$A$-numerical radius inequalities for semi-Hilbertian space operators

TL;DR

This paper establishes new inequalities and characterizations for the $A$-numerical radius of operators in semi-Hilbertian spaces, providing bounds and formulas that extend classical results to this generalized setting.

Contribution

It introduces a novel characterization of the $A$-numerical radius using $A$-adjoint operators and derives new bounds for it in semi-Hilbertian spaces.

Findings

Characterization of $w_A(T)$ via supremum over $A$-semi-norms.

Derived upper and lower bounds for $w_A(T)$ involving $A$-operator semi-norm.

Provided bounds for $A$-numerical radius of operator commutators, anticommutators, and products.

Abstract

Let $A$ be a positive bounded operator on a Hilbert space $\big(\mathcal{H}, \langle \cdot, \cdot\rangle \big)$. The semi-inner product ${\langle x, y\rangle}_A := \langle Ax, y\rangle$, $x, y\in\mathcal{H}$ induces a semi-norm ${\|\cdot\|}_A$ on $\mathcal{H}$. Let ${\|T\|}_A$ and $w_A(T)$ denote the $A$-operator semi-norm and the $A$-numerical radius of an operator $T$ in semi-Hilbertian space $\big(\mathcal{H}, {\|\cdot\|}_A\big)$, respectively. In this paper, we prove the following characterization of $w_A(T)$ \begin{align*} w_A(T) = \displaystyle{\sup_{\alpha^2 + \beta^2 = 1}} {\left\|\alpha \frac{T + T^{\sharp_A}}{2} + \beta \frac{T - T^{\sharp_A}}{2i}\right\|}_A, \end{align*} where $T^{\sharp_A}$ is a distinguished $A$-adjoint operator of $T$. We then apply it to find upper and lower bounds for $w_A(T)$. In particular, we show that \begin{align*} \frac{1}{2}{\|T\|}_A \leq \max\Big\{\sqrt{1 - {|\cos|}^2_AT}, \frac{\sqrt{2}}{2}\Big\}w_A(T)\leq w_A(T), \end{align*} where ${|\cos|}_AT$ denotes the $A$-cosine of angle of $T$. Some upper bounds for the $A$-numerical radius of commutators, anticommutators, and products of semi-Hilbertian space operators are also given.

Equations

$$w_A(T)=\underset{{\alpha }^2+{\beta }^2=1}{\sup }{\Vert \alpha \frac{T+T^{{♯}_A}}{2}+\beta \frac{T-T^{{♯}_A}}{2i}\Vert }_A,$$

$$\displaystyle\frac{1}{2}{\|T\|}_{A}\leq\max\Big{\{}\sqrt{1-{|\cos|}^{2}_{A}T},\frac{\sqrt{2}}{2}\Big{\}}w_{A}(T)\leq w_{A}(T),$$

$$\displaystyle{\|T\|}_{A}:=\displaystyle{\sup_{x\in\overline{\mathcal{R}(A)},x\neq 0}}\frac{{\|Tx\|}_{A}}{{\|x\|}_{A}}=\inf\Big{\{}c>0:\,\,{\|Tx\|}_{A}\leq c{\|x\|}_{A},x\in\mathcal{H}\Big{\}}<\infty.$$

$$\displaystyle{\|T\|}_{A}=\sup\Big{\{}|{\langle Tx,y\rangle}_{A}|:\,\,x,y\in\overline{\mathcal{R}(A)},{\|x\|}_{A}={\|y\|}_{A}=1\Big{\}}.$$

$$\displaystyle{\|T\|}_{A}=\sup\Big{\{}{\langle Tx,x\rangle}_{A}:\,\,x\in\mathcal{H},{\|x\|}_{A}=1\Big{\}}.$$

$${\Vert T^{{♯}_A}T\Vert }_A={\Vert T{T}^{{♯}_A}\Vert }_A={\Vert T\Vert }_A^2={\Vert T^{{♯}_A}\Vert }_A^2.$$

$$\displaystyle w(T)=\sup\Big{\{}|\langle Tx,x\rangle|:\,\,x\in\mathcal{H},\|x\|=1\Big{\}}.$$

$$\frac{1}{2}\Vert T\Vert \le w(T)\le \Vert T\Vert .$$

$$\displaystyle w_{A}(T)=\sup\Big{\{}\big{|}{\langle Tx,x\rangle}_{A}\big{|}:\,\,x\in\mathcal{H},\,{\|x\|}_{A}=1\Big{\}},$$

$$\displaystyle c_{A}(T)=\inf\Big{\{}|{\langle Tx,x\rangle}_{A}|:\,\,x\in\mathcal{H},{\|x\|}_{A}=1\Big{\}},$$

