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 T ∈ B(H) is said to be k-quasi-class A(s, t) if $${T}^{*k}\left({\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}-{\left|{T}^{*}\right|}^{2t}\right){T}^{k}\ge 0,$$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 T ∈ B(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 $\frac{1}{2}$-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 ||T^{2}x|| ≥ ||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 ${\left({T}^{*}{\left|T\right|}^{2k}T\right)}^{\frac{1}{k+1}}\ge {\left|T\right|}^{2}$. 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 ${\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}\ge {\left|{T}^{*}\right|}^{2t}$. 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 $\left|T\right|={\left({T}^{*}T\right)}^{\frac{1}{2}}$. For s, t > 0, the generalized Aluthge transformation T̃_{s,t} of T, is T̃_{s,t} = |T|^{s}U|T|^{t}. If $s=t=\frac{1}{2}$, then T̃_{s,t} is called the Aluthge transformation of T denoted by T̃ [2]. The following equalities are relations between T and its transformation T̃_{s,t}$$\begin{array}{c}{\tilde{T}}_{s,t}{\left|T\right|}^{s}={\left|T\right|}^{s}U{\left|T\right|}^{t}{\left|T\right|}^{s}={\left|T\right|}^{s}T,\\ U{\left|T\right|}^{t}{\tilde{T}}_{s,t}=U{\left|T\right|}^{t}{\left|T\right|}^{s}U{\left|T\right|}^{t}=TU{\left|T\right|}^{t}.\end{array}$$
Aluthge and Wang [3] introduced ω-hyponormal operators defined as follows: An operator T is said to be ω-hyponormal if |T̃| ≥ |T| ≥ |T̃^{*}|. 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 $${\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{s+t}}\ge {\left|{T}^{*}\right|}^{2t}$$and $${\left|T\right|}^{2s}\ge {\left({\left|T\right|}^{s}{\left|{T}^{*}\right|}^{2t}{\left|T\right|}^{s}\right)}^{\frac{s}{s+t}}.$$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 $${\left|{\tilde{T}}_{s,t}\right|}^{\frac{2t}{s+t}}\ge {\left|T\right|}^{2t}\hspace{0.17em}\hspace{0.17em}\text{and}\hspace{0.17em}\hspace{0.17em}{\left|T\right|}^{2s}\ge {\left|{\tilde{T}}^{*}\right|}^{\frac{2s}{s+t}}.$$
An operator T is ω-hyponormal if and only if T belongs to the class $\omega A\left(\frac{1}{2},\frac{1}{2}\right)$ [39]. An operator T is said to be of class ωF (s, t, q) for s, t > 0 and q ≥ 1 if $${\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{q}}\ge {\left|{T}^{*}\right|}^{\frac{2(s+t)}{q}}$$and $${\left|T\right|}^{2(s+t)\left(1-\frac{1}{q}\right)}\ge {\left({\left|T\right|}^{s}{\left|{T}^{*}\right|}^{2t}{\left|T\right|}^{s}\right)}^{1-\frac{1}{q}}.$$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 $\omega F\left(s,t,\frac{s+t}{t}\right)$. We have the following inclusion of classes from [39] for all s > 0, t > 0: $$A(s,t)\supseteq \omega A(s,t)\supseteq AI(s,t).$$
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 T ∈ B(H) is said to be k-quasi-class A(s, t), or in k_{QC}A(s, t) if ${T}^{*k}\left({\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}-{\left|{T}^{*}\right|}^{2t}\right){T}^{k}\ge 0$, where s > 0, t > 0 and k is a natural number.
