Some New Estimates for the Berezin Number of Hilbert Space Operators

Abstract

In this paper, we have developed new estimates of some estimates involving the Berezin norm and Berezin number of bounded linear operators defined on a reproducing kernel Hilbert space ${\mathcal{H}}_{\mathrm{\Omega }}$. The uniqueness or novelty of this article consists of new estimates of Berezin numbers for different types of operators. These estimates improve the upper bounds of the Berezin numbers obtained by other similar papers. We give several upper bounds for ${\mathbf{ber}}^r\left(S^{*}T\right)$, where $T,S\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ and $r\ge 1$. We also present an estimation of ${\mathbf{ber}}^{2r}\left({\sum }_{i=1}^d{T}_i\right)$ where $T_i\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$, $i=\overline{1,d}$ and $r\ge 1$. Some of the obtained inequalities represent improvements to earlier ones. In this work, the ideas and methodologies presented may serve as a starting point for future investigation in this field.

1. Introduction

In the literature related to operator theory, the Berezin norm and Berezin number of an operator have been studied for their many applications in engineering, quantum computing, quantum mechanics, numerical analysis, differential equations, etc. To characterize the Berezin number and the Berezin norm, we first present some concepts and properties of bounded linear operators on a Hilbert space.

Let $\mathcal{H}$ be a complex Hilbert space, endowed with the inner product $⟨·,·⟩$ and associated norm $\parallel ·\parallel$. Let $\mathbb{B}\left(\mathcal{H}\right)$ denote the $C^{*}$-algebra of all bounded linear operators on $\mathcal{H}$. An operator $T\in \mathbb{B}\left(\mathcal{H}\right)$ is called positive if $⟨Tx,x⟩\ge 0$ for all $x\in \mathcal{H}$, and then we write $T\ge 0$. If a bounded linear operator T on $\mathcal{H}$ is positive, then there exists a unique positive bounded linear operator denoted by $T^{1/2}$ such that $T={\left(T^{1/2}\right)}^2$. Furthermore, the absolute value of T, denoted by $\left|T\right|$, is defined by $\left|T\right|={\left(T^{*}T\right)}^{1/2}$. We remark that $\left|T\right|\ge 0$. For $T\in \mathbb{B}\left(\mathcal{H}\right)$, we have the following numerical values: the operator norm $\parallel T\parallel$ given by $\parallel T\parallel =sup\left\{\parallel Tx\parallel ;x\in \mathcal{H},\parallel x\parallel =1\right\}$ and the numerical radius of the operator T defined by $\omega \left(T\right)=sup\left\{\left|⟨Tx,x⟩\right|;x\in \mathcal{H},\parallel x\parallel =1\right\}$. It is easy to see that $\omega \left(T\right)\le \parallel T\parallel$. If T is a normal operator, i.e., $T^{*}T=T{T}^{*}$, then $\omega \left(T\right)=\parallel T\parallel$. The operator norm and the numerical radius norm are equivalent, because $\frac{\parallel T\parallel }{2}\le \omega \left(T\right)\le \parallel T\parallel$ for every $T\in \mathbb{B}\left(\mathcal{H}\right)$. We also have $\omega \left(T\right)\le \omega \left(\right|T\left|\right)$. Recent contributions concerning numerical radius inequalities of Hilbert space operators can be found in [1] and references therein. Properties of some operators on Hilbert spaces can be found in [2,3].

Let $\mathrm{\Omega }$ be a non-empty set and $\mathcal{F}\left(\mathrm{\Omega }\right)$ be the set of all functions from $\mathrm{\Omega }$ to $\mathbb{C}$, where $\mathbb{C}$ is the field of all complex numbers, a set ${\mathcal{H}}_{\mathrm{\Omega }}$ included in $\mathcal{F}\left(\mathrm{\Omega }\right)$ is called a reproducing kernel Hilbert space (RKHS for short) on $\mathrm{\Omega }$ if ${\mathcal{H}}_{\mathrm{\Omega }}$ is a Hilbert space (with identity $I_{\mathrm{\Omega }}$) and for every $\lambda \in \mathrm{\Omega }$, the linear evaluation functional $E_{\lambda }:{\mathcal{H}}_{\mathrm{\Omega }}\to \mathbb{C}$ given by $E_{\lambda }\left(f\right)=f\left(\lambda \right)$ is bounded. Using the Riesz representation theorem, we show that for each $\lambda \in \mathrm{\Omega }$, there exists a unique vector $k_{\lambda }\in {\mathcal{H}}_{\mathrm{\Omega }}$ such that $f\left(\lambda \right)=E_{\lambda }\left(f\right)=⟨f,k_{\lambda }⟩$ for all $f\in {\mathcal{H}}_{\mathrm{\Omega }}$. Here, the function $k_{\lambda }$ is called the reproducing kernel for the element $\lambda$ and the set $\{k_{\lambda };\lambda \in \mathrm{\Omega }\}$ is called the reproducing kernel of ${\mathcal{H}}_{\mathrm{\Omega }}.$ We denote by ${\widehat{k}}_{\lambda }=\frac{k_{\lambda }}{\parallel k_{\lambda }\parallel }$, for $\lambda \in \mathrm{\Omega }$, the normalized reproducing kernel of ${\mathcal{H}}_{\mathrm{\Omega }}$. Note that the set $\{{\widehat{k}}_{\lambda }:\lambda \in \mathrm{\Omega }\}$ is a total set in ${\mathcal{H}}_{\mathrm{\Omega }}$. For $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$, the Berezin symbol (or Berezin transform) of T, which was first introduced by Berezin [4,5], is the bounded function $\tilde{T}:\mathrm{\Omega }\to \mathbb{C}$ defined by $\tilde{T}\left(\lambda \right)=⟨T\widehat{k_{\lambda }},\widehat{k_{\lambda }}⟩$. If the operator T is selfadjoint ($T=T^{*}$) then $\tilde{T}\left(\lambda \right)\in \mathbb{R}$, and if the operator T is positive then $\tilde{T}\left(\lambda \right)\ge 0$.

The Berezin symbol has been investigated in detail for the Toeplitz and Hankel operators on the Hardy and Bergman spaces. It is widely applied in various areas of analysis and uniquely determines an operator (i.e., for all $\lambda \in \mathrm{\Omega },\tilde{T}\left(\lambda \right)=\tilde{S}\left(\lambda \right)$ implies $T=S$). For further information about the Berezin symbol we refer the reader to [6,7,8,9] and references therein.

