Weyl-type theorems and k-quasi-M-hyponormal operators

Abstract

In this paper, we show that if E is the Riesz idempotent for a non-zeroisolated point λ of the spectrum of ak-quasi-M-hyponormal operator T, then E isself-adjoint, and $R(E)=N(T-\lambda )=N{(T-\lambda )}^{\ast }$. Also, we obtain that Weyl-type theorems hold foralgebraically k-quasi-M-hyponormal operators.

MSC: 47B20, 47A10.

1 Introduction

Let T be a bounded linear operator on a complex Hilbert space H,write it for $T\in B(H)$, take a complex number λ in ℂ, and,henceforth, shorten $T-\lambda I$ to $T-\lambda$. One of recent trends in operator theory is studying naturalextensions of normal operators. We introduce some of these operators as follows.

T is said to be a hyponormal operator if $T^{\ast }T\ge T{T}^{\ast }$;

T is M-hyponormal [1] if there exists a real positive number M such that

$$M^2{(T-\lambda )}^{\ast }(T-\lambda )\ge (T-\lambda ){(T-\lambda )}^{\ast }\text{for all }\lambda \in \mathbb{C};$$

T is quasi-M-hyponormal [2] if there exits a real positive number M such that

$$T^{\ast }(M^2{(T-\lambda )}^{\ast }(T-\lambda ))T\ge T^{\ast }(T-\lambda ){(T-\lambda )}^{\ast }T\text{for all }\lambda \in \mathbb{C};$$

T is k-quasi-M-hyponormal [3] if there exists a real positive number M such that

$$T^{\ast k}(M^2{(T-\lambda )}^{\ast }(T-\lambda ))T^k\ge T^{\ast k}(T-\lambda ){(T-\lambda )}^{\ast }{T}^k\text{for all }\lambda \in \mathbb{C},$$

where k is a natural number.

It is clear that $\text{hyponormal}\Rightarrow M\text{-hyponormal}\Rightarrow k\text{-quasi-}M\text{-hyponormal}$.

We give the following example to indicate that there exists an M-hyponormaloperator, which is not hyponormal.

Example 1.1 Consider the unilateral weighted shift operator as aninfinite-dimensional Hilbert space operator. Recall that given a bounded sequence ofpositive numbers α: ${\alpha }_1,{\alpha }_2,{\alpha }_3,\dots$ (called weights), the unilateral weighted shift $W_{\alpha }$ associated with α is the operator on $H=l_2$ defined by $W_{\alpha }{e}_n:={\alpha }_n{e}_{n+1}$ for all $n\ge 1$, where ${\{e_n\}}_{n=1}^{\mathrm{\infty }}$ is the canonical orthogonal basis for $l_2$. It is well known that $W_{\alpha }$ is hyponormal if and only if α is monotonicallyincreasing. Also, $W_{\alpha }$ is M-hyponormal if and only if α iseventually increasing. Hence, if we take the weights α such that ${\alpha }_1=2$, ${\alpha }_2=1$, ${\alpha }_3=2$, ${\alpha }_3={\alpha }_4=\cdots$ , then $W_{\alpha }$ is an M-hyponormal operator, but it is nothyponormal.

Next, we give a 2-quasi-M-hyponormal operator, which is notM-hyponormal.

Example 1.2 Let $T=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ defined on ${\mathbb{C}}^2$. Then by simple calculations, we see that T is a2-quasi-M-hyponormal operator, but is not M-hyponormal.

If $T\in B(H)$, we shall write $N(T)$ and $R(T)$ for the null space and the range space of T. Also, let $\alpha (T):=dimN(T)$, $\beta (T):=dimN(T^{\ast })$, $\sigma (T)$ and $iso\sigma (T)$ for the spectrum and the isolated points of the spectrum ofT, respectively.

Let $\lambda \in iso\sigma (T)$. The Riesz idempotent E of T with respect toλ is defined by $E=\frac{1}{2\pi i}{\int }_{\partial D}{(\mu -T)}^{-1}d\mu$, where D is a closed disk, centered at λ,which contains no other points of $\sigma (T)$. It is well known that the Riesz idempotent satisfies $E^2=E$, $ET=TE$, $\sigma (T{|}_{R(E)})=\{\lambda \}$, and $N(T-\lambda )\subseteq R(E)$. Stampfli [4] showed that if T satisfies the growth condition $G_1$, then E is self-adjoint and $R(E)=N(T-\lambda )$. Recently, Chō and Tanahashi [5] obtained an improvement of Stampfli’s result top-hyponormal operators or log-hyponormal operators. Furthermore, Chōand Han extended it to M-hyponormal operators as follows.

