Let $H$ be an arbitrary complex Hilbert space and $T$ be a bounded linear operator on $H$. We denote the $*$-algebra of all bounded linear operators on $H$ by $B(H)$.

An operator $T$ is $*$-paranormal if $||T^{2}x||\geq||T^{*}x||^{2}$ for unit vector $x.$ $*$-paranormal operators have been studied by many researchers, see [5, 9, 11], etc. An operator $T$ is said to be class $*$- $A$ if $|T^{2}|\geq |T^{*}|^{2}$, where $|T|=(T^{*}T)^{\frac{1}{2}}$. As an easy extension of class $*$- $A$ operators, an operator $T$ is said to be quasi- $*$- $A$ if $T^{*}|T^{2}|T\geq T^{*}|T^{*}|^{2}T$ in [17]. Moreover, for $k>0$, an operator $T$ belongs to $*$- $A(k)$ if $(T^{*}|T|^{2k}T)^{\frac{1}{k+1}}\geq|T^{*}|^{2}.$ An operator $T$ is absolute- $*$- $k$-paranormal if $||T^{*}x||^{k+1}\leq|||T|^{k}Tx||||x||^{k}$ for every $x\in H.$ Particularly an operator $T$ is a class $*$- $A$(resp. $*$-paranormal) operator if and only if $T$ is a $*$- $A(1)$(resp. absolute- $*$- $1$-paranormal) operator.

In this note we extend $*$- $A(k)$(resp. $*$-paranormal) operators and quasi- $*$- $A$ operators to a new class of operators called quasi- $*$- $A(k)$(resp.quasi-absolute- $*$- $k$-paranormal) operators, and study their spectral properties.

Definition 1.1  Let $T\in B(H)$.

(ⅰ) For each $k>0$, $T$ belongs to quasi- $*$- $A(k)$ if

$ T^{*}(T^{*}|T|^{2k}T)^{\frac{1}{k+1}}T\geq|T|^{4}. $

(ⅱ) For each $k>0$, $T$ belongs to quasi-absolute- $*$- $k$-paranormal if

$ ||T^{*}Tx||^{k+1}\leq|||T|^{k}T^{2}x||||Tx||^{k}\mbox{for every} x\in H. $

It's known that quasi- $*$- $A(1)$ operator is quasi- $*$- $A$, and a quasi- $*$- $A(k)$ operator is quasi-absolute- $*$- $k$-paranormal (see Lemma 2.2), then we have the following implications:

$*$- $A(k)$ $\Rightarrow$ quasi- $*$- $A(k)$ $\Rightarrow$ quasi-absolute- $*$- $k$-paranormal.

By simple calculation, we have the following lemma as appeared in [22].

Lemma 1.2  Let $K=\oplus_{n=1}^{+\infty} H_{n}$, where $H_{n}\cong H$. For given positive operators $A$ and $B$ on $H$, define the operator $ T_{A, B}$ on $K$ as follows:

$ T_{A, B}= \left(\begin {array}{llllllll} 0&0&0&0&0&0&\cdots\\ A&0&0&0&0&0&\cdots \\ 0&B&0& 0& 0& 0&\cdots\\ 0&0&B&0&0&0&\cdots\\ 0&0&0&B&0&0&\cdots\\ 0&0&0&0&B&0&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{array}\right). $

Then the following assertions hold:

(ⅰ) $T_{A, B}$ belongs to $*$- $A(k)$ if and only if $B^{2}\geq A^{2}.$

(ⅱ) $T_{A, B}$ belongs to quasi- $*$- $A(k)$ if and only if $AB^{2}A\geq A^{4}.$

The following example provides an operator which is quasi- $*$- $A(k)$ but not $*$- $A(k)$.

Example 1.3  A non- $*$- $A(k)$ and quasi- $*$- $A(k)$ operator.

Take $A$ and $B$ as

$ A=\left(\begin {array}{lll} 1&0\\ 0&0\\ \end{array}\right) B= \left(\begin {array}{lll} 1&1\\ 1&1\\ \end{array}\right). $

Then

$ B^{2}-A^{2}= \left(\begin {array}{lll} 1&2\\ 2&2\\ \end{array}\right)\ngeqslant 0. $

Hence $ T_{A, B}$ is not a $*$- $A(k)$ operator.

