Generalized Cauchy–Schwarz Inequalities and A-Numerical Radius Applications

Abstract

The purpose of this research paper is to introduce new Cauchy–Schwarz inequalities that are valid in semi-Hilbert spaces, which are generalizations of Hilbert spaces. We demonstrate how these new inequalities can be employed to derive novel A-numerical radius inequalities, where A denotes a positive semidefinite operator in a complex Hilbert space. Some of our novel A-numerical radius inequalities expand upon the existing literature on numerical radius inequalities with Hilbert space operators, which are important tools in functional analysis. We use techniques from semi-Hilbert space theory to prove our results and highlight some applications of our findings.

1. Introduction and Preliminaries

The Cauchy–Schwarz inequality is a fundamental inequality in mathematics, with applications in many areas such as linear algebra, analysis, and probability theory. In the context of Hilbert spaces, the Cauchy–Schwarz inequality takes on a particularly elegant form, which has led to its extensive use in functional analysis and related fields.

Semi-Hilbert spaces are mathematical spaces that extend the concept of Hilbert spaces, which are commonly studied in analysis and linear algebra. Unlike Hilbert spaces, semi-Hilbert spaces allow for more flexible inner products that do not need to satisfy all the usual requirements. This flexibility makes semi-Hilbert spaces useful for investigating various mathematical and physical problems involving unbounded operators, singularities, or non-local interactions.

In the field of operator theory, semi-Hilbert spaces have attracted significant interest. They can be constructed by starting with a positive semidefinite sesquilinear form and then building a space based on this form. Recent research has made notable contributions to this area. Noteworthy works exploring operator theory in semi-Hilbert spaces include references [1,2,3,4,5,6,7,8,9], as well as the additional references provided therein. These resources provide valuable insights and advancements in understanding semi-Hilbert spaces and their applications in operator theory.

In this paper, we establish new types of Cauchy–Schwarz inequalities in the context of semi-Hilbert spaces and apply them to derive novel A-numerical radius inequalities, where A is a positive semidefinite operator in a complex Hilbert space.

To set the foundation for our findings, we initially present certain symbols and remind readers of some commonly acknowledged facts. Our analysis involves a complex Hilbert space denoted as $\mathcal{X}$, which comes equipped with an inner product $\langle ·,·\rangle$ and a corresponding norm $\parallel ·\parallel$. The $C^{*}$-algebra, which comprises all bounded linear operators on $\mathcal{X}$ along with the identity operator $I_{\mathcal{X}}$ (or simply I), is denoted as $\mathcal{L}\left(\mathcal{X}\right)$ throughout this paper. When we refer to an “operator”, we are specifically referring to a bounded linear operator that acts on $\mathcal{X}$.

For any operator S, we use $ker\left(S\right)$ and $\mathrm{ran}\left(S\right)$ to respectively denote its nullspace and range. The orthogonal projection onto any closed linear subspace $\mathrm{\Omega }$ of $\mathcal{X}$ is denoted by $P_{\mathrm{\Omega }}$. We say that an operator S is positive if $\langle Sy,y\rangle \ge 0$ for all $y\in \mathcal{X}$, and we write $S\ge 0$ to indicate this property. When $S\ge 0$, we introduce the notation $S^{1/2}$ for the unique positive bounded linear operator satisfying $S={\left(S^{1/2}\right)}^2$.

The absolute value of an operator $S\in \mathcal{L}\left(\mathcal{X}\right)$ is defined as $\left|S\right|={\left(S^{*}S\right)}^{1/2}$, which is always non-negative. We define the Moore–Penrose inverse of an operator S as $S^{†}$, which is the unique linear extension of ${\tilde{S}}^{-1}$ to $\mathcal{D}\left(S^{†}\right):=\mathrm{ran}\left(S\right)\oplus \mathrm{ran}{\left(S\right)}^{\perp }$ such that $ker\left(S^{†}\right)=\mathrm{ran}{\left(S\right)}^{\perp }$. Here, $\tilde{S}$ is the isomorphism $\tilde{S}{:=S|}_{{ker\left(S\right)}^{\perp }}:{ker\left(S\right)}^{\perp }⟶\mathrm{ran}\left(S\right)$. The operator $S^{†}$ is obtained as the unique solution to the set of four equations known as the Moore–Penrose equations:

$$SYS=S,YSY=Y,YS=P_{ker{\left(S\right)}^{\perp }}\mathrm{and}SY=P_{\overline{\mathrm{ran}\left(S\right)}}{|}_{\mathrm{ran}\left(S\right)\oplus \mathrm{ran}{\left(S\right)}^{\perp }}.$$

For more details on the Moore–Penrose inverse, interested readers can refer to [10].

From this point forward, the following assumptions will be made: $A\ge 0$ and $A\ne 0$. We can use A to define a positive semidefinite sesquilinear form on $\mathcal{X}\times \mathcal{X}$, denoted by ${\langle ·,·\rangle }_A$, given by ${\langle x_1,x_2\rangle }_A:=\langle A{x}_1,x_2\rangle$ for all $x_1,x_2\in \mathcal{X}$. The seminorm induced by ${\langle ·,·\rangle }_A$ is ${\parallel y\parallel }_A=\sqrt{{\langle y,y\rangle }_A}=\parallel A^{1/2}y\parallel$ for every $y\in \mathcal{X}$. It is worth mentioning that ${\parallel ·\parallel }_A$ serves as a norm on $\mathcal{X}$ if and only if the operator A is injective. Moreover, the pair $(\mathcal{X},{\parallel ·\parallel }_A)$ forms a semi-Hilbert space that is complete if and only if $\mathrm{ran}\left(A\right)$ is a closed subspace in $\mathcal{X}$. Finally, when $A=I$, we get ${\langle x_1,x_2\rangle }_A=\langle x_1,x_2\rangle$ and $\parallel x_1{\parallel }_A=\parallel x_1\parallel$ for all $x_1,x_2\in \mathcal{X}$. The set of all unit vectors in $\mathcal{X}$ is given by

$$S_A^1:=\left\{\zeta \in {\mathcal{X};\parallel \zeta \parallel }_A=1\right\}.$$

The numerical radius is a crucial concept in matrix analysis and operator theory. In recent times, there have been various extensions of the numerical radius, one of which is the A-numerical radius of an operator $Q\in \mathcal{L}\left(\mathcal{X}\right)$. The definition of the A-numerical radius of Q was first introduced by Saddi in [11], and it can be expressed as:

$$\begin{array}{c}\hfill {\omega }_A\left(Q\right)=sup\left\{{|\langle Qy,y\rangle }_A|;y\in S_A^1\right\}.\end{array}$$

This concept has recently received significant attention, with several papers exploring its properties and applications, including [1,2,3] and the references therein. As established in [12], it is common knowledge that ${\omega }_A\left(Q\right)$ can be infinite for certain operators $Q\in \mathcal{L}\left(\mathcal{X}\right)$. Therefore, in order to ensure that ${\omega }_A\left(Q\right)$ is well defined and finite, we must revisit the notion of A-adjoint operators presented in [13].