The Berezin set and the Berezin number of an operator T are, respectively, defined by

$$\begin{array}{c}\hfill \mathbf{Ber}\left(T\right):=\left\{\tilde{T}\left(\lambda \right);\lambda \in \mathrm{\Omega }\right\}\mathrm{and}\mathbf{ber}\left(T\right):=\underset{\lambda \in \mathrm{\Omega }}{sup}|\tilde{T}\left(\lambda \right)|=\underset{\lambda \in \mathrm{\Omega }}{sup}|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩|.\end{array}$$

Through some simple calculations we get $0\le \mathbf{ber}\left(T\right)\le \omega \left(T\right)\le \parallel T\parallel$, for all $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right).$ Karaev [10] showed that $\frac{\parallel T\parallel }{2}\le \mathbf{ber}\left(T\right)$ does not hold for every $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$.

It is easy to see that $\mathbf{ber}\left(I_{\mathrm{\Omega }}\right)=1$ and $\left|⟨T{k}_{\lambda },k_{\lambda }⟩\right|\le \mathbf{ber}\left(T\right){\parallel k_{\lambda }\parallel }^2$, for any reproducing kernel $k_{\lambda }$. Here $I_{\mathrm{\Omega }}$ denotes the identity operator on ${\mathcal{H}}_{\mathrm{\Omega }}$.

We remark that for every $T,S\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$, we have the following properties:

(i)

$\mathbf{ber}\left(\alpha T\right)=\left|\alpha \right|\mathbf{ber}\left(T\right)$ for all $\alpha \in \mathbb{C}$,

(ii)

$\mathbf{ber}(T+S)\le \mathbf{ber}\left(T\right)+\mathbf{ber}\left(S\right)$.

Furthermore, it can be checked that $\mathbf{ber}\left(T\right)=0$ if and only if T is equal to the zero function of ${\mathcal{H}}_{\mathrm{\Omega }}$. Hence, using the above considerations regarding the Berezin number, it follows that ber(·) is a norm on $\mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right).$ Similarly for $\mathbf{ber}\left(T\right)$, as was proven in [11]: $c\left(T\right):=inf\left\{|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩|:\lambda \in \mathrm{\Omega }\right\}$.

For $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$, the Berezin norm of T is given by

$${\parallel T\parallel }_{\mathbf{ber}}:=sup\left\{|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\mu }⟩|;\lambda ,\mu \in \mathrm{\Omega }\right\},$$

where ${\widehat{k}}_{\lambda }$ and ${\widehat{k}}_{\mu }$ are two normalized reproducing kernels of the space ${\mathcal{H}}_{\mathrm{\Omega }}$ (see [12,13]). We note that ${\parallel ·\parallel }_{\mathbf{ber}}$ does not, in general, imply the submultiplicativity property. The equality ${\parallel T\parallel }_{\mathbf{ber}}={sup}_{\lambda \in \mathrm{\Omega }}\parallel T{\widehat{k}}_{\lambda }\parallel$ may not hold in general for $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ (see [14]) as well. An important observation is that

$$\mathbf{ber}\left(T\right)\le {\parallel T\parallel }_{\mathbf{ber}}\le \parallel T\parallel ,\forall T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right).$$

(1)

It should be mentioned here that inequality (1) is strict in general. However, Bhunia et al. proved in [15] that if $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ is a positive operator, then

$$\mathbf{ber}\left(T\right)={\parallel T\parallel }_{\mathbf{ber}}.$$

(2)

Remark 1. 

It is crucial to note that (2) may not be true, in general, for selfadjoint operators (see [15]).

The following inequality, which was recently proven in (1), provides a refinement of the inequality $\mathbf{ber}\left(T\right)\le \parallel T\parallel$.

$${\mathbf{ber}}^2\left(T\right)\le \frac{1}{2}{\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}},\forall T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right).$$

(3)

The uniqueness or novelty of this article consists of new estimates of the Berezin number and Berezin norm of different types of bounded linear operators acting on reproducing kernel Hilbert space (RKHS). These estimates improve the upper bounds of the Berezin numbers obtained by other similar articles.

The present article is organized in the following manner: In Section 2, some lemmas that are required to establish our main results are collected. In Section 3, we present our main results, which include several Berezin number and norm inequalities of operators. In particular, we establish two estimations of ${\mathbf{ber}}^{2r}\left({\sum }_{i=1}^d{T}_i\right)$ where $T_i\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ for all $i\in \{1,\dots ,d\}$ and $r\ge 1$ with $d\in {\mathbb{N}}^{*}$. Here ${\mathbb{N}}^{*}$ denotes the set of all positive integers. Some of the obtained bounds improve on the earlier ones.

2. Useful Lemmas

In this section, we collect some well-known useful lemmas, which will be used repeatedly to reach our goal in this present paper. In all that follows, $\mathcal{H}$ stands for a complex Hilbert space with inner product $⟨·,·⟩$ and associated norm $\parallel ·\parallel$.

The first lemma was proven by Kittaneh and Moradi in [16] and provides a refinement of the well-known Cauchy–Schwarz inequality.

Lemma 1. 

Let $x,y\in \mathcal{H}$. Then

$$\begin{array}{c}\hfill {\left|⟨x,y⟩\right|}^2\le \left|⟨x,y⟩\right|∥x∥∥y∥+\frac{1}{2}\left({∥x∥}^2{∥y∥}^2-{\left|⟨x,y⟩\right|}^2\right)\le {∥x∥}^2{∥y∥}^2.\end{array}$$

Remark 2. 

It follows from Lemma 1 that

$$\begin{array}{c}\hfill |⟨x,y⟩{|}^2\le \frac{1}{3}{\parallel x\parallel }^2{\parallel y\parallel }^2+\frac{2}{3}|⟨x,y⟩|\parallel x\parallel \parallel y\parallel ,\end{array}$$

(4)

for all $x,y\in \mathcal{H}$.