$${|\cos |}_{A}T=\inf \left\{\frac{|{⟨Tx,x⟩}_A|}{{\Vert Tx\Vert }_A{\Vert x\Vert }_A}:x\in \mathcal{H},{\Vert Tx\Vert }_A{\Vert x\Vert }_A\ne 0\right\},$$

$${|\sin |}_{A}T=\sqrt{1-{|\cos |}_A^{2}T}.$$

$$w_A(T)=\underset{{\alpha }^2+{\beta }^2=1}{\sup }{\Vert \alpha \frac{T+T^{{♯}_A}}{2}+\beta \frac{T-T^{{♯}_A}}{2i}\Vert }_A.$$

$$\displaystyle w_{A}(T)\leq\frac{\sqrt{2}}{2}\sqrt{{\big{\|}TT^{\sharp_{A}}+T^{\sharp_{A}}T\big{\|}}_{A}}\leq{\|T\|}_{A},$$

$$\frac{1}{2}{\Vert T\Vert }_A\le \sqrt{\frac{w_A^2(T)}{2}+\frac{w_A(T)}{2}\sqrt{w_A^2(T)-c_A^2(T)}}\le w_A(T),$$

$$\displaystyle\frac{1}{2}{\|T\|}_{A}\leq\max\Big{\{}{|\sin|}_{A}T,\frac{\sqrt{2}}{2}\Big{\}}w_{A}(T)\leq w_{A}(T).$$

$$\displaystyle w_{A}(TS)\leq w_{A}(T){\|S\|}_{A}+\frac{1}{2}w_{A}\Big{(}(TS)^{\sharp_{A}}\pm T^{\sharp_{A}}S\Big{)}\leq 2w_{A}(T){\|S\|}_{A}.$$

$$\displaystyle w_{A}(TS^{\sharp_{A}}\pm ST^{\sharp_{A}})\leq{\big{\|}T^{\sharp_{A}}T+SS^{\sharp_{A}}\big{\|}}_{A}.$$

$$w_A(T)={\Vert T\Vert }_A.$$

$$\displaystyle\big{|}{\langle Tx,x\rangle}_{A}\big{|}\leq{\|Tx\|}_{A}{\|x\|}_{A}\leq{\|T\|}_{A},$$

$${⟨T(y+z),y+z⟩}_A={⟨Ty,y⟩}_A+2\text{mbox}Re{⟨Ty,z⟩}_A+{⟨Tz,z⟩}_A$$

$${⟨T(y-z),y-z⟩}_A={⟨Ty,y⟩}_A-2\text{mbox}Re{⟨Ty,z⟩}_A+{⟨Tz,z⟩}_A.$$

$$\displaystyle\mbox{Re}{\langle Ty,z\rangle}_{A}=\frac{1}{4}\Big{(}{\langle T(y+z),y+z\rangle}_{A}-{\langle T(y-z),y-z\rangle}_{A}\Big{)}.$$

$$\displaystyle\big{|}\mbox{Re}{\langle Ty,z\rangle}_{A}\big{|}\leq\frac{w_{A}(T)}{4}\big{(}{\|y+z\|}^{2}_{A}+{\|y-z\|}^{2}_{A}\big{)}.$$

$$\displaystyle\big{|}\mbox{Re}{\langle Ty,z\rangle}_{A}\big{|}\leq\frac{w_{A}(T)}{4}\big{(}2{\|y\|}^{2}_{A}+2{\|z\|}^{2}_{A}\big{)}=w_{A}(T).$$

$$\displaystyle 0\leq{\|T\|}_{A}^{2}-w_{A}^{2}(T)\leq\inf_{\gamma\in\mathbb{C}}\Big{\{}{\|T+\gamma I\|}_{A}^{2}-c_{A}^{2}(T+\gamma I)\Big{\}}.$$

$$\displaystyle{\|Tx\|}^{2}_{A}-\big{|}{\langle Tx,x\rangle}_{A}\big{|}^{2}={\|Tx-\gamma x\|}^{2}_{A}-\big{|}{\langle Tx-\gamma x,x\rangle}_{A}\big{|}^{2}\qquad(\gamma\in\mathbb{C}),$$

$$\displaystyle{\|Tx\|}^{2}_{A}-\big{|}{\langle Tx,x\rangle}_{A}\big{|}^{2}\leq{\|T-\gamma I\|}^{2}_{A}-c^{2}_{A}(T-\gamma I)\qquad(\gamma\in\mathbb{C}).$$

$$\displaystyle{\|Tx\|}^{2}_{A}-\big{|}{\langle Tx,x\rangle}_{A}\big{|}^{2}\leq\inf_{\gamma\in\mathbb{C}}\Big{\{}{\|T-\gamma I\|}^{2}_{A}-c^{2}_{A}(T-\gamma I)\Big{\}}.$$

$$w_A\left(\frac{e^{i\theta }T+(e^{i\theta }T{)}^{{♯}_A}}{2}\right)={\Vert \frac{e^{i\theta }T+(e^{i\theta }T{)}^{{♯}_A}}{2}\Vert }_A.$$

