$\mathcal{A}$-Joint Numerical Radius Bounds for Operator Pairs in Semi-Hilbert Spaces with Applications

Abstract

Consider a complex Hilbert space $\left(\mathbb{X},⟨·,·⟩\right)$ equipped with a positive (semidefinite) bounded linear operator $\mathcal{A}$ on $\mathbb{X}$. The $\mathcal{A}$-joint numerical radius for two $\mathcal{A}$-bounded operators $\mathcal{T}$ and $\mathcal{S}$ is defined as ${\omega }_{\mathcal{A},e}(\mathcal{T},\mathcal{S})={sup}_{{\parallel x\parallel }_{\mathcal{A}}=1}\sqrt{{\left|{⟨\mathcal{T}x,x⟩}_{\mathcal{A}}\right|}^2+{\left|{⟨\mathcal{S}x,x⟩}_{\mathcal{A}}\right|}^2}.$ Among other results, in this study we demonstrate that ${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\le 2^{1-{\displaystyle \frac{1}{r}}}{\left[max\left\{{∥\mathcal{T}∥}_{\mathcal{A}}^{2r},{∥\mathcal{S}∥}_{\mathcal{A}}^{2r}\right\}+{\omega }_{\mathcal{A}}^r\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{r}}}$ for $r\ge 1$ and that ${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\ge \frac{1}{2}{\omega }_{\mathcal{A}}\left({\mathcal{T}}^2+{\mathcal{S}}^2\right)+\frac{1}{2}max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{S}+\mathcal{T}\right),{\omega }_{\mathcal{A}}\left(\mathcal{S}-\mathcal{T}\right)\right\}\left|{\omega }_{\mathcal{A}}\left(\mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\mathcal{T}\right)\right|$. Additionally, we provide several inequalities related to the $\mathcal{A}$-numerical radius and $\mathcal{A}$-seminorm.

1. Introduction

In recent years, the study of operators on semi-Hilbert spaces has experienced a significant resurgence of interest driven by their ability to generalize classical Hilbert spaces and provide a versatile framework for operator theory, as evidenced by works such as [1,2,3,4,5,6,7,8,9,10,11,12,13]. Semi-Hilbert spaces are equipped with a semi-inner product induced by a positive operator $\mathcal{A}$, offering a robust and promising branch of functional analysis with applications in quantum theory and beyond. This framework enables the study of operators under modified inner products, facilitating the analysis of physical observables in quantum systems where traditional self-adjointness constraints are restrictive [8,10,11,12]. Specifically, physical states in a quantum system are represented on a Hilbert space $\mathbb{X}$ with an inner product $\langle ·,·\rangle$. Non-Hermitian quantum theory introduces a new inner product, defined as ${\langle {\xi }_1,{\xi }_2\rangle }_{\mathcal{A}}=\langle \mathcal{A}{\xi }_1,{\xi }_2\rangle$ for all ${\xi }_1,{\xi }_2\in \mathbb{X}$, relative to which operators are self-adjoint, thereby broadening the scope of quantum theory [13]. In quasi-Hermitian and pseudo-Hermitian quantum theory [8,10,11], $\mathcal{A}$ is invertible, self-adjoint, and positive, while in indefinite metric quantum theory [12] $\mathcal{A}$ is unitary and self-adjoint but not necessarily positive. This flexibility makes semi-Hilbert spaces critical for advancing operator theory and its applications in functional analysis, spectral theory, and quantum mechanics.

Mathematical inequalities, particularly norm and numerical radius inequalities, have received considerable attention in this context [9,14,15,16,17,18,19]. These inequalities provide insights into the geometric and spectral properties of operators, with significant implications for functional analysis. For instance, ref. [1] established metric properties of projections in semi-Hilbertian spaces, while [2] developed a framework for partial isometries that enhanced the understanding of operator behavior. The foundational theory of $\mathcal{A}$-normal operators was introduced in [20], while [3,4] respectively explored joint normality and closed operators in semi-Hilbertian spaces, offering results tied to the $\mathcal{A}$-joint numerical radius. Inequalities for the Euclidean operator radius of operator pairs in Hilbert spaces were derived in [17], inspiring generalizations to semi-Hilbert spaces in [16], which extended seminorm and numerical radius inequalities. Further advancements include bounds for the $\mathcal{A}$-joint numerical radius and $\mathcal{A}$-numerical radius of operator matrices [6,7], refinements of $\mathcal{A}$-numerical radius inequalities [21,22,23], and studies on $2\times 2$ operator matrices [9,24]. Orthogonality and norm-parallelism in the $\mathcal{A}$-numerical radius were examined in [25], while [15,26,27,28,29] provided additional bounds and generalizations, including $\mathcal{A}$-translatable radii. Structured operator matrices were addressed in [18,19,30], complementing these results. The works in [31,32,33] further solidify the importance of semi-Hilbert spaces in studying operator inequalities.

In this paper, we establish novel inequalities for the $\mathcal{A}$-joint numerical radius and $\mathcal{A}$-numerical radius of operators in a semi-Hilbert space, building on frameworks from [1,3,7,17]. We also investigate connections between these quantities and the $\mathcal{A}$-seminorm, providing refined bounds and applications to operator theory. Our results leverage the generalized structure of semi-Hilbert spaces to extend classical operator theory, offering new tools for spectral analysis and quantum theory applications.

Prior to presenting our findings, we clarify notations and concepts. Let $\left(\mathbb{X},\langle ·,·\rangle \right)$ be a complex Hilbert space with norm $\parallel ·\parallel$ induced by $\langle ·,·\rangle$. The $C^{*}$-algebra of bounded linear operators on $\mathbb{X}$ with identity I is denoted by $\mathbf{L}\left(\mathbb{X}\right)$. The sets $\mathbb{N}$ and ${\mathbb{N}}^{*}$ represent the non-negative and positive integers, respectively. Throughout this paper, an operator refers to an element of $\mathbf{L}\left(\mathbb{X}\right)$.

An operator $\mathcal{T}\in \mathbf{L}\left(\mathbb{X}\right)$ is called positive, denoted $\mathcal{T}\ge 0$, if $\langle \mathcal{T}x,x\rangle \ge 0$ for all $x\in \mathbb{X}$ and $n\in {\mathbb{N}}^{*}$. For a positive operator $\mathcal{T}$, its square root is denoted by ${\mathcal{T}}^{\frac{1}{2}}$. Henceforth, we assume that $\mathcal{A}$ is a nonzero positive operator inducing the semi-inner product

$${\langle ·,·\rangle }_{\mathcal{A}}:\mathbb{X}\times \mathbb{X}⟶\mathbb{C},(\mathcal{T},\mathcal{S})⟼{\langle \mathcal{T},\mathcal{S}\rangle }_{\mathcal{A}}:=\langle \mathcal{A}\mathcal{T},\mathcal{S}\rangle .$$

Observe that $(\mathbb{X},\parallel ·{\parallel }_{\mathcal{A}})$ is called a semi-Hilbert space. Here, ${\parallel ·\parallel }_{\mathcal{A}}$ denotes the seminorm induced by the semi-inner product ${\langle ·,·\rangle }_{\mathcal{A}}$, defined by ${\parallel \xi \parallel }_{\mathcal{A}}=\sqrt{{\langle \xi ,\xi \rangle }_{\mathcal{A}}}$ for all $\xi \in \mathbb{X}$. It should be noted that $(\mathbb{X},\parallel ·{\parallel }_{\mathcal{A}})$ is, in general, neither a normed space nor a complete one. However, it becomes a Hilbert space if and only if $\mathcal{A}$ is injective and $\overline{\mathrm{Ran}\left(\mathcal{A}\right)}=\mathrm{Ran}\left(\mathcal{A}\right)$. Here, $\overline{\mathrm{Ran}\left(\mathcal{A}\right)}$ denotes the closure of $\mathrm{Ran}\left(\mathcal{A}\right)$ with respect to the topology induced by the norm $\parallel ·\parallel$.

To illustrate, consider the complex field $\mathbb{C}$ and let ${\ell }^2$ be the set of all sequences $\xi =\left\{{\xi }_i\right\}$ (${\xi }_i\in \mathbb{C}$ for all $i\in {\mathbb{N}}^{*}$) satisfying ${\sum }_{m\in \mathbb{N}}{\left|{\xi }_m\right|}^2<\infty$. The inner product on ${\ell }^2$ is defined by

$$\langle \xi ,\nu \rangle =\sum _{m\in {\mathbb{N}}^{*}}{\xi }_m\overline{{\nu }_m}\mathrm{for}\mathrm{all}\xi =\left\{{\xi }_i\right\},\nu =\left\{{\nu }_i\right\}\in {\ell }^2.$$

It is well known that $({\ell }^2,\langle ·,·\rangle )$ is a Hilbert space. We define the diagonal operator $\mathcal{A}$ on ${\ell }^2$ by

$$\mathcal{A}{e}_1=0,\mathrm{and}\mathcal{A}{e}_i={\displaystyle \frac{e_i}{i}},\mathrm{for}\mathrm{all}i\ge 2,$$

where $\left\{e_i\right\}$ is the canonical basis of ${\ell }^2$. It can be easily verified that $\mathcal{A}$ is not injective and that $\overline{\mathrm{Ran}\left(\mathcal{A}\right)}\ne \mathrm{Ran}\left(\mathcal{A}\right)$. Therefore, the semi-Hilbert space $({\ell }^2,{\langle ·,·\rangle }_{\mathcal{A}})$ is neither normed nor complete.

An operator $\mathcal{M}\in \mathbf{L}\left(\mathbb{X}\right)$ is called an $\mathcal{A}$-adjoint of $\mathcal{T}$ if

$${\langle \mathcal{T}\xi ,\nu \rangle }_{\mathcal{A}}={\langle \xi ,\mathcal{M}\nu \rangle }_{\mathcal{A}}\mathrm{for}\mathrm{all}\xi ,\nu \in \mathbb{X},$$

or equivalently, if $\mathcal{A}\mathcal{M}={\mathcal{T}}^{*}\mathcal{A}$. It is important to note that not every operator $\mathcal{T}\in \mathbf{L}\left(\mathbb{X}\right)$ admits an $\mathcal{A}$-adjoint. Moreover, even when such an operator $\mathcal{M}$ exists, it is not necessarily unique. This non-uniqueness introduces several difficulties in the study of operators on semi-Hilbert spaces.

In this framework, the celebrated Douglas range inclusion theorem [34] plays a crucial role. In essence, the Douglas theorem asserts that the operator equation $\mathcal{T}X=\mathcal{S}$ admits a solution $\mathcal{Y}\in \mathbf{L}\left(\mathbb{X}\right)$ if and only if $\mathrm{Ran}\left(\mathcal{S}\right)\subseteq \mathrm{Ran}\left(\mathcal{T}\right)$. Equivalently, there exists a constant $\kappa >0$ such that

$$\parallel {\mathcal{S}}^{*}\xi \parallel \le \kappa \parallel {\mathcal{T}}^{*}\xi \parallel \mathrm{for}\mathrm{all}\xi \in \mathbb{X}.$$

By applying the Douglas theorem, it is possible to identify from among the operators $\mathcal{T}$ that possess $\mathcal{A}$-adjoints a distinguished operator denoted by ${\mathcal{T}}^{{♯}_{\mathcal{A}}}$. If ${\mathcal{A}}^{†}$ stands for the Moore–Penrose pseudo-inverse of $\mathcal{A}$, then (see [2,32])

$${\mathcal{T}}^{{♯}_{\mathcal{A}}}={\mathcal{A}}^{†}{\mathcal{T}}^{*}\mathcal{A}.$$

The operator ${\mathcal{T}}^{{♯}_{\mathcal{A}}}$ shares certain properties with ${\mathcal{T}}^{*}$, but not all. For instance, the equality ${\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^{{♯}_{\mathcal{A}}}=\mathcal{T}$ does not generally hold. However, it is valid precisely when $\mathrm{Ran}\left(\mathcal{T}\right)\subseteq \overline{\mathrm{Ran}\left(\mathcal{A}\right)}$. Furthermore, we have

$${\left({\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^{{♯}_{\mathcal{A}}}\right)}^{{♯}_{\mathcal{A}}}={\mathcal{T}}^{{♯}_{\mathcal{A}}}.$$

(1)

