KYUNGPOOK Math. J. 2019; 59(3): 415-431
Spectral Properties of k-quasi-class A(s,t) Operators
Salah Mecheri, Naim Latif Braha?
Taibah University, College of Science Department of Mathematics, P. O. Box 20003 Al Madinah Al Munawarah, Saudi Arabia
e-mail : mecherisalah@hotmail.com
Research Institute Ilirias, Rruga Janina, No-2, ferizaj, 70000, Kosovo Department of Mathematics and Computer Sciences, University of Prishtina, Avenue Mother Teresa, No-4, Prishtine, 10000, Kosova
e-mail : nbraha@yahoo.com
* Corresponding Author.
Received: March 13, 2016; Revised: October 11, 2018; Accepted: October 16, 2018; Published online: September 23, 2019.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

In this paper we introduce a new class of operators which will be called the class of k-quasi-class A(s, t) operators. An operator TB(H) is said to be k-quasi-class A(s, t) if $T*k((|T*|t|T|2s|T*|t)1t+s−|T*|2t)Tk≥0,$where s > 0, t > 0 and k is a natural number. We show that an algebraically k-quasi-class A(s, t) operator T is polaroid, has Bishop’s property β and we prove that Weyl type theorems for k-quasi-class A(s, t) operators. In particular, we prove that if T* is algebraically k-quasi-class A(s, t), then the generalized a-Weyl’s theorem holds for T. Using these results we show that T* satisfies generalized the Weyl’s theorem if and only if T satisfies the generalized Weyl’s theorem if and only if T satisfies Weyl’s theorem. We also examine the hyperinvariant subspace problem for k-quasi-class A(s, t) operators.

Introduction
roduction

Let B(H) be the algebra of all bounded linear operators acting on infinite dimensional separable complex Hilbert space H. An operator TB(H) is said to be p-hyponormal, for p ∈ (0, 1], if (T*T)p ≥ (TT*)p [2]. A 1-hyponormal operator is simply called hyponormal and $12$-hyponormal is simply called semi-hyponormal. An invertible operator T is said to be log-hyponormal if log |T| ≥ log |T*| [37]. An operator T is said to be paranormal if ||T2x|| ≥ ||Tx||2. It is known [22] that p-hyponormal and log-hyponormal operators are paranormal. An operator T belongs to class A(k) for k > 0 if $(T*|T|2kT)1k+1≥|T|2$. When k = 1 we say that T belongs to class A. Furuta et al. [19] showed that every class A operator is paranormal. As a further generalization of class A(k), Fujii et al. [20] introduced the class A(s, t): T belongs to the class A(s, t) for s > 0 and t > 0 if $(|T*|t|T|2s|T*|t)1t+s≥|T*|2t$. Class AI(s, t) is the class of all invertible class A(s, t) operators for s > 0 and t > 0. Fujii et al [20] showed several properties of class A(s, t) and class AI(s, t) as extensions of the properties of class A(k) shown in [20]. They also showed that T is log-hyponormal if and only if T belongs to class AI(s, t) for all s, t > 0. It is known [40] that class A(k, 1) equals class A(k).

Let T be an operator with polar decomposition T = U|T|, where $|T|=(T*T)12$. For s, t > 0, the generalized Aluthge transformation s,t of T, is s,t = |T|sU|T|t. If $s=t=12$, then s,t is called the Aluthge transformation of T denoted by [2]. The following equalities are relations between T and its transformation s,t$T˜s,t|T|s=|T|sU|T|t|T|s=|T|sT,U|T|tT˜s,t=U|T|t|T|sU|T|t=TU|T|t.$

Aluthge and Wang [3] introduced ω-hyponormal operators defined as follows: An operator T is said to be ω-hyponormal if || ≥ |T| ≥ |*|. In order to generalize the class of ω-hyponormal operators, Ito [22] introduced the class ωA(s, t). An operator T belongs to class ωA(s, t) for s, t > 0 if $(|T*|t|T|2s|T*|t)1s+t≥|T*|2t$and $|T|2s≥(|T|s|T*|2t|T|s)ss+t.$Ito [22] showed that ωA(s, t) operators can be characterized via generalized Aluthge transformation as follows:

