@ Applied General Topology c© Universidad Politécnica de Valencia Volume 4, No. 2, 2003 pp. 301–316 Bounded point evaluations for cyclic Hilbert space operators A. Bourhim ∗ Dedicated to Professor S. Naimpally on the occasion of his 70th birthday. Abstract. In this talk, to be given at a conference at Seconda Università degli Studi di Napoli in September 2001, we shall describe the set of analytic bounded point evaluations for an arbitrary cyclic bounded linear operator T on a Hilbert space H and shall answer some questions due to L. R. Williams. 2000 AMS Classification: Primary 47A10; Secondary 47B20. Keywords: cyclic operator, bounded point evaluation, single-valued extension property, Bishop’s property (β). 1. Introduction. Throughout this paper, L(H) will denote the algebra of all linear bounded operators on an infinite–dimensional separable complex Hilbert space H. Let T ∈ L(H) be a cyclic operator on H with cyclic vector x ∈ H i.e., the linear subspace {p(T)x : p polynomial} is dense in H. A complex number λ ∈ C is said to be a bounded point evaluation for T if there is a positive constant M such that for every polynomial p, |p(λ)| ≤ M‖p(T)x‖; equivalently, if λ induces a continuous linear functional on H which maps p(T)x to p(λ) for every polynomial p. Therefore, it follows from the Riesz Represen- tation Theorem that a complex number λ ∈ C is a bounded point evaluation for T if and only if there is a unique vector k(λ) ∈H such that (1.1) p(λ) = 〈p(T)x,k(λ)〉 for every polynomial p. The set of all bounded point evaluations for T will be denoted by B(T). A point λ ∈ B(T) is called an analytic bounded point evaluation for T if there is ∗This research is supported in part by the Abdus Salam ICTP, Trieste, Italy. 302 A. Bourhim an open neighborhood O of λ contained in B(T) such that for every y ∈H, the complex function ŷ defined on B(T) by ŷ(λ) := 〈y,k(λ)〉, is analytic on O. The set of all analytic bounded point evaluations for T will be denoted by Ba(T). An operator T ∈L(H) is said to be subnormal if it has a normal extension, i.e., if there is a normal operator N on a Hilbert space K, containing H, such that H is a closed invariant subspace of N and the restriction N|H coincides with T . The operator T is said to be hyponormal if ‖T∗x‖ ≤ ‖Tx‖ for every x ∈H, where T∗ denotes the adjoint of T . Note that every subnormal operator is hyponormal, with the converse false (see [8] and also Example 5.6). Using Bran’s theorem [5] and the maximum modulus principle for analytic functions, Tavan T. Trent proved in [20] that for every cyclic subnormal operator T ∈ L(H), we have Ba(T) = Γ(T)\σap(T). Here, σap(T) denotes the approximate point spectrum of T , that is the set of complex numbers λ for which there is a sequence (xn)n of elements of the unit sphere of H such that lim n→+∞ ‖(T − λ)xn‖ = 0, and Γ(T) denotes the compression spectrum of T , that is the set of complex numbers λ such that the range of (T − λ) is not dense in H. More informations about bounded point evaluations for cyclic subnormal operators can be found in [8]. In [21], L. R. Williams followed Trent’s method to shown that (1.2) Γ(T)\σap(T) ⊂ Ba(T) for every cyclic operator T ∈L(H) and posed the following question. Question 1.1. Let T ∈L(H) be a cyclic operator. Is Ba(T) = Γ(T)\σap(T)? Note that, in general, the basic spectral properties of subnormal operators remain valid for hyponormal operators. Thus we pose the following weaker question. Question 1.2. Is Ba(T) = Γ(T)\σap(T) for