On the other hand,

$ A(B^{2}-A^{2})A= \left(\begin {array}{lll} 1&0\\ 0&0\\ \end{array}\right)\left(\begin {array}{lll} 1&2\\ 2&2\\ \end{array}\right)\left(\begin {array}{lll} 1&0\\ 0&0\\ \end{array}\right)=\left(\begin {array}{lll} 1&0\\ 0&0\\ \end{array}\right)\geq0. $

Thus $ T_{A, B}$ is a quasi- $*$- $A(k)$ operator.

Consider unilateral weighted shift operator as an infinite dimensional Hilbert space operator. Recall that given a bounded sequence of positive numbers $\alpha : \alpha_{1}, \alpha_{2}, \alpha_{3}, \cdots$(called weights), the unilateral weighted shift $W_{\alpha}$ associated with $\alpha$ is the operator on $H =l_{ 2}$ defined by $W_{\alpha} e_{n} := \alpha_{n}e_{n+1}$ for all $n\geq1$, where $\{e_{n}\}_{n=1}^{\infty}$ is the canonical orthogonal basis for $l_{2}$. Straightforward calculations show that $W_{\alpha}$ is quasi- $*$- $A(k)$ if and only if

$ W_{\alpha}=\left(\begin {array}{lllllll} 0&0&0&0&0&\cdots\\ \alpha_{1}&0&0&0&0&\cdots\\ 0&\alpha_{2}&0&0&0&\cdots\\ 0&0&\alpha_{3}&0&0&\cdots\\ 0&0&0&\alpha_{4}&0&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\ \end{array}\right), $

where

$ (\alpha_{i+1}\alpha_{i+2}^{k})^{\frac{1}{k+1}}\geq \alpha_{i}\ \ (i=1, 2, 3, \cdots). $

The following examples show that quasi- $*$- $A$ operator and quasi- $*$- $A(2)$ operator are independent.

Example 1.4  A non-quasi- $*$- $A$ and quasi- $*$- $A(2)$ operator.

Let $T$ be a unilateral weighted shift operator with weighted sequence $(\alpha_{i})$, given $\alpha_{1}=3, \alpha_{2}=1, \alpha_{3}=8, \alpha_{3}=\alpha_{4}=\alpha_{5}=\cdots .$ Simple calculations show that $T$ is quasi- $*$- $A(2)$ and a non-quasi- $*$- $A$ operator.

Example 1.5  A non-quasi- $*$- $A(2)$ and quasi- $*$- $A$ operator.

Let $T$ be a unilateral weighted shift operator with weighted sequence $(\alpha_{i})$, given $\alpha_{1}=1, \alpha_{2}=\frac{1}{2}, \alpha_{3}=2, \alpha_{4}=\frac{1}{8}, \alpha_{5}=64, \alpha_{5}=\alpha_{6}=\cdots .$ Simple calculations show that $T$ is quasi- $*$- $A$ but not a quasi- $*$- $A(2)$ operator.

In the sequel, let $\sigma(T)$, $\sigma_{a}(T)$, $\sigma_{p}(T)$, $\sigma_{ea}(T)$, $\sigma_{jp}(T)$, $\sigma_{ja}(T)$ for the spectrum of $T$, the approximate point spectrum of $T$, the point spectrum of $T$, the essential approximate point spectrum of $T$, the joint point spectrum of $T$, the joint approximate point spectrum of $T$, respectively. $\lambda\in\sigma_{p}(T)$ if there is a nonzero $x\in H $ such that $(T-\lambda)x=0$. If in addition, $(T^{*}-\overline{\lambda})x=0$, then $\lambda\in\sigma_{jp}(T)$. Analogously, $\lambda\in\sigma_{a}(T)$ if there is a sequence $\{x_{n}\}$ of unit vectors in $ H $ such that $(T-\lambda)x_{n}\rightarrow 0$. If in addition, $(T^{*}-\overline{\lambda})x_{n}\rightarrow 0$, then $\lambda\in\sigma_{ja}(T)$.

