����������� ������ 54 (2002), 71–81

UDK 517.984.3

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WEYL’S THEOREM FOR A GENERALIZED DERIVATION AND AN ELEMENTARY OPERATOR

B. P. Duggal

Abstract. For a, b ∈ B(H), B(H) the algebra of operators on a complex infinite dimensional Hilbert space H, the generalized derivation δ_{ab} ∈ B(B(H)) and the elementary operator △_{ab} ∈ B(B(H)) are defined by δ_{ab}(x) = ax − xb and △_{ab}(x) = axb − x. Let d_{ab} = δ_{ab} or △_{ab}. It is proved that if a, b^{∗} are hyponormal, then f(d_{ab}) satisfies (generalized) Weyl’s theorem for each function f analytic on a neighbourhood of σ(d_{ab}).

1. Introduction

Let B(H) denote the algebra of operators (i.e., bounded linear transformations) on a complex infinite dimensional Hilbert space H. For a, b ∈ B(H), let δ_{ab}: B(H) → B(H) and △_{ab}: B(H) → B(H) denotes the generalized derivation δ_{ab}(x) = ax − xb and the elementary operator △_{ab}(x) = axb − x. Let d_{ab} = δ_{ab} or △_{ab}. The following implications hold for a general bounded linear operator t on a normed linear space V, in particular for t = d_{ab}:

t^{−1}(0) ⊥ t(V) ⟹ t^{−1}(0) ∩ clt(V) = {0}
⟹ t^{−1}(0) ∩ t(V) = {0} ⟺ asc(t) ≤ 1

[6, page 25]. Here asc(t) denotes the ascent of t, clt(V) denote the closure of the range of t and t^{−1}(0) ⊥ t(V) denotes that the kernel of t is orthogonal to the range of t in the sense of G. Birkhoff. Recall that if M, N are linear subspaces of a normed linear space V, then M ⊥ N in the sense of Birkhoff if ∥m∥ ≤ ∥m+n∥ for all m ∈ M and n ∈ N. This concept of orthogonality is not symmetric, i.e., M ⊥ N does not imply N ⊥ M, but the concept does agree with the usual concept of orthogonality in the case in which V = H. The range-kernel orthogonality of d_{ab} has been considered by a number of authors (see [1,6,10,15,22,23] for further references). A sufficient condition guaranteeing d_{ab}^{−1}(0) ⊥ d_{ab}(B(H)) is that d_{ab}^{−1}(0) ⊆ d_{a^{∗}b^{∗}}^{−1}(0) [10, Theorem (i)]. The class of operators a, b^{∗} ∈ B(H) such that d_{ab}^{−1}(0) ⊆ d_{a^{∗}b^{∗}}^{−1}(0) is large, and includes in particular the class of hyponormal a and b^{∗} [9,20]. If a, b^{∗} ∈ B(H) are hyponormal, then (d_{ab} − λ)^{−1}(0) ⊆ (d_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0) and asc(d_{ab} − λ) ≤ 1 for all complex numbers λ. This implies that d_{ab} has the single-valued extension property and hence satisfies Browder’s theorem [8].

AMS Subject Classification: 47B47, 47B20, 47A53.

Communicated at the 5th International Symposium on Mathematical Analysis and its Applications, Niška banja, Yugoslavia, October, 2–6, 2002.

A detailed study of the spectral properties of the operator d_{ab} has been carried out in a series of papers by L. A. Fialkow (of which [14] is an earlier sample). Our aim here is a modest one. We show that if a, b^{∗} ∈ B(H) are hyponormal, and f is a function which is analytic on a neighbourhood of the spectrum of d_{ab}, then f(d_{ab}) satisfies Weyl’s theorem. Indeed more is true: f(d_{ab}) satisfies the generalized Weyl’s theorem. Problems of this type seem not to have previously been considered.

The plan of this paper is as follows. We use the remainder of this section to introduce some of our notation and terminology. (Any additional notation or terminology will be introduced as and when required.) Section 2 will be devoted to proving some complementary results, amongst them that d_{ab}, a, b^{∗} ∈ B(H) hyponormal, is isoloid and that the range of d_{ab} −λ is closed for each isolated point λ of the spectrum of d_{ab}. We shall prove Weyl’s theorems for d_{ab} in Section 3.

We shall denote the spectrum, the point spectrum and the set of isolated points of the spectrum of a Banach space operator t ∈ B(V) by σ(t), σ_{p}(t) and isoσ(t), respectively. The range, the kernel and the restriction to an invariant subspace M of t will be denoted by t(V) (or, ran(t)), t^{−1}(0) (or, ker t) and t_{|M}, respectively. The operator t is a quasi-affinity if it is injective and has dense range, and t is said to be isoloid if there is implication λ ∈ isoσ(t) ⟹ λ ∈ σ_{p}(t). Recall that the ascent asc(t) of an operator t is the smallest non-negative integer n such that t^{−n}(0) = t^{−(n+1)}(0).

Let V be a (complex) Banach space, and let U be an open subset of the complex plane C. Let O(U, V) denote the Frechet space of V -valued analytic functions from U. The operator t ∈ B(V) is said to satisfy Bishop’s condition (β) if, for each open subset U of C, the operator t_{|U} given by (t_{|U}f)(λ) := (t − λ)f(λ) is injective and has dense range in O(U, V) for each f ∈ O(U, V) and all λ ∈ U. For a closed subset F of C, let V_{t}(F) denote the analytic spectral manifold

V_{t}(F) = {v ∈ V : (t − λ)f(λ) = v has an analytic solution f: C \ F → V }.