An operator T belongs to the class ωA(s, t) for s, t > 0 if and only if $|T˜s,t|2ts+t≥|T|2t and |T|2s≥|T˜*|2ss+t.$

An operator T is ω-hyponormal if and only if T belongs to the class $ωA(12,12)$ [39]. An operator T is said to be of class ωF (s, t, q) for s, t > 0 and q ≥ 1 if $(|T*|t|T|2s|T*|t)1q≥|T*|2(s+t)q$and $|T|2(s+t)(1−1q)≥(|T|s|T*|2t|T|s)1−1q.$Because q and (1 − q−1)−1 with q > 1 are conjugate exponents, it is clear that the class ωA(s, t) equals the class $ωF(s,t,s+tt)$. We have the following inclusion of classes from [39] for all s > 0, t > 0: $A(s,t)⊇ωA(s,t)⊇AI(s,t).$

In order to generalize the class A(s, t) we introduce the class of k-quasi-class A(s, t) operators defined as follows:

### Definition 1.1

An operator TB(H) is said to be k-quasi-class A(s, t), or in kQCA(s, t) if $T*k((|T*|t|T|2s|T*|t)1t+s−|T*|2t)Tk≥0$, where s > 0, t > 0 and k is a natural number.

We have $A(s,t)⊇ωA(s,t)⊇AI(s,t)⊇kQCA(s,t)$for all s > 0, t > 0. The connection between k-quasi-class A(s, t) operators and the classes of operators mentioned above is given in Figure 1.

If TB(ℋ), we shall write ker(T) and R(T) for the null space and the range of T, respectively. Also, let σ(T) and σa(T) denote the spectrum and the approximate point spectrum of T, respectively. An operator T is called Fredholm if R(T) is closed, α(T) = dim ker(T) < ∞ and β(T) = dim ℋ/R(T) < ∞. Moreover if i(T) = α(T) − β(T) = 0, then T is called Weyl. The essential spectrum σe(T) and the Weyl spectrum σW (T) are defined by $σe(T)={λ∈ℂ:T−λ is not Fredholm}$and $σW(A)={λ∈ℂ:A−λ is not Weyl},$respectively. It is known that σe(T) ⊂ σW (T) ⊂ σe(T) ∪ acc σ(T) where we write acc K for the set of all accumulation points of K ⊂ ℂ. If we write iso K = Kacc K, then we let $π00(T)={λ∈iso σ(T):0<α(T−λ)<∞}.$We say that Weyl’s theorem holds for T if $σ(T)σW(T)=π00(T).$

In [38], H. Weyl proved that Weyl’s theorem holds for hermitian operators. Weyl’s theorem has been extended from hermitian operators to hyponormal operators [14], algebraically hyponormal operators [21], p-hyponormal operators [13] and quasi-*-class A [36].

More generally, M. Berkani investigated the generalized Weyl’s theorem which extends Weyl’s theorem, and proved that the generalized Weyl’s theorem holds for hyponormal operators ([6, 7, 8]). In a recent paper [25] the author showed that the generalized Weyl’s theorem holds for (p, k)-quasi-hyponormal operators.

M. Berkani also investigated B-Fredholm theory as follows (see [4, 6, 7, 8]). An operator T is called B-Fredholm if there exists n ∈ ℕ such that R(Tn) is closed and the induced operator $T[n]:R(Tn)∋x→Tx∈R(Tn)$is Fredholm, i.e., R(T[n]) = R(Tn+1) is closed, α(T[n]) = dim N(T[n]) < ∞ and β(T[n]) = dim R(Tn)/R(T[n]) < ∞. Similarly, a B-Fredholm operator T is called B-Weyl if i(T[n]) = 0. The following results is due to M. Berkani and M. Sarih [8].