Definition 1

([13]). An operator $Q\in \mathcal{L}\left(\mathcal{X}\right)$ is said to be an A-adjoint operator of $S\in \mathcal{L}\left(\mathcal{X}\right)$ if the identity ${\langle S{x}_1,x_2\rangle }_A={\langle x_1,Q{x}_2\rangle }_A$ holds for every $x_1,x_2\in \mathcal{X}$. In other words, Q is the solution of the operator equation $AY=S^{*}A$.

To investigate this equation, we can use a theorem developed by Douglas [14]. This theorem states that the operator equation $TY=S$ has a solution $Y\in \mathcal{L}\left(\mathcal{X}\right)$ if and only if $\mathrm{ran}\left(S\right)\subseteq \mathrm{ran}\left(T\right)$, which is equivalent to the existence of a positive number $\beta$ such that $\parallel S^{*}\xi \parallel \le \beta \parallel T^{*}\xi \parallel$ for all $\xi \in \mathcal{X}$. Furthermore, the same theorem developed by Douglas shows that, if $TY=S$ has more than one solution, then there exists only one solution, denoted by R, that satisfies $\mathrm{ran}\left(R\right)\subseteq \overline{\mathrm{ran}\left(T\right)}$. Such a solution R is referred to as the reduced solution of the equation $TY=S$. However, it is not guaranteed that an A-adjoint operator exists or is unique for a given operator T.

Let us define the sets ${\mathcal{L}}_A\left(\mathcal{X}\right)$ and ${\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$ as the sets of all operators that have A-adjoints and $A^{1/2}$-adjoints, respectively. Based on Douglas’s theorem, we can derive the following equivalences:

$${\mathcal{L}}_A\left(\mathcal{X}\right)=\left\{Q\in \mathcal{L}\left(\mathcal{X}\right);\mathrm{ran}\left(Q^{*}A\right)\subseteq \mathrm{ran}\left(A\right)\right\}$$

and

$${\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)=\left\{Q\in \mathcal{L}\left(\mathcal{X}\right);\exists \mu >0\mathrm{such}\mathrm{that}{\parallel Qy\parallel }_A\le \mu {\parallel y\parallel }_A,\forall y\in \mathcal{X}\right\}.$$

It is worth noting that ${\mathcal{L}}_A\left(\mathcal{X}\right)$ and ${\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$ are two subalgebras of $\mathcal{L}\left(\mathcal{X}\right)$, but they are not necessarily closed or dense in $\mathcal{L}\left(\mathcal{X}\right)$. Furthermore, we can show that ${\mathcal{L}}_A\left(\mathcal{X}\right)\subseteq {\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$ using the reference [12].

An operator is considered A-bounded if it is a member of the set ${\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$. The set ${\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$ is equipped with the seminorm ${\parallel Q\parallel }_A$, which is defined as follows:

$${\displaystyle {\parallel Q\parallel }_A:=\underset{\begin{array}{c}y\in \overline{\mathrm{ran}\left(A\right)},\\ x\ne 0\end{array}}{sup}\frac{{\parallel Qy\parallel }_A}{{\parallel y\parallel }_A}=\underset{y\in S_A^1}{sup}{\parallel Qy\parallel }_A<\infty .}$$

This seminorm is also given by

$$\begin{array}{c}\hfill {\parallel Q\parallel }_A=sup\left\{\left|{\langle Q{x}_1,x_2\rangle }_A\right|;x_1,x_2\in S_A^1\right\},\end{array}$$

for $T\in {\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$, as shown in [15]. This definition is well established in the literature and has been used extensively in the study of bounded linear operators in A-weighted spaces. It can be shown that ${\parallel Q\parallel }_A=0$ if and only if $AQ=0$ for any $Q\in {\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$. Moreover, for all $Q\in {\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$ and $y\in \mathcal{X}$, we have ${\parallel Qy\parallel }_A\le {\parallel Q\parallel }_A{\parallel y\parallel }_A$, which implies the inequality $\parallel Q_1{Q}_2{\parallel }_A\le \parallel Q_1{\parallel }_A{\parallel Q_2\parallel }_A$ for all $Q_1,Q_2\in {\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$.

It is noteworthy that the seminorms ${\parallel ·\parallel }_A$ and ${\omega }_A(·)$ are equivalent on ${\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$. Specifically, for any $Q\in {\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$, we have (see [16])

$$\frac{1}{2}{\parallel Q\parallel }_A\le {\omega }_A\left(Q\right)\le {\parallel Q\parallel }_A.$$

(1)

Moreover, it was demonstrated in [12] that

$${\omega }_A\left(Q^n\right)\le {\omega }_A^n\left(Q\right),$$

(2)

for any positive integer n and $Q\in {\mathcal{L}}_{A^{1/2}}\left(\mathcal{X}\right)$.

Let us remember that an operator $Q\in \mathcal{L}\left(\mathcal{X}\right)$ is referred to as being A-selfadjoint if $AQ$ is selfadjoint. It is clear that if Q is A-selfadjoint, then $Q\in {\mathcal{L}}_A\left(\mathcal{X}\right)$. If an operator Q satisfies $AQ\ge 0$, it is referred to as A-positive, denoted as $Q{\ge }_{A}0$. It is important to highlight that in the context of a complex Hilbert space $\mathcal{X}$, an A-positive operator is also considered A-selfadjoint. In [12], it was demonstrated that, for any A-selfadjoint operator Q (especially if $Q{\ge }_{A}0$), the following equality holds:

$${\parallel Q\parallel }_A={\omega }_A\left(Q\right).$$

(3)

Suppose $Q\in {\mathcal{L}}_A\left(\mathcal{X}\right)$. The solution to the equation $AY=Q^{*}A$ can be reduced and denoted as $Q^{{*}_A}$. This reduced solution can be expressed as $A^{†}{Q}^{*}A$. Moreover, if $Q\in {\mathcal{L}}_A\left(\mathcal{X}\right)$, then $Q^{{*}_A}\in {\mathcal{L}}_A\left(\mathcal{X}\right)$ and we have the properties ${\left(Q^{{*}_A}\right)}^{{*}_A}=P_{\overline{\mathrm{ran}\left(A\right)}}Q{P}_{\overline{\mathrm{ran}\left(A\right)}}$ and ${\left({\left(Q^{{*}_A}\right)}^{{*}_A}\right)}^{{*}_A}=Q^{{*}_A}$. For detailed proofs and additional related results, refer to [10,13,17] and the references therein.