An operator $\mathcal{T}\in \mathbf{L}\left(\mathbb{X}\right)$ is said to be $\mathcal{A}$-selfadjoint if $\mathcal{A}\mathcal{T}$ is selfadjoint. It was shown in [35], Lemma 1, that if $\mathcal{T}\in \mathbf{L}\left(\mathbb{X}\right)$ is an $\mathcal{A}$-selfadjoint operator, then ${\mathcal{T}}^{{♯}_{\mathcal{A}}}$ is also $\mathcal{A}$-selfadjoint and

$${\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^{{♯}_{\mathcal{A}}}={\mathcal{T}}^{{♯}_{\mathcal{A}}}.$$

(2)

Let ${\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and ${\mathbf{L}}_{\sqrt{\mathcal{A}}}\left(\mathbb{X}\right)$ denote the sets of all operators that admit $\mathcal{A}$-adjoints and $\sqrt{\mathcal{A}}$-adjoints, respectively. Using the Douglas theorem, it follows immediately that

$$\begin{array}{c}\hfill {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)=\left\{\mathcal{T}\in \mathbf{L}\left(\mathbb{X}\right);\mathrm{Ran}\left({\mathcal{T}}^{*}\mathcal{A}\right)\subseteq \mathrm{Ran}\left(\mathcal{A}\right)\right\}\end{array}$$

and

$${\mathbf{L}}_{\sqrt{\mathcal{A}}}\left(\mathbb{X}\right)=\left\{\mathcal{T}\in {\mathbf{L}\left(\mathbb{X}\right);\exists \kappa >0\mathrm{such}\mathrm{that}\parallel \mathcal{T}x\parallel }_{\mathcal{A}}\le \kappa {\parallel x\parallel }_{\mathcal{A}},\forall x\in \mathbb{X}\right\}.$$

Operators in ${\mathbf{L}}_{\sqrt{\mathcal{A}}}\left(\mathbb{X}\right)$ are called $\mathcal{A}$-bounded. It is also worth noting that both ${\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and ${\mathbf{L}}_{\sqrt{\mathcal{A}}}\left(\mathbb{X}\right)$ are subalgebras of $\mathbf{L}\left(\mathbb{X}\right)$, which are in general neither closed nor dense in the $C^{*}$-algebra $\mathbf{L}\left(\mathbb{X}\right)$. Moreover, the proper inclusions

$${\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)\subseteq {\mathbf{L}}_{\sqrt{\mathcal{A}}}\left(\mathbb{X}\right)\subseteq \mathbf{L}\left(\mathbb{X}\right)$$

hold.

The $\mathcal{A}$-joint numerical radius of two $\mathcal{A}$-bounded operators $\mathcal{T}$ and $\mathcal{S}$ is defined in [36] as

$${\omega }_{\mathcal{A},e}(\mathcal{T},\mathcal{S})=\underset{{\parallel x\parallel }_{\mathcal{A}}=1}{sup}\sqrt{{\left|{\langle \mathcal{T}x,x\rangle }_{\mathcal{A}}\right|}^2+{\left|{\langle \mathcal{S}x,x\rangle }_{\mathcal{A}}\right|}^2}.$$

In the same work [36], the third author established the following upper bounds:

$${\omega }_{\mathcal{A},e}(\mathcal{T},\mathcal{S})\le \sqrt{{∥\mathcal{T}∥}_{\mathcal{A}}^4+{∥\mathcal{S}∥}_{\mathcal{A}}^4+2{\omega }_{\mathcal{A}}^2\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)}\le {∥\mathcal{T}∥}_{\mathcal{A}}^2+{∥\mathcal{S}∥}_{\mathcal{A}}^2,$$

$${\omega }_{\mathcal{A},e}(\mathcal{T},\mathcal{S})\le {\left[{\omega }_{\mathcal{A}}\left({\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)}^2+{\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)}^2\right)+2{\omega }_{\mathcal{A}}^2\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{\frac{1}{4}},$$

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A},e}(\mathcal{T},\mathcal{S})& \le {\displaystyle \frac{\sqrt{2}}{2}}\sqrt{{\parallel {\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\parallel }_{\mathcal{A}}+{\parallel {\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\parallel }_{\mathcal{A}}+2{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)}\hfill \\ \hfill & \le \sqrt{{\parallel \mathcal{T}\parallel }_{\mathcal{A}}^2+{\parallel \mathcal{S}\parallel }_{\mathcal{A}}^2+{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)},\hfill \end{array}$$

and

$${\omega }_{\mathcal{A},e}(\mathcal{T},\mathcal{S})\le \sqrt{max\left({\parallel \mathcal{T}\parallel }_{\mathcal{A}}^2,{\parallel \mathcal{S}\parallel }_{\mathcal{A}}^2\right)+{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)}.$$

(3)

Furthermore, the inequality in (3) is sharp.

The following upper bound also holds:

$${\omega }_{\mathcal{A},e}(\mathcal{T},\mathcal{S})\le \sqrt{max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}\right),{\omega }_{\mathcal{A}}\left(\mathcal{S}\right)\right\}\sqrt{{\parallel {\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\parallel }_{\mathcal{A}}+2{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)}}.$$

(4)

Additional lower and upper bounds were derived in [36]:

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\sqrt{2}}{2}}max\left\{{\omega }_{\mathcal{A}}(\mathcal{T}+\mathcal{S}),{\omega }_{\mathcal{A}}(\mathcal{T}-\mathcal{S})\right\}& \le & {\omega }_{\mathcal{A},\mathrm{e}}(\mathcal{T},\mathcal{S})\hfill \\ & \le & {\displaystyle \frac{\sqrt{2}}{2}}\sqrt{{\omega }_{\mathcal{A}}^2(\mathcal{T}+\mathcal{S})+{\omega }_{\mathcal{A}}^2(\mathcal{T}-\mathcal{S})}.\hfill \end{array}$$

The constant ${\displaystyle \frac{\sqrt{2}}{2}}$ is sharp in both inequalities.

Additionally, the following bounds were recalled:

$${\displaystyle \frac{\sqrt{2}}{2}}\sqrt{{\omega }_{\mathcal{A}}({\mathcal{T}}^2+{\mathcal{S}}^2)}\le {\omega }_{\mathcal{A},e}(\mathcal{T},\mathcal{S})\le \sqrt{{\parallel {\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\parallel }_{\mathcal{A}}},$$

(5)

with ${\displaystyle \frac{\sqrt{2}}{2}}$ being a sharp constant.

In [36], the author also derived several inequalities involving the $\mathcal{A}$-Davis–Wielandt radius and the $\mathcal{A}$-numerical radii of $\mathcal{A}$-bounded operators.

2. Upper Bounds for $\mathcal{A}$-Euclidean Numerical Radius

R.P. Boas in 1941 [37], along with R. Bellman independently in 1944 [38], established a generalization of Bessel’s inequality; for vectors $x,y_1,\dots ,y_n$ in an inner product space $\left(\mathbb{X};〈·,·〉\right)$, the following inequality holds:

$$\sum _{i=1}^n{\left|〈x,y_i〉\right|}^2\le {∥x∥}^2\left[\underset{1\le i\le n}{max}{∥y_i∥}^2+{\left(\sum _{1\le i\ne j\le n}{\left|〈y_i,y_j〉\right|}^2\right)}^{\frac{1}{2}}\right],$$

(6)

When $n=2$, $y_1=y$, and $y_2=z$ in (6), we obtain

$${\left|〈x,y〉\right|}^2+{\left|〈x,z〉\right|}^2\le {∥x∥}^2\left[max\left\{{∥y∥}^2,{∥z∥}^2\right\}+\sqrt{2}\left|〈y,z〉\right|\right]$$

(7)

for all $x,y,z\in \mathbb{X}$.

In 2006, the second author [17], Equation (2.26) derived a sharper inequality than (7):

$${\left|〈x,y〉\right|}^2+{\left|〈x,z〉\right|}^2\le {∥x∥}^2\left[max\left\{{∥y∥}^2,{∥z∥}^2\right\}+\left|〈y,z〉\right|\right]$$

(8)

for all $x,y,z\in \mathbb{X}$.

In 2004, the second author [39] also obtained a result related to (6):

$$\sum _{i=1}^n{\left|〈x,y_i〉\right|}^2\le ∥x∥\underset{1\le i\le n}{max}\left|〈x,y_i〉\right|{\left\{\sum _{i=1}^n{∥y_i∥}^2+\sum _{1\le i\ne j\le n}\left|〈y_i,y_j〉\right|\right\}}^{\frac{1}{2}}$$

(9)

for any $x,y_1,\dots ,y_n$ in the inner product space $\left(\mathbb{X};〈·,·〉\right)$. Setting $n=2$, $y_1=y$, and $y_2=z$ in (9), we get

$$\begin{array}{cc}\hfill & {\left|〈x,y〉\right|}^2+{\left|〈x,z〉\right|}^2\hfill \\ \hfill & \le \sqrt{2}∥x∥max\left\{\left|〈x,y〉\right|,\left|〈x,z〉\right|\right\}{\left\{{\displaystyle \frac{{∥y∥}^2+{∥z∥}^2}{2}}+\left|〈y,z〉\right|\right\}}^{\frac{1}{2}}\hfill \end{array}$$

(10)

for all $x,y,z\in \mathbb{X}$.

We require the following result for the $\mathcal{A}$-inner product and norm.

Lemma 1.

For all $u,v,w\in \mathbb{X}$, we have

$${\left|{〈u,v〉}_{\mathcal{A}}\right|}^2+{\left|{〈u,w〉}_{\mathcal{A}}\right|}^2\le {∥u∥}_{\mathcal{A}}^2\left[max\left\{{∥v∥}_{\mathcal{A}}^2,{∥w∥}_{\mathcal{A}}^2\right\}+\left|{〈v,w〉}_{\mathcal{A}}\right|\right].$$

(11)

Proof. 

By setting $x={\mathcal{A}}^{\frac{1}{2}}u$, $y={\mathcal{A}}^{\frac{1}{2}}v$, and $z={\mathcal{A}}^{\frac{1}{2}}w$ in (8), we obtain

$$\begin{array}{cc}\hfill & {\left|〈{\mathcal{A}}^{\frac{1}{2}}u,{\mathcal{A}}^{\frac{1}{2}}v〉\right|}^2+{\left|〈{\mathcal{A}}^{\frac{1}{2}}u,{\mathcal{A}}^{\frac{1}{2}}w〉\right|}^2\hfill \\ \hfill & \le {∥{\mathcal{A}}^{\frac{1}{2}}u∥}^2\left[max\left\{{∥{\mathcal{A}}^{\frac{1}{2}}v∥}^2,{∥{\mathcal{A}}^{\frac{1}{2}}w∥}^2\right\}+\left|〈{\mathcal{A}}^{\frac{1}{2}}v,{\mathcal{A}}^{\frac{1}{2}}w〉\right|\right],\hfill \end{array}$$

which simplifies to

$${\left|〈\mathcal{A}u,v〉\right|}^2+{\left|〈\mathcal{A}u,w〉\right|}^2\le 〈\mathcal{A}u,u〉\left[max\left\{〈\mathcal{A}v,v〉,〈\mathcal{A}w,w〉\right\}+\left|〈\mathcal{A}v,w〉\right|\right]$$

for all $u,v,w\in \mathbb{X}$, which is equivalent to (11). □

Our first result is as follows:

Theorem 1.

For all $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $r\ge 1$, the following inequalities hold:

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\le 2^{1-{\displaystyle \frac{1}{r}}}{\left[max\left\{{∥\mathcal{T}∥}_{\mathcal{A}}^{2r},{∥\mathcal{S}∥}_{\mathcal{A}}^{2r}\right\}+{\omega }_{\mathcal{A}}^r\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{\frac{1}{r}}$$

(12)

and

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \le 2^{1-\frac{1}{r}}{\left[{\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^r+{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^r}{2}}+{\omega }_{\mathcal{A}}^r\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{\frac{1}{r}}.\hfill \end{array}$$

(13)

Proof. 

From (11), we obtain

$${\left|{〈x,y〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,z〉}_{\mathcal{A}}\right|}^2\le 2{∥x∥}_{\mathcal{A}}^2\left[{\displaystyle \frac{max\left\{{∥y∥}_{\mathcal{A}}^2,{∥z∥}_{\mathcal{A}}^2\right\}+\left|{〈y,z〉}_{\mathcal{A}}\right|}{2}}\right]$$

for all $x,y,z\in \mathbb{X}$.