### Proposition 1.1

Let T ∈ B(ℋ).

• If R(Tn) is closed and T[n]is Fredholm, then R(Tm) is closed and T[m]is Fredholm for every mn. Moreover, ind T[m] = ind T[n](= ind T).

• An operator T is B-Fredholm (B-Weyl) if and only if there exist T-invariant subspaces ℳandsuch thatwhere T|ℳ is Fredholm (Weyl) andis nilpotent.

The B-Weyl spectrum σBW (T) is defined by $σBW(T)={λ∈ℂ:T−λ is not B-Weyl}⊂σW(T).$We say that generalized Weyl’s theorem holds for T if $σ(T)σBW(T)=E(T)$where E(T) denotes the set of all isolated points of the spectrum which are eigenvalues (no restriction on multiplicity). Note that, if the generalized Weyl’s theorem holds for T, then so does Weyl’s theorem [7]. Recently in [6] M. Berkani and A. Arroud showed that if T is hyponormal, then generalized Weyl’s theorem holds for T.

We define $T∈SF+−$ if R(T) is closed, dim ker(T) < ∞ and ind T ≤ 0. Let $π00a(T)$ denote the set of all isolated points λ of σa(T) with 0 < dim ker(Tλ) < ∞. Let $σSF+−(T)={λ|T−λ∉SF+−}⊂σW(T)$. We say that a-Weyl’s theorem holds for T if $σa(T)σSF+−(T)=π00a(T).$V. Rakočević [33, Corollary 2.5] proved that if a-Weyl’s theorem holds for T, then Weyl’s theorem holds for T.

We define $T∈SBF+−$ if there exists a positive integer n such that R(Tn) is closed, T[n] : R(Tn) ∋ xTxR(Tn) is upper semi-Fredholm (i.e., R(T[n]) = R(Tn+1) is closed, dim ker(T[n]) = dim ker(T) ∩ R(Tn) < ∞) and 0 ≥ ind T[n](= ind T) [8]. We define $σSBF+−(T)={λ|T−λ∉SBF+−}⊂σSF+−(T)$. Let Ea(T) denote the set of all isolated points λ of σa(T) with 0 < dim ker(Tλ). We say that generalized a-Weyl’s theorem holds for T if $σa(T)σSBF+−(T)=Ea(T).$M. Berkani and J.J. Koliha [7] proved that if generalized a-Weyl’s theorem holds for T, then a-Weyl’s theorem holds for T.

An operator TB(H) is said to have the single-valued extension property (or SVEP) if for every open subset G of ℂ and any analytic function f : GH such that (Tz)f(z) ≡ 0 on G, we have f(z) ≡ 0 on G. For TB(H) and xH, the set ρT (x) is defined to consist of elements z0 ∈ ℂ such that there exists an analytic function f(z) defined in a neighborhood of z0, with values in H, which verifies (Tz)f(z) = x, and it is called the local resolvent set of T at x. We denote the complement of ρT (x) by σT (x), called the local spectrum of T at x, and define the local spectral subspace of T, HT (F) = {xH : σT (x) ⊂ F} for each subset F of ℂ. An operator TB(H) is said to have the property (β) if for every open subset G of ℂ and every sequence fn : GH of H-valued analytic functions such that (Tz)fn(z) converges uniformly to 0 in norm on compact subsets of G, fn(z) converges uniformly to 0 in norm on compact subsets of G. An operator TB(H) is said to have Dunford’s property (C) if HT (F) is closed for each closed subset F of ℂ. It is well known that $Property (β)⇒Dunford's property(C)⇒SVEP.$

SVEP is possessed by many important classes of operators such as hyponormal operators and decomposable operators. The interested reader is referred to [26, 27, 28, 29, 30] for more details. A closed subspace of H is said to be hyperinvariant for T if it is invariant under every operator in the commutant {T}′ of T. Knowledge of the hyperinvariant subspaces of T gives information on the structure of the commutant of T. The question of whether every operator on H has an hyperinvariant subspace is one of the most difficult problems in operator theory. Some partial solutions on this problem were given in the literature. Perhaps the most elegant results on the hyperinvariant subspace problem are the affirmative answers for non-scalar normal operators and for non-zero compact operators. It is known that every operator which commutes with a (nonzero) compact operator has a (proper closed) hyperinvariant subspace [31].