We have $$A(s,t)\supseteq \omega A(s,t)\supseteq AI(s,t)\supseteq {k}_{QC}A(s,t)$$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 T ∈ B(ℋ), we shall write ker(T) and R(T) for the null space and the range of T, respectively. Also, let σ(T) and σ_{a}(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 σ_{e}(T) and the Weyl spectrum σ_{W} (T) are defined by $${\sigma}_{e}(T)=\left\{\lambda \in \u2102:T-\lambda \hspace{0.17em}\text{is}\hspace{0.17em}\text{not}\hspace{0.17em}\text{Fredholm}\right\}$$and $${\sigma}_{W}(A)=\left\{\lambda \in \u2102:A-\lambda \hspace{0.17em}\text{is}\hspace{0.17em}\text{not}\hspace{0.17em}\text{Weyl}\right\},$$respectively. It is known that σ_{e}(T) ⊂ σ_{W} (T) ⊂ σ_{e}(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 $${\pi}_{00}(T)=\left\{\lambda \in \text{iso}\hspace{0.17em}\sigma (T):0<\alpha (T-\lambda )<\infty \right\}.$$We say that Weyl’s theorem holds for T if $$\sigma (T){\sigma}_{W}(T)={\pi}_{00}(T).$$
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(T^{n}) is closed and the induced operator $${T}_{[n]}:R({T}^{n})\ni x\to Tx\in R({T}^{n})$$is Fredholm, i.e., R(T_{[n]}) = R(T^{n+1}) is closed, α(T_{[n]}) = dim N(T_{[n]}) < ∞ and β(T_{[n]}) = dim R(T^{n})/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(T^{n}) is closed and T_{[n]}is Fredholm, then R(T^{m}) is closed and T_{[m]}is Fredholm for every m ≥ n. 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 σ_{BW} (T) is defined by $${\sigma}_{BW}(T)=\left\{\lambda \in \u2102:T-\lambda \hspace{0.17em}\text{is}\hspace{0.17em}\text{not}\hspace{0.17em}\text{B-Weyl}\right\}\subset {\sigma}_{W}(T).$$We say that generalized Weyl’s theorem holds for T if $$\sigma (T){\sigma}_{BW}(T)=E(T)$$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 $T\in {\mathit{SF}}_{+}^{-}$ if R(T) is closed, dim ker(T) < ∞ and ind T ≤ 0. Let ${\pi}_{00}^{a}(T)$ denote the set of all isolated points λ of σ_{a}(T) with 0 < dim ker(T − λ) < ∞. Let ${\sigma}_{{\mathit{SF}}_{+}^{-}}(T)=\left\{\lambda |T-\lambda \notin {\mathit{SF}}_{+}^{-}\right\}\subset {\sigma}_{W}(T)$. We say that a-Weyl’s theorem holds for T if $${\sigma}_{a}(T){\sigma}_{{\mathit{SF}}_{+}^{-}}(T)={\pi}_{00}^{a}(T).$$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 $T\in {\mathit{SBF}}_{+}^{-}$ if there exists a positive integer n such that R(T^{n}) is closed, T_{[n]} : R(T^{n}) ∋ x → Tx ∈ R(T^{n}) is upper semi-Fredholm (i.e., R(T_{[n]}) = R(T^{n+1}) is closed, dim ker(T_{[n]}) = dim ker(T) ∩ R(T^{n}) < ∞) and 0 ≥ ind T_{[n]}(= ind T) [8]. We define ${\sigma}_{{\mathit{SBF}}_{+}^{-}}(T)=\left\{\lambda |T-\lambda \notin {\mathit{SBF}}_{+}^{-}\right\}\subset {\sigma}_{{\mathit{SF}}_{+}^{-}}(T)$. Let E^{a}(T) denote the set of all isolated points λ of σ_{a}(T) with 0 < dim ker(T − λ). We say that generalized a-Weyl’s theorem holds for T if $${\sigma}_{a}(T){\sigma}_{SB{F}_{+}^{-}}(T)={E}^{a}(T).