Clearly, $\sigma_{jp}(T)\subseteq\sigma_{p}(T)$, $\sigma_{ja}(T)\subseteq\sigma_{a}(T)$. In general, $\sigma_{jp}(T)\neq\sigma_{p}(T)$, $\sigma_{ja}(T)\neq\sigma_{a}(T)$.

Recently, it was shown that, for some nonnormal operator $T$, the nonzero points of its point spectrum and joint point spectrum are identical, the nonzero points of its approximate point spectrum and joint approximate point spectrum are identical[3, 7, 19-21]. In this section, we will extend that result to quasi- $*$- $A(k)$ for $0 < k\leq 1$.

To prove the inclusion relation between quasi- $*$- $A(k)$ operator and quasi-absolute- $*$- $k$-paranormal operator, we need the following lemma.

Lemma 2.1[10]  Let $A$ be a positive linear operator on a Hilbert space $H$. Then

(ⅰ) ( $A^{\lambda}x, x)\geq (Ax, x)^{\lambda}$ for any $\lambda>1$ and $||x||=1$.

(ⅱ) ( $A^{\lambda}x, x)\leq(Ax, x)^{\lambda}$ for any $0\leq \lambda\leq 1$ and $||x||=1$.

Lemma 2.2  For each $k>0$, every quasi- $*$- $A(k)$ operator is a quasi-absolute- $*$-k-paranormal operator.

Proof  Suppose that $T$ belongs to quasi- $*$- $A(k)$ for $k>0$, i.e.,

$ T^{*}|T^{*}|^{2}T\leq T^{*}(T^{*}|T|^{2k}T)^{\frac{1}{k+1}}T. $

Then, for every $x\in H$,

$ \begin{align*} || T^{*}Tx||^{2(k+1)}&=(T^{*}|T^{*}|^{2}Tx, x)^{(k+1)}\\ &\leq(T^{*}(T^{*}|T|^{2k}T)^{^{\frac{1}{k+1}}}Tx, x)^{(k+1)}\\ &=((T^{*}|T|^{2k}T)^{^{\frac{1}{k+1}}}Tx, Tx)^{(k+1)}\\ &\leq(T^{*}|T|^{2k}T^{2}x, Tx)|| Tx||^{^{2k}}\\ &=|||T|^{k}T^{2}x||^{2}|| Tx||^{^{2k}}. \end{align*} $

Therefore,

$ || T^{*}Tx||^{k+1}\leq|||T|^{k}T^{2}x|||| Tx||^{k} \ \mbox{for every }x\in H, $

that is, $T$ is quasi-absolute- $*$- $k$-paranormal for $k>0$.

Lemma 2.3[6]  Let $H$ be a complex Hilbert space. Then there exists a Hilbert space $K$ such that $H\subset K$ and a map $\varphi : B(H) \rightarrow B(K) $ such that

(ⅰ) $\varphi$ is a faithful $*$-representation of the algebra $B(H)$ on $K$;

(ⅱ) $\varphi(A)\geq 0$ for any $A \geq 0$ in $B(H)$;

(ⅲ) $\sigma_{a}(T)=\sigma_{a}(\varphi(T))=\sigma_{p}(\varphi(T))$ for any $T \in B(H)$.

Lemma 2.4[21]  Let $\varphi : B(H) \rightarrow B(K)$ be Berberian's faithful $*$-representation. Then $\sigma_{ja}(T)=\sigma_{jp}(\varphi(T))$.

Lemma 2.5  Let $T$ be a quasi- $*$- $A(k)$ operator for $0 < k\leq 1$ and $\lambda\neq 0$. Then $Tx=\lambda x$ implies $T^{*}x=\bar{\lambda}x$.

