R, S, RS and SR, in particularwe show several results concerning the quasi-nilpotent parts and the analyticcores of these operators. It should be noted that these results are establishedin a more general framework, assuming that only one of the operator equationsin (1) holds.Supported in part by MICINN (Spain), Grant MTM2013-45643.3738 p. aiena, m. gonzálezWe shall denote by X a complex infinite dimensional Banach space. Givena bounded linear operator T ∈ L(X), the local resolvent set of T at a pointx ∈ X is defined as the union of all open subsets U of C such that there existsan analytic function f : U → X satisfying(λI − T )f(λ) = x for all λ ∈ U . (2)The local spectrum σT (x) of T at x is the set defined by σT (x) := C \ ρT (x).Obviously, σT (x) ⊆ σ(T ), where σ(T ) denotes the spectrum of T .The following result shows that σT (Tx) and σT (x) may differ only at 0. Itwas proved in [7] for operators satisfying the SVEP.Lemma 1.1. For every T ∈ L(X) and x ∈ X we haveσT (Tx) ⊆ σT (x) ⊆ σT (Tx) ∪ {0}. (3)Moreover, if T is injective then σT (Tx) = σT (x) for all x ∈ X.Proof. Take S = T and R = I in [6, Proposition 3.1 ].For every subset F of C, the local spectral subspace of T at F is the setXT (F) := {x ∈ X : σT (x) ⊆ F}.It is easily seen from the definition thatXT (F) is a linear subspace T -invariantof X. Furthermore, for every closed F ⊆ C we have(λI − T )XT (F) = XT (F) for all λ ∈ C \ F . (4)See [9, Proposition 1.2.16].An operator T ∈ L(X) is said to have the single valued extension propertyat λo ∈ C (abbreviated SVEP at λo), if for every open disc Dλo centered atλo the only analytic function f : Dλo → X which satisfies the equation(λI − T )f(λ) = 0 (5)is the function f ≡ 0. An operator T ∈ L(X) is said to have the SVEP if Thas the SVEP at every point λ ∈ C. Clearly, the SVEP is inherited by therestrictions to invariant subspaces.A variant of XT (F) which is more useful for operators without SVEP isthe glocal spectral subspace XT (F). For an operator T ∈ L(X) and a closedlocal spectral theory for operators R and S 39subset F of C, we define XT (F) as the set of all x ∈ X for which there existsan analytic function f : C \ F → X which satisfies(λI − T )f(λ) = x for all λ ∈ C \ F .Clearly XT (F) ⊆ XT (F) for every closed F ⊆ C. Moreover T has SVEP ifand only ifXT (F) = XT (F) for all closed subsets F ⊆ C.See [9, Proposition 3.3.2]. Note that XT (F) and XT (F) are not closed ingeneral.Given a closed subspace Z of X and T ∈ L(X), we denote by T |Z therestriction of T to Z.Lemma 1.2. [2, Lemmas 2.3 and 2.4] Let F be a closed subset of C andT ∈ L(X).(1) If 0 ∈ F and Tx ∈ XT (F) then x ∈ XT (F).(2) Suppose T has SVEP, Z := XT (F) is closed, and A := T |XT (F). ThenXT (K) = ZA(K) for all closed K ⊆ F .Lemma 1.3. Suppose that T has SVEP and F is a closed subset of Csuch that 0 /∈ F . If XT (F ∪ {0}) is closed then XT (F) is closed.Proof. Set Z := XT (F ∪{0}) and S := T |Z. By [9, Proposition 1.2.20] wehave σ(S) ⊆ F ∪{0}. We suppose first that 0 /∈ σ(S). Then σ(S) ⊆ F , henceZ = ZS(F). By Lemma 1.2 we have ZS(F) = XT (F), so XT (F) is closed.For the case 0 ∈ σ(S), we set F0 := σ(S) ∩ F . Then σ(S) = F0 ∪ {0}. Since0 ∈ σ(S), by Lemma 1.2 we have Z = ZS(F0)⊕ ZS({0}) andZS(F0) = ZS(σ(S) ∩ F) = ZS(F) = XT (F),hence XT (F) is closed.2. Operator equation RSR = R2Operators S,R ∈ L(X) satisfying the operator equations RSR = R2 andSRS = S2 were studied first in [12], and more recently in [10], [11], [8], andother papers. An easy example of operators for which these equations hold is40 p. aiena, m. gonzálezgiven in the case that R = PQ and S = QP , where P,Q ∈ L(X) are idem-potents. A remarkable result of Vidav [12, Theorem 2] shows that if R,S areself-ajoint operators on a Hilbert space then the equations (1) hold if and onlyif there exists an (uniquely determined) idempotent P such that R = PP ∗and S = P ∗P , where P ∗ is the adjoint of P .The operators R, S, SR and RS for which the equations (1) hold sharemany spectral properties ([10], [11]), and local spectral properties as decom-posability, property (β) and SVEP ([8]). In this section we consider the per-manence of property (C), property (Q) in this context.It is easily seen that if 0 /∈ σ(R) ∩ σ(S) then R = S = I, so this case istrivial. Thus we shall assume that 0 ∈ σ(R) ∩ σ(S). Evidently, the operatorequation RSR = R2 implies(SR)2 = SR2 and (RS)2 = R2S.Lemma 2.1. Suppose that R,S ∈ L(X) satisfy RSR = R2. Then forevery x ∈ X we haveσR(Rx) ⊆ σSR(x) and σSR(SRx) ⊆ σR(x). (6)Proof. For the first inclusion, suppose that λ0 /∈ σSR(x). Then there existsan open neighborhood U0 of λ0 and an analytic function f : U0 → X suchthat(λI − SR)f(λ) = x for all λ ∈ U0.From this it follows thatRx = R(λI − SR)f(λ) = (λR−RSR)f(λ)= (λR−R2)f(λ) = (λI −R)(Rf)(λ),for all λ ∈ U0. Since Rf : U0 → X is analytic we get λ0 /∈ σR(Rx).For the second inclusion, let λ0 /∈ σR(x). Then there exists an openneighborhood U0 of λ0 and an analytic function f : U0 → X such that(λI −R)f(λ) = x for all λ ∈ U0.Consequently,SRx = SR(λI −R)f(λ) = (λSR− SR2)f(λ)= (λSR− (SR)2)f(λ) = (λI − SR)(SRf)(λ),for all λ ∈ U0, and since (SR)f is analytic we obtain λ0 /∈ σSR(SRx).local spectral theory for operators R and S 41Theorem 2.2. Let S,R ∈ L(X) satisfy RSR = R2, and let F be a closedsubset of C with 0 ∈ F . Then XR(F) is closed if and only if so is XSR(F).Proof. Suppose thatXR(F) is closed and let (xn) be a sequence ofXSR(F)which converges to x ∈ X. We need to show that x ∈ XSR(F). For everyn ∈ N we have σSR(xn) ⊆ F and hence, by Lemma 2.1, we have σR(Rxn) ⊆ F ,i.e. Rxn ∈ XR(F). Since 0 ∈ F , by Lemma 1.2 we have xn ∈ XR(F), andsince XR(F) is closed, x ∈ XR(F), i.e. σR(x) ⊆ F . Now from Lemma 2.1 wederive σSR(SRx) ⊆ F , and this implies SRx ∈ XSR(F). Again by Lemma 1.2,we obtain x ∈ XSR(F), thus XSR(F) is closed.Conversely, suppose that XSR(F) is closed and let (xn) be a sequenceof XR(F) which converges to x ∈ X. Then σR(xn) ⊆ F for every n ∈ N,hence σSR(SRxn) ⊆ F , i.e. SRxn ∈ XSR(F) by Lemma 2.1. But 0 ∈ F , so,by Lemma 1.2, xn ∈ XSR(F). Since XSR(F) is closed, x ∈ XSR(F), henceσSR(x) ⊆ F . Now from Lemma 2.1 we obtain σR(Rx) ⊆ F , i.e. Rx ∈ XR(F),and the condition 0 ∈ F implies x ∈ XR(F).The following