The classical Schwarz inequality for positive operators is given below:

$${\left|⟨Tx,y⟩\right|}^2\le ⟨Tx,x⟩⟨Ty,y⟩,$$

(5)

for any positive operator $T\in \mathbb{B}\left(\mathcal{H}\right)$ and for any vectors $x,y\in \mathcal{H}$. Kato [17] established a companion of the Schwarz inequality (5), which asserts:

$${|⟨Tx,y⟩|}^2\le {⟨|T|}^{2\theta }x,x⟩⟨|T^{*}{|}^{2(1-\theta )}y,y⟩,$$

(6)

for every operator $T\in \mathbb{B}\left(\mathcal{H}\right)$, for any vectors $x,y\in \mathcal{H}$, and $\theta \in [0,1]$. For $\theta =\frac{1}{2}$ we obtain a result attributed to Halmos [18] (pp. 75–76), thus

$$\begin{array}{c}\hfill \left|⟨Tx,x⟩\right|\le \sqrt{⟨\left|T\right|x,x⟩⟨\left|T^{*}\right|x,x⟩},\end{array}$$

(7)

for every $T\in \mathbb{B}\left(\mathcal{H}\right)$ and for all $x,y\in \mathcal{H}$.

The inequality in the following lemma deals with positive operators and is known as the McCarthy inequality.

Lemma 2 

([19], Theorem 1.4). Let $T\in \mathbb{B}\left(\mathcal{H}\right)$ be a positive operator and $x\in \mathcal{H}$ be such that $\parallel x\parallel =1$. Then, for every $r\ge 1$ we have

$$\begin{array}{c}\hfill {⟨Tx,x⟩}^r\le ⟨T^{r}x,x⟩.\end{array}$$

The next lemma is stated as follows.

Lemma 3 

([16]). Let $x,y,e\in \mathcal{H}$ be such that $\parallel e\parallel =1$. Then

$$\begin{array}{cc}\hfill 12|⟨x,e⟩⟨e,y⟩{|}^2& \le {\parallel x\parallel }^2{\parallel y\parallel }^2+|⟨x,y⟩{|}^2+2\parallel x\parallel \parallel y\parallel |⟨x,y⟩|\hfill \\ \hfill & +4\left|⟨x,e⟩⟨e,y⟩\right|\left(\parallel x\parallel \parallel y\parallel +|⟨x,y⟩|\right).\hfill \end{array}$$

The interesting inequality in the following lemma is proved by Buzano in [20].

Lemma 4. 

Let $x,y,e\in \mathcal{H}$ be such that $\parallel e\parallel =1$. Then

$$|⟨x,e⟩⟨e,y⟩|\le \frac{1}{2}\left(\parallel x\parallel \parallel y\parallel +\left|⟨x,y⟩\right|\right).$$

The following lemma is known as Bohr’s inequality.

Lemma 5 

([21]). Let $a_k$ be a positive real number for every $k\in \{1,2,\dots ,d\}$. Then,

$${\left(\sum _{k=1}^d{a}_k\right)}^r\le d^{r-1}\sum _{k=1}^d{a}_k^r,\forall r\ge 1.$$

In the next lemma, we recall an important inequality that has been recently proven by Omidvar et al. in [22].

Lemma 6. 

Let $x,y,e\in \mathcal{H}$ be such that $\parallel e\parallel =1$. Then

$$|⟨x,e⟩{|}^2|⟨e,y⟩{|}^2\le \frac{3}{4}{\parallel x\parallel }^2{\parallel y\parallel }^2+\frac{1}{4}\parallel x\parallel \parallel y\parallel \left|⟨x,y⟩\right|.$$

3. Main Results

In this section, $({\mathcal{H}}_{\mathrm{\Omega }},⟨·,·⟩)$ denotes an RKHS on a set $\mathrm{\Omega }$ with associated norm $\parallel ·\parallel$.

Our first result in this paper reads as follows:

Theorem 1. 

Let $T,S\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then, for all $r\ge 1$ we have

$${\mathit{ber}}^r\left(S^{*}T\right)\le \frac{1}{2}{\parallel \left|T\right|}^{2r}+{\left|S\right|}^{2r}{\parallel }_{\mathit{ber}}.$$

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of the space ${\mathcal{H}}_{\mathrm{\Omega }}$. Further, let $r\ge 1$. By using the Cauchy–Schwarz inequality together with the arithmetic-geometric mean inequality, we see that

$$\begin{array}{cc}\hfill \left|⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|& =\left|⟨T{\widehat{k}}_{\lambda },S{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & \le \parallel T{\widehat{k}}_{\lambda }\parallel \parallel S{\widehat{k}}_{\lambda }\parallel \hfill \\ \hfill & =\sqrt{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\sqrt{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\hfill \\ \hfill & \le \frac{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}{2}\hfill \\ \hfill & \le {\left(\frac{{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r+{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r}{2}\right)}^{\frac{1}{r}},\hfill \end{array}$$

where the last inequality follows by applying the convexity of the function $t\mapsto t^r$. Further, by applying Lemma 2, we get

$$\begin{array}{cc}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^r& \le \frac{{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r+{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r}{2}\hfill \\ \hfill & \le \frac{⟨\left({\left|T\right|}^{2r}+{\left|S\right|}^{2r}\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}{2}\hfill \\ \hfill & \le \frac{1}{2}\mathbf{ber}\left({\left|T\right|}^{2r}+{\left|S\right|}^{2r}\right)\hfill \\ \hfill & =\frac{1}{2}{\parallel \left|T\right|}^{2r}+{\left|S\right|}^{2r}{\parallel }_{\mathbf{ber}},\hfill \end{array}$$

where the last equality follows by using (2) since ${\left|T\right|}^{2r}+{\left|S\right|}^{2r}\ge 0$. Hence, we deduce that

$$\begin{array}{c}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^r\le \frac{1}{2}{\parallel \left|T\right|}^{2r}+{\left|S\right|}^{2r}{\parallel }_{\mathbf{ber}},\end{array}$$

for all $\lambda \in \mathrm{\Omega }$. Therefore, by taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the above inequality, we get the desired result. □

Our next result is stated as follows.

Theorem 2. 

