E extracta mathematicae Vol. 31, Núm. 1, 37 – 46 (2016) Local Spectral Theory for Operators R and S Satisfying RSR = R2 Pietro Aiena, Manuel González Dipartimento di Metodi e Modelli Matematici, Facoltà di Ingegneria, Università di Palermo (Italia), paiena@unipa.it Departamento de Matemáticas, 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 ∈ L(Y, X) and S ∈ 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 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 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. gonzález We shall denote by X a complex infinite dimensional Banach space. Given a bounded linear operator T ∈ L(X), the local resolvent set of T at a point x ∈ X is defined as the union of all open subsets U of C such that there exists an 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. It was proved in [7] for operators satisfying the SVEP. Lemma 1.1. For every T ∈ L(X) and x ∈ X we have σT (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 set XT (F) := {x ∈ X : σT (x) ⊆ F}. It is easily seen from the definition that XT (F) is a linear subspace T-invariant of 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 property at λo ∈ C (abbreviated SVEP at λo), if for every open disc Dλo centered at λo 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 T has the SVEP at every point λ ∈ 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 ∈ L(X) and a closed local spectral theory for operators R and S 39 subset F of C, we define XT (F) as the set of all x ∈ X for which there exists an 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 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 ∈ L(X), we denote by T |Z 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 ∈ 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). 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 /∈ 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] we have σ(S) ⊆ F ∪ {0}. We suppose first that 0 /∈ σ(S). Then σ(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 ∈ σ(S), we set F0 := σ(S) ∩ F. Then σ(S) = F0 ∪ {0}. Since 0 ∈ σ(S), by Lemma 1.2 we have Z = ZS(F0) ⊕ ZS({0}) and ZS(F0) = ZS(σ(S) ∩ F) = ZS(F) = XT (F), hence XT (F) is closed. 2. Operator equation RSR = R2 Operators S, R ∈ L(X) satisfying the operator equations RSR = R2 and SRS = S2 were studied first 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. gonzález given 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 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 (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 is trivial. Thus we shall assume that 0 ∈ σ(R) ∩ σ(S). Evidently, the operator equation RSR = R2 implies (SR)2 = SR2 and (RS)2 = R2S. Lemma 2.1. Suppose that R, S ∈ L(X) satisfy RSR = R2. Then for every x ∈ X we have σR(Rx) ⊆ σSR(x) and σSR(SRx) ⊆ σR(x). (6) Proof. For the first inclusion, suppose that λ0 /∈ σSR(x). Then there exists an open neighborhood U0 of λ0 and an analytic function f : U0 → X such that (λI − SR)f(λ) = x for all λ ∈ U0. From this it follows that Rx = 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 open neighborhood 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 41 Theorem 2.2. Let S, R ∈ L(X) satisfy RSR = R2, and let F be a closed subset of C with 0 ∈ F. Then XR(F) is closed if and only if so is XSR(F). Proof. Suppose that XR(F) is closed and let (xn) be a sequence of XSR(F) which converges to x ∈ X. We need to show that x ∈ XSR(F). For every n ∈ 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), and since XR(F) is closed, x ∈ XR(F), i.e. σR(x) ⊆ F. Now from Lemma 2.1 we derive 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 sequence of 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), hence σSR(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 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 λ0 if an only if so has RS. Suppose that R has SVEP at λ0 and let f : U0 → X be an analytic function on an open neighborhood U0 of λ0 for which (λI − RS)f(λ) ≡ 0 on U0. Then RSf(λ) = λf(λ) and 0 = 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 that RSf(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 an 42 p. aiena, m. gonzález analytic function on an open neighborhood U0 of λ0 such that (λI−R)f(λ) ≡ 0 on U0. Then R 2f(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 has SVEP at λ0. 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 /∈ F. Theorem 2.4. Let F be a closed subset of C such that 0 /∈ F. Suppose that 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 by assumption 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 by Lemma 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 set F ⊆ 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 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 ∈ F then XSR(F) is closed, by Theorem 2.2, while in the case where 0 /∈ F we have that XR(F ∪ {0}) 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 /∈ F then XSR(F ∪ {0}) 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 := {λ}. The glocal spectral subspace XT ({λ}) coincides with the quasi-nilpotent part H0(λI − T) of λI − T defined by H0(λI − T) := {x ∈ X : lim sup n→∞ ∥(λI − T)nx∥1/n = 0}. See [1, Theorem 2.20]. In general H0(λI − T) is not closed, but it coincides with 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 have H0(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), if T(X) is closed and the kernel ker T is finite-dimensional, and T is said to be lower 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. gonzález Proof. Suppose that R, S ∈ Φ+(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(λI − R) = XR({λ}) is closed for every λ ∈ C, and H0(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 2.4 implies 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 have XSR({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). 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 ∈ L(X) is defined [1, Definition 1.20] as the set 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 45 The 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, or equivalently K(RS) is closed. Proof. (1) Suppose λ ̸= 0 and K(λI − R) closed. Let (xn) be a sequence of 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 and K(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) again by 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) be a 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 have λ /∈ σSR(xn), so xn ∈ K(λI − SR), and hence x ∈ K(λI − SR), since the last set is closed. This implies that λ /∈ σSR(x), and hence λ /∈ σR(Rx), again by Lemma 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 if K(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. 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