One can verify that the operators $Q^{{*}_A}Q{\ge }_{A}0$ and $Q{Q}^{{*}_A}{\ge }_{A}0$. Additionally, for any operator $Q\in {\mathcal{L}}_A\left(\mathcal{X}\right)$, the following equalities hold (see Proposition 2.3 in [10]):

$$\begin{array}{c}\hfill {\parallel Q^{{*}_A}Q\parallel }_A={\parallel Q{Q}^{{*}_A}\parallel }_A={\parallel Q\parallel }_A^2={\parallel Q^{{*}_A}\parallel }_A^2.\end{array}$$

(4)

The term “A-normal” is used to describe an operator $Q\in {\mathcal{L}}_A\left(\mathcal{X}\right)$ that satisfies $Q{Q}^{{*}_A}=Q^{{*}_A}Q$ (refer to [11]). It should be noted that, while all selfadjoint operators are normal, an A-selfadjoint operator may not necessarily be A-normal (see Example 4 in [12]).

Many authors have recently demonstrated various improvements to the inequalities shown in Equation (1). These can be found in studies such as [1,16], as well as other references mentioned therein. Specifically, it has been demonstrated in [16] that, for $Q\in {\mathcal{L}}_A\left(\mathcal{X}\right)$, the following inequality

$$\begin{array}{c}\hfill \frac{1}{2}\sqrt{\parallel Q^{{*}_A}Q+Q{Q}^{{*}_A}{\parallel }_A}\le {\omega }_A\left(Q\right)\le \frac{\sqrt{2}}{2}\sqrt{\parallel Q^{{*}_A}Q+Q{Q}^{{*}_A}{\parallel }_A},\end{array}$$

(5)

holds. When $A=I$ in Equation (5), the resulting inequalities are the well established ones that were proven by Kittaneh in Theorem 1 in [18].

Conde et al. in [19] established important numerical radius upper bounds. Specifically, for operators $T,S\in {\mathcal{L}}_A\left(\mathcal{X}\right)$ and a positive integer n, the following inequalities hold:

$${\omega }_A^n\left(T{S}^{{*}_A}\right)\le \frac{1}{2}\parallel {\left(T{T}^{{*}_A}\right)}^n+{\left(S{S}^{{*}_A}\right)}^n{\parallel }_A.$$

These results can be further improved by replacing T with $T^{{*}_A}$ and S with $S^{{*}_A}$, yielding the following inequalities:

$${\omega }_A^n\left(S^{{*}_A}T\right)\le \frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^n+{\left(S^{{*}_A}S\right)}^n{\parallel }_A.$$

(6)

In this paper, we present new types of Cauchy–Schwarz inequalities within the framework of semi-Hilbert spaces and employ them to derive innovative A-numerical radius inequalities. Notably, several of our findings expand upon the established body of knowledge concerning the classical numerical radius inequalities of Hilbert space operators. The inspiration for our investigation comes from recent works in this area [20,21], which have highlighted the importance of developing new mathematical tools for studying inner product spaces. To establish our results, we employ techniques rooted in semi-Hilbert space theory, which provides a more general framework for studying inner product spaces. Our findings have important implications in various branches of mathematics, including functional analysis and operator theory. We demonstrate the versatility of the new Cauchy–Schwarz inequalities and show how they can be used to derive innovative A-numerical radius inequalities, where A represents a positive semidefinite operator in a complex Hilbert space.

Overall, our research contributes to the ongoing study of inner product spaces and provides new insights and tools for various areas of mathematics.

2. Main Results

In this section, we present our main results. To prove our first main result, we require three lemmas. We will establish the first one, while the second and third are quoted from the references [19,20], respectively. Let us start by presenting the first lemma, which concerns a refined version of the Cauchy–Schwarz inequality in the context of semi-Hilbert spaces.

Lemma 1.

Consider $x,y\in \mathcal{X}$ and $\epsilon \in [0,1]$. Then, we have:

$$\left|{\langle x,y\rangle }_A\right|\le \sqrt{{\epsilon \parallel x\parallel }_A^2{\parallel y\parallel }_A^2+(1-\epsilon )|{\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A}\le {\parallel x\parallel }_A{\parallel y\parallel }_A.$$

Proof. 

Let $x,y\in \mathcal{X}$. Starting from the classical Cauchy–Schwarz inequality, we observe that

$$\left|{\langle x,y\rangle }_A\right|\le {\parallel x\parallel }_A{\parallel y\parallel }_A.$$

(7)

This leads us to the conclusion that

$$\left|{\langle x,y\rangle }_A\right|\le {\epsilon \parallel x\parallel }_A{\parallel y\parallel }_A+(1-\epsilon )\left|{\langle x,y\rangle }_A\right|\le {\parallel x\parallel }_A{\parallel y\parallel }_A.$$

By multiplying both sides by ${\parallel x\parallel }_A{\parallel y\parallel }_A$, we get

$$|{\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A\le {\epsilon \parallel x\parallel }_A^2{\parallel y\parallel }_A^2+(1-\epsilon )|{\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A\le {\parallel x\parallel }_A^2{\parallel y\parallel }_A^2.$$

Using inequality (7) again, we can conclude that

$$\begin{array}{cc}\hfill |{\langle x,y\rangle }_A{|}^2& \le |{\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A\hfill \\ \hfill & \le {\epsilon \parallel x\parallel }_A^2{\parallel y\parallel }_A^2+(1-\epsilon )|{\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A\le {\parallel x\parallel }_A^2{\parallel y\parallel }_A^2.\hfill \end{array}$$

Therefore, we have demonstrated the validity of the inequality presented in the statement. □

Lemma 2

([20]). Suppose that $Q_1$ and $Q_2$ are both A-positive operators in $\mathcal{L}\left(\mathcal{X}\right)$. Then,

$${∥{\displaystyle \frac{Q_1+Q_2}{2}}∥}_A^n\le {∥{\displaystyle \frac{Q_1^n+Q_2^n}{2}}∥}_A,\forall n\in {\mathbb{N}}^{*}.$$

Lemma 3

([19]). Suppose that T is an operator in $\mathcal{L}\left(\mathcal{X}\right)$ such that $T{\ge }_{A}0$. Then, for any $n\in {\mathbb{N}}^{*}$ and $y\in S_A^1$, the inequality

$${〈Ty,y〉}_A^n\le {〈T^{n}y,y〉}_A,$$

holds.

