#  $(\;\alpha,\beta)$-$A$-NORMAL OPERATORS IN SEMI-HILBERTIAN SPACES

Abdelkader Benali [1] and Ould Ahmed Mahmoud Sid Ahmed [2]

[1] Faculty of science, Mathematics Department,University of Hassiba
Benbouali, Chlef Algeria. B.P. 151 Hay Essalem, chlef 02000, Algeria.
<EMAIL_ADDRESS>
[2] Mathematics Department, College of Science. Aljouf University
Aljouf 2014. Saudi Arabia
<EMAIL_ADDRESS>

###### Abstract

Let $\mathcal{H}$ be a Hilbert space and let $A$ be a positive bounded
operator on $\mathcal{H}$. The semi-inner product $\langle
u\;|\;v\rangle_{A}:=\langle Au\;|\;v\rangle,\;\;u,v\in\mathcal{H}$ induces a
semi-norm $\left\|.\;\right\|_{A}$ on $\mathcal{H}.$ This makes $\mathcal{H}$
into a semi-Hilbertian space. In this paper we introduce and prove some
proprieties of $(\alpha,\beta)$-normal operators according to semi-Hilbertian
space structures. Furthermore we state various inequalities between the
$A$-operator norm and $A$-numerical radius of $(\alpha,\beta)$-normal
operators in semi Hilbertian spaces.

Keywords. Semi-Hilbertian space, $A$-selfadjoint operators,$A$-normal
operators, $A$-positive operators, $(\alpha,\beta)$-normal operators.

Mathematics Subject Classification: Primary 46C05, Secondary 47A05.

##  1 INTRODUCTION AND PRELIMINARIES RESULTS

One of the most important subclasses of the algebra of all bounded linear
operators acting on Hilbert space, the class of normal operators
($TT^{*}=T^{*}T$).They have been the object of some intensive studies. The
theory of these operators was investigated in $[5]$ and $[20]$.

This class has been generalized, in some sense, to the larger sets of so-
called quasinormal, hyponormal, isometry, partial isometry, $m$-isometries
operators on Hilbert spaces.

Recently, these classes of operators have been generalized by many authors
when an additional semi-inner product is considered (see $[2,3,4,17,18,21]$ )
and other papers.

In this framework, we show that many results from $[7,8,12]$ remain true if we
consider an additional semi-inner product defined by a positive semi-definite
operator A. We are interested to introducing a new concept of normality in
semi-Hilbertian spaces.
The contents of the paper are the following. In Section 1, we give notation
and results about the concept of $A$-adjoint operators that will be useful in
the sequel. In Section 2 we introduce the new concept of normality of
operators in semi-Hilbertian space
$(\mathcal{H},\;\langle.\;|\;.\rangle_{A})$, called
$(\alpha,\beta)$-$A$-normality and we investigate various structural
properties of this class of operators. In Section 3,we state various
inequalities between the $A$-operator norm and $A$-numerical radius of
$(\alpha,\beta)$-$A$-normal operators.

We start by introducing some notations. The symbol $\mathcal{H}$ stands for a
complex Hilbert space with inner product $\langle.\;|\;.\;\rangle$ and norm
$\|.\|$ . We denote by $\mathcal{B}(\mathcal{H})$ the Banach algebra of all
bounded linear operators on $\mathcal{H}$, $I=I_{\mathcal{H}}$ being the
identity operator. $\mathcal{B}(\mathcal{H})^{+}$ is the cone of positive
(semi-definite) operators, i.e.,
$\mathcal{B}(\mathcal{H})^{+}=\\{A\in\mathcal{B}(\mathcal{H}):\langle
Au,\;|\;u\rangle\geq 0,\;\forall\;u\in\mathcal{H}\;\\}$. For every
$T\in\mathcal{L}(\mathcal{H})$ its range is denoted by $\mathcal{R}(T)$, its
null space by $\mathcal{N}(T)$ and its adjoint by $T^{*}$. If
$\mathcal{M}\subset\mathcal{H}$ is a closed subspace, $P_{\mathcal{M}}$ is the
orthogonal projection onto $\mathcal{M}$. The subspace $\mathcal{M}$ is
invariant for $T$ if $T\mathcal{M}\subset\mathcal{M}.$ We shall denote the set
of all complex numbers and the complex conjugate of a complex number $\lambda$
by $\mathbb{C}$ and $\overline{\lambda}$, respectively. The closure of
$\mathcal{R}(T)$ will be denoted by $\overline{\mathcal{R}(T)}$, and we shall
henceforth shorten $T-\lambda I$ by $T-\lambda.$ In addition, if
$T,S\in\mathcal{B}(\mathcal{H})$ then $T\geq S$ means that $T-S\geq 0.$ .
Any $A\in\mathcal{B}(\mathcal{H})^{+}$ defines a positive semi-definite
sesquilinear form, denoted by

$\langle.\;|\;.\rangle_{A}:\mathcal{B}(\mathcal{H})\times\mathcal{B}(\mathcal{H})\rightarrow\mathbb{C},\;\langle
u\;|\;v\rangle_{A}=\langle Au\;|\;v\rangle.$

We remark that $\langle u\;|\;v\rangle_{A}=\langle
A^{\frac{1}{2}}u\;|\;A^{\frac{1}{2}}v\rangle.$ The semi-norm induced by
$\langle.|.\rangle_{A}$, which is denoted by $\|.\|_{A}$, is given by
$\|u\|_{A}=\langle u\;|\;u\rangle_{A}^{\frac{1}{2}}=\|A^{\frac{1}{2}}u\|.$
This makes $\mathcal{H}$ into a semi-Hilbertian space. Observe that
$\|u\|_{A}=0$ if and only if $u\in\mathcal{N}(A).$ Then $\|.\|_{A}$ is a norm
if and only if $A$ is an injective operator, and the semi-normed space
$(\mathcal{B}(\mathcal{H}),\|.\|_{A})$ is complete if and only if
$\mathcal{R}(A)$ is closed. Moreover $\langle\;|\;\rangle_{A}$ induced a
seminorm on a certain subspace of $\mathcal{B}(\mathcal{H}),$ namely, on the
subset of all $T\in\mathcal{B}(\mathcal{H})$ for witch there exists a constant
$c>0$ such that $\|Tu\|_{A}\leq c\|u\|_{A}$ for every $u\in\mathcal{H}$ ($T$
is called $A$-bounded). For this operators it holds

$\|T\|_{A}=\sup_{u\in\overline{\mathcal{R}(A)}\\\
,u\not=0}\frac{\|Tu\|_{A}}{\|u\|_{A}}<\infty.$

It is straightforward that

$\|T\|_{A}=\sup\\{|\langle
Tu\;|\;v\rangle_{A}|:u,v\in\mathcal{H}\;\hbox{and}\;\;\|u\|_{A}\leq
1,\|v\|_{A}\leq 1\;\\}.$

######  Definition 1.1.

$([2])$ For $T\in\mathcal{B}(\mathcal{H}),$ an operator
$S\in\mathcal{B}(\mathcal{H})$ is called an $A$-adjoint of $T$ if for every
$u,v\in\mathcal{H}$

$\langle Tu\;|\;v\rangle_{A}=\langle u\;|\;Sv\rangle_{A},$

i.e., $AS=T^{*}A.$

If $T$ is an $A$-adjoint of itself, then $T$ is called an $A$-selfadjoint
operator $\big{(}AT=T^{*}A\big{)}.$

It is possible that an operator $T$ does not have an $A$-adjoint, and if $S$
is an $A$-adjoint of $T$ we may find many $A$-adjoints; In fact, in $AR=0$ for
some $R\in\mathcal{B}(\mathcal{H})$,then $S+R$ is an $A$-adjoint of $T$. The
set of all $A$-bounded operators which admit an $A$-adjoint is denoted by
$\mathcal{B}_{A}(\mathcal{H}).$ By Douglas Theorem $(\hbox{see}\;[6,\;10]\;)$
we have that

$\mathcal{B}_{A}(\mathcal{H})=\big{\\{}\;T\in\mathcal{B}(\mathcal{H})\;/\;\mathcal{R}(T^{*}A)\subset\mathcal{R}(A)\;\big{\\}}.$

If $T\in\mathcal{B}_{A}(\mathcal{H})$, then there exists a distinguished
$A$-adjoint operator of $T$, namely,the reduced solution of equation
$AX=T^{*}A$, i.e., $A^{\dagger}T^{*}A$. This operator is denoted by
$T^{\sharp}$. Therefore, $T^{\sharp}=A^{\dagger}T^{*}A$ and

$AT^{\sharp}=T^{*}A,\;\mathcal{R}(T^{\sharp})\subset\overline{\mathcal{R}(A)}\;\;\hbox{and}\;\;\mathcal{N}(T^{\sharp})=\mathcal{N}(T^{*}A).$

Note that in which $A^{\dagger}$ is the Moore-Penrose inverse of $A$. For more
details see $[2,3,4]$.

In the next proposition we collect some properties of $T^{\sharp}$ and its
relationship with the seminorm $\|\;.\;\|_{A}.$ For the proof see $[2,3,4]$ .

######  Proposition 1.1.

Let $T\in\mathcal{B}_{A}({\mathcal{H}})$. Then the following statements hold.

(1)
$T^{\sharp}\in\mathcal{B}_{A}({\mathcal{H}}),(T^{\sharp})^{\sharp}=P_{\overline{R(A)}}TP_{\overline{R(A)}}$
and $(T^{\sharp})^{\sharp})^{\sharp}=T^{\sharp}.$

(2) If $S\in\mathcal{B}_{A}(\mathcal{H})$ then
$TS\in\mathcal{B}_{A}({\mathcal{H}})$ and
$(TS)^{\sharp}=S^{\sharp}T^{\sharp}.$

(3) $T^{\sharp}T$ and $TT^{\sharp}$ are $A$-selfadjoint.

(4)
$\|T\|_{A}=\|T^{\sharp}\|_{A}=\|T^{\sharp}T\|^{\frac{1}{2}}_{A}=\|TT^{\sharp}\|^{\frac{1}{2}}_{A}$.