result is inspired by [8, Theorem 2.1].Lemma 2.3. Let S,R ∈ L(X) be such that RSR = R2 and one of theoperators R, SR, RS has SVEP. Then all of them have SVEP. Additionally,if SRS = S2 and one of R, S, SR, RS has SVEP then all of them have SVEP.Proof. By [6, Proposition 2.1], SR has SVEP if and only if RS has SVEP.So it is enough to prove that R has SVEP at λ0 if an only if so has RS.Suppose that R has SVEP at λ0 and let f : U0 → X be an analytic functionon an open neighborhood U0 of λ0 for which (λI −RS)f(λ) ≡ 0 on U0. ThenRSf(λ) = λf(λ) and0 = RS(λI −RS)f(λ) = (λRS − (RS)2)f(λ) = (λRS − (R2S)f(λ)= (λI −R)RSf(λ).The SVEP of R at λ0 implies thatRSf(λ) = λf(λ) = 0 for all λ ∈ U0.Hence f ≡ 0 on U0, and we conclude that RS has SVEP at λ0.Conversely, suppose that RS has SVEP at λ0 and let f : U0 → X be an42 p. aiena, m. gonzálezanalytic function on an open neighborhood U0 of λ0 such that (λI−R)f(λ) ≡ 0on U0. Then R2f(λ) = λRf(λ) = λ2f(λ) for all λ ∈ U0. Moreover,0 = RS(λI −R)f(λ) = λRSf(λ)−R2f(λ) = λRSf(λ)− λ2f(λ)= (λI −RS)(−λf(λ)),and since RS has SVEP at λ0 we have λf(λ) ≡ 0, hence f(λ) ≡ 0, so R hasSVEP at λ0.The second assertion is clear, if SRS = S2, just interchanging R and S inthe argument above, the SVEP fo S holds if and only if SR, or equivalentlyRS, has SVEP.We now consider the result of Theorem 2.2 when 0 /∈ F .Theorem 2.4. Let F be a closed subset of C such that 0 /∈ F . Supposethat R,S ∈ L(X) satisfy RSR = R2 and R has SVEP. Then we have(1) If XR(F ∪ {0}) is closed then XSR(F) is closed.(2) If XSR(F ∪ {0}) is closed then XR(F) is closed.Proof. (1) Let us denote F1 := F ∪ {0}. The set F1 is closed, and byassumption XR(F1) is closed. Since 0 ∈ F1 then XSR(F1) is closed, by The-orem 2.2. Moreover, the SVEP for R is equivalent to the SVEP for SR byLemma 2.3. Then XSR(F) is closed by Lemma 1.3.(2) The argument is similar: if XSR(F ∪ {0}) is closed then XR(F ∪ {0})by Theorem 2.2, and since R has SVEP, XR(F) is closed by Lemma 1.3.Definition 2.5. An operator T ∈ L(X) is said to have Dunford’s prop-erty (C) (abbreviated property (C)) if XT (F ) is closed for every closed setF ⊆ C.It should be noted that Dunford property (C) implies SVEP.Theorem 2.6. Suppose that S,R ∈ L(X) satisfy RSR = R2, and any oneof the operators R, SR, RS, has property (C).Then all of them have property(C). If, additionally, SRS = S2 and one of R, S, RS, SR has property (C),then all of them have property (C).Proof. Since property (C) implies SVEP, all the operators have SVEP byLemma 2.3. Moreover the equivalence of property (C) for SR and RS haslocal spectral theory for operators R and S 43been proved in [2] (see also [13]). So it is enough to prove that R has property(C) if an only if so has RS.Suppose that R has property (C) and let F be a closed set. If 0 ∈ Fthen XSR(F) is closed, by Theorem 2.2, while in the case where 0 /∈ F wehave that XR(F ∪ {0}) is closed, and hence, by part (1) of Theorem 2.4, theSVEP for R ensures that also in this case XSR(F) is closed. Therefore, SRhas property (C).Conversely, suppose that SR has property (C). For every closed subset Fcontaining 0, XR(F) is closed by Theorem 2.2. If 0 /∈ F then XSR(F ∪ {0})is closed, hence XR(F) is closed by part (2) of Theorem 2.4 and we concludethat R has property (C).If additionally, SRS = S2 then, by interchanging S with R, the sameargument above proves the second assertion, so the proof is complete.Next we consider the case when F is a singleton set, say F := {λ}.The glocal spectral subspace XT ({λ}) coincides with the quasi-nilpotent partH0(λI − T ) of λI − T defined byH0(λI − T ) := {x ∈ X : lim supn→∞∥(λI − T )nx∥1/n = 0}.See [1, Theorem 2.20]. In general H0(λI − T ) is not closed, but it coincideswith the kernel of a power of λI − T in some cases [3, Theorem 2.2].Definition 2.7. An operator T ∈ L(X) is said to have the property (Q)if H0(λI − T ) is closed for every λ ∈ C.It is known that if H0(λI − T ) is closed then T has SVEP at λ ([4]), thus,property (C) ⇒ property (Q) ⇒ SVEP.Therefore, for operators T having property (Q) we haveH0(λI−T )=XT ({λ}).In [13, Corollary 3.8] it was observed that if R ∈ L(Y,X) and S ∈ L(X,Y )are both injective then RS has property (Q) precisely when SR has property(Q).Recall that T ∈ L(X) is said to be upper semi-Fredholm, T ∈ Φ+(X), ifT (X) is closed and the kernel ker T is finite-dimensional, and T is said to belower semi-Fredholm, T ∈ Φ−(X), if the range T (X) has finite codimension.Theorem 2.8. Let R,S ∈ L(X) satisfying RSR = R2, and R,S ∈ Φ+(X)or R,S ∈ Φ−(X). Then R has property (Q) if and only if so has SR.44 p. aiena, m. gonzálezProof. Suppose that R,S ∈ Φ+(X) and R has property (Q). Then Rhas SVEP and, by Lemma 2.3, also SR has SVEP. Consequently, the localand glocal spectral subspaces relative to the a closed set coincide for R andSR. By assumption H0(λI − R) = XR({λ}) is closed for every λ ∈ C, andH0(SR) = XSR({0}) is closed by Theorem 2.2. Let 0 ̸= λ ∈ C. By [9,Proposition 3.3.1, part (f)]XR({λ} ∪ {0}) = XR({λ}) +XR({0}) = H0(λI −R) +H0(R).Since R ∈ Φ+(X) the SVEP at 0 implies that H0(R) is finite-dimensional, see[1, Theorem 3.18 ], so XR({λ} ∪ {0}) is closed. Then part (1) of Theorem 2.4implies that H0(λI − SR) = XSR({λ}) is closed, hence SR has property (Q).Conversely, suppose that SR has property (Q). If λ = 0 then H0(SR) =XRS({0}) is closed by assumption, and H0(R) = XR({0}) is closed by Theo-rem 2.2. In the case λ ̸= 0 we haveXSR({λ} ∪ {0}) = XSR({λ}) +XSR({0}) = H0(λI − SR) +H0(SR).Since SR has SVEP and SR ∈ Φ+(X), H0(SR) is finite dimensional by [1,Theorem 3.18]. So XSR({λ} ∪ {0}) is closed. By part (2) of Theorem 2.4,XR({λ}) = H0(λI −R) is closed. Therefore R has property (Q).The proof in the case where R,S ∈ Φ−(X) is analogous.Corollary 2.9. Let S,R ∈ L(X) satisfy the operator equations (1). Ifone of the operators R, S, RS and SR is bounded below and has property(Q), then all of them have property (Q).Proof. Note that all the operators R, S, RS, and SR are injective whenone of them is injective [8, Lemma 2.3], and the same is true for being uppersemi-Fredholm [8, Theorem 2.5]. Hence, if one of the operators is boundedbelow, then all of them are bounded below.By Theorem 2.8 property (Q) for R and for SR are equivalent. So the sameis true for S and RS, and also for RS and SR since R and S are injective.The analytical core K(T ) of T ∈ L(X) is defined [1, Definition 1.20] as theset of all λ ∈ C for which there exists a constant δ > 0 and a sequence (un)in X such that x = u0, and Tun+1 = un and ∥un∥ ≤ δn∥x∥ for each n ∈ N.The following characterization can be found in [1, Theorem 2.18]:K(T ) = XT (C \ {0}) = {x ∈ X : 0 /∈ σT (x)}.local spectral theory for operators R and S 45The analytical core of T is an invariant subspace and, in general, is not closed.Theorem 2.10. Suppose that R,S ∈ L(X) satisfy RSR = R2.