every cyclic hyponormal operator T ∈L(H)? In this paper, we shall explain more about bounded point evaluations for cyclic Hilbert space operators from the point of view of local spectral theory and shall answer the above questions. In Section 2, we give a complete description of the set of analytic bounded point evaluations for arbitrary cyclic operators and derive some consequences from it. In Section 3, we give a necessary and sufficient condition for unilateral weighted shift operators to satisfy Trent’s re- sult, and exhibit some operators which provide a negative answer to Question 1.1. In Section 4, we generalize a result of L. Yang which allows us to give a positive answer to Question 1.2. As a corollary, we get that two quasisimilar cyclic hyponormal operators have equal approximate point spectra; this result is a generalization of Theorem 4 of [16]. In Section 5, we show that if T ∈L(H) is a cyclic operator for which the span of the eigenvectors of T∗ associated with Bounded point evaluations for . . . 303 a connected component of Ba(T) is dense in H, then T is without eigenvalues and has a connected spectrum. Some related examples are given. Before going further, we need to introduce some notations and recall some basic notions concerning local spectral theory; we refer to the monographs [7] and [13] for further informations. For an operator T ∈ L(H), we denote as usual by σ(T) := {λ ∈ C : T −λ is not invertible}, ρ(T) := C\σ(T), σp(T) := {λ ∈ C : T − λ is not injective}, ker T , and ranT the spectrum, the resolvent set, the point spectrum, the kernel, and the range of T , respectively. For a subset M of H, we use cl(M), and ∨ M to denote the closure of M, and the closed linear subspace generated by M, respectively. For a subset F of C, we use F, and Fr(F) to denote the complex conjugates of the points in F, and the boundary of F, respectively. For an open subset U of C, we let O(U,H) denote the space of analytic H−valued functions on U. It is a Fréchet space when equipped with the topology of uniform convergence on compact subsets of U and the space H may be viewed as simply the constants in O(U,H). One says that an operator T ∈ L(H) possesses Bishop’s property (β) if for each open subset U of C, the multiplication operator TU : O(U,H) −→O(U,H), f 7−→ (T −z)f is injective with closed range. M. Putinar [14] has shown that hyponormal operators have Bishop’s property (β). Recall that T is said to have the single– valued extension property provided that, for every open subset U of C, the only analytic H−valued solution f of the equation (T −λ)f(λ) = 0, (λ ∈ U), is the identically zero function f ≡ 0 on U. Equivalently, if for every open subset U of C, the mapping TU is injective. A localized version of this property dates back to J. K. Finch [10] and can be defined as follows (see [1], and [10]). An operator T ∈ L(H) is said to have the single–valued extension property at a point λ ∈ C if for every open disk U centered at λ, the mapping TU is injective. Let <(T) := { λ ∈ C : T does not have the single-valued extension property at λ } . Obviously, <(T) is an open subset contained in σp(T) and is empty precisely when T has the single–valued extension property. The local resolvent set of T at a vector y ∈ H, denoted by ρ T (y), is the union of all open subsets U of C for which y ∈ ranTU . The local spectrum of T at y is σT (y) := C\ρT (y); it is a closed subset of σ(T). In the sequel, T ∈ L(H) will be a cyclic operator with cyclic vector x ∈ H; and for λ ∈ B(T), k(λ) will denote the vector of H as defined in (1.1). 