(5) $\|S\|_{A}=\|T^{\sharp}\|_{A}$ for every $S\in\mathcal{B}({\mathcal{H}})$
which is an $A$-adjoint of $T.$

(6) If $S\in\mathcal{B}_{A}({\mathcal{H}})$ then $\|TS\|_{A}=\|ST\|_{A}.$

We recapitulate very briefly the following definitions. For more details, the
interested reader is referred to $[2,4,21]$ and the references therein.

######  Definition 1.2.

Any operator $T\in\mathcal{B}_{A}(\mathcal{H})$ is called

(1) $A$-normal if $TT^{\sharp}=T^{\sharp}T.$

(2) $A$-isometry if $T^{\sharp}T=P_{\overline{\mathcal{R}(A)}}.$

(3) $A$-unitary if $T^{\sharp}T=TT^{\sharp}=P_{\overline{\mathcal{R}(A)}}.$

In $[21]$, the $A$-spectral radius of an operator
$T\in\mathcal{B}(\mathcal{H})$, denoted $r_{A}(T)$ is defined as

$r_{A}(T)=\limsup_{n\longrightarrow\infty}\left\|T^{n}\right\|_{A}^{\frac{1}{n}}$

and the $A$-numerical radius of an operator $T\in\mathcal{B}(\mathcal{H})$,
denoted by $\omega_{A}(T)$ is defined as

$w_{A}(T)=\sup\big{\\{}\;\left|\langle
Tu\;|\;u\rangle_{A}\right|,\;\;u\in\mathcal{H}\;,\left\|u\right\|_{A}=1\;\;\big{\\}}.$

It is a generalization of the concept of numerical radius of an operator.
Clearly, $\omega_{A}$ defines a seminorm on $\mathcal{B}(\mathcal{H})$.
Furthermore, for every $u\in\mathcal{H},$

$\left|\left\langle
Tu|\;u\right\rangle_{A}\right|\leq\omega_{A}(T)\left\|u\right\|_{A}^{2}.$

######  Remark 1.1.

If $T\in\mathcal{B}_{A}(\mathcal{H})$ is $A$-selfadjoint,then
$\left\|T\right\|_{A}=w_{A}(T)$ $(\hbox{see}\;[21])$.

######  Theorem 1.1.

$([21],\;\;\hbox{ Theorem 3.1}\;)$

A necessary and sufficient condition for an operator
$T\in\mathcal{B}_{A}(\mathcal{H})$ to be $A$-normal is that

(1) $\mathcal{R}(TT^{\sharp})\subset\overline{\mathcal{R}(A)}$ and

(2) $\|T^{\sharp}Tu\|_{A}=\|TT^{\sharp}u\|_{A}$ for all $u\in\mathcal{H}$.

##  2  PROPERTIES OF $(\alpha,\beta)$-$A$-NORMAL OPERATORS

In this section we define the class of $(\alpha,\beta)$-$A$-normal operators
according to semi-Hilbertian space structures and we give some their
proprieties.

Let $(\alpha,\beta)\in\mathbb{R}^{2}$ such that $0\leq\alpha\leq 1\leq\beta,$
an operator $T\in\mathcal{B}(\mathcal{H})$ is said to be
$(\alpha,\beta)$-normal $[7,19]$ if

$\alpha^{2}T^{*}T\leq TT^{*}\leq\beta^{2}T^{*}T,$

which is equivalent to the condition

$\alpha\|Tu\|\leq\|T^{*}u\|\leq\beta\|Tu\|$

for all $u\in\mathcal{H}.$ For $\alpha=1=\beta$ is a normal operator. For
$\alpha=1,$ we observe from the left inequality that $T^{*}$ is hyponormal and
for $\beta=1,$ from the right inequality we obtain that $T$ is hyponormal. In
recent work, Senthilkumar $[22]$ introduced $p$-$(\alpha,\beta)$-normal
operators as a generalization of $(\alpha,\beta)$\- normal operators . An
operator $T\in\mathcal{B}(\mathcal{H})$ is said to be $p$-$(\alpha,\beta)$\-
normal operators for $0<p\leq 1$ if

$\alpha^{2}(T^{*}T)^{p}\leq(TT^{*})^{p}\leq\beta^{2}(T^{*}T)^{p},0\leq\alpha\leq
1\leq\beta.$

When $p=1,$ this coincide with $(\alpha,\beta)$-normal operators.

Now we are going to consider an extension of the notion of $(\alpha,\beta)$
-normal operators, similar to those extensions of the notion of normality to
$A$-normality and hyponormality to $A$-hyponormality (see $[18,21]$).

######  Definition 2.1.

$([18])$ Let $A\in\mathcal{B}(\mathcal{H})^{+}$ and
$T\in\mathcal{B}(\mathcal{H})$. We say that $T$ is an $A$-positive if
$AT\in\mathcal{B}(\mathcal{H})^{+}$ which is equivalent to the condition

$\langle Tu\;|\;u\rangle_{A}\geq 0\;\;\forall\;u\;\in\mathcal{H}.$

We note $T\geq_{A}0.$

######  Definition 2.2.

$([18])$ An operator $T\in\mathcal{B}_{A}(\mathcal{H})$ is said to be
$A$-hyponormal if $T^{\sharp}T-TT^{\sharp}$ is $A$-positive i.e.,
$T^{\sharp}T-TT^{\sharp}\geq_{A}0.$

######  Proposition 2.1.

$([18])$ Let $T\in\mathcal{B}_{A}(\mathcal{H})$. Then $T$ is $A$-hyponormal if
and only if

$\|Tu\|_{A}\geq\|T^{\sharp}u\|_{A}\;\;\hbox{for all}\;\;u\in\mathcal{H}.$

As a generalization of $A$-normal and $A$-hyponormal operators, we introduce
$(\alpha,\beta)$-$A$-normal operators.

######  Definition 2.3.

An operator $T\in\mathcal{B}_{A}(\mathcal{H})$ is said to be
$(\alpha,\beta)$-$A$-normal for $0\leq\alpha\leq 1\leq\beta,$ if

$\beta^{2}T^{\sharp}T\geq_{A}TT^{\sharp}\geq_{A}\alpha^{2}T^{\sharp}T,$

which is equivalent to the condition

$\beta\|Tu\|_{A}\geq\|T^{\sharp}u\|_{A}\geq\alpha\|Tu\|_{A},\;\hbox{for
all}\;u\in\mathcal{H}.$

When $A=I$ (the identity operator), this coincide with $(\alpha,\beta)$-normal
operator.

  * •

For $\alpha=1=\beta$ is a $A$ normal operator.

  * •

For $\beta=1,$ we observe from the right inequality that $T$ is
$A$-hyponormal.

  * •

For $\alpha=1,$ and $\mathcal{N}(A)$ is invariant subspace for $T$ from the
right inequality we obtain that $T^{\sharp}$ is $A$-hyponormal.

######  Remark 2.1.

(1) Every $A$-normal operator is $(\alpha,\beta)$-$A$-normal operator.

(2) If $A$ is injective,then $(1,1)$-$A$-normal is $A$-normal operator.

(3) If $\mathcal{R}(TT^{\sharp})\subset\mathcal{R}(A)$,then $(1,1)$-$A$-normal
is $A$-normal operator.

We give an example of $(\alpha,\beta)$-$A$-normal operator which is neither
$A$-normal nor $A$-hyponormal.

######  Example 2.1.

Let $A=\left(\begin{array}[]{cc}1&0\\\ 0&2\\\ \end{array}\right)$ and
$T=\left(\begin{array}[]{ccc}1&2\\\ 0&1\\\
\end{array}\right)\in\mathcal{B}(\mathbb{C}^{2}).$ It easy to check that

$A\geq
0,\mathcal{R}(T^{*}A)\subset\mathcal{R}(A)\hbox{and}\;\;T^{\sharp}=\left(\begin{array}[]{ccc}1&0\\\
1&1\\\
\end{array}\right),T^{\sharp}T\not=TT^{\sharp}\hbox{and}\;\;\|Tu\|_{A}\not\geq\|T^{\sharp}u\|_{A}.$

$T$ is neither $A$-normal nor $A$-hyponormal. Moreover

$10T^{\sharp}T\geq_{A}TT^{\sharp}\geq_{A}\frac{1}{6}T^{\sharp}T.$

So $T$ is $(\displaystyle\frac{1}{\sqrt{6}},\sqrt{10})$-$A$-normal operator.

The following theorem gives a necessary and sufficient conditions that an
operator to be $(\alpha,\beta)$-$A$-normal. It is similar to
$[14,\;\hbox{Theorem}\;2.3].$

######  Theorem 2.1.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ and $(\alpha,\beta)\in\mathbb{R}^{2}$
such that $0\leq\alpha\leq 1\leq\beta$. Then $T$ is
$(\alpha,\beta)$-$A$-normal if and only if the following conditions are
satisfied

$\left\\{\begin{array}[]{lll}\lambda^{2}TT^{\sharp}+2\alpha^{2}\lambda
T^{\sharp}T+TT^{\sharp}\geq_{A}0,\;\;\;\;\hbox{for
all}\;\;\lambda\in\mathbb{R}\quad\quad\quad(1)\\\ \\\ \hbox{and}\\\ \\\
\lambda^{2}T^{\sharp}T+2\lambda
TT^{\sharp}+\beta^{4}T^{\sharp}T\geq_{A}0,\;\;\;\;\hbox{for
all}\;\;\lambda\in\mathbb{R}\quad\quad\quad(2).\end{array}\right.$

###### Proof.

Assume that the conditions (1) and (2) are satisfied and prove that $T$ is
$(\alpha,\beta)$-$A$-normal.