Let $T,S\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ be positive operators and $\theta \in [0,1]$. Then

$$\begin{array}{c}\hfill {∥\theta T+(1-\theta )S∥}_{\mathit{ber}}^r\le {∥\theta T^r+(1-\theta )S^r∥}_{\mathit{ber}},\forall r\ge 1.\end{array}$$

(8)

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of the space ${\mathcal{H}}_{\mathrm{\Omega }}$. By using the convexity of the function $h\left(t\right)=t^r$ with $r\ge 1$, we see that

$$\begin{array}{cc}\hfill {\left(⟨\left(\theta T+(1-\theta )S\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)}^r& ={\left(\theta ⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+(1-\theta )⟨S{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)}^r\hfill \\ \hfill & \stackrel{Power-Mean}{\le }\theta {⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r+(1-\theta ){⟨S{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r\hfill \\ \hfill & \stackrel{McCarthy}{\le }\theta ⟨T^r{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+(1-\theta )⟨S^r{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & =⟨\left(\theta T^r+(1-\theta )S^r\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & \le \mathbf{ber}\left(\theta T^r+(1-\theta )S^r\right)={∥\theta T^r+(1-\theta )S^r∥}_{\mathbf{ber}},\hfill \end{array}$$

because the operator $\theta T^r+(1-\theta )S^r$ is positive, when T and S are the positive operators and $\theta \in [0,1]$. Hence,

$$⟨(\theta T+(1-\theta )S){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }{⟩}^r\le {∥\theta T^r+(1-\theta )S^r∥}_{\mathbf{ber}}.$$

Hence, by taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the above inequality, we deduce that

$${\mathbf{ber}}^r\left(\theta T+(1-\theta )S\right)\le {∥\theta T^r+(1-\theta )S^r∥}_{\mathbf{ber}}.$$

This completes the proof by using (2) since $\theta T+(1-\theta )S\ge 0$. □

The following corollary is an immediate consequence of Theorem 2.

Corollary 1. 

Let $T,S\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ be positive operators. Then

$${∥{\displaystyle \frac{T+S}{2}}∥}_{\mathit{ber}}^r\le {∥{\displaystyle \frac{T^r+S^r}{2}}∥}_{\mathit{ber}},\forall r\ge 1.$$

(9)

Proof. 

Using $\theta =\frac{1}{2}$ in inequality (8), we obtain the relation of the statement. □

The following lemma is useful in proving our next result.

Lemma 7. 

Let $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then

$$\begin{array}{c}\hfill \mathit{ber}\left(T\right)\le \frac{1}{2}\parallel \left|T\right|+|T^{*}|{\parallel }_{\mathit{ber}}.\end{array}$$

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of ${\mathcal{H}}_{\mathrm{\Omega }}$. By applying inequality (7) together with the arithmetic-geometric mean inequality, one observes that

$$\begin{array}{cc}\hfill \left|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|& \le \sqrt{⟨\left|T\right|{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\sqrt{⟨\left|T^{*}\right|{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\hfill \\ \hfill & \le \frac{1}{2}\left(⟨\left|T\right|{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+⟨\left|T^{*}\right|{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)\hfill \\ \hfill & =\frac{1}{2}⟨\left(\left|T\right|+\left|T^{*}\right|\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & \le \frac{1}{2}\mathbf{ber}\left(\left|T\right|+\left|T^{*}\right|\right)\hfill \\ \hfill & =\frac{1}{2}\parallel \left|T\right|+\left|T^{*}\right|{\parallel }_{\mathbf{ber}},\hfill \end{array}$$

where the last equality follows from (2) since $\left|T\right|+\left|T^{*}\right|\ge 0$. Hence

$$\begin{array}{c}\hfill \left|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\le \frac{1}{2}\parallel \left|T\right|+\left|T^{*}\right|{\parallel }_{\mathbf{ber}}.\end{array}$$

Therefore, by taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the above inequality, we obtain the desired result. □

Now, we can prove the following result that provides an improvement in inequality (3).

Theorem 3. 

Let $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then

$$\begin{array}{cc}\hfill {\mathit{ber}}^2\left(T\right)& \le \frac{1}{6}{\parallel \left|T\right|}^2+|T^{*}{|}^2{\parallel }_{\mathit{ber}}+\frac{1}{3}\mathit{ber}\left(T\right)\parallel \left|T\right|+|T^{*}|{\parallel }_{\mathit{ber}}\hfill \\ \hfill & \le \frac{1}{2}{\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathit{ber}}.\hfill \end{array}$$

Proof. 

By using (3) together with Lemma 7, we see that

$$\begin{array}{cc}\hfill {\mathbf{ber}}^2\left(T\right)& =\frac{1}{3}{\mathbf{ber}}^2\left(T\right)+\frac{2}{3}{\mathbf{ber}}^2\left(T\right)\hfill \\ \hfill & \le \frac{1}{6}{\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}}+\frac{2}{3}\mathbf{ber}\left(T\right)\mathbf{ber}\left(T\right)\hfill \\ \hfill & \le \frac{1}{6}{\parallel \left|T\right|}^2+|T^{*}{|}^2{\parallel }_{\mathbf{ber}}+\frac{1}{3}\mathbf{ber}\left(T\right)\parallel \left|T\right|+|T^{*}|{\parallel }_{\mathbf{ber}}.\hfill \end{array}$$

This proves the first inequality in Theorem 6. On the other hand, by applying Lemma 7, we see that

$$\begin{array}{cc}\hfill & \frac{1}{6}{\parallel \left|T\right|}^2+|T^{*}{|}^2{\parallel }_{\mathbf{ber}}+\frac{1}{3}\mathbf{ber}\left(T\right)\parallel \left|T\right|+|T^{*}|{\parallel }_{\mathbf{ber}}\hfill \\ \hfill & \le \frac{1}{6}{\parallel \left|T\right|}^2+|T^{*}{|}^2{\parallel }_{\mathbf{ber}}+\frac{1}{6}\parallel \left|T\right|+|T^{*}|{\parallel }_{\mathbf{ber}}^2\hfill \\ \hfill & =\frac{1}{6}{\parallel \left|T\right|}^2+{|T^{*}|}^2{\parallel }_{\mathbf{ber}}+\frac{1}{6}\parallel \frac{2\left|T\right|+2|T^{*}|}{2}{\parallel }_{\mathbf{ber}}^2\hfill \\ \hfill & \le \frac{1}{6}{\parallel \left|T\right|}^2+|T^{*}{|}^2{\parallel }_{\mathbf{ber}}+\frac{1}{12}{\parallel 4\left|T\right|}^2+4{|T^{*}|}^2{\parallel }_{\mathbf{ber}}\left(\mathrm{by}\left(9\right)\right)\hfill \\ \hfill & =\frac{1}{6}{\parallel \left|T\right|}^2+|T^{*}{|}^2{\parallel }_{\mathbf{ber}}+\frac{1}{3}{\parallel \left|T\right|}^2+|T^{*}{|}^2{\parallel }_{\mathbf{ber}}=\frac{1}{2}{\parallel \left|T\right|}^2+{|T^{*}|}^2{\parallel }_{\mathbf{ber}}.\hfill \end{array}$$

Hence, the proof is complete. □

In the next result, we establish an upper bound involving ${\mathbf{ber}}^2\left(S^{*}T\right)$, which provides an improvement of the inequality in Theorem 1 for $r=2$.