2. Description of B(T) and Ba(T). The proofs of Proposition 2.1 and Lemma 2.2 are similar to the ones for the cyclic subnormal operators (see [2], and [8]); we include them for completeness. A complete description of B(T) is given by the following result. 304 A. Bourhim Proposition 2.1. Let λ ∈ C; the following statements are equivalent. (i) λ ∈ B(T). (ii) λ ∈ Γ(T). (iii) ker ( (T −λ)∗ ) is one dimensional. Proof. First, note that if (T − λ)∗u = 0 for some u ∈ H, then for every polynomial p, we have (2.3) 〈p(T)x,u〉 = p(λ)〈x,u〉. Next, we mention that it suffices to prove the implications (ii)⇒(i) and (i)⇒(iii), since the implication (iii)⇒(ii) can be deduced trivially from the fact that Γ(T) = σp(T∗). Let λ ∈ B(T); it is clear that k(λ) 6= 0 since 〈x,k(λ)〉 = 1. On the other hand, for every polynomial p, we have 0 = 〈(T −λ)p(T)x,k(λ)〉 = 〈p(T)x, (T −λ)∗k(λ)〉. Since x is a cyclic vector for T , we have (T −λ)∗k(λ) = 0. Hence, λ ∈ σp(T∗). Now, let u ∈H be such that (T −λ)∗u = 0. It follows then from equation (2.3) that u = 〈x , u〉k(λ). Therefore, (i)⇒(iii). Let λ ∈ Γ(T). Then there is a non-zero vector u ∈H such that (T−λ)∗u = 0. Since x is a cyclic vector of T , it follows from equation (2.3) that 〈x,u〉 6= 0. Hence, p(λ) = 〈p(T)x, u 〈x,u〉 〉 for every polynomial p. Therefore, λ ∈ B(T) and k(λ) = u 〈x,u〉 , which proves that (ii)⇒(i). � Lemma 2.2. An open subset O of C is contained in Ba(T) if and only if it is contained in B(T) and the function λ 7−→ ‖k(λ)‖ is bounded on compact subsets of O. Proof. Assume that O ⊂ Ba(T) and let K be a compact subset of O. For every y ∈H, the function ŷ is analytic on O; in particular, sup λ∈K |〈y,k(λ)〉| < +∞. So it follows from the Uniform Boundedness Principle that sup λ∈K ‖k(λ)‖ < +∞. Conversely, suppose that O ⊂ B(T) and the function λ 7−→ ‖k(λ)‖ is bounded on compact subsets of O. Let y ∈ H; then there is a sequence of polynomials (pn)n such that lim n→+∞ ‖pn(T)x−y‖ = 0. And so, for every com- pact subset K of O, it follows from the Cauchy-Schwartz inequality that, sup λ∈K |pn(λ) − ŷ(λ)| ≤ sup λ∈K ‖k(λ)‖‖pn(T)x−y‖. Hence, the function ŷ is analytic on O. Therefore, O ⊂ Ba(T). � The following gives a complete description of Ba(T). Bounded point evaluations for . . . 305 Theorem 2.3. Ba(T) = <(T∗). Proof. Suppose that λ ∈ <(T∗). Then there is a non-zero analytic H–valued function φ on some open disk V centered at λ such that (T −µ)φ(µ) = 0 for all µ ∈V. Using the fact that a non-zero analytic H–valued function has isolated zeros, one can assume that the function φ has no zero in V. Hence, V ⊂ σp(T∗) = B(T). As before, it follows from (2.3) that k(µ) = φ(µ) 〈x,φ(µ)〉 for every µ ∈V. Therefore, the function k : V → H is continuous; in particular, the function µ 7−→‖k(µ)‖ is bounded on compact subsets of V. By Lemma 2.2, V ⊂ Ba(T). Thus <(T∗) ⊂ Ba(T). Conversely, set O = Ba(T) and consider the following H–valued function φ defined on O by φ(λ) := k(λ), λ ∈ O. First, we show that φ is analytic on O. Indeed, for every y ∈H and for every λ0 ∈ O, we have lim λ→λ0 〈φ(λ),y〉−〈φ(λ0),y〉 λ−λ0 = lim λ→λ0 〈k(λ),y〉−〈k(λ0),y〉 λ−λ0 = lim λ→λ0 ŷ(λ) − ŷ(λ0) λ−λ0 = ŷ′(λ0). Hence, for every y ∈ H, the function λ 7−→ 〈φ(λ),y〉 is differentiable on O; therefore, φ is analytic on O. On the other hand, φ has no zeros on O and satisfies the following equation (T∗ −λ)φ(λ) = 0 for every λ ∈ O. This gives O = Ba(T) ⊂<(T∗), and the proof is complete. � Corollary 2.4. The following holds: Ba(T) = {λ ∈ Γ(T) : σ T∗−λ (k(λ)) = ∅} = {λ ∈ Γ(T) : σ T∗ (k(λ)) = ∅}. Proof. Since, for every λ ∈ Ba(T), λ is a simple eigenvalue of T∗ with corre- sponding eigenvector k(λ), the proof follows by combining Theorem 2.3 and Theorem 1.9 of [1]. � Remark 2.5. (i) Note that Proposition 2.1 and Proposition 2.3 show that both B(T) and Ba(T) are independent of the choice of cyclic vector for T (see Proposition 1.4 of [21]). (ii) Using Theorem 2.3 and Theorem 2.6 of [1], one can easily prove (1.2). 