New Approach to Generalized Berezin Norms and Rigorous Operator Bounds

Abstract

Let $\left({\mathfrak{H}}_{\mathsf{\Theta }},⟨·,·⟩\right)$ be a reproducing kernel Hilbert space over a non-empty set $\mathsf{\Theta }$, and let $\mathrm{A}$ be a non-zero positive operator on ${\mathfrak{H}}_{\mathsf{\Theta }}$. This operator induces a semi-inner product given by ${⟨\xi ,\eta ⟩}_{\mathrm{A}}=⟨\mathrm{A}\xi ,\eta ⟩$ for all $\xi ,\eta \in {\mathfrak{H}}_{\mathsf{\Theta }}$, with the associated seminorm ${\parallel \xi \parallel }_{\mathrm{A}}=\sqrt{{⟨\xi ,\xi ⟩}_{\mathrm{A}}}$. The $\mathrm{A}$-normalized Berezin number and the $\mathrm{A}$-normalized Berezin norm of an $\mathrm{A}$-bounded linear operator $\mathrm{C}$ on ${\mathfrak{H}}_{\mathsf{\Theta }}$ are defined by ${\displaystyle {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)=\underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\left|{⟨\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}⟩}_{\mathrm{A}}\right|}$ and ${\displaystyle {\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}=\underset{\gamma ,\delta \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\left|{⟨\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}⟩}_{\mathrm{A}}\right|}$, where ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}={\displaystyle \frac{{\mathrm{x}}_{\gamma }}{\parallel {\mathrm{x}}_{\gamma }{\parallel }_{\mathrm{A}}}}$ and ${\mathsf{\Theta }}_{\mathrm{A}}=\{\gamma \in \mathsf{\Theta }:\parallel {\mathrm{x}}_{\gamma }{\parallel }_{\mathrm{A}}\ne 0\}$. The primary aim of this paper is to establish new sharp bounds and inequalities involving these two quantities and related operator-theoretic notions. In doing so, we propose a novel method to address the challenges of operator bounds. Furthermore, we revisit recent results on generalized Berezin norms, in particular those of Huban’s work in 2022. We show that some of these results rely on the incorrect assumption that the $\mathrm{A}$-Berezin number coincides with the $\mathrm{A}$-Berezin norm for $\mathrm{A}$-selfadjoint operators. By providing corrected arguments and employing tools such as the $\mathrm{A}$-Cartesian decomposition and the generalized Buzano inequality, we develop a consistent and rigorous framework for the study of generalized Berezin symbols in semi-Hilbertian spaces.

1. Introduction and Preliminaries

Mathematical inequalities play an important role in the development of operator theory and functional analysis. They provide useful tools for estimating operator norms and describing the behavior of linear operators [1,2,3]. Specifically, these techniques help in obtaining sharp bounds for spectral quantities within the intricate framework of semi-Hilbertian spaces, providing foundational support for numerous analytical results [4,5,6].

A large part of this study concerns the numerical radius, which often gives more precise information than the usual operator norm [7,8,9]. In addition to abstract Hilbert spaces, operator inequalities have been widely studied in reproducing kernel Hilbert spaces (RKHSs) [10,11,12]. These spaces provide a natural setting that connects operator theory to function theory, and they appear in several applications, such as mathematical physics, machine learning, and integral equations [13,14,15].

Additionally, functional analysis has been extended to semi-Hilbertian spaces. These are spaces equipped with a semi-inner product induced by a non-zero positive operator [16,17,18]. This generalization has led to further studies of partial isometries and operator ranges [19,20,21], as well as developments in spectral theory [22,23,24].

Motivated by these directions, several authors have extended classical operator inequalities to the semi-Hilbertian setting, with particular interest in the $\mathrm{A}$-numerical radius [25,26,27]. More recently, attention has turned to generalized Berezin symbols and related quantities. In this context, the $\mathrm{A}$-Berezin number and the $\mathrm{A}$-Berezin norm have attracted growing interest, leading to new inequalities in semi-Hilbertian spaces [28,29,30].

In this section, we present the main definitions and basic results concerning reproducing kernels, Berezin symbols, and semi-Hilbertian spaces, which will be used in the sequel.

1.1. Reproducing Kernel Hilbert Spaces and Berezin Symbols

Let $\mathsf{\Theta }$ be a non-empty set and $\mathcal{F}\left(\mathsf{\Theta }\right)$ be the set of all functions from $\mathsf{\Theta }$ to $\mathbb{C}$. A set ${\mathfrak{H}}_{\mathsf{\Theta }}\subseteq \mathcal{F}\left(\mathsf{\Theta }\right)$ is called a reproducing kernel Hilbert space (RKHS) on $\mathsf{\Theta }$ if ${\mathfrak{H}}_{\mathsf{\Theta }}$ is a Hilbert space and, for every $\gamma \in \mathsf{\Theta }$, the evaluation map $E_{\gamma }:{\mathfrak{H}}_{\mathsf{\Theta }}\to \mathbb{C}$, defined by $E_{\gamma }\left(f\right)=f\left(\gamma \right)$, is bounded. By the Riesz representation theorem, for each $\gamma \in \mathsf{\Theta }$, there exists a unique vector ${\mathrm{x}}_{\gamma }\in {\mathfrak{H}}_{\mathsf{\Theta }}$ such that $f\left(\gamma \right)=\langle f,{\mathrm{x}}_{\gamma }\rangle$ for every $f\in {\mathfrak{H}}_{\mathsf{\Theta }}$. The map $\mathrm{x}:\mathsf{\Theta }\times \mathsf{\Theta }\to \mathbb{C}$ defined by $\mathrm{x}(\xi ,\gamma )={\mathrm{x}}_{\gamma }\left(\xi \right)=\langle {\mathrm{x}}_{\gamma },{\mathrm{x}}_{\xi }\rangle$ is called the reproducing kernel function. Familiar examples include the Hardy space $H^2\left(\mathbb{D}\right)$ equipped with the classical Szegő kernel [12,14]. The existence and uniqueness of such spaces are guaranteed by the classical Moore–Aronszajn theorem [10].

For $\gamma \in \mathsf{\Theta }$, let ${\widehat{\mathrm{x}}}_{\gamma }={\displaystyle \frac{{\mathrm{x}}_{\gamma }}{\parallel {\mathrm{x}}_{\gamma }\parallel }}$ denote the normalized reproducing kernel. The set $\{{\widehat{\mathrm{x}}}_{\gamma };\gamma \in \mathsf{\Theta }\}$ is total in ${\mathfrak{H}}_{\mathsf{\Theta }}$. For a bounded linear operator $\mathrm{C}\in \mathcal{L}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$, the Berezin symbol ${\mathrm{C}}^{◊}$, initially introduced by F. A. Berezin [31,32], is defined on $\mathsf{\Theta }$ by ${\mathrm{C}}^{◊}\left(\gamma \right)=\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma },{\widehat{\mathrm{x}}}_{\gamma }\rangle$. This transform is an exceptionally powerful tool. On essential functional spaces (such as Bergman, Hardy, and Fock spaces), an operator is uniquely determined by its Berezin transform [13,15].

We formally define the Berezin set of an operator $\mathrm{C}$ as the collection of its Berezin symbols over the entire domain:

$$\mathbf{Ber}\left(\mathrm{C}\right):=\left\{{\mathrm{C}}^{◊}\left(\gamma \right);\gamma \in \mathsf{\Theta }\right\}.$$

The Berezin number of $\mathrm{C}$ is given by

$${\displaystyle \mathbf{ber}\left(\mathrm{C}\right):=\underset{\gamma \in \mathsf{\Theta }}{\sup }\left|{\mathrm{C}}^{◊}\left(\gamma \right)\right|=\underset{\gamma \in \mathsf{\Theta }}{\sup }|\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma },{\widehat{\mathrm{x}}}_{\gamma }\rangle |.}$$

The Berezin norm and the modified Berezin norm are defined, respectively, by

$${\displaystyle {\parallel \mathrm{C}\parallel }_{\mathbf{ber}}:=\underset{\gamma ,\delta \in \mathsf{\Theta }}{\sup }|\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma },{\widehat{\mathrm{x}}}_{\delta }\rangle |,{\parallel \mathrm{C}\parallel }_{\widetilde{\mathbf{ber}}}:=\underset{\gamma \in \mathsf{\Theta }}{\sup }\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }\parallel .}$$

The definitions and basic properties of these four quantities, namely, the Berezin set, Berezin number, Berezin norm, and modified Berezin norm, can be found in [33,34,35]. These quantities define norms on $\mathcal{L}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$ and satisfy

$$\mathbf{ber}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{\mathbf{ber}}\le {\parallel \mathrm{C}\parallel }_{\widetilde{\mathbf{ber}}}\le \parallel \mathrm{C}\parallel ,\forall \mathrm{C}\in \mathcal{L}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right).$$

(1)

In general, the inequalities in (1) are strict. Moreover, the Berezin norm is not submultiplicative, even for positive operators, and the Berezin number does not necessarily satisfy the power inequality $\mathbf{ber}\left({\mathrm{C}}^n\right)\le {\mathbf{ber}}^n\left(\mathrm{C}\right)$; see [29,36].

It is a standard result that $\mathbf{ber}\left(\mathrm{C}\right)=\mathbf{ber}\left({\mathrm{C}}^{*}\right)$ and ${\parallel \mathrm{C}\parallel }_{\mathbf{ber}}={\parallel {\mathrm{C}}^{*}\parallel }_{\mathbf{ber}}$; however, the equality ${∥\mathrm{C}∥}_{\widetilde{\mathbf{ber}}}={∥{\mathrm{C}}^{*}∥}_{\widetilde{\mathbf{ber}}}$ fails in general. Furthermore, since $\mathbf{Ber}\left(\mathrm{C}\right)\subseteq W\left(\mathrm{C}\right)$, we naturally deduce $\mathbf{ber}\left(\mathrm{C}\right)\le \omega \left(\mathrm{C}\right)$, where $W\left(\mathrm{C}\right)$ and $\omega \left(\mathrm{C}\right)$ are the classical numerical range and numerical radius, respectively, [3,5,8].