Theorem 4. 

Let $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then

$$\begin{array}{cc}\hfill {\mathit{ber}}^2\left(S^{*}T\right)& \le \frac{1}{6}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathit{ber}}+\frac{1}{3}\mathit{ber}\left(S^{*}T\right){\parallel \left|T\right|}^2+{\left|S\right|}^2{\parallel }_{\mathit{ber}}\hfill \\ \hfill & \le \frac{1}{2}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathit{ber}}.\hfill \end{array}$$

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of the space ${\mathcal{H}}_{\mathrm{\Omega }}$. By replacing x and y with $T{\widehat{k}}_{\lambda }$ and $S{\widehat{k}}_{\lambda }$, respectively, in (4) and then using the arithmeticgeometric mean inequality, we see that

$$\begin{array}{cc}\hfill & |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2\hfill \\ \hfill & \le \frac{1}{3}\parallel T{\widehat{k}}_{\lambda }{\parallel }^2\parallel S{\widehat{k}}_{\lambda }{\parallel }^2+\frac{2}{3}\left|⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\parallel T{\widehat{k}}_{\lambda }\parallel \parallel S{\widehat{k}}_{\lambda }\parallel \hfill \\ \hfill & =\frac{1}{3}⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\frac{2}{3}\left|⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\sqrt{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\sqrt{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\hfill \\ \hfill & \le \frac{1}{6}\left({⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2+{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2\right)+\frac{1}{3}\mathbf{ber}\left(S^{*}T\right)⟨\left({\left|T\right|}^2+{\left|S\right|}^2\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & \le \frac{1}{6}\left({⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2+{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2\right)+\frac{1}{3}\mathbf{ber}\left(S^{*}T\right)\mathbf{ber}\left({\left|T\right|}^2+{\left|S\right|}^2\right)\hfill \\ \hfill & =\frac{1}{6}\left({⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2+{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2\right)+\frac{1}{3}\mathbf{ber}\left(S^{*}T\right){\parallel \left|T\right|}^2+{\left|S\right|}^2{\parallel }_{\mathbf{ber}},\hfill \end{array}$$

where the last equality follows by applying (2) since ${\left|T\right|}^2+{\left|S\right|}^2\ge 0$. Moreover, by applying Lemma 2, we conclude that

$$\begin{array}{cc}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2& \le \frac{1}{6}⟨\left({\left|T\right|}^4+{\left|S\right|}^4\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\frac{1}{3}\mathbf{ber}\left(S^{*}T\right){\parallel \left|T\right|}^2+{\left|S\right|}^2{\parallel }_{\mathbf{ber}}\hfill \\ \hfill & \le \frac{1}{6}\mathbf{ber}\left({\left|T\right|}^4+{\left|S\right|}^4\right)+\frac{1}{3}\mathbf{ber}\left(S^{*}T\right){\parallel \left|T\right|}^2+{\left|S\right|}^2{\parallel }_{\mathbf{ber}}.\hfill \end{array}$$

Since ${\left|T\right|}^4+{\left|S\right|}^4\ge 0$, then (2) we have

$$\mathbf{ber}\left({\left|T\right|}^4+{\left|S\right|}^4\right)={\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}},$$

whence

$$\begin{array}{cc}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2& \le \frac{1}{6}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{3}\mathbf{ber}\left(S^{*}T\right){\parallel \left|T\right|}^2+{\left|S\right|}^2{\parallel }_{\mathbf{ber}}.\hfill \end{array}$$

Therefore, by taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the above inequality, we get the first inequality in Theorem 4. Now, by applying Theorem 1 for $r=2$, we see that

$$\begin{array}{cc}\hfill & \frac{1}{6}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{3}\mathbf{ber}\left(S^{*}T\right){\parallel \left|T\right|}^2+{\left|S\right|}^2{\parallel }_{\mathbf{ber}}\hfill \\ \hfill & \le \frac{1}{6}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{6}{\parallel \left|T\right|}^2+{\left|S\right|}^2{\parallel }_{\mathbf{ber}}^2\hfill \\ \hfill & =\frac{1}{6}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{6}\parallel \frac{{2\left|T\right|}^2+2{\left|S\right|}^2}{2}{\parallel }_{\mathbf{ber}}^2\hfill \\ \hfill & \le \frac{1}{6}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{12}{\parallel 4\left|T\right|}^4+4{\left|S\right|}^4{\parallel }_{\mathbf{ber}}\left(\mathrm{by}\left(9\right)\right)\hfill \\ \hfill & =\frac{1}{6}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{3}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}=\frac{1}{2}{\parallel \left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}.\hfill \end{array}$$

This completes the proof. □

Our next result reads as follows.

Theorem 5. 