Raising both sides to the power $r\ge 1$ and leveraging the convexity of the power function for $r\ge 1$, we get

$$\begin{array}{cc}\hfill & {\left({\left|{〈x,y〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,z〉}_{\mathcal{A}}\right|}^2\right)}^r\hfill \\ \hfill & \le 2^r{∥x∥}_{\mathcal{A}}^{2r}{\left[{\displaystyle \frac{max\left\{{∥y∥}_{\mathcal{A}}^2,{∥z∥}_{\mathcal{A}}^2\right\}+\left|{〈y,z〉}_{\mathcal{A}}\right|}{2}}\right]}^r\hfill \\ \hfill & \le 2^r{∥x∥}_{\mathcal{A}}^{2r}{\displaystyle \frac{{\left(max\left\{{∥y∥}_{\mathcal{A}}^2,{∥z∥}_{\mathcal{A}}^2\right\}\right)}^r+{\left|{〈y,z〉}_{\mathcal{A}}\right|}^r}{2}}\hfill \\ \hfill & =2^{r-1}{∥x∥}_{\mathcal{A}}^{2r}\left[max\left\{{∥y∥}_{\mathcal{A}}^{2r},{∥z∥}_{\mathcal{A}}^{2r}\right\}+{\left|{〈y,z〉}_{\mathcal{A}}\right|}^r\right]\hfill \end{array}$$

(14)

for all $x,y,z\in \mathbb{X}$.

Substituting $y=\mathcal{T}x$, $z=\mathcal{S}x$ with $x\in \mathbb{X}$ and ${∥x∥}_{\mathcal{A}}=1$ into (14), we obtain

$$\begin{array}{cc}\hfill & {\left({\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^r\hfill \\ \hfill & \le 2^{r-1}\left[{\left(max\left\{{∥\mathcal{T}x∥}_{\mathcal{A}}^2,{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right\}\right)}^r+{\left|{〈\mathcal{T}x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^r\right]\hfill \\ \hfill & =2^{r-1}\left[{\left(max\left\{{∥\mathcal{T}x∥}_{\mathcal{A}}^2,{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right\}\right)}^r+{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^r\right].\hfill \end{array}$$

(15)

Taking the supremum over ${∥x∥}_{\mathcal{A}}=1$, we get

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^{2r}\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & =\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left({\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^r\hfill \\ \hfill & \le 2^{r-1}\underset{{∥x∥}_{\mathcal{A}}=1}{sup}\left[{\left(max\left\{{∥\mathcal{T}x∥}_{\mathcal{A}}^2,{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right\}\right)}^r+{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^r\right]\hfill \\ \hfill & \le 2^{r-1}\left[{\left(\underset{{∥x∥}_{\mathcal{A}}=1}{sup}max\left\{{∥\mathcal{T}x∥}_{\mathcal{A}}^2,{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right\}\right)}^r\right.\hfill \\ \hfill & \left.+\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^r\right]\hfill \\ \hfill & =2^{r-1}\left[{\left(max\left\{\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{∥\mathcal{T}x∥}_{\mathcal{A}}^2,\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right\}\right)}^r\right.\hfill \\ \hfill & \left.+\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^r\right]\hfill \\ \hfill & =2^{r-1}\left[{\left(max\left\{{∥\mathcal{T}∥}_{\mathcal{A}}^2,{∥\mathcal{S}∥}_{\mathcal{A}}^2\right\}\right)}^r+{\omega }_{\mathcal{A}}^r\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]\hfill \\ \hfill & =2^{r-1}\left[max\left\{{∥\mathcal{T}∥}_{\mathcal{A}}^{2r},{∥\mathcal{S}∥}_{\mathcal{A}}^{2r}\right\}+{\omega }_{\mathcal{A}}^r\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right],\hfill \end{array}$$

which establishes (12).

Additionally, observe that

$$\begin{array}{cc}\hfill & max\left\{{∥\mathcal{T}x∥}_{\mathcal{A}}^2,{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right\}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left({∥\mathcal{T}x∥}_{\mathcal{A}}^2+{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right)+{\displaystyle \frac{1}{2}}\left|{∥\mathcal{T}x∥}_{\mathcal{A}}^2-{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right|\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left({〈{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}+{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}x,x〉}_{\mathcal{A}}\right)+{\displaystyle \frac{1}{2}}\left|{〈{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}-{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}x,x〉}_{\mathcal{A}}\right|\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left({〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}+\left|{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}\right|\right).\hfill \end{array}$$

Using the convexity of the power function for exponents greater than 1, we have

$$\begin{array}{cc}\hfill & {\left(max\left\{{∥\mathcal{T}x∥}_{\mathcal{A}}^2,{∥\mathcal{S}x∥}_{\mathcal{A}}^2\right\}\right)}^r\hfill \\ \hfill & ={\left({\displaystyle \frac{{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}+\left|{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}\right|}{2}}\right)}^r\hfill \\ \hfill & \le {\displaystyle \frac{{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}^r+{\left|{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}\right|}^r}{2}}.\hfill \end{array}$$

From (15), we obtain

$$\begin{array}{cc}\hfill & {\left({\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^r\hfill \\ \hfill & \le 2^{r-1}\left[{\displaystyle \frac{{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}^r+{\left|{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}\right|}^r}{2}}\right.\hfill \\ \hfill & \left.+{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^r\right]\hfill \end{array}$$

(16)

for $x\in \mathbb{X}$ with ${∥x∥}_{\mathcal{A}}=1$.

Taking the supremum in (16) over ${∥x∥}_{\mathcal{A}}=1$, we get

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^{2r}\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & =\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left({\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^r\hfill \\ \hfill & \le 2^{r-1}\underset{{∥x∥}_{\mathcal{A}}=1}{sup}\left[{\displaystyle \frac{{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}^r+{\left|{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}\right|}^r}{2}}\right.\hfill \\ \hfill & \left.+{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^r\right]\hfill \\ \hfill & \le 2^{r-1}\left[{\displaystyle \frac{1}{2}}\left[\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}^r\right.\right.\hfill \\ \hfill & \left.\left.+\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left|{〈\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)x,x〉}_{\mathcal{A}}\right|}^r\right]+\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^r\right]\hfill \\ \hfill & =2^{r-1}\left[{\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^r+{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^r}{2}}+{\omega }_{\mathcal{A}}^r\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right],\hfill \end{array}$$

which establishes (13). □

Corollary 1.

For all $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $r\ge 1$, the following inequalities hold:

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le \left[{\displaystyle \frac{1}{2}}max\left\{{∥\mathcal{T}+\mathcal{S}∥}_{\mathcal{A}}^{2r},{∥\mathcal{T}-\mathcal{S}∥}_{\mathcal{A}}^{2r}\right\}\right.\hfill \\ \hfill & {\left.+{\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}^r\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right)\right]}^{\frac{1}{r}}\hfill \end{array}$$

(17)

and

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le \left[2^{r-2}\left({∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^r+{∥{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^r\right)\right.\hfill \\ \hfill & {\left.+{\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}^r\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right)\right]}^{{\displaystyle \frac{1}{r}}}.\hfill \end{array}$$

(18)

Proof. 

We first note that by the parallelogram identity for the $\mathcal{A}$-inner product, we have

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T}+\mathcal{S},\mathcal{T}-\mathcal{S}\right)\hfill \\ \hfill & =\underset{{∥x∥}_{\mathcal{A}}=1}{sup}\left({\left|{〈x,\left(\mathcal{T}+\mathcal{S}\right)x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\left(\mathcal{T}-\mathcal{S}\right)x〉}_{\mathcal{A}}\right|}^2\right)\hfill \\ \hfill & =2\underset{{∥x∥}_{\mathcal{A}}=1}{sup}\left({\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)=2{\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \end{array}$$

(19)

for $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$.

Applying (12) to the pair $\left(\mathcal{T}+\mathcal{S},\mathcal{T}-\mathcal{S}\right)$, we obtain

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T}+\mathcal{S},\mathcal{T}-\mathcal{S}\right)\hfill \\ \hfill & \le 2^{1-{\displaystyle \frac{1}{r}}}{\left[max\left\{{∥\mathcal{T}+\mathcal{S}∥}_{\mathcal{A}}^{2r},{∥\mathcal{T}-\mathcal{S}∥}_{\mathcal{A}}^{2r}\right\}+{\omega }_{\mathcal{A}}^r\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right)\right]}^{\frac{1}{r}}.\hfill \end{array}$$

Using (19), this implies (17).

Next, we observe that

$$\begin{array}{cc}\hfill & {\left(\mathcal{T}+\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}+\mathcal{S}\right)+{\left(\mathcal{T}-\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}-\mathcal{S}\right)\hfill \\ \hfill & =\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)+\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-\mathcal{S}\right)\hfill \\ \hfill & ={\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\hfill \\ \hfill & =2\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & {\left(\mathcal{T}+\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}+\mathcal{S}\right)-{\left(\mathcal{T}-\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}-\mathcal{S}\right)\hfill \\ \hfill & =\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)-\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-\mathcal{S}\right)\hfill \\ \hfill & ={\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}-\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)\hfill \\ \hfill & =2\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right).\hfill \end{array}$$

Applying (13), we obtain

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T}+\mathcal{S},\mathcal{T}-\mathcal{S}\right)\hfill \\ \hfill & \le 2^{1-{\displaystyle \frac{1}{r}}}\left[{\displaystyle \frac{{∥2\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)∥}_{\mathcal{A}}^r+{∥2\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}\right)∥}_{\mathcal{A}}^r}{2}}\right.\hfill \\ \hfill & {\left.+{\omega }_{\mathcal{A}}^r\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right)\right]}^{\frac{1}{r}}\hfill \\ \hfill & =2^{1-\frac{1}{r}}\left[2^{r-1}\left({∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^r+{∥{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^r\right)\right.\hfill \\ \hfill & {\left.+{\omega }_{\mathcal{A}}^r\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right)\right]}^{\frac{1}{r}}.\hfill \end{array}$$

Using (19), this implies (18). □

Remark 1.

Setting $r=1$ in Theorem 1, we obtain the inequalities

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\le max\left\{{∥\mathcal{T}∥}_{\mathcal{A}}^2,{∥\mathcal{S}∥}_{\mathcal{A}}^2\right\}+{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right),$$

(20)

which correspond to inequality (3) from the Introduction, along with

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le {\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}+{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}}{2}}\hfill \\ \hfill & +{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right).\hfill \end{array}$$

For $r=2$, we derive

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\le \sqrt{2}{\left[max\left\{{∥\mathcal{T}∥}_{\mathcal{A}}^4,{∥\mathcal{S}∥}_{\mathcal{A}}^4\right\}+{\omega }_{\mathcal{A}}^2\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{2}}}$$

and

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \le \sqrt{2}{\left[{\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^2+{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^2}{2}}+{\omega }_{\mathcal{A}}^2\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Similarly, setting $r=1$ in Corollary 1, we obtain

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}max\left\{{∥\mathcal{T}+\mathcal{S}∥}_{\mathcal{A}}^2,{∥\mathcal{T}-\mathcal{S}∥}_{\mathcal{A}}^2\right\}+{\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le {\displaystyle \frac{1}{2}}\left({∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}+{∥{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right).\hfill \end{array}$$

For $r=2$, we get

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le \left[{\displaystyle \frac{1}{2}}max\left\{{∥\mathcal{T}+\mathcal{S}∥}_{\mathcal{A}}^4,{∥\mathcal{T}-\mathcal{S}∥}_{\mathcal{A}}^4\right\}\right.\hfill \\ \hfill & {\left.+{\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}^2\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right)\right]}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le \left[{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^2+{∥{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^2\right.\hfill \\ \hfill & {\left.+{\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}^2\left(\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-{\mathcal{S}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}+\mathcal{S}\right)\right)\right]}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

We now require the following vector inequality for the $\mathcal{A}$-inner product and $\mathcal{A}$-seminorm.

Lemma 2.