(1) If 0 ̸= λ ∈ C, then K(λI−R) is closed if and only K(λI−SR) is closed,or equivalently K(λI −RS) is closed.(2) If R is injective, then K(R) is closed if and only K(SR) is closed, orequivalently K(RS) is closed.Proof. (1) Suppose λ ̸= 0 and K(λI − R) closed. Let (xn) be a sequenceof K(λI − SR) which converges to x ∈ X. Then λ /∈ σSR(xn) and hence,by Lemma 2.1, λ /∈ σR(Rxn), thus Rxn ∈ K(λI − R). Since Rxn → Rx andK(λI − R) is closed, it then follows that Rx ∈ K(λI − R), i.e., λ /∈ σR(Rx).Since λ ̸= 0, by Lemma 1.1 we have λ /∈ σR(x), hence λ /∈ σSR(SRx) againby Lemma 2.1. By Lemma 1.1 this implies λ /∈ σSR(x). Therefore x ∈K(λI − SR), and consequently, K(λI − SR) is closed.Conversely, suppose that λ ̸= 0 and K(λI − SR) is closed. Let (xn) bea sequence of K(λI − R) which converges to x ∈ X. Then λ /∈ σR(xn)and, by Lemma 2.1, we have λ /∈ σSR(SRxn). By Lemma 1.1 then we haveλ /∈ σSR(xn), so xn ∈ K(λI − SR), and hence x ∈ K(λI − SR), since the lastset is closed. This implies that λ /∈ σSR(x), and hence λ /∈ σR(Rx), again byLemma 2.1. By Lemma 1.1 we have λ /∈ σR(x), so x ∈ K(λI−R). Therefore,K(λI − R) is closed. The equivalence K(λI − SR) is closed if and only ifK(λI −RS) is closed was proved in [13, Corollary 3.3].(2) The proof is analogous to that of part (1) applying Lemma 1.1.Corollary 2.11. Suppose RSR = R2, SRS = S2 and λ ̸= 0. Then thefollowing statements are equivalent:(1) K(λI −R) is closed;(2) K(λI − SR) is closed;(3) K(λI −RS) is closed;(4) K(λI − S) is closed.When R is injective, the equivalence also holds for λ = 0.Proof. The equivalence of (3) and (4) follows from Theorem 2.10, inter-changing R and S. Since, as noted in the proof of Corollary 2.9, the injectivityof R is equivalent to the injectivity of S, the equivalence of (1) and (4) alsoholds for λ = 0.46 p. aiena, m. gonzálezReferences[1] P. Aiena, “Fredholm and Local Spectral Theory, with Application to Multi-pliers”, Kluwer Acad. Publishers, Dordrecht, 2004.[2] P. Aiena, M. González, On the Dunford property (C) for bounded linearoperators RS and SR, Integral Equations Operator Theory 70 (4) (2011),561 – 568.[3] P. Aiena, M. Chō, M. González, Polaroid type operators under quasi-affinities, J. Math. Anal. Appl. 371 (2) (2010), 485 – 495.[4] P. Aiena, M.L. Colasante, M. González, Operators which have aclosed quasi-nilpotent part, Proc. Amer. Math. Soc. 130 (9) (2002), 2701 –2710.[5] B. Barnes, Common operator properties of the linear operators RS and SR,Proc. Amer. Math. Soc. 126 (4) (1998), 1055 – 1061.[6] C. Benhida, E.H. Zerouali, Local spectral theory of linear operators RSand SR Integral Equations Operator Theory 54 (1) (2006), 1 – 8.[7] T. Bermúdez, M. González, A. Martinón, Stability of the local spec-trum Proc. Amer. Math. Soc. 125 (2) (1997), 417 – 425.[8] B.P. Duggal, Operator equations ABA = A2 and BAB = B2, Funct. Anal.Approx. Comput. 3 (1) (2011), 9 – 18.[9] K.B. Laursen, M.M. Neumann, “An Introduction to Local SpectralTheory”, London Mathematical Society Monographs, New Series, 20, TheClarendon Press, New York, 2000.[10] C. Schmoeger, On the operator equations ABA = A2 and BAB = B2, Publ.Inst. Math, (Beograd) (N.S.) 78(92) (2005), 127 – 133.[11] C. Schmoeger, Common spectral properties of linear operators A and Bsuch that ABA = A2 and BAB = B2, Publ. Inst. Math. (Beograd) (N.S.)79(93) (2006), 109 – 114.[12] I. Vidav, On idempotent operators in a Hilbert space, Publ. Inst. Math.(Beograd) (N.S.) 4(18), (1964), 157 – 163.[13] Q. Zeng, H. Zhong, Common properties of bounded linear operators ACand BA: Local spectral theory, J. Math. Anal. Appl. 414 (2) (2014), 553 –560.

English

Local spectral theory for operators R and S satisfying RSR = R^2

Institutional Repository University of Extremadura

E extracta mathematicae Vol. 31, Num. 1, 37 { 46 (2016)
Local Spectral Theory for Operators R and S
Satisfying RSR = R2
Pietro Aiena, Manuel Gonzalez
Dipartimento di Metodi e Modelli Matematici, Facolta di Ingegneria, Universita di
Palermo (Italia), paiena@unipa.it
Departamento de Matematicas, Facultad de Ciencias, Universidad de Cantabria,
E-39071 Santander (Spain), manuel.gonzalez@unican.es
Received February 5, 2016
Abstract : We study some local spectral properties for bounded operators R, S, RS and SR
in the case that R and S satisfy the operator equation RSR = R2. Among other results, we
prove that S, R, SR and RS share Dunford's property (C) when RSR = R2 and SRS = S2.
Key words: Local spectral subspace, Dunford's property (C), operator equation.
AMS Subject Class. (2010): 47A10, 47A11, 47A53, 47A55.
1. Introduction and preliminaries
The equivalence of Dunford's property (C) for products RS and SR of
operators R 2 L(Y;X) and S 2 L(X;Y ), X and Y Banach spaces, has been
studied in [2]. As noted in [13] the proof of Theorem 2.5 in [2] contains a gap
which was lled up in [13, Theorem 2.7]. In [2] it was also studied property
(C) for operators R;S 2 L(X) which satisfy the operator equations
RSR = R2 and SRS = S2: (1)
A similar gap exists in the proof of Theorem 3.3 in [2], which states the
equivalence of property (C) for R, S, RS and SR, when R, S satisfy (1).
In this paper we give a correct proof of this result and we prove further
results concerning the local spectral theory of R, S, RS and SR, in particular
we show several results concerning the quasi-nilpotent parts and the analytic
cores of these operators. It should be noted that these results are established
in a more general framework, assuming that only one of the operator equations
in (1) holds.
Supported in part by MICINN (Spain), Grant MTM2013-45643.