306 A. Bourhim 3. Resolution of question 1.1. The weighted shift operators have proven to be an interesting rich collec- tion of operators providing examples and counterexamples to illustrate many properties of operators. The Allen Shields’s excellent survey [18] contains their basic facts and properties concerning their spectral theory (see also [2]). Throughout this section, S will denote a unilateral weighted shift operator on H with a positive bounded weight sequence (ωn)n≥0, that is Sen = ωnen+1, n ≥ 0, where (en)n≥0 is an orthonormal basis of H. Let (βn)n≥0 be the following sequence given by: βn =   ω0...ωn−1 if n > 0 1 if n = 0 Set r1(S) = lim n→∞ [ inf k≥0 βn+k βk ] 1 n , r2(S) = lim inf n→∞ [ βn ] 1 n and r(S) = lim n→∞ [ sup k≥0 βn+k βk ] 1 n ; and note that, r1(S) ≤ r2(S) ≤ r(S) ≤‖S‖. The following gives a necessary and sufficient condition for the weighted shift S to answer affirmatively Question 1.1. Theorem 3.1. The following are equivalent. (i) Ba(S) = Γ(S)\σap(S). (ii) r1(S) = r2(S). Proof. Since σp(S∗) ⊂{λ ∈ C : |λ| ≤ r2(S)} (see Theorem 9 of [18]), we have <(S∗) ⊂ O := {λ ∈ C : |λ| < r2(S)}. Conversely, consider the following non-zero analytic H–valued function φ de- fined on O by φ(λ) := +∞∑ n=0 λn βn en. It is easy to see that (S∗ −λ)φ(λ) = 0 for every λ ∈ O. Hence, O = {λ ∈ C : |λ| < r2(S)}⊂<(S∗). Therefore, <(S∗) = {λ ∈ C : |λ| < r2(S)}; by Theorem 2.3, (3.4) Ba(S) = {λ ∈ C : |λ| < r2(S)}. On the other hand, by [18, Theorems 4 and 6], the spectrum and the approxi- mate point spectrum of S are given, respectively, by σ(S) = {λ ∈ C : |λ| ≤ r(S)} and σap(S) = {λ ∈ C : r1(S) ≤ |λ| ≤ r(S)}. Bounded point evaluations for . . . 307 Hence, (3.5) Γ(S)\σap(S) = σ(S)\σap(S) = {λ ∈ C : |λ| < r1(S)}. And so the proof follows from (3.4) and (3.5). � Now, to give a counterexample to Question 1.1, we only need to produce a weight sequence (ωn)n≥0 for which the corresponding weighted shift S satisfies r1(S) < r2(S). Example 3.2. Let (Ck)k≥0 be the sequence of successive disjoint segments covering the set of non-negative integers N such that each segment Ck contains k2 + k elements. Let R > 1 be a real number and let k ∈ N; for every n ∈ Ck we set, ωn =   R if n lies in the first k2 terms of Ck 1 otherwise Hence, the unilateral weighted shift S corresponding to the weight (ωn)n≥0 is bounded and satisfies ‖S‖ = R and r1(S) = 1. On the other hand for every n ≥ 3, there is k(n) ≥ 2 such that n ∈ Ck(n). Hence, R k(n)−1∑ s=1 s2 ≤ βn and n ≤ k(n)∑ s=1 (s2 + s). Therefore, R (2k(n)−1)(k(n)−1) 2(k(n)+1)(k(n)+2) ≤ [ βn ] 1 n . Since lim n→+∞ k(n) = +∞, it follows that, R ≤ lim inf n→+∞ [ βn ] 1 n . We deduce that r1(S) = 1 and r2(S) = ‖S‖ = R. Thus, Γ(S)\σap(S) = {λ ∈ C : |λ| < 1}$ Ba(S) = {λ ∈ C : |λ| < R}. The original idea of this construction is due to W. C. Ridge [17]. For other example see [3], where the authors constructed a unilateral weighted shift S for which Γ(S)\σap(S) = ∅ and Ba(S) = {λ ∈ C : |λ| < 1}. 