In this paper, we show that an algebraically k-quasi-class A(s, t) operator T is polaroid and we prove that Weyl type theorems hold for k-quasi-class A(s, t) operators. Especially we prove that if T* is an algebraically k-quasi-class A(s, t), then generalized a-Weyl’s theorem holds for T. Using these results we show that T* satisfies generalized Weyl’s theorem if and only if T satisfies generalized Weyl’s theorem if and only if T satisfies Weyl’s theorem. We also examine the hyperinvariant subspace problem for k-quasi-class A(s, t) operators.

Main Results
n Results

### Proposition 2.1

Let T be k-quasi-class A(s, t) operator for$t=12$and s > 0. Then the restriction T|to an invariant subspaceof T is also k-quasi-class A(s, t).

Proof

Let P be the projection of H onto M. Thus we can represent T as the following matrix with respect to the decomposition MM, $T=(ABOC).$

Put A = T|M and we have $(AOOO)=TP=PTP.$Since T is k-quasi-class A(s, t), we have $PT*k((|T*|t|T|2s|T*|t)1t+s−|T*|2t)TkP≥0.$We remark $PT*k|T*|2tTkP=PT*kP|T*|2tPTkP=(A*k(|A*|2t+|B*|2t)AkOOO)=(A*k|A*|2tAkOOO)+(A*k|B*|2tAkOOO)≥(A*k|A*|2tAkOOO)$and by Hansen inequality, we have $PT*k(|T*|t|T|2s|T*|t)1t+sTkP=PT*kP(|T*|t|T|2s|T*|t)1t+sPTkP≤PT*k(P|T*|t|T|2s|T*|tP)1t+sTkP=PT*k(P|T*|tP|T|2sP|T*|tP)1t+sTkP≤PT*k(P|T*|tP(PTT*P)sP|T*|tP)1t+sTkP=(A*k(|A*|t|A|2s|A*|t)1t+sAkOOO).$Thus, $(A*k(|A*|t|A|2s|A*|t)1t+sAkOOO)≥PT*k(|T*|t|T|2s|T*|t)1t+sTkP≥PT*k|T*|2tTkP≥(A*k|A*|2tAkOOO).$Hence, A is k-quasi-class A(s, t) operator on M.

### Lemma 2.1

([39]) Let A, B and C be positive operators, 0 < p and 0 < r ≤ 1. If$(Br2ApBr2)rp+r≥Br$and BC, then$(Cr2ApCr2)rp+r≥Cr$.

### Lemma 2.2

Let T be k-quasi-class A(s, t) operator for$t=12$and s > 0. Let R(Tk) be not dense. Decompose$T=(T1T20T3) on ℋ=R(Tk)¯⊕ker(T*k)$where$R(Tk)¯$is closure of R(Tk). Then T1is in A(s, t), $T3k=0$, and σ(T) = σ(T1) ∪ {0}.

Proof

Suppose TB(H) is k–quasi-class A(s, t) for $t=12$ and s > 0. If R(Tk) is not dense and T has representation $T=(T1T20T3)$on H = [R(Tk)] ⊕ [ker(T*k)], then T1 = TP = PTP, where P is the projection onto R(Tk).