1.2. Operators in Semi-Hilbert Spaces

Throughout this section, let $\mathfrak{H}$ be a complex Hilbert space with inner product $\langle ·,·\rangle$ and associated norm $\parallel ·\parallel$. We use $\mathcal{L}\left(\mathfrak{H}\right)$ to denote the $C^{*}$-algebra of all bounded linear operators from $\mathfrak{H}$ to itself. It is crucial to mention that all operators in this work are assumed to be bounded and linear. For $\mathrm{C}\in \mathcal{L}\left(\mathfrak{H}\right)$, its range and its null space are denoted by $\mathbf{R}\left(\mathrm{C}\right)$ and $\ker \left(\mathrm{C}\right)$, respectively. Furthermore, ${\mathrm{C}}^{*}$ stands for the adjoint of $\mathrm{C}$. Let $\mathcal{S}$ be any linear subspace of $\mathfrak{H}$. We use $\overline{\mathcal{S}}$ to denote its closure with respect to the topology generated by $\parallel ·\parallel$. If $\mathcal{M}$ is a closed subspace of $\mathfrak{H}$, then $P_{\mathcal{M}}$ stands for the orthogonal projection onto $\mathcal{M}$.

Let $\mathrm{A}\in \mathcal{L}\left(\mathfrak{H}\right)$ be a non-zero positive operator. The semi-inner product induced by $\mathrm{A}$ is given by

$${\langle \xi ,\eta \rangle }_{\mathrm{A}}=\langle \mathrm{A}\xi ,\eta \rangle =\langle \sqrt{\mathrm{A}}\xi ,\sqrt{\mathrm{A}}\eta \rangle ,\forall \xi ,\eta \in \mathfrak{H}.$$

Here, $\sqrt{\mathrm{A}}$ denotes the square root of $\mathrm{A}$, that is, the unique positive operator such that ${\left(\sqrt{\mathrm{A}}\right)}^2=\mathrm{A}$. The corresponding seminorm induced by ${\langle ·,·\rangle }_{\mathrm{A}}$ is defined as ${\parallel \xi \parallel }_{\mathrm{A}}=\parallel \sqrt{\mathrm{A}}\xi \parallel$. Notably, ${\parallel \xi \parallel }_{\mathrm{A}}=0$ if and only if $\xi \in \ker \left(\mathrm{A}\right)$, making it a true norm if and only if $\mathrm{A}$ is injective. Here, $\ker \left(\mathrm{A}\right)$ denotes the null space of $\mathrm{A}$.

We consider the space ${\mathcal{L}}^{\mathrm{A}}\left(\mathfrak{H}\right)$ containing all operators $\mathrm{C}\in \mathcal{L}\left(\mathfrak{H}\right)$ for which there exists a constant $k>0$ such that ${\parallel \mathrm{C}\xi \parallel }_{\mathrm{A}}\le k{\parallel \xi \parallel }_{\mathrm{A}}$ for all $\xi \in \overline{\mathbf{R}\left(\mathrm{A}\right)}$. For $\mathrm{C}\in {\mathcal{L}}^{\mathrm{A}}\left(\mathfrak{H}\right)$, the $\mathrm{A}$-operator seminorm is given by

$${\displaystyle {\parallel \mathrm{C}\parallel }_{\mathrm{A}}=\underset{\begin{array}{c}\xi \in \overline{\mathbf{R}\left(\mathrm{A}\right)},\\ {\parallel \xi \parallel }_{\mathrm{A}}=1\end{array}}{\sup }{\parallel \mathrm{C}\xi \parallel }_{\mathrm{A}}<\infty .}$$

An operator $\mathrm{D}\in \mathcal{L}\left(\mathfrak{H}\right)$ is called an $\mathrm{A}$-adjoint of $\mathrm{C}$ if ${\langle \mathrm{C}\xi ,\eta \rangle }_{\mathrm{A}}={\langle \xi ,\mathrm{D}\eta \rangle }_{\mathrm{A}}$ for all $\xi ,\eta \in \mathfrak{H}$ [16]; this is equivalent to solving the operator equation $\mathrm{A}\mathrm{Y}={\mathrm{C}}^{*}\mathrm{A}$. The existence of such solutions is governed by Douglas’s theorem:

Theorem 1

(Douglas [18]). Let $\mathrm{C},\mathrm{D}\in \mathcal{L}\left(\mathfrak{H}\right)$. The following statements are equivalent:

(i)

$\mathbf{R}\left(\mathrm{D}\right)\subseteq \mathbf{R}\left(\mathrm{C}\right)$;

(ii)

$\mathrm{C}{\mathrm{D}}_1=\mathrm{D}$ for some ${\mathrm{D}}_1\in \mathcal{L}\left(\mathfrak{H}\right)$;

(iii)

There exists $a_1>0$ such that $\parallel {\mathrm{D}}^{*}\xi \parallel \le a_1\parallel {\mathrm{C}}^{*}\xi \parallel$ for all $\xi \in \mathfrak{H}$.

If any of these equivalent conditions holds, there is a unique reduced solution ${\mathrm{D}}_2\in \mathcal{L}\left(\mathfrak{H}\right)$ of $\mathrm{C}\mathrm{Y}=\mathrm{D}$ such that $\mathbf{R}\left({\mathrm{D}}_2\right)\subseteq \overline{\mathbf{R}\left({\mathrm{C}}^{*}\right)}$.

Let ${\mathcal{L}}_{\mathrm{A}}\left(\mathfrak{H}\right)$ and ${\mathcal{L}}_{\sqrt{\mathrm{A}}}\left(\mathfrak{H}\right)$ denote the collections of operators admitting an $\mathrm{A}$-adjoint and $\sqrt{\mathrm{A}}$-adjoint, respectively. By Theorem 1, ${\mathcal{L}}_{\mathrm{A}}\left(\mathfrak{H}\right)=\{\mathrm{C}\in \mathcal{L}\left(\mathfrak{H}\right):\mathbf{R}\left({\mathrm{C}}^{*}\mathrm{A}\right)\subseteq \mathbf{R}\left(\mathrm{A}\right)\}$, and operators in ${\mathcal{L}}_{\sqrt{\mathrm{A}}}\left(\mathfrak{H}\right)$ are explicitly referred to as $\mathrm{A}$-bounded. For $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left(\mathfrak{H}\right)$, the reduced solution to $\mathrm{A}\mathrm{Y}={\mathrm{C}}^{*}\mathrm{A}$ is denoted by ${\mathrm{C}}^{{♯}_{\mathrm{A}}}={\mathrm{A}}^{†}{\mathrm{C}}^{*}\mathrm{A}$. Importantly, ${\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)}^{{♯}_{\mathrm{A}}}=P_{\overline{\mathbf{R}\left(\mathrm{A}\right)}}\mathrm{C}{P}_{\overline{\mathbf{R}\left(\mathrm{A}\right)}}$, which implies the absorption property ${\left({\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)}^{{♯}_{\mathrm{A}}}\right)}^{{♯}_{\mathrm{A}}}={\mathrm{C}}^{{♯}_{\mathrm{A}}}$.

An operator $\mathrm{C}$ is called $\mathrm{A}$-selfadjoint if $\mathrm{A}\mathrm{C}$ is selfadjoint. This condition implies that $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left(\mathfrak{H}\right)$, but it does not necessarily guarantee that ${\mathrm{C}}^{{♯}_{\mathrm{A}}}=\mathrm{C}$. However, the equality $\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}{\xi \parallel }_{\mathrm{A}}={\parallel \mathrm{C}\xi \parallel }_{\mathrm{A}}$ holds for all $\xi \in \mathfrak{H}$ [25].

For $\mathrm{C}\in {\mathcal{L}}_{\sqrt{\mathrm{A}}}\left(\mathfrak{H}\right)$, the $\mathrm{A}$-numerical range and $\mathrm{A}$-numerical radius [21,22], respectively, are defined as follows:

$$W_{\mathrm{A}}\left(\mathrm{C}\right)=\left\{{\langle \mathrm{C}\xi ,\xi \rangle }_{\mathrm{A}};\xi \in \mathfrak{H},{\parallel \xi \parallel }_{\mathrm{A}}=1\right\},{\omega }_{\mathrm{A}}\left(\mathrm{C}\right)=\sup \left\{|a_2|;a_2\in W_{\mathrm{A}}\left(\mathrm{C}\right)\right\}.$$

The $\mathrm{A}$-numerical radius is equivalent to the $\mathrm{A}$-operator seminorm via the bounds ${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\mathrm{A}}\le {\omega }_{\mathrm{A}}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{\mathrm{A}}$.

To prove some of the main results, we first recall the generalized Buzano inequality established by Saddi in [21].

Lemma 1.

Let $\xi ,\eta ,\zeta \in \mathfrak{H}$ be such that ${\parallel \zeta \parallel }_{\mathrm{A}}=1$. Then

$$|{\langle \xi ,\zeta \rangle }_{\mathrm{A}}{\langle \zeta ,\eta \rangle }_{\mathrm{A}}|\le {\displaystyle \frac{1}{2}}\left(\left|{\langle \xi ,\eta \rangle }_{\mathrm{A}}\right|+{\parallel \xi \parallel }_{\mathrm{A}}{\parallel \eta \parallel }_{\mathrm{A}}\right).$$

(2)

1.3. The $\mathrm{A}$-Normalized Berezin Framework

When the Berezin number is extended to the semi-Hilbertian setting, two distinct normalization methods naturally arise.