$$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 T ∈ B(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 : G → H such that (T − z)f(z) ≡ 0 on G, we have f(z) ≡ 0 on G. For T ∈ B(H) and x ∈ H, the set ρ_{T} (x) is defined to consist of elements z_{0} ∈ ℂ such that there exists an analytic function f(z) defined in a neighborhood of z_{0}, with values in H, which verifies (T − z)f(z) = x, and it is called the local resolvent set of T at x. We denote the complement of ρ_{T} (x) by σ_{T} (x), called the local spectrum of T at x, and define the local spectral subspace of T, H_{T} (F) = {x ∈ H : σ_{T} (x) ⊂ F} for each subset F of ℂ. An operator T ∈ B(H) is said to have the property (β) if for every open subset G of ℂ and every sequence f_{n} : G → H of H-valued analytic functions such that (T − z)f_{n}(z) converges uniformly to 0 in norm on compact subsets of G, f_{n}(z) converges uniformly to 0 in norm on compact subsets of G. An operator T ∈ B(H) is said to have Dunford’s property (C) if H_{T} (F) is closed for each closed subset F of ℂ. It is well known that $$\text{Property}\hspace{0.17em}(\beta )\Rightarrow \text{Dunford's}\hspace{0.17em}\text{property}(C)\Rightarrow \text{SVEP}.$$
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$t=\frac{1}{2}$and s > 0. Then the restriction T|_{ℳ}to an invariant subspace ℳ of 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 M ⊕ M^{⊥}, $$T=\left(\begin{array}{cc}A& B\\ O& C\end{array}\right).$$
Put A = T|_{M} and we have $$\left(\begin{array}{cc}A& O\\ O& O\end{array}\right)=TP=PTP.$$Since T is k-quasi-class A(s, t), we have $$P{T}^{*k}\left({\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}-{\left|{T}^{*}\right|}^{2t}\right){T}^{k}P\ge 0.$$We remark $$\begin{array}{ll}P{T}^{*k}{\left|{T}^{*}\right|}^{2t}{T}^{k}P\hfill & =P{T}^{*k}P{\left|{T}^{*}\right|}^{2t}P{T}^{k}P\hfill \\ \hfill & =\left(\begin{array}{cc}{A}^{*k}\left({\left|{A}^{*}\right|}^{2t}+{\left|{B}^{*}\right|}^{2t}\right){A}^{k}& O\\ O& O\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{A}^{*k}{\left|{A}^{*}\right|}^{2t}{A}^{k}& O\\ O& O\end{array}\right)+\left(\begin{array}{cc}{A}^{*k}{\left|{B}^{*}\right|}^{2t}{A}^{k}& O\\ O& O\end{array}\right)\hfill \\ \hfill & \ge \left(\begin{array}{cc}{A}^{*k}{\left|{A}^{*}\right|}^{2t}{A}^{k}& O\\ O& O\end{array}\right)\hfill \end{array}$$and by Hansen inequality, we have $$\begin{array}{cc}P{T}^{*k}{\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}{T}^{k}P& =P{T}^{*k}P{\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}P{T}^{k}P\\ & \le P{T}^{*k}{\left(P{\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}P\right)}^{\frac{1}{t+s}}{T}^{k}P\\ & =P{T}^{*k}{\left(P{\left|{T}^{*}\right|}^{t}P{\left|T\right|}^{2s}P{\left|{T}^{*}\right|}^{t}P\right)}^{\frac{1}{t+s}}{T}^{k}P\\ & \le P{T}^{*k}{\left(P{\left|{T}^{*}\right|}^{t}P{\left(PT{T}^{*}P\right)}^{s}P{\left|{T}^{*}\right|}^{t}P\right)}^{\frac{1}{t+s}}{T}^{k}P\\ & =\left(\begin{array}{cc}{A}^{*k}{\left({\left|{A}^{*}\right|}^{t}{\left|A\right|}^{2s}{\left|{A}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}{A}^{k}& O\\ O& O\end{array}\right).\end{array}$$Thus, $$\begin{array}{cc}\left(\begin{array}{cc}{A}^{*k}{\left({\left|{A}^{*}\right|}^{t}{\left|A\right|}^{2s}{\left|{A}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}{A}^{k}& O\\ O& O\end{array}\right)& \ge P{T}^{*k}{\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{1}{t+s}}{T}^{k}P\\ & \ge P{T}^{*k}{\left|{T}^{*}\right|}^{2t}{T}^{k}P\\ & \ge \left(\begin{array}{cc}{A}^{*k}{\left|{A}^{*}\right|}^{2t}{A}^{k}& O\\ O& O\end{array}\right).\end{array}$$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${\left({B}^{\frac{r}{2}}{A}^{p}{B}^{\frac{r}{2}}\right)}^{\frac{r}{p+r}}\ge {B}^{r}$and B ≥ C, then${\left({C}^{\frac{r}{2}}{A}^{p}{C}^{\frac{r}{2}}\right)}^{\frac{r}{p+r}}\ge {C}^{r}$.