By Hansen’s inequality, we have $|T1|2s=(P|T|2P)2≥P|T|2sP$and $|T*|2=TT*≥TPT*=|T1|2$T is k–quasi-class A(s, t) for $t=12$ and s > 0 if and only if $P(|T*|t|T|2s|T*|t)ts+tP≥P|T*|2tP$By Hansen’s inequality, (2.3) becomes, $(P|T*|t|T|2s|T*|tP)ts+t≥P|T*|2tP$ which implies, $(P|T*|tP|T|2sP|T*|tP)ts+t≥P|T*|2tP$Since $t=12$, the above inequality can be written as $((P|T*|P)t|T|2s(P|T*|P)t)ts+t≥(P|T*|P)2t$Then from (2.2) and applying Lemma 2.2, (2.4) becomes $((P|T1*|P)t|T|2s(P|T1*|P)t)ts+t≥(P|T1*|P)2t$Since $|T1*|tP=P|T1*|t=|T1*|t$, the above inequality changes to $(|T1*|t|T|2s|T1*|t)ts+t≥|T1*|2t$Then by (2.1) we have the following: $(|T1*|t|T1|2s|T1*|t)ts+t≥|T1*|2t.$Thus, T1 is in A(s, t) for $t=12$ and s > 0.

Let x ∈ ker(T*k). Then by simple calculations we have Tkx = 0 and so $T3kx=0$. Further, we have $〈T3kx2,y2〉=〈Tk(I−P)x,(I−P)y〉=〈(I−P)x,T*k(I−P)y〉=0,$for any $x=(x1x2)$, $y=(y1y2)∈H$. Thus $T3*k=0$. Since σ(T3) = {0}, we have σ(T) = σ(T1) ∪ {0}.

### Lemma 2.3

Let T be k-quasi-class A(s, t) for$t=12$and 0 < s < 1. Then T has Bishop’s property β. Hence T has SVEP.

Proof

Suppose TB(H) be k–quasi-class A(s, t) for $t=12$ and s ∈ (0, 1]. If R(Tk) is dense, then T is class A(s, t) for $t=12$ and s ∈ (0, 1], so T has Bishop’s property β [35]. Suppose that R(Tk) is not dense. From Lemma 2.2, we write the matrix representation of T on H = [R(Tk)] ⊕ [kerT*k] as follows. $T=(T1T20T3).$where T1 = T | R(Tk) is in A(s, t) for $t=12$ and s > 0, and T3 is nilpotent.

Let fn(µ) be analytic on with (Tµ)fn(µ) → 0 uniformly on every compact subset of . Then $(T1−μT20T3−μ)(fn1(μ)fn2(μ))=((T1−μ)fn1(μ)+T2fn2(μ)(T3−μ)fn2(μ))→0$Since T3 is nilpotent, T3 satisfies Bishop property β. Thus, fn2(µ) → 0 uniformly on every compact subset of . Thus, (T1µ)fn1 → 0. Since T1 is in A(s, t) with $t=12$and s ∈ (0, 1], T1 satisfies Bishop property (β) [35]. Thus, fn1(µ) → 0 and so T has Bishop’s property β.

### Corollary 2.1

Let TB(H) be a k-quasi-class A(s, t) with thick spectra, then T has a nontrivial invariant subspace

Proof

Since T has Bishop’s property, it suffices to apply [18].

An operator T is said to be polaroid if points in iso(σ(T)) are poles of the resolvent of T. Recall that if T is algebraically polaroid, i.e., p(T) is polaroid for some non-constant polynomial p, then T is polaroid [15]. We will show that an algebraically k-quasi-class A(s, t) operator is polaroid. For this we need the following lemmas.

### Lemma 2.4

Assume that T is quasinilpotent k-quasi-class A(s, t). Then T is nilpotent.

Proof

Assume that T is k-quasi-class A(s, t). We consider two cases:

• Case 1: Suppose Tk has dense range. It follows that T is class A(s, t). Since a class A(s, t) operator is normaloid [20, 39], every quasinilpotent class A(s, t) operator is the zero operator. Hence T is nilpotent.

• Case 2: Suppose Tk does not have dense range. Then by Lemma 2.1, T can be represented on $ℋ=R(Tk)¯⊕kerT*k$ as $T=(T1T20T3).$So T1 is in A(s, t), $T3k=0$ and σ(T) = σ(T1) ∪ {0}. Since T is quasinilpotent, σ(T) = {0}. But σ(T) = σ(T1) ∪ {0}. Since T1 is in A(s, t), T1 = 0, so T is nilpotent.