For all $u,v,w\in \mathbb{X}$, we have

$$\begin{array}{cc}\hfill {\left|{〈u,v〉}_{\mathcal{A}}\right|}^2+{\left|{〈u,w〉}_{\mathcal{A}}\right|}^2& \le \sqrt{2}{∥u∥}_{\mathcal{A}}max\left\{\left|{〈u,v〉}_{\mathcal{A}}\right|,\left|{〈u,w〉}_{\mathcal{A}}\right|\right\}\hfill \\ \hfill & \times {\left\{{\displaystyle \frac{{∥v∥}_{\mathcal{A}}^2+{∥w∥}_{\mathcal{A}}^2}{2}}+\left|{〈v,w〉}_{\mathcal{A}}\right|\right\}}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(21)

The proof follows by applying (10) with $x={\mathcal{A}}^{\frac{1}{2}}u$, $y={\mathcal{A}}^{\frac{1}{2}}v$, and $z={\mathcal{A}}^{\frac{1}{2}}w$.

Furthermore, we have the following theorem.

Theorem 2.

For all $\mathcal{T},\mathcal{S}\in \mathbf{L}\left(\mathbb{X}\right)$ and $r\ge 2$, the following inequalities hold:

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le 2^{1-{\displaystyle \frac{1}{r}}}max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}\right),{\omega }_{\mathcal{A}}\left(\mathcal{S}\right)\right\}\hfill \\ \hfill & \times {\left\{{∥{\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}}{2}}∥}_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}+{\omega }_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right\}}^{{\displaystyle \frac{1}{r}}}.\hfill \end{array}$$

(22)

Additionally, we have

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \le max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}+\mathcal{S}\right),{\omega }_{\mathcal{A}}\left(\mathcal{T}-\mathcal{S}\right)\right\}\hfill \\ \hfill & \times {\left\{{\displaystyle \frac{1}{2}}{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}+{\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}\left({\left(\mathcal{T}-\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}+\mathcal{S}\right)\right)\right\}}^{{\displaystyle \frac{1}{r}}}.\hfill \end{array}$$

(23)

Proof. 

By setting $v=\mathcal{T}x$, $w=\mathcal{S}x$ with $x\in \mathbb{X}$ and ${∥x∥}_{\mathcal{A}}=1$ in (21), we obtain

$$\begin{array}{cc}\hfill & {\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\hfill \\ \hfill & \le \sqrt{2}max\left\{\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|,\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|\right\}{\left\{{\displaystyle \frac{{∥\mathcal{T}x∥}_{\mathcal{A}}^2+{∥\mathcal{S}x∥}_{\mathcal{A}}^2}{2}}+\left|{〈\mathcal{T}x,\mathcal{S}x〉}_{\mathcal{A}}\right|\right\}}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & =\sqrt{2}max\left\{\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|,\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|\right\}{\left\{{〈\left({\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}}{2}}\right)x,x〉}_{\mathcal{A}}+\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|\right\}}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Raising both sides to the power $r\ge 2$ and applying the convexity of the power function, we get

$$\begin{array}{cc}\hfill & {\left({\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^r\hfill \\ \hfill & \le 2^{{\displaystyle \frac{r}{2}}}max\left\{{\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^r,{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^r\right\}{\left\{{〈\left({\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}}{2}}\right)x,x〉}_{\mathcal{A}}+\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|\right\}}^{{\displaystyle \frac{r}{2}}}\hfill \\ \hfill & =2^{{\displaystyle \frac{r}{2}}}max\left\{{\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^r,{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^r\right\}{2}^{{\displaystyle \frac{r}{2}}}{\left\{{\displaystyle \frac{{〈\left(\frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}}{2}\right)x,x〉}_{\mathcal{A}}+\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}{2}}\right\}}^{{\displaystyle \frac{r}{2}}}\hfill \\ \hfill & \le 2^{r}max\left\{{\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^r,{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^r\right\}{\displaystyle \frac{{〈\left(\frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}}{2}\right)x,x〉}_{\mathcal{A}}^{\frac{r}{2}}+{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^{\frac{r}{2}}}{2}}\hfill \\ \hfill & =2^{r-1}max\left\{{\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^r,{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^r\right\}\left[{〈\left({\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}}{2}}\right)x,x〉}_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}+{\left|{〈{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^{{\displaystyle \frac{r}{2}}}\right]\hfill \end{array}$$

(24)

for all $x\in \mathbb{X}$ with ${∥x∥}_{\mathcal{A}}=1$.

Taking the supremum in (24) over ${∥x∥}_{\mathcal{A}}=1$, we obtain (22).

Now, applying (22) to the pair $\left(\mathcal{T}+\mathcal{S},\mathcal{T}-\mathcal{S}\right)$, we have

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T}+\mathcal{S},\mathcal{T}-\mathcal{S}\right)\hfill \\ \hfill & \le 2^{1-{\displaystyle \frac{1}{r}}}max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}+\mathcal{S}\right),{\omega }_{\mathcal{A}}\left(\mathcal{T}-\mathcal{S}\right)\right\}\hfill \\ \hfill & \times \left[{∥{\displaystyle \frac{{\left(\mathcal{T}+\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}+\mathcal{S}\right)+{\left(\mathcal{T}-\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}-\mathcal{S}\right)}{2}}∥}_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}\right.\hfill \\ \hfill & {\left.+{\omega }_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}\left({\left(\mathcal{T}-\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}+\mathcal{S}\right)\right)\right]}^{\frac{1}{r}},\hfill \end{array}$$

which by the equality in (19) yields (23). □

Remark 2.

If we take $r=2$ in (22), then we get

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le \sqrt{2}max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}\right),{\omega }_{\mathcal{A}}\left(\mathcal{S}\right)\right\}{\left\{{∥{\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}}{2}}∥}_{\mathcal{A}}+{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{*}\mathcal{T}\right)\right\}}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

This is the result obtained in (4) from the Introduction.

For $r=4$, we obtain

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le 2\sqrt{2}max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}\right),{\omega }_{\mathcal{A}}\left(\mathcal{S}\right)\right\}{\left\{{∥{\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}}{2}}∥}_{\mathcal{A}}^2+{\omega }^2\left({\mathcal{S}}^{*}\mathcal{T}\right)\right\}}^{{\displaystyle \frac{1}{4}}}.\hfill \end{array}$$

If we take $r=2$ in (23), then we get

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}+\mathcal{S}\right),{\omega }_{\mathcal{A}}\left(\mathcal{T}-\mathcal{S}\right)\right\}\hfill \\ \hfill & \times {\left\{{\displaystyle \frac{1}{2}}{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}+{\displaystyle \frac{1}{2}}\omega \left({\left(\mathcal{T}-\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}+\mathcal{S}\right)\right)\right\}}^{{\displaystyle \frac{1}{2}}},\hfill \end{array}$$

while for $r=4$ we obtain

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \le max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}+\mathcal{S}\right),{\omega }_{\mathcal{A}}\left(\mathcal{T}-\mathcal{S}\right)\right\}\hfill \\ \hfill & \times {\left\{{\displaystyle \frac{1}{2}}{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^2+{\displaystyle \frac{1}{2}}{\omega }^2\left({\left(\mathcal{T}-\mathcal{S}\right)}^{{♯}_{\mathcal{A}}}\left(\mathcal{T}+\mathcal{S}\right)\right)\right\}}^{{\displaystyle \frac{1}{4}}}.\hfill \end{array}$$

Theorem 3.

For all $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \mathbb{C}$ with $\left|\alpha +\beta \right|=1$ and $\left|\gamma +\delta \right|=1$, we have

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \le {\omega }_{e,\mathcal{A}}\left(\alpha \mathcal{T},\gamma \mathcal{S}\right)+{\omega }_{e,\mathcal{A}}\left(\beta \mathcal{T},\delta \mathcal{S}\right)\hfill \\ \hfill & \le {∥{\left|\alpha \right|}^2{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\left|\gamma \right|}^2{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}}+{∥{\left|\beta \right|}^2{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\left|\delta \right|}^2{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

(25)

and

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \le {\omega }_{e,\mathcal{A}}\left(\alpha \mathcal{T},\gamma \mathcal{S}\right)+{\omega }_{e,\mathcal{A}}\left(\beta \mathcal{T},\delta \mathcal{S}\right)\hfill \\ \hfill & \le {\left[max\left\{{\left|\alpha \right|}^2{∥\mathcal{T}∥}_{\mathcal{A}}^2,{\left|\gamma \right|}^2{∥\mathcal{S}∥}_{\mathcal{A}}^2\right\}+\left|\alpha \right|\left|\gamma \right|{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\left[max\left\{{\left|\beta \right|}^2{∥\mathcal{T}∥}_{\mathcal{A}}^2,{\left|\delta \right|}^2{∥\mathcal{S}∥}_{\mathcal{A}}^2\right\}+\left|\beta \right|\left|\delta \right|{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(26)

Proof. 

We use the elementary inequality of Minkowski type:

$$\sqrt{{\left|a+b\right|}^2+{\left|c+d\right|}^2}\le \sqrt{{\left|a\right|}^2+{\left|c\right|}^2}+\sqrt{{\left|b\right|}^2+{\left|d\right|}^2}$$

which holds for the complex numbers $a,b,c,d.$

Becaude $\left|\overline{\alpha }+\overline{\beta }\right|=1$ and $\left|\overline{\gamma }+\overline{\delta }\right|=1$, we have

$$\begin{array}{cc}\hfill & {\left({\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & ={\left({\left|\overline{\alpha }{〈x,\mathcal{T}x〉}_{\mathcal{A}}+\overline{\beta }{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|\overline{\gamma }{〈x,\mathcal{S}x〉}_{\mathcal{A}}+\overline{\delta }{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & ={\left({\left|{〈x,\alpha \mathcal{T}x〉}_{\mathcal{A}}+{〈x,\beta \mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\gamma \mathcal{S}x〉}_{\mathcal{A}}+〈x,\delta \mathcal{S}x〉\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le {\left({\left|{〈x,\alpha \mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\gamma \mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}+{\left({\left|{〈x,\beta \mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\delta \mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

for all $x\in \mathbb{X}$, ${∥x∥}_{\mathcal{A}}=1$.

If we take the supremum over $x\in \mathbb{X}$, ${∥x∥}_{\mathcal{A}}=1,$ then we get

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}\left(\mathcal{T},\mathcal{S}\right)& =\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left({\left|{〈x,\mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le \underset{{∥x∥}_{\mathcal{A}}=1}{sup}\left[{\left({\left|{〈x,\alpha \mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\gamma \mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\right.\hfill \\ \hfill & \left.+{\left({\left|{〈x,\beta \mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\delta \mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\right]\hfill \\ \hfill & \le \underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left({\left|{〈x,\alpha \mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\gamma \mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +\underset{{∥x∥}_{\mathcal{A}}=1}{sup}{\left({\left|{〈x,\beta \mathcal{T}x〉}_{\mathcal{A}}\right|}^2+{\left|{〈x,\delta \mathcal{S}x〉}_{\mathcal{A}}\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & ={\omega }_{e,\mathcal{A}}\left(\alpha \mathcal{T},\gamma \mathcal{S}\right)+{\omega }_{e,\mathcal{A}}\left(\beta \mathcal{T},\delta \mathcal{S}\right),\hfill \end{array}$$

which proves the first part of (25).

We know that the following inequality is true for two operators $\mathcal{X}$, $\mathcal{Y}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$; see [36], Theorem 2.27:

$${\omega }_{e,\mathcal{A}}\left(\mathcal{X},\mathcal{Y}\right)\le {∥{\mathcal{X}}^{{♯}_{\mathcal{A}}}\mathcal{X}+{\mathcal{Y}}^{{♯}_{\mathcal{A}}}\mathcal{Y}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}}.$$

Using this inequality, we have

$${\omega }_{e,\mathcal{A}}\left(\alpha \mathcal{T},\gamma \mathcal{S}\right)\le {∥{\left|\alpha \right|}^2{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\left|\gamma \right|}^2{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}}$$

and

$${\omega }_{e,\mathcal{A}}\left(\beta \mathcal{T},\delta \mathcal{S}\right)\le {∥{\left|\beta \right|}^2{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\left|\delta \right|}^2{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}},$$

which proves (25).