By utilizing the aforementioned lemmas, we can present the following result, which offers an improvement to inequality (6) for the case where $n=2$.

Theorem 1.

Let $T,S\in {\mathcal{L}}_A\left(\mathcal{X}\right)$. Then, for any $\epsilon \in [0,1]$, we have

$$\begin{array}{cc}\hfill {\omega }_A^2\left(S^{{*}_A}T\right)& \le \frac{(1-\epsilon )}{2}{\omega }_A\left(S^{{*}_A}T\right)\parallel T^{{*}_A}T+S^{{*}_A}S{\parallel }_A+\frac{\epsilon }{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A\hfill \\ \hfill & \le \frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2{\parallel }_A.\hfill \end{array}$$

Proof. 

Let $\epsilon \in [0,1]$. Assuming $\xi \in S_A^1$, we can apply Lemma 1 by setting $x=T\xi$ and $y=S\xi$ to obtain:

$$\begin{array}{cc}\hfill & |{\langle T\xi ,S\xi \rangle }_A{|}^2\hfill \\ \hfill & \le {(1-\epsilon )\parallel T\xi \parallel }_A{\parallel S\xi \parallel }_A\left|{\langle T\xi ,S\xi \rangle }_A\right|+{\epsilon \parallel T\xi \parallel }_A^2{\parallel S\xi \parallel }_A^2\hfill \\ \hfill & =(1-\epsilon )\sqrt{{\langle T^{{*}_A}T\xi ,\xi \rangle }_A}\sqrt{{\langle S^{{*}_A}S\xi ,\xi \rangle }_A}\left|{\langle S^{{*}_A}T\xi ,\xi \rangle }_A\right|+\epsilon {\langle T^{{*}_A}T\xi ,\xi \rangle }_A{\langle S^{{*}_A}S\xi ,\xi \rangle }_A\hfill \\ \hfill & \le \frac{(1-\epsilon )}{2}{\omega }_A\left(S^{{*}_A}T\right){\langle \left(T^{{*}_A}T+S^{{*}_A}S\right)\xi ,\xi \rangle }_A+\frac{\epsilon }{2}\left({\langle T^{{*}_A}T\xi ,\xi \rangle }_A^2+{\langle S^{{*}_A}S\xi ,\xi \rangle }_A^2\right).\hfill \end{array}$$

In the final inequality, we have employed the arithmetic–geometric mean inequality. Furthermore, as both $T^{{*}_A}T{\ge }_{A}0$ and $S^{{*}_A}S{\ge }_{A}0$, we can utilize Lemma 3 to deduce that:

$$\begin{array}{cc}\hfill & |{\langle S^{{*}_A}T\xi ,\xi \rangle }_A{|}^2\hfill \\ \hfill & \le \frac{(1-\epsilon )}{2}{\omega }_A\left(S^{{*}_A}T\right){\langle \left(T^{{*}_A}T+S^{{*}_A}S\right)\xi ,\xi \rangle }_A+\frac{\epsilon }{2}{\langle \left({\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2\right)\xi ,\xi \rangle }_A\hfill \\ \hfill & \le \frac{(1-\epsilon )}{2}{\omega }_A\left(S^{{*}_A}T\right){\omega }_A\left(T^{{*}_A}T+S^{{*}_A}S\right)+\frac{\epsilon }{2}{\omega }_A\left({\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2\right).\hfill \end{array}$$

Given that $T^{{*}_A}T+S^{{*}_A}S{\ge }_{A}0$ and ${\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2{\ge }_{A}0$, we can apply Equation (3) to conclude that:

$$|{\langle S^{{*}_A}T\xi ,\xi \rangle }_A{|}^2\le \frac{(1-\epsilon )}{2}{\omega }_A\left(S^{{*}_A}T\right)\parallel T^{{*}_A}T+S^{{*}_A}S{\parallel }_A+\frac{\epsilon }{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A.$$

Therefore, the first inequality in Theorem 1 can be obtained by taking the supremum over all $\xi \in S_A^1$.

Now, by applying inequality (6) for $n=2$, we see that

$$\begin{array}{cc}\hfill {\omega }_A^2\left(S^{{*}_A}T\right)& \le \frac{(1-\epsilon )}{2}{\omega }_A\left(S^{{*}_A}T\right)\parallel T^{{*}_A}T+S^{{*}_A}S{\parallel }_A+\frac{\epsilon }{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A\hfill \\ \hfill & \le \frac{(1-\epsilon )}{4}\parallel T^{{*}_A}T+S^{{*}_A}S{\parallel }_A^2+\frac{\epsilon }{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A\hfill \\ \hfill & =(1-\epsilon ){∥\frac{T^{{*}_A}T+S^{{*}_A}S}{2}∥}_A^2+\frac{\epsilon }{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A\hfill \\ \hfill & \le (1-\epsilon ){∥\frac{{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2}{2}∥}_A+\frac{\epsilon }{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A\hfill \\ \hfill & =\frac{(1-\epsilon )}{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A+\frac{\epsilon }{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A.\hfill \end{array}$$

Here, we used Lemma 2 to obtain the last inequality since $T^{{*}_A}T$ and $S^{{*}_A}S$ are A-positive operators. This establishes the second inequality in Theorem 1, thereby completing the proof. □

Remark 1.

The statement from Theorem 1 implies that, given any T and S belonging to ${\mathcal{L}}_A\left(\mathcal{X}\right)$ and a non-negative value of ν, it holds true that:

$$\begin{array}{cc}\hfill {\omega }_A^2\left(S^{{*}_A}T\right)& \le \frac{1}{2\nu +2}{\omega }_A\left(S^{{*}_A}T\right)\parallel T^{{*}_A}T+S^{{*}_A}S{\parallel }_A+\frac{\nu }{2\nu +2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2∥}_A\hfill \\ \hfill & \le \frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(S^{{*}_A}S\right)}^2{\parallel }_A.\hfill \end{array}$$

Remark 2.