Full text

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where $T^{\sharp_{A}}$ is a distinguished $A$-adjoint operator of $T$. We then apply it to find upper and lower bounds for $w_{A}(T)$. In particular, we show that

[TABLE]

where ${|\cos|}_{A}T$ denotes the $A$-cosine of angle of $T$. Some upper bounds for the $A$-numerical radius of commutators, anticommutators, and products of semi-Hilbertian space operators are also given.

1. Introduction and preliminaries

Let $\mathbb{B}(\mathcal{H})$ denote the $C^{\ast}$-algebra of all bounded linear operators on a complex Hilbert space \big{(}\mathcal{H},\langle\cdot,\cdot\rangle\big{)} and the corresponding norm $\|\cdot\|$. Let the symbol $I$ stand for the identity operator on $\mathcal{H}$. If $T\in\mathbb{B}(\mathcal{H})$, then we denote by $\mathcal{R}(T)$ the range of $T$, and by $\overline{\mathcal{R}(T)}$ the norm closure of $\mathcal{R}(T)$. Throughout this paper, we assume that $A\in\mathbb{B}(\mathcal{H})$ is a positive operator and that $P$ is the orthogonal projection onto $\overline{\mathcal{R}(A)}$. Recall that $A$ is called positive, denoted by $A\geq 0$, if $\langle Ax,x\rangle\geq 0$ for all $x\in\mathcal{H}$. Such an operator $A$ induces a positive semidefinite sesquilinear form ${\langle\cdot,\cdot\rangle}_{A}:\,\mathcal{H}\times\mathcal{H}\rightarrow\mathbb{C}$ defined by ${\langle x,y\rangle}_{A}=\langle Ax,y\rangle$,  $x,y\in\mathcal{H}$. Denote by ${\|\cdot\|}_{A}$ the seminorm induced by ${\langle\cdot,\cdot\rangle}_{A}$, that is, ${\|x\|}_{A}=\sqrt{{\langle x,x\rangle}_{A}}$ for every $x\in\mathcal{H}$. It can be easily seen that ${\|\cdot\|}_{A}$ is a norm if and only if $A$ is an injective operator, and that $(\mathcal{H},{\|\cdot\|}_{A})$ is a complete space if and only if $\mathcal{R}(A)$ is closed in $\mathcal{H}$. The semi-inner product ${\langle\cdot,\cdot\rangle}_{A}$ induces a semi-norm on a certain subspace of $\mathbb{B}(\mathcal{H})$. Namely, given $T\in\mathbb{B}(\mathcal{H})$, if there exists $c>0$ such that ${\|Tx\|}_{A}\leq c{\|x\|}_{A}$ for all $x\in\overline{\mathcal{R}(A)}$, then it holds that

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We set \mathbb{B}^{A}(\mathcal{H}):=\Big{\{}T\in\mathbb{B}(\mathcal{H}):\,\,{\|T\|}_{A}<\infty\Big{\}}. It can be seen that $\mathbb{B}^{A}(\mathcal{H})$ is not generally a subalgebra of $\mathbb{B}(\mathcal{H})$ and ${\|T\|}_{A}=0$ if and only if $ATA=0$. In addition, for $T\in\mathbb{B}^{A}(\mathcal{H})$, we have

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An operator $T$ is called $A$-positive if $AT\geq 0$. Note that if $T$ is $A$-positive, then

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For $T\in\mathbb{B}(\mathcal{H})$, an operator $R\in\mathbb{B}(\mathcal{H})$ is called an $A$-adjoint of $T$ if for every $x,y\in\mathcal{H}$, we have ${\langle Tx,y\rangle}_{A}={\langle x,Ry\rangle}_{A}$, i.e., $AR=T^{*}A$. The existence of an $A$-adjoint operator is not guaranteed. In fact, an operator $T\in\mathbb{B}(\mathcal{H})$ may admit none, one or many $A$-adjoints. The set of all operators which admit $A$-adjoints is denoted by $\mathbb{B}_{A}(\mathcal{H})$. Note that $\mathbb{B}_{A}(\mathcal{H})$ is a subalgebra of $\mathbb{B}(\mathcal{H})$ which is neither closed nor dense in $\mathbb{B}(\mathcal{H})$. Moreover, the following inclusions $\mathbb{B}_{A}(\mathcal{H})\subseteq\mathbb{B}^{A}(\mathcal{H})\subseteq\mathbb{B}(\mathcal{H})$ hold with equality if $A$ is injective and has a closed range.