37
38 p. aiena, m. gonzalez
We shall denote by X a complex innite dimensional Banach space. Given
a bounded linear operator T 2 L(X), the local resolvent set of T at a point
x 2 X is dened as the union of all open subsets U of C such that there exists
an analytic function f : U ! X satisfying
(I   T )f() = x for all  2 U : (2)
The local spectrum T (x) of T at x is the set dened by T (x) := C n T (x).
Obviously, T (x)  (T ), where (T ) denotes the spectrum of T .
The following result shows that T (Tx) and T (x) may dier only at 0. It
was proved in [7] for operators satisfying the SVEP.
Lemma 1.1. For every T 2 L(X) and x 2 X we have
T (Tx)  T (x)  T (Tx) [ f0g: (3)
Moreover, if T is injective then T (Tx) = T (x) for all x 2 X.
Proof. Take S = T and R = I in [6, Proposition 3.1 ].
For every subset F of C, the local spectral subspace of T at F is the set
XT (F) := fx 2 X : T (x)  Fg:
It is easily seen from the denition thatXT (F) is a linear subspace T -invariant
of X. Furthermore, for every closed F  C we have
(I   T )XT (F) = XT (F) for all  2 C n F : (4)
See [9, Proposition 1.2.16].
An operator T 2 L(X) is said to have the single valued extension property
at o 2 C (abbreviated SVEP at o), if for every open disc Do centered at
o the only analytic function f : Do ! X which satises the equation
(I   T )f() = 0 (5)
is the function f  0. An operator T 2 L(X) is said to have the SVEP if T
has the SVEP at every point  2 C. Clearly, the SVEP is inherited by the
restrictions to invariant subspaces.
A variant of XT (F) which is more useful for operators without SVEP is
the glocal spectral subspace XT (F). For an operator T 2 L(X) and a closed
local spectral theory for operators R and S 39
subset F of C, we dene XT (F) as the set of all x 2 X for which there exists
an analytic function f : C n F ! X which satises
(I   T )f() = x for all  2 C n F :
Clearly XT (F)  XT (F) for every closed F  C. Moreover T has SVEP if
and only if
XT (F) = XT (F) for all closed subsets F  C:
See [9, Proposition 3.3.2]. Note that XT (F) and XT (F) are not closed in
general.
Given a closed subspace Z of X and T 2 L(X), we denote by T jZ the
restriction of T to Z.
Lemma 1.2. [2, Lemmas 2.3 and 2.4] Let F be a closed subset of C and
T 2 L(X).
(1) If 0 2 F and Tx 2 XT (F) then x 2 XT (F).
(2) Suppose T has SVEP, Z := XT (F) is closed, and A := T jXT (F). Then
XT (K) = ZA(K) for all closed K  F .
Lemma 1.3. Suppose that T has SVEP and F is a closed subset of C
such that 0 =2 F . If XT (F [ f0g) is closed then XT (F) is closed.
Proof. Set Z := XT (F [f0g) and S := T jZ. By [9, Proposition 1.2.20] we
have (S)  F [f0g. We suppose rst that 0 =2 (S). Then (S)  F , hence
Z = ZS(F). By Lemma 1.2 we have ZS(F) = XT (F), so XT (F) is closed.
For the case 0 2 (S), we set F0 := (S) \ F . Then (S) = F0 [ f0g. Since
0 2 (S), by Lemma 1.2 we have Z = ZS(F0) ZS(f0g) and
ZS(F0) = ZS((S) \ F) = ZS(F) = XT (F);
hence XT (F) is closed.
2. Operator equation RSR = R2
Operators S;R 2 L(X) satisfying the operator equations RSR = R2 and
SRS = S2 were studied rst in [12], and more recently in [10], [11], [8], and
other papers. An easy example of operators for which these equations hold is
40 p. aiena, m. gonzalez
given in the case that R = PQ and S = QP , where P;Q 2 L(X) are idem-
potents. A remarkable result of Vidav [12, Theorem 2] shows that if R;S are
self-ajoint operators on a Hilbert space then the equations (1) hold if and only
if there exists an (uniquely determined) idempotent P such that R = PP 
and S = P P , where P  is the adjoint of P .
The operators R, S, SR and RS for which the equations (1) hold share
many spectral properties ([10], [11]), and local spectral properties as decom-
posability, property () and SVEP ([8]). In this section we consider the per-
manence of property (C), property (Q) in this context.
It is easily seen that if 0 =2 (R) \ (S) then R = S = I, so this case is
trivial. Thus we shall assume that 0 2 (R) \ (S). Evidently, the operator
equation RSR = R2 implies
(SR)2 = SR2 and (RS)2 = R2S:
Lemma 2.1. Suppose that R;S 2 L(X) satisfy RSR = R2. Then for
every x 2 X we have
R(Rx)  SR(x) and SR(SRx)  R(x): (6)
Proof. For the rst inclusion, suppose that 0 =2 SR(x). Then there exists
an open neighborhood U0 of 0 and an analytic function f : U0 ! X such
that
(I   SR)f() = x for all  2 U0:
From this it follows that
Rx = R(I   SR)f() = (R RSR)f()
= (R R2)f() = (I  R)(Rf)();
for all  2 U0: Since Rf : U0 ! X is analytic we get 0 =2 R(Rx).
For the second inclusion, let 0 =2 R(x). Then there exists an open
neighborhood U0 of 0 and an analytic function f : U0 ! X such that
(I  R)f() = x for all  2 U0:
Consequently,
SRx = SR(I  R)f() = (SR  SR2)f()
= (SR  (SR)2)f() = (I   SR)(SRf)();
for all  2 U0, and since (SR)f is analytic we obtain 0 =2 SR(SRx).
local spectral theory for operators R and S 41
Theorem 2.2. Let S;R 2 L(X) satisfy RSR = R2, and let F be a closed
subset of C with 0 2 F . Then XR(F) is closed if and only if so is XSR(F).
Proof. Suppose thatXR(F) is closed and let (xn) be a sequence ofXSR(F)
which converges to x 2 X. We need to show that x 2 XSR(F). For every
n 2 N we have SR(xn)  F and hence, by Lemma 2.1, we have R(Rxn)  F ,
i.e. Rxn 2 XR(F). Since 0 2 F , by Lemma 1.2 we have xn 2 XR(F), and
since XR(F) is closed, x 2 XR(F), i.e. R(x)  F . Now from Lemma 2.1 we
derive SR(SRx)  F , and this implies SRx 2 XSR(F). Again by Lemma 1.2,
we obtain x 2 XSR(F), thus XSR(F) is closed.
Conversely, suppose that XSR(F) is closed and let (xn) be a sequence
of XR(F) which converges to x 2 X. Then R(xn)  F for every n 2 N,
hence SR(SRxn)  F , i.e. SRxn 2 XSR(F) by Lemma 2.1. But 0 2 F , so,
by Lemma 1.2, xn 2 XSR(F). Since XSR(F) is closed, x 2 XSR(F), hence
SR(x)  F . Now from Lemma 2.1 we obtain R(Rx)  F , i.e. Rx 2 XR(F),
and the condition 0 2 F implies x 2 XR(F).
The following result is inspired by [8, Theorem 2.1].
Lemma 2.3. Let S;R 2 L(X) be such that RSR = R2 and one of the
operators R, SR, RS has SVEP. Then all of them have SVEP. Additionally,
if SRS = S2 and one of R, S, SR, RS has SVEP then all of them have SVEP.
Proof. By [6, Proposition 2.1], SR has SVEP if and only if RS has SVEP.
So it is enough to prove that R has SVEP at 0 if an only if so has RS.