### Lemma 2.5

Let TB(H) be an algebraically k-quasi-class A(s, t) operator, and σ(T) = {µ0}, then Tµ0is nilpotent.

Proof

Assume p(T) is k-quasi-class A(s, t) for some nonconstant polynomial p(z). Since σ(p(T)) = p(σ(T)) = {p(µ0)}, the operator p(T) − p(µ0) is nilpotent by Lemma 2.4. Let $p(z)−p(μ0)=a(z−μ0)k0(z−μ1)k1⋯(z−μt)kt,$where µjµs for js. Then $0={p(T)−p(μ0)}m=am(T−μ0)mk0(T−μ1)mk1⋯(T−μt)mkt$and hence (Tµ0)mk0 = 0.

In the following theorem we will prove that an algebraically k-quasi-class A(s, t) operator is polaroid.

### Theorem 2.1

Let T be an algebraically k-quasi-class A(s, t) operator. Then T is polaroid.

Proof

If T is an algebraically k-quasi-class A(s, t) operator. Then p(T) is a k-quasi-class A(s, t) operator for some nonconstant polynomial p. Let µiso(σ(T)) and let Eµ be the Riesz idempotent associated to µ defined by $E:=12πi∫∂D(μI−T)−1dμ,$where D is a closed disk centered at µ which contains no other points of the spectrum of T. Then T can be represented as follows $(T100T2),$where σ(T1) = {µ} and σ(T2) = σ(T2) {µ}. Since T1 is an algebraically k-quasi-class A(s, t) operator by Proposition 2.1 and σ(T1) = µ, it follows from Lemma 2.4 that T1λI is nilpotent. Therefore T1µ has finite ascent and descent. On the other hand, since T2µI is invertible, it has finite ascent and descent. Therefore TµI has finite ascent and descent. Therefore µ is a pole of the resolvent of T. Now if µiso(σ(T)), then µπ(T). Thus iso(σ(T)) ∈ π(T), where π(T) denote the set of poles of the resolvent of T. Hence T is polaroid.

### Corollary 2.2

A k-quasi-class A(s, t) operator is isoloid.

If a Banach space operator T has SVEP (everywhere), the single-valued extension property, then T and T* satisfy Browder’s (equivalently, generalized Browder’s) theorem and a-Browder’s (equivalently, generalized a-Browder’s) theorem. A sufficient condition for an operator T satisfying Browder’s (generalized Browder’s) theorem to satisfy Weyl’s (resp., generalized Weyl’s) theorem is that T is polaroid. Observe that if TB(H) has SVEP, then σ(T) = σa(T*). Hence, if T has SVEP and is polaroid, then T* satisfies generalized a-Weyl’s (so also, a-Weyl’s) theorem [5].

### Theorem 2.2

Let TB(H).

• If T*is an algebraically k-quasi-class A(s, t) operator, then generalized a-Weyl’s theorem holds for T.

• If T is an algebraically k-quasi-class A(s, t) operator, then generalized a-Weyl’s theorem holds for T*.

Proof

(i) It is well known that T is polaroid if and only if T* is polaroid [5, Theorem 2.11]. Now since an algebraically k-quasi-class A(s, t) operator is polaroid by Lemma 2.2 and has SVEP by Theorem 2.1, [5, Theorem 3.10] gives us the result of the theorem. For (ii) we can also apply [5, Theorem 3.10].

### Corollary 2.3

If T is algebraically k-quasi-class A(s, t), then the following statements are equivalent.

• generalized Weyl’s theorem holds for T*.

• generalized Weyl’s theorem holds for T.

• Weyl’s theorem holds for T.

• Weyl’s theorem holds for T*.