It is possible to use the following better inequality instead of Lemma 1 to get the corresponding result from Theorem 1:

$$\begin{array}{cc}\hfill {|\langle x,y\rangle }_A|& \le {\parallel x\parallel }_A^{\frac{1+\epsilon }{2}}{\parallel y\parallel }_A^{\frac{1+\epsilon }{2}}{\left|{\langle x,y\rangle }_A\right|}^{\frac{1-\epsilon }{2}}\hfill \\ \hfill & \le \sqrt{{\epsilon \parallel x\parallel }_A^2{\parallel y\parallel }_A^2{+(1-\epsilon )|\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A}\hfill \\ \hfill & \le \left(\frac{1+\epsilon }{2}\right){\parallel x\parallel }_A{\parallel y\parallel }_A+\left(\frac{1-\epsilon }{2}\right)\left|{\langle x,y\rangle }_A\right|\hfill \\ \hfill & \le {\parallel x\parallel }_A{\parallel y\parallel }_A.\hfill \end{array}$$

The inequalities above are due to the following: if $0<\alpha \le \beta$, then we have

$$\alpha \le {\alpha }^{\frac{1+\epsilon }{2}}{\beta }^{\frac{1-\epsilon }{2}}\le \sqrt{\epsilon {\alpha }^2+(1-\epsilon )\alpha \beta }\le \left(\frac{1+\epsilon }{2}\right)\alpha +\left(\frac{1-\epsilon }{2}\right)\beta \le \beta .$$

for $0\le \epsilon \le 1$. Since the second term ${\alpha }^{\frac{1+\epsilon }{2}}{\beta }^{\frac{1-\epsilon }{2}}$ and the fourth one $\left(\frac{1+\epsilon }{2}\right)\alpha +\left(\frac{1-\epsilon }{2}\right)\beta$ are respectively the geometric mean and arithmetic mean of α and β with a weight $\frac{1+\epsilon }{2}$, their appearance is quite natural (i.e., we judged that they may not be so interesting for the readers), so we adopted the third term $\sqrt{\epsilon {\alpha }^2+(1-\epsilon )\alpha \beta }$ to obtain the bound of the numerical radius in Theorem 1. It is notable that we have

$$\alpha \le \sqrt{\alpha \beta }=\sqrt{\alpha \beta }\le \frac{\alpha +\beta }{2}\le \beta$$

for a special case $\epsilon =0$. In this case ($\epsilon =0$), two bounds ${\alpha }^{\frac{1+\epsilon }{2}}{\beta }^{\frac{1-\epsilon }{2}}$ and $\sqrt{\epsilon {\alpha }^2+(1-\epsilon )\alpha \beta }$ acquire the same value $\sqrt{\alpha \beta }$.

In [11], Saddi established an interesting inequality that generalizes the well-known Buzano inequality. Specifically, the inequality is given by

$${|\langle x,e\rangle }_A{\langle e,y\rangle }_A|\le \frac{1}{2}\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right),$$

(8)

where $x,y\in \mathcal{X}$ and $e\in S_A^1$.

Considering inequality (8), we can derive a valuable lemma as follows.

Lemma 4.

Consider $x,y\in \mathcal{X}$, $e\in S_A^1$ and $\epsilon \in [0,1]$. Then we have:

$${|\langle x,e\rangle }_A{\langle e,y\rangle }_A|\le \sqrt{\frac{\epsilon +1}{2}{\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+\frac{(1-\epsilon )}{2}|{\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A}.$$

(9)

Proof. 

Suppose that $\epsilon$ is a scalar in the interval $[0,1]$. Let x, y, and e be vectors in $\mathcal{X}$ with $e\in S_A^1$. By using inequality (8), along with the convexity of the function $t\mapsto t^2$, we can obtain the following:

$$\begin{array}{cc}\hfill |{\langle x,e\rangle }_A{\langle e,y\rangle }_A{|}^2& \le \frac{1}{4}{\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right)}^2\hfill \\ \hfill & \le \frac{1}{2}\left({\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+|{\langle x,y\rangle }_A{|}^2\right).\hfill \end{array}$$

Moreover, we can use Lemma 1 to derive the following inequality:

$$\begin{array}{cc}\hfill |{\langle x,e\rangle }_A{\langle e,y\rangle }_A{|}^2& \le \frac{1}{2}\left({\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+{\epsilon \parallel x\parallel }_A^2{\parallel y\parallel }_A^2+(1-\epsilon )|{\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A\right)\hfill \\ \hfill & =\frac{\epsilon +1}{2}{\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+\frac{(1-\epsilon )}{2}|{\langle x,y\rangle }_A{|\parallel x\parallel }_A{\parallel y\parallel }_A.\hfill \end{array}$$

This inequality allows us to obtain the desired result. □

Based on the previous lemma, we can now demonstrate the next result.

Theorem 2.

Let $T\in {\mathcal{L}}_A\left(\mathcal{X}\right)$. Then, for any $\epsilon \in [0,1]$, we have

$$\begin{array}{cc}\hfill {\omega }_A^4\left(T\right)& \le \frac{\epsilon +1}{4}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{(1-\epsilon )}{4}{\omega }_A\left(T^2\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A\hfill \\ \hfill & \le \frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A.\hfill \end{array}$$

Proof. 

Let $\xi \in S_A^1$ and $\epsilon \in [0,1]$. We can apply Lemma 4 by setting $e=\xi$, $x=T\xi$ and $y=T^{{*}_A}\xi$, which gives us:

$$\begin{array}{cc}\hfill \left|{\langle T\xi ,\xi \rangle }_A{|}^2\right|{\langle \xi ,T^{{*}_A}\xi \rangle }_A{|}^2& \le \frac{\epsilon +1}{2}{\parallel T\xi \parallel }_A^2\parallel T^{{*}_A}{\xi \parallel }_A^2+\frac{(1-\epsilon )}{2}|{\langle T\xi ,T^{{*}_A}\xi \rangle }_A{|\parallel T\xi \parallel }_A{\parallel T^{{*}_A}\xi \parallel }_A.\hfill \end{array}$$

Using the same arguments as in the proof of Theorem 1, we can conclude that

$$\begin{array}{cc}\hfill & |{\langle T\xi ,\xi \rangle }_A{|}^4\hfill \\ \hfill & \le \frac{\epsilon +1}{2}{\parallel T\xi \parallel }_A^2\parallel T^{{*}_A}{\xi \parallel }_A^2+\frac{(1-\epsilon )}{2}|{\langle T^2\xi ,\xi \rangle }_A{|\parallel T\xi \parallel }_A{\parallel T^{{*}_A}\xi \parallel }_A\hfill \\ \hfill & =\frac{\epsilon +1}{2}{\langle T^{{*}_A}T\xi ,\xi \rangle }_A{\langle T{T}^{{*}_A}\xi ,\xi \rangle }_A+\frac{(1-\epsilon )}{2}\left|{\langle T^2\xi ,\xi \rangle }_A\right|\sqrt{{\langle T^{{*}_A}T\xi ,\xi \rangle }_A}\sqrt{{\langle T{T}^{{*}_A}\xi ,\xi \rangle }_A}\hfill \\ \hfill & \le \frac{\epsilon +1}{4}(\langle \left(T^{{*}_A}{T\xi ,\xi \rangle }_A^2+{\langle T{T}^{{*}_A}\xi ,\xi \rangle }_A^2\right)+\frac{(1-\epsilon )}{4}{\omega }_A\left(T^2\right){\langle \left(T^{{*}_A}T+T{T}^{{*}_A}\right)\xi ,\xi \rangle }_A\hfill \\ \hfill & =\frac{\epsilon +1}{4}\left({\langle \left[{\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2\right]\xi ,\xi \rangle }_A\right)+\frac{(1-\epsilon )}{4}{\omega }_A\left(T^2\right){\langle \left(T^{{*}_A}T+T{T}^{{*}_A}\right)\xi ,\xi \rangle }_A.\hfill \end{array}$$