By (20), we also have

$${\omega }_{e,\mathcal{A}}\left(\alpha \mathcal{T},\gamma \mathcal{S}\right)\le {\left[max\left\{{\left|\alpha \right|}^2{∥\mathcal{T}∥}_{\mathcal{A}}^2,{\left|\gamma \right|}^2{∥\mathcal{S}∥}_{\mathcal{A}}^2\right\}+\left|\alpha \right|\left|\gamma \right|{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{2}}}$$

and

$${\omega }_{e,\mathcal{A}}\left(\beta \mathcal{T},\delta \mathcal{S}\right)\le {\left[max\left\{{\left|\beta \right|}^2{∥\mathcal{T}∥}_{\mathcal{A}}^2,{\left|\delta \right|}^2{∥\mathcal{S}∥}_{\mathcal{A}}^2\right\}+\left|\beta \right|\left|\delta \right|{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{2}}},$$

which proves (26). □

Remark 3.

If we take $\alpha ,\beta \in \mathbb{C}$ with $\left|\alpha +\beta \right|=1$ and choose $\gamma =$β, $\delta =\alpha$ in Theorem 3, then we get

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}\left(\mathcal{T},\mathcal{S}\right)& \le {\omega }_{e,\mathcal{A}}\left(\alpha \mathcal{T},\beta \mathcal{S}\right)+{\omega }_{e,\mathcal{A}}\left(\beta \mathcal{T},\alpha \mathcal{S}\right)\hfill \\ \hfill & \le {∥{\left|\alpha \right|}^2{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\left|\beta \right|}^2{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}}+{∥{\left|\beta \right|}^2{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{\left|\alpha \right|}^2{\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{S}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}\left(\mathcal{T},\mathcal{S}\right)& \le {\omega }_{e,\mathcal{A}}\left(\alpha \mathcal{T},\beta \mathcal{S}\right)+{\omega }_{e,\mathcal{A}}\left(\beta \mathcal{T},\alpha \mathcal{S}\right)\hfill \\ \hfill & \le {\left[max\left\{{\left|\alpha \right|}^2{∥\mathcal{T}∥}_{\mathcal{A}}^2,{\left|\beta \right|}^2{∥\mathcal{S}∥}_{\mathcal{A}}^2\right\}+\left|\alpha \right|\left|\beta \right|{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\left[max\left\{{\left|\beta \right|}^2{∥\mathcal{T}∥}_{\mathcal{A}}^2,{\left|\alpha \right|}^2{∥\mathcal{S}∥}_{\mathcal{A}}^2\right\}+\left|\alpha \right|\left|\beta \right|{\omega }_{\mathcal{A}}\left({\mathcal{S}}^{{♯}_{\mathcal{A}}}\mathcal{T}\right)\right]}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

3. Lower Bounds

We start with the following main result.

Theorem 4.

For $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta ,\lambda ,\mu \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2={\left|\lambda \right|}^2+{\left|\mu \right|}^2=1,$ we have

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\lambda }^2\right){\mathcal{T}}^2+\left({\eta }^2+{\mu }^2\right){\mathcal{S}}^2+\left(\zeta \eta +\lambda \mu \right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\lambda }^2\right){\mathcal{T}}^2+\left({\eta }^2+{\mu }^2\right){\mathcal{S}}^2-\left(\zeta \eta +\lambda \mu \right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left(\zeta +\lambda \right)\mathcal{T}+\left(\eta +\mu \right)\mathcal{S}\right],{\omega }_{\mathcal{A}}\left[\left(\zeta -\lambda \right)\mathcal{T}+\left(\eta -\mu \right)\mathcal{S}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\right|\hfill \end{array}$$

(27)

and

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left[{\left(\zeta +\eta \right)}^2+{\left(\lambda +\mu \right)}^2\right]{\mathcal{T}}^2+\left[{\left(\zeta -\eta \right)}^2+{\left(\lambda -\mu \right)}^2\right]{\mathcal{S}}^2\right.\right.\hfill \\ \hfill & +\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right),\hfill \\ \hfill & {\omega }_{\mathcal{A}}\left[\left[{\left(\zeta -\eta \right)}^2+{\left(\lambda -\mu \right)}^2\right]{\mathcal{T}}^2+\left[{\left(\zeta +\eta \right)}^2+{\left(\lambda +\mu \right)}^2\right]{\mathcal{S}}^2\right.\hfill \\ \hfill & \left.+\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left(\zeta +\lambda +\eta +\mu \right)\mathcal{T}+\left(\zeta +\lambda -\eta -\mu \right)\mathcal{S}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left(\zeta +\eta -\lambda -\mu \right)\mathcal{T}+\left(\zeta -\eta -\lambda +\mu \right)\mathcal{S}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\left(\zeta +\eta \right)\mathcal{T}+\left(\zeta -\eta \right)\mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\left(\lambda +\mu \right)\mathcal{T}+\left(\lambda -\mu \right)\mathcal{S}\right)\right|.\hfill \end{array}$$

(28)

Proof. 

By using the Cauchy–Schwarz inequality for ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1,$ we have

$$\begin{array}{cc}\hfill & {\left|{〈\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^2+{\left|{〈\mathcal{S}x,x〉}_{\mathcal{A}}\right|}^2\hfill \\ \hfill & =\left({\left|\zeta \right|}^2+{\left|\eta \right|}^2\right)\left({\left|{〈\mathcal{T}x,x〉}_{\mathcal{A}}\right|}^2+{\left|{〈\mathcal{S}x,x〉}_{\mathcal{A}}\right|}^2\right)\hfill \\ \hfill & \ge {\left(\left|\zeta \right|\left|{〈\mathcal{T}x,x〉}_{\mathcal{A}}\right|+\left|\eta \right|\left|{〈\mathcal{S}x,x〉}_{\mathcal{A}}\right|\right)}^2\hfill \\ \hfill & ={\left(\left|{〈\zeta \mathcal{T}x,x〉}_{\mathcal{A}}\right|+\left|{〈\eta \mathcal{S}x,x〉}_{\mathcal{A}}\right|\right)}^2\hfill \\ \hfill & \ge {\left|{〈\zeta \mathcal{T}x,x〉}_{\mathcal{A}}+{〈\eta \mathcal{S}x,x〉}_{\mathcal{A}}\right|}^2={\left|{〈\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)x,x〉}_{\mathcal{A}}\right|}^2\hfill \end{array}$$

for all $x\in \mathbb{X}.$

If we take the supremum over $x\in \mathbb{X}$, ${∥x∥}_{\mathcal{A}}=1,$ then we get

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\ge {\omega }_{\mathcal{A}}^2\left(\zeta \mathcal{T}+\eta \mathcal{S}\right).$$

If ${\left|\lambda \right|}^2+{\left|\mu \right|}^2=1,$ then we also have

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\ge {\omega }_{\mathcal{A}}^2\left(\lambda \mathcal{T}+\mu \mathcal{S}\right).$$

Therefore,

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \ge max\left\{{\omega }_{\mathcal{A}}^2\left(\zeta \mathcal{T}+\eta \mathcal{S}\right),{\omega }_{\mathcal{A}}^2\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\right\}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[{\omega }_{\mathcal{A}}^2\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)+{\omega }_{\mathcal{A}}^2\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left|{\omega }_{\mathcal{A}}^2\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)-{\omega }_{\mathcal{A}}^2\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\right|\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[{\omega }_{\mathcal{A}}^2\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)+{\omega }_{\mathcal{A}}^2\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left[{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)+{\omega }_{\mathcal{A}}\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\right]\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\right|.\hfill \end{array}$$

(29)

Now, observing that

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}^2\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)+{\omega }_{\mathcal{A}}^2\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\hfill \\ \hfill & \ge {\omega }_{\mathcal{A}}\left[{\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)}^2\right]+{\omega }_{\mathcal{A}}\left[{\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)}^2\right]\hfill \\ \hfill & \ge max\left\{{\omega }_{\mathcal{A}}\left[{\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)}^2+{\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)}^2\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[{\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)}^2-{\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)}^2\right]\right\}\hfill \\ \hfill & =max\left\{{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\lambda }^2\right){\mathcal{T}}^2+\left(\zeta \eta +\lambda \mu \right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)+\left({\eta }^2+{\mu }^2\right){\mathcal{S}}^2\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\lambda }^2\right){\mathcal{T}}^2-\left(\zeta \eta +\lambda \mu \right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)+\left({\eta }^2+{\mu }^2\right){\mathcal{S}}^2\right]\right\}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)+{\omega }_{\mathcal{A}}\left(\lambda \mathcal{T}+\mu \mathcal{S}\right)\hfill \\ \hfill & \ge max\left\{{\omega }_{\mathcal{A}}\left[\left(\zeta +\lambda \right)\mathcal{T}+\left(\eta +\mu \right)\mathcal{S}\right],{\omega }_{\mathcal{A}}\left[\left(\zeta -\lambda \right)\mathcal{T}+\left(\eta -\mu \right)\mathcal{S}\right]\right\},\hfill \end{array}$$

by (29) we derive (27).

Now, if we replace $\mathcal{T}$ with $\mathcal{S}+\mathcal{D}$ and $\mathcal{S}$ with $\mathcal{S}-\mathcal{D},$ we get

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{S}+\mathcal{D},\mathcal{S}-\mathcal{D}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\lambda }^2\right){\left(\mathcal{S}+\mathcal{D}\right)}^2+\left({\eta }^2+{\mu }^2\right){\left(\mathcal{S}-\mathcal{D}\right)}^2\right.\right.\hfill \\ \hfill & +\left(\zeta \eta +\lambda \mu \right)\left(\left(\mathcal{S}+\mathcal{D}\right)\left(\mathcal{S}-\mathcal{D}\right)+\left(\mathcal{S}-\mathcal{D}\right)\left(\mathcal{S}+\mathcal{D}\right)\right),\hfill \\ \hfill & {\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\lambda }^2\right){\left(\mathcal{S}+\mathcal{D}\right)}^2+\left({\eta }^2+{\mu }^2\right){\left(\mathcal{S}-\mathcal{D}\right)}^2\right.\hfill \\ \hfill & \left.\left.-\left(\zeta \eta +\lambda \mu \right)\left(\left(\mathcal{S}+\mathcal{D}\right)\left(\mathcal{S}-\mathcal{D}\right)+\left(\mathcal{S}-\mathcal{D}\right)\left(\mathcal{S}+\mathcal{D}\right)\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left(\zeta +\lambda \right)\left(\mathcal{S}+\mathcal{D}\right)+\left(\eta +\mu \right)\left(\mathcal{S}-\mathcal{D}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left(\zeta -\lambda \right)\left(\mathcal{S}+\mathcal{D}\right)+\left(\eta -\mu \right)\left(\mathcal{S}-\mathcal{D}\right)\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \left(\mathcal{S}+\mathcal{D}\right)+\eta \left(\mathcal{S}-\mathcal{D}\right)\right)-\omega \left(\lambda \left(\mathcal{S}+\mathcal{D}\right)+\mu \left(\mathcal{S}-\mathcal{D}\right)\right)\right|.\hfill \end{array}$$

(30)