Lemma 2.2
Let T be k-quasi-class A(s, t) operator for$t=\frac{1}{2}$and s > 0. Let R(T^{k}) be not dense. Decompose$$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{on}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathscr{H}=\overline{R({T}^{k})}\oplus \text{ker}\left({T}^{*k}\right)$$where$\overline{R({T}^{k})}$is closure of R(T^{k}). Then T_{1}is in A(s, t), ${T}_{3}^{k}=0$, and σ(T) = σ(T_{1}) ∪ {0}.
Proof
Suppose T ∈ B(H) is k–quasi-class A(s, t) for $t=\frac{1}{2}$ and s > 0. If R(T^{k}) is not dense and T has representation $$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right)$$on H = [R(T^{k})] ⊕ [ker(T^{*k})], then T_{1} = TP = PTP, where P is the projection onto R(T^{k}).
By Hansen’s inequality, we have $${\left|{T}_{1}\right|}^{2s}={\left(P{\left|T\right|}^{2}P\right)}^{2}\ge P{\left|T\right|}^{2s}P$$and $${\left|{T}^{*}\right|}^{2}=T{T}^{*}\ge TP{T}^{*}={\left|{T}_{1}\right|}^{2}$$T is k–quasi-class A(s, t) for $t=\frac{1}{2}$ and s > 0 if and only if $$P{\left({\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}\right)}^{\frac{t}{s+t}}P\ge P{\left|{T}^{*}\right|}^{2t}P$$By Hansen’s inequality, (2.3) becomes, ${\left(P{\left|{T}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}^{*}\right|}^{t}P\right)}^{\frac{t}{s+t}}\ge P{\left|{T}^{*}\right|}^{2t}P$ which implies, $${\left(P{\left|{T}^{*}\right|}^{t}P{\left|T\right|}^{2s}P{\left|{T}^{*}\right|}^{t}P\right)}^{\frac{t}{s+t}}\ge P{\left|{T}^{*}\right|}^{2t}P$$Since $t=\frac{1}{2}$, the above inequality can be written as $${\left({\left(P\left|{T}^{*}\right|P\right)}^{t}{\left|T\right|}^{2s}{\left(P\left|{T}^{*}\right|P\right)}^{t}\right)}^{\frac{t}{s+t}}\ge {\left(P\left|{T}^{*}\right|P\right)}^{2t}$$Then from (2.2) and applying Lemma 2.2, (2.4) becomes $${\left({\left(P\left|{T}_{1}^{*}\right|P\right)}^{t}{\left|T\right|}^{2s}{\left(P\left|{T}_{1}^{*}\right|P\right)}^{t}\right)}^{\frac{t}{s+t}}\ge {\left(P\left|{T}_{1}^{*}\right|P\right)}^{2t}$$Since ${\left|{T}_{1}^{*}\right|}^{t}P=P{\left|{T}_{1}^{*}\right|}^{t}={\left|{T}_{1}^{*}\right|}^{t}$, the above inequality changes to $${\left({\left|{T}_{1}^{*}\right|}^{t}{\left|T\right|}^{2s}{\left|{T}_{1}^{*}\right|}^{t}\right)}^{\frac{t}{s+t}}\ge {\left|{T}_{1}^{*}\right|}^{2t}$$Then by (2.1) we have the following: $${\left({\left|{T}_{1}^{*}\right|}^{t}{\left|{T}_{1}\right|}^{2s}{\left|{T}_{1}^{*}\right|}^{t}\right)}^{\frac{t}{s+t}}\ge {\left|{T}_{1}^{*}\right|}^{2t}.$$Thus, T_{1} is in A(s, t) for $t=\frac{1}{2}$ and s > 0.
Let x ∈ ker(T^{*}^{k}). Then by simple calculations we have T^{k}x = 0 and so ${T}_{3}^{k}x=0$. Further, we have $$\langle {T}_{3}^{k}{x}_{2},{y}_{2}\rangle =\langle {T}^{k}(I-P)x,(I-P)y\rangle =\langle (I-P)x,{T}^{*k}(I-P)y\rangle =0,$$for any $x=\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)$, $y=\left(\begin{array}{c}{y}_{1}\\ {y}_{2}\end{array}\right)\in H$. Thus ${T}_{3}^{*k}=0$. Since σ(T_{3}) = {0}, we have σ(T) = σ(T_{1}) ∪ {0}.