The first approach, introduced by Conde, Feki, and Kittaneh [29], utilizes the standard Hilbert-normalized kernel ${\widehat{\mathrm{x}}}_{\gamma }={\displaystyle \frac{{\mathrm{x}}_{\gamma }}{\parallel {\mathrm{x}}_{\gamma }\parallel }}$. In this context, the $\mathrm{A}$-Berezin number is defined as

$${\displaystyle {\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)=\underset{\gamma \in \mathsf{\Theta }}{\sup }\frac{|{\langle \mathrm{C}{\mathrm{x}}_{\gamma },{\mathrm{x}}_{\gamma }\rangle }_{\mathrm{A}}|}{\parallel {\mathrm{x}}_{\gamma }{\parallel }^2}=\underset{\gamma \in \mathsf{\Theta }}{\sup }\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma },{\widehat{\mathrm{x}}}_{\gamma }\rangle }_{\mathrm{A}}\right|.}$$

While this definition is well posed and naturally avoids division by zero when ${\mathrm{x}}_{\gamma }\in \ker \left(\mathrm{A}\right)$, it presents certain geometric limitations. Specifically, many semi-Hilbertian inequalities (such as the generalized Buzano inequality (2)) require unit vectors with respect to the $\mathrm{A}$-seminorm (i.e., vectors $\zeta$ satisfying ${\parallel \zeta \parallel }_{\mathrm{A}}=1$). Since ${\parallel \widehat{\mathrm{x}}\gamma \parallel }_{\mathrm{A}}\ne 1$ in general, classical semi-Hilbertian bounds cannot be directly applied using this normalization.

To circumvent this limitation, an alternative $\mathrm{A}$-normalized approach can be employed. Inspired by [37], we define the following set:

$${\mathsf{\Theta }}_{\mathrm{A}}=\{\gamma \in \mathsf{\Theta }:\parallel {\mathrm{x}}_{\gamma }{\parallel }_{\mathrm{A}}\ne 0\}.$$

We first note that the set ${\mathsf{\Theta }}_{\mathrm{A}}$ is non-empty. If it were empty, then $\mathrm{A}{\mathrm{x}}_{\delta }=0$ would be true for all $\delta \in \mathsf{\Theta }$. Since the linear span of ${\left\{{\mathrm{x}}_{\delta }\right\}}_{\delta \in \mathsf{\Theta }}$ is dense in ${\mathfrak{H}}_{\mathsf{\Theta }}$, this would imply that $\mathrm{A}=\mathbf{0}$, which contradicts our initial assumption that $\mathrm{A}$ is a non-zero positive operator. Consequently, the $\mathrm{A}$-normalization is always well defined.

Definition 1.

Let $\mathrm{C}\in {\mathcal{L}}_{\sqrt{\mathrm{A}}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. For $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$, we define the $\mathrm{A}$-normalized kernel as ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}={\displaystyle \frac{{\mathrm{x}}_{\gamma }}{\parallel {\mathrm{x}}_{\gamma }{\parallel }_{\mathrm{A}}}}$. We formalize the $\mathrm{A}$-Berezin definitions as follows:

(i)

The $\mathrm{A}$-Berezin symbol of $\mathrm{C}$ at γ is ${\mathrm{C}}_{\mathrm{A}}^{◊}\left(\gamma \right)={\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}$.

(ii)

The $\mathrm{A}$-Berezin range of $\mathrm{C}$ is ${\mathit{Ber}}_{\mathrm{A}}\left(\mathrm{C}\right):=\left\{{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}};\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}\right\}$.

(iii)

The $\mathrm{A}$-Berezin number of $\mathrm{C}$ is ${\displaystyle {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right):=\underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}$.

(iv)

The $\mathrm{A}$-Berezin seminorm of $\mathrm{C}$ is ${\displaystyle {\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}:=\underset{\gamma ,\delta \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}$.

(v)

The $\mathrm{A}$-Berezin radius of $\mathrm{C}$ is ${\displaystyle {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right):=\underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}}$.

Remark 1.

Let $\mathrm{C}\in {\mathcal{L}}_{\sqrt{\mathrm{A}}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$.

(i)

Since ${\parallel {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}=1$ for all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$, it immediately follows that ${\mathit{Ber}}_{\mathrm{A}}\left(\mathrm{C}\right)\subseteq W_{\mathrm{A}}\left(\mathrm{C}\right),$ and therefore,

$${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\omega }_{\mathrm{A}}\left(\mathrm{C}\right).$$

(ii)

By applying the Cauchy–Schwarz inequality for the semi-inner product, we obtain the following sequence of inequalities:

$${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}\le {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{\mathrm{A}}.$$

(3)

(iii)

The quantities ${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)$ and ${\mathit{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)$ are generally incomparable. For a counterexample, consider ${\mathfrak{H}}_{\mathsf{\Theta }}={\mathbb{C}}^2$ equipped with the canonical basis as reproducing kernels, and let $\mathrm{A}=\mathrm{diag}(2,{\displaystyle \frac{1}{2}})$. In this case, ${\mathsf{\Theta }}_{\mathrm{A}}=\{1,2\}$. If we define ${\mathrm{C}}_1=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ and ${\mathrm{C}}_2=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$, we observe that

$${\mathit{ber}}_{\mathrm{A}}\left({\mathrm{C}}_1\right)=2>1={\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}_1\right),{\mathit{ber}}_{\mathrm{A}}\left({\mathrm{C}}_2\right)=\frac{1}{2}<1={\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}_2\right).$$

For the remainder of this paper, we focus solely on the $\mathrm{A}$-normalized Berezin number ${\mathbf{b}}_{\mathrm{A}}$. To ensure a solid theoretical foundation for our results, we conclude this section by addressing and clarifying certain methodological and notational discrepancies found in recent literature, particularly in the work of Huban [38].

In [38], an attempt was made to introduce the modified $\mathrm{A}$-Berezin norm for an operator. However, the initial definition contained a typographical inconsistency, being presented as ${\displaystyle {\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}:=\underset{\gamma \in \mathsf{\Theta }}{\sup }{\parallel \mathrm{A}\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }\parallel }_{{\mathfrak{H}}_{\mathsf{\Theta }}}}$, where ${\widehat{\mathrm{x}}}_{\gamma }={\displaystyle \frac{{\mathrm{x}}_{\gamma }}{\parallel {\mathrm{x}}_{\gamma }\parallel }}$. In the subsequent proofs of [38], however, the quantity actually utilized is as follows:

$${\displaystyle {\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}=\underset{\gamma \in \mathsf{\Theta }}{\sup }\parallel \sqrt{\mathrm{A}}\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }{\parallel }_{{\mathfrak{H}}_{\mathsf{\Theta }}}=\underset{\gamma \in \mathsf{\Theta }}{\sup }{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }\parallel }_{\mathrm{A}}.}$$

(4)

It is worth noting that when $\mathrm{A}=I_{{\mathfrak{H}}_{\mathsf{\Theta }}}$, this reduces to the standard modified Berezin norm ${\parallel \mathrm{C}\parallel }_{\widetilde{\mathbf{ber}}}$ as defined in [29]. Furthermore, the prerequisite for $\mathrm{C}$ to belong to the class of $\mathrm{A}$-bounded operators, ${\mathcal{L}}_{\sqrt{\mathrm{A}}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$, was not explicitly stated in [38], although this assumption is mathematically necessary to guarantee the finiteness of the supremum in (4).

Additionally, a technical ambiguity arises in some of the proofs in [38] regarding mathematical types. For instance, in the proof of Lemma 2.1 in [38], the operator norm notation is applied to vectors in the Hilbert space, resulting in expressions such as $\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }{\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}$ and $\parallel {\widehat{\mathrm{x}}}_{\gamma }\pm {\widehat{\mathrm{x}}}_{\delta }{\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}$. This overlap in notation between the vector semi-norm ${\parallel ·\parallel }_{\mathrm{A}}$ and an operator semi-norm can lead to algebraic manipulations that are not formally justified.

To further clarify this point, for any operator $\mathrm{C}\in {\mathcal{L}}_{\sqrt{\mathrm{A}}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$, the following inequalities hold legitimately:

$${\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}\le {\parallel \mathrm{C}\parallel }_{\mathrm{A}}.$$

However, the proofs in [38] critically rely on the following operator properties for $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$:

(1)

${\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}={\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}$.

(2)

${\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)={\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}$ for every $\mathrm{A}$-selfadjoint operator $\mathrm{C}$.

(3)

${\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}={\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}={\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}^2$.

It is important to note that these properties do not hold in general. As shown in [29], these equalities may fail even in the classical case where $\mathrm{A}$ is the identity operator.

Motivated by these observations, one of the main goals of this paper is to correct the proofs and clarify the bounds presented by Huban [38]. In addition, we establish several new inequalities and upper bounds for the $\mathrm{A}$-normalized Berezin number, the $\mathrm{A}$-normalized Berezin norm and related operator-theoretic notions. Using tools such as the $\mathrm{A}$-Cartesian decomposition and the generalized Buzano inequality, we aim to develop a clear and consistent framework for the study of Berezin-type quantities in semi-Hilbertian spaces.

2. Main Results

In this section, we present our main results, in which we derive several operator inequalities for the $\mathrm{A}$-normalized Berezin number ${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)$. As mentioned previously, it is incorrect to treat the set of normalized kernels $\{{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}};\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}\}$ as a linear subspace and then apply the polarization identity. This approach is not valid and may lead to incorrect results. To avoid this issue, we work directly with pointwise inequalities on the elements of ${\mathsf{\Theta }}_{\mathrm{A}}$.

Finally, we note that the equality ${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)={\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}$ does not hold in general for $\mathrm{A}$-selfadjoint operators. However, it does hold for the class of $\mathrm{A}$-positive operators.

Lemma 2.

Let $\mathrm{C}\in {\mathcal{L}}_{\sqrt{\mathrm{A}}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$ be an $\mathrm{A}$-positive operator. Then,

$${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)={\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}.$$

(5)

Proof. 