If $T\in\mathbb{B}_{A}(\mathcal{H})$, the reduced solution of the equation $AX=T^{*}A$ is a distinguished $A$-adjoint operator of $T$, which is denoted by $T^{\sharp_{A}}$; see [25]. Note that, $T^{\sharp_{A}}=A^{{\dagger}}T^{*}A$ in which $A^{{\dagger}}$ is the Moore–Penrose inverse of $A$. It is useful that if $T\in\mathbb{B}_{A}(\mathcal{H})$, then $AT^{\sharp_{A}}=T^{*}A$. An operator $T\in\mathbb{B}(\mathcal{H})$ is said to be $A$-selfadjoint if $AT$ is selfadjoint, i.e., $AT=T^{*}A$. Observe that if $T$ is $A$-selfadjoint, then $T\in\mathbb{B}_{A}(\mathcal{H})$. However it does not hold, in general, that $T=T^{\sharp_{A}}$. For example, consider the operators $A=\begin{bmatrix}1&1\\ 1&1\end{bmatrix}$ and $T=\begin{bmatrix}2&2\\ 0&0\end{bmatrix}$. Then simple computations show that $T$ is $A$-selfadjoint and $T^{\sharp_{A}}=\begin{bmatrix}1&1\\ 1&1\end{bmatrix}\neq T$. More precisely, if $T\in\mathbb{B}_{A}(\mathcal{H})$, then $T=T^{\sharp_{A}}$ if and only if $T$ is $A$-selfadjoint and $\mathcal{R}(T)\subseteq\overline{\mathcal{R}(A)}$. Notice that if $T\in\mathbb{B}_{A}(\mathcal{H})$, then $T^{\sharp_{A}}\in\mathbb{B}_{A}(\mathcal{H})$, $(T^{\sharp_{A}})^{\sharp_{A}}=PTP$ and \big{(}(T^{\sharp_{A}})^{\sharp_{A}}\big{)}^{\sharp_{A}}=T^{\sharp_{A}}. In addition, $T^{\sharp_{A}}T$, $TT^{\sharp_{A}}$ are $A$-selfadjoint and $A$-positive and so we have

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Furthermore, if $T,S\in\mathbb{B}_{A}(\mathcal{H})$, then $(TS)^{\sharp_{A}}=S^{\sharp_{A}}T^{\sharp_{A}}$, ${\|TS\|}_{A}\leq{\|T\|}_{A}{\|S\|}_{A}$ and ${\|Tx\|}_{A}\leq{\|T\|}_{A}{\|x\|}_{A}$ for all $x\in\mathcal{H}$.

For proofs and more facts about this class of operators, we refer the reader to [4, 5] and their references.

In recent years, several results covering some classes of operators on a complex Hilbert space \big{(}\mathcal{H},\langle\cdot,\cdot\rangle\big{)} are extended to \big{(}\mathcal{H},{\langle\cdot,\cdot\rangle}_{A}\big{)} (see, e.g., [5, 7, 8, 13, 14, 18, 24, 30, 31, 35]).

The numerical radius of $T\in\mathbb{B}(\mathcal{H})$ is defined by

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This concept is useful in studying linear operators and has attracted the attention of many authors in the last few decades (e.g., see [11, 17, 22, 26, 36], and their references).

It is well known that $w(\cdot)$ defines a norm on $\mathbb{B}(\mathcal{H})$ such that for all $T\in\mathbb{B}(\mathcal{H})$,

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The inequalities in (1.1) are sharp. The first inequality becomes an equality if $T^{2}=0$. The second inequality becomes an equality if $T$ is normal.

For more material about the numerical radius and other results on numerical radius inequality, see, e.g., [12, 17, 22, 23], and the references therein.

Some interesting numerical radius inequalities improving inequalities (1.1) have been obtained by several mathematicians (see, e.g., [1, 2, 10, 16, 21, 23, 32, 33]).

Motivated by theoretical study and applications, there have been many generalizations of the numerical radius (e.g., see [3, 6, 9, 15, 17, 19, 20, 26, 27, 28, 29, 34]). One of these generalizations is the $A$-numerical radius of an operator $T\in\mathbb{B}(\mathcal{H})$ defined by

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see, e.g., [8].