In fact we have by using elementary properties of real quadratic forms

$\displaystyle\lambda^{2}TT^{\sharp}+2\alpha^{2}\lambda
T^{\sharp}T+TT^{\sharp}\geq_{A}0$ $\displaystyle\Leftrightarrow$
$\displaystyle\left\langle(\lambda^{2}TT^{\sharp}+2\alpha^{2}\lambda
T^{\sharp}T+TT^{\sharp})u\;|\;u\right\rangle\geq_{A}0,\;\;\forall\;u\in\mathcal{H},\;\forall\;\lambda\in\mathbb{R}$
$\displaystyle\Leftrightarrow$
$\displaystyle\lambda^{2}\left\|T^{\sharp}u\right\|_{A}^{2}+2\alpha^{2}\lambda\left\|Tu\right\|_{A}^{2}+\left\|T^{\sharp}u\right\|_{A}^{2}\geq_{A}0,\;\;\forall\;u\in\mathcal{H},\;\forall\;\lambda\in\mathbb{R}$
$\displaystyle\Leftrightarrow$
$\displaystyle\alpha\left\|Tu\right\|_{A}\leq\left\|T^{\sharp}u\right\|_{A},\;\;\forall\;u\in\mathcal{H}.$

Similarly

$\displaystyle\lambda^{2}T^{\sharp}T+2\lambda
TT^{\sharp}+\beta^{4}T^{\sharp}T\geq_{A}0$ $\displaystyle\Leftrightarrow$
$\displaystyle\left\langle(\lambda^{2}T^{\sharp}T+2\lambda
TT^{\sharp}+\beta^{4}T^{\sharp}T)u\;|\;u\right\rangle\geq_{A}0,\;\;\forall\;u\in\mathcal{H},\;\forall\;\lambda\in\mathbb{R}$
$\displaystyle\Leftrightarrow$
$\displaystyle\lambda^{2}\left\|Tu\right\|_{A}^{2}+2\lambda\left\|T^{\sharp}u\right\|_{A}^{2}+\beta^{4}\left\|Tu\right\|_{A}^{2}\geq_{A}0,\;\;\forall\;u\in\mathcal{H},\;\forall\;\lambda\in\mathbb{R}$
$\displaystyle\Leftrightarrow$
$\displaystyle\left\|T^{\sharp}u\right\|_{A}\leq\beta\left\|Tu\right\|_{A},\;\;\forall\;u\in\mathcal{H}.$

Consequently

$\alpha\left\|Tu\right\|_{A}\leq\left\|T^{\sharp}u\right\|_{A}\leq\beta\left\|Tu\right\|_{A},\;\;\forall\;u\in\mathcal{H}.$

So $T$ is $(\alpha,\beta)$-$A$-normal as desired.

The proof of the converse seems obvious.

∎

######  Proposition 2.2.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ such that $\mathcal{N}(A)$ is invariant
subspace for $T$ and let $(\alpha,\beta)\in\mathbb{R}^{2}$ such that
$0<\alpha\leq 1\leq\beta$ . Then $T$ is an $(\alpha,\beta)$-$A$-normal if and
only if $T^{\sharp}$ is
$(\displaystyle\frac{1}{\beta},\frac{1}{\alpha})$-$A$-normal operator.

###### Proof.

First assume that $T$ is $(\alpha,\beta)$-$A$-normal operator. We have for all
$u\in\mathcal{H}$

$\alpha\left\|Tu\right\|_{A}\leq\left\|T^{\sharp}u\right\|_{A}\leq\beta\left\|Tu\right\|_{A}.$

It follows that

$\frac{1}{\beta}\left\|T^{\sharp}u\right\|_{A}\leq\left\|Tu\right\|_{A}\;\;\hbox{and}\;\left\|Tu\right\|_{A}\leq\frac{1}{\alpha}\left\|T^{\sharp}u\right\|_{A}.$

On the other hand,since $\mathcal{N}(A)$ is invariant subspace for $T$ we
observe that $TP_{\overline{\mathcal{R}(A)}}=P_{\overline{\mathcal{R}(A)}}T$
and $AP_{\overline{\mathcal{R}(A)}}=P_{\overline{\mathcal{R}(A)}}A=A$ and it
follows that

$\left\|\big{(}T^{\sharp}\big{)}^{\sharp}u\right\|_{A}=\left\|P_{\overline{\mathcal{R}(A)}}TP_{\overline{\mathcal{R}(A)}}u\right\|_{A}=\left\|Tu\right\|_{A}.$

Consequently

$\frac{1}{\beta}\left\|T^{\sharp}u\right\|_{A}\leq\left\|\big{(}T^{\sharp}\big{)}^{\sharp}u\right\|_{A}\leq\frac{1}{\alpha}\left\|T^{\sharp}u\right\|_{A}$

for all $u\in\mathcal{H}.$ Therefore $T^{\sharp}$ is
$(\displaystyle\frac{1}{\beta},\frac{1}{\alpha})$-$A$-normal operator.

Conversely assume that $T^{\sharp}$ is
$(\displaystyle\frac{1}{\beta},\frac{1}{\alpha})$-$A$-normal operator. We have

$\frac{1}{\beta}\left\|T^{\sharp}u\right\|_{A}\leq\left\|\big{(}T^{\sharp}\big{)}^{\sharp}u\right\|_{A}\leq\frac{1}{\alpha}\left\|T^{\sharp}u\right\|_{A}$

for all $u\in\mathcal{H},$ and from which it follows that

$\frac{1}{\beta}\left\|T^{\sharp}u\right\|_{A}\leq\left\|Tu\right\|_{A}\leq\frac{1}{\alpha}\left\|T^{\sharp}u\right\|_{A}$

for all $u\in\mathcal{H}.$ Hence

$\alpha\left\|Tu\right\|_{A}\leq\left\|T^{\sharp}u\right\|_{A}\leq\beta\left\|Tu\right\|_{A},\;\forall\;u\in\mathcal{H}.$

This completes the proof. ∎

The following corollary is a immediate consequence of Proposition 2.2.

######  Corollary 2.1.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ such that $\mathcal{N}(A)$ is invariant
subspace for $T$ and let $(\alpha,\beta)\in\mathbb{R}^{2}$ such that
$0<\alpha\leq 1\leq\beta$ and $\alpha\beta=1$. Then $T$ is an
$(\alpha,\beta)$-$A$-normal if and only if $T^{\sharp}$ is
$(\alpha,\beta)$-$A$-normal operator.

######  Remark 2.2.

$(\alpha,\beta)$-$A$-normality is not translation invariant, more precisely,
there exists an operator $T\in\mathcal{B}_{A}(\mathcal{H})$ that $T$ is
$(\alpha,\beta)$-$A$-normal,but $T+\lambda$ is not $(\alpha,\beta)$-$A$-normal
for some $\lambda\in\mathbb{C}.$ The following example shows that such
operators exist:

######  Example 2.2.

Consider the operators $A=\left(\begin{array}[]{cc}1&0\\\ 0&2\\\
\end{array}\right)$ ,$T=\left(\begin{array}[]{ccc}1&2\\\ 0&1\\\
\end{array}\right)$ and $S=T+I=\left(\begin{array}[]{ccc}2&2\\\ 0&2\\\
\end{array}\right)\in\mathcal{B}(\mathbb{R}^{2})$. It is easily to check that
$T$ is $(\displaystyle\frac{1}{\sqrt{6}},\sqrt{10})$-$A$-normal, but $S$ is
not $(\displaystyle\frac{1}{\sqrt{6}},\sqrt{10})$-$A$-normal. So
$(\alpha,\beta)$-$A$-normality is not translation-invariant.

Similarly to $[12]$, we define the following quantities

$\mu_{A}^{1}(T)=\inf\bigg{\\{}\;\frac{Re\left\langle
Tu\;|\;u\right\rangle_{A}}{\|Tu\|_{A}},\;\|u\|_{A}=1,\;\;Tu\notin\mathcal{N}(A^{\frac{1}{2}})\;\bigg{\\}}$

and

$\mu_{A}^{2}(T)=\sup\bigg{\\{}\;\frac{Re\left\langle
Tu\;|\;u\right\rangle_{A}}{\|Tu\|_{A}},\;\|u\|_{A}=1,\;\;Tu\notin\mathcal{N}(A^{\frac{1}{2}})\;\bigg{\\}}.$

A.Saddi $[21,\;\hbox{Corollary}\;3.2\;]$ have shown that if $T$ is $A$-normal
operator such that $\mathcal{N}(A)$ is invariant subspace for $T$, then
$T-\lambda$ is $A$-normal. In $[18,\;\hbox{Theorem}\;2.7\;]$ the authors
proved this property for $A$-hyponormal operators. In the following theorem we
extend these results to $(\alpha,\beta)$-$A$-normal operators. This is a
generalization of $[12,\;\hbox{ Theorem}\;2.1].$

######  Theorem 2.2.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ such that $\mathcal{N}(A)$ is invariant
subspace for $T$ and $0\leq\alpha\leq 1\leq\beta$. The following statements
hold.

(1) If $T$ is $(\alpha,\beta)$-$A$-normal, then $\lambda T$ is
$(\alpha,\beta)$-$A$-normal for $\lambda\in\mathbb{C}$ .

(2) If $T$ is $(\alpha,\beta)$-$A$-normal, then $T+\lambda$ for
$\lambda\in\mathbb{C}$ is $(\alpha,\beta)$-$A$-normal, if one of the following
conditions holds:

(i) $\mu_{A}^{1}(\overline{\lambda}T)\geq 0$

(ii) $\mu_{A}^{1}(\overline{\lambda}T)<0,$
$|\lambda|^{2}+2|\lambda|\|T\|_{A}\mu_{A}^{1}(\overline{\lambda}T)>0.$

###### Proof.

(1) Since $\mathcal{N}(A)$ is invariant subspace for $T$ we observe that
$TP_{\overline{\mathcal{R}(A)}}=P_{\overline{\mathcal{R}(A)}}T$ and
$AP_{\overline{\mathcal{R}(A)}}=P_{\overline{\mathcal{R}(A)}}A=A.$ Let $T$ be
$(\alpha,\beta)$-$A$-normal then

$\displaystyle\beta^{2}T^{\sharp}T\geq_{A}TT^{\sharp}\geq_{A}\alpha^{2}T^{\sharp}T$
$\displaystyle\Leftrightarrow$
$\displaystyle\beta^{2}|\lambda|^{2}T^{\sharp}T\geq_{A}|\lambda|^{2}TT^{\sharp}\geq_{A}|\lambda|^{2}\alpha^{2}T^{\sharp}T$
$\displaystyle\Leftrightarrow$ $\displaystyle
A\overline{\lambda}T^{\sharp}\lambda T\geq A\lambda
T\overline{\lambda}T^{\sharp}\geq\alpha^{2}A\overline{\lambda}T^{\sharp}\lambda
T$ $\displaystyle\Leftrightarrow$ $\displaystyle
AP_{\overline{\mathcal{R}(A)}}\overline{\lambda}T^{\sharp}\lambda T\geq
AP_{\overline{\mathcal{R}(A)}}\lambda
T\overline{\lambda}T^{\sharp}\geq\alpha^{2}AP_{\overline{\mathcal{R}(A)}}\overline{\lambda}T^{\sharp}\lambda
T$ $\displaystyle\Leftrightarrow$ $\displaystyle\beta^{2}A(\lambda
T)^{\sharp}(\lambda T)\geq A(\lambda T)(\lambda
T)^{\sharp}\geq\alpha^{2}A(\lambda T)^{\sharp}(\lambda T)$
$\displaystyle\Leftrightarrow$ $\displaystyle\beta^{2}(\lambda
T)^{\sharp}(\lambda T)\geq_{A}(\lambda T)(\lambda
T)^{\sharp}\geq_{A}\alpha^{2}(\lambda T)^{\sharp}(\lambda T).$

Therefore $\lambda T$ is $(\alpha,\beta)$-$A$-normal operator.