The inequality ${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}$ holds by definition. To prove the reverse inequality, let $\gamma ,\delta \in {\mathsf{\Theta }}_{\mathrm{A}}$ and ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}$ be the corresponding $\mathrm{A}$-normalized reproducing kernels. Since $\mathrm{C}$ is $\mathrm{A}$-positive (i.e., $\mathrm{A}\mathrm{C}\ge 0$), the mapping $(\xi ,\eta )\mapsto {\langle \mathrm{C}\xi ,\eta \rangle }_{\mathrm{A}}=\langle \mathrm{A}\mathrm{C}\xi ,\eta \rangle$ defines a positive semi-inner product on ${\mathfrak{H}}_{\mathsf{\Theta }}$. By the Cauchy–Schwarz inequality, we obtain

$$\begin{array}{cc}\hfill |{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2& \le {\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}\rangle }_{\mathrm{A}}\hfill \\ \hfill & \le {\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right).\hfill \end{array}$$

Taking the supremum over all $\gamma ,\delta \in {\mathsf{\Theta }}_{\mathrm{A}}$, we deduce that ${\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}^2\le {\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)$. This proves the inequality (5). □

Remark 2. (1) If $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$, then ${\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}$ and $\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}$ are $\mathrm{A}$-positive operators. Hence, by Lemma 2, we have

$${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)=\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\parallel }_{{\mathbf{b}}_{\mathrm{A}}}={\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)\mathit{and}{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\right)={\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}={\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right).$$

(6)

(2) Bhunia et al. [7] provided an example (for the case $\mathrm{A}=I_{{\mathfrak{H}}_{\mathsf{\Theta }}}$) demonstrating that the equality (5) does not hold even for selfadjoint operators. It is also important to note that the equality ${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)={\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)$ may fail even for $\mathrm{A}$-positive operators (see [29]).

For $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$, we utilize the $\mathrm{A}$-Cartesian decomposition $\mathrm{C}={\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)+i{\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)$, where ${\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)={\displaystyle \frac{\mathrm{C}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}}{2}}$ and ${\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)={\displaystyle \frac{\mathrm{C}-{\mathrm{C}}^{{♯}_{\mathrm{A}}}}{2i}}$. Both ${\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)$ and ${\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)$ are $\mathrm{A}$-selfadjoint. Consequently, for any $\xi \in {\mathfrak{H}}_{\mathsf{\Theta }}$, the quantities ${\langle {\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)\xi ,\xi \rangle }_{\mathrm{A}}$ and ${\langle {\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\xi ,\xi \rangle }_{\mathrm{A}}$ are purely real.

We are now ready to prove the following theorem, which provides a mathematically rigorous improvement of a result proposed by Huban in [38].

Theorem 2.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\displaystyle \frac{\sqrt{2}}{2}}\sqrt{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\parallel }_{{\mathbf{b}}_{\mathrm{A}}}}\le {\displaystyle \frac{\sqrt{2}}{2}}\sqrt{{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)+{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)}\le {∥\mathrm{C}∥}_{\mathrm{A}}.$$

(7)

Proof. 

Since $\mathrm{C}={\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)+i{\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)$, a straightforward calculation shows that

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}\left(\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\right)={\mathrm{Re}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+{\mathrm{Im}}_{\mathrm{A}}^2\left(\mathrm{C}\right).\end{array}$$

(8)

Let $\delta \in {\mathsf{\Theta }}_{\mathrm{A}}$, and let ${\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}$ be the corresponding $\mathrm{A}$-normalized reproducing kernel of the space ${\mathfrak{H}}_{\mathsf{\Theta }}$. By applying the Cauchy–Schwarz inequality, we can see that

$$\begin{array}{cc}\hfill |{〈\mathrm{C}{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}〉}_{\mathrm{A}}{|}^2& =|{〈{\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right){\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}〉}_{\mathrm{A}}{|}^2+|{〈{\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right){\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}〉}_{\mathrm{A}}{|}^2\hfill \\ \hfill & \le {\parallel {\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right){\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2+{\parallel {\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right){\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\hfill \\ \hfill & ={〈{\mathrm{Re}}_{\mathrm{A}}^2\left(\mathrm{C}\right){\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}〉}_{\mathrm{A}}+{〈{\mathrm{Im}}_{\mathrm{A}}^2\left(\mathrm{C}\right){\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}〉}_{\mathrm{A}}\hfill \\ \hfill & ={〈\left({\mathrm{Re}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+{\mathrm{Im}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\right){\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}〉}_{\mathrm{A}}\hfill \\ \hfill & \le {\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+{\mathrm{Im}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\right)\hfill \\ \hfill & ={\parallel {\mathrm{Re}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+{\mathrm{Im}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\parallel }_{{\mathbf{b}}_{\mathrm{A}}}.\hfill \end{array}$$

Note that we have used Lemma 2 in the final equality because ${\mathrm{Re}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+{\mathrm{Im}}_{\mathrm{A}}^2\left(\mathrm{C}\right){\ge }_{\mathrm{A}}0$. Thus, taking (8) into consideration, we obtain

$$|{〈\mathrm{C}{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}〉}_{\mathrm{A}}{|}^2\le {\displaystyle \frac{1}{2}}{\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}.$$

Taking the supremum over all $\delta \in {\mathsf{\Theta }}_{\mathrm{A}}$ yields the first inequality in (7). Finally, applying (6) and the fundamental bounds ${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{\mathrm{A}}$ presented in (3) completes the proof. □

Remark 3.

Kittaneh [39] proved that for every $\mathrm{C}\in \mathcal{L}\left(\mathfrak{H}\right)$, the following bounds hold for the classical numerical radius:

$${\displaystyle \frac{1}{2}}\sqrt{\parallel {\mathrm{C}}^{*}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{*}\parallel }\le \omega \left(\mathrm{C}\right)\le {\displaystyle \frac{\sqrt{2}}{2}}\sqrt{\parallel {\mathrm{C}}^{*}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{*}\parallel }.$$

However, it should be noted that the analogous lower bound does not hold in general in the context of the Berezin number, even when $\mathrm{A}=I_{{\mathfrak{H}}_{\mathsf{\Theta }}}$ (see [29]). Consequently, the corresponding inequality for the $\mathrm{A}$-Berezin number,

$${\displaystyle \frac{1}{2}}\sqrt{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\parallel }_{{\mathbf{b}}_{\mathrm{A}}}}\le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right),$$

fails to hold for some $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$.

To further refine this upper bound, we can incorporate the square of the operator, ${\mathrm{C}}^2$. A similar attempt was made in [38] (Theorem 2.10), but the proof was invalidated by the incorrect application of the norm-number equality for $\mathrm{A}$-selfadjoint operators. We provide a definitive, corrected bound by rigorously applying the generalized Buzano inequality pointwise.

Theorem 3.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$${\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\le {\displaystyle \frac{1}{2}}{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)+{\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}.$$

Proof. 

Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ be the corresponding $\mathrm{A}$-normalized reproducing kernel. By setting $\xi =\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$, $\eta ={\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$, and $\zeta ={\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ in the generalized Buzano inequality (Lemma 1), we obtain the following:

$$\begin{array}{cc}\hfill |{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2& =\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\langle {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left(\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|+{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left|{\langle {\mathrm{C}}^2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|+{\displaystyle \frac{1}{4}}\left(\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2+{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left|{\langle {\mathrm{C}}^2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|+{\displaystyle \frac{1}{4}}{〈({\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}〉}_{\mathrm{A}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)+{\displaystyle \frac{1}{4}}{\mathbf{b}}_{\mathrm{A}}(\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}).\hfill \end{array}$$

Here, we utilized Young’s inequality (or the AM–GM inequality applied to $a^2$ and $b^2$) $ab\le {\displaystyle \frac{1}{2}}(a^2+b^2)$ for $a,b\ge 0$, alongside the identities ${\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}={\langle {\mathrm{C}}^2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}$ and $\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2={\langle {\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}$. Furthermore, since ${\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}$ is an $\mathrm{A}$-positive operator, Lemma 2 guarantees that

$$\begin{array}{c}\hfill |{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2\le {\displaystyle \frac{1}{2}}{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)+{\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}.\end{array}$$

Taking the supremum over all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ in the above inequality directly yields the desired result. □

As a direct consequence of Theorem 3, we state the next corollary.

Corollary 1.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\displaystyle \frac{1}{2}}\sqrt{\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}+2{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)}.$$

Remark 4.

The upper bound established in Theorem 3 provides a refinement over the bound obtained in Theorem 2. To mathematically justify this refinement, observe that for any $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and its corresponding $\mathrm{A}$-normalized reproducing kernel ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$, the Cauchy–Schwarz inequality and Young’s inequality $ab\le {\displaystyle \frac{1}{2}}(a^2+b^2)$ imply the following:

$$\begin{array}{cc}\hfill \left|{\langle {\mathrm{C}}^2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|& =\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|\hfill \\ \hfill & \le {\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left(\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2+{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{〈({\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}〉}_{\mathrm{A}}.\hfill \end{array}$$

Taking the supremum over $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$, and using Lemma 2, we obtain

$${\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)\le {\displaystyle \frac{1}{2}}{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}.$$

Incorporating this bound into the result of Theorem 3 gives the following:

$$\begin{array}{cc}\hfill {\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)& \le {\displaystyle \frac{1}{2}}{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)+{\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}+{\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}\mathrm{C}+\mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}\parallel }_{{\mathbf{b}}_{\mathrm{A}}},\hfill \end{array}$$

which precisely recovers the upper bound from Theorem 2. This deduction rigorously proves that Theorem 3 guarantees an inequality that is always at least as sharp as that in Theorem 2.

Remark 5.

It can be seen that Theorem 3 provides the following estimate:

$${\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\le {\displaystyle \frac{1}{2}}{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)+{\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\mathrm{A}}^2.$$

Remark 6.