The spaces V_{t}(F) are t-invariant (generally, non-closed) manifolds of V. If, for every closed F ⊆ C, V_{t}(F) is closed, then t is said to satisfy Dunford’s property (C). Condition (β) implies property (C), which in turn implies that the operator t_{|U} is injective for every open U ⊆ C. This last property is the single-valued extension property, shortened henceforth to SVEP. (Thus t has SVEP if, for every v ∈ V, (t − λ)f(u) = v has a unique solution f : U → V on U ⊆ C.) We shall denote the set of natural numbers by N.

A Banach space operator t ∈ B(V) is said to be Fredholm if t(V) is closed, and both t^{−1}(0) and V \ clt(V) are finite dimensional. The Fredholm index ind(t) of t is defined by ind(t) = dim(t^{−1}(0)) − dim(V \ t(V)). The operator t is Weyl if it is Fredholm of index 0, and it is Browder if it is Fredholm and both asc(t) and dsc(t) are finite [11]. The (Fredholm) essential spectrum σ_{e}(t), the Weyl spectrum σ_{w}(t) and the Browder spectrum σ_{b}(t) of t are defined by

σ_{e}(t) = {λ ∈ C : t − λ is not Fredholm},
σ_{w}(t) = {λ ∈ C : t − λ is not Weyl},
σ_{b}(t) = {λ ∈ C : t − λ is not Browder}.

Evidently, σ_{e}(t) ⊆ σ_{w}(t) ⊆ σ_{b}(t) ⊆ σ_{e}(t) ∪ acc σ(t). In general, the spectral mapping theorem holds for σ_{b}(t) but fails for σ_{w}(t) [12,13]. Let σ_{o}(t) denote the set of Riesz points of t, and let σ_{oo}(t) = {λ ∈ isoσ(t) : 0 < dim(t − λ)^{−1}(0) < ∞}. Then isoσ(t) \ σ_{e}(t) = isoσ(t) \ σ_{w}(t) = σ_{o}(t) ⊆ σ_{oo}(t). We say that t satisfies Weyl’s theorem (resp., Browder’s theorem) if

σ(t) \ σ_{w}(t) = σ_{oo}(t) (resp., σ(t) \ σ_{w}(t) = σ_{o}(t)).

2. Complementary results

We prove in this section that if a, b^{∗} ∈ B(H) are hyponormal, then d_{ab} is isoloid and ran(d_{ab} − λ) is closed for each λ ∈ σ_{oo}(d_{ab}). But we start by working towards proving that d_{ab} has SVEP. Throughout the following, we write t − λ for the operator t − λI. The operators of “left multiplication by a” and “right multiplication by b” shall be denoted by L_{a} and R_{b}, respectively.

Lemma 2.1. Let a, b ∈ B(H) be normal. If there exists a quasi- affinity x ∈ △_{ab}^{−1}(0), then b is invertible and x ∈ δ_{ab^{−1}}^{−1}(0).

Proof. The operators a and b being normal, it follows from an application of the Putnam-Fuglede theorem for normal operators that △_{ab}^{−1}(0) = △_{a^{∗}b^{∗}}^{−1}(0) [9, Corollary 3]. Let the quasiaffinity x ∈ △_{ab}^{−1}(0) have the polar decomposition x = u|x| (where u is unitary). Since △_{ab}(x) = 0 = △_{a^{∗}b^{∗}}(x), it follows from

b|x|^{2} = bx^{∗}(axb) = (bx^{∗}a)xb = |x|^{2}b

that |x|^{2}, and so also |x|, ∈ δ_{bb}^{−1}(0). Hence, since △_{ab}(x) = △_{ab}(u|x|) = △_{ab}(u)|x| = 0, u ∈ △_{ab}^{−1}(0). Let h ∈ H. Then

△_{ab}(u)h = 0 ⟹ u^{∗}aubh = h ⟹ ∥h∥ = ∥u^{∗}aubh∥ ≤ ∥a∥∥bh∥,

i.e., b is bounded below. It is clear from △_{a^{∗}b^{∗}}(x) = a^{∗}xb^{∗} − x = 0 that b^{∗} is injective. Hence b is invertible and x ∈ δ_{ab^{−1}}^{−1}(0).

The following lemma is proved in [20] for the case in which d = δ; for the case in which d = △ a proof follows from the argument of the proof of [9, Lemma 4].

Lemma 2.2. If a, b ∈ B(H) are normal, then d_{ab}^{−2}(0) = d_{ab}^{−1}(0).

The ascent of the operator d_{ab} (indeed, any operator) equals 0 if and only if d_{ab} is injective, and then {0} = d_{ab}^{−1}(0) ⊆ d_{a^{∗}b^{∗}}^{−1}(0) trivially. The following proposition says that non-injective d_{ab} − λ satisfying (d_{ab} − λ)^{−1}(0) ⊆ (d_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0) have ascent one.

Proposition 2.3. If a, b ∈ B(H) and λ ∈ C are such that (d_{ab} − λ)^{−1}(0) ⊆ (d_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0) , then asc(d_{ab} − λ) ≤ 1.