Let $T_i\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ for all $i\in \{1,2,\dots ,d\}$ with $d\in {\mathbb{N}}^{*}$. Then, for every $\theta \in [0,1]$ and $r\ge 1$, we have

$$\begin{array}{cc}\hfill {\mathit{ber}}^{2r}\left(\sum _{i=1}^d{T}_i\right)& \le \frac{d^{2r-1}}{4}\parallel \sum _{i=1}^d\left(|T_i{|}^{4r\theta }+{\left|T_i^{*}\right|}^{4r(1-\theta )}\right){\parallel }_{\mathit{ber}}+\frac{d^{2r-1}}{2}\sum _{i=1}^d\mathit{ber}\left(|T_i{|}^{2r\theta }{\left|T_i^{*}\right|}^{2r(1-\theta )}\right).\hfill \end{array}$$

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of the space ${\mathcal{H}}_{\mathrm{\Omega }}$. By using Lemmas 2 and 5, and inequality (7), we see that

$$\begin{array}{cc}\hfill {\left|⟨\left(\sum _{i=1}^d{T}_i\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|}^{2r}& \le {\left(\sum _{i=1}^d|⟨T_i{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩|\right)}^{2r}\hfill \\ \hfill & \le d^{2r-1}\sum _{i=1}^d|⟨T_i{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^{2r}\hfill \\ \hfill & \stackrel{Kato}{\le }{d}^{2r-1}\sum _{i=1}^d{⟨|T_i{|}^{2\theta }{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r{⟨|T_i^{*}{|}^{2(1-\theta )}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r\hfill \\ \hfill & \stackrel{McCarthy}{\le }{d}^{2r-1}\sum _{i=1}^d⟨|T_i{|}^{2r\theta }{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨|T_i^{*}{|}^{2r(1-\theta )}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & =d^{2r-1}\sum _{i=1}^d⟨|T_i{|}^{2r\theta }{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨{\widehat{k}}_{\lambda },{\left|T_i^{*}\right|}^{2r(1-\theta )}{\widehat{k}}_{\lambda }⟩.\hfill \end{array}$$

Moreover, by letting $x=|T_i{|}^{2r\theta }{\widehat{k}}_{\lambda }$, $y=|T_i^{*}{|}^{2r(1-\theta )}{\widehat{k}}_{\lambda }$ and $e={\widehat{k}}_{\lambda }$ in Lemma 4 and then applying the arithmetic-geometric mean inequality, we obtain

$$\begin{array}{cc}\hfill & {\left|⟨\left(\sum _{i=1}^d{T}_i\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|}^{2r}\hfill \\ \hfill & \le \frac{d^{2r-1}}{2}\sum _{i=1}^d\left(\parallel |T_i{|}^{2r\theta }{\widehat{k}}_{\lambda }\parallel \parallel {\left|T_i^{*}\right|}^{2r(1-\theta )}{\widehat{k}}_{\lambda }\parallel +\left|⟨|T_i{|}^{2r\theta }{\widehat{k}}_{\lambda },{\left|T_i^{*}\right|}^{2r(1-\theta )}{\widehat{k}}_{\lambda }⟩\right|\right)\hfill \\ \hfill & \le \frac{d^{2r-1}}{2}\sum _{i=1}^d\left(\frac{1}{2}\left(\parallel |T_i{|}^{2r\theta }{\widehat{k}}_{\lambda }{\parallel }^2+\parallel {\left|T_i^{*}\right|}^{2r(1-\theta )}{\widehat{k}}_{\lambda }{\parallel }^2\right)+\left|⟨|T_i{|}^{2r\theta }{\left|T_i^{*}\right|}^{2r(1-\theta )}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\right)\hfill \\ \hfill & =\frac{d^{2r-1}}{4}\sum _{i=1}^d⟨\left(|T_i{|}^{4r\theta }+{\left|T_i^{*}\right|}^{4r(1-\theta )}\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\frac{d^{2r-1}}{2}\sum _{i=1}^d\left|⟨|T_i{|}^{2r\theta }{\left|T_i^{*}\right|}^{2r(1-\theta )}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & =\frac{d^{2r-1}}{4}⟨\left[\sum _{i=1}^d\left(\right|T_i{|}^{4r\theta }+|T_i^{*}{|}^{4r(1-\theta )})\right]{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\frac{d^{2r-1}}{2}\sum _{i=1}^d\left|⟨|T_i{|}^{2r\theta }{\left|T_i^{*}\right|}^{2r(1-\theta )}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & \le \frac{d^{2r-1}}{4}\mathbf{ber}\left(\sum _{i=1}^d\left(|T_i{|}^{4r\theta }+{\left|T_i^{*}\right|}^{4r(1-\theta )}\right)\right)+\frac{d^{2r-1}}{2}\sum _{i=1}^d\mathbf{ber}\left(|T_i{|}^{2r\theta }{\left|T_i^{*}\right|}^{2r(1-\theta )}\right).\hfill \end{array}$$

Since, ${\sum }_{i=1}^d\left(|T_i{|}^{4r\theta }+{\left|T_i^{*}\right|}^{4r(1-\theta )}\right)\ge 0$, then by (2), we deduce that

$$\begin{array}{cc}\hfill & {\left|⟨\left(\sum _{i=1}^d{T}_i\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|}^{2r}\hfill \\ \hfill & \le \frac{d^{2r-1}}{4}\parallel \sum _{i=1}^d\left(|T_i{|}^{4r\theta }+{\left|T_i^{*}\right|}^{4r(1-\theta )}\right){\parallel }_{\mathbf{ber}}+\frac{d^{2r-1}}{2}\sum _{i=1}^d\mathbf{ber}\left(|T_i{|}^{2r\theta }{\left|T_i^{*}\right|}^{2r(1-\theta )}\right).\hfill \end{array}$$

Taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the last inequality yields the desired result. □

Remark 3. 

Substituting $\theta =\frac{1}{2}$ into the inequality of Theorem 5, we obtain the inequality given by Bhunia et al. in [12], thus

$${\mathit{ber}}^{2r}\left(\sum _{i=1}^d{T}_i\right)\le \frac{d^{2r-1}}{4}\parallel \sum _{i=1}^d\left(|T_i{|}^{2r}+{\left|T_i^{*}\right|}^{2r}\right){\parallel }_{\mathit{ber}}+\frac{d^{2r-1}}{2}\sum _{i=1}^d\mathit{ber}\left(|T_i{|}^r{\left|T_i^{*}\right|}^r\right).$$

For $d=1$ and $r=1$ in the inequality in Theorem 5, we deduce the following inequality:

$${\mathit{ber}}^2\left(T\right)\le \frac{1}{4}\parallel {\left|T\right|}^{4\theta }+{\left|T^{*}\right|}^{4(1-\theta )}{\parallel }_{\mathit{ber}}+\frac{1}{2}\mathit{ber}\left({\left|T\right|}^{2\theta }{\left|T^{*}\right|}^{2(1-\theta )}\right),$$

(10)

for all $\theta \in [0,1]$ and $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. If we take $\theta =\frac{1}{2}$ in inequality (10), then we find

$${\mathit{ber}}^2\left(T\right)\le \frac{1}{4}{\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathit{ber}}+\frac{1}{2}\mathit{ber}\left(\left|T\right||T^{*}|\right)$$

for every $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$.