Proposition 1.3 [[6], Theorem 4]

Let T be an M-hyponormal operator, and let λ be an isolated point of$\sigma (T)$. If E is the Riesz idempotent for λ, then E is self-adjoint, and$R(E)=N(T-\lambda )=N{(T-\lambda )}^{\ast }$.

2 Isolated point of spectrum of k-quasi-M-hyponormaloperators

Lemma 2.1 Let T be a k-quasi-M-hyponormal operator. If$0\ne \lambda \in \mathbb{C}$, and assume that$\sigma (T)=\{\lambda \}$, then$T=\lambda I$.

Proof If $\lambda \ne 0$ and $\sigma (T)=\{\lambda \}$, then T is invertible, so T is anM-hyponormal operator, and hence, $T=\lambda I$ by [6]. □

Lemma 2.2 Let T be a k-quasi-M-hyponormal operator and$0\ne \lambda \in \mathbb{C}$. Then$Tx=\lambda x$implies that$T^{\ast }x=\overline{\lambda }x$.

Proof Suppose that $Tx=\lambda x$. Since T is a k-quasi-M-hyponormaloperator, $M\parallel (T-\alpha )T^{k}y\parallel \ge \parallel {(T-\alpha )}^{\ast }{T}^{k}y\parallel$ for all vectors $y\in H$ and $\alpha \in \mathbb{C}$. In particular, $M\parallel (T-\lambda )T^{k}x\parallel \ge \parallel {(T-\lambda )}^{\ast }{T}^{k}x\parallel$. Since $Tx=\lambda x$, $0=M{|\lambda |}^k\parallel (T-\lambda )x\parallel =M\parallel (T-\lambda )T^{k}x\parallel \ge \parallel {(T-\lambda )}^{\ast }{T}^{k}x\parallel ={|\lambda |}^k\parallel {(T-\lambda )}^{\ast }x\parallel$. $|\lambda |\ne 0$, therefore $\parallel {(T-\lambda )}^{\ast }x\parallel =0$. □

Theorem 2.3 Let T be a k-quasi-M-hyponormal operator, and let λ be a non-zero isolated point of$\sigma (T)$. Then the Riesz idempotent E for λ is self-adjoint, and

$$R(E)=N(T-\lambda )=N{(T-\lambda )}^{\ast }.$$

Proof We can derive the result from Lemma 2.2, [[3], Theorem 2.5] and [[7], Lemma 5.2]. □

3 Weyl-type theorems of algebraically k-quasi-M-hyponormaloperators

We say that T is an algebraically k-quasi-M-hyponormaloperator if there exists a nonconstant complex polynomial p such that $p(T)$ is a k-quasi-M-hyponormal operator. From thedefinition above, T is an algebraicallyk-quasi-M-hyponormal operator, then so is $T-\lambda$ for each $\lambda \in \mathbb{C}$.

An operator T is called Fredholm if $R(T)$ is closed, and both $N(T)$ and $N(T^{\ast })$ are finite-dimensional. The index of a Fredholm operatorT is given by $i(T)=\alpha (T)-\beta (T)$. An operator T is called Weyl if it is Fredholm of indexzero. The Weyl spectrum of T[8] is defined by $w(T):=\{\lambda \in \mathbb{C}:T-\lambda \text{ is not Weyl}\}$. Following [9], we say that Weyl’s theorem holds for T if $\sigma (T)\mathrm{\setminus }w(T)={\pi }_{00}(T)$, where ${\pi }_{00}(T):=\{\lambda \in iso\sigma (T):0<\alpha (T-\lambda )<\mathrm{\infty }\}$.

More generally, Berkani investigated the B-Fredholm theory (see [10–12]). We define $T\in SB{F}_{+}^{-}(H)$ if there exists a positive integer n such that $R(T^n)$ is closed, $T_{[n]}:R(T^n)\ni x\to Tx\in R(T^n)$ is upper semi-Fredholm (i.e., $R(T_{[n]})=R(T^{n+1})$ is closed, $dimN(T_{[n]})=dimN(T)\cap R(T^n)<\mathrm{\infty }$) and $i(T_{[n]})\le 0$[12]. We define ${\sigma }_{SB{F}_{+}^{-}}(T)=\{\lambda \in \mathbb{C}:T-\lambda \notin SB{F}_{+}^{-}(H)\}$. Let $E^a(T)$ denote the set of all isolated points λ of ${\sigma }_a(T)$ with $0<\alpha (T-\lambda )$. We say that generalized a-Weyl’s theorem holds forT if ${\sigma }_a(T)\setminus {\sigma }_{SB{F}_{+}^{-}}(T)=E^a(T)$.