We show that Corollary 2.8 and Theorem 2.11 in [38] are incorrect. Corollary 2.8 in [38] claims that

$${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\widetilde{{\mathit{ber}}_{\mathrm{A}}}}\le {\mathit{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{\widetilde{{\mathit{ber}}_{\mathrm{A}}}}.$$

Theorem 2.11 in [38] claims that

$${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\widetilde{{\mathit{ber}}_{\mathrm{A}}}}\le {\displaystyle \frac{1}{2}}\sqrt{\parallel \mathrm{C}{\mathrm{C}}^{{♯}_{\mathrm{A}}}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\mathrm{C}\parallel }_{\widetilde{{\mathit{ber}}_{\mathrm{A}}}}+2{\mathbf{c}}_{{\mathit{ber}}_{\mathrm{A}}}\left({\mathrm{C}}^2\right)}\le {\mathit{ber}}_{\mathrm{A}}\left(\mathrm{C}\right).$$

Since both results rely on the false lower bound ${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\widetilde{{\mathit{ber}}_{\mathrm{A}}}}\le {\mathit{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)$, they are invalid. To prove this, we use the nilpotent counterexample. Let $\mathrm{C}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $\mathrm{A}=I_{{\mathfrak{H}}_{\mathsf{\Theta }}}$ on ${\mathbb{C}}^2$. Since $\mathrm{C}$ is nilpotent, it follows that ${\mathit{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)=0$ and ${\parallel \mathrm{C}\parallel }_{\widetilde{{\mathit{ber}}_{\mathrm{A}}}}=1$. By using these values in Corollary 2.8, we obtain ${\displaystyle \frac{1}{2}}\le 0$, which is absurd. For Theorem 2.11, since ${\mathrm{C}}^2=\mathbf{0}$, then ${\mathbf{c}}_{{\mathit{ber}}_{\mathrm{A}}}\left({\mathrm{C}}^2\right)=0$. By using $\mathrm{C}{\mathrm{C}}^{*}+{\mathrm{C}}^{*}\mathrm{C}=I_{{\mathfrak{H}}_{\mathsf{\Theta }}}$, we get $\parallel \mathrm{C}{\mathrm{C}}^{*}+{\mathrm{C}}^{*}{\mathrm{C}\parallel }_{\widetilde{{\mathit{ber}}_{\mathrm{A}}}}=1$. By using these values in Theorem 2.11, we get ${\displaystyle \frac{1}{2}}\le 0$, which is false. Corollary 1 successfully resolves this by providing the correct analogous upper bound.

Theorems 2.5 and 2.6 in [38] attempted to express the Berezin number via the supremum of the Berezin norm over various combinations of the Cartesian parts. In the following theorem, we establish a mathematically sound version formulating the problem solely in terms of the $\mathrm{A}$-Berezin number.

Theorem 4.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$${\displaystyle {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)=\underset{\theta \in \mathbb{R}}{\sup }{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}\left(e^{i\theta }\mathrm{C}\right)\right)=\underset{\begin{array}{c}{a}_2,a_3\in \mathbb{R}\\ a_2^2+a_3^2=1\end{array}}{\sup }{\mathbf{b}}_{\mathrm{A}}\left(a_2{\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)+a_3{\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\right).}$$

Proof. 

For any $z\in \mathbb{C}$, where ℜ denotes the real part, its modulus can be expressed as ${\displaystyle \left|z\right|=\underset{\theta \in \mathbb{R}}{\sup }\Re \left(e^{i\theta }z\right)}$. Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ be the corresponding $\mathrm{A}$-normalized reproducing kernel. Applying this identity to the complex number $z={\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}$, we obtain

$$\begin{array}{cc}\hfill \left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|& {\displaystyle =\underset{\theta \in \mathbb{R}}{\sup }\Re \left(e^{i\theta }{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right)}\hfill \\ \hfill & {\displaystyle =\underset{\theta \in \mathbb{R}}{\sup }{〈\left(\frac{e^{i\theta }\mathrm{C}+{\left(e^{i\theta }\mathrm{C}\right)}^{{♯}_{\mathrm{A}}}}{2}\right){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}〉}_{\mathrm{A}}}\hfill \\ \hfill & {\displaystyle =\underset{\theta \in \mathbb{R}}{\sup }{\langle {\mathrm{Re}}_{\mathrm{A}}\left(e^{i\theta }\mathrm{C}\right){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}.}\hfill \end{array}$$

Taking the supremum over all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and noting that the symmetry ${\mathrm{Re}}_{\mathrm{A}}\left(e^{i(\theta +\pi )}\mathrm{C}\right)=-{\mathrm{Re}}_{\mathrm{A}}\left(e^{i\theta }\mathrm{C}\right)$ allows us to place the absolute value inside the supremum over $\theta$, yielding ${\displaystyle {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)=\underset{\theta \in \mathbb{R}}{\sup }{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}\left(e^{i\theta }\mathrm{C}\right)\right)}$. Substituting $e^{i\theta }=\cos \theta +i\sin \theta$ and parameterizing the unit circle via real numbers $a_2=\cos \theta$ and $a_3=-\sin \theta$ directly yields the second equality. □

Inspired by [27], we deduce the following corollary as a consequence of Theorem 4.

Corollary 2.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$$\begin{array}{c}\hfill {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le \sqrt{{\mathbf{b}}_{\mathrm{A}}^2\left({\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)+{\mathbf{b}}_{\mathrm{A}}^2\left({\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)}\le \sqrt{2}\max \left\{{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)\right),{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)\right\}.\end{array}$$

Proof. 

By Theorem 4 and the subadditivity of the $\mathrm{A}$-Berezin number, we have

$$\begin{array}{cc}\hfill {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)& {\displaystyle =\underset{\begin{array}{c}{a}_2,a_3\in \mathbb{R}\\ a_2^2+a_3^2=1\end{array}}{\sup }{\mathbf{b}}_{\mathrm{A}}\left(a_2{\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)+a_3{\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)}\hfill \\ \hfill & {\displaystyle \le \underset{\begin{array}{c}{a}_2,a_3\in \mathbb{R}\\ a_2^2+a_3^2=1\end{array}}{\sup }\left(|a_2|{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)+\left|a_3\right|{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)\right).}\hfill \end{array}$$

By the classical Cauchy–Schwarz inequality, we immediately obtain

$${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le \sqrt{{\mathbf{b}}_{\mathrm{A}}^2\left({\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)+{\mathbf{b}}_{\mathrm{A}}^2\left({\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)}.$$

To establish the second inequality, we utilize the elementary algebraic fact that $x^2+y^2\le 2\max \{x^2,y^2\}$ for any real numbers $x,y$. □

Our next corollary establishes a foundational lower bound for ${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)$ by relating it to the Berezin numbers of its Cartesian components. This provides a mathematically sound alternative to previously published attempts (e.g., [38] (Corollary 2.7)) that relied on the invalid equality assumption that the $\mathrm{A}$-Berezin number equals the $\mathrm{A}$-Berezin norm for $\mathrm{A}$-selfadjoint operators.

Corollary 3.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$$\max \left\{{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)\right),{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)\right\}\le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right).$$

Proof. 

This result follows immediately from Theorem 4. By taking $(a_2,a_3)=(1,0)$ and $(a_2,a_3)=(0,1)$, we directly obtain ${\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)\le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)$ and ${\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Im}}_{\mathrm{A}}\left(\mathrm{C}\right)\right)\le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)$, respectively. □

In the next result, we establish an exact integral representation for the quantity ${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)$, expressed in terms of convex combinations of the operator $\mathrm{C}$ and its $\mathrm{A}$-adjoint ${\mathrm{C}}^{{♯}_{\mathrm{A}}}$ along rotated directions. We point out that a similar integral formula was recently obtained in [40] for the quantity ${\mathbf{ber}}_{\mathrm{A}}$. However, since the two normalization approaches, ${\mathbf{ber}}_{\mathrm{A}}$ and ${\mathbf{b}}_{\mathrm{A}}$, are generally not comparable (see Remark 1), the derivation of an analogous result for ${\mathbf{b}}_{\mathrm{A}}$ is not a straightforward consequence of the existing literature and therefore deserves separate consideration. Moreover, the proof presented here relies on different and more direct techniques, highlighting a new approach to this type of representation.

Theorem 5.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$${\displaystyle {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)=\underset{\beta \in \mathbb{R}}{\sup }{\int }_0^1{\mathbf{b}}_{\mathrm{A}}\left(\tau e^{i\beta }\mathrm{C}+(1-\tau ){\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)d\tau .}$$

(9)

Proof. 

For any $\beta \in \mathbb{R}$ and $\tau \in [0,1]$, we have

$$\begin{array}{cc}\hfill {\mathbf{b}}_{\mathrm{A}}\left(\tau e^{i\beta }\mathrm{C}+(1-\tau ){\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)& \le \tau {\mathbf{b}}_{\mathrm{A}}\left(e^{i\beta }\mathrm{C}\right)+(1-\tau ){\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)\hfill \\ \hfill & =\tau {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)+(1-\tau ){\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right).\hfill \end{array}$$

This gives

$${\displaystyle {\int }_0^1{\mathbf{b}}_{\mathrm{A}}\left(\tau e^{i\beta }\mathrm{C}+(1-\tau ){\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)d\tau \le {\int }_0^1{\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)d\tau ={\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right).}$$

Taking the supremum over $\beta \in \mathbb{R}$ immediately gives

$${\displaystyle \underset{\beta \in \mathbb{R}}{\sup }{\int }_0^1{\mathbf{b}}_{\mathrm{A}}\left(\tau e^{i\beta }\mathrm{C}+(1-\tau ){\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)d\tau \le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right).}$$

(10)

For any $\beta \in \mathbb{R}$, since ${\mathbf{b}}_{\mathrm{A}}$ is a seminorm, we obtain

$$\begin{array}{cc}\hfill {\displaystyle {\int }_0^1{\mathbf{b}}_{\mathrm{A}}\left(\tau e^{i\beta }\mathrm{C}+(1-\tau ){\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)d\tau }& \ge {\mathbf{b}}_{\mathrm{A}}\left({\displaystyle {\int }_0^1\left(\tau e^{i\beta }\mathrm{C}+(1-\tau ){\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)d\tau }\right)\hfill \\ \hfill & ={\mathbf{b}}_{\mathrm{A}}\left({\left[{\displaystyle \frac{{\tau }^2}{2}}{e}^{i\beta }\mathrm{C}+\left(\tau -{\displaystyle \frac{{\tau }^2}{2}}\right){\mathrm{C}}^{{♯}_{\mathrm{A}}}\right]}_0^1\right)\hfill \\ \hfill & ={\mathbf{b}}_{\mathrm{A}}\left({\displaystyle \frac{e^{i\beta }\mathrm{C}+{\mathrm{C}}^{{♯}_{\mathrm{A}}}}{2}}\right)\hfill \\ \hfill & ={\mathbf{b}}_{\mathrm{A}}\left(e^{i{\displaystyle \frac{\beta }{2}}}{\mathrm{Re}}_{\mathrm{A}}\left(e^{i{\displaystyle \frac{\beta }{2}}}\mathrm{C}\right)\right)={\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}\left(e^{i{\displaystyle \frac{\beta }{2}}}\mathrm{C}\right)\right).\hfill \end{array}$$

Taking the supremum over all $\beta \in \mathbb{R}$ on both sides of the inequality sequence and invoking Theorem 4, we deduce the following:

$${\displaystyle \underset{\beta \in \mathbb{R}}{\sup }{\int }_0^1{\mathbf{b}}_{\mathrm{A}}\left(\tau e^{i\beta }\mathrm{C}+(1-\tau ){\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)d\tau \ge \underset{\beta \in \mathbb{R}}{\sup }{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{Re}}_{\mathrm{A}}\left(e^{i\frac{\beta }{2}}\mathrm{C}\right)\right)={\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right).}$$

(11)

Combining (10) and (11) yields (9). □

In Corollary 2.2 of [38], the bounds for the difference between the Berezin norm and Berezin number were incorrectly derived. To provide a rigorous counterpart by properly analyzing the pointwise $\mathrm{A}$-variance, we first introduce the $\mathrm{A}$-normalized Berezin Crawford number.