Proof

If T is algebraically k-quasi-class A(s, t), then T is polaroid. Now, observe for polaroid operators T satisfying generalized Weyl’s theorem, $E(T)=π(T)=π(T*)=E(T*),$where π(T) is the set of poles of the resolvent of T. Hence, for a polaroid operator T, T* satisfies generalized Weyl’s theorem if and only if T satisfies generalized Weyl’s theorem if and only if T satisfies Weyl’s theorem if and only if T* satisfies Weyl’s theorem.

### Remark 2.1

• Recall [5] that if T is polaroid, then T satisfies generalized Weyl’s theorem (resp. generalized a-Weyl’s) theorem if and only if T satisfies Weyl’s theorem (resp. a-Weyl’s theorem). Hence if T is an algebraically k-quasi-class A(s, t) operator, the above equivalences hold.

• Let f(z) be an analytic function on σ(T). If T is polaroid, then f(T) is polaroid too [5].

• If T* is algebraically k-quasi-paranormal, then f(T) satisfies generalized a-Weyl’s theorem. Indeed, since T* is polaroid, the result holds by [5, Theorem 3.12]

• If T is algebraically k-quasi-class A(s, t), then f(T*) satisfies generalized a-Weyl’s theorem. Indeed, since T is polaroid, the result holds by [5, Theorem 3.12].

Since dA,B is polaroid and has SVEP, we get the following theorems for dA,B

### Theorem 2.3

If A, B*are k-quasi-class A(s, t) operators, then the following statements are equivalent.

• generalized Weyl’s theorem holds for$dA,B*$.

• generalized Weyl’s theorem holds for dA,B.

• Weyl’s theorem holds for dA,B.

### Theorem 2.4

Let A, BB(H).

• If A, B*are k-quasi-class A(s, t) operators, then generalized a-Weyl’s theorem holds for dA,B.

• If A, B*are k-quasi-class A(s, t) operators, then generalized a-Weyl’s theorem holds for$dA,B*$.

### Theorem 2.5

If A, B*are k-quasi-class A(s, t) operators, then

• Weyl’s theorem holds for dA,Bif and only if generalized Weyl’s theorem holds for dA,B.

• a-Weyl’s theorem holds for dA,Bif and only if generalized a-Weyl’s theorem holds for d2A,B.

### Corollary 2.4

Let A, BB(H).

• If A, B*are k-quasi-class A(s, t) operators, then Weyl’s theorem, a-Weyl’s theorem, generalized Weyl’s theorem and generalized a-Weyl’s theorem hold for dA,Band are equivalent.

• If A, B*are k-quasi-class A(s, t) operators, then Weyl’s theorem, a-Weyl’s theorem, generalized Weyl’s theorem and generalized a-Weyl’s theorem hold for$dA,B*$and are equivalent.

Let fHol(σ(T)), where Hol(σ(T)) is the space of all functions that are analytic in an open neighborhoods of σ(T). If dA,B is polaroid, then f(dA,B) is also polaroid [5]. Thus we have

### Theorem 2.6

Let A, B*B(H)

• If A, B*are k-quasi-class A(s, t) operators, then f(dA,B) satisfies Weyl’s theorem, a-Weyl’s theorem, generalized Weyl’s theorem and generalized a-Weyl’s theorem.

• If A, B*are k-quasi-class A(s, t) operators, then$f(dA,B*)$satisfies Weyl’s theorem, a-Weyl’s theorem, generalized Weyl’s theorem and generalized a-Weyl’s theorem.

Corollary 2.16 may be extended as follows.

### Theorem 2.7

Let A, BB(H).

• If A, B*are k-quasi-class A(s, t) operators, then Weyl’s theorem, a-Weyl’s theorem, generalized Weyl’s theorem and generalized a-Weyl’s theorem hold for f(dA,B) and are equivalent.

• If A, B*are k-quasi-class A(s, t) operators, then Weyl’s theorem, a-Weyl’s theorem, generalized Weyl’s theorem and generalized a-Weyl’s theorem hold for$f(dA,B*)$and are equivalent.