Suppose that R has SVEP at 0 and let f : U0 ! X be an analytic function
on an open neighborhood U0 of 0 for which (I  RS)f()  0 on U0. Then
RSf() = f() and
0 = RS(I  RS)f() = (RS   (RS)2)f() = (RS   (R2S)f()
= (I  R)RSf():
The SVEP of R at 0 implies that
RSf() = f() = 0 for all  2 U0:
Hence f  0 on U0, and we conclude that RS has SVEP at 0.
Conversely, suppose that RS has SVEP at 0 and let f : U0 ! X be an
42 p. aiena, m. gonzalez
analytic function on an open neighborhood U0 of 0 such that (I R)f()  0
on U0. Then R
2f() = Rf() = 2f() for all  2 U0. Moreover,
0 = RS(I  R)f() = RSf() R2f() = RSf()  2f()
= (I  RS)( f());
and since RS has SVEP at 0 we have f()  0, hence f()  0, so R has
SVEP at 0.
The second assertion is clear, if SRS = S2, just interchanging R and S in
the argument above, the SVEP fo S holds if and only if SR, or equivalently
RS, has SVEP.
We now consider the result of Theorem 2.2 when 0 =2 F .
Theorem 2.4. Let F be a closed subset of C such that 0 =2 F . Suppose
that R;S 2 L(X) satisfy RSR = R2 and R has SVEP. Then we have
(1) If XR(F [ f0g) is closed then XSR(F) is closed.
(2) If XSR(F [ f0g) is closed then XR(F) is closed.
Proof. (1) Let us denote F1 := F [ f0g. The set F1 is closed, and by
assumption XR(F1) is closed. Since 0 2 F1 then XSR(F1) is closed, by The-
orem 2.2. Moreover, the SVEP for R is equivalent to the SVEP for SR by
Lemma 2.3. Then XSR(F) is closed by Lemma 1.3.
(2) The argument is similar: if XSR(F [ f0g) is closed then XR(F [ f0g)
by Theorem 2.2, and since R has SVEP, XR(F) is closed by Lemma 1.3.
Definition 2.5. An operator T 2 L(X) is said to have Dunford's prop-
erty (C) (abbreviated property (C)) if XT (F ) is closed for every closed set
F  C.
It should be noted that Dunford property (C) implies SVEP.
Theorem 2.6. Suppose that S;R 2 L(X) satisfy RSR = R2, and any one
of the operators R, SR, RS, has property (C).Then all of them have property
(C). If, additionally, SRS = S2 and one of R, S, RS, SR has property (C),
then all of them have property (C).
Proof. Since property (C) implies SVEP, all the operators have SVEP by
Lemma 2.3. Moreover the equivalence of property (C) for SR and RS has
local spectral theory for operators R and S 43
been proved in [2] (see also [13]). So it is enough to prove that R has property
(C) if an only if so has RS.
Suppose that R has property (C) and let F be a closed set. If 0 2 F
then XSR(F) is closed, by Theorem 2.2, while in the case where 0 =2 F we
have that XR(F [ f0g) is closed, and hence, by part (1) of Theorem 2.4, the
SVEP for R ensures that also in this case XSR(F) is closed. Therefore, SR
has property (C).
Conversely, suppose that SR has property (C). For every closed subset F
containing 0, XR(F) is closed by Theorem 2.2. If 0 =2 F then XSR(F [ f0g)
is closed, hence XR(F) is closed by part (2) of Theorem 2.4 and we conclude
that R has property (C).
If additionally, SRS = S2 then, by interchanging S with R, the same
argument above proves the second assertion, so the proof is complete.
Next we consider the case when F is a singleton set, say F := fg.
The glocal spectral subspace XT (fg) coincides with the quasi-nilpotent part
H0(I   T ) of I   T dened by
H0(I   T ) := fx 2 X : lim sup
n!1
k(I   T )nxk1=n = 0g:
See [1, Theorem 2.20]. In general H0(I   T ) is not closed, but it coincides
with the kernel of a power of I   T in some cases [3, Theorem 2.2].
Definition 2.7. An operator T 2 L(X) is said to have the property (Q)
if H0(I   T ) is closed for every  2 C.
It is known that if H0(I   T ) is closed then T has SVEP at  ([4]), thus,
property (C)) property (Q)) SVEP:
Therefore, for operators T having property (Q) we haveH0(I T )=XT (fg).
In [13, Corollary 3.8] it was observed that if R 2 L(Y;X) and S 2 L(X;Y )
are both injective then RS has property (Q) precisely when SR has property
(Q).
Recall that T 2 L(X) is said to be upper semi-Fredholm, T 2 +(X), if
T (X) is closed and the kernel ker T is nite-dimensional, and T is said to be
lower semi-Fredholm, T 2  (X), if the range T (X) has nite codimension.
Theorem 2.8. Let R;S 2 L(X) satisfying RSR = R2, and R;S 2 +(X)
or R;S 2  (X). Then R has property (Q) if and only if so has SR.
44 p. aiena, m. gonzalez
Proof. Suppose that R;S 2 +(X) and R has property (Q). Then R
has SVEP and, by Lemma 2.3, also SR has SVEP. Consequently, the local
and glocal spectral subspaces relative to the a closed set coincide for R and
SR. By assumption H0(I   R) = XR(fg) is closed for every  2 C, and
H0(SR) = XSR(f0g) is closed by Theorem 2.2. Let 0 6=  2 C. By [9,
Proposition 3.3.1, part (f)]
XR(fg [ f0g) = XR(fg) +XR(f0g) = H0(I  R) +H0(R):
Since R 2 +(X) the SVEP at 0 implies that H0(R) is nite-dimensional, see
[1, Theorem 3.18 ], so XR(fg [ f0g) is closed. Then part (1) of Theorem 2.4
implies that H0(I   SR) = XSR(fg) is closed, hence SR has property (Q).
Conversely, suppose that SR has property (Q). If  = 0 then H0(SR) =
XRS(f0g) is closed by assumption, and H0(R) = XR(f0g) is closed by Theo-
rem 2.2. In the case  6= 0 we have
XSR(fg [ f0g) = XSR(fg) +XSR(f0g) = H0(I   SR) +H0(SR):
Since SR has SVEP and SR 2 +(X), H0(SR) is nite dimensional by [1,
Theorem 3.18]. So XSR(fg [ f0g) is closed. By part (2) of Theorem 2.4,
XR(fg) = H0(I  R) is closed. Therefore R has property (Q).
The proof in the case where R;S 2  (X) is analogous.
Corollary 2.9. Let S;R 2 L(X) satisfy the operator equations (1). If
one of the operators R, S, RS and SR is bounded below and has property
(Q), then all of them have property (Q).
Proof. Note that all the operators R, S, RS, and SR are injective when
one of them is injective [8, Lemma 2.3], and the same is true for being upper
semi-Fredholm [8, Theorem 2.5]. Hence, if one of the operators is bounded
below, then all of them are bounded below.
By Theorem 2.8 property (Q) for R and for SR are equivalent. So the same
is true for S and RS, and also for RS and SR since R and S are injective.
The analytical core K(T ) of T 2 L(X) is dened [1, Denition 1.20] as the
set of all  2 C for which there exists a constant  > 0 and a sequence (un)
in X such that x = u0, and Tun+1 = un and kunk  nkxk for each n 2 N.
The following characterization can be found in [1, Theorem 2.18]:
K(T ) = XT (C n f0g) = fx 2 X : 0 =2 T (x)g:
local spectral theory for operators R and S 45
The analytical core of T is an invariant subspace and, in general, is not closed.
Theorem 2.10. Suppose that R;S 2 L(X) satisfy RSR = R2.
(1) If 0 6=  2 C, then K(I R) is closed if and only K(I SR) is closed,
or equivalently K(I  RS) is closed.
(2) If R is injective, then K(R) is closed if and only K(SR) is closed, or
equivalently K(RS) is closed.