Now, following by the Crawford number and the cosine of angle of an operator $T\in\mathbb{B}(\mathcal{H})$ introduced by Gustafson and Rao in [17], we introduce the following notations

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and

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The paper is organized as follows.

In Section 2, inspired by the numerical radius inequalities of bounded linear operators in [1], [2], [21], [23], [33] and by using some ideas of them, we first state a useful characterization of the $A$-numerical radius for $T\in\mathbb{B}_{A}(\mathcal{H})$ as follows:

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This expression was motivated by [23, Theorem 2.1]. We then apply it to find upper and lower bounds for the $A$-numerical radius of semi-Hilbertian space operators. Particularly, for $T\in\mathbb{B}_{A}(\mathcal{H})$ we prove that

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[TABLE]

and

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In Section 3, some upper bounds for the $A$-numerical radius of products of semi-Hilbertian space operators are given. In particular, for $T,S\in\mathbb{B}_{A}(\mathcal{H})$ we show that

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In the last section we present some upper bounds for the $A$-numerical radius of commutators and anticommutators of semi-Hilbertian space operators. Particularly, for $T,S\in\mathbb{B}_{A}(\mathcal{H})$ we prove that

[TABLE]

Our results generalize recent numerical radius inequalities of bounded linear operators due to Kittaneh et al. [1, 2, 12, 22, 23, 33].

2. Upper and lower bounds of the $A$-numerical radius of operators

We start our work with the following lemmas. To establish the first lemma we use some ideas of [17, Theorem 1.3-1].

Lemma 2.1**.**

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

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Proof.

Let $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$. By the Cauchy-Schwarz inequality, we have

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and hence w_{A}(T)=\sup\Big{\{}\big{|}{\langle Tx,x\rangle}_{A}\big{|}:\,\,x\in\mathcal{H},\,{\|x\|}_{A}=1\Big{\}}\leq{\|T\|}_{A}.

Moreover, since $T$ is an $A$-selfadjoint operator, for every $y,z\in\mathcal{H}$ such that ${\|y\|}_{A}={\|z\|}_{A}=1$ we have

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and

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Consequently, we deduce

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So, we obtain

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Then it follows from parallelogram law that

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Now, consider the polar decomposition {\langle Ty,z\rangle}_{A}=e^{i\theta}\big{|}{\langle Ty,z\rangle}_{A}\big{|} with $\theta\in\mathbb{R}$. By replacing $z$ by $e^{i\theta}z$ in (2.1), we get \big{|}{\langle Ty,z\rangle}_{A}\big{|}=\mbox{Re}{\langle Ty,e^{i\theta}z\rangle}_{A}\leq w_{A}(T). From this it follows that {\|T\|}_{A}=\sup\Big{\{}\big{|}{\langle Ty,z\rangle}_{A}\big{|}:\,\,y,z\in\mathcal{H},\,{\|y\|}_{A}={\|z\|}_{A}=1\Big{\}}\leq w_{A}(T) and consequently $w_{A}(T)={\|T\|}_{A}$. ∎

Remark 2.2*.*

Note that for an arbitrary operator $T$ of $\mathbb{B}_{A}(\mathcal{H})$, we have

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Indeed, if $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$, then simple computations show that

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whence

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Thus

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Taking the supremum in the above inequality over $x\in\mathcal{H}$, ${\|x\|}_{A}=1$, we deduce the desired inequality.

The second lemma is stated as follows.

Lemma 2.3**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. For every $\theta\in\mathbb{R}$,

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Proof.

Let $\theta\in\mathbb{R}$. We have \Big{(}\frac{(e^{i\theta}T)^{\sharp_{A}}+\big{(}(e^{i\theta}T)^{\sharp_{A}}\big{)}^{\sharp_{A}}}{2}\Big{)}^{\sharp_{A}}=\frac{\big{(}(e^{i\theta}T)^{\sharp_{A}}\big{)}^{\sharp_{A}}+(e^{i\theta}T)^{\sharp_{A}}}{2}. Hence \frac{(e^{i\theta}T)^{\sharp_{A}}+\big{(}(e^{i\theta}T)^{\sharp_{A}}\big{)}^{\sharp_{A}}}{2} is an $A$-selfadjoint operator. So, by Lemma 2.1 we get

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Since $w_{A}(R^{\sharp_{A}})=w_{A}(R)$ and ${\|R^{\sharp_{A}}\|}_{A}={\|R\|}_{A}$ for every $R\in\mathbb{B}_{A}(\mathcal{H})$, from (2.2) it follows that

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∎

We now state the third lemma, which will be used to prove Theorem 2.5.