From this, we can draw the inference that:

$$|{\langle T\xi ,\xi \rangle }_A{|}^4\le \frac{\epsilon +1}{4}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{(1-\epsilon )}{4}{\omega }_A\left(T^2\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A.$$

We can arrive at the first inequality in Theorem 2 by taking the supremum of all $\xi$ in $S_A^1$.

It is now evident that:

$$\begin{array}{cc}\hfill & {\omega }_A^4\left(T\right)\hfill \\ \hfill & \le \frac{\epsilon +1}{4}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{(1-\epsilon )}{4}{\omega }_A\left(T^2\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A\hfill \\ \hfill & \le \frac{\epsilon +1}{4}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{(1-\epsilon )}{4}{\omega }_A^2\left(T\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A,\left(\mathrm{by}\left(2\right)\right)\hfill \\ \hfill & \le \frac{\epsilon +1}{4}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{(1-\epsilon )}{8}\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A^2,\left(\mathrm{by}\left(5\right)\right)\hfill \\ \hfill & \le \frac{\epsilon +1}{4}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{(1-\epsilon )}{2}{∥\frac{T^{{*}_A}T+T{T}^{{*}_A}}{2}∥}_A^2\hfill \\ \hfill & \le \frac{\epsilon +1}{4}{∥{\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2∥}_A+\frac{(1-\epsilon )}{2}{\parallel \frac{{\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2}{2}\parallel }_A\hfill \\ \hfill & =\frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A.\hfill \end{array}$$

□

The following presents an enhanced version of Theorem 2.

Proposition 1.

Let $T\in {\mathcal{L}}_A\left(\mathcal{X}\right)$. Then, for any $\epsilon \in [0,1]$, we have

$$\begin{array}{cc}\hfill {\omega }_A^4\left(T\right)& \le \left(\frac{1+\epsilon }{4}\right)\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\left(\frac{1-\epsilon }{2}\right){\omega }_A^2\left(T^2\right)\hfill \\ \hfill & \le \left(\frac{1+\epsilon }{4}\right)\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\left(\frac{1-\epsilon }{4}\right){\omega }_A\left(T^2\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A\hfill \\ \hfill & \le \frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A.\hfill \end{array}$$

Proof. 

We have the following inequalities

$$\begin{array}{cc}\hfill |{\langle x,e\rangle }_A{\langle e,y\rangle }_A{|}^2& \le \frac{1}{4}{\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right)}^2\hfill \\ \hfill & \le \frac{1}{2}\left({\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+|{\langle x,y\rangle }_A{|}^2\right)\hfill \\ \hfill & \le \frac{1}{2}\left({\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+{\epsilon \parallel x\parallel }_A^2{\parallel y\parallel }_A^2+(1-\epsilon )|{\langle x,y\rangle }_A{|}^2\right)\hfill \\ \hfill & =\left(\frac{1+\epsilon }{2}\right){\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+\left(\frac{1-\epsilon }{2}\right)|{\langle x,y\rangle }_A{|}^2,\hfill \end{array}$$

(10)

since $0\le |{\langle x,y\rangle }_A{|}^2\le {\parallel x\parallel }_A^2{\parallel y\parallel }_A^2$. By setting $e=\xi$, $x=T\xi$ and $y=T^{{*}_A}\xi$ in the above, we have

$$\begin{array}{cc}\hfill {\omega }_A^4\left(T\right)& \le \left(\frac{1+\epsilon }{4}\right)\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\left(\frac{1-\epsilon }{2}\right){\omega }_A^2\left(T^2\right)\hfill \\ \hfill & \le \left(\frac{1+\epsilon }{4}\right)\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\left(\frac{1-\epsilon }{4}\right){\omega }_A\left(T^2\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A\hfill \\ \hfill & \le \frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A.\hfill \end{array}$$

The second inequality above is due to inequalities (2) and (5). □

Remark 3.

Since the inequalities in Equation (10) give a better bound than Lemma 4, Proposition 1 gives an improvement of Theorem 2.

This result follows directly from Theorem 2 by setting $\epsilon =\frac{1}{2}$.

Corollary 1.

Let $T\in {\mathcal{L}}_A\left(\mathcal{X}\right)$. Then,

$$\begin{array}{cc}\hfill {\omega }_A^4\left(T\right)& \le \frac{3}{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{1}{8}{\omega }_A\left(T^2\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A.\hfill \end{array}$$

The following lemma plays a pivotal role in establishing our forthcoming result.

Lemma 5.

For $x,y\in \mathcal{X}$, $e\in S_A^1$, and $\epsilon \in [0,1]$, the following inequality holds:

$$\begin{array}{cc}\hfill |{\langle x,e\rangle }_A{\langle e,y\rangle }_A{|}^2& \le \frac{\epsilon }{4}|{\langle x,y\rangle }_A{|}^2+\frac{1-\epsilon }{2}|{\langle x,e\rangle }_A{\langle e,y\rangle }_A|\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right)\hfill \\ \hfill & +\frac{\epsilon }{4}{\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+\frac{\epsilon }{2}{\parallel x\parallel }_A{\parallel y\parallel }_A\left|{\langle x,y\rangle }_A\right|.\hfill \end{array}$$

(11)

Proof. 