Definition 2.

Let $\mathrm{C}\in {\mathcal{L}}_{\sqrt{\mathrm{A}}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. The $\mathrm{A}$-normalized Berezin Crawford number of $\mathrm{C}$ is defined as

$${\mathbf{c}}_{{\mathit{ber}}_{\mathrm{A}}}\left(\mathrm{C}\right):=\underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\inf }\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|.$$

Theorem 6.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then, for any $\zeta \in \mathbb{C}$,

$${\displaystyle {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)-{\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\le \underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\left(\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2\right)\le {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}})-{\mathbf{c}}_{{\mathit{ber}}_{\mathrm{A}}}^2(\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}).}$$

Proof. 

Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$, ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ be the corresponding $\mathrm{A}$-normalized reproducing kernel, and let $\zeta \in \mathbb{C}$. We observe that

$$\begin{array}{cc}\hfill {\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2-{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2& =\parallel (\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}+\zeta {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-|{\langle (\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}+\zeta {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2\hfill \\ \hfill & =\parallel (\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-|{\langle (\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2\hfill \\ \hfill & {\displaystyle \le \underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }{\parallel (\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2-\underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\inf }|{\langle (\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2}\hfill \\ \hfill & ={\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}})-{\mathbf{c}}_{{\mathbf{ber}}_{\mathrm{A}}}^2(\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}).\hfill \end{array}$$

Therefore, we deduce that

$$\begin{array}{cc}\hfill {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)-{\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)& {\displaystyle \le \underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\left(\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2\right)}\hfill \\ \hfill & \le {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}})-{\mathbf{c}}_{{\mathbf{ber}}_{\mathrm{A}}}^2(\mathrm{C}-\zeta I_{{\mathfrak{H}}_{\mathsf{\Theta }}}).\hfill \end{array}$$

This immediately completes the proof. □

Our next theorem parameterizes bounds using real and imaginary translations.

Theorem 7.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then, for every $\zeta \in [0,1]$ and $\tau \in \mathbb{R}$,

$${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)\le [{(1-\zeta )}^2+{\zeta }^2]{\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+\zeta {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(\tau I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C})+(1-\zeta ){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(i\tau I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C}).$$

Proof. 

We employ a generalized version of Dragomir’s inequality [1] for the $\mathrm{A}$-semi-inner product. More precisely, it is not difficult to show that for any $a_1,a_2\in {\mathfrak{H}}_{\mathsf{\Theta }}$, $\tau \in \mathbb{R}$, and $\zeta \in [0,1]$,

$$\begin{array}{cc}\hfill & \parallel a_1{\parallel }_{\mathrm{A}}^2{\parallel a_2\parallel }_{\mathrm{A}}^2\hfill \\ \hfill & \le \left[\zeta \parallel \tau a_2-a_1{\parallel }_{\mathrm{A}}^2+(1-\zeta ){\parallel i\tau a_2-a_1\parallel }_{\mathrm{A}}^2\right]{\parallel a_2\parallel }_{\mathrm{A}}^2+{\left[(1-\zeta )\Im {\langle a_1,a_2\rangle }_{\mathrm{A}}+\zeta \Re {\langle a_1,a_2\rangle }_{\mathrm{A}}\right]}^2\hfill \\ \hfill & \le \left[\zeta \parallel \tau a_2-a_1{\parallel }_{\mathrm{A}}^2+(1-\zeta ){\parallel i\tau a_2-a_1\parallel }_{\mathrm{A}}^2\right]{\parallel a_2\parallel }_{\mathrm{A}}^2+[{(1-\zeta )}^2+{\zeta }^2]|{\langle a_1,a_2\rangle }_{\mathrm{A}}{|}^2.\hfill \end{array}$$

Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ be the corresponding $\mathrm{A}$-normalized reproducing kernel. Substituting $a_1=\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ and $a_2={\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ (noting that $\parallel a_2{\parallel }_{\mathrm{A}}=1$) yields

$$\begin{array}{cc}\hfill {\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2& \le \zeta \parallel \tau {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2+(1-\zeta ){\parallel i\tau {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2+[{(1-\zeta )}^2+{\zeta }^2]|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2\hfill \\ \hfill & \le \zeta {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(\tau I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C})+(1-\zeta ){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(i\tau I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C})+[{(1-\zeta )}^2+{\zeta }^2]{\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right).\hfill \end{array}$$

Taking the supremum over all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ establishes the result. □

By evaluating $\zeta$ at specific values ($\zeta =0,1,{\displaystyle \frac{1}{2}}$), we immediately obtain the following constraints.

Corollary 4.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$$\begin{array}{cc}\hfill {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)& \le {\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+\underset{\tau \in \mathbb{R}}{\inf }{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(\tau I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C}),\hfill \end{array}$$

(12)

$$\begin{array}{ccc} & \hfill {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)& \le {\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+\underset{\tau \in \mathbb{R}}{\inf }{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(i\tau I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C}),\hfill \\ & \hfill {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)& \le {\displaystyle \frac{1}{2}}{\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)+{\displaystyle \frac{1}{2}}\underset{\tau \in \mathbb{R}}{\inf }\left[{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(\tau I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C})+{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(i\tau I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C})\right].\hfill \end{array}$$

(13)

We conclude this section with an upper bound for the $\mathrm{A}$-variance using an elementary reverse Cauchy–Schwarz inequality.

Proposition 1.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then, for any $z\in \mathbb{C}$,

$${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)-{\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\le {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-z\mathrm{C}).$$

Proof. 

It can be checked that for any $a_1,a_2\in {\mathfrak{H}}_{\mathsf{\Theta }}$ and $z\in \mathbb{C}$, we have

$$\parallel a_1{\parallel }_{\mathrm{A}}^2\parallel a_2{\parallel }_{\mathrm{A}}^2-|{\langle a_1,a_2\rangle }_{\mathrm{A}}{|}^2\le \parallel a_2{\parallel }_{\mathrm{A}}^2{\parallel a_1-z{a}_2\parallel }_{\mathrm{A}}^2.$$

Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ be the corresponding $\mathrm{A}$-normalized reproducing kernel. By setting $a_1={\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ and $a_2=\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$, where $\parallel a_1{\parallel }_{\mathrm{A}}=1$, we obtain the following:

$$\begin{array}{cc}\hfill \parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2& \le {\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2{\parallel {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-z\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\hfill \\ \hfill & \le {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(I_{{\mathfrak{H}}_{\mathsf{\Theta }}}-z\mathrm{C}).\hfill \end{array}$$

Furthermore, we have

$${\displaystyle {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)-{\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\le \underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }(\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2).}$$

This immediately proves the desired result. □

Corollary 5.

If there exists ${\tau }_0\in \mathbb{R}$ such that ${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2({\tau }_0{I}_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C})\le \beta$ or ${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(i{\tau }_0{I}_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C})\le \beta$, then

$${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)-{\mathbf{b}}_{\mathrm{A}}^2\left(\mathrm{C}\right)\le \beta .$$

Here, $\beta \ge 0$ geometrically bounds the maximal $\mathrm{A}$-variance of $\mathrm{C}$:

$${\displaystyle \underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\left(\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2\right)\le \underset{z\in \{{\tau }_0,i{\tau }_0\}}{\min }{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^2(z{I}_{{\mathfrak{H}}_{\mathsf{\Theta }}}-\mathrm{C})\le \beta .}$$

Proof. 

Substituting $\tau ={\tau }_0$ into either inequality (12) or (13) immediately establishes the bound. □

We now extend our pointwise framework to establish parameterized inequalities and higher-power bounds.

Theorem 8.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then, for every $r\ge 1$ and $\zeta \in [0,1]$,

$${\mathbf{b}}_{\mathrm{A}}^{2r}\left(\mathrm{C}\right)\le {\displaystyle \frac{\zeta }{2}}{\mathbf{b}}_{\mathrm{A}}^r\left({\mathrm{C}}^2\right)+\left(1-{\displaystyle \frac{\zeta }{2}}\right){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^r\left(\mathrm{C}\right){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^r\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\right).$$

Proof. 