Proof. We consider the cases d_{ab} = δ_{ab} and d_{ab} = △_{ab} separately.

Case d_{ab} = δ_{ab}. Let x ∈ (δ_{ab} − λ)^{−1}(0). Then the hypothesis (δ_{ab} − λ)^{−1}(0) ⊆ (δ_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0) ⟹ ax − x(b + λ) = 0 = a^{∗}x − x(b^{∗} + \overline{λ}) ⟹ \overline{ranx} reduces a and ker^{⊥}x reduces b+λ. Since x ∈ (δ_{ab} −λ)^{−1}(0) ⟹ ax and x(b+λ) ∈ (δ_{ab} −λ)^{−1}(0),

a^{∗}ax = ax(b + λ)^{∗} = aa^{∗}x

and

x(b + λ)^{∗}(b + λ) = a^{∗}x(b + λ) = x(b + λ)(b + λ)^{∗}.

Hence a_{1} = a|\overline{ranx} and b_{1} = (b + λ)|_{ker^{⊥}x} are normal operators.

Suppose now that y ∈ (δ_{ab} − λ)^{−2}(0). Set (δ_{ab} − λ)y = x, let x_{1} : ker^{⊥}x → \overline{ranx} be the quasi-affinity defined by setting x_{1}h = xh for each h ∈ H and let y : ker^{⊥}x ⊕ kerx → \overline{ranx} ⊕ \overline{ranx}^{⊥} have the matrix representation y = [y_{ij}]_{i,j=1}^{2}. Then 0 = δ_{ab}(x) = δ_{a_{1}b_{1}}(x_{1}) ⊕ 0 = δ_{a_{1}b_{1}}^{2}(y_{11}) ⊕ 0. The operators a_{1} and b_{1} being normal, it follows from Lemma (2.2) that δ_{a_{1}b_{1}}(y_{11}) = 0. Hence x = δ_{a_{1}b_{1}}(y_{11})⊕0 = 0 ⟹ (δ_{ab} − λ)(y) = 0 ⟹ asc(δ_{ab} − λ) ≤ 1.

Case d_{ab} = △_{ab}. The proof is split into the cases λ = −1 and λ ≠ −1.

If λ = −1, then y ∈ (△_{ab} − λ)^{−2}(0) ⟹ ayb ∈ (△_{ab} − λ)^{−1}(0) ⟹ |a|^{2}y|b^{∗}|^{2} = 0. If both |a| and |b^{∗}| are injective, then y = 0 (and we are done). If only one of |a| and |b^{∗}| is injective, say |a|, then y|b^{∗}|^{2} = 0. Letting |b^{∗}| = 0 ⊕ b_{2}, b_{2} invertible, and y = [y_{ij}]_{i,j=1}^{2} it then follows that y_{12} = y_{22} = 0. Hence y|b^{∗}| = 0, which implies that |a|y|b^{∗}| = 0 ⟹ ayb = 0. Finally, if both |a| and |b^{∗}| are not injective, then upon letting |a| = 0 ⊕ a_{2}, |b^{∗}| = 0 ⊕ b_{2} and y = [y_{ij}]_{i,j=1}^{2} it follows that y_{22} = 0 ⟹ |a|y|b^{∗}| = 0 ⟹ ayb = 0. In either case asc(△_{ab} − λ) ≤ 1.

If λ ≠ −1, then △_{ab} − λ = (1 + λ)△_{cb} and (△_{ab} − λ)^{−1}(0) ⊆ (△_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0) ⟺ △_{cb}^{−1}(0) ⊆ △_{c^{∗}b^{∗}}^{−1}(0), where we have set \frac{1}{1+λ}a = c. Let x ∈ △_{cb}^{−1}(0). Then cxb − x = 0 = c^{∗}xb^{∗} − x ⟹ \overline{ranx} reduces c, ker^{⊥}x reduces b and b|_{ker^{⊥}x} is invertible (see the proof of Lemma 2.1). Obviously, x ∈ △_{cb}^{−1}(0) ⟹ cx and xb ∈ △_{cb}^{−1}(0). Since △_{cb}^{−1}(0) ⊆ △_{c^{∗}b^{∗}}^{−1}(0),

△_{c^{∗}b^{∗}}(cx) = 0 = c△_{c^{∗}b^{∗}}(x) and △_{c^{∗}b^{∗}}(xb) = 0 = △_{c^{∗}b^{∗}}(x)b,

which implies that c_{1} = c|\overline{ranx} and b_{1} = b|_{ker^{⊥}x} are normal operators. Assume now that y ∈ △_{cb}^{−2}(0). Set △_{cb}(y) = x, let y : ker^{⊥}x ⊕ kerx → \overline{ranx} ⊕ \overline{ranx}^{⊥} have the matrix representation y = [y_{ij}]_{i,j=1}^{2}, and define the quasi-affinity x_{1} : ker^{⊥}x → \overline{ranx} as above. Then

0 = △_{cb}(x) = △_{c_{1}b_{1}}(x_{1}) ⊕ 0 = △_{c_{1}b_{1}}^{2}(y_{11}) ⊕ 0.

The operators c_{1} and b_{1} being normal, it follows (from an application of Lemma (2.2)) that △_{c_{1}b_{1}}(y_{11}) = 0. Hence x = △_{c_{1}b_{1}}(y_{11})⊕0 = 0 ⟹ (△_{ab} −λ)(y) = 0 ⟹ asc(△_{ab} − λ) ≤ 1.