Proof  Let $\lambda\neq0$ and suppose $x\in N(T-\lambda)$, we get $Tx=\lambda x$. Since $T$ is quasi- $*$- $A(k)$, for every unit vector $x \in H$, $|| T^{*}Tx||^{k+1}\leq|||T|^{k}T^{2}x|| || Tx||^{k}, $ then

$ \begin{align*} || T^{*}x||^{k+1}|\lambda|^{k+1}&\leq |\lambda|^{k+2}|||T|^{k}x||=|\lambda|^{k+2}(|T|^{2k}x, x)^{\frac{1}{2}}\\ &\leq|\lambda|^{k+2}(|T|^{2}x, x)^{\frac{k}{2}}=|\lambda|^{k+2}|| Tx||^{k}\\ &=|\lambda|^{2k+2}. \end{align*} $

Hence, for all $|| x||= 1$, $|| T^{*}x||^{2}\leq |\lambda|^{2}, $and then

$ \begin{align} || T^{*}x-\overline{\lambda}x||&=((T-\lambda)^{*}x, (T-\lambda)^{*}x)\nonumber\\ &=|| T^{*}x||^{2}-(x, \overline{\lambda}Tx)-(\overline{\lambda}Tx, x) +|\lambda|^{2}|| x||^{2}\nonumber\\ &\leq|\lambda|^{2}-|\lambda|^{2}-|\lambda|^{2} +|\lambda|^{2}\nonumber\\ &=0\nonumber. \end{align} $

Therefore, we have $|| T^{*}x-\overline{\lambda}x||=0$, that is, $T^{*}x=\bar{\lambda}x$.

Remark 2.6  The condition $``\lambda\neq0"$ cannot be omitted in Lemma 2.5. In fact, Example 1.3 shows that $ T_{A, B}$ is a quasi- $*$- $A(k)$ operator, however for the vector $x=(0, 0, 1, -1, 0, 0, \cdots)$, $T_{A, B}(x)=0$, but $T_{A, B}^{*}(x)\neq0$. Therefore, the relation $N(T_{A, B})\subseteq N(T_{A, B}^{*})$ does not always hold.

Theorem 2.7  Let $T\in B(H)$ be quasi- $*$- $A(k)$ for $0 < k\leq 1$. Then

(ⅰ) $\sigma_{jp}(T)\backslash\{0\}=\sigma_{p}(T)\backslash\{0\}$;

(ⅱ) If $(T-\lambda)x=0, (T-\mu)y=0$, and $\lambda\neq\mu$, then we have $ < x, y > =0;$

(ⅲ) $\sigma_{ja}(T)\backslash\{0\}=\sigma_{a}(T)\backslash\{0\}$.

Proof  (ⅰ) Clearly by Lemma 2.5.

(ⅱ) Without loss of generality, we assume $\mu\neq0$. Then $(T -\mu)^{*}y =0$ by Lemma 2.5. Thus we have $\mu < x, y > = < x, T^{*}y > = < Tx, y > =\lambda < x, y > .$ Since $\lambda\neq\mu$, $ < x, y > =0.$

(ⅲ) Let $\varphi$: $B(H)\rightarrow B(K)$ be Berberian's faithful $*$-representation of Lemma 2.3. In the following, we shall show that $\varphi(T)$ is also a quasi- $*$- $A(k)$ operator.

In fact, since $T$ is a quasi- $*$- $A(k)$ operator, by Lemma 2.3, we have

$ \begin{align*} &(\varphi(T))^{*}[((\varphi(T))^{*}|\varphi (T)|^{2k}\varphi(T))^{\frac{1}{k+1}}-|(\varphi(T))^{*}|^{2}]\varphi (T)\\ &=\varphi(T^{*}[(T^{*}|T|^{2k}T)^{\frac{1}{k+1}}-|T^{*}|^{2}]T)\geq 0. \end{align*} $

Then

$ \begin{align*} \sigma_{a}(T)\backslash\{0\}&=\sigma_{a}(\varphi(T))\backslash\{0\}\ \ \mbox{by Lemma 2.3} \\ &=\sigma_{p}(\varphi(T))\backslash\{0\}\ \ \mbox{by Lemma 2.3} \\ &=\sigma_{jp}(\varphi(T))\backslash\{0\} \ \ \mbox{by(ⅰ)} \\ &=\sigma_{ja}(T)\backslash\{0\}\ \ \mbox{by Lemma 2.4}. \end{align*} $

Recall that $T\in B(H)$ is said to have finite ascent if $N(T^{n})= N(T^{n+1})$ for some positive integer $n$, where $N(T)$ for the null space of $T$.