Observe that

$$\begin{array}{cc}\hfill & \left({\zeta }^2+{\lambda }^2\right){\left(\mathcal{S}+\mathcal{D}\right)}^2+\left({\eta }^2+{\mu }^2\right){\left(\mathcal{S}-\mathcal{D}\right)}^2\hfill \\ \hfill & =\left({\zeta }^2+{\lambda }^2\right)\left({\mathcal{S}}^2+{\mathcal{D}}^2+\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right)+\left({\eta }^2+{\mu }^2\right)\left({\mathcal{S}}^2+{\mathcal{D}}^2-\mathcal{S}\mathcal{D}-\mathcal{D}\mathcal{S}\right)\hfill \\ \hfill & =\left({\zeta }^2+{\lambda }^2+{\eta }^2+{\mu }^2\right)\left({\mathcal{S}}^2+{\mathcal{D}}^2\right)+\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \left(\mathcal{S}+\mathcal{D}\right)\left(\mathcal{S}-\mathcal{D}\right)+\left(\mathcal{S}-\mathcal{D}\right)\left(\mathcal{S}+\mathcal{D}\right)\hfill \\ \hfill & ={\mathcal{S}}^2-{\mathcal{D}}^2+\mathcal{D}\mathcal{S}-\mathcal{S}\mathcal{D}+{\mathcal{S}}^2-{\mathcal{D}}^2=\mathcal{D}\mathcal{S}+\mathcal{S}\mathcal{D}\hfill \\ \hfill & =2\left({\mathcal{S}}^2-{\mathcal{D}}^2\right);\hfill \end{array}$$

then,

$$\begin{array}{cc}\hfill & \left({\zeta }^2+{\lambda }^2\right){\left(\mathcal{S}+\mathcal{D}\right)}^2+\left({\eta }^2+{\mu }^2\right){\left(\mathcal{S}-\mathcal{D}\right)}^2\hfill \\ \hfill & +\left(\zeta \eta +\lambda \mu \right)\left(\left(\mathcal{S}+\mathcal{D}\right)\left(\mathcal{S}-\mathcal{D}\right)+\left(\mathcal{S}-\mathcal{D}\right)\left(\mathcal{S}+\mathcal{D}\right)\right)\hfill \\ \hfill & =\left({\zeta }^2+{\lambda }^2+{\eta }^2+{\mu }^2\right)\left({\mathcal{S}}^2+{\mathcal{D}}^2\right)+\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right)\hfill \\ \hfill & +2\left(\zeta \eta +\lambda \mu \right)\left({\mathcal{S}}^2-{\mathcal{D}}^2\right)\hfill \\ \hfill & =\left({\zeta }^2+{\lambda }^2+{\eta }^2+{\mu }^2+2\left(\zeta \eta +\lambda \mu \right)\right){\mathcal{S}}^2\hfill \\ \hfill & +\left({\zeta }^2+{\lambda }^2+{\eta }^2+{\mu }^2-2\left(\zeta \eta +\lambda \mu \right)\right){\mathcal{D}}^2\hfill \\ \hfill & +\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right)\hfill \\ \hfill & =\left[{\left(\zeta +\eta \right)}^2+{\left(\lambda +\mu \right)}^2\right]{\mathcal{S}}^2+\left[{\left(\zeta -\eta \right)}^2+{\left(\lambda -\mu \right)}^2\right]{\mathcal{D}}^2\hfill \\ \hfill & +\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \left({\zeta }^2+{\lambda }^2\right){\left(\mathcal{S}+\mathcal{D}\right)}^2+\left({\eta }^2+{\mu }^2\right){\left(\mathcal{S}-\mathcal{D}\right)}^2\hfill \\ \hfill & -\left(\zeta \eta +\lambda \mu \right)\left(\left(\mathcal{S}+\mathcal{D}\right)\left(\mathcal{S}-\mathcal{D}\right)+\left(\mathcal{S}-\mathcal{D}\right)\left(\mathcal{S}+\mathcal{D}\right)\right)\hfill \\ \hfill & =\left({\zeta }^2+{\lambda }^2+{\eta }^2+{\mu }^2\right)\left({\mathcal{S}}^2+{\mathcal{D}}^2\right)+\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right)\hfill \\ \hfill & -2\left(\zeta \eta +\lambda \mu \right)\left({\mathcal{S}}^2-{\mathcal{D}}^2\right)\hfill \\ \hfill & =\left({\zeta }^2+{\lambda }^2+{\eta }^2+{\mu }^2-2\left(\zeta \eta +\lambda \mu \right)\right){\mathcal{S}}^2\hfill \\ \hfill & +\left({\zeta }^2+{\lambda }^2+{\eta }^2+{\mu }^2+2\left(\zeta \eta +\lambda \mu \right)\right){\mathcal{D}}^2\hfill \\ \hfill & +\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right)\hfill \\ \hfill & =\left[{\left(\zeta -\eta \right)}^2+{\left(\lambda -\mu \right)}^2\right]{\mathcal{S}}^2+\left[{\left(\zeta +\eta \right)}^2+{\left(\lambda +\mu \right)}^2\right]{\mathcal{D}}^2\hfill \\ \hfill & +\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right).\hfill \end{array}$$

Additionally,

$$\left(\zeta +\lambda \right)\left(\mathcal{S}+\mathcal{D}\right)+\left(\eta +\mu \right)\left(\mathcal{S}-\mathcal{D}\right)=\left(\zeta +\lambda +\eta +\mu \right)\mathcal{S}+\left(\zeta +\lambda -\eta -\mu \right)\mathcal{D}$$

and

$$\left(\zeta -\lambda \right)\left(\mathcal{S}+\mathcal{D}\right)+\left(\eta -\mu \right)\left(\mathcal{S}-\mathcal{D}\right)=\left(\zeta +\eta -\lambda -\mu \right)\mathcal{S}+\left(\zeta -\lambda -\eta +\mu \right)\mathcal{D}.$$

Moreover,

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{S}+\mathcal{D},\mathcal{S}-\mathcal{D}\right)=2{\omega }_{e,\mathcal{A}}^2\left(\mathcal{S},\mathcal{D}\right),$$

and by (30) we get

$$\begin{array}{cc}\hfill & 2{\omega }_{e,\mathcal{A}}^2\left(\mathcal{S},\mathcal{D}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left[{\left(\zeta +\eta \right)}^2+{\left(\lambda +\mu \right)}^2\right]{\mathcal{S}}^2+\left[{\left(\zeta -\eta \right)}^2+{\left(\lambda -\mu \right)}^2\right]{\mathcal{D}}^2\right.\right.\hfill \\ \hfill & +\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right),\hfill \\ \hfill & {\omega }_{\mathcal{A}}\left[\left[{\left(\zeta -\eta \right)}^2+{\left(\lambda -\mu \right)}^2\right]{\mathcal{S}}^2+\left[{\left(\zeta +\eta \right)}^2+{\left(\lambda +\mu \right)}^2\right]{\mathcal{D}}^2\right.\hfill \\ \hfill & \left.+\left({\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2\right)\left(\mathcal{S}\mathcal{D}+\mathcal{D}\mathcal{S}\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left(\zeta +\lambda +\eta +\mu \right)\mathcal{S}+\left(\zeta +\lambda -\eta -\mu \right)\mathcal{D}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left(\zeta +\eta -\lambda -\mu \right)\mathcal{S}+\left(\zeta -\lambda -\eta +\mu \right)\mathcal{D}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\left(\zeta +\eta \right)\mathcal{S}+\left(\zeta -\eta \right)\mathcal{D}\right)-\omega \left(\left(\lambda +\mu \right)\mathcal{S}+\left(\lambda -\mu \right)\mathcal{D}\right)\right|.\hfill \end{array}$$

By replacing $\mathcal{S}$ with $\mathcal{T}$ and $\mathcal{D}$ with $\mathcal{S},$ the inequality in (28) is obtained. □

Remark 4.

By setting $\zeta =0$, $\eta =1$, $\lambda =1$, and $\mu =0$ in (27), we obtain

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \ge {\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}\left({\mathcal{T}}^2+{\mathcal{S}}^2\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{S}+\mathcal{T}\right),{\omega }_{\mathcal{A}}\left(\mathcal{S}-\mathcal{T}\right)\right\}\left|{\omega }_{\mathcal{A}}\left(\mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\mathcal{T}\right)\right|,\hfill \end{array}$$

(31)

while from (28) we get

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \ge {\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}\left({\mathcal{T}}^2+{\mathcal{S}}^2\right)+{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left(\mathcal{T}\right),{\omega }_{\mathcal{A}}\left(\mathcal{S}\right)\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\mathcal{T}-\mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\mathcal{T}+\mathcal{S}\right)\right|.\hfill \end{array}$$

(32)

Both inequalities (31) and (32) refine the inequality

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\ge {\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}\left({\mathcal{T}}^2+{\mathcal{S}}^2\right)$$

established in (5) from the Introduction.

Remark 5.

By setting $\lambda =\zeta$ and $\mu =\eta$ in (27), we obtain

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\ge max\left\{{\omega }_{\mathcal{A}}\left[{\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)}^2\right],{\omega }_{\mathcal{A}}\left[{\left(\zeta \mathcal{T}-\eta \mathcal{S}\right)}^2\right]\right\}$$

for $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta =\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, this yields

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[{\left(\mathcal{T}+\mathcal{S}\right)}^2\right],{\omega }_{\mathcal{A}}\left[{\left(\mathcal{T}-\mathcal{S}\right)}^2\right]\right\}.$$

(33)

Setting $\lambda =\eta$ and $\mu =\zeta$ in (27), we get

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\eta }^2\right)\left({\mathcal{T}}^2+{\mathcal{S}}^2\right)+2\zeta \eta \left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\eta }^2\right)\left({\mathcal{T}}^2+{\mathcal{S}}^2\right)-2\zeta \eta \left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{\left|\zeta +\eta \right|{\omega }_{\mathcal{A}}\left(\mathcal{T}+\mathcal{S}\right),\left|\zeta -\eta \right|{\omega }_{\mathcal{A}}\left(\mathcal{T}-\mathcal{S}\right)\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\eta \mathcal{T}+\zeta \mathcal{S}\right)\right|\hfill \end{array}$$

(34)

for $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta =\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, this again yields (33).

Setting $\lambda =i\zeta$ and $\mu =-i\eta$ in (27), we obtain

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \ge \left|\zeta \eta \right|{\omega }_{\mathcal{A}}\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\zeta \left(1+i\right)\mathcal{T}+\eta \left(1-i\right)\mathcal{S}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\zeta \left(1-i\right)\mathcal{T}+\eta \left(1+i\right)\mathcal{S}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}-\eta \mathcal{S}\right)\right|\hfill \end{array}$$

(35)

for $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta =\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, we get

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \ge {\displaystyle \frac{1}{2}}{\omega }_{\mathcal{A}}\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left(1+i\right)\mathcal{T}+\left(1-i\right)\mathcal{S}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left(1-i\right)\mathcal{T}+\left(1+i\right)\mathcal{S}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\mathcal{T}+\mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\mathcal{T}-\mathcal{S}\right)\right|.\hfill \end{array}$$

(36)

Setting $\lambda =i\eta$ and $\mu =-i\zeta$ in (27), we obtain

$$\begin{array}{cc}\hfill & {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left({\eta }^2-{\zeta }^2\right)\left({\mathcal{S}}^2-{\mathcal{T}}^2\right)+2\zeta \eta \left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left({\eta }^2-{\zeta }^2\right)\left({\mathcal{S}}^2-{\mathcal{T}}^2\right)-2\zeta \eta \left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left(\zeta +i\eta \right)\mathcal{T}+\left(\eta -i\zeta \right)\mathcal{S}\right],{\omega }_{\mathcal{A}}\left[\left(\zeta -i\eta \right)\mathcal{T}+\left(\eta +i\zeta \right)\mathcal{S}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta \mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(\eta \mathcal{T}-\zeta \mathcal{S}\right)\right|\hfill \end{array}$$

for $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta ={\displaystyle \frac{\sqrt{2}}{2}}i$ and $\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, we get

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[{\left(\mathcal{S}+i\mathcal{T}\right)}^2\right],{\omega }_{\mathcal{A}}\left[{\left(\mathcal{S}-i\mathcal{T}\right)}^2\right]\right\}.$$

By choosing $\zeta =sin\theta$, $\eta =cos\theta$, $\theta \in \mathbb{R}$, we can also state the trigonometric inequalities

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)\ge max\left\{{\omega }_{\mathcal{A}}\left[{\left(sin\theta \mathcal{T}+cos\theta \mathcal{S}\right)}^2\right],{\omega }_{\mathcal{A}}\left[{\left(sin\theta \mathcal{T}-cos\theta \mathcal{S}\right)}^2\right]\right\},$$

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[{\mathcal{T}}^2+{\mathcal{S}}^2+sin\left(2\theta \right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[{\mathcal{T}}^2+{\mathcal{S}}^2-sin\left(2\theta \right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{\left|sin\theta +cos\theta \right|{\omega }_{\mathcal{A}}\left(\mathcal{T}+\mathcal{S}\right),\left|sin\theta -cos\theta \right|{\omega }_{\mathcal{A}}\left(\mathcal{T}-\mathcal{S}\right)\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(sin\theta \mathcal{T}+cos\theta \mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(cos\theta \mathcal{T}+sin\theta \mathcal{S}\right)\right|,\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \ge {\displaystyle \frac{1}{2}}\left|sin\left(2\theta \right)\right|{\omega }_{\mathcal{A}}\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[sin\theta \left(1+i\right)\mathcal{T}+cos\theta \left(1-i\right)\mathcal{S}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[sin\theta \left(1-i\right)\mathcal{T}+cos\theta \left(1+i\right)\mathcal{S}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(sin\theta \mathcal{T}+cos\theta \mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(sin\theta \mathcal{T}-cos\theta \mathcal{S}\right)\right|,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},\mathcal{S}\right)& \ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[cos\left(2\theta \right)\left({\mathcal{S}}^2-{\mathcal{T}}^2\right)+sin\left(2\theta \right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[cos\left(2\theta \right)\left({\mathcal{S}}^2-{\mathcal{T}}^2\right)-sin\left(2\theta \right)\left(\mathcal{T}\mathcal{S}+\mathcal{S}\mathcal{T}\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left(sin\theta +icos\theta \right)\mathcal{T}+\left(cos\theta -isin\theta \right)\mathcal{S}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left(sin\theta -icos\theta \right)\mathcal{T}+\left(cos\theta +isin\theta \right)\mathcal{S}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(sin\theta \mathcal{T}+cos\theta \mathcal{S}\right)-{\omega }_{\mathcal{A}}\left(cos\theta \mathcal{T}-sin\theta \mathcal{S}\right)\right|\hfill \end{array}$$

for $\mathcal{T},\mathcal{S}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\theta \in \mathbb{R}$.