If t ∈ B(H) is hyponormal, then so are the operators λt and t + λ for every λ ∈ C. Since the inclusion d_{ab}^{−1}(0) ⊆ d_{a^{∗}b^{∗}}^{−1}(0) holds for hyponormal a, b^{∗} ∈ B(H) [9,20], it follows that

(δ_{ab} − λ)^{−1}(0) = δ_{a(b+λ)}^{−1}(0) ⊆ δ_{a^{∗}(b+λ)^{∗}}^{−1}(0) = (δ_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0).

Again, if λ ≠ −1, then

(△_{ab} − λ)^{−1}(0) = (L_{\frac{1}{1+λ} a}R_{b} − 1)^{−1}(0) ⊆ (L_{\frac{1}{1+\overline{λ}} a^{∗}}R_{b^{∗}} − 1)^{−1}(0) = (△_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0);

and if λ = −1, then

(△_{ab} − λ)^{−1}(0) = (L_{a}R_{b})^{−1}(0) ⊆ (L_{a^{∗}}R_{b^{∗}})^{−1}(0) = (△_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0).

Corollary 2.4. If a, b^{∗} ∈ B(H) are hyponormal, then asc(d_{ab} − λ) ≤ 1 for all λ ∈ C. In particular, d_{ab} has SVEP.

Proof. Since (d_{ab} − λ)^{−1}(0) ⊆ (d_{a^{∗}b^{∗}} − \overline{λ})^{−1}(0) for all λ ∈ C, Proposition 2.3 applies. The finite ascent property of (d_{ab} − λ) implies SVEP [17].

Remarks 2.5. (i) The asymmetric hypotheses on a and b in Corollary 2.4 are not surprising; for the record the corollary fails if a and b are hyponormal (even, subnormal). Specifically, take u to be the (forward) unilateral shift and let

x = [ \begin{matrix}0 & 0 \\ 1 − uu^{∗} & 0 \end{matrix} ] (on H ⊕ H). Choose a = b = u ⊕ 0 in the case in which d = δ

,and a = u ⊕ I and b = (I + u) ⊕ 0 in the case in which d = △. Then x ∈ d_{ab}^{−1}(0), but x ∉ d_{a^{∗}b^{∗}}^{−1}(0).

(ii) More is true in Corollary 2.4 in the case in which d = δ. The hypothesis a, b^{∗} ∈ B(H) are hyponormal implies that a, b^{∗} satisfy Bishop’s condition (β) [17]. Hence δ_{ab} satisfies condition (C) [17, Theorem 3.6.10, page 277] (which implies that δ_{ab} has SVEP). Denoting B(H) by V and δ_{ab} by t, this implies that E_{t}(F) is closed for all closed sets F ⊆ C if and only if E_{t}(F) = V_{t}(F) [17, Proposition 1.4.13], where the algebraic spectral subspace E_{t}(F) is the largest subspace of V on which all restrictions of t − λ, λ ∈ C \ F, are surjective. (We note here that E_{t}(F) = ∩_{λ∉F,n∈N}(t − λ)^{n} for all subsets F of C, because of the finite ascent property of t.)

(iii) Does △_{ab}, a and b^{∗} ∈ B(H) hyponormal, satisfy condition (C)?

For the remainder of this section we assume that a, b^{∗} ∈ B(H) are hyponormal.

Theorem 2.6. (d_{ab} − λ) has closed range for each λ ∈ isoσ(d_{ab}).

Proof. Before proceeding with the proof proper let us recall that if t ∈ B(H) is hyponormal, then: (i) t − λ is hyponormal; (ii) t quasi-nilpotent implies t = 0; (iii) the isolated points of σ(t) are poles of order one of the resolvent of t; and (iv) the eigenvalues of t are normal eigenvalues. Let λ ∈ isoσ(d_{ab}).

The case d_{ab} = △_{a,b}. We divide the proof into the cases λ = −1 and λ ≠ −1. Let Φ_{ab} = L_{a}R_{b}. If λ = −1, then 0 ∈ isoσ(Φ_{ab}). Since σ(Φ_{ab}) = ∪{σ(za) : z ∈ σ(b)} (this well known fact follows from [11, Theorem 3.2]), we must have that either 0 ∈ isoσ(b) or 0 ∈ isoσ(a). Suppose that 0 ∈ isoσ(b). (The other case is similarly dealt with.) Then 0 can not be a limit point of σ(a). For if 0 is a limit point of σ(a), then there exists a sequence {α_{n}} ∈ σ(a) such that α_{n} → 0 ∈ σ(a). Choosing a non-zero z ∈ σ(b) we then have a sequence {zα_{n}} ∈ σ(Φ_{ab}) such that zα_{n} → 0, which contradicts the fact that 0 ∈ isoσ(Φ_{ab}). (We remark here that such a choice of z is always possible, for if not then σ(b) = {0} and b is the zero operator.) The conclusion that 0 can not be a limit point of σ(a) implies that either 0 ∉ σ(a) or 0 ∈ isoσ(a). If 0 ∉ σ(a), then a is invertible and ran(Φ_{ab}) is closed whenever ran(Φ_{Ib}) is closed. Notice that 0 ∈ isoσ(b) ⟹ 0 ∈ isoσ(b^{∗}). Since b^{∗} is hyponormal, ker(b^{∗}) reduces b and b = 0 ⊕ b_{2} with respect to the decomposition H = ker(b^{∗}) ⊕ ker^{⊥}(b^{∗}) = H_{1} ⊕ H_{2}, say, of H. Clearly, the operator b_{2} = b|_{H_{2}} is invertible. Let x : H_{1}⊕H_{2} → H_{1}⊕H_{2} have the matrix representation x = [x_{ij}]_{i,j=1}^{2}.