In the next result, we prove an estimate for ${\mathbf{ber}}^4\left(T\right)$.

Theorem 6. 

Let $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then,

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

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of ${\mathcal{H}}_{\mathrm{\Omega }}$. By putting $e={\widehat{k}}_{\lambda }$ and then replacing x and y by $T{\widehat{k}}_{\lambda }$ and $T^{*}{\widehat{k}}_{\lambda }$, respectively, in Lemma 3 we see that

$$\begin{array}{cc}\hfill & 12|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4\hfill \\ \hfill & \le \parallel T{\widehat{k}}_{\lambda }{\parallel }^2\parallel T^{*}{\widehat{k}}_{\lambda }{\parallel }^2+\left|⟨T{\widehat{k}}_{\lambda },T^{*}{\widehat{k}}_{\lambda }⟩{|}^2+2\parallel T{\widehat{k}}_{\lambda }\parallel \parallel T^{*}{\widehat{k}}_{\lambda }\parallel |⟨T{\widehat{k}}_{\lambda },T^{*}{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & +4|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2\left(\parallel T{\widehat{k}}_{\lambda }\parallel \parallel T^{*}{\widehat{k}}_{\lambda }\parallel +|⟨T{\widehat{k}}_{\lambda },T^{*}{\widehat{k}}_{\lambda }⟩|\right)\hfill \\ \hfill & ={⟨\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+2\sqrt{{⟨\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\left|⟨T^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & +4|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2\left(\sqrt{{⟨\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}+\left|⟨T^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\right)+|⟨T^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2.\hfill \end{array}$$

Further, by applying the arithmetic-geometric mean inequality, we get

$$\begin{array}{cc}\hfill & 12|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4\hfill \\ \hfill & \le \frac{1}{2}\left({⟨\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }{⟩}^2+⟨|T^{*}{{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2\right)+\left|⟨T^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\left({⟨\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)\hfill \\ \hfill & +2|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2\left({⟨\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }{⟩}^2+⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }{⟩}^2+2\left|⟨T^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\right)+|⟨T^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2\hfill \\ \hfill & \le \frac{1}{2}\left(⟨\left[{\left|T\right|}^4+{\left|T^{*}\right|}^4\right]{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)+\mathbf{ber}\left(T^2\right)\left(⟨\left[{\left|T\right|}^2+{\left|T^{*}\right|}^2\right]{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)\hfill \\ \hfill & +2{\mathbf{ber}}^2\left(T\right)\left(⟨\left({\left|T\right|}^2+{\left|T^{*}\right|}^2\right),{\widehat{k}}_{\lambda }⟩+2\mathbf{ber}\left(T^2\right)\right)+{\mathbf{ber}}^2\left(T^2\right),\hfill \end{array}$$

Since the operators ${\left|T\right|}^2$ and $|T^{*}{|}^2$ are positive, then by applying Lemma 2, we see that

$$\begin{array}{cc}\hfill 12|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4& \le \frac{1}{2}\left(⟨\left[{\left|T\right|}^4+{\left|T^{*}\right|}^4\right]{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)+\mathbf{ber}\left(T^2\right)\left(⟨\left[{\left|T\right|}^2+{\left|T^{*}\right|}^2\right]{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)\hfill \\ \hfill & +2{\mathbf{ber}}^2\left(T\right)\left(⟨\left({\left|T\right|}^2+{\left|T^{*}\right|}^2\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+2\mathbf{ber}\left(T^2\right)\right)+{\mathbf{ber}}^2\left(T^2\right)\hfill \\ \hfill & \le \frac{1}{2}\mathbf{ber}\left({\left|T\right|}^4+{\left|T^{*}\right|}^4\right)+\mathbf{ber}\left(T^2\right)\mathbf{ber}\left({\left|T\right|}^2+{\left|T^{*}\right|}^2\right)\hfill \\ \hfill & +2{\mathbf{ber}}^2\left(T\right)\left(\mathbf{ber}\left({\left|T\right|}^2+{\left|T^{*}\right|}^2\right)+2\mathbf{ber}\left(T^2\right)\right)+{\mathbf{ber}}^2\left(T^2\right)\hfill \\ \hfill & =\frac{1}{2}{\parallel \left|T\right|}^4+|T^{*}{|}^4{\parallel }_{\mathbf{ber}}+\mathbf{ber}\left(T^2\right){\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}}\hfill \\ \hfill & +2{\mathbf{ber}}^2\left(T\right)\left({\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}}+2\mathbf{ber}\left(T^2\right)\right)+{\mathbf{ber}}^2\left(T^2\right),\hfill \end{array}$$

where the last inequality follows by applying (2) since the operators ${\left|T\right|}^4+{\left|T^{*}\right|}^4$ and ${\left|T\right|}^2+{\left|T^{*}\right|}^2$ are positive. Thus,

$$\begin{array}{cc}\hfill 12|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4& \le \frac{1}{2}{\parallel \left|T\right|}^4+|T^{*}{|}^4{\parallel }_{\mathbf{ber}}+\mathbf{ber}\left(T^2\right){\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}}\hfill \\ \hfill & +2{\mathbf{ber}}^2\left(T\right)\left({\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}}+2\mathbf{ber}\left(T^2\right)\right)+{\mathbf{ber}}^2\left(T^2\right)\hfill \\ \hfill & =\frac{1}{2}{\parallel \left|T\right|}^4+{\left|T^{*}\right|}^4{\parallel }_{\mathbf{ber}}+{\mathbf{ber}}^2\left(T^2\right)+4{\mathbf{ber}}^2\left(T\right)\mathbf{ber}\left(T^2\right)\hfill \\ \hfill & +{\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}}\left(\mathbf{ber}\left(T^2\right)+2{\mathbf{ber}}^2\left(T\right)\right),\hfill \end{array}$$

whence

$$\begin{array}{cc}\hfill |⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4& \le \frac{1}{24}{\parallel \left|T\right|}^4+{\left|T^{*}\right|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{12}{\mathbf{ber}}^2\left(T^2\right)+\frac{1}{3}{\mathbf{ber}}^2\left(T\right)\mathbf{ber}\left(T^2\right)\hfill \\ \hfill & +\frac{1}{12}{\parallel \left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}}\left(\mathbf{ber}\left(T^2\right)+2{\mathbf{ber}}^2\left(T\right)\right).\hfill \end{array}$$

This proves the desired inequality by taking the supremum over $\lambda \in \mathrm{\Omega }$ in the last inequality. □

Our next theorem is stated as follows.