4. Resolution of question 1.2. If T possesses Bishop’s property (β), then we obtain the following. Theorem 4.1. If T possesses Bishop’s property (β), then the following are equivalent. (i) Ba(T) = Γ(T)\σap(T). (ii) Ba(T) ∩σp(T) = ∅. 308 A. Bourhim Proof. If Ba(T) = Γ(T)\σap(T) then it is clear that Ba(T) ∩ σp(T) = ∅ since σp(T) ⊂ σap(T). Conversely, suppose that Ba(T) ∩ σp(T) = ∅. Since Γ(T)\σap(T) ⊂ Ba(T) ⊂ Γ(T), it suffices to prove that Ba(T) ∩ σap(T) = ∅. Suppose that there is λ ∈ Ba(T) ∩σap(T). It then follows that ran(T −λ) is not closed. Let y ∈ cl ( ran(T −λ) ) \ran(T −λ); then y 6∈ ran(T −λ) and 〈y,k(λ)〉 = 0. Therefore, there is a sequence of polynomials (pn)n vanishing at λ such that lim n→+∞ ‖pn(T)x−y‖ = 0. Define on U := Ba(T) the following analytic H–valued functions f and fn by f(µ) = y− ŷ(µ)x and fn(µ) = pn(T)x−pn(µ)x, n ≥ 0. Since f(λ) = y 6∈ ran(T −λ), then f 6∈ ran(TU ). On the other hand, it is easy to see that fn ∈ ran(TU ) for every n ≥ 0. Now, let K be a compact subset of U; we have sup µ∈K ‖fn(µ) −f(µ)‖ ≤ ‖pn(T)x−y‖ + sup µ∈K ‖ [ pn(µ) − ŷ(µ) ] x‖ ≤ ‖pn(T)x−y‖ + ‖x‖ sup µ∈K |pn(µ) − ŷ(µ)| ≤ [ 1 + ‖x‖ sup µ∈K ‖k(µ)‖ ] ‖pn(T)x−y‖. Therefore, fn −→ f in O(U,H). Thus ran(TU ) is not closed which contradicts the fact that T possess Bishop’s property (β). The proof is complete. � The following is immediate (see Theorem 3.1 of [23]). Corollary 4.2. Suppose that T possesses Bishop’s property (β). If σp(T) = ∅, then Ba(T) = Γ(T)\σap(T). The following gives a positive answer to Question 1.2. Theorem 4.3. If T is hyponormal, then Ba(T) = Γ(T)\σap(T). Proof. Since T is hyponormal, then for every λ ∈ C, we have T −λ is hyponor- mal i.e., (4.6) ‖(T −λ)∗y‖≤‖(T −λ)y‖, ∀y ∈H. Now, let λ ∈ Ba(T) and suppose that there is y ∈ H such that Ty = λy. For every µ ∈ Ba(T), we have, λŷ(µ) = 〈Ty , k(µ)〉 = 〈y , T∗k(µ)〉 = 〈y , µk(µ)〉 = µŷ(µ). Hence, the analytic function ŷ is identically zero on Ba(T). On the other hand, it follows from (4.6) and Proposition 2.1 that y = αk(λ) for some α ∈ C. Therefore, ŷ(λ) = α‖k(λ)‖2 = 0 gives y = 0. Thus, Ba(T) ∩ σp(T) = ∅. By Theorem 4.1, the proof is complete. � Bounded point evaluations for . . . 309 In [16], M. Raphael has shown that two quasisimilar cyclic subnormal op- erators have equal approximate point spectra. The following generalizes M. Raphael’s result to cyclic hyponormal operators. Suppose that H1 and H2 are Hilbert spaces. Recall that two operators T1 ∈ L(H1) and T2 ∈ L(H2) are said to be quasisimilar if there exist two bounded linear transformations X : H1 → H2 and Y : H2 → H1 injectives and having dense range such that XT1 = T2X and T1Y = Y T2. Corollary 4.4. Suppose that H1 and H2 are Hilbert spaces, and let T1 ∈L(H1) and T2 ∈ L(H2). If T1 and T2 are quasisimilar cyclic hyponormal operators, then σap(T1) = σap(T2). Proof. In view of Theorem 1.5 of [21] and Theorem 4.3, we have Γ(T1)\σap(T1) = Γ(T2)\σap(T2). Hence, σ(T1)\σap(T1) = σ(T2)\σap(T2). Since σ(T1) = σ(T2) (see Theorem 2 of [6], and also Corollary 2.2 of [23]), we deduce that σap(T1) = σap(T2). � Remark 4.5. Theorem 4.3 and Corollary 4.4 can be extended with no extra effort to the class of operators satisfying the following. (i) T possesses Bishop’s property (β). (ii) ker(T −λ) ⊂ ker(T −λ)∗ for every λ ∈ C. Immediate other examples of operators satisfying (i) and (ii) are provided by M–hyponormal and p–hyponormal operators (see [14] and [9]). 