Then Φ_{Ib}(x) = [ \begin{matrix}0 & x_{12}b_{2} \\ 0 & x_{22}b_{2} \end{matrix} ]. The operator b_{2} being invertible, Φ_{Ib_{2}} is invertible,

and hence ran(Φ_{Ib}) (and so also ran(Φ_{ab})) is closed. Now let 0 ∈ isoσ(a). Then a = 0 ⊕ a_{2} with respect to the decomposition H = ker(a) ⊕ ker^{⊥}(a) = H′_{1} ⊕ H′_{2}, say, of H, where the operator a_{2} = a|_{H′_{2}} is invertible. Let x : H_{1} ⊕ H_{2} → H′_{1} ⊕ H′_{2}

have the matrix representation x = [x_{ij}]_{i,j=1}^{2}. Then Φ_{ab}(x) = [ \begin{matrix}0 & 0 \\ 0 & a_{2}x_{22}b_{2} \end{matrix} ]. The

operator Φ_{a_{2}b_{2}} being invertible, ran(Φ_{a_{2}b_{2}}) ( and so also ran(Φ_{ab})) is closed. This leaves us with the case λ ≠ −1, which we consider next.

If λ ≠ −1, then (△_{ab}−λ)(x) = axb−(1+λ)x, and it follows from [11, Theorem 3.2] that

σ(△_{ab} − λ) = ⋃{σ(−(1 + λ) + za) : z ∈ σ(b)}.

If λ ∈ isoσ(△_{ab}), then 0 ∈ isoσ(△_{ab} − λ). There exists a finite set {β_{1}, β_{2},…, β_{n}} of distinct non-zero values of z ∈ isoσ(b), and corresponding to these values of z a finite set {α_{1}, α_{2},…, α_{n}} of distinct non-zero values α_{i} ∈ isoσ(a) such that α_{i}β_{i} = 1 + λ for all 1 ≤ i ≤ n. Let

H_{1} = ⋁_{i=1}^{n}ker(b−β_{i})^{∗}, H′_{1} = ⋁_{i=1}^{n}ker(a−α_{i}), H_{2} = H⊖H_{1} and H′_{2} = H⊖H′_{1}.

Then a and b have the direct sum decompositions a = a_{1} ⊕ a_{2} and b = b_{1} ⊕ b_{2}, where a_{1} = a|_{H′_{1}} and b_{1} = b|_{H_{1}} are normal operators with finite spectrum, b_{1} is invertible, a_{2} = a|_{H′_{2}}, b_{2} = b|_{H_{2}}, and σ(a_{1}) ∩ σ(a_{2}) = ∅ = σ(b_{1}) ∩ σ(b_{2}). Let x : H_{1} ⊕ H_{2} → H′_{1} ⊕ H′_{2} have the matrix representation x = [x_{ij}]_{i,j=1}^{2}. Then

(△_{ab} − λ)x = [ \begin{matrix}(△_{a_{1}b_{1}} − λ)x_{11} & (△_{a_{1}b_{2}} − λ)x_{12} \\ (△_{a_{2}b_{1}} − λ)x_{21} & (△_{a_{2}b_{2}} − λ)x_{22} \end{matrix} ] ,

where 0 ∉ σ(△_{a_{i}b_{j}} − λ) for all 1 ≤ i, j ≤ 2 such that i, j ≠ 1. To prove that ran(△_{ab} − λ) is closed it thus remains to prove that ran(△_{a_{1}b_{1}} − λ) is closed. This follows from [1, (2.4) Theorem], since b_{1} invertible implies (△_{a_{1}b_{1}} − λ)x_{11} = (a_{1}x_{11} − x_{11}(1 + λ)b_{1}^{−1})b_{1} = δ_{a_{1}((1+λ)b_{1}^{−1})}(x_{11}b_{1}), where the normal operators a_{1} and (1 + λ)b_{1}^{−1} have finite spectrum.

The case d_{ab} = δ_{ab}. Let λ ∈ isoσ(δ_{ab}). Then 0 ∈ isoσ(δ_{ab} − λ), where σ(δ_{ab} − λ) = σ(a) − σ(b + λ) [11]. Hence σ(a) ∩ σ(b + λ) consists of points which are isolated in both σ(a) and σ(b + λ). In particular, σ(a) ∩ σ(b + λ) does not contain any limit points of σ(a) ∪ σ(b + λ). There exists a finite set S = {α_{1}, α_{2},…, α_{n}} of distinct values α_{i} such that S = σ(a) ∩ σ(b + λ) and each α_{i}, 1 ≤ i ≤ n, is an isolated point of both σ(a) and σ(b + λ). Let

H_{1} = ⋁_{i=1}^{n}ker(b−α_{i})^{∗}, H′_{1} = ⋁_{i=1}^{n}ker(a−α_{i}), H_{2} = H⊖H_{1} and H′_{2} = H⊖H′_{1}.