Let $x,y\in \mathcal{X}$, $e\in S_A^1$, and $\epsilon \in [0,1]$. By utilizing inequality (8), we obtain

$$\begin{array}{cc}\hfill & |{\langle x,e\rangle }_A{\langle e,y\rangle }_A{|}^2\hfill \\ \hfill & \le \frac{1}{2}\left|{\langle x,e\rangle }_A{\langle e,y\rangle }_A\right|\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right)\hfill \\ \hfill & =\frac{\epsilon }{2}\left|{\langle x,e\rangle }_A{\langle e,y\rangle }_A\right|\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right)+\frac{1-\epsilon }{2}\left|{\langle x,e\rangle }_A{\langle e,y\rangle }_A\right|\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right)\hfill \\ \hfill & \le \frac{\epsilon }{4}{\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right)}^2+\frac{1-\epsilon }{2}\left|{\langle x,e\rangle }_A{\langle e,y\rangle }_A\right|\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right).\hfill \end{array}$$

Therefore, we have obtained the desired result. □

Remark 4.

The upper bound of $|{\langle x,e\rangle }_A{\langle e,y\rangle }_A{|}^2$ in Equation (11) is better than in Equation (10) because we have

$$\begin{array}{cc}\hfill & \frac{\epsilon }{4}|{\langle x,y\rangle }_A{|}^2+\frac{1-\epsilon }{2}|{\langle x,e\rangle }_A{\langle e,y\rangle }_A|\left({\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|\right)\hfill \\ \hfill & +\frac{\epsilon }{4}{\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+\frac{\epsilon }{2}{\parallel x\parallel }_A{\parallel y\parallel }_A\left|{\langle x,y\rangle }_A\right|\hfill \\ \hfill & \le {\left(\frac{{\parallel x\parallel }_A{\parallel y\parallel }_A+\left|{\langle x,y\rangle }_A\right|}{2}\right)}^2\hfill \\ \hfill & \le \left(\frac{1+\epsilon }{2}\right){\parallel x\parallel }_A^2{\parallel y\parallel }_A^2+\left(\frac{1-\epsilon }{2}\right)|{\langle x,y\rangle }_A{|}^2.\hfill \end{array}$$

This is because the upper bound in (11) still uses $\left|{\langle x,e\rangle }_A{\langle e,y\rangle }_A\right|$, while the one in (10) does not use it.

With the help of inequality (11), we can now demonstrate the following result.

Theorem 3.

Consider $T\in {\mathcal{L}}_A\left(\mathcal{X}\right)$ and $\epsilon \in [0,1]$. We have the following inequalities:

$$\begin{array}{cc}\hfill {\omega }_A^4\left(T\right)& \le \frac{\epsilon }{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\left[\frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\right]\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A\hfill \\ \hfill & +\frac{\epsilon }{4}{\omega }_A^2\left(T^2\right)+\frac{1-\epsilon }{2}{\omega }_A\left(T^2\right){\omega }_A^2\left(T\right)\hfill \\ \hfill & \le \frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A.\hfill \end{array}$$

Proof. 

Let $\xi \in S_A^1$. If we replace e by $\xi$, x by $T\xi$, and y by $T^{{*}_A}\xi$ in the above inequality (Equation (11)), we obtain

$$\begin{array}{cc}\hfill & {|\langle T\xi ,\xi \rangle }_A{|}^4\hfill \\ \hfill & {=|\langle T\xi ,\xi \rangle }_A{\langle x,T^{{*}_A}\xi \rangle }_A{|}^2\hfill \\ \hfill & \le \frac{\epsilon }{4}|{\langle T\xi ,T^{{*}_A}\xi \rangle }_A{|}^2+\frac{1-\epsilon }{2}\left|{\langle T\xi ,\xi \rangle }_A{\langle x,T^{{*}_A}\xi \rangle }_A\right|\left({\parallel T\xi \parallel }_A\parallel T^{{*}_A}{\xi \parallel }_A+\left|{\langle T\xi ,T^{{*}_A}\xi \rangle }_A\right|\right)\hfill \\ \hfill & +\frac{\epsilon }{4}{\parallel T\xi \parallel }_A^2\parallel T^{{*}_A}{\xi \parallel }_A^2+\frac{\epsilon }{2}{\parallel T\xi \parallel }_A\parallel T^{{*}_A}{\xi \parallel }_A\left|{\langle T\xi ,T^{{*}_A}\xi \rangle }_A\right|\hfill \\ \hfill & \le \frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{2}{\omega }_A^2\left(T\right)\left({\parallel T\xi \parallel }_A{\parallel T^{{*}_A}\xi \parallel }_A+{\omega }_A\left(T^2\right)\right)\hfill \\ \hfill & +\frac{\epsilon }{4}{\parallel T\xi \parallel }_A^2\parallel T^{{*}_A}{\xi \parallel }_A^2+\frac{\epsilon }{2}{\parallel T\xi \parallel }_A{\parallel T^{{*}_A}\xi \parallel }_A{\omega }_A\left(T^2\right).\hfill \end{array}$$

Furthermore, through the application of the arithmetic–geometric mean inequality, we can derive

$$\begin{array}{cc}\hfill & {|\langle T\xi ,\xi \rangle }_A{|}^4\hfill \\ \hfill & \le \frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\left({\parallel T\xi \parallel }_A^2+{\parallel T^{{*}_A}\xi \parallel }_A^2+2{\omega }_A\left(T^2\right)\right)\hfill \\ \hfill & +\frac{\epsilon }{8}\left({\parallel T\xi \parallel }_A^4+{\parallel T^{{*}_A}\xi \parallel }_A^4\right)+\frac{\epsilon }{2}{\omega }_A\left(T^2\right)\left({\parallel T\xi \parallel }_A^2+{\parallel T^{{*}_A}\xi \parallel }_A^2\right)\hfill \\ \hfill & =\frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\left({\langle (T^{{*}_A}T+T{T}^{{*}_A})\xi ,\xi \rangle }_A+2{\omega }_A\left(T^2\right)\right)\hfill \\ \hfill & +\frac{\epsilon }{8}\left({\langle \left(T^{{*}_A}T\right)\xi ,\xi \rangle }_A^2+{\langle \left(T{T}^{{*}_A}\right)\xi ,\xi \rangle }_A^2\right)+\frac{\epsilon }{2}{\omega }_A\left(T^2\right)\left({\parallel T\xi \parallel }_A^2+{\parallel T^{{*}_A}\xi \parallel }_A^2\right)\hfill \end{array}$$