5. examples and comments. In this section, we start by proving the following result that we will need throughout. Proposition 5.1. If H = ∨ {k(λ) : λ ∈ Ba(T)}, then σp(T) = ∅. Moreover, if H = ∨ {k(λ) : λ ∈ G} for some connected component G of Ba(T), then the following hold. (i) cl(G) ⊂ σ T (y) for every non-zero element y ∈H. (ii) For every y ∈H, σ T (y) is a connected set. (iii) σ(T) is a connected set. Proof. Suppose that H = ∨ {k(λ) : λ ∈ Ba(T)}, and Ty = λy for some λ ∈ C and y ∈H. For every µ ∈ Ba(T)\{λ}, we have, λŷ(µ) = 〈Ty,k(µ)〉 = 〈y,T∗k(µ)〉 = µŷ(µ). Hence, the analytic function ŷ is identically zero on Ba(T); this means that 〈y,k(λ)〉 = 0 for every λ ∈ Ba(T). Since H = ∨ {k(λ) : λ ∈ Ba(T)}, we have 310 A. Bourhim y = 0. Thus σp(T) = ∅. Now, suppose that H = ∨ {k(λ) : λ ∈ G} for some connected component G of Ba(T). (i) Suppose that there is an element y ∈H such that G∩ρ T (y) 6= ∅. Then there is an analytic H–valued function f such that (T −λ)f(λ) = y for λ ∈ V, where V = G∩ρ T (y). Hence, for every λ ∈ V , we have ŷ(λ) = 〈y,k(λ)〉 = 〈(T −λ)f(λ),k(λ)〉 = 〈f(λ), (T −λ)∗k(λ)〉 = 0. Therefore, ŷ ≡ 0 on G. Since H = ∨ {k(λ) : λ ∈ G}, it follows that y = 0. Thus the first statement (i) holds. (ii) Suppose that there is a non-zero element y ∈ H for which σ T (y) is disconnected, then σ T (y) = σ1 ∪ σ2, where σ1 and σ2 are two non-empty disjoint compact subsets of C. Since σp(T) = ∅, we note that T has the single–valued extension property. And so, using the local version of Riesz’s functional calculus, one shows that there are two non-zero elements y1 and y2 of H such that y = y1 + y2 and σT (yi) ⊂ σi, i = 1, 2 which contradicts (i). (iii) By (i), we have ⋂ y∈H\{0} σ T (y) 6= ∅, and σ T (y) is connected for every y ∈H\{0}. Since T has the single–valued extension property, we have σ(T) =⋃ y∈H\{0} σ T (y) (see Proposition 1.3.2 of [13]). The result follows. � Corollary 5.2. Suppose that T possesses Bishop’s property (β). If H =∨ {k(λ) : λ ∈ Ba(T)}, then Ba(T) = Γ(T)\σap(T). We shall also need the following result due to L. R. Williams [21]. Recall that T is said to be pure if T does not have a non-zero reducing subspace M for which T is normal when restricted to M. Theorem 5.3. If T is a pure cyclic hyponormal operator so that σap(T) has zero planar Lebesgue measure, then H = ∨ {k(λ) : λ ∈ Ba(T)}. Example 5.4. Let D be the open unit disk in the complex plane C, and let H = L2a(D) be the Bergman space, consisting of those analytic functions on D that are square integrable on D with respect to area measure. It is a Hilbert space when equipped with the inner product given by the formula 〈f,g〉 = ∫ D f(z)g(z)dm(z), f, g ∈ L2a(D), Bounded point evaluations for . . . 311 where dm denotes planar area measure, normalized so that D has total mass 1. The Bergman operator S for D is the operator multiplication by z on L2a(D); i.e., (Sf)(z) = zf(z) for f ∈ L2a(D) and z ∈ D. It is a pure cyclic subnormal operator, with cyclic vector the constant function 1, with σ(S) = cl(D) and σap(S) = {λ ∈ C : |λ| = 1} (see [8]). In particular, Ba(S) = σ(S)\σap(S) = D 6= ∅. By Theorem 5.3, we have L2a(D) = ∨ {k(λ) : λ ∈ Ba(S)}. This fact can be proved without using Theorem 5.3 and the vectors k(λ) can be given explicitely. To do this, fix λ ∈ D and consider the power series expansion (5.7) k(λ)(z) = +∞∑ n=0 anz n. Fixing a non-negative integer n, and taking the polynomial p(z) = zn, we have λn = 〈p(S)1,k(λ)〉 = 〈zn,k(λ)〉 = ∫ D zn [ +∞∑ i=0 aizi ] dm(z) = π an n + 1 . So, by (5.7), k(λ) should be given by the formula (5.8) k(λ)(z) = 1 π +∞∑ n=0 (n + 1)(λz)n = 1 π (1 −λz)−2. The above