Then, upon defining the normal operators a_{1} and b_{1} as before and letting x : H_{1} ⊕ H_{2} → H′_{1} ⊕ H′_{2} have the matrix representation x = [x_{ij}]_{i,j=1}^{2}, it is seen that

(δ_{ab} − λ)x = [ \begin{matrix}(δ_{a_{1}b_{1}} − λ)x_{11} & (δ_{a_{1}b_{2}} − λ)x_{12} \\ (δ_{a_{2}b_{1}} − λ)x_{21} & (δ_{a_{2}b_{2}} − λ)x_{22} \end{matrix} ] ,

where a_{2} = a|_{H′_{2}}, b_{2} = b|_{H_{2}} and σ(a_{i}) ∩ σ(b_{j} + λ) = ∅ for all 1 ≤ i, j ≤ 2 such that i, j ≠ 1 (so that 0 ∉ σ(δ_{ab} − λ) for all 1 ≤ i, j ≤ 2 such that i, j ≠ 1). That ran(δ_{ab} − λ) is closed now follows (see the proof above).

As earlier stated, hyponormal operators are isoloid. The following theorem says that d_{ab} retains this property in the case in which a, b^{∗} are hyponormal.

Theorem 2.7. d_{ab} is isoloid.

Proof. If λ ∈ isoσ(d_{ab}), then 0 ∈ isoσ(d_{ab} − λ). Let P denote the spectral projection of d_{ab} − λ at 0. Then

0 ≠ P (B(H)) = \{x ∈ B(H) : lim_{n→∞} ∥(d_{ab} − λ)^{n}x∥^{\frac{1}{n}} = 0\}.

We prove that

P (B(H)) = {x ∈ B(H) : (d_{ab} − λ)x = 0}.

Let d = δ. Then upon arguing as in the proof of Theorem (2.6) it is seen that there exist decompositions H = H_{1} ⊕ H_{2} and H = H′_{1} ⊕ H′_{2} such that x : H_{1} ⊕ H_{2} → H′_{1} ⊕ H′_{2} has the representation x = [x_{ij}]_{i,j=1}^{2} and

(δ_{ab} − λ)x = [(δ_{a_{i}b_{j}} − λ)x_{ij}]_{i,j=1}^{2},

where a_{1} = a_{|H′_{1}} and b_{1} = b_{|H_{1}} are normal operators with finite spectrum, and where 0 ∉ σ(δ_{a_{i}b_{j}} − λ) for all 1 ≤ i, j ≤ 2 such that i, j ≠ 1. Since

\{∥(δ_{a_{i}b_{j}} − λ)^{n}x_{ij}∥\}^{\frac{1}{n}} ≤ ∥(δ_{ab} − λ)^{n}x∥^{\frac{1}{n}} → 0

as n → ∞ implies that (δ_{a_{i}b_{j}} − λ) is quasi-nilpotent for all 1 ≤ i, j ≤ 2, it follows that x_{ij} = 0 for all 1 ≤ i, j ≤ 2 such that i, j ≠ 1. Consequently,

P (B(H)) = \{x = x_{11} ⊕ 0 ∈ B(H) : lim_{n→∞} ∥(δ_{a_{1}b_{1}} − λ)^{n}x_{11}∥^{\frac{1}{n}} = 0\}.

The operators a_{1} and b_{1} + λ in δ_{a_{1}(b_{1}+λ)} = δ_{a_{1}b_{1}} − λ being normal,

lim_{n→∞} ∥(δ_{a_{1}b_{1}} − λ)^{n}x_{11}∥^{\frac{1}{n}} ⇔ (δ_{a_{1}b_{1}} − λ)x_{11} = 0

[20, Lemma 2], i.e., if and only if 0 is an eigenvalue of δ_{a_{1}b_{1}} − λ. Hence

P (B(H)) = {x = x_{11} ⊕ 0 ∈ B(H) : (δ_{ab} − λ)x_{11} ⊕ 0 = 0}.

Now let d = △, and let 0 ∈ isoσ(△_{ab}). If λ ≠ −1, then (it follows from Theorem 2.6 that)

(△_{ab} − λ)x = [(△_{a_{i}b_{j}} − λ)x_{ij}]_{i,j=1}^{2},

where a_{1} and b_{1} are normal invertible operators with finite spectrum, and where 0 ∉ σ(△_{a_{i}b_{j}} − λ) for all 1 ≤ i, j ≤ 2 such that i, j ≠ 1. Consequently,

P (B(H)) = \{x = x_{11} ⊕ 0 ∈ B(H) : lim_{n→∞} ∥(△_{a_{1}b_{1}} − λ)^{n}x_{11}∥^{\frac{1}{n}} = 0\}.

The operator b_{1} being invertible normal,

∥(δ_{a_{1}b_{1}^{−1}} − λ)^{n}x_{11}∥^{\frac{1}{n}} = ∥(△_{a_{1}b_{1}} − λ)^{n}x_{11})b_{1}^{−n}∥^{\frac{1}{n}} ≤ ∥b_{1}^{−1}∥∥((△_{a_{1}b_{1}} − λ)^{n}x_{11}∥^{\frac{1}{n}} → 0

as n → ∞. This implies that

(δ_{a_{1}b_{1}^{−1}} − λ)x_{11} = 0 ⇔ (△_{a_{1}b_{1}} − λ)x_{11} = 0

and hence that

P (B(H)) = {x = x_{11} ⊕ 0 ∈ B(H) : (△_{ab} − λ)(x_{11} ⊕ 0) = 0}

in the case in which λ ≠ −1. Arguing similarly it is seen that if λ = −1, then

P (B(H)) = \{x = [ \begin{matrix}x_{11} & 0 \\ x_{21} & 0 \end{matrix} ] or [ \begin{matrix}x_{11} & x_{12} \\ x_{21} & 0 \end{matrix} ] ∈ B(H) :

(△_{ab} − λ)x = Φ_{Ib}(x) or Φ_{ab}(x) = 0}.