Theorem 2.8  If $T$ is quasi- $*$- $A(k)$ for $0 < k\leq1$, then $T-\lambda$ has finite ascent for each $\lambda \in{\mathbb C}$.

Proof  If $\lambda\neq0$, then $N(T-\lambda)\subseteq N(T^{*}-\overline{\lambda}$) by Lemma 2.5, thus $N(T-\lambda) = N(T-\lambda)^{2}$. If $\lambda=0$, let $x \in N(T^{2})$, since $T$ is quasi- $*$- $A(k)$, then

$ || T^{*}Tx||^{k+1} \leq |||T|^{k}T^{2}x|| || Tx||^{k}\ \ \mbox{for every } x\in H. $

We have

$ \begin{align*} || |T|^{2}x||^{k+1}&=|| T^{*}Tx||^{k+1}\leq |||T|^{k}T^{2}x|| || Tx||^{k}\\ &=(|T|^{2k}T^{2}x, T^{2}x)^{\frac{1}{2}}||Tx||^{k}\\ &\leq (|T|^{2}T^{2}x, T^{2}x)^{\frac{k}{2}}||Tx||^{k}||T^{2}x||^{(1-k)}\\ &=||T^{3}x||^{k}||Tx||^{k}||T^{2}x||^{(1-k)}\mbox{for every } x\in H. \end{align*} $

Hence $|T|^{2}x=0$ implies $Tx=0$. This shows that $T-\lambda$ has finite ascent for each $\lambda \in{\mathbb C}$.

Recall that $T\in B(H)$ has the single valued extension property (abbrev. SVEP), if for every open set $U$ of ${\mathbb C}$, the only analytic solution $f$: $U\rightarrow H$ of the equation

$ (T-\lambda)f(\lambda)=0 $

for all $\lambda \in U $ is the zero function on $U$.

Theorem 2.9  If $T$ is quasi- $*$- $A(k)$ for $0 < k\leq1$, then $T$ has SVEP.

Proof  Clearly by Theorem 2.8 and [14, Proposition 1.8].

As a simple consequence of the preceding result, we obtain

Corollary 2.10  If $T$ is quasi- $*$- $A(k)$ for $0 < k\leq1$, then

(ⅰ) $\sigma_{ea}(f(T))= f(\sigma_{ea}(T))$ for every $f \in H(\sigma(T))$, where $H(\sigma(T))$ is the space of functions analytic on an open neighborhood of $\sigma(T)$;

(ⅱ) $T$ obeys $a$-Browder's theorem, that is $\sigma_{ea}(T)=\sigma_{ab}(T)$, where $\sigma_{ab}(T) := \cap\{ \sigma_{a}(T+K): TK=KT \ \mbox{and}\ K\ \mbox{is a compact operator}\}$;

(ⅲ) $a$-Browder's theorem holds for $f(T)$ for every $f \in H(\sigma(T))$.

Proof  Note that $T$ has SVEP, Corollary 2.10 follows by [1].

Given non-zero $T$, $S$ $\in B(H)$, let $T\otimes S $ denote the tensor product on the product space $H \otimes H$. The operation of taking tensor products $T\otimes S $ preserves many properties of $T$, $S$ $\in B(H)$, but by no means all of them. The normaloid property is invariant under tensor products [16, p.623], $T\otimes S $ is normal if and only if $T$ and $S$ are normal [12, 18], however, there exist paranormal operators $T$ and $S$ such that $T\otimes S $ is not paranormal [4]. Duggal [8] showed that for non-zero $T$, $S$ $\in B(H)$, $T\otimes S \in H(p)$ if and only if $T$, $S$ $\in H(p)$. Recently, this result was extended to class $*$- $A$ operators and class $A$ operators in [9, 13], respectively.

In this section we consider the tensor products of $*$- $A(k)$ operators. The following key lemma is due to J. Stochel.

Lemma 3.1[18, Proposition 2.2]  Let $A_{1}, A_{2} \in B(H)$, $B_{1}, B_{2} \in B(K)$ be non-negative operators. If $A_{1}$ and $B_{1}$ are non-zero, then the following assertions are equivalent:

(ⅰ) $A_{1} \otimes B_{1} \leq A_{2 }\otimes B_{2}$.