Lemma 2.3
Let T be k-quasi-class A(s, t) for$t=\frac{1}{2}$and 0 < s < 1. Then T has Bishop’s property β. Hence T has SVEP.
Proof
Suppose T ∈ B(H) be k–quasi-class A(s, t) for $t=\frac{1}{2}$ and s ∈ (0, 1]. If R(T^{k}) is dense, then T is class A(s, t) for $t=\frac{1}{2}$ and s ∈ (0, 1], so T has Bishop’s property β [35]. Suppose that R(T^{k}) is not dense. From Lemma 2.2, we write the matrix representation of T on H = [R(T^{k})] ⊕ [kerT^{*k}] as follows. $$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right).$$where T_{1} = T | R(T^{k}) is in A(s, t) for $t=\frac{1}{2}$ and s > 0, and T_{3} is nilpotent.
Let f_{n}(µ) be analytic on with (T − µ)f_{n}(µ) → 0 uniformly on every compact subset of . Then $$\left(\begin{array}{cc}{T}_{1}-\mu & {T}_{2}\\ 0& {T}_{3}-\mu \end{array}\right)\left(\begin{array}{c}{f}_{{n}_{1}}(\mu )\\ {f}_{{n}_{2}}(\mu )\end{array}\right)=\left(\begin{array}{c}({T}_{1}-\mu ){f}_{{n}_{1}}(\mu )+{T}_{2}{f}_{{n}_{2}}(\mu )\\ ({T}_{3}-\mu ){f}_{{n}_{2}}(\mu )\end{array}\right)\to 0$$Since T_{3} is nilpotent, T_{3} satisfies Bishop property β. Thus, f_{n2}(µ) → 0 uniformly on every compact subset of . Thus, (T_{1} − µ)f_{n1} → 0. Since T_{1} is in A(s, t) with $t=\frac{1}{2}$and s ∈ (0, 1], T_{1} satisfies Bishop property (β) [35]. Thus, f_{n1}(µ) → 0 and so T has Bishop’s property β.
Corollary 2.1
Let T ∈ B(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 T^{k} 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 T^{k} does not have dense range. Then by Lemma 2.1, T can be represented on $\mathscr{H}=\overline{R({T}^{k})}\oplus \text{ker}{T}^{*k}$ as $$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right).$$So T_{1} is in A(s, t), ${T}_{3}^{k}=0$ and σ(T) = σ(T_{1}) ∪ {0}. Since T is quasinilpotent, σ(T) = {0}. But σ(T) = σ(T_{1}) ∪ {0}. Since T_{1} is in A(s, t), T_{1} = 0, so T is nilpotent.
Lemma 2.5
Let T ∈ B(H) be an algebraically k-quasi-class A(s, t) operator, and σ(T) = {µ_{0}}, then T − µ_{0}is 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(µ_{0})}, the operator p(T) − p(µ_{0}) is nilpotent by Lemma 2.4. Let $$p(z)-p({\mu}_{0})=a{\left(z-{\mu}_{0}\right)}^{{k}_{0}}{\left(z-{\mu}_{1}\right)}^{{k}_{1}}\cdots {\left(z-{\mu}_{t}\right)}^{{k}_{t}},$$where µ_{j} ≠ µ_{s} for j ≠ s. Then $$0={\left\{p(T)-p({\mu}_{0})\right\}}^{m}={a}^{m}{\left(T-{\mu}_{0}\right)}^{m}{k}_{0}{\left(T-{\mu}_{1}\right)}^{m{k}_{1}}\cdots {\left(T-{\mu}_{t}\right)}^{m{k}_{t}}$$and hence (T − µ_{0})^{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 µ ∈ iso(σ(T)) and let E_{µ} be the Riesz idempotent associated to µ defined by $$E:=\frac{1}{2\pi i}{\int}_{\partial D}{\left(\mu I-T\right)}^{-1}d\mu ,$$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 $$\left(\begin{array}{cc}{T}_{1}& 0\\ 0& {T}_{2}\end{array}\right),$$where σ(T_{1}) = {µ} and σ(T_{2}) = σ(T_{2}) {µ}. Since T_{1} is an algebraically k-quasi-class A(s, t) operator by Proposition 2.1 and σ(T_{1}) = µ, it follows from Lemma 2.4 that T_{1} − λI is nilpotent. Therefore T_{1} − µ has finite ascent and descent. On the other hand, since T_{2} − µI is invertible, it has finite ascent and descent. Therefore T − µI has finite ascent and descent. Therefore µ is a pole of the resolvent of T. Now if µ ∈ iso(σ(T)), then µ ∈ π(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 T ∈ B(H) has SVEP, then σ(T) = σ_{a}(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 T ∈ B(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, $$E(T)=\pi (T)=\pi ({T}^{*})=E({T}^{*}),$$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 d_{A,B} is polaroid and has SVEP, we get the following theorems for d_{A,B}
Theorem 2.3
If A, B^{*}are k-quasi-class A(s, t) operators, then the following statements are equivalent.