(2) Assume that $T$ is $(\alpha,\beta)$-$A$-normal and the condition $(i)$
holds. We need to prove that

$\displaystyle\left\\{\begin{array}[]{c}\alpha^{2}\left\langle\big{(}T+\lambda\big{)}^{\sharp}\big{(}T+\lambda\big{)}u\;|\;u\right\rangle_{A}\leq\left\langle\big{(}T+\lambda\big{)}\big{(}T+\lambda\big{)}^{\sharp}u\;|\;u\right\rangle_{A}\\\
\\\
\left\langle\big{(}T+\lambda\big{)}\big{(}T+\lambda\big{)}^{\sharp}u\;|\;u\right\rangle_{A}\leq\beta^{2}\left\langle\big{(}T+\lambda\big{)}^{\sharp}\big{(}T+\lambda\big{)}u\;|\;u\right\rangle_{A}.\end{array}\right.$
(2.4)

In order To verify $(2.1)$ we have

$\displaystyle\alpha^{2}\left\langle\big{(}T+\lambda\big{)}^{\sharp}\big{(}T+\lambda\big{)}u\;|\;u\right\rangle_{A}$
$\displaystyle=$ $\displaystyle\alpha^{2}\bigg{\\{}\left\langle
T^{\sharp}Tu\;|u\right\rangle_{A}+\left\langle{\lambda}T^{\sharp}u\;|\;u\right\rangle_{A}+\left\langle\overline{\lambda}P_{\overline{\mathcal{R}(A)}}Tu\;|\;u\right\rangle_{A}$
$\displaystyle\quad\quad+|\lambda|^{2}\left\langle
P_{\overline{\mathcal{R}(A)}}u\;|\;u\right\rangle_{A}\bigg{\\}}$
$\displaystyle=$ $\displaystyle\alpha^{2}\bigg{\\{}\left\langle
T^{\sharp}Tu\;|u\right\rangle_{A}+2Re\left\langle\overline{\lambda}Tu\;|\;u\right\rangle_{A}+|\lambda|^{2}\left\|u\right\|_{A}^{2}\bigg{\\}}$
$\displaystyle\leq$ $\displaystyle\alpha^{2}\left\langle
T^{\sharp}Tu\;|u\right\rangle_{A}+\alpha^{2}\bigg{\\{}2Re\left\langle\overline{\lambda}Tu\;|\;u\right\rangle_{A}+|\lambda|^{2}\left\|u\right\|_{A}^{2}\bigg{\\}}.$

The condition $(i)$ implies that
$2Re\left\langle\overline{\lambda}Tu\;|\;u\right\rangle_{A}\geq 0$ and it
follows that

$\displaystyle\alpha^{2}\left\langle\big{(}T+\lambda\big{)}^{\sharp}\big{(}T+\lambda\big{)}u\;|\;u\right\rangle_{A}$
$\displaystyle\leq$ $\displaystyle\bigg{\\{}\left\langle
TT^{\sharp}u\;|u\right\rangle_{A}+2Re\left\langle\overline{\lambda}Tu\;|\;u\right\rangle_{A}+|\lambda|^{2}\left\|u\right\|_{A}^{2}\bigg{\\}}$
$\displaystyle=$
$\displaystyle\left\langle\big{(}T+\lambda\big{)}\big{(}T+\lambda\big{)}^{\sharp}u\;|\;u\right\rangle_{A}$
$\displaystyle=$ $\displaystyle\bigg{\\{}\left\langle
TT^{\sharp}u\;|u\right\rangle_{A}+2Re\left\langle\overline{\lambda}Tu\;|\;u\right\rangle_{A}+|\lambda|^{2}\left\|u\right\|_{A}^{2}\bigg{\\}}$
$\displaystyle\leq$
$\displaystyle\beta^{2}\left\langle\big{(}T+\lambda\big{)}^{\sharp}\big{(}T+\lambda\big{)}u\;|\;u\right\rangle_{A}$

and hence $T+\lambda$ is $(\alpha,\beta)$-$A$-normal. On the other hand if the
condition (ii) is satisfied then we have for $\lambda\not=0$

$\displaystyle|\lambda|^{2}+2|\lambda|\left\|T\right\|_{A}\mu_{A}^{1}(\overline{\lambda}T)$
$\displaystyle=$
$\displaystyle|\lambda|^{2}+2|\lambda|\bigg{(}\sup_{\|u\|_{A}=1}\|Tu\|_{A}\bigg{)}\bigg{(}\inf\bigg{\\{}\;\frac{Re\left\langle\overline{\lambda}Tu\;|\;u\right\rangle_{A}}{|\lambda|\|Tu\|_{A}},\;\|u\|_{A}=1,\;\;Tu\notin\mathcal{N}(A^{\frac{1}{2}})\;\bigg{\\}}\bigg{)}$
$\displaystyle\leq$
$\displaystyle|\lambda|^{2}+2\inf_{\|u\|_{A}=1}Re\left\langle\overline{\lambda}Tu\;|\;u\right\rangle_{A}$
$\displaystyle\leq$
$\displaystyle|\lambda|^{2}+2Re\left\langle\overline{\lambda}Tu\;|\;u\right\rangle_{A}.$

A similar argument used as above shows that $T+\lambda$ is
$(\alpha,\beta)$-$A$-normal.

∎

######  Corollary 2.2.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ be an $(\alpha,\beta)$-$A$-normal
operator. The following statement hold

(1) If $\mu_{A}^{1}(T)\geq 0$ then $T+\lambda$ is $(\alpha,\beta)$-$A$-normal
for every $\lambda>0.$

(1) If $\mu_{A}^{2}(T)\leq 0$ then $T+\lambda$ is $(\alpha,\beta)$-$A$-normal
for every $\lambda<0.$

###### Proof.

(1) For every $\lambda>0$ we have
$\mu_{A}^{1}(\overline{\lambda}T)=\mu_{A}^{1}(\lambda T)=\mu_{A}^{1}(T)\geq
0$.By using Theorem 2.2 (i) we have that $T+\lambda$ is an
$(\alpha,\beta)$-$A$-normal.

(2) For every $\lambda<0$ we have
$\mu_{A}^{1}(\overline{\lambda}T)=-\mu_{A}^{2}(T)\geq 0$.By using Theorem 2.2
(ii) we have that $T+\lambda$ is an $(\alpha,\beta)$-$A$-normal. ∎

######  Lemma 2.1.

$([18],\;\hbox{Lemma}\;2.1\;)$ Let $T,S\in\mathcal{B}(\mathcal{H})$ such that
$T\geq_{A}S$ and let $R\in\mathcal{B}_{A}(\mathcal{H}).$ Then the following
properties hold

(1) $R^{\sharp}TR\geq_{A}R^{\sharp}SR.$

(2) $RTR^{\sharp}\geq_{A}RSR^{\sharp}.$

(3) If $R$ is $A$-selfadjoint then $RTR\geq_{A}RSR.$

######  Proposition 2.3.

Let $T,V\in\mathcal{B}_{A}(\mathcal{H})$ such that $\mathcal{N}(A)$ is
invariant subspace for both $T$ and $V$.If $T$ is an
$(\alpha,\beta)$-$A$-normal $(0\leq\alpha\leq 1\leq\beta)$ and $V$ is an
$A$-isometry , then $VTV^{\sharp}$ is an $(\alpha,\beta)$-$A$-normal operator.