Remark 2.8. It is clear from the proof of Theorem 2.7 that if 0 ∈ isoσ(d_{ab}), then P (B(H)) = d_{ab}^{−1}(0). Since B(H) = P (B(H)) ⊕ P^{−1}(0) and P^{−1}(0) ⊂ d_{ab}(B(H)), it follows from [16, Theorem 3.4] that 0 is a pole of order one of the resolvent of d_{ab} and B(H) = d_{ab}^{−1}(0) ⊕ d_{ab}(B(H)).

The descent of the operator t, dsc(t), is the smallest non-negative integer n such that ran(t^{n}) = ran(t^{n+1}). The operator t is said to be Drazin invertible if there is an operator s and an n ∈ N such that

t^{n}st = t^{n}, sts = s and st = ts.

It is known that t is Drazin invertible if and only if both asc(t) and dsc(t) are finite (and this is equivalent to the existence of a decomposition t = t_{0} ⊕ t_{1}, where t_{0} is nilpotent and t_{1} is invertible) [18]. The following theorem relates the Drazin invertibility of d_{ab} to the finiteness of a subset of σ(d_{ab}). But before that we recall (once again) from [2, Theorem 3.3] that if s, t ∈ B(H) are normal, then cl(δ_{st}(B(H)))⊕δ_{st}^{−1}(0) = δ_{st}(B(H))⊕δ_{st}^{−1}(0) = B(H) if and only if the set σ(s)∩σ(t) is isolated in σ(δ_{st}). Since σ(δ_{ab}) = σ(a) − σ(b) (and σ(△_{ab}) = ∪{σ(−1 + za) : z ∈ σ(b)}), 0 ∈ isoσ(δ_{ab}) (resp., 0 ∈ isoσ(△_{ab}))if and only if the set σ(a) ∩ σ(b) is isolated in σ(δ_{ab}) (resp., the set {αβ : α ∈ σ(a), β ∈ σ(b), and αβ = 1} is isolated in σ(△_{ab}).)

Theorem 2.9. δ_{ab} (resp., △_{ab}) is Drazin invertible if and only if the set {σ(a) ∩ σ(b)} is isolated in σ(δ_{ab})(resp., the set {α ∈ σ(a) : α^{−1} ∈ σ(b)} is isolated in σ(△_{ab})).

Proof. We prove the case in which d = δ; the other case is similarly proved. If δ_{ab} is Drazin invertible, then both asc(δ_{ab}) and dsc(δ_{ab}) are finite. Since asc(δ_{ab}) ≤ 1 by Corollary (2.4), it follows from [21, Theorem V.6.2] that asc(δ_{ab}) = dsc(δ_{ab}) ≤ 1 and δ_{ab}(B(H)) ⊕ δ_{ab}^{−1}(0) = B(H). Hence 0 ∈ isoσ(δ_{ab}), which implies that {σ(a) ∩ σ(b)} is isolated in σ(δ_{ab}).

Conversely, {σ(a) ∩ σ(b)} isolated in σ(δ_{ab}) ⟹ 0 ∈ isoσ(δ_{ab}). (Clearly, 0 ∉ σ(δ_{ab}) ⟹ Drazin invertibility, trivially.) By Remark 2.8, 0 is a pole of order 1 of δ_{ab} and δ_{ab}(B(H)) ⊕ δ_{ab}^{−1}(0) = B(H). Hence asc(δ_{ab}) = dsc(δ_{ab}) ≤ 1 [17, Proposition 4.10.6] and δ_{ab} is Drazin invertible.

Note that the Drazin invertibility of d_{ab} implies the existence of a projection p and a bijection c on B(H) such that d_{ab} = pc = cp (see [17, Proposition 4.10.7]).

3. Weyl’s Theorem

The implication Weyl’s theorem ⟹ Browder’s theorem holds, but the reverse implication is in general false. SVEP ⟹ Browder’s theorem [8], but this implication fails if one replaces “Browder’s theorem” by “Weyl’s theorem” [7]. Let V = B(H) and let (as before) a, b^{∗} ∈ B(H) be hyponormal. Then the SVEP of d_{ab} ⟹ Browder’s theorem holds for d_{ab}. Recall from [7, Theorem 2.5] that if an operator t on a Banach space has SVEP, then t satisfies Weyl’s theorem ⟺ ran(t − λ) is closed for every λ ∈ σ_{oo}(t). Hence, in view of Theorem (2.6), d_{ab} satisfies Weyl’s theorem. More is true.

Theorem 3.1. If f is analytic on a neighbourhood of σ(d_{ab}), then f(d_{ab}) satisfies Weyl’s theorem.