4. Applications to a Single Operator

For $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$, setting $\mathcal{T}={\mathcal{T}}^{{♯}_{\mathcal{A}}}$ and $\mathcal{S}={\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^{{♯}_{\mathcal{A}}}$, we have

$${\omega }_{e,\mathcal{A}}^2\left(\mathcal{T},{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)=2{\omega }_{\mathcal{A}}^2\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)=2{\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right).$$

By applying Theorem 1 combined with (1) and the properties that ${\omega }_{\mathcal{A}}\left({\mathcal{X}}^{{♯}_{\mathcal{A}}}\right)={\omega }_{\mathcal{A}}\left(\mathcal{X}\right)$ and $\parallel {\mathcal{X}}^{{♯}_{\mathcal{A}}}{\parallel }_{\mathcal{A}}={\parallel \mathcal{X}\parallel }_{\mathcal{A}}$ for all $\mathcal{X}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$, we obtain for $r\ge 1$ that

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\le {\displaystyle \frac{1}{2^{\frac{1}{r}}}}{\left[{∥\mathcal{T}∥}_{\mathcal{A}}^{2r}+{\omega }_{\mathcal{A}}^r\left({\mathcal{T}}^2\right)\right]}^{{\displaystyle \frac{1}{r}}}$$

and

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2^{\frac{1}{r}}}}{\left[{\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^r+{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^r}{2}}+{\omega }_{\mathcal{A}}^r\left({\mathcal{T}}^2\right)\right]}^{{\displaystyle \frac{1}{r}}}.\hfill \end{array}$$

For $r=1$, we derive

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\le {\displaystyle \frac{1}{2}}\left[{∥\mathcal{T}∥}_{\mathcal{A}}^2+{\omega }_{\mathcal{A}}\left({\mathcal{T}}^2\right)\right],$$

(37)

which was established in [36], Corollary 2.16 and

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}+{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}}{2}}+{\omega }_{\mathcal{A}}\left({\mathcal{T}}^2\right)\right].\hfill \end{array}$$

For $r=2$, we obtain

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\le {\displaystyle \frac{\sqrt{2}}{2}}{\left[{∥\mathcal{T}∥}_{\mathcal{A}}^4+{\omega }_{\mathcal{A}}^2\left({\mathcal{T}}^2\right)\right]}^{\frac{1}{2}}$$

and

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{\sqrt{2}}{2}}{\left[{\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^2+{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}-\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^2}{2}}+{\omega }_{\mathcal{A}}^2\left({\mathcal{T}}^2\right)\right]}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Similarly, by setting $\mathcal{T}={\mathcal{T}}^{{♯}_{\mathcal{A}}}$ and $\mathcal{S}={\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^{{♯}_{\mathcal{A}}}$ in Theorem 2, we obtain for $r\ge 2$ that

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\le {\displaystyle \frac{1}{4^{\frac{1}{r}}}}{\left\{{∥{\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}∥}_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}+{\omega }_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}\left({\mathcal{T}}^2\right)\right\}}^{2/r}.$$

For $r=2$, we obtain

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\le {\displaystyle \frac{1}{2}}\left\{{∥{\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}∥}_{\mathcal{A}}+{\omega }_{\mathcal{A}}\left({\mathcal{T}}^2\right)\right\},$$

(38)

while for $r=4$ we get

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\le {\displaystyle \frac{\sqrt{2}}{2}}{\left\{{∥{\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}∥}_{\mathcal{A}}^2+{\omega }_{\mathcal{A}}^2\left({\mathcal{T}}^2\right)\right\}}^{\frac{1}{2}}.$$

Because

$${∥{\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}∥}_{\mathcal{A}}\le {\displaystyle \frac{1}{2}}\left({∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}∥}_{\mathcal{A}}+{∥\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}\right)={∥\mathcal{T}∥}_{\mathcal{A}}^2,$$

it follows that (38) refines (37).

Now, let $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and consider its $\mathcal{A}$-Cartesian decomposition $\mathcal{T}=\mathcal{B}+i\mathcal{C}$, where $\mathcal{B}$ and $\mathcal{C}$ are $\mathcal{A}$-selfadjoint operators defined by

$$\mathcal{B}={\mathfrak{R}}_{\mathcal{A}}\left(\mathcal{T}\right):={\displaystyle \frac{1}{2}}\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\mathrm{and}\mathcal{C}={\mathfrak{I}}_{\mathcal{A}}\left(\mathcal{T}\right):={\displaystyle \frac{1}{2i}}\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right).$$

It can be seen that

$${\omega }_{e,\mathcal{A}}^2\left({\mathcal{B}}^{{♯}_{\mathcal{A}}},{\mathcal{C}}^{{♯}_{\mathcal{A}}}\right)={\omega }_{\mathcal{A}}^2\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)={\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right),$$

(39)

$${\left({\mathcal{B}}^{{♯}_{\mathcal{A}}}\right)}^2+{\left({\mathcal{C}}^{{♯}_{\mathcal{A}}}\right)}^2={\left({\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}\right)}^{{♯}_{\mathcal{A}}},$$

(40)

and

$${\left({\mathcal{B}}^{{♯}_{\mathcal{A}}}\right)}^2-{\left({\mathcal{C}}^{{♯}_{\mathcal{A}}}\right)}^2={\left({\displaystyle \frac{{\mathcal{T}}^2+{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2}{2}}\right)}^{{♯}_{\mathcal{A}}}.$$

(41)

By applying Theorem 1 to the $\mathcal{A}$-Cartesian decomposition of ${\mathcal{T}}^{{♯}_{\mathcal{A}}}$, combined with (2), (39), and the properties that ${\omega }_{\mathcal{A}}\left({\mathcal{X}}^{{♯}_{\mathcal{A}}}\right)={\omega }_{\mathcal{A}}\left(\mathcal{X}\right)$ and $\parallel {\mathcal{X}}^{{♯}_{\mathcal{A}}}{\parallel }_{\mathcal{A}}={\parallel \mathcal{X}\parallel }_{\mathcal{A}}$ for all $\mathcal{X}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$, we obtain for $r\ge 1$ that

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\le {\displaystyle \frac{1}{2^{1+\frac{1}{r}}}}\left[max\left\{{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^{2r},{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^{2r}\right\}\right.\hfill \\ \hfill & {\left.+{\omega }_{\mathcal{A}}^r\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right)\right]}^{{\displaystyle \frac{1}{r}}}.\hfill \end{array}$$

(42)

Further, by taking into account (40) and (41), we get

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{1}{2^{\frac{1}{r}}}}\left[{\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^r+{∥{\mathcal{T}}^2+{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2∥}_{\mathcal{A}}^r}{2}}\right.\hfill \\ \hfill & {\left.+{\displaystyle \frac{1}{2^r}}{\omega }_{\mathcal{A}}^r\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right)\right]}^{{\displaystyle \frac{1}{r}}}.\hfill \end{array}$$

(43)

By setting $r=1$ in (42) and (43), we obtain results analogous to those in ([17], Equation (2.29)) and ([36], Corollary 2.18), i.e., we have

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{1}{4}}max\left\{{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^2,{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^2\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}{\omega }_{\mathcal{A}}\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}+{∥{\mathcal{T}}^2+{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2∥}_{\mathcal{A}}}{4}}+{\displaystyle \frac{1}{4}}{\omega }_{\mathcal{A}}\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right).\hfill \end{array}$$

For $r=2$ in (42) and (43), we obtain

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{\sqrt{2}}{4}}{\left[max\left\{{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^4,{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^4\right\}+{\omega }_{\mathcal{A}}^2\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right)\right]}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{\sqrt{2}}{2}}{\left[{\displaystyle \frac{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^2+{∥{\mathcal{T}}^2+{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2∥}_{\mathcal{A}}^2}{2}}+{\displaystyle \frac{1}{4}}{\omega }_{\mathcal{A}}^2\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right)\right]}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Similarly, applying Theorem 2 to the $\mathcal{A}$-Cartesian decomposition of $\mathcal{T}$, for $r\ge 2$ we obtain

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{1}{2^{1+\frac{1}{r}}}}max\left\{{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}},{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}\right\}\hfill \\ \hfill & \times {\left\{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}+{\omega }_{\mathcal{A}}^{{\displaystyle \frac{r}{2}}}\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right)\right\}}^{{\displaystyle \frac{1}{r}}}.\hfill \end{array}$$

For $r=2$, we obtain

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{\sqrt{2}}{4}}max\left\{{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}},{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}\right\}\hfill \\ \hfill & \times {\left\{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}+{\omega }_{\mathcal{A}}\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right)\right\}}^{{\displaystyle \frac{1}{2}}},\hfill \end{array}$$

while for $r=4$ we get

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \le {\displaystyle \frac{\sqrt[4]{8}}{4}}max\left\{{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}},{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}\right\}\hfill \\ \hfill & \times {\left\{{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^2+{\omega }_{\mathcal{A}}^2\left(\left(\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\left(\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right)\right\}}^{{\displaystyle \frac{1}{4}}}.\hfill \end{array}$$

For $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$, by proceeding as above and setting $\mathcal{T}={\mathcal{T}}^{{♯}_{\mathcal{A}}}$ and $\mathcal{S}={\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^{{♯}_{\mathcal{A}}}$ in Theorem 3, we obtain for $t\in [0,1]$ that

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}\left(\mathcal{T}\right)\hfill \\ \hfill & \le {∥{\displaystyle \frac{{(1-t)}^2{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+t^2\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}}+{∥{\displaystyle \frac{t^2{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+{(1-t)}^2\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}∥}_{\mathcal{A}}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

and

$${\omega }_{\mathcal{A}}\left(\mathcal{T}\right)\le \sqrt{2}{\left[{\left({\displaystyle \frac{1}{2}}+\left|t-{\displaystyle \frac{1}{2}}\right|\right)}^2{∥\mathcal{T}∥}_{\mathcal{A}}^2+(1-t)t{\omega }_{\mathcal{A}}\left({\mathcal{T}}^2\right)\right]}^{\frac{1}{2}}$$

(44)

for all $t\in [0,1]$.

Setting $t={\displaystyle \frac{1}{2}}$ in (44), we recover (37).

Now, by setting $\mathcal{T}=\mathcal{T}$ and $\mathcal{S}={\mathcal{T}}^{{♯}_{\mathcal{A}}}$ in Theorem 4, we obtain

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\lambda }^2\right){\mathcal{T}}^2+\left({\eta }^2+{\mu }^2\right){\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2+\left(\zeta \eta +\lambda \mu \right)\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\lambda }^2\right){\mathcal{T}}^2+\left({\eta }^2+{\mu }^2\right){\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2-\left(\zeta \eta +\lambda \mu \right)\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left(\zeta +\lambda \right)\mathcal{T}+\left(\eta +\mu \right){\mathcal{T}}^{{♯}_{\mathcal{A}}}\right],{\omega }_{\mathcal{A}}\left[\left(\zeta -\lambda \right)\mathcal{T}+\left(\eta -\mu \right){\mathcal{T}}^{{♯}_{\mathcal{A}}}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)-{\omega }_{\mathcal{A}}\left(\lambda \mathcal{T}+\mu {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right|\hfill \end{array}$$

and for $\zeta ,\eta ,\lambda ,\mu \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2={\left|\lambda \right|}^2+{\left|\mu \right|}^2=1$, we have

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\ge {\displaystyle \frac{1}{8}}max\{& {\omega }_{\mathcal{A}}\left[a{\mathcal{T}}^2+b{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2+c({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}})\right],\hfill \\ \hfill & {\omega }_{\mathcal{A}}\left[b{\mathcal{T}}^2+a{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2+c({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}})\right]\}\hfill \\ \hfill +{\displaystyle \frac{1}{8}}max\{& {\omega }_{\mathcal{A}}\left[d\mathcal{T}+e{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right],{\omega }_{\mathcal{A}}\left[f\mathcal{T}+g{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right]\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(h\mathcal{T}+i{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)-{\omega }_{\mathcal{A}}\left(j\mathcal{T}+k{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right|,\hfill \end{array}$$

where $a={(\zeta +\eta )}^2+{(\lambda +\mu )}^2$, $b={(\zeta -\eta )}^2+{(\lambda -\mu )}^2$, $c={\zeta }^2+{\lambda }^2-{\eta }^2-{\mu }^2$, $d=\zeta +\lambda +\eta +\mu$, $e=\zeta +\lambda -\eta -\mu$, $f=\zeta +\eta -\lambda -\mu$, $g=\zeta -\eta -\lambda +\mu$, $h=\zeta +\eta$, $i=\zeta -\eta$, $j=\lambda +\mu$, and $k=\lambda -\mu$.