###### Proof.

Assume that
$\beta^{2}T^{\sharp}T\geq_{A}TT^{\sharp}\geq_{A}\alpha^{2}T^{\sharp}V$ and
$V^{\sharp}V=P_{\overline{\mathcal{R}(A)}}.$ This implies

$\displaystyle\beta^{2}\big{(}VTV^{\sharp}\big{)}^{\sharp}\big{(}VTV^{\sharp}\big{)}$
$\displaystyle=$
$\displaystyle\beta^{2}\bigg{(}\big{(}V^{\sharp})^{\sharp}T^{\sharp}V^{\sharp}VTV^{\sharp}\bigg{)}$
$\displaystyle=$
$\displaystyle\beta^{2}\bigg{(}P_{\overline{\mathcal{R}(A)}}VP_{\overline{\mathcal{R}(A)}}T^{\sharp}P_{\overline{\mathcal{R}(A)}}TV^{\sharp}\bigg{)}$
$\displaystyle=$
$\displaystyle\beta^{2}\bigg{(}VP_{\overline{\mathcal{R}(A)}}T^{\sharp}T\big{(}VP_{\overline{\mathcal{R}(A)}}\big{)}^{\sharp}\bigg{)}$
$\displaystyle\geq_{A}$ $\displaystyle
VP_{\overline{\mathcal{R}(A)}}TT^{\sharp}\big{(}VP_{\overline{\mathcal{R}(A)}}\big{)}^{\sharp}\;\quad\quad(\hbox{by
Lemma 2.1})$ $\displaystyle\geq_{A}$
$\displaystyle\big{(}VTV^{\sharp}\big{)}\big{(}VTV^{\sharp}\big{)}^{\sharp}.$

Similarly, we have

$\displaystyle\big{(}VTV^{\sharp}\big{)}\big{(}VTV^{\sharp}\big{)}^{\sharp}$
$\displaystyle=$ $\displaystyle
VP_{\overline{\mathcal{R}(A)}}TT^{\sharp}\big{(}VP_{\overline{\mathcal{R}(A)}}\big{)}^{\sharp}$
$\displaystyle\geq_{A}$
$\displaystyle\alpha^{2}VP_{\overline{\mathcal{R}(A)}}T^{\sharp}T\big{(}VP_{\overline{\mathcal{R}(A)}}\big{)}^{\sharp}\;\quad\quad(\hbox{by
Lemma 2.1})$ $\displaystyle\geq_{A}$
$\displaystyle\alpha^{2}\big{(}VTV^{\sharp}\big{)}^{\sharp}\big{(}VTV^{\sharp}\big{)}.$

The conclusion holds. ∎

######  Proposition 2.4.

Let $T,S\in\mathcal{B}_{A}(\mathcal{H})$ such that $T$ is
$(\alpha,\beta)$-$A$-normal and $S$ is $A$-selfadjoint. If
$T^{\sharp}S=ST^{\sharp}$ then $TS$ is $(\alpha,\beta)$-$A$-normal.

###### Proof.

Since $T$ is $(\alpha,\beta)$-$A$-normal we have for $u\in\mathcal{H}$

$\alpha\left\|TSu\right\|_{A}\leq\left\|T^{\sharp}Su\right\|_{A}\leq\beta\left\|TSu\right\|_{A}.$

On the other hand

$\displaystyle\left\|T^{\sharp}Su\right\|_{A}^{2}=\left\langle
T^{\sharp}Su\;|\;T^{\sharp}Su\right\rangle_{A}=\left\langle
AST^{\sharp}u\;|\;ST^{\sharp}u\right\rangle=\left\langle(TS)^{\sharp}u\;|\;(TS)^{\sharp}u\right\rangle_{A}=\left\|(TS)^{\sharp}u\right\|_{A}^{2}.$

This implies

$\alpha\left\|TSu\right\|_{A}\leq\left\|(TS)^{\sharp}u\right\|_{A}\leq\beta\left\|TSu\right\|_{A}.$

∎

######  Proposition 2.5.

Let $T,S\in\mathcal{B}_{A}(\mathcal{H})$ such that $T$ is
$(\alpha,\beta)$-$A$-normal and $S$ is $A$-unitary. If $TS=ST$ and
$\mathcal{N}(A)$ is invariant subspace for $T$ then $TS$ is
$(\alpha,\beta)$-$A$-normal.

###### Proof.

Since $\mathcal{N}(A)$ is invariant subspace for $T$ we observe that
$TP_{\overline{\mathcal{R}(A)}}=P_{\overline{\mathcal{R}(A)}}T$ and
$T^{\sharp}P_{\overline{\mathcal{R}(A)}}=P_{\overline{\mathcal{R}(A)}}T^{\sharp}.$
Let $S$ be $A$-unitary then
$S^{\sharp}S=SS^{\sharp}=P_{\overline{\mathcal{R}(A)}}.$

Now it is easy to see that

$\beta^{2}\bigg{(}(TS)^{\sharp}(TS)\bigg{)}=\beta^{2}\bigg{(}T^{\sharp}S^{\sharp}ST\bigg{)}=\beta^{2}\bigg{(}T^{\sharp}P_{\overline{\mathcal{R}(A)}}T\bigg{)}=\beta^{2}\bigg{(}P_{\overline{\mathcal{R}(A)}}T^{\sharp}TP_{\overline{\mathcal{R}(A)}}\bigg{)}.$

By using the fact that $T$ is $(\alpha,\beta)$-$A$-normal,it follows
immediately from Lemma 2.1 that

$\beta^{2}\bigg{(}(TS)^{\sharp}(TS)\bigg{)}\geq_{A}\underbrace{\bigg{(}P_{\overline{\mathcal{R}(A)}}TT^{\sharp}P_{\overline{\mathcal{R}(A)}}\bigg{)}}_{(1)}\geq_{A}\alpha^{2}\underbrace{(P_{\overline{\mathcal{R}(A)}}T^{\sharp}TP_{\overline{\mathcal{R}(A)}}\bigg{)}}_{(2)}.$

Notice that (1) gives

$(P_{\overline{\mathcal{R}(A)}}TT^{\sharp}P_{\overline{\mathcal{R}(A)}}=TP_{\overline{\mathcal{R}(A)}}T^{\sharp}=TSS^{\sharp}T^{\sharp}=TS(TS)^{\sharp}$

and similarly (2) gives

$(P_{\overline{\mathcal{R}(A)}}T^{\sharp}TP_{\overline{\mathcal{R}(A)}}=T^{\sharp}P_{\overline{\mathcal{R}(A)}}T=T^{\sharp}S^{\sharp}ST=(TS)^{\sharp}(TS).$

So

$\beta^{2}(TS)^{\sharp}(TS)\geq_{A}TS(TS)^{\sharp}\geq_{A}\alpha^{2}(TS)^{\sharp}(TS).$

Hence $TS$ is $(\alpha,\beta)$-$A$-normal operator. ∎

The following example proves that even if $T$ and $S$ are
$(\alpha,\beta)$-$A$–normal operators, their product $TS$ is not in general
$(\alpha,\beta)$-$A$-normal operator.

######  Example 2.3.

(1) Consider $T=\left(\begin{array}[]{ccc}0&0&1\\\ 0&1&0\\\ 1&0&0\\\
\end{array}\right)$ and $S=\left(\begin{array}[]{ccc}-1&0&0\\\ 0&-1&0\\\
0&0&-1\\\ \end{array}\right)$ which are $(\alpha,\beta)$-$I_{3}$-normal and
their product is $(\alpha,\beta)$-$I_{3}$-normal.

(2) Consider $T=\left(\begin{array}[]{cc}1&0\\\ 1&1\\\ \end{array}\right)$ and
$S=\left(\begin{array}[]{cc}-1&0\\\ 0&-1\\\ \end{array}\right)$ which are
$(\alpha,\beta)$-$I_{2}$-normal whereas their product
$TS=\left(\begin{array}[]{cc}-1&0\\\ -1&-1\\\ \end{array}\right)$ is not
$(\alpha,\beta)$-$I_{2}$-normal.

######  Theorem 2.3.

Let $T,S\in\mathcal{B}_{A}(\mathcal{H})$ such that $T$ is
$(\alpha,\beta)$-$A$-normal $(0\leq\alpha\leq 1\leq\beta)$ and $S$ is
$(\alpha^{\prime},\beta^{\prime})$ -$A$-normal $(0\leq\alpha^{\prime}\leq
1\leq\beta^{\prime}).$ Then the following statements hold:

(1) If $T^{\sharp}S=ST^{\sharp}$,then $TS$ is
$(\alpha\alpha^{\prime},\beta\beta^{\prime})$-$A$-normal operator.

(2) If $S^{\sharp}T=TS^{\sharp}$,then $ST$ is
$(\alpha\alpha^{\prime},\beta\beta^{\prime})$-$A$-normal operator.

###### Proof.

(1) Since $T$ is $(\alpha,\beta)$-$A$-normal and $S$ is
$(\alpha^{\prime},\beta^{\prime})$\- $A$-normal, it follows that for all
$u\in\mathcal{H}$

$\displaystyle\alpha\alpha^{\prime}\left\|TSu\right\|_{A}\leq\alpha^{\prime}\left\|T^{\sharp}Su\right\|_{A}$
$\displaystyle=$ $\displaystyle\alpha^{\prime}\left\|ST^{\sharp}u\right\|_{A}$
$\displaystyle\leq$ $\displaystyle\left\|S^{\sharp}T^{\sharp}u\right\|_{A}$
$\displaystyle=$ $\displaystyle\left\|(TS)^{\sharp}u\right\|_{A}$
$\displaystyle\leq$
$\displaystyle\beta^{\prime}\left\|ST^{\sharp}u\right\|_{A}$ $\displaystyle=$
$\displaystyle\beta^{\prime}\left\|T^{\sharp}Su\right\|_{A}$
$\displaystyle\leq$ $\displaystyle\beta\beta^{\prime}\left\|TSu\right\|_{A}.$

It follows that

$\alpha\alpha^{\prime}\left\|STu\right\|_{A}\leq\left\|(TS)^{\sharp}u\right\|_{A}\leq\beta\beta^{\prime}\left\|TSu\right\|_{A}.$

The proof of the second assertion is completed in much the same way as the
first assertion. ∎

The following example shows that the power of $(\alpha,\beta)$-$A$-normal
operator not necessarily an $(\alpha,\beta)$-$A$-normal.

######  Example 2.4.

Let $A=\left(\begin{array}[]{cc}1&0\\\ 0&2\\\ \end{array}\right)$ and
$T=\left(\begin{array}[]{ccc}1&2\\\ 0&1\\\
\end{array}\right)\in\mathcal{B}(\mathbb{C}^{2}).$ By Example 2.1,$T$ is
$(\displaystyle\frac{1}{\sqrt{6}},\sqrt{10})$-$A$-normal.However by direct
computation one can show that $T^{2}$ is is neither
$(\displaystyle\frac{1}{\sqrt{6}},\sqrt{10})$-$A$-normal nor
$(\displaystyle\frac{1}{6},10)$-$A$-normal. But it is
$(\displaystyle\frac{1}{36},100)$-$A$-normal i.e., $T^{2}$ is
$\big{(}\big{(}\displaystyle\frac{1}{\sqrt{6}}\big{)}^{2^{2}},(\sqrt{10})^{2^{2}}\big{)}$-$A$-normal.

Question. If $T\in\mathcal{B}_{A}(\mathcal{H})$ which is
$(\alpha,\beta)$-$A$-normal operator, is that true $T^{n}$ is
$(\alpha^{n^{2}},\beta^{n^{2}})$-$A$-normal operator?