infinite sum is evaluated by letting ω = λz and noting that +∞∑ n=0 (n + 1)ωn = [ +∞∑ n=0 ωn+1 ]′ = [ ω 1 −ω ]′ = 1 (1 −ω)2 . Now, let f ∈ L2a(D) with a power series expansion f(z) = +∞∑ n=0 anz n, z ∈ D. It is easy to verify that the monomials (zn)n≥0 form an orthogonal basis for L2a(D), hence the partial sums of the above series converge to f in L 2 a(D). Thus, for every λ ∈ D, we have f̂(λ) = 〈f,k(λ)〉 = lim n→+∞ 〈 n∑ i=0 aiz i,k(λ)〉 = lim n→+∞ n∑ i=0 aiλ i = f(λ). So, if f is orthogonal to ∨ {k(λ) : λ ∈ Ba(S)}, then f must be identically zero. Therefore, L2a(D) = ∨ {k(λ) : λ ∈ Ba(S)}. 312 A. Bourhim The following example shows that ∨ {k(λ) : λ ∈ Ba(S)} needs not always be equal to H. It also shows that the purity of T is necessary condition in Theorem 5.3. Example 5.5. Let δ be the σ–finite Dirac measure on C at 2; i.e., for every subset A of C, δ(A) =   1 if 2 ∈ A 0 otherwise The normal operator Nδ : L2(δ) → L2(δ) defined by Nδf(z) = zf(z) for all f ∈ L2(δ), is cyclic and satisfies σ(Nδ) = σap(Nδ) = σp(Nδ) = {2}. Now, let H = L2a(D) ⊕L2(δ) and let T = S ⊕Nδ, where S : L2a(D) → L2a(D) is the Bergman operator for D. By Proposition 1-viii of [12], T is cyclic subnormal operator with cyclic vector 1 ⊕ 1. On the other hand, we have, σ(T) = cl(D) ∪{2}, σap(T) = {λ ∈ C : |λ| = 1}∪{2} and σp(T) = {2}. In particular, Ba(T) = σ(T)\σap(T) = D 6= ∅. Since σp(T) 6= ∅, it then follows from Proposition 5.1 that∨ {k(λ) : λ ∈ Ba(T)}$H. Note that all the examples considered in the present paper are given either by weighted shift operators or by subnormal operators. So, it would be interesting to give other examples. We begin by recalling that an operator R ∈L(H) is said to be Fredholm if ranR is closed, and ker R and ker R∗ are finite dimensional. The essential spectrum σe(R) of R is the set of all λ ∈ C such that R − λ is not Fredholm; in fact, σe(R) is exactly the spectrum σ ( π(R) ) in the Calkin algebra L(H)/C of π(R), where π is the natural quotient map of L(H) onto L(H)/C; here, C denotes the ideal of all compact operators in L(H). Note that if R ∈L(H) is a cyclic operator such that σp(R) = ∅, then σap(R) = σe(R) = {λ ∈ C : ran ( R−λ ) is not closed }. Example 5.6. Let S be the unweighted shift on H i.e., Sen = en+1 for every non-negative integer n, where (en)n≥0 is an orthonormal basis of H. It is shown in [11] that the operator T := S∗ + 2S is hyponormal, but T 2 is not. Therefore, T is not subnormal operator since every power of a subnormal operator is subnormal. On the other hand, it is easy to see that T is a cyclic operator. Indeed, by induction, one can show that ek = pk(T)e0 for every k ≥ 0, where (5.9)   p0(z) = 1, p1(z) = 12z pk+1(z) = 12 [ zpk(z) −pk−1(z) ] , ∀k ≥ 1. Bounded point evaluations for . . . 313 Hence, e0 is a cyclic vector for the operator T . Moreover, the following prop- erties hold. (i) T is a pure cyclic hyponormal operator. (ii) σap(T) = {a + ib ∈ C : (a3 ) 2 + b2 = 1}. (iii) σ(T) = {a + ib ∈ C : (a 3 )2 + b2 ≤ 1}. Hence, T is not a weighted shift operator since its spectrum is not a disc. (iv) Ba(T) = {a + ib ∈ C : (a3 ) 2 + b2 < 1} and for every λ ∈ Ba(T), k(λ) = +∞∑ n=0 pn(λ)en, where the polynomials pn are given by (5.9). (v) σ T (y) = σ(T) for every non-zero element y ∈H. (i) Let M be a reducing subspace of T . Since S = 2T −T∗ 3 , then M is a reducing subspace of S. And so M = {0} or M = H. Hence T is pure operator. (ii) Since, σ ( π(S) ) = σe(S) = T, where T is the unit circle of C, and π(S) is a normal element in the