Setting $(\mathcal{T},\mathcal{S})=(\mathcal{T},{\mathcal{T}}^{{♯}_{\mathcal{A}}})$ and $\lambda =\zeta$, $\mu =\eta$ in (27), we obtain

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[{\left(\zeta \mathcal{T}+\eta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2\right],{\omega }_{\mathcal{A}}\left[{\left(\zeta \mathcal{T}-\eta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2\right]\right\}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta =\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\ge {\displaystyle \frac{1}{4}}max\left\{{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^2,{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}^2\right\}.$$

Setting $(\mathcal{T},\mathcal{S})=(\mathcal{T},{\mathcal{T}}^{{♯}_{\mathcal{A}}})$ and $\lambda =\eta$, $\mu =\zeta$ in (27), we obtain

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\eta }^2\right)\left({\mathcal{T}}^2+{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2\right)+2\zeta \eta \left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\eta }^2\right)\left({\mathcal{T}}^2+{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2\right)-2\zeta \eta \left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{\left|\zeta +\eta \right|{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}},\left|\zeta -\eta \right|{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)-{\omega }_{\mathcal{A}}\left(\eta \mathcal{T}+\zeta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right|\hfill \end{array}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$.

For $\zeta =0$ and $\eta =1$, we get

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\ge {\displaystyle \frac{1}{4}}{\omega }_{\mathcal{A}}\left[{\mathcal{T}}^2+{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2\right]={\displaystyle \frac{1}{4}}{∥{\mathcal{T}}^2+{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2∥}_{\mathcal{A}}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$.

Further, from (35) with $(\mathcal{T},\mathcal{S})=(\mathcal{T},{\mathcal{T}}^{{♯}_{\mathcal{A}}})$, we obtain

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \ge {\displaystyle \frac{1}{2}}\left|\zeta \eta \right|{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\zeta \left(1+i\right)\mathcal{T}+\eta \left(1-i\right){\mathcal{T}}^{{♯}_{\mathcal{A}}}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\zeta \left(1-i\right)\mathcal{T}+\eta \left(1+i\right){\mathcal{T}}^{{♯}_{\mathcal{A}}}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)-{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}-\eta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right|\hfill \end{array}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta =\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, we get

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \ge {\displaystyle \frac{1}{4}}{∥{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}\hfill \\ \hfill & +{\displaystyle \frac{1}{8}}max\left\{{\omega }_{\mathcal{A}}\left[\left(1+i\right)\mathcal{T}+\left(1-i\right){\mathcal{T}}^{{♯}_{\mathcal{A}}}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left(1-i\right)\mathcal{T}+\left(1+i\right){\mathcal{T}}^{{♯}_{\mathcal{A}}}\right]\right\}\hfill \\ \hfill & \times \left|{∥\mathcal{T}+{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}-{∥\mathcal{T}-{\mathcal{T}}^{{♯}_{\mathcal{A}}}∥}_{\mathcal{A}}\right|\hfill \end{array}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$.

Moreover, setting $(\mathcal{T},\mathcal{S})=(\mathcal{T},{\mathcal{T}}^{{♯}_{\mathcal{A}}})$ in (36), we obtain

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left({\eta }^2-{\zeta }^2\right)\left({\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2-{\mathcal{T}}^2\right)+2\zeta \eta \left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left({\eta }^2-{\zeta }^2\right)\left({\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2-{\mathcal{T}}^2\right)-2\zeta \eta \left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left(\zeta +i\eta \right)\mathcal{T}+\left(\eta -i\zeta \right){\mathcal{T}}^{{♯}_{\mathcal{A}}}\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left(\zeta -i\eta \right)\mathcal{T}+\left(\eta +i\zeta \right){\mathcal{T}}^{{♯}_{\mathcal{A}}}\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta \mathcal{T}+\eta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)-{\omega }_{\mathcal{A}}\left(\eta \mathcal{T}-\zeta {\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)\right|\hfill \end{array}$$

(45)

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta ={\displaystyle \frac{\sqrt{2}}{2}}i$ and $\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, we get

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \ge {\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}+i\mathcal{T}\right)}^2\right],{\omega }_{\mathcal{A}}\left[{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}-i\mathcal{T}\right)}^2\right]\right\}.\hfill \end{array}$$

Additionally, setting $\zeta =0$ and $\eta =1$ in (45), we obtain

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\ge {\displaystyle \frac{1}{4}}{\omega }_{\mathcal{A}}\left[{\left({\mathcal{T}}^{{♯}_{\mathcal{A}}}\right)}^2-{\mathcal{T}}^2\right].$$

If we apply the inequality in (33) with $\mathcal{T}={\Re }_{\mathcal{A}}\left(\mathcal{T}\right)$ and $\mathcal{S}={\Im }_{\mathcal{A}}\left(\mathcal{T}\right)$, we get

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\hfill \\ \hfill & \ge max\left\{{\omega }_{\mathcal{A}}\left[{\left(\zeta {\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+\eta {\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right)}^2\right],{\omega }_{\mathcal{A}}\left[{\left(\zeta {\Re }_{\mathcal{A}}\left(\mathcal{T}\right)-\eta {\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right)}^2\right]\right\}\hfill \end{array}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta =\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$${\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\ge {\displaystyle \frac{1}{2}}max\left\{{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}^2,{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)-{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}^2\right\}.$$

From (34) with $\mathcal{T}={\Re }_{\mathcal{A}}\left(\mathcal{T}\right)$ and $\mathcal{S}={\Im }_{\mathcal{A}}\left(\mathcal{T}\right)$, we also have

$$\begin{array}{cc}\hfill & {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\eta }^2\right)\left({\Re }_{\mathcal{A}}^2\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}^2\left(\mathcal{T}\right)\right)+2\zeta \eta \left({\Re }_{\mathcal{A}}\left(\mathcal{T}\right){\Im }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right){\Re }_{\mathcal{A}}\left(\mathcal{T}\right)\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left({\zeta }^2+{\eta }^2\right)\left({\Re }_{\mathcal{A}}^2\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}^2\left(\mathcal{T}\right)\right)-2\zeta \eta \left({\Re }_{\mathcal{A}}\left(\mathcal{T}\right){\Im }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right){\Re }_{\mathcal{A}}\left(\mathcal{T}\right)\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{\left|\zeta +\eta \right|{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}},\left|\zeta -\eta \right|{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)-{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta {\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+\eta {\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right)-{\omega }_{\mathcal{A}}\left(\eta {\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+\zeta {\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right)\right|\hfill \end{array}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$.

Setting $\zeta =0$ and $\eta =1$ in this inequality, we obtain

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \ge {\displaystyle \frac{1}{2}}{∥{\Re }_{\mathcal{A}}^2\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}^2\left(\mathcal{T}\right)∥}_{\mathcal{A}}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}},{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)-{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}\right\}\hfill \\ \hfill & \times \left|{∥{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}-{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}\right|\hfill \end{array}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$.

Because ${\Re }_{\mathcal{A}}^2\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}^2\left(\mathcal{T}\right)={\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}$, we have

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \ge {\displaystyle \frac{1}{4}}{∥{\displaystyle \frac{{\mathcal{T}}^{{♯}_{\mathcal{A}}}\mathcal{T}+\mathcal{T}{\mathcal{T}}^{{♯}_{\mathcal{A}}}}{2}}∥}_{\mathcal{A}}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}},{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)-{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}\right\}\hfill \\ \hfill & \times \left|{∥{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}-{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}\right|.\hfill \end{array}$$

Further, setting $\mathcal{T}={\Re }_{\mathcal{A}}\left(\mathcal{T}\right)$ and $\mathcal{S}={\Im }_{\mathcal{A}}\left(\mathcal{T}\right)$ in (35), we obtain

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \ge \left|\zeta \eta \right|{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right){\Im }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right){\Re }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{{\omega }_{\mathcal{A}}\left[\zeta \left(1+i\right){\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+\eta \left(1-i\right){\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\zeta \left(1-i\right){\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+\eta \left(1+i\right){\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right]\right\}\hfill \\ \hfill & \times \left|{\omega }_{\mathcal{A}}\left(\zeta {\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+\eta {\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right)-{\omega }_{\mathcal{A}}\left(\zeta {\Re }_{\mathcal{A}}\left(\mathcal{T}\right)-\eta {\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right)\right|\hfill \end{array}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$ and $\zeta ,\eta \in \mathbb{C}$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$. For $\zeta =\eta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$$\begin{array}{cc}\hfill {\omega }_{\mathcal{A}}^2\left(\mathcal{T}\right)& \ge {\displaystyle \frac{1}{2}}{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right){\Im }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right){\Re }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{{\omega }_{\mathcal{A}}\left[\left(1+i\right){\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+\left(1-i\right){\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right],\right.\hfill \\ \hfill & \left.{\omega }_{\mathcal{A}}\left[\left(1-i\right){\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+\left(1+i\right){\Im }_{\mathcal{A}}\left(\mathcal{T}\right)\right]\right\}\hfill \\ \hfill & \times \left|{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)+{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}-{∥{\Re }_{\mathcal{A}}\left(\mathcal{T}\right)-{\Im }_{\mathcal{A}}\left(\mathcal{T}\right)∥}_{\mathcal{A}}\right|\hfill \end{array}$$

for $\mathcal{T}\in {\mathbf{L}}_{\mathcal{A}}\left(\mathbb{X}\right)$.

5. Conclusions

In this paper, we have established a series of new inequalities for the $\mathcal{A}$-joint numerical radius of operators in the generalized setting of semi-Hilbert spaces. Our primary contributions include the derivation of novel upper and lower bounds for ${\omega }_{e,\mathcal{A}}(\mathcal{T},\mathcal{S})$, which refine and extend several known results in operator theory. Key findings include power-dependent inequalities that offer tighter estimations by connecting the $\mathcal{A}$-joint numerical radius with the $\mathcal{A}$-seminorms and individual $\mathcal{A}$-numerical radii of related operators. Furthermore, the application of our main theorems to a single operator and its $\mathcal{A}$-adjoint has yielded several new bounds for the standard $\mathcal{A}$-numerical radius, providing deeper insights into the properties of operators.

The results presented herein not only advance the study of operator inequalities but also open avenues for further investigation. This work may serve as a foundational starting point for future projects. A natural and significant extension would be the generalization of the $\mathcal{A}$-joint numerical radius to higher powers. Investigating and proving novel inequalities for the generalized radius, potentially of the form

$${\omega }_{\mathcal{A},e,\rho }(\mathcal{T},\mathcal{S})=\underset{{\parallel \xi \parallel }_{\mathcal{A}}=1}{sup}{\left({\left|{\langle \mathcal{T}\xi ,\xi \rangle }_{\mathcal{A}}\right|}^{\rho }+{\left|{\langle \mathcal{S}\xi ,\xi \rangle }_{\mathcal{A}}\right|}^{\rho }\right)}^{{\displaystyle \frac{1}{\rho }}}$$

for $\rho \ge 1$, could lead to a rich new class of inequalities and a more comprehensive understanding of operator behavior. We believe that these potential directions will continue to enrich the theory of operators on semi-Hilbert spaces.