######  Remark 2.3.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ ,then

(1) If $T$ is $A$-normal, then $r_{A}(T)=\left\|T\right\|_{A}$
(see,$[21,\hbox{Corollary}\;3.2]$).

(2) If $T$ is $A$-hyponormal, then $r_{A}(T)=\left\|T\right\|_{A}$
(see,$[18,\hbox{Theorem}\;2.6]$).

The following theorem presents a generalization of these results to
$(\alpha,\beta)$-$A$-normal. Our inspiration cames from
$[12,\;\hbox{Theorem}\;2.5]$.

######  Theorem 2.4.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ be an $(\alpha,\beta)$-$A$-normal such
that $T^{2^{n}}$ is $(\alpha,\beta)$-$A$-normal for every
$n\in\mathbb{N}$,too. Then, we have

$\frac{1}{\beta}\left\|T\right\|_{A}\leq r_{A}(T)\leq\left\|T\right\|_{A}.$

###### Proof.

It is we know that if $T\in\mathcal{B}_{A}(\mathcal{H})$ then

$\left\|T^{\sharp}T\right\|_{A}=\left\|TT^{\sharp}\right\|_{A}=\left\|T\right\|_{A}^{2}$

and if $T$ is $A$-selfadjoint then

$\left\|T^{2}\right\|_{A}=\left\|T\right\|_{A}^{2}.$

From the definition of $(\alpha,\beta)$-$A$-normal operator and Lemma 2.1.1 we
deduce that

$\beta^{2}\big{(}T^{\sharp}\big{)}^{2}T^{2}\geq_{A}\big{(}T^{\sharp}T\big{)}^{2}\geq_{A}\alpha^{2}\big{(}T^{\sharp}\big{)}^{2}T^{2}$

and so

$\sup_{\|u\|_{A}=1}\left\langle\big{(}T^{\sharp}\big{)}^{2}T^{2}u\;|\;u\right\rangle_{A}\geq\frac{1}{\beta^{2}}\sup_{\|u\|_{A}=1}\left\langle\big{(}T^{\sharp}T\big{)}^{2}u\;|\;u\right\rangle_{A}.$

Hence

$\left\|\big{(}T^{\sharp}\big{)}^{2}T^{2}\right\|_{A}^{2}\geq\frac{1}{\beta^{2}}\left\|\big{(}T^{\sharp}T\big{)}^{2}\right\|_{A}^{2}=\frac{1}{\beta^{2}}\left\|T\right\|_{A}^{4}.$

Now using a mathematical induction, we observe that for every positive integer
number $n$,

$\left\|\big{(}T^{\sharp}\big{)}^{2^{n}}T^{2^{n}}\right\|_{A}\geq\frac{1}{\beta^{2^{n+1}-2}}\left\|T\right\|_{A}^{2^{n+1}}.$

We have

$\displaystyle r_{A}(T)^{2}=r_{A}(T^{\sharp})r_{A}(T)$ $\displaystyle=$
$\displaystyle\lim\sup_{n\longrightarrow\infty}\left\|\big{(}T^{\sharp}\big{)}^{2^{n}}\right\|_{A}^{\frac{1}{2^{n}}}\lim\sup_{n\longrightarrow\infty}\left\|T^{2^{n}}\right\|_{A}^{\frac{1}{2^{n}}}$
$\displaystyle\geq$
$\displaystyle\lim_{n\longrightarrow\infty}\bigg{(}\left\|\big{(}T^{\sharp}\big{)}^{2^{n}}\right\|_{A}\left\|T^{2^{n}}\right\|_{A}\bigg{)}^{\frac{1}{2^{n}}}$
$\displaystyle\geq$
$\displaystyle\lim_{n\longrightarrow\infty}\bigg{(}\left\|\big{(}T^{\sharp}\big{)}^{2^{n}}T^{2^{n}}\right\|_{A}\bigg{)}^{\frac{1}{2^{n}}}$
$\displaystyle\geq$
$\displaystyle\frac{1}{\beta^{2}}\left\|T\right\|_{A}^{2}.$

Therefore, we get

$\frac{1}{\beta}\left\|T\right\|_{A}\leq r_{A}(T)\leq\left\|T\right\|_{A}.$

This completes the proof. ∎

Let $\mathcal{H}\overline{\otimes}\mathcal{H}$ denote the completion, endowed
with a reasonable uniform crose-norm, of the algebraic tensor product
$\mathcal{H}{\otimes}\mathcal{H}$ of $\mathcal{H}$ with $\mathcal{H}$. Given
non-zero $T,S\in\mathcal{B}(\mathcal{H}),$ let $T\otimes
S\in\mathcal{B}(\mathcal{H}\overline{\otimes}\mathcal{H})$ denote the tensor
product on the Hilbert space $\mathcal{H}\overline{\otimes}\mathcal{H}$, when
$T\otimes S$ is defined as follows

$\langle T\otimes
S(\xi_{1}\otimes\eta_{1})|\;(\xi_{2}\otimes\eta_{2})\rangle=\langle
T\xi_{1}|\;\xi_{2}\rangle\langle S\eta_{1}|\;\eta_{2}\rangle.$

The operation of taking tensor products $T\otimes S$ preserves many properties
of $T,S\in\mathcal{B}(\mathcal{H})$, but by no means all of them. Thus,
whereas $T\otimes S$ is normal if and only if $T$ and $S$ are normal $[15]$,
there exist paranormal operators $T$ and $S$ such that $T\otimes S$ is not
paranormal $[1]$. In $[9]$, Duggal showed that if for non-zero
$T,S\in\mathcal{B}(\mathcal{H}),T\otimes S$ is $p$-hyponormal if and only if
$T$ and $S$ are $p$-hyponormal. Thus result was extended to $p$-quasi-
hyponormal operators in $[16]$ .

Recall that for $T\in\mathcal{B}_{A}(\mathcal{H})$ and
$S\in\mathcal{B}_{B}(\mathcal{H})$, $T\otimes S$ is
$(\alpha,\beta)$-$(A\otimes B)$-normal operator with $0\leq\alpha\leq
1\leq\beta$, if

$\beta^{2}\big{(}T\otimes S\big{)}^{\sharp}\big{(}T\otimes
S\big{)}\geq_{A\otimes B}\big{(}T\otimes S\big{)}\big{(}T\otimes
S\big{)}^{\sharp}\geq_{A\otimes B}\alpha^{2}\big{(}T\otimes
S\big{)}^{\sharp}\big{(}T\otimes S\big{)}$

or equivalently

$\alpha\left\|\big{(}T\otimes S\big{)}\big{(}u\otimes
v\big{)}\right\|_{A\otimes B}\leq\left\|\big{(}T\otimes
S\big{)}^{\sharp}\big{(}u\otimes v\big{)}\right\|_{A\otimes
B}\leq\beta\left\|\big{(}T\otimes S\big{)}\big{(}u\otimes
v\big{)}\right\|_{A\otimes B},$

for all $u,v\in\mathcal{H}.$

######  Lemma 2.2.

$(\;[18],\;\hbox{Lemma}\;3.1\;)$
Let $T_{k},S_{k}\in\mathcal{B}(\mathcal{H}),\;k=1,2$ and Let
$A,B\in\mathcal{B}(\mathcal{H})^{+},$ such that $T_{1}\geq_{A}T_{2}\geq_{A}0$
and $S_{1}\geq_{B}S_{2}\geq_{B}0$, then

$\big{(}T_{1}\otimes S_{1}\big{)}\geq_{A\otimes B}\big{(}T_{2}\otimes
S_{2}\big{)}\geq_{A\otimes B}0.$

######  Proposition 2.6.

$(\;[18],\;\hbox{Proposition}\;3.2\;)$ Let
$T_{1},T_{2},S_{1},S_{2}\in\mathcal{B}(\mathcal{H})$ and let
$A,B\in\mathcal{B}(\mathcal{H})^{+}$ such that $T_{k}$ is $A$\- positive and
$S_{k}$ is $B$-positive for $k=1,2$. If $T_{1}\not=0$ and $S_{1}\not=0$,then
the following conditions are equivalents

(1) $T_{2}\otimes S_{2}\geq_{A\otimes B}T_{1}\otimes S_{1}$

(2) there exists $d>0$ such that $dT_{2}\geq_{A}T_{1}$ and
$d^{-1}S_{2}\geq_{B}S_{1}.$

The following theorem gives a necessary and sufficient condition for $T\otimes
S$ to be $(\alpha,\beta)$\- -$A\otimes B$-normal operator when $T$ and $S$ are
both nonzero operators.

######  Proposition 2.7.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ and let
$S\in\mathcal{B}_{B}(\mathcal{H})$ with $T\neq 0$ and $S\neq 0.$ Let
$(\alpha,\beta)\in\mathbb{R}^{2}$ and
$(\alpha^{\prime},\beta^{\prime})\in\mathbb{R}^{2}$ such that
$0\leq\alpha,\;\alpha^{\prime}\leq 1$ and $1\leq\beta,\;1\leq\beta^{\prime}.$
The following properties hold:

(1) If $T$ is an $(\alpha,\beta)$-$A$ normal and $S$ is an
$(\alpha^{\prime},\beta^{\prime})$-$B$-normal ,then $T\otimes S$ is a
$(\alpha\alpha^{\prime},\beta\beta^{\prime})$-$A\otimes B$-normal operator.

(2) If $T\otimes S$ is an $(\alpha,\beta)$-$A\otimes B$-normal, then there
exist two constants $d>0$ and $d_{0}>0$ such that $T$ is
$\displaystyle\big{(}\sqrt{d_{0}^{-1}}\alpha,\sqrt{{d}}\beta\big{)}$-$A$-normal
and $S$ is
$\big{(}\sqrt{d_{0}},\displaystyle\frac{1}{\sqrt{d}}\big{)}$-$B$-normal
operator.