We know that Weyl’s theorem holds for hermitian operators [13], which have been extended to hyponormal operators [14], algebraically hyponormal operators by [15], algebraically M-hyponormal operators [6] and algebraically quasi-M-hyponormal operators [2], respectively. In this section, we obtain that generalizeda-Weyl’s theorems hold for algebraicallyk-quasi-M-hyponormal operators.

Lemma 3.1[3]

Let$T\in B(H)$be a k-quasi-M-hyponormal operator, let the rangeof$T^k$be not dense and

$$T=\left(\begin{array}{cc}{T}_1& T_2\\ 0& T_3\end{array}\right)\mathit{\text{on}}H=\overline{R\left(T^k\right)}\oplus N\left(T^{\ast k}\right).$$

Then$T_1$is M-hyponormal, $T_3^k=0$and$\sigma (T)=\sigma (T_1)\cup \{0\}$.

Theorem 3.2 Let T be a quasinilpotent algebraically k-quasi-M-hyponormal operator. Then T is nilpotent.

Proof We first assume that T is ak-quasi-M-hyponormal operator. Consider two cases, Case I: If therange of $T^k$ has dense range, then it is an M-hyponormal operator.Hence, by [[6], Lemma 8], T is nilpotent. Case II: If T does nothave dense range, then by Lemma 3.1, we can represent T as the uppertriangular matrix

$$T=\left(\begin{array}{cc}{T}_1& T_2\\ 0& T_3\end{array}\right)\text{on}H=\overline{R\left(T^k\right)}\oplus N\left(T^{\ast k}\right),$$

where $T_1:=T|\overline{R(T^k)}$ is an M-hyponormal operator. Since T isquasinilpotent, $\sigma (T)=\{0\}$. But $\sigma (T)=\sigma (T_1)\cup \{0\}$, hence, $\sigma (T_1)=\{0\}$. Since $T_1$ is an M-hyponormal operator, $T_1=0$. Since $T_3^k=0$, simple computation shows that

$$T^{k+1}=\left(\begin{array}{cc}0& T_2{T}_3^k\\ 0& T_3^{k+1}\end{array}\right)=0.$$

Now, suppose that T is an algebraicallyk-quasi-M-hyponormal operator. Then there exists a nonconstantpolynomial p such that $p(T)$ is a k-quasi-M-hyponormal operator. If $p(T)$ has dense range, then $p(T)$ is an M-hyponormal operator. Thus T is analgebraically M-hyponormal operator. It follows from [[6], Lemma 8] that it is nilpotent. If ${(p(T))}^k$ does not have a dense range, then by Lemma 3.1, we canrepresent $p(T)$ as the upper triangular matrix

$$p(T)=\left(\begin{array}{cc}A& B\\ 0& C\end{array}\right)\text{on}H=\overline{R\left({(p(T))}^k\right)}\oplus N\left({(p(T))}^{\ast k}\right),$$

where $A:=p(T)|\overline{R({(p(T))}^k)}$ is an M-hyponormal operator. Since $\sigma (T)=\{0\}$ and $\sigma (p(T))=p(\sigma (T))=\{p(0)\}$, the operator $p(T)-p(0)$ is quasinilpotent. But $\sigma (p(T))=\sigma (A)\cup \{0\}$, thus $\sigma (A)\cup \{0\}=\{p(0)\}$. So $p(0)=0$, and hence, $p(T)$ is quasinilpotent. Since $p(T)$ is a k-quasi-M-hyponormal operator, by theprevious argument $p(T)$ is nilpotent. On the other hand, since $p(0)=0$, $p(z)=c{z}^m(z-{\lambda }_1)(z-{\lambda }_2)\cdots (z-{\lambda }_n)$ for some natural number m. $p(T)=c{T}^m(T-{\lambda }_1)(T-{\lambda }_2)\cdots (T-{\lambda }_n)$. $p(T)$ is nilpotent, therefore, T isnilpotent. □

Recall that an operator T is said to be isoloid if every isolated point of $\sigma (T)$ is an eigenvalue of T and polaroid if every isolatedpoint of $\sigma (T)$ is a pole of the resolvent of T. In general, ifT is polaroid, then it is isoloid. However, the converse is not true.In [6], it is showed that every algebraically M-hyponormal operator isisoloid, we can prove more.

Theorem 3.3 Let T be an algebraically k-quasi-M-hyponormal operator. Then T is polaroid.