Proof. (1) Suppose  6= 0 and K(I   R) closed. Let (xn) be a sequence
of K(I   SR) which converges to x 2 X. Then  =2 SR(xn) and hence,
by Lemma 2.1,  =2 R(Rxn), thus Rxn 2 K(I   R): Since Rxn ! Rx and
K(I   R) is closed, it then follows that Rx 2 K(I   R), i.e.,  =2 R(Rx).
Since  6= 0, by Lemma 1.1 we have  =2 R(x), hence  =2 SR(SRx) again
by Lemma 2.1. By Lemma 1.1 this implies  =2 SR(x). Therefore x 2
K(I   SR), and consequently, K(I   SR) is closed.
Conversely, suppose that  6= 0 and K(I   SR) is closed. Let (xn) be
a sequence of K(I   R) which converges to x 2 X. Then  =2 R(xn)
and, by Lemma 2.1, we have  =2 SR(SRxn). By Lemma 1.1 then we have
 =2 SR(xn), so xn 2 K(I   SR), and hence x 2 K(I   SR), since the last
set is closed. This implies that  =2 SR(x), and hence  =2 R(Rx), again by
Lemma 2.1. By Lemma 1.1 we have  =2 R(x), so x 2 K(I R). Therefore,
K(I   R) is closed. The equivalence K(I   SR) is closed if and only if
K(I  RS) is closed was proved in [13, Corollary 3.3].
(2) The proof is analogous to that of part (1) applying Lemma 1.1.
Corollary 2.11. Suppose RSR = R2, SRS = S2 and  6= 0. Then the
following statements are equivalent:
(1) K(I  R) is closed;
(2) K(I   SR) is closed;
(3) K(I  RS) is closed;
(4) K(I   S) is closed.
When R is injective, the equivalence also holds for  = 0.
Proof. The equivalence of (3) and (4) follows from Theorem 2.10, inter-
changing R and S. Since, as noted in the proof of Corollary 2.9, the injectivity
of R is equivalent to the injectivity of S, the equivalence of (1) and (4) also
holds for  = 0.
46 p. aiena, m. gonzalez
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UCrea

E extracta mathematicae Vol. 31, Num. 1, 37 { 46 (2016)Local Spectral Theory for Operators R and SSatisfying RSR = R2Pietro Aiena, Manuel GonzalezDipartimento di Metodi e Modelli Matematici, Facolta di Ingegneria, Universita diPalermo (Italia), paiena@unipa.itDepartamento de Matematicas, Facultad de Ciencias, Universidad de Cantabria,E-39071 Santander (Spain), manuel.gonzalez@unican.esReceived February 5, 2016Abstract : We study some local spectral properties for bounded operators R, S, RS and SRin the case that R and S satisfy the operator equation RSR = R2. Among other results, weprove that S, R, SR and RS share Dunford's property (C) when RSR = R2 and SRS = S2.Key words: Local spectral subspace, Dunford's property (C), operator equation.AMS Subject Class. (2010): 47A10, 47A11, 47A53, 47A55.1. Introduction and preliminariesThe equivalence of Dunford's property (C) for products RS and SR ofoperators R 2 L(Y;X) and S 2 L(X;Y ), X and Y Banach spaces, has beenstudied in [2]. As noted in [13] the proof of Theorem 2.5 in [2] contains a gapwhich was lled up in [13, Theorem 2.7]. In [2] it was also studied property(C) for operators R;S 2 L(X) which satisfy the operator equationsRSR = R2 and SRS = S2: (1)A similar gap exists in the proof of Theorem 3.3 in [2], which states theequivalence of property (C) for R, S, RS and SR, when R, S satisfy (1).In this paper we give a correct proof of this result and we prove furtherresults concerning the local spectral theory of R, S, RS and SR, in particularwe show several results concerning the quasi-nilpotent parts and the analyticcores of these operators. It should be noted that these results are establishedin a more general framework, assuming that only one of the operator equationsin (1) holds.Supported in part by MICINN (Spain), Grant MTM2013-45643.3738 p. aiena, m. gonzalezWe shall denote by X a complex innite dimensional Banach space. Givena bounded linear operator T 2 L(X), the local resolvent set of T at a pointx 2 X is dened as the union of all open subsets U of C such that there existsan analytic function f : U ! X satisfying(I   T )f() = x for all  2 U : (2)The local spectrum T (x) of T at x is the set dened by T (x) := C n T (x).Obviously, T (x)  (T ), where (T ) denotes the spectrum of T .The following result shows that T (Tx) and T (x) may dier only at 0. Itwas proved in [7] for operators satisfying the SVEP.Lemma 1.1. For every T 2 L(X) and x 2 X we haveT (Tx)  T (x)  T (Tx) [ f0g: (3)Moreover, if T is injective then T (Tx) = T (x) for all x 2 X.Proof. Take S = T and R = I in [6, Proposition 3.1 ].For every subset F of C, the local spectral subspace of T at F is the setXT (F) := fx 2 X : T (x)  Fg:It is easily seen from the denition thatXT (F) is a linear subspace T -invariantof X. Furthermore, for every closed F  C we have(I   T )XT (F) = XT (F) for all  2 C n F : (4)See [9, Proposition 1.2.16].An operator T 2 L(X) is said to have the single valued extension propertyat o 2 C (abbreviated SVEP at o), if for every open disc Do centered ato the only analytic function f : Do ! X which satises the equation(I   T )f() = 0 (5)is the function f  0. An operator T 2 L(X) is said to have the SVEP if Thas the SVEP at every point  2 C. Clearly, the SVEP is inherited by therestrictions to invariant subspaces.A variant of XT (F) which is more useful for operators without SVEP isthe glocal spectral subspace XT (F). For an operator T 2 L(X) and a closedlocal spectral theory for operators R and S 39subset F of C, we dene XT (F) as the set of all x 2 X for which there existsan analytic function f : C n F ! X which satises(I   T )f() = x for all  2 C n F :Clearly XT (F)  XT (F) for every closed F  C. Moreover T has SVEP ifand only ifXT (F) = XT (F) for all closed subsets F  C:See [9, Proposition 3.3.2]. Note that XT (F) and XT (F) are not closed ingeneral.Given a closed subspace Z of X and T 2 L(X), we denote by T jZ therestriction of T to Z.Lemma 1.2. [2, Lemmas 2.3 and 2.4] Let F be a closed subset of C andT 2 L(X).(1) If 0 2 F and Tx 2 XT (F) then x 2 XT (F).