(ⅱ) There exists $c > 0$ for which $A_{1} \leq cA_{2}$ and $B_{1} \leq c^{-1}B_{2}$.

Theorem 3.2  Let $T$, $S$ $\in B(H)$ be non-zero operators. Then $T\otimes S$ is a $*$- $A(k)$ operator if and only if $T$ and $S$ are $*$- $A(k)$ operators.

Proof  By simple calculation we have

$ \begin{align*}&T \otimes S \mbox{ is a} *\mbox{-}A(k) \mbox{ operator}\\ &\Leftrightarrow[(T \otimes S)^{*}|T \otimes S|^{2k}(T \otimes S)]^{\frac{1}{k+1}}\geq |(T \otimes S)^{*}|^{2}\\ &\Leftrightarrow[(T \otimes S)^{*}|T \otimes S|^{2k}(T \otimes S)]^{\frac{1}{k+1}}- |T^{*}|^{2} \otimes |S^{*}|^{2} \geq 0\\ &\Leftrightarrow [(T^{*} \otimes S^{*})(|T|^{2k} \otimes |S|^{2k})(T \otimes S)]^\frac{1}{k+1} - |T^{*}|^{2} \otimes|S^{*}|^{2} \geq 0\\ &\Leftrightarrow (T^{*}|T|^{2k}T)^{\frac{1}{k+1}}\otimes(S^{*}|S|^{2k}S)^{\frac{1}{k+1}}-|T^{*}|^{2} \otimes |S^{*}|^{2} \geq 0\\ &\Leftrightarrow |T^{*}|^{2} \otimes [(S^{*}|S|^{2k}S)^{\frac{1}{k+1}}-|S^{*}|^{2}]+[(T^{*}|T|^{2k}T)^{\frac{1}{k+1}} -|T^{*}|^{2}] \otimes (S^{*}|S|^{2k}S)^{\frac{1}{k+1}} \geq 0. \end{align*} $

Thus the sufficiency is easily proved. Conversely, suppose that $T\otimes S$ belongs to $*$- $A(k)$. Without loss of generality, it is enough to show that $T$ belongs to $*$- $A(k)$. Since $T\otimes S$ is $*$- $A(k)$, we obtain

$ (T^{*}|T|^{2k}T)^{\frac{1}{k+1}}\otimes(S^{*}|S|^{2k}S)^{\frac{1}{k+1}}\geq |(T \otimes S)^{*}|^{2}. $

Therefore, by Lemma 3.1, there exists a positive real number $l$ for which

$ l(T^{*}|T|^{2k}T)^{\frac{1}{k+1}}\geq |T^{*}|^{2} $

and

$ l^{-1}(S^{*}|S|^{2k}S)^{\frac{1}{k+1}}\geq |S^{*}|^{2}. $

Consequently, for arbitrary $x, y \in H$, using Lemma 2.1 we have

$ \begin{align*} || T||^{2}&=|| T^{*}||^{2}=\sup\{(|T^{*}|^{2}x, x):|| x||=1\}\\ &\leq \sup\{(l(T^{*}|T|^{2k}T)^{\frac{1}{k+1}}x, x):|| x||=1\}\\ &\leq l\sup\{((T^{*}|T|^{2k}T)x, x)^{\frac{1}{1+k}}:|| x||=1\}\\ &\leq l|| T^{*}|T|^{2k}T||^{\frac{1}{1+k}}\\ &\leq l|| T||^{2} \end{align*} $

and

$ \begin{align*} || S||^{2}&=|| S^{*}||^{2}=\sup\{(|S^{*}|^{2}y, y):|| y||=1\}\\ &\leq \sup\{[l^{-1}(S^{*}|S|^{2k}S)^{\frac{1}{k+1}}y, y]:|| y||=1\}\\ &\leq l^{-1}\sup\{[(S^{*}|S|^{2k}S)y, y]^{\frac{1}{1+k}}:|| y||=1\}\\ &\leq l^{-1}|| S^{*}|S|^{2k}S||^{\frac{1}{1+k}}\\ &\leq l^{-1}|| S||^{2}. \end{align*} $

Clearly, we must have $l = 1$, and hence $T$ is a $*$- $A(k)$ operator.