Lemma 2.4**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$ and $x\in\mathcal{H}$. Then

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Proof.

Let $\theta\in\mathbb{R}$. We have

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Thus

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From this it follows that

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whence

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Now, if $|{\langle Tx,x\rangle}_{A}|=0$, then from (2.3) we obtain \left|{\Big{\langle}\frac{e^{i\theta}T+(e^{i\theta}T)^{\sharp_{A}}}{2}x,x\Big{\rangle}}_{A}\right|=0 and so

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If $|{\langle Tx,x\rangle}_{A}|\neq 0$, then we put $e^{i\theta_{0}}=\frac{{\langle x,Tx\rangle}_{A}}{|{\langle Tx,x\rangle}_{A}|}$. Therefore, by (2.3), we obtain

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From (2.4) and (2.5) it follows that

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∎

Now, we are in a position to state a useful characterization of the $A$-numerical radius for semi-Hilbertian space operators.

Theorem 2.5**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

Let $\theta\in\mathbb{R}$. By Lemma 2.3 it follows that

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Therefore, by Lemma 2.4 we conclude that

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∎

Here we present one of the main results of this section.

Theorem 2.6**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then for $\alpha,\beta\in\mathbb{R}$,

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Proof.

Let $\theta\in\mathbb{R}$. Put $\alpha=\cos\theta$ and $\beta=-\sin\theta$. We have

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Therefore

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and hence, by Theorem 2.5, we obtain

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∎

As a consequence of Theorem 2.6, we have the following result.

Corollary 2.7**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

By setting $(\alpha,\beta)=(1,0)$ and $(\alpha,\beta)=(0,1)$ in Theorem 2.6, the result follows. ∎

The following result is another consequence of Theorem 2.6.

Corollary 2.8**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

Clearly, $w_{A}(T)\leq{\|T\|}_{A}$. On the other hand, by using Corollary 2.7, we get

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Hence $\frac{1}{2}{\|T\|}_{A}\leq w_{A}(T)$. ∎

Remark 2.9*.*

Corollary 2.8 has recently been proved by Baklouti et al. in [8]. Our approach here is different from theirs.

In the following theorem, we give a improvement of the second inequality in (2.6).

Theorem 2.10**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

Put $M:=\frac{T^{\sharp_{A}}+(T^{\sharp_{A}})^{\sharp_{A}}}{2}$ and $N:=\frac{T^{\sharp_{A}}-(T^{\sharp_{A}})^{\sharp_{A}}}{2i}$. Then $T^{\sharp_{A}}=M+iN$. Also, simple computations show that

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Since ${\|R^{\sharp_{A}}\|}_{A}={\|R\|}_{A}$ for every $R\in\mathbb{B}_{A}(\mathcal{H})$, hence

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Now, let $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$. We have

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Hence

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or equivalently,

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Further, since ${\|TT^{\sharp_{A}}\|}_{A}={\|T^{\sharp_{A}}T\|}_{A}={\|T\|}^{2}_{A}$, by the triangle inequality we obtain

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Thus

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By (2.8) and (2.9) we deduce the desired result. ∎

Next, we present another improvement of the second inequality in (2.6).

Theorem 2.11**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

By Theorem 2.5, we have

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which proves the desired inequalities. ∎

In the following theorem, we establish an improvement of the first inequality in (2.6).

Theorem 2.12**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

Let $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$. Suppose that \big{|}{\langle T^{\sharp_{A}}T^{\sharp_{A}}x,x\rangle}_{A}\big{|}=e^{-2i\theta}{\langle T^{\sharp_{A}}T^{\sharp_{A}}x,x\rangle}_{A} for some real number $\theta$. Then, we have

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Thus

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So, by Theorem 2.5, we obtain

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From this it follows that

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Taking the supremum over $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$ in the above inequality we get

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Furthermore, since $T^{\sharp_{A}}T$ is an $A$-positive operator, from {\big{\|}TT^{\sharp_{A}}+T^{\sharp_{A}}T\big{\|}}_{A}\geq{\|TT^{\sharp_{A}}\|}_{A}={\|T\|}^{2}_{A} it follows that

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Now, by (2.11) and (2.12) we conclude that

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∎

Now, we give another improvement of the first inequality in (2.6).