By taking into account a refinement of the Cauchy–Schwarz inequality [41] (Corollary 2.5), we can prove that for any $a_1,a_2,a_3\in {\mathfrak{H}}_{\mathsf{\Theta }}$ with $\parallel a_3{\parallel }_{\mathrm{A}}=1$ and $\zeta \in [0,1]$, we have

$$\left|{\langle a_1,a_3\rangle }_{\mathrm{A}}{\langle a_3,a_2\rangle }_{\mathrm{A}}\right|\le {\displaystyle \frac{\zeta }{2}}\left|{\langle a_1,a_2\rangle }_{\mathrm{A}}\right|+\left(1-{\displaystyle \frac{\zeta }{2}}\right)\parallel a_1{\parallel }_{\mathrm{A}}{\parallel a_2\parallel }_{\mathrm{A}}.$$

(14)

Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$. Setting $a_3={\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$, $a_1=\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$, and $a_2={\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ in (14) yields the following:

$$\begin{array}{cc}\hfill |{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2& =\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\langle {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|\hfill \\ \hfill & \le {\displaystyle \frac{\zeta }{2}}\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|+\left(1-{\displaystyle \frac{\zeta }{2}}\right)\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\hfill \\ \hfill & ={\displaystyle \frac{\zeta }{2}}\left|{\langle {\mathrm{C}}^2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|+\left(1-{\displaystyle \frac{\zeta }{2}}\right)\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}.\hfill \end{array}$$

Since $f\left(x\right)=x^r$ is convex and increasing on $[0,\infty )$ for $r\ge 1$, applying it to both sides and using Jensen’s inequality gives

$$\begin{array}{cc}\hfill |{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^{2r}& \le {\displaystyle \frac{\zeta }{2}}|{\langle {\mathrm{C}}^2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^r+\left(1-{\displaystyle \frac{\zeta }{2}}\right)\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^r{\parallel {\mathrm{C}}^{{♯}_{\mathrm{A}}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^r\hfill \\ \hfill & \le {\displaystyle \frac{\zeta }{2}}{\mathbf{b}}_{\mathrm{A}}^r\left({\mathrm{C}}^2\right)+\left(1-{\displaystyle \frac{\zeta }{2}}\right){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^r\left(\mathrm{C}\right){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}^r\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\right).\hfill \end{array}$$

Taking the supremum over all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ provides the desired inequality. □

Remark 7.

By fixing $r=1$ and $\zeta =1$, Theorem 8 provides the refined estimate:

$${\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\left({\displaystyle \frac{1}{2}}{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)+{\displaystyle \frac{1}{2}}{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right){\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left({\mathrm{C}}^{{♯}_{\mathrm{A}}}\right)\right)}^{{\displaystyle \frac{1}{2}}}\le {\left({\displaystyle \frac{1}{2}}{\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}^2\right)+{\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\mathrm{A}}^2\right)}^{{\displaystyle \frac{1}{2}}}.$$

In Theorem 2.12 of [38], an upper bound for the Berezin number was proposed using the Crawford number ${\mathbf{c}}_{{\mathbf{ber}}_{\mathrm{A}}}\left(\mathrm{C}\right)$. The author’s proof incorrectly assumed that for the $\mathrm{A}$-Cartesian parts ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$, the bounds $\parallel {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}\le {\mathbf{b}}_{\mathrm{A}}\left({\mathrm{C}}_1\right)\le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)$ hold. This sequence of inequalities is false because the Berezin number does not bound the pointwise norm of an $\mathrm{A}$-selfadjoint operator evaluation vector.

To rectify this mathematically, we employ the $\mathrm{A}$-normalized Berezin operator radius ${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)$. To bound the Cartesian components, we define

$${\displaystyle M_{\mathrm{A}}\left(\mathrm{C}\right):=\underset{\theta \in \mathbb{R}}{\sup }{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left({\mathrm{Re}}_{\mathrm{A}}\left(e^{i\theta }\mathrm{C}\right)\right).}$$

Before establishing our main theorem for this section, we formally define $\mathrm{A}$-orthogonality and state the corresponding Pythagorean theorem in semi-Hilbertian spaces.

Definition 3.

Two vectors $\xi ,\eta \in {\mathfrak{H}}_{\mathsf{\Theta }}$ are said to be $\mathrm{A}$-orthogonal, written $\xi {\perp }_{\mathrm{A}}\eta$, if ${\langle \xi ,\eta \rangle }_{\mathrm{A}}=0$.

The next lemma provides the Pythagorean identity for $\mathrm{A}$-orthogonal vectors.

Lemma 3.

Let $\xi ,\eta \in {\mathfrak{H}}_{\mathsf{\Theta }}$ with $\xi {\perp }_{\mathrm{A}}\eta$. Then,

$${\parallel \xi +\eta \parallel }_{\mathrm{A}}^2={\parallel \xi \parallel }_{\mathrm{A}}^2+{\parallel \eta \parallel }_{\mathrm{A}}^2.$$

Proof. 

We see that

$$\begin{array}{cc}\hfill {\parallel \xi +\eta \parallel }_{\mathrm{A}}^2& ={\langle \xi +\eta ,\xi +\eta \rangle }_{\mathrm{A}}\hfill \\ \hfill & ={\parallel \xi \parallel }_{\mathrm{A}}^2+2\Re {\langle \xi ,\eta \rangle }_{\mathrm{A}}+{\parallel \eta \parallel }_{\mathrm{A}}^2\hfill \\ \hfill & ={\parallel \xi \parallel }_{\mathrm{A}}^2+{\parallel \eta \parallel }_{\mathrm{A}}^2,\hfill \end{array}$$

where the last equality follows from ${\langle \xi ,\eta \rangle }_{\mathrm{A}}=0$. □

Theorem 9.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$. Then,

$${\displaystyle \frac{1}{2}}{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)\le \sqrt{{\displaystyle \frac{M_{\mathrm{A}}^2\left(\mathrm{C}\right)}{2}}+{\displaystyle \frac{M_{\mathrm{A}}\left(\mathrm{C}\right)}{2}}\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\mathbf{c}}_{{\mathbf{ber}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)}}\le M_{\mathrm{A}}\left(\mathrm{C}\right).$$

Proof. 

Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ be the $\mathrm{A}$-normalized reproducing kernel. Write

$${\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}=r{e}^{-i\theta },r=\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|.$$

Set ${\mathrm{C}}_1={\mathrm{Re}}_{\mathrm{A}}\left(e^{i\theta }\mathrm{C}\right)$ and ${\mathrm{C}}_2={\mathrm{Im}}_{\mathrm{A}}\left(e^{i\theta }\mathrm{C}\right)$, so $e^{i\theta }\mathrm{C}={\mathrm{C}}_1+i{\mathrm{C}}_2$. Since ${\mathrm{C}}_1,{\mathrm{C}}_2$ are $\mathrm{A}$-selfadjoint,

$${\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}+i{\langle {\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}=r.$$

Hence,

$${\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}=r,{\langle {\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}=0.$$

By observing that $\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}={\parallel e^{i\theta }\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}$, we can see that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2& ={\displaystyle \frac{1}{4}}{\parallel ({\mathrm{C}}_1+i{\mathrm{C}}_2){\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\hfill \\ \hfill & ={\displaystyle \frac{1}{4}}{\parallel \left({\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}+i{\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\right)+{\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2.\hfill \end{array}$$

Since ${\langle {\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}=0$, the vector $({\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}+i{\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}})$ is $\mathrm{A}$-orthogonal to ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$. Thus, applying Lemma 3 and the triangle inequality yields the following:

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2& ={\displaystyle \frac{1}{4}}\left(\parallel {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}+i{\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2+{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}\left({\left(\parallel {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}+{\parallel {\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\right)}^2+{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2\right).\hfill \end{array}$$

Using the identity

$${\parallel {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2={\parallel {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2-{\langle {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}^2,$$

we obtain the following:

$${\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\le {\displaystyle \frac{1}{4}}\left({\left(\sqrt{\parallel {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2}+{\parallel {\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\right)}^2+{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2\right).$$

By definition, ${\parallel {\mathrm{C}}_1{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\le {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left({\mathrm{C}}_1\right)\le M_{\mathrm{A}}\left(\mathrm{C}\right)$ and ${\parallel {\mathrm{C}}_2{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\le {\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left({\mathrm{C}}_2\right)\le M_{\mathrm{A}}\left(\mathrm{C}\right)$. The map $t\mapsto {(\sqrt{t^2-a^2}+t)}^2$ is monotonically increasing with respect to t for $t\ge a\ge 0$, allowing us to deduce that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2& \le {\displaystyle \frac{1}{4}}\left({\left(\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2}+M_{\mathrm{A}}\left(\mathrm{C}\right)\right)}^2+{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2\right)\hfill \\ \hfill & ={\displaystyle \frac{M_{\mathrm{A}}^2\left(\mathrm{C}\right)}{2}}+{\displaystyle \frac{M_{\mathrm{A}}\left(\mathrm{C}\right)}{2}}\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2}.\hfill \\ \hfill & \le {\displaystyle \frac{M_{\mathrm{A}}^2\left(\mathrm{C}\right)}{2}}+{\displaystyle \frac{M_{\mathrm{A}}\left(\mathrm{C}\right)}{2}}\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\mathbf{c}}_{{\mathbf{ber}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)}.\hfill \end{array}$$

Taking the square root and subsequently taking the supremum over all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ gives

$${\displaystyle \frac{1}{2}}{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)\le \sqrt{{\displaystyle \frac{M_{\mathrm{A}}^2\left(\mathrm{C}\right)}{2}}+{\displaystyle \frac{M_{\mathrm{A}}\left(\mathrm{C}\right)}{2}}\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\mathbf{c}}_{{\mathbf{ber}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)}}.$$

The rightmost inequality follows trivially, since $\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\mathbf{c}}_{{\mathbf{ber}}_{\mathrm{A}}}^2\left(\mathrm{C}\right)}\le M_{\mathrm{A}}\left(\mathrm{C}\right)$. Hence, the proof is complete. □

Classically, $\frac{1}{2}\parallel \mathrm{C}\parallel \le \omega \left(\mathrm{C}\right)$. In Theorem 2.13 of [38], the analogous bound

$$\frac{1}{2}{\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}\le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)$$

was assumed, but it is false in general, even for $\mathrm{A}=I_{{\mathfrak{H}}_{\mathsf{\Theta }}}$ (see [29]).