(2) Suppose T has SVEP, Z := XT (F) is closed, and A := T jXT (F). ThenXT (K) = ZA(K) for all closed K  F .Lemma 1.3. Suppose that T has SVEP and F is a closed subset of Csuch that 0 =2 F . If XT (F [ f0g) is closed then XT (F) is closed.Proof. Set Z := XT (F [f0g) and S := T jZ. By [9, Proposition 1.2.20] wehave (S)  F [f0g. We suppose rst that 0 =2 (S). Then (S)  F , henceZ = ZS(F). By Lemma 1.2 we have ZS(F) = XT (F), so XT (F) is closed.For the case 0 2 (S), we set F0 := (S) \ F . Then (S) = F0 [ f0g. Since0 2 (S), by Lemma 1.2 we have Z = ZS(F0) ZS(f0g) andZS(F0) = ZS((S) \ F) = ZS(F) = XT (F);hence XT (F) is closed.2. Operator equation RSR = R2Operators S;R 2 L(X) satisfying the operator equations RSR = R2 andSRS = S2 were studied rst in [12], and more recently in [10], [11], [8], andother papers. An easy example of operators for which these equations hold is40 p. aiena, m. gonzalezgiven in the case that R = PQ and S = QP , where P;Q 2 L(X) are idem-potents. A remarkable result of Vidav [12, Theorem 2] shows that if R;S areself-ajoint operators on a Hilbert space then the equations (1) hold if and onlyif there exists an (uniquely determined) idempotent P such that R = PP and S = P P , where P  is the adjoint of P .The operators R, S, SR and RS for which the equations (1) hold sharemany spectral properties ([10], [11]), and local spectral properties as decom-posability, property () and SVEP ([8]). In this section we consider the per-manence of property (C), property (Q) in this context.It is easily seen that if 0 =2 (R) \ (S) then R = S = I, so this case istrivial. Thus we shall assume that 0 2 (R) \ (S). Evidently, the operatorequation RSR = R2 implies(SR)2 = SR2 and (RS)2 = R2S:Lemma 2.1. Suppose that R;S 2 L(X) satisfy RSR = R2. Then forevery x 2 X we haveR(Rx)  SR(x) and SR(SRx)  R(x): (6)Proof. For the rst inclusion, suppose that 0 =2 SR(x). Then there existsan open neighborhood U0 of 0 and an analytic function f : U0 ! X suchthat(I   SR)f() = x for all  2 U0:From this it follows thatRx = R(I   SR)f() = (R RSR)f()= (R R2)f() = (I  R)(Rf)();for all  2 U0: Since Rf : U0 ! X is analytic we get 0 =2 R(Rx).For the second inclusion, let 0 =2 R(x). Then there exists an openneighborhood U0 of 0 and an analytic function f : U0 ! X such that(I  R)f() = x for all  2 U0:Consequently,SRx = SR(I  R)f() = (SR  SR2)f()= (SR  (SR)2)f() = (I   SR)(SRf)();for all  2 U0, and since (SR)f is analytic we obtain 0 =2 SR(SRx).local spectral theory for operators R and S 41Theorem 2.2. Let S;R 2 L(X) satisfy RSR = R2, and let F be a closedsubset of C with 0 2 F . Then XR(F) is closed if and only if so is XSR(F).Proof. Suppose thatXR(F) is closed and let (xn) be a sequence ofXSR(F)which converges to x 2 X. We need to show that x 2 XSR(F). For everyn 2 N we have SR(xn)  F and hence, by Lemma 2.1, we have R(Rxn)  F ,i.e. Rxn 2 XR(F). Since 0 2 F , by Lemma 1.2 we have xn 2 XR(F), andsince XR(F) is closed, x 2 XR(F), i.e. R(x)  F . Now from Lemma 2.1 wederive SR(SRx)  F , and this implies SRx 2 XSR(F). Again by Lemma 1.2,we obtain x 2 XSR(F), thus XSR(F) is closed.Conversely, suppose that XSR(F) is closed and let (xn) be a sequenceof XR(F) which converges to x 2 X. Then R(xn)  F for every n 2 N,hence SR(SRxn)  F , i.e. SRxn 2 XSR(F) by Lemma 2.1. But 0 2 F , so,by Lemma 1.2, xn 2 XSR(F). Since XSR(F) is closed, x 2 XSR(F), henceSR(x)  F . Now from Lemma 2.1 we obtain R(Rx)  F , i.e. Rx 2 XR(F),and the condition 0 2 F implies x 2 XR(F).The following result is inspired by [8, Theorem 2.1].Lemma 2.3. Let S;R 2 L(X) be such that RSR = R2 and one of theoperators R, SR, RS has SVEP. Then all of them have SVEP. Additionally,if SRS = S2 and one of R, S, SR, RS has SVEP then all of them have SVEP.Proof. By [6, Proposition 2.1], SR has SVEP if and only if RS has SVEP.So it is enough to prove that R has SVEP at 0 if an only if so has RS.Suppose that R has SVEP at 0 and let f : U0 ! X be an analytic functionon an open neighborhood U0 of 0 for which (I  RS)f()  0 on U0. ThenRSf() = f() and0 = RS(I  RS)f() = (RS   (RS)2)f() = (RS   (R2S)f()= (I  R)RSf():The SVEP of R at 0 implies thatRSf() = f() = 0 for all  2 U0:Hence f  0 on U0, and we conclude that RS has SVEP at 0.Conversely, suppose that RS has SVEP at 0 and let f : U0 ! X be an42 p. aiena, m. gonzalezanalytic function on an open neighborhood U0 of 0 such that (I R)f()  0on U0. Then R2f() = Rf() = 2f() for all  2 U0. Moreover,0 = RS(I  R)f() = RSf() R2f() = RSf()  2f()= (I  RS)( f());and since RS has SVEP at 0 we have f()  0, hence f()  0, so R hasSVEP at 0.The second assertion is clear, if SRS = S2, just interchanging R and S inthe argument above, the SVEP fo S holds if and only if SR, or equivalentlyRS, has SVEP.We now consider the result of Theorem 2.2 when 0 =2 F .Theorem 2.4. Let F be a closed subset of C such that 0 =2 F . Supposethat R;S 2 L(X) satisfy RSR = R2 and R has SVEP. Then we have(1) If XR(F [ f0g) is closed then XSR(F) is closed.(2) If XSR(F [ f0g) is closed then XR(F) is closed.Proof. (1) Let us denote F1 := F [ f0g. The set F1 is closed, and byassumption XR(F1) is closed. Since 0 2 F1 then XSR(F1) is closed, by The-orem 2.2. Moreover, the SVEP for R is equivalent to the SVEP for SR byLemma 2.3. Then XSR(F) is closed by Lemma 1.3.(2) The argument is similar: if XSR(F [ f0g) is closed then XR(F [ f0g)by Theorem 2.2, and since R has SVEP, XR(F) is closed by Lemma 1.3.Definition 2.5. An operator T 2 L(X) is said to have Dunford's prop-erty (C) (abbreviated property (C)) if XT (F ) is closed for every closed setF  C.It should be noted that Dunford property (C) implies SVEP.Theorem 2.6. Suppose that S;R 2 L(X) satisfy RSR = R2, and any oneof the operators R, SR, RS, has property (C).Then all of them have property(C). If, additionally, SRS = S2 and one of R, S, RS, SR has property (C),then all of them have property (C).Proof. Since property (C) implies SVEP, all the operators have SVEP byLemma 2.3. Moreover the equivalence of property (C) for SR and RS haslocal spectral theory for operators R and S 43been proved in [2] (see also [13]). So it is enough to prove that R has property(C) if an only if so has RS.Suppose that R has property (C) and let F be a closed set. If 0 2 Fthen XSR(F) is closed, by Theorem 2.2, while in the case where 0 =2 F wehave that XR(F [ f0g) is closed, and hence, by part (1) of Theorem 2.4, theSVEP for R ensures that also in this case XSR(F) is closed. Therefore, SRhas property (C).Conversely, suppose that SR has property (C). For every closed subset Fcontaining 0, XR(F) is closed by Theorem 2.2. If 0 =2 F then XSR(F [ f0g)is closed, hence XR(F) is closed by part (2) of Theorem 2.4 and we concludethat R has property (C).If additionally, SRS = S2 then, by interchanging S with R, the sameargument above proves the second assertion, so the proof is complete.Next we consider the case when F is a singleton set, say F := fg.The glocal spectral subspace XT (fg) coincides with the quasi-nilpotent partH0(I   T ) of I   T dened byH0(I   T ) := fx 2 X : lim supn!1k(I   T )nxk1=n = 0g:See [1, Theorem 2.20]. In general H0(I   T ) is not closed, but it coincideswith the kernel of a power of I   T in some cases [3, Theorem 2.2].Definition 2.7. An operator T 2 L(X) is said to have the property (Q)if H0(I   T ) is closed for every  2 C.It is known that if H0(I   T ) is