Theorem 2.13**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

Clearly, $\sqrt{\frac{w^{2}_{A}(T)}{2}+\frac{w_{A}(T)}{2}\sqrt{w^{2}_{A}(T)-c^{2}_{A}(T)}}\leq w_{A}(T)$. Now, let $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$. Suppose that \big{|}{\langle Tx,x\rangle}_{A}\big{|}=e^{i\theta}{\langle Tx,x\rangle}_{A} for some real number $\theta$. Put $M:=\frac{e^{i\theta}T+(e^{i\theta}T)^{\sharp_{A}}}{2}$ and $N:=\frac{e^{i\theta}T-(e^{i\theta}T)^{\sharp_{A}}}{2i}$. Then $M+iN=e^{i\theta}T$ and

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It follows from ${\langle Nx,x\rangle}_{A}=\mbox{Im}{\langle e^{i\theta}Tx,x\rangle}_{A}\in\mathbb{R}$ that

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So, we have

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Hence

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which implies

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Taking the supremum over $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$ in the above inequality we get

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∎

We end this section with a considerable improvement of the first inequality in (2.6).

Theorem 2.14**.**

Let $T\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

Obviously,

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Furthermore, by (2.13) we have

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and hence

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From this it follows that

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We consider two cases.

Case 1. ${\|Tx\|}^{2}_{A}-2w^{2}_{A}(T)\leq 0$. Then we reach that ${\|Tx\|}_{A}\leq\sqrt{2}w_{A}(T)$ and so

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Case 2. ${\|Tx\|}^{2}_{A}-2w^{2}_{A}(T)>0$. By (2.14) it follows that

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Thus

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This yields

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and hence

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Now, by (2.15) and (2.16) we obtain

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∎

3. Upper bounds for the $A$-numerical radius of products of operators

In this section, we derive upper bounds for the $A$-numerical radius of products of semi-Hilbertian space operators. Since for every $T,S\in\mathbb{B}_{A}(\mathcal{H})$ we have ${\|TS\|}_{A}\leq{\|T\|}_{A}{\|S\|}_{A}$, by the inequalities of (2.6) we obtain

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In the following theorems, we improve the inequalities 3.1. To achieve our goal, we need the following lemma.

Lemma 3.1**.**

Let $T,S\in\mathbb{B}_{A}(\mathcal{H})$. Then

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Proof.

Let $\theta\in\mathbb{R}$. Since \big{(}(R^{\sharp_{A}})^{\sharp_{A}}\big{)}^{\sharp_{A}}=R^{\sharp_{A}} for every $R\in\mathbb{B}_{A}(\mathcal{H})$, we have

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Therefore, by Theorem 2.5 and (3), we obtain

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Hence

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Since $w_{A}(R^{\sharp_{A}})=w_{A}(R)$ for every $R\in\mathbb{B}_{A}(\mathcal{H})$, from (3.3) we obtain

[TABLE]

Finally, by replacing $S$ by $-iS$ in (3.4), we reach that

[TABLE]

∎

In the next theorem, we give a new upper bound for the $A$-numerical radius of products of semi-Hilbertian space operators.

Theorem 3.2**.**

Let $T,S\in\mathbb{B}_{A}(\mathcal{H})$. Then

[TABLE]

Proof.

The second inequality follows from Lemma 3.1. It is therefore enough to prove the first inequality. Let $\theta\in\mathbb{R}$. By Lemma 2.3 we have

[TABLE]

Thus

[TABLE]

Now, by replacing $S$ by $-iS$ in (3.5), we conclude that

[TABLE]

By (3.5) and (3.6) we deduce the desired result. ∎

As an immediate consequence of the preceding theorem, we have the following result.

Corollary 3.3**.**

Let $T,S\in\mathbb{B}_{A}(\mathcal{H})$. If $(TS)^{\sharp_{A}}=T^{\sharp_{A}}S$, then

[TABLE]

In the following, for $R\in\mathbb{B}_{A}(\mathcal{H})$, let $d_{A}(R)$ denote the $A$-numerical radius distance of $R$ from the scalar operators, that is,

[TABLE]

Next, we present a improvement of the second inequality in (3.1).

Theorem 3.4**.**

Let $T,S\in\mathbb{B}_{A}(\mathcal{H})$. Then

[TABLE]

Proof.

Using a compactness argument, let $\zeta_{0}\in\mathbb{C}$ such that $d_{A}(T)=w_{A}(T+\zeta_{0}I)$. If $\zeta_{0}=0$, then by the inequalities of (2.6) we get

[TABLE]

Hence, we may assume that $\zeta_{0}\neq 0$. Put $\zeta=\frac{\overline{\zeta_{0}}}{|\zeta_{0}|}$. Then

[TABLE]

Hence

[TABLE]

Since $d_{A}(T)\leq w_{A}(T)$, from (3.7) and (3.8) it follows that

[TABLE]

By a similar argument we have

[TABLE]

Now, by (3.9) and (3.10) we obtain the desired inequalities. ∎

We finish this section by another upper bound for the $A$-numerical radius of products of semi-Hilbertian space operators.