Calkin algebra, it follows from the Spectral Mapping Theorem that σ ( π(T) ) = {λ + 2λ : λ ∈ T} = {a + ib ∈ C : ( a 3 )2 + b2 = 1}. Since σap(T) = σe(T) = σ ( π(T) ) , the desired result holds. (iii) Note that the operator T is a Toeplitz operator Tφ with associated function φ = z + 2z. It then follows from Theorem 5 of [19] that { a + ib ∈ C : (a 3 )2 + b2 > 1 } ⊂ ρ(T), and either{ a + ib ∈ C : (a 3 )2 + b2 < 1 } ⊂ σ(T) or { a + ib ∈ C : (a 3 )2 + b2 < 1 } ⊂ ρ(T). In view of Theorem 1 of [15], the last inclusion is impossible since T is not normal. Hence, σ(T) = {a + ib ∈ C : (a 3 )2 + b2 ≤ 1}. (iv) By Theorem 4.3, we have Ba(T) = σ(T)\σap(T) = { a + ib ∈ C : (a 3 )2 + b2 < 1 } . 314 A. Bourhim Now, let λ ∈ Ba(T); then we have k(λ) = +∞∑ n=0 〈k(λ),en〉en = +∞∑ n=0 〈k(λ),pn(T)e0〉en = +∞∑ n=0 pn(λ)en. (v) By Theorem 5.3, we have H = ∨ {k(λ) : λ ∈ Ba(T)}. And so, by Proposition 5.1, σ(T) = cl ( Ba(T) ) ⊂ σ T (y) for every non-zero element y ∈H. The statement (v) is proved. The following example shows that H = ∨ {k(λ) : λ ∈ Ba(T)} is not sufficient condition for T to have a connected spectrum, and gives a negative answer to Question B of [21] (see also [22]). The following lemma is needed. Lemma 5.7. Suppose that R ∈ L(H). If σ R (y) = σ(R) for every non-zero element y ∈H, then σ(R) is connected. Proof. The proof is similar to the one of Proposition 5.1-(ii). � Example 5.8. In considering the operator T given in Example 5.6, we let T̃ := (T − 4) ⊕ (T + 4) ∈L(H̃), where H̃ = H⊕H. If we set G+ = { a+ib ∈ C : (a + 4 3 )2 +b2 < 1 } and G− = { a+ib ∈ C : (a− 4 3 )2 +b2 < 1 } , then the following hold. (i) T̃ is a pure cyclic hyponormal operator with cyclic vector x = e0 ⊕e0. (ii) σap(T̃) = Fr(G−) ∪ Fr(G+). Hence, by Theorem 5.3, H̃ = ∨ {k T̃ (λ) : λ ∈ Ba(T̃)}. (iii) σ(T̃) = cl(G−) ∪ cl(G+). Hence, σ(T̃) is disconnected; by Lemma 5.7, there is a non-zero element y ∈ H̃ for which σ T̃ (y) $ σ(T̃). (iv) Ba(T̃) = G− ∪G+. What we need to show here is that x = e0 ⊕e0 is a cyclic vector for T̃ . Indeed, let B+ = { λ ∈ C : |λ + 4| < 7 2 } and B− = { λ ∈ C : |λ− 4| < 7 2 } . Bounded point evaluations for . . . 315 Observe that B− ∩ B+ = ∅, and consider the following analytic function on B := B− ∪B+ f(z) =   1 if z ∈ B+ 0 if z ∈ B−. Since C\B is connected, Runge’s theorem shows that there is a sequence of polynomials (pn)n which converges uniformly to f on compact subsets of B. Hence, pn(T̃)x → f(T̃)x = e0 ⊕ 0. Since e0 is cyclic vector for T , for every y ∈H, we have y ⊕ 0 ∈ ∨{ p(T̃)x : p polynomial } . Similarly, for every y ∈H, we have 0 ⊕y ∈ ∨{ p(T̃)x : p polynomial } . Therefore, H̃ = ∨{ p(T̃)x : p polynomial } . So, x = e0 ⊕e0 is cyclic vector for T̃ . Acknowledgements. Part of this material is contained in the author’s Ph.D thesis, [4], written at the Mohammed V university, Rabat-Morocco. The author expresses his gratitude to Professor O. El-Fallah for his encourage- ment and helpful discussion. 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Williams, Subdecomposable operators and rationally invariant subspaces, Operator theory: Adv. Appl., 115 (2000), 297-309. [23] L. Yang, Hyponormal and subdecomposable operators, J. Functional Anal. 112 (1993), 204- 217. Received November 2001 Revised September 2002 A. Bourhim The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy E-mail address : bourhim@ictp.trieste.it Current address: A. Bourhim Département de Mathématiques, Université Mohamed V, B.P. 1014, Rabat, Morocco E-mail address : abourhim@fsr.ac.ma