###### Proof.

Assume that $T$ is an $(\alpha,\beta)$-$A$ normal and $S$ is an
$(\alpha^{\prime},\beta^{\prime})$-$B$-normal. By assumptions we have

$\beta^{2}T^{\sharp}T\geq_{A}TT^{\sharp}\geq_{A}\alpha^{2}T^{\sharp}T$

and

$\beta^{\prime 2}S^{\sharp}S\geq_{B}SS^{\sharp}\geq_{B}\alpha^{\prime
2}S^{\sharp}S.$

It follows from the inequalities above and Lemma 2.2 that

$\beta^{2}\beta^{\prime 2}T^{\sharp}T\otimes S^{\sharp}S\geq_{A\otimes
B}TT^{\sharp}\otimes SS^{\sharp}\geq_{A\otimes B}\alpha^{2}\alpha^{\prime
2}T^{\sharp}T\otimes S^{\sharp}S$

and so

$(\beta\beta^{\prime})^{2}\big{(}T\otimes S\big{)}^{\sharp}\big{(}T\otimes
S\big{)}\geq_{A\otimes B}\big{(}T\otimes S\big{)}\big{(}T\otimes
S\big{)}^{\sharp}\geq_{A\otimes B}(\alpha\alpha^{\prime})^{2}\big{(}T\otimes
S\big{)}^{\sharp}\big{(}T\otimes S\big{)}.$

Hence,$T\otimes S$ is a
$(\alpha\alpha^{\prime},\beta\beta^{\prime})$-$A\otimes B$-normal operator.

Conversely assume that $T\otimes S$ is a $(\alpha,\beta)$-$A\otimes B$-normal
operator.

We have

$\beta^{2}T^{\sharp}T\otimes S^{\sharp}S\geq_{A\otimes B}TT^{\sharp}\otimes
SS^{\sharp}\geq_{A\otimes B}\alpha^{2}T^{\sharp}T\otimes S^{\sharp}S.$

So

$\beta^{2}T^{\sharp}T\otimes S^{\sharp}S\geq_{A\otimes B}TT^{\sharp}\otimes
SS^{\sharp}$ (2.5)

and

$TT^{\sharp}\otimes SS^{\sharp}\geq_{A\otimes B}\alpha^{2}T^{\sharp}T\otimes
S^{\sharp}S$ (2.6)

We deduce from inequality $(2.2)$ and Proposition 2.6 that there exists a
constant $d>0$ such that

$\left\\{\begin{array}[]{lll}d\beta^{2}T^{\sharp}T\geq_{A}TT^{\sharp}\\\ \\\
\hbox{and}\\\ \\\ d^{-1}S^{\sharp}S\geq_{B}SS^{\sharp}\\\ \end{array}\right.$

$d\beta^{2}\sup_{\left\|u\right\|_{A}=1}\left\langle
T^{\sharp}Tu\;|\;u\right\rangle_{A}\geq\sup_{\left\|u\right\|_{A}=1}\left\langle
TT^{\sharp}u\;|\;u\right\rangle_{A}$

and so

$d\beta^{2}\left\|T^{\sharp}T\right\|_{A}\geq\left\|TT^{\sharp}\right\|_{A}.$

Thus, $d\beta^{2}\geq 1.$ Similarly, we obtain $d^{-1}\geq 1.$

On the other had by inequality $(2.3)$ and we can find a constant $d_{0}>0$
satisfies

$\left\\{\begin{array}[]{lll}d_{0}TT^{\sharp}\geq_{A}\alpha^{2}T^{\sharp}T\\\
\\\ \hbox{and}\\\ \\\ d_{0}^{-1}SS^{\sharp}\geq_{B}S^{\sharp}S\\\
\end{array}\right.$

It easily to see that

$\sqrt{d_{0}^{-1}}\alpha\leq 1\;\;\hbox{and}\;d_{0}\leq 1.$

Consequently we have

$\big{(}\sqrt{d}\beta\big{)}^{2}T^{\sharp}T\geq_{A}TT^{\sharp}\geq_{A}\big{(}\sqrt{d_{0}^{-1}}\alpha\big{)}^{2}T^{\sharp}T$

and

$\big{(}\frac{1}{\sqrt{d}}\big{)}^{2}S^{\sharp}S\geq_{B}SS^{\sharp}\geq_{B}\big{(}\sqrt{d_{0}}\big{)}^{2}S^{\sharp}S.$

This proof is completes. ∎

######  Theorem 2.5.

Let $T,S\in\mathcal{B}_{A}(\mathcal{H})$ such that $T$is
$(\alpha,\beta)$-$A$-normal and $S$ is $(\alpha^{\prime},\beta^{\prime})$-$A$
-normal operators with $0\leq\alpha\leq 1\leq\beta$ and
$0\leq\alpha^{\prime}\leq 1\leq\beta^{\prime}$. The following statements hold:

(1) If $T^{\sharp}S=ST^{\sharp}$, then $TS\otimes T$ is
$(\alpha^{2}\alpha^{\prime},\beta^{2}\beta^{\prime})$-$(A\otimes A)$-normal
operator and $TS\otimes S$ is

$(\alpha\alpha^{\prime 2},\beta\beta^{\prime 2})$-$A\otimes A$-normal operator

(2) If $S^{\sharp}T=TS^{\sharp}$,then $ST\otimes T$ is
$(\alpha^{\prime}\alpha^{2},\beta^{\prime}\beta^{2})$-$(A\otimes A)$-normal
operator and $ST\otimes S$ is

$(\alpha^{\prime 2}\alpha,\beta^{\prime 2}\beta)$-$A\otimes A$-normal
operator.

###### Proof.

The proof is an immediate consequence of Theorem 2.3 and Proposition 2.7. ∎

##  3  INEQUALITIES INVOLVING $A$-OPERATOR NORMS AND $A$-NUMERICAL RADIUS OF
$(\alpha,\beta)$-$A$-NORMAL OPERATORS

Drogomir and Moslehian $[7]$ have given various inequalities between the
operator norm and the numerical radius of $(\alpha,\beta)$-normal operators in
Hilbert spaces.

Motivated by this work, we will extended some of these inequalities to
$A$-operator norm and $A$-numerical radius $\omega_{A}$ of
$(\alpha,\beta)$-$A$-normal in semi-Hilbertian spaces by employing some known
results for vectors in inner product spaces. We start with the following lemma
reproduced from $[13]$.

######  Lemma 3.1.

Let $r\in\mathbb{R}$ and $u,v\in\mathcal{H}$ such that
$\left\|u\right\|_{A}\geq\left\|v\right\|_{A}$ and $u,v\notin\mathcal{N}(A)$
,then the following inequalities hold

$\left\|u\right\|_{A}^{2r}+\left\|v\right\|_{A}^{2r}-2\left\|u\right\|_{A}^{r}\left\|v\right\|_{A}^{r}\frac{Re\left\langle
u\;|\;v\right\rangle_{A}}{\left\|u\right\|_{A}\left\|v\right\|_{A}}\leq\left\\{\begin{array}[]{lll}r^{2}\left\|u\right\|_{A}^{2r-2}\left\|u-v\right\|_{A}^{2}\;\;\hbox{if}\;\;r\geq
1\\\ \\\ \hbox{and}\\\ \\\
\left\|v\right\|_{A}^{2r-2}\left\|u-v\right\|_{A}^{2}\;\;\hbox{if}\;\;r<1.\end{array}\right.$
(3.1)

######  Theorem 3.1.

$T\in\mathcal{B}_{A}(\mathcal{H})$ be an $(\alpha,\beta)$-$A$-normal operator.
Then

$\bigg{(}\alpha^{2r}+\beta^{2r}\bigg{)}\left\|T\right\|_{A}^{2}\leq\left\\{\begin{array}[]{lll}2\beta^{r}\omega_{A}(T^{2})+\beta^{2r-2}\left\|\beta
T-T^{\sharp}\right\|_{A}^{2}\;\;\hbox{if}\;\;r\geq 1\\\ \\\ \hbox{and}\\\ \\\
2\beta^{r}\omega_{A}(T^{2})+\left\|\beta
T-T^{\sharp}\right\|_{A}^{2}\;\;\hbox{if}\;\;r<1.\end{array}\right.$ (3.2)

###### Proof.

Firstly , assume that $r\geq 1$ and let $u\in\mathcal{H}$ with
$\left\|u\right\|_{A}=1$. Since $T$ is $(\alpha,\beta)$-$A$-normal

$\alpha^{2}\left\|Tu\right\|_{A}^{2}\leq\left\|T^{\sharp}u\right\|_{A}^{2}\leq\beta^{2}\left\|Tu\right\|_{A}^{2}$

we have

$\bigg{(}\alpha^{2r}+\beta^{2r}\bigg{)}\left\|Tu\right\|_{A}^{2r}\leq\beta^{2r}\left\|Tu\right\|_{A}^{2r}+\left\|T^{\sharp}u\right\|_{A}^{2r}.$

Applying Lemma 3.1 with the choices $u_{0}=\beta\ Tu$ and $v_{0}=T^{\sharp}u$
we get

$\left\|\beta
Tu\right\|_{A}^{2r}+\left\|T^{\sharp}u\right\|_{A}^{2r}-2\left\|\beta
Tu\right\|_{A}^{r-1}\left\|T^{\sharp}u\right\|_{A}^{r-1}Re\left\langle\beta
Tu\;|\;T^{\sharp}u\right\rangle_{A}\leq r^{2}\left\|\beta
Tu\right\|_{A}^{2r-2}\left\|\beta Tu-T^{\sharp}u\right\|_{A}^{2}.$ (3.3)

From which , it follows that

$\displaystyle\bigg{(}\alpha^{2r}+\beta^{2r}\bigg{)}\left\|Tu\right\|_{A}^{2r}$
$\displaystyle\leq$ $\displaystyle 2\left\|\beta
Tu\right\|_{A}^{r-1}\left\|T^{\sharp}u\right\|_{A}^{r-1}\left|\left\langle\beta
T^{2}u\;|\;u\right\rangle_{A}\right|$ (3.4) $\displaystyle+r^{2}\left\|\beta
Tu\right\|_{A}^{2r-2}\left\|\beta Tu-T^{\sharp}u\right\|_{A}^{2}.$

Taking the supremum in $(3.4)$ over $u\in\mathcal{H},\|u\|_{A}=1$ and using
the fact that