If A, B^{*}are k-quasi-class A(s, t) operators, then generalized a-Weyl’s theorem holds for d_{A,B}.
If A, B^{*}are k-quasi-class A(s, t) operators, then generalized a-Weyl’s theorem holds for${d}_{A,B}^{*}$.
Theorem 2.5
If A, B^{*}are k-quasi-class A(s, t) operators, then
Weyl’s theorem holds for d_{A,B}if and only if generalized Weyl’s theorem holds for d_{A,B}.
a-Weyl’s theorem holds for d_{A,B}if and only if generalized a-Weyl’s theorem holds for d2_{A,B}.
Corollary 2.4
Let A, B ∈ B(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 d_{A,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${d}_{A,B}^{*}$and are equivalent.
Let f ∈ Hol(σ(T)), where Hol(σ(T)) is the space of all functions that are analytic in an open neighborhoods of σ(T). If d_{A,B} is polaroid, then f(d_{A,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(d_{A,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$f\left({d}_{A,B}^{*}\right)$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, B ∈ B(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(d_{A,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$f\left({d}_{A,B}^{*}\right)$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 d_{A,B} generalize a recent result on generalized Weyl’s theorem for d_{A,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$$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right)\in B\left({H}_{1}\oplus {H}_{2}\right).$$If T has bishop’s property β and there exists a non zero x ∈ H_{1} ⊕ H_{2}such that${\sigma}_{T}(x)\underset{\ne}{\subset}\sigma (T)$. Then T has a nontrivial hyperinvariant subspace.
Proof
Assume that $$\mathcal{M}=\left\{y\in H\oplus H:{\sigma}_{T}(y)\subseteq {\sigma}_{T}(x)\right\},$$that is, ℳ = H_{T} (σ_{T} (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 = H_{1} ⊕ H_{2}. Since T has the single extension property, we get $\sigma (T)=\cup \left\{{\sigma}_{T}(y):y\in {H}_{1}\oplus {H}_{2}\right\}\subseteq {\sigma}_{T}(x)\underset{\ne}{\subset}\sigma (T)$ 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$$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right)\in B\left(H\oplus H\right).$$be a k-quasi-class A(s, t) operator. If there exists a nonzero x ∈ H ⊕ H such that${\sigma}_{T}(x)\underset{\ne}{\subset}\sigma (T)$, then T has a nontrivial hyperinvariant subspace.
It is remarkable that simply knowing when solutions to AX − XB = 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, T ∈ B(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: $$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right).$$Thus we have the following theorem.
Theorem 3.2
Let$$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right),\hspace{0.17em}\hspace{0.17em}S=\left(\begin{array}{cc}{S}_{1}& {S}_{2}\\ {S}_{3}& {S}_{4}\end{array}\right).$$If the spectrum of T_{1}and T_{2}are disjoint and ST = TS, then S_{4} = 0.
Proof
It is clear that if ST = TS, then T_{3}S_{4} = S_{4}T_{1}. Hence it follows from the Rosemblum theorem that S_{4} = 0.
Thus ℳ is a nontrivial hyperinvariant subspace for T if the spectrum of T_{1} and the spectrum of T_{3} are disjoint. In particular we have.
Corollary 3.2
Let$$T=\left(\begin{array}{cc}{T}_{1}& {T}_{2}\\ 0& {T}_{3}\end{array}\right)$$be k-quasi-class A(s, t). If 0 ∉ σ(T_{1}), then T has a non trivial hyperinvariant subspace.
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