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Weyltype theorems and kquasiMhyponormal operators
Journal of Inequalities and Applicationsvolume 2013, Article number: 446 (2013)
Abstract
In this paper, we show that if E is the Riesz idempotent for a nonzeroisolated point λ of the spectrum of akquasiMhyponormal operator T, then E isselfadjoint, and $R(E)=N(T\lambda )=N{(T\lambda )}^{\ast}$. Also, we obtain that Weyltype theorems hold foralgebraically kquasiMhyponormal 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 Mhyponormal [1] if there exists a real positive number M such that
T is quasiMhyponormal [2] if there exits a real positive number M such that
T is kquasiMhyponormal [3] if there exists a real positive number M such that
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 Mhyponormaloperator, which is not hyponormal.
Example 1.1 Consider the unilateral weighted shift operator as aninfinitedimensional 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 Mhyponormal 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 Mhyponormal operator, but it is nothyponormal.
Next, we give a 2quasiMhyponormal operator, which is notMhyponormal.
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 a2quasiMhyponormal operator, but is not Mhyponormal.
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}\phantom{\rule{0.2em}{0ex}}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 selfadjoint and $R(E)=N(T\lambda )$. Recently, Chō and Tanahashi [5] obtained an improvement of Stampfli’s result tophyponormal operators or loghyponormal operators. Furthermore, Chōand Han extended it to Mhyponormal operators as follows.
Proposition 1.3 [[6], Theorem 4]
Let T be an Mhyponormal operator, and let λ be an isolated point of$\sigma (T)$. If E is the Riesz idempotent for λ, then E is selfadjoint, and$R(E)=N(T\lambda )=N{(T\lambda )}^{\ast}$.
2 Isolated point of spectrum of kquasiMhyponormaloperators
Lemma 2.1 Let T be a kquasiMhyponormal 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 anMhyponormal operator, and hence, $T=\lambda I$ by [6]. □
Lemma 2.2 Let T be a kquasiMhyponormal 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 kquasiMhyponormaloperator, $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 kquasiMhyponormal operator, and let λ be a nonzero isolated point of$\sigma (T)$. Then the Riesz idempotent E for λ is selfadjoint, and
Proof We can derive the result from Lemma 2.2, [[3], Theorem 2.5] and [[7], Lemma 5.2]. □
3 Weyltype theorems of algebraically kquasiMhyponormaloperators
We say that T is an algebraically kquasiMhyponormaloperator if there exists a nonconstant complex polynomial p such that $p(T)$ is a kquasiMhyponormal operator. From thedefinition above, T is an algebraicallykquasiMhyponormal 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 finitedimensional. 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 BFredholm 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 semiFredholm (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 aWeyl’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 Mhyponormal operators [6] and algebraically quasiMhyponormal operators [2], respectively. In this section, we obtain that generalizedaWeyl’s theorems hold for algebraicallykquasiMhyponormal operators.
Lemma 3.1[3]
Let$T\in B(H)$be a kquasiMhyponormal operator, let the rangeof${T}^{k}$be not dense and
Then${T}_{1}$is Mhyponormal, ${T}_{3}^{k}=0$and$\sigma (T)=\sigma ({T}_{1})\cup \{0\}$.
Theorem 3.2 Let T be a quasinilpotent algebraically kquasiMhyponormal operator. Then T is nilpotent.
Proof We first assume that T is akquasiMhyponormal operator. Consider two cases, Case I: If therange of ${T}^{k}$ has dense range, then it is an Mhyponormal 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
where ${T}_{1}:=T\overline{R({T}^{k})}$ is an Mhyponormal 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 Mhyponormal operator, ${T}_{1}=0$. Since ${T}_{3}^{k}=0$, simple computation shows that
Now, suppose that T is an algebraicallykquasiMhyponormal operator. Then there exists a nonconstantpolynomial p such that $p(T)$ is a kquasiMhyponormal operator. If $p(T)$ has dense range, then $p(T)$ is an Mhyponormal operator. Thus T is analgebraically Mhyponormal 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
where $A:=p(T)\overline{R({(p(T))}^{k})}$ is an Mhyponormal 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 kquasiMhyponormal 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 Mhyponormal operator isisoloid, we can prove more.
Theorem 3.3 Let T be an algebraically kquasiMhyponormal operator. Then T is polaroid.
Proof Suppose that T is an algebraicallykquasiMhyponormal operator. Then $p(T)$ is a kquasiMhyponormal 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}\phantom{\rule{0.2em}{0ex}}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:
where $\sigma ({T}_{1})=\{\lambda \}$ and $\sigma ({T}_{2})=\sigma (T)\mathrm{\setminus}\{\lambda \}$. Since ${T}_{1}$ is an algebraically kquasiMhyponormaloperator, 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 kquasiMhyponormal 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
is a zero function on U.
Theorem 3.5 Let T be an algebraically kquasiMhyponormal operator. Then T has SVEP.
Proof Suppose that T is an algebraicallykquasiMhyponormal operator. Then $p(T)$ is a kquasiMhyponormal 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 kquasiMhyponormal operator. ThenWeyl’s theorem holds for$f(T)$for every$f\in H(\sigma (T))$.
Proof Firstly, suppose that T is an algebraicallykquasiMhyponormal 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 kquasiMhyponormaloperator. 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 kquasiMhyponormal operator, then $f(T)$ satisfies a generalized aWeyl’s theorem.

(ii)
If T is an algebraically kquasiMhyponormal operator, then $f({T}^{\ast})$ satisfies a generalized aWeyl’s theorem.
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Acknowledgements
We wish to thank the referee for careful reading and valuable comments for theorigin draft. This work is supported by the Basic Science and TechnologicalFrontier Project of Henan Province (No. 132300410261). This work is partiallysupported by the National Natural Science Foundation of China (No. 11271112,11201126).
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Both authors have given equal contributions in this paper.
Fei Zuo and Hongliang Zuo contributed equally to this work.
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Keywords
 kquasiMhyponormal operators
 Riesz idempotent operators
 Weyltype theorems