$\sup_{\left\|u\right\|_{A}=1}\left|\left\langle
T^{2}u\;|\;u\right\rangle\rangle_{A}\right|=\omega_{A}(T^{2})$

we get

$\bigg{(}\alpha^{2r}+\beta^{2r}\bigg{)}\left\|T\right\|_{A}^{2r}\leq
2\beta^{r}\left\|T\right\|_{A}^{2r-2}\omega_{A}(T^{2})+r^{2}\beta^{2r-2}\left\|T\right\|_{A}^{2r-2}\left\|\beta
T-T^{\sharp}\right\|_{A}^{2}.$

So

$\bigg{(}\alpha^{2r}+\beta^{2r}\bigg{)}\left\|T\right\|_{A}^{2}\leq
2\beta^{2r}\omega_{A}(T^{2})+r^{2}\beta^{2r-2}\left\|\beta
T-T^{\sharp}\right\|_{A}^{2}$

which is the first inequality in $(3.2).$

By employing a similar argument to that used in the first inequality in
$(3.1)$ , gives the second inequality of $(3.2).$

∎

######  Theorem 3.2.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ be an $A(\alpha.\beta)$-normal
operator. Then

$\omega_{A}(T)^{2}\leq\frac{1}{2}\bigg{(}\beta\left\|T\right\|_{A}^{2}+\omega_{A}(T^{2})\bigg{)}.$
(3.5)

###### Proof.

Since for all $u,v$ and $e\in\mathcal{H}$

$\left|\left\langle u|\;v\right\rangle_{A}-\left\langle
u|\;e\right\rangle_{A}\left\langle
e|\;v\right\rangle_{A}\right|\geq\left|\left\langle
u|\;e\right\rangle_{A}\left\langle
e|\;v\right\rangle_{A}\right|-\left|\left\langle
u|\;v\right\rangle_{A}\right|$

we have by applying the inequalities reproduced from $[11]$

$\left\|u\right\|_{A}\left\|v\right\|_{A}\geq\left|\left\langle
u|\;v\right\rangle_{A}-\left\langle u|\;e\right\rangle_{A}\left\langle
e|\;v\right\rangle_{A}\right|+\left|\left\langle
u|\;e\right\rangle_{A}\left\langle
e|\;v\right\rangle_{A}\right|\geq\left|\left\langle
u|\;v\right\rangle_{A}\right|$

that

$\left|\left\langle u\;|\;e\right\rangle_{A}\right|\left|\left\langle
e\;|\;v\right\rangle_{A}\right|\leq\frac{1}{2}\bigg{(}\left\|u\right\|_{A}\left\|v\right\|_{A}+\left|\left\langle
u\;|\;v\right\rangle_{A}\right|\bigg{)}$ (3.6)

for all $u,v,e\in\mathcal{H}$ with $\left\|e\right\|_{A}=1.$

Let $x\in\mathcal{H}$ with $\left\|x\right\|_{A}=1$ and choosing in $(3.6)$
$u=Tx$ , $v=T^{\sharp}x$ and $e=x$ we get

$\left|\left\langle Tx\;|\;x\right\rangle_{A}\right|\left|\left\langle
x\;|\;T^{\sharp}x\right\rangle_{A}\right|\leq\frac{1}{2}\bigg{(}\left\|Tx\right\|_{A}\left\|T^{\sharp}x\right\|_{A}+\left|\left\langle
Tx\;|\;T^{\sharp}x\right\rangle_{A}\right|\bigg{)}.$ (3.7)

Since $T$ is $(\alpha,\beta)$-$A$-normal, it follows that

$\left|\left\langle
Tx\;|\;x\right\rangle_{A}\right|^{2}\leq\frac{1}{2}\bigg{(}\beta\left\|Tx\right\|_{A}^{2}+\left|\left\langle
T^{2}x\;|\;x\right\rangle_{A}\right|\bigg{)}.$ (3.8)

Tanking the supremum over $x\in\mathcal{H}$ $\left\|x\right\|_{A}=1,$ we get
the desired inequality in $(3.5).$ ∎

######  Theorem 3.3.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ be an $(\alpha,\beta)$-$A$-normal
operator and $\lambda\in\mathbb{C}$. Then

$\alpha\left\|T\right\|_{A}^{2}\leq\omega_{A}(T^{2})+\frac{2\beta\left\|T-\lambda
T^{\sharp}\right\|_{A}^{2}}{\big{(}1+|\lambda|\alpha\big{)}^{2}}.$ (3.9)

###### Proof.

For $\lambda=0$, the inequality (3.9) is obvious. Assume that
$\lambda\not=0$.From the following inequality $[13]$

$\frac{1}{2}\bigg{(}\left\|u\right\|+\left\|v\right\|\bigg{)}\left\|\frac{u}{\left\|u\right\|}-\frac{v}{\left\|v\right\|}\right\|\leq\left\|u-v\right\|,\;;\;u,v\in\mathcal{H}-\\{0\\}$

which is well known in the literature as the Dunkl-Williams inequality, it
follows that

$\frac{1}{2}\big{(}\left\|u\right\|_{A}+\left\|v\right\|_{A}\big{)}\left\|\frac{u}{\left\|u\right\|_{A}}-\frac{v}{\left\|v\right\|_{A}}\right\|_{A}\leq\left\|u-v\right\|_{A}\;\;\hbox{for
all }\;\;u,v\in\mathcal{H}\;/\;u,v\notin\mathcal{N}(A).$

A simple computation shows that

$\left\|\frac{u}{\left\|u\right\|_{A}}-\frac{v}{\left\|v\right\|_{A}}\right\|_{A}^{2}=2-2\frac{Re\left\langle
u\;|\;v\right\rangle_{A}}{\left\|u\right\|_{A}\left\|v\right\|_{A}}\leq\frac{4\left\|u-v\right\|_{A}^{2}}{\big{(}\left\|u\right\|_{A}+\left\|v\right\|_{A}\big{)}^{2}}$

which shows that

$\frac{\left\|u\right\|_{A}\left\|v\right\|_{A}-\left|\left\langle
u\;|\;v\right\rangle_{A}\right|}{\left\|u\right\|_{A}\left\|v\right\|_{A}}\leq\frac{2\left\|u-v\right\|_{A}^{2}}{\big{(}\left\|u\right\|_{A}+\left\|v\right\|_{A}\big{)}^{2}},\;\;\hbox{for
all }\;\;u,v\in\mathcal{H}\;/\;u,v\notin\mathcal{N}(A)\;\;$

and so

$\left\|u\right\|_{A}\left\|v\right\|_{A}\leq\left|\left\langle
u\;|\;v\right\rangle_{A}\right|+\frac{2\left\|u\right\|_{A}\left\|v\right\|_{A}}{\big{(}\left\|u\right\|_{A}+\left\|v\right\|_{A}\big{)}^{2}}\left\|u-v\right\|_{A}^{2}.$

Let $x\in\mathcal{H}$ with $\left\|x\right\|_{A}=1$ and consider $u=Tx$ and
$v=\lambda T^{\sharp}x$ with
$x\notin\mathcal{N}(A^{\frac{1}{2}}T)=\mathcal{N}(A^{\frac{1}{2}}T^{\sharp})$
we obtain

$\left\|Tx\right\|_{A}\left\|\lambda
T^{\sharp}x\right\|_{A}\leq\left|\left\langle Tx\;|\;\lambda
T^{\sharp}x\right\rangle_{A}\right|+\frac{2\left\|Tx\right\|_{A}\left\|\lambda
T^{\sharp}x\right\|_{A}}{\big{(}\left\|Tx\right\|_{A}+\left\|\lambda
T^{\sharp}x\right\|_{A}\big{)}^{2}}\left\|Tx-\lambda
T^{\sharp}x\right\|_{A}^{2}.$

Since $T$ being $(\alpha,\beta)$-$A$-normal operator, we get

$\alpha\left\|Tx\right\|_{A}^{2}\leq\left|\left\langle
T^{2}x\;|\;x\right\rangle_{A}\right|+\frac{2\beta\left\|Tx\right\|_{A}^{2}}{\big{(}\left\|Tx\right\|_{A}+\alpha|\lambda|\left\|Tx\right\|_{A}\big{)}^{2}}\left\|Tx-\lambda
T^{\sharp}x\right\|_{A}^{2}.$

Tanking the supremum over $x\in\mathcal{H};$ $\left\|x\right\|_{A}=1,$ we get
the desired inequality in $(3.9).$

∎

######  Theorem 3.4.

Let $T\in\mathcal{B}_{A}(\mathcal{H})$ be an $(\alpha,\beta)$-$A$-normal
operator and $\lambda\in\mathbb{C}$. Then

$\bigg{[}\alpha^{2}-\big{(}\frac{1}{|\lambda|}+\beta\big{)}^{2}\bigg{]}\left\|T\right\|_{A}^{4}\leq\omega_{A}(T^{2}).$
(3.10)

###### Proof.

We apply the following inequality inspired from $[{8}]$

$0\leq\left\|u\right\|_{A}^{2}\left\|v\right\|_{A}^{2}-\left|\left\langle
u\;|\;v\right\rangle_{A}\right|\leq\frac{1}{|\lambda|^{2}}\left\|u\right\|_{A}^{2}\left\|u-v\right\|_{A}^{2}$
(3.11)

for all $u,v\in\mathcal{H}$ and $\lambda\in\mathbb{C}\;,\lambda\not=0.$

Let $x\in\mathcal{H}$ and set $u=Tx$ and $v=T^{\sharp}x$ in $(3.11)$ we get

$\alpha^{2}\left\|Tx\right\|_{A}^{4}\leq\left|\left\langle
T^{2}x\;|\;x\right\rangle_{A}\right|^{2}+\frac{1}{|\lambda|^{2}}\left\|Tx\right\|_{A}^{2}\big{(}1+|\lambda|\beta\big{)}^{2}\left\|Tx\right\|_{A}^{2}.$
(3.12)

Tanking the supremum over $x\in\mathcal{H}$ $\left\|x\right\|_{A}=1,$ we get
the desired inequality in $(3.10).$ ∎

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