Proof Suppose that T is an algebraicallyk-quasi-M-hyponormal operator. Then $p(T)$ is a k-quasi-M-hyponormal operator for somenonconstant polynomial p. Let $\lambda \in iso\sigma (T)$ and $E_{\lambda }$ be the Riesz idempotent associated to λ defined by $E_{\lambda }:=\frac{1}{2\pi i}{\int }_{\partial D}{(\mu -T)}^{-1}d\mu$, where D is a closed disk of center λ,which contains no other point of $\sigma (T)$. We can represent T as the direct sum in the followingform:

$$T=\left(\begin{array}{cc}{T}_1& 0\\ 0& T_2\end{array}\right),$$

where $\sigma (T_1)=\{\lambda \}$ and $\sigma (T_2)=\sigma (T)\mathrm{\setminus }\{\lambda \}$. Since $T_1$ is an algebraically k-quasi-M-hyponormaloperator, so is $T_1-\lambda$. But $\sigma (T_1-\lambda )=\{0\}$, it follows from Theorem 3.2 that $T_1-\lambda$ is nilpotent, thus $T_1-\lambda$ has finite ascent and descent. On the other hand, since $T_2-\lambda$ is invertible, clearly, it has finite ascent and descent. $T-\lambda$ has finite ascent and descent, and hence, λ is apole of the resolvent of T, therefore, T ispolaroid. □

Corollary 3.4 Let T be an algebraically k-quasi-M-hyponormal operator. Then T is isoloid.

We say that T has the single valued extension property (abbreviated SVEP)if, for every open set U of ℂ, the only analytic solution f: $U\to H$ of the equation

$$(T-\lambda )f(\lambda )=0\text{for all }\lambda \in U$$

is a zero function on U.

Theorem 3.5 Let T be an algebraically k-quasi-M-hyponormal operator. Then T has SVEP.

Proof Suppose that T is an algebraicallyk-quasi-M-hyponormal operator. Then $p(T)$ is a k-quasi-M-hyponormal operator for somenonconstant complex polynomial p, and hence, $p(T)$ has SVEP by [[3], Theorem 2.1]. Therefore, T has SVEP by [[16], Theorem 3.3.9]. □

In the following theorem, $H(\sigma (T))$ denotes the space of functions analytic in an open neighborhood of $\sigma (T)$.

Theorem 3.6 Let T or$T^{\ast }$be an algebraically k-quasi-M-hyponormal operator. ThenWeyl’s theorem holds for$f(T)$for every$f\in H(\sigma (T))$.

Proof Firstly, suppose that T is an algebraicallyk-quasi-M-hyponormal operator. We first show that Weyl’stheorem holds for T. Using the fact [[17], Theorem 2.2] that if T is polaroid, then Weyl’stheorem holds for T if and only if T has SVEP at points of $\lambda \in \sigma (T)\setminus w(T)$. We have that T is polaroid by Theorem 3.3, andT has SVEP by Theorem 3.5. Hence, T satisfiesWeyl’s theorem.

Next, suppose that $T^{\ast }$ is an algebraically k-quasi-M-hyponormaloperator. Now we show that Weyl’s theorem holds for T. We use thefact [[18], Theorem 3.1] that if T or $T^{\ast }$ has SVEP, then Weyl’s theorem holds for T if andonly if ${\pi }_{00}(T)=p_{00}(T)$. Since $T^{\ast }$ has SVEP, it is sufficient to show that ${\pi }_{00}(T)=p_{00}(T)$. $p_{00}(T)\subseteq {\pi }_{00}(T)$ is clear, so we only need to prove ${\pi }_{00}(T)\subseteq p_{00}(T)$. Let $\lambda \in {\pi }_{00}(T)$. Then λ is an isolated point of $\sigma (T)$. Hence, λ is a pole of the resolvent of T,since T is polaroid by Theorem 3.3, that is, $p(\lambda -T)=q(\lambda -T)<\mathrm{\infty }$. By assumption, we have $\alpha (\lambda -T)<\mathrm{\infty }$, so $\beta (\lambda -T)<\mathrm{\infty }$. Hence, we conclude that $\lambda \in p_{00}(T)$. Therefore, Weyl’s theorem holds for T.

Finally, we can derive the result by Theorem 3.5 and [[17], Theorem 2.4]. □

Following [[19], Theorem 3.12], we obtain the following result.

Theorem 3.7 Let f be an analytic function on$\sigma (T)$, and f is not constant on each connected component of the open set U containing$\sigma (T)$.

• (i) If $T^{\ast }$ is an algebraically k-quasi-M-hyponormal operator, then $f(T)$ satisfies a generalized a-Weyl’s theorem.

• (ii) If T is an algebraically k-quasi-M-hyponormal operator, then $f(T^{\ast })$ satisfies a generalized a-Weyl’s theorem.