### Remark 2.2

• Since a quasi-class A operator is k-quasi-class A(s, t), hence all Weyl’s theorems (generalized or not) hold for algebraically quasi-class A operators and are equivalent by Theorem 2.8, Corollary 2.9 and Remark 2.10. This subsumes and extends [1, Theorem 2.4 and Theorem 3.3].

• Since a class A(s, t) operator is k-quasi-class A(s, t), hence Theorem 2.8, Corollary 2.9 and Remark 2.10 generalize the results on Weyl type theorems for class A(s, t) operators proved in [11, 41].

• Our results on Weyl type theorems for dA,B generalize a recent result on generalized Weyl’s theorem for dA,B when A and B* are class A operators [24]. Also, since a quasi-class A operators is a k-quasi-class A(s, t), hence all Weyl’s theorems (generalized or not) hold for algebraically quasi-class A and algebraically w-hyponormal operators and are equivalent by Theorem 2.15, Theorem 2.17 and Theorem 2.18. This subsumes and extends [12, 16, 17]

Hyperinvariant Subspaces
Subspaces

The purpose of this section is to make a beginning on the hyperinvariant subspace problem for another class of operators closely related to the normal operators namely, the class of k-quasi-paranormal operators.

### Theorem 3.1

Let$T=(T1T20T3)∈B(H1⊕H2).$If T has bishop’s property β and there exists a non zero xH1H2such that$σT(x)⊂≠σ(T)$. Then T has a nontrivial hyperinvariant subspace.

Proof

Assume that $ℳ={y∈H⊕H:σT(y)⊆σT(x)},$that is, ℳ = HT (σT (x)). Since T has Bishop’s property β, hence T has Dunford’s property (C). It follows from [10] that ℳ is a T-hyperinvariant subspace. Since x ∈ ℳ, we have ℳ ≠ {0}. Now, set ℳ = H1H2. Since T has the single extension property, we get $σ(T)=∪{σT(y):y∈H1⊕H2}⊆σT(x)⊂≠σ(T)$ from [23]. This is a contradiction. Hence ℳ is a nontrivial T-hyperinvariant subspace.

Since a k-quasi-class A(s, t) operator has property β by applying Lemma 2.1 and Theorem 3.1, we get the following corollary.

### Corollary 3.1

Let$T=(T1T20T3)∈B(H⊕H).$be a k-quasi-class A(s, t) operator. If there exists a nonzero xHH such that$σT(x)⊂≠σ(T)$, then T has a nontrivial hyperinvariant subspace.

It is remarkable that simply knowing when solutions to AXXB = Y exist gives striking results on many topics, including similarity, commutativity, hyperinvariant subspaces, spectral operators and differential equations for details (see [9, 32]). Some of these are discussed below.

### Lemma 3.1

([34, Rosemblum Theorem]) Let S, TB(H). If the spectrum of S and T are disjoint, then the operator equation TX − XS = R has a unique solution for every operator X.

It is known that an invariant subspace for an operator T may not be hyperinvariant. However, a sufficient condition that an invariant subspace be hyperinvariant can be derived from the Rosemblum theorem. If a subspace ℳ of the Hilbert space H is invariant under T, then with respect to the orthogonal decomposition H = ℳ ⊕ ℳ, then the operator T can be decomposed as follows: $T=(T1T20T3).$Thus we have the following theorem.

### Theorem 3.2

Let$T=(T1T20T3), S=(S1S2S3S4).$If the spectrum of T1and T2are disjoint and ST = TS, then S4 = 0.

Proof

It is clear that if ST = TS, then T3S4 = S4T1. Hence it follows from the Rosemblum theorem that S4 = 0.

Thus ℳ is a nontrivial hyperinvariant subspace for T if the spectrum of T1 and the spectrum of T3 are disjoint. In particular we have.

### Corollary 3.2

Let$T=(T1T20T3)$be k-quasi-class A(s, t). If 0 ∉ σ(T1), then T has a non trivial hyperinvariant subspace.

Figures
Fig. 1. Relationship of classes of operators
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