closed then T has SVEP at  ([4]), thus,property (C)) property (Q)) SVEP:Therefore, for operators T having property (Q) we haveH0(I T )=XT (fg).In [13, Corollary 3.8] it was observed that if R 2 L(Y;X) and S 2 L(X;Y )are both injective then RS has property (Q) precisely when SR has property(Q).Recall that T 2 L(X) is said to be upper semi-Fredholm, T 2 +(X), ifT (X) is closed and the kernel ker T is nite-dimensional, and T is said to belower semi-Fredholm, T 2  (X), if the range T (X) has nite codimension.Theorem 2.8. Let R;S 2 L(X) satisfying RSR = R2, and R;S 2 +(X)or R;S 2  (X). Then R has property (Q) if and only if so has SR.44 p. aiena, m. gonzalezProof. Suppose that R;S 2 +(X) and R has property (Q). Then Rhas SVEP and, by Lemma 2.3, also SR has SVEP. Consequently, the localand glocal spectral subspaces relative to the a closed set coincide for R andSR. By assumption H0(I   R) = XR(fg) is closed for every  2 C, andH0(SR) = XSR(f0g) is closed by Theorem 2.2. Let 0 6=  2 C. By [9,Proposition 3.3.1, part (f)]XR(fg [ f0g) = XR(fg) +XR(f0g) = H0(I  R) +H0(R):Since R 2 +(X) the SVEP at 0 implies that H0(R) is nite-dimensional, see[1, Theorem 3.18 ], so XR(fg [ f0g) is closed. Then part (1) of Theorem 2.4implies that H0(I   SR) = XSR(fg) is closed, hence SR has property (Q).Conversely, suppose that SR has property (Q). If  = 0 then H0(SR) =XRS(f0g) is closed by assumption, and H0(R) = XR(f0g) is closed by Theo-rem 2.2. In the case  6= 0 we haveXSR(fg [ f0g) = XSR(fg) +XSR(f0g) = H0(I   SR) +H0(SR):Since SR has SVEP and SR 2 +(X), H0(SR) is nite dimensional by [1,Theorem 3.18]. So XSR(fg [ f0g) is closed. By part (2) of Theorem 2.4,XR(fg) = H0(I  R) is closed. Therefore R has property (Q).The proof in the case where R;S 2  (X) is analogous.Corollary 2.9. Let S;R 2 L(X) satisfy the operator equations (1). Ifone of the operators R, S, RS and SR is bounded below and has property(Q), then all of them have property (Q).Proof. Note that all the operators R, S, RS, and SR are injective whenone of them is injective [8, Lemma 2.3], and the same is true for being uppersemi-Fredholm [8, Theorem 2.5]. Hence, if one of the operators is boundedbelow, then all of them are bounded below.By Theorem 2.8 property (Q) for R and for SR are equivalent. So the sameis true for S and RS, and also for RS and SR since R and S are injective.The analytical core K(T ) of T 2 L(X) is dened [1, Denition 1.20] as theset of all  2 C for which there exists a constant  > 0 and a sequence (un)in X such that x = u0, and Tun+1 = un and kunk  nkxk for each n 2 N.The following characterization can be found in [1, Theorem 2.18]:K(T ) = XT (C n f0g) = fx 2 X : 0 =2 T (x)g:local spectral theory for operators R and S 45The analytical core of T is an invariant subspace and, in general, is not closed.Theorem 2.10. Suppose that R;S 2 L(X) satisfy RSR = R2.(1) If 0 6=  2 C, then K(I R) is closed if and only K(I SR) is closed,or equivalently K(I  RS) is closed.(2) If R is injective, then K(R) is closed if and only K(SR) is closed, orequivalently K(RS) is closed.Proof. (1) Suppose  6= 0 and K(I   R) closed. Let (xn) be a sequenceof K(I   SR) which converges to x 2 X. Then  =2 SR(xn) and hence,by Lemma 2.1,  =2 R(Rxn), thus Rxn 2 K(I   R): Since Rxn ! Rx andK(I   R) is closed, it then follows that Rx 2 K(I   R), i.e.,  =2 R(Rx).Since  6= 0, by Lemma 1.1 we have  =2 R(x), hence  =2 SR(SRx) againby Lemma 2.1. By Lemma 1.1 this implies  =2 SR(x). Therefore x 2K(I   SR), and consequently, K(I   SR) is closed.Conversely, suppose that  6= 0 and K(I   SR) is closed. Let (xn) bea sequence of K(I   R) which converges to x 2 X. Then  =2 R(xn)and, by Lemma 2.1, we have  =2 SR(SRxn). By Lemma 1.1 then we have =2 SR(xn), so xn 2 K(I   SR), and hence x 2 K(I   SR), since the lastset is closed. This implies that  =2 SR(x), and hence  =2 R(Rx), again byLemma 2.1. By Lemma 1.1 we have  =2 R(x), so x 2 K(I R). Therefore,K(I   R) is closed. The equivalence K(I   SR) is closed if and only ifK(I  RS) is closed was proved in [13, Corollary 3.3].(2) The proof is analogous to that of part (1) applying Lemma 1.1.Corollary 2.11. Suppose RSR = R2, SRS = S2 and  6= 0. Then thefollowing statements are equivalent:(1) K(I  R) is closed;(2) K(I   SR) is closed;(3) K(I  RS) is closed;(4) K(I   S) is closed.When R is injective, the equivalence also holds for  = 0.Proof. The equivalence of (3) and (4) follows from Theorem 2.10, inter-changing R and S. Since, as noted in the proof of Corollary 2.9, the injectivityof R is equivalent to the injectivity of S, the equivalence of (1) and (4) alsoholds for  = 0.46 p. aiena, m. gonzalezReferences[1] P. Aiena, \Fredholm and Local Spectral Theory, with Application to Multi-pliers", Kluwer Acad. Publishers, Dordrecht, 2004.[2] P. Aiena, M. Gonzalez, On the Dunford property (C) for bounded linearoperators RS and SR, Integral Equations Operator Theory 70 (4) (2011),561 { 568.[3] P. Aiena, M. Cho, M. Gonzalez, Polaroid type operators under quasi-anities, J. Math. Anal. Appl. 371 (2) (2010), 485 { 495.[4] P. Aiena, M.L. Colasante, M. Gonzalez, Operators which have aclosed quasi-nilpotent part, Proc. Amer. Math. Soc. 130 (9) (2002), 2701 {2710.[5] B. Barnes, Common operator properties of the linear operators RS and SR,Proc. Amer. Math. Soc. 126 (4) (1998), 1055 { 1061.[6] C. Benhida, E.H. Zerouali, Local spectral theory of linear operators RSand SR Integral Equations Operator Theory 54 (1) (2006), 1 { 8.[7] T. Bermudez, M. Gonzalez, A. Martinon, Stability of the local spec-trum Proc. Amer. Math. Soc. 125 (2) (1997), 417 { 425.[8] B.P. Duggal, Operator equations ABA = A2 and BAB = B2, Funct. Anal.Approx. Comput. 3 (1) (2011), 9 { 18.[9] K.B. Laursen, M.M. Neumann, \An Introduction to Local SpectralTheory", London Mathematical Society Monographs, New Series, 20, TheClarendon Press, New York, 2000.[10] C. Schmoeger, On the operator equations ABA = A2 and BAB = B2, Publ.Inst. Math, (Beograd) (N.S.) 78(92) (2005), 127 { 133.[11] C. Schmoeger, Common spectral properties of linear operators A and Bsuch that ABA = A2 and BAB = B2, Publ. Inst. Math. (Beograd) (N.S.)79(93) (2006), 109 { 114.[12] I. Vidav, On idempotent operators in a Hilbert space, Publ. Inst. Math.(Beograd) (N.S.) 4(18), (1964), 157 { 163.[13] Q. Zeng, H. Zhong, Common properties of bounded linear operators ACand BA: Local spectral theory, J. Math. Anal. Appl. 414 (2) (2014), 553 {560.

Local spectral theory for operators R and S satisfying RSR =R2

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