Proof. SVEP being stable under the functional calculus [17], d_{ab} has SVEP ⟹ f(d_{ab}) has SVEP for each f analytic in a neighbourhood of σ(d_{ab}) ⟹ σ_{b}(f(d_{ab})) = σ_{w}(f(d_{ab})) [13]. Since the spectral mapping theorem holds for σ_{b}, we have

σ_{w}(f(d_{ab})) = σ_{b}(f(d_{ab})) = f(σ_{b}(d_{ab})) = f(σ_{w}(d_{ab})).

To complete the proof we have to show that f(σ_{w}(d_{ab})) = σ(f(d_{ab})) \ σ_{oo}(f(d_{ab})): this follows from Theorem 2.7 and a limit argument applied to [19, Proposition 1].

Remark 3.2. Browder’s theorem is transmitted to and from dual operators, but the same does not in general hold for Weyl’ theorem [13]. It is known that if T ∈ B(H) is hyponormal, then both T and T^{∗} satisfy Weyl’s theorem. A formal dual of the operator d_{ab} may be defined by d_{a^{∗}b^{∗}}. Does d_{a^{∗}b^{∗}} satisfy Browder’s theorem? Notice that σ(d_{a^{∗}b^{∗}}) = \overline{σ(d_{ab})} and λ ∈ isoσ(d_{a^{∗}b^{∗}}) ⟹ λ ∈ isoσ(d_{ab}) ⟹ λ ∈ σ_{oo}(d_{ab}) ⟹ \overline{λ} ∈ σ_{oo}(d_{a^{∗}b^{∗}}). Since Weyl’s theorem holds for d_{ab},

σ(d_{a^{∗}b^{∗}}) \ σ_{oo}(d_{a^{∗}b^{∗}}) = \overline{σ(d_{ab}) \ σ_{oo}(d_{ab})} = \overline{σ_{w}(d_{ab})}.

Does \overline{σ_{w}(d_{ab})} = σ_{w}(d_{a^{∗}b^{∗}})?

Generalized Weyl’s Theorem. An operator t ∈ B(V) is said to be generalized Fredholm, or B-Fredholm, if there is an n ∈ N for which the induced operator t_{n} : t^{n}(V) → t^{n}(V) is Fredholm in the usual sense, and generalized Weyl, or “B-Weyl”, if in addition t_{n} has index zero. The generalized Weyl spectrum σ_{Bw}(t) of t is defined to be the set {λ ∈ C : (t − λ) is not generalized Weyl}, and we say that t satisfies generalized Weyl’s theorem (resp., generalized Browder’s theorem) if σ_{Bw}(t) = σ(t) \ E(t) (resp., σ_{Bw}(t) = σ(t) \ Π(t)), where E(t) = {λ ∈ isoσ(t) : λ is an eigenvalue of t} and Π(t) is the set of poles of t. (See [3,4,5] for further information.) The implication t satisfies genarlized Weyl’s theorem ⟹ t satisfies Weyl’s theorem holds, but the reverse implication in general fails [5, Example 4.1]. Operators d_{ab}, recall a, b^{∗} are hyponormal, satisfy generalized Weyl’s theorem.

Theorem 3.3. σ_{Bw}(d_{ab}) = σ(d_{ab}) \ E(d_{ab}). Furthermore, if f is analytic on a neighbourhood of σ(d_{ab}), then f(d_{ab}) satisfies generalized Weyl’s theorem.

Proof. Let λ ∈ σ(d_{ab}) \ σ_{Bw}(d_{ab}). Since d_{ab} has SVEP, it follows upon arguing as in the proof of [5, Theorem 3.12] and an application of Theorem (2.7) that λ ∈ isoσ(d_{ab}) = E(d_{ab}). Conversely, if λ ∈ E(d_{ab}), then d_{ab} − λ is Fredholm of index 0 ( by Theorems (2.6) and (2.7)). Hence d_{ab} satisfies generalized Weyl’s theorem.

Now let f be as in the statement of the theorem, and let σ_{D}(d_{ab}) = {λ ∈ C : (d_{ab} − λ) is not Drazin invertible } denote the Drazin spectrum of d_{ab}. Then σ_{D}(f(d_{ab})) = f(σ_{D}(d_{ab})) [3, Corollary 2.4]. Also, since d_{ab} and f(d_{ab}) have SVEP, σ_{D}(d_{ab}) = σ_{Bw}(d_{ab}) and σ_{D}(f(d_{ab})) = σ_{Bw}(f(d_{ab})) [5, Theorem 3.12]. Hence

f(σ_{Bw}(d_{ab})) = f(σ(d_{ab}) \ E(d_{ab})) = σ_{Bw}(f(d_{ab})).

The isoloid property of σ(d_{ab}), Theorem 2.7, now implies that

σ_{Bw}(f(d_{ab})) = σ(f(d_{ab})) \ E(f(d_{ab}))

[4, Lemma 2.9], and the proof is complete.

Acknowledgement. My thanks are due to Prof. Robin Harte for his many incisive comments during the preparation of this note. My thanks are also due to a referee for his thorough reading of the manuscript (and pointing out a number of typographical errors). It is my great pleasure to thank the organisers of the conference MAA5, especially Professors V. Rakočević and D. -Dor - dević, for their hospitality.

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(received 23.10.2002)

Department of Mathematics, College of Science, UAEU, P. O. Box 17551, Al Ain, Arab Emirates E-mail: bpduggal@uaeu.ac.ae.