Classically, $\frac{1}{2}\parallel \mathrm{C}\parallel \le \omega \left(\mathrm{C}\right)$. In Theorem 2.13 of [38], the analogous bound

$${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}\le {\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)$$

was assumed, but it is false in general. To prove this, let ${\mathfrak{H}}_{\mathsf{\Theta }}={\mathbb{C}}^2$ equipped with the standard basis $\{e_1,e_2\}$ as reproducing kernels. Let $\mathrm{A}=I_{{\mathfrak{H}}_{\mathsf{\Theta }}}$ and $\mathrm{C}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Since $\mathrm{A}=I_{{\mathfrak{H}}_{\mathsf{\Theta }}}$, the two normalization approaches coincide. Since $\mathrm{C}$ is nilpotent, $\mathrm{C}{e}_1=0$ and $\mathrm{C}{e}_2=e_1$. By using the definition of the $\mathrm{A}$-Berezin number, we obtain ${\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)=0$. By using the definition of the modified $\mathrm{A}$-Berezin norm, we obtain ${\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}=1$. The assumed bound then implies ${\displaystyle \frac{1}{2}}\le 0$, which is absurd. This clearly justifies the necessity of using the $\mathrm{A}$-Berezin radius ${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)$ for our estimates.

Therefore, we work with pointwise estimates via the $\mathrm{A}$-Berezin radius ${\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)$. We introduce the active-domain $\mathrm{A}$-cosine and $\mathrm{A}$-sine. To avoid division by zero, we impose the restriction $\xi \notin \ker \left(\mathrm{A}\right)$, since ${\parallel \xi \parallel }_{\mathrm{A}}=0$ iff $\xi \in \ker \left(\mathrm{A}\right)$.

Definition 4.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$ such that $\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\notin \ker \left(\mathrm{A}\right)$ for all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$. The active-domain $\mathrm{A}$-cosine and $\mathrm{A}$-sine of $\mathrm{C}$, respectively, are defined as

$$\begin{array}{cc}\hfill {\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)& :=\underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\inf }{\displaystyle \frac{|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}|}{{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}}},\hfill \\ \hfill {\mathrm{Sin}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)& :=\underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\sqrt{1-{\displaystyle \frac{|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2}{{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2}}}.\hfill \end{array}$$

With these trigonometric quantities established, we formulate the following exact geometric bounds.

Theorem 10.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$ such that $\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\notin \ker \left(\mathrm{A}\right)$ for all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$. Then, the $\mathrm{A}$-Berezin number satisfies the following lower bound:

$$\begin{array}{cc}\hfill {\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)·{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)& \le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right).\hfill \end{array}$$

(15)

Furthermore, we have

$$\begin{array}{cc}\hfill \underset{\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}& \le {\mathrm{Sin}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)·{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right).\hfill \end{array}$$

(16)

Proof. 

Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ be the $\mathrm{A}$-normalized reproducing kernel. From Definition 4, we infer that

$$\begin{array}{cc}\hfill |{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}|& \ge {\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right){\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}.\hfill \end{array}$$

Taking the supremum over all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ directly yields the inequality (15).

To prove (16), we write $\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ as follows:

$$\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}={\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}+\left(\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\right).$$

It can be seen that $\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ is $\mathrm{A}$-orthogonal to ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$.

Since $\parallel {\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}=1$, by applying Lemma 3, we see that

$$\begin{array}{cc}\hfill {\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}-{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2& =\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2\hfill \\ \hfill & =\left(1-{\displaystyle \frac{|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2}{{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2}}\right){\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\hfill \\ \hfill & \le {\left({\mathrm{Sin}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2.\hfill \end{array}$$

Taking square roots and then the supremum over $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ gives (16). □

In connection with the previous estimates, we must explicitly mention that Theorem 2.13 from Huban [38] is mathematically incorrect. The theorem claims that for any operator $\mathrm{C}$, the following inequality holds:

$${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}\le \max \left\{{\mathrm{Sin}}_{\mathrm{A}}\left(\mathrm{C}\right),{\displaystyle \frac{\sqrt{2}}{2}}\right\}{\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right).$$

Since this chain implies ${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{\widetilde{{\mathbf{ber}}_{\mathrm{A}}}}\le {\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)$, it is false, as previously proved.

To provide a correct alternative to Theorem 2.13, we replace the quantity ${\mathbf{ber}}_{\mathrm{A}}\left(\mathrm{C}\right)$ with $M_{\mathrm{A}}\left(\mathrm{C}\right)$ and use our robust $\mathrm{A}$-normalized quantities. We present the corrected theorem below.

Theorem 11.

Let $\mathrm{C}\in {\mathcal{L}}_{\mathrm{A}}\left({\mathfrak{H}}_{\mathsf{\Theta }}\right)$ such that $\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\notin \ker \left(\mathrm{A}\right)$ for all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$. Then,

$${\displaystyle \frac{1}{2}}{\mathbf{N}}_{{\mathbf{b}}_{\mathrm{A}}}\left(\mathrm{C}\right)\le \max \left\{{\mathrm{Sin}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right),{\displaystyle \frac{\sqrt{2}}{2}}\right\}{M}_{\mathrm{A}}\left(\mathrm{C}\right).$$

Proof. 

Let $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$ and let ${\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}$ be the $\mathrm{A}$-normalized reproducing kernel. By using the proof of Theorem 9, we obtain

$${\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\le {\displaystyle \frac{M_{\mathrm{A}}^2\left(\mathrm{C}\right)}{2}}+{\displaystyle \frac{M_{\mathrm{A}}\left(\mathrm{C}\right)}{2}}\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|}^2}.$$

Since $\left|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\rangle }_{\mathrm{A}}\right|\ge {\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right){\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}$ by Definition 4, we can deduce the following:

$${\displaystyle \frac{1}{4}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\le {\displaystyle \frac{M_{\mathrm{A}}^2\left(\mathrm{C}\right)}{2}}+{\displaystyle \frac{M_{\mathrm{A}}\left(\mathrm{C}\right)}{2}}\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\left({\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2}.$$

By using algebraic rearrangement, we obtain

$$\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-2M_{\mathrm{A}}^2\left(\mathrm{C}\right)\le 2M_{\mathrm{A}}\left(\mathrm{C}\right)\sqrt{M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\left({\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2}.$$

We consider two cases.

Case 1. Assume $\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-2M_{\mathrm{A}}^2\left(\mathrm{C}\right)\le 0$. Since $\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}\le \sqrt{2}{M}_{\mathrm{A}}\left(\mathrm{C}\right)$, we obtain

$${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\le {\displaystyle \frac{\sqrt{2}}{2}}{M}_{\mathrm{A}}\left(\mathrm{C}\right).$$

Case 2. Assume $\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2-2M_{\mathrm{A}}^2\left(\mathrm{C}\right)>0$. By squaring both sides of the rearranged inequality, we obtain

$$\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^4-4M_{\mathrm{A}}^2\left(\mathrm{C}\right){\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2+4M_{\mathrm{A}}^4\left(\mathrm{C}\right)\le 4M_{\mathrm{A}}^2\left(\mathrm{C}\right)\left(M_{\mathrm{A}}^2\left(\mathrm{C}\right)-{\left({\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\right).$$

This simplifies to

$$\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^4-4M_{\mathrm{A}}^2\left(\mathrm{C}\right)\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2+4M_{\mathrm{A}}^2\left(\mathrm{C}\right){\left({\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}^2\le 0.$$

Since $\mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\notin \ker \left(\mathrm{A}\right)$, it follows that $\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}>0$. By dividing by $\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2$, we obtain

$$\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2\le 4M_{\mathrm{A}}^2\left(\mathrm{C}\right)\left(1-{\left({\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2\right).$$

Since ${\displaystyle {\left({\mathrm{Sin}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2=\underset{\delta \in {\mathsf{\Theta }}_{\mathrm{A}}}{\sup }\left(1-\frac{|{\langle \mathrm{C}{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}},{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}\rangle }_{\mathrm{A}}{|}^2}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\delta }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2}\right)=1-{\left({\mathrm{Cos}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2}$, we can infer the following:

$$\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}{\parallel }_{\mathrm{A}}^2\le 4M_{\mathrm{A}}^2\left(\mathrm{C}\right){\left({\mathrm{Sin}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)\right)}^2.$$

By taking the square root, we obtain ${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\le {\mathrm{Sin}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right)M_{\mathrm{A}}\left(\mathrm{C}\right)$.

Combining both cases, we deduce that for any $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$,

$${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}{\widehat{\mathrm{x}}}_{\gamma }^{\mathrm{A}}\parallel }_{\mathrm{A}}\le \max \left\{{\mathrm{Sin}}_{\mathrm{A}}^{\mathbf{b}}\left(\mathrm{C}\right),{\displaystyle \frac{\sqrt{2}}{2}}\right\}{M}_{\mathrm{A}}\left(\mathrm{C}\right).$$

By taking the supremum over all $\gamma \in {\mathsf{\Theta }}_{\mathrm{A}}$, we obtain the final result. □

Our next aim is to respectfully demonstrate that Corollary 2.8 and Theorem 2.11 in [38] are also incorrect.

Corollary 2.8 in [38] claims that

$${\displaystyle \frac{1}{2}}{\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}\le {\mathbf{b}}_{\mathrm{A}}\left(\mathrm{C}\right)\le {\parallel \mathrm{C}\parallel }_{{\mathbf{b}}_{\mathrm{A}}}.$$

This formula corrects the error in Theorem 2.11.

3. Conclusions

In this paper, we have established new, sharp operator bounds and inequalities involving the $\mathrm{A}$-normalized Berezin number and the $\mathrm{A}$-normalized Berezin norm in semi-Hilbertian spaces. We have rigorously corrected several inconsistencies present in the recent literature by properly utilizing the $\mathrm{A}$-Cartesian decomposition and the generalized Buzano inequality. Correcting the research in this direction is of great importance, as it builds a strong and reliable mathematical foundation for the field.

Furthermore, extending these novel abstract operator techniques to concrete physical models and differential equations represents a highly promising direction for future research. For instance, exploring applications within specific reproducing kernel Hilbert spaces could yield significant insights, as demonstrated in recent applied studies [37].

Overall, the consistent mathematical framework developed herein provides a solid foundation for further investigations into generalized Berezin symbols and their associated geometric properties. We hope that this work will serve as a strong starting point to inspire other elegant research papers in the future.