Theorem 3.5**.**

Let $T,S\in\mathbb{B}_{A}(\mathcal{H})$. Then

[TABLE]

Proof.

The fact that $d_{A}(R)\leq w_{A}(R)$ holds for every $R\in\mathbb{B}_{A}(\mathcal{H})$ implies that the third desired inequality.

Now, let $\zeta_{0},\xi_{0}\in\mathbb{C}$ such that $d_{A}(T)=w_{A}(T+\zeta_{0}I)$ and $d_{A}(S)=w_{A}(S+\xi_{0}I)$. As in the proof of Theorem 3.4 we may assume that $\zeta_{0}\xi_{0}\neq 0$. Put $\zeta=\frac{\overline{\zeta_{0}}}{|\zeta_{0}|}$ and $\xi=\frac{\overline{\xi_{0}}}{|\xi_{0}|}$. Therefore, we have

[TABLE]

∎

4. Upper bounds for the $A$-numerical radius of commutators, and anticommutators of operators

In this section, we present some upper bounds for the $A$-numerical radius of commutators, and anticommutators of semi-Hilbertian space operators. To achieve the first main result in this section, we need the following lemma.

Lemma 4.1**.**

Let $R\in\mathbb{B}_{A}(\mathcal{H})$. Then

[TABLE]

Proof.

Observe that, from $d_{A}(R)\leq w_{A}(R)$ we have 2\big{(}w^{2}_{A}(R)+d^{2}_{A}(R)\big{)}\leq 4w^{2}_{A}(R). It is therefore enough to prove the first inequality. Let $\zeta_{0}\in\mathbb{C}$ such that $d_{A}(R)=w_{A}(R+\zeta_{0}I)$. If $\zeta_{0}=0$, then by employing Corollary 2.7 we have

[TABLE]

If $\zeta_{0}\neq 0$, then put $\zeta=\frac{\overline{\zeta_{0}}}{|\zeta_{0}|}$. A simple computation together with Corollary 2.7 gives

[TABLE]

∎

The following result may be stated as well.

Theorem 4.2**.**

Let $T,S\in\mathbb{B}_{A}(\mathcal{H})$. Then

[TABLE]

Proof.

Clearly,

[TABLE]

Now, let $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$. By the Cauchy–Schwarz inequality, we have

[TABLE]

Thus

[TABLE]

Taking the supremum over $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$ in the above inequality we get

[TABLE]

From (4.1) and Lemma 4.1 it follows that

[TABLE]

whence

[TABLE]

Similarly,

[TABLE]

Hence by (4) and (4) we deduce the desired result. ∎

As a consequence of Lemma 4.1 and Theorem 4.2, we have the following result.

Corollary 4.3**.**

Let $T,S\in\mathbb{B}_{A}(\mathcal{H})$. Then

[TABLE]

For the second main result in this section, we need the following lemma that is interesting on its own right.

Lemma 4.4**.**

For $T,S,R\in\mathbb{B}_{A}(\mathcal{H})$ the following statements hold.

• (i)

w_{A}\Big{(}TRT^{\sharp_{A}}\Big{)}\leq{\|T\|}^{2}_{A}w_{A}(R).

• (ii)

w_{A}\Big{(}SRT^{\sharp_{A}}\Big{)}\leq\frac{1}{2}{\big{\|}TT^{\sharp_{A}}+SS^{\sharp_{A}}\big{\|}}_{A}{\|R\|}_{A}.

Proof.

(i) Let $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$. We have

[TABLE]

Now, by taking the supremum over all $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$ we conclude that

[TABLE]

(ii) Let $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$. We have

[TABLE]

which, by taking the supremum over $x\in\mathcal{H}$, ${\|x\|}_{A}=1$, implies that

[TABLE]

∎

Finally, we present the following result.

Theorem 4.5**.**

Let $T,S\in\mathbb{B}_{A}(\mathcal{H})$. Then

[TABLE]

Proof.

Let $\theta\in\mathbb{R}$. We have

[TABLE]

Thus

[TABLE]

and so,

[TABLE]

Then, by Theorem 2.5, we get

[TABLE]

Finally, by replacing $T$ by $iT$ in (4.4), we obtain

[TABLE]

and the proof is completed. ∎

Acknowledgments. The author expresses his gratitude to the referee for his/hers comments towards an improved final version of the paper. He would also like to thank Professor M. S. Moslehian for his helpful suggestions. This work was supported by a grant from Shanghai Municipal Science and Technology Commission (18590745200).