As a result, this leads to

$$\begin{array}{cc}\hfill & {|\langle T\xi ,\xi \rangle }_A{|}^4\hfill \\ \hfill & \le \frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\left({\langle (T^{{*}_A}T+T{T}^{{*}_A})\xi ,\xi \rangle }_A+2{\omega }_A\left(T^2\right)\right)\hfill \\ \hfill & +\frac{\epsilon }{8}\left({\langle \left(T^{{*}_A}T\right)\xi ,\xi \rangle }_A^2+{\langle \left(T{T}^{{*}_A}\right)\xi ,\xi \rangle }_A^2\right)+\frac{\epsilon }{2}{\omega }_A\left(T^2\right){\langle (T^{{*}_A}T+T{T}^{{*}_A})\xi ,\xi \rangle }_A\hfill \\ \hfill & \le \frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\left({\omega }_A\left(T^{{*}_A}T+T{T}^{{*}_A}\right)+2{\omega }_A\left(T^2\right)\right)\hfill \\ \hfill & +\frac{\epsilon }{8}\left({\langle {\left(T^{{*}_A}T\right)}^2\xi ,\xi \rangle }_A+{\langle {\left(T{T}^{{*}_A}\right)}^2\xi ,\xi \rangle }_A\right)+\frac{\epsilon }{2}{\omega }_A\left(T^2\right){\omega }_A\left(T^{{*}_A}T+T{T}^{{*}_A}\right),\hfill \end{array}$$

where the last inequality is derived by utilizing Lemma 3. Therefore, we deduce that

$$\begin{array}{cc}\hfill {|\langle T\xi ,\xi \rangle }_A{|}^4& \le \frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\left({\omega }_A\left(T^{{*}_A}T+T{T}^{{*}_A}\right)+2{\omega }_A\left(T^2\right)\right)\hfill \\ \hfill & +\frac{\epsilon }{8}{\omega }_A\left({\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2\right)+\frac{\epsilon }{2}{\omega }_A\left(T^2\right){\omega }_A\left(T^{{*}_A}T+T{T}^{{*}_A}\right),\hfill \end{array}$$

Consequently, by considering the supremum over $\xi \in S_A^1$ in the last inequality, we can derive the following inequality:

$$\begin{array}{cc}\hfill {\omega }_A^4\left(T\right)& \le \frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\left({\omega }_A\left(T^{{*}_A}T+T{T}^{{*}_A}\right)+2{\omega }_A\left(T^2\right)\right)\hfill \\ \hfill & +\frac{\epsilon }{8}{\omega }_A\left({\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2\right)+\frac{\epsilon }{2}{\omega }_A\left(T^2\right){\omega }_A\left(T^{{*}_A}T+T{T}^{{*}_A}\right),\hfill \end{array}$$

On the other hand, since $T^{{*}_A}T+T{T}^{{*}_A}{\ge }_{A}0$ and ${\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\ge }_{A}0$, we can apply Equation (3) to derive the following inequality:

$$\begin{array}{cc}\hfill {\omega }_A^4\left(T\right)& \le \frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\left(\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A+2{\omega }_A\left(T^2\right)\right)\hfill \\ \hfill & +\frac{\epsilon }{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{\epsilon }{2}{\omega }_A\left(T^2\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A.\hfill \end{array}$$

This inequality directly proves the first inequality in Theorem 3.

To derive the second inequality in Theorem 3, we utilize Equation (2) in conjunction with the second inequality stated in Equation (5). This allows us to obtain:

$$\begin{array}{cc}\hfill & {\omega }_A^4\left(T\right)\hfill \\ \hfill & \le \frac{\epsilon }{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\left[\frac{\epsilon }{4}{\omega }_A\left(T^2\right)+\frac{1-\epsilon }{4}{\omega }_A^2\left(T\right)\right]\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A\hfill \\ \hfill & +\frac{\epsilon }{4}{\omega }_A^2\left(T^2\right)+\frac{1-\epsilon }{2}{\omega }_A\left(T^2\right){\omega }_A^2\left(T\right)\hfill \\ \hfill & \le \frac{\epsilon }{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{1}{4}{\omega }_A^2\left(T\right)\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A+\frac{2-\epsilon }{4}{\omega }_A^4\left(T\right)\hfill \\ \hfill & \le \frac{\epsilon }{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{1}{8}\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A^2+\frac{2-\epsilon }{16}\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A^2\hfill \\ \hfill & =\frac{\epsilon }{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{1}{2}{∥\frac{T^{{*}_A}T+T{T}^{{*}_A}}{2}∥}_A^2+\frac{2-\epsilon }{4}{∥\frac{T^{{*}_A}T+T{T}^{{*}_A}}{2}∥}_A^2\hfill \\ \hfill & \le \frac{\epsilon }{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{1}{4}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A\hfill \\ \hfill & +\frac{2-\epsilon }{8}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A\hfill \\ \hfill & =\frac{1}{2}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A.\hfill \end{array}$$

It is worth noting that, in the last inequality, we employed Lemma 2 due to the A-positive nature of the operators $T^{{*}_A}T$ and $T{T}^{{*}_A}$. This concludes the proof. □

The corollary presented below is a direct result of Theorem 3 where we set $\epsilon =\frac{1}{3}$. This corollary highlights a discovery mentioned in [20].

Corollary 2.

For any $T\in {\mathcal{L}}_A\left(\mathcal{X}\right)$, the following inequalities hold:

$$\begin{array}{cc}\hfill {\omega }_A^4\left(T\right)& \le \frac{1}{24}\parallel {\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2{\parallel }_A+\frac{1}{12}\parallel T^{{*}_A}T+T{T}^{{*}_A}{\parallel }_A\left({\omega }_A\left(T^2\right)+2{\omega }_A^2\left(T\right)\right)\hfill \\ \hfill & +\frac{1}{12}{\omega }_A^2\left(T^2\right)+\frac{1}{3}{\omega }_A^2\left(T\right){\omega }_A\left(T^2\right)\hfill \\ \hfill & \le \frac{1}{2}{∥{\left(T^{{*}_A}T\right)}^2+{\left(T{T}^{{*}_A}\right)}^2∥}_A.\hfill \end{array}$$

3. Conclusions

In conclusion, this research paper introduced new Cauchy–Schwarz inequalities for use in semi-Hilbert spaces, which are extensions of Hilbert spaces. These new inequalities were used to create innovative A-numerical radius inequalities, where A is a positive semidefinite operator in a complex Hilbert space. These findings are important tools in functional analysis and build upon the existing literature on numerical radius inequalities with Hilbert space operators.

We used techniques from semi-Hilbert space theory to prove the validity of these new inequalities and demonstrated their potential applications in operator theory and functional analysis. Our work provides a foundation for future research, extending the theoretical framework beyond traditional Hilbert spaces and inspiring the development of new mathematical tools and techniques.

In summary, this research contributes to the advancement of mathematical theory by introducing new Cauchy–Schwarz inequalities and deriving novel A-numerical radius inequalities in semi-Hilbert spaces. We anticipate that these findings will serve as a valuable starting point for future research and stimulate further investigations in functional analysis and related areas of study.