Theorem 7. 

Let $T,S\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then, for all $r\ge 1$ and $\theta \in [0,1]$, we have

$${\mathit{ber}}^{2r}\left(S^{*}T\right)\le {\parallel \theta \left|T\right|}^{2r}+{(1-\theta )\left|S\right|}^{2r}{\parallel }_{\mathit{ber}}{\parallel (1-\theta )\left|T\right|}^{2r}+\theta {\left|S\right|}^{2r}{\parallel }_{\mathit{ber}}.$$

(11)

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of the space ${\mathcal{H}}_{\mathrm{\Omega }}$. Further, let $r\ge 1$. By using the Cauchy–Schwarz inequality, we deduce that

$$\begin{array}{cc}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^{2r}& =|⟨T{\widehat{k}}_{\lambda },S{\widehat{k}}_{\lambda }⟩{|}^{2r}\hfill \\ \hfill & \le {(\parallel T{\widehat{k}}_{\lambda }{\parallel }^2\parallel S{\widehat{k}}_{\lambda }{\parallel }^2)}^r\hfill \\ \hfill & ={\left(⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩|\right)}^r\hfill \\ \hfill & ={\left({⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^{\theta }{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^{1-\theta }\right)}^r{\left({⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^{1-\theta }{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^{\theta }\right)}^r.\hfill \end{array}$$

Now, by applying Young’s inequality and the convexity of the function $t\mapsto t^r$, we have

$$a^{\theta }{b}^{1-\theta }\le \theta a+(1-\theta )b\le {\left(\theta a^r+(1-\theta )b^r\right)}^{\frac{1}{r}},$$

(12)

for every $a,b>0$, $r\ge 1$ and $\theta \in [0,1]$. Therefore, by using (12) together with Lemma 2, we get

$$\begin{array}{cc}\hfill & |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^{2r}\hfill \\ \hfill & =\left(\theta {⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r+(1-\theta ){⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r\right)\left((1-\theta ){⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r+\theta {⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r\right)\hfill \\ \hfill & =\left(\theta ⟨{\left|T\right|}^{2r}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+(1-\theta )⟨{\left|S\right|}^{2r}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)\left((1-\theta )⟨{\left|T\right|}^{2r}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\theta ⟨{\left|S\right|}^{2r}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)\hfill \\ \hfill & =⟨\left({\theta \left|T\right|}^{2r}+(1-\theta ){\left|S\right|}^{2r}\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨\left({(1-\theta )\left|T\right|}^{2r}+\theta {\left|S\right|}^{2r}\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & \le \mathbf{ber}\left({\theta \left|T\right|}^{2r}+(1-\theta ){\left|S\right|}^{2r}\right)\mathbf{ber}\left({(1-\theta )\left|T\right|}^{2r}+\theta {\left|S\right|}^{2r}\right).\hfill \end{array}$$

Therefore, by taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the above inequality and taking into account that ${\theta \left|T\right|}^{2r}+(1-\theta ){\left|S\right|}^{2r}\ge 0$ and ${(1-\theta )\left|T\right|}^{2r}+\theta {\left|S\right|}^{2r}\ge 0$, we obtain

$$\begin{array}{cc}\hfill {\mathbf{ber}}^{2r}\left(S^{*}T\right)& \le \mathbf{ber}\left({\theta \left|T\right|}^{2r}+(1-\theta ){\left|S\right|}^{2r}\right)\mathbf{ber}\left({(1-\theta )\left|T\right|}^{2r}+\theta {\left|S\right|}^{2r}\right)\hfill \\ \hfill & ={\parallel \theta \left|T\right|}^{2r}+{(1-\theta )\left|S\right|}^{2r}{\parallel }_{\mathbf{ber}}{\parallel (1-\theta )\left|T\right|}^{2r}+\theta {\left|S\right|}^{2r}{\parallel }_{\mathbf{ber}},\hfill \end{array}$$

for all $\theta \in [0,1]$ and $r\ge 1$. Consequently, we obtain the inequality of the statement. □

Remark 4. 

By letting $\theta =\frac{1}{2}$ in (11), we get the inequality in Theorem 1.

The following corollary is an immediate consequence of Theorem 7.

Corollary 2. 

Let $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then, for all $r\ge 1$ and $\theta \in [0,1]$, we have

$${\mathit{ber}}^{2r}\left(T^2\right)\le {\parallel \theta \left|T\right|}^{2r}+(1-\theta )|T^{*}{|}^{2r}{\parallel }_{\mathit{ber}}{\parallel (1-\theta )\left|T\right|}^{2r}+\theta {\left|T^{*}\right|}^{2r}{\parallel }_{\mathit{ber}}.$$

(13)

Proof. 

For $S=T^{*}$ in inequality (11), we deduce the inequality from the statement. □

Remark 5. 

For $\theta =\frac{1}{2}$ in the relation (13), we deduce the following inequality:

$$\mathit{ber}\left(T^2\right)\le \parallel \frac{{\left|T\right|}^{2r}+{\left|T^{*}\right|}^{2r}}{2}{\parallel }_{\mathit{ber}}^{\frac{1}{r}},$$

for all $r\ge 1$ and $T,S\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$.

In the following theorem, we establish a new Berezin number inequality that refines an earlier result by Bhunia et al. in [12].

Theorem 8. 

Let $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$, with $T\ne 0$. Then, for all $r\ge 1$ and $\theta \in [0,1]$, we have

$$\begin{array}{c}\hfill {\mathit{ber}}^{2r}\left(T\right)+\frac{{min}^r\{\theta ,1-\theta \}}{4^r{max}^r{\{\parallel |T|}^2{\parallel }_{\mathit{ber}},\parallel |T^{*}{|}^2{\parallel }_{\mathit{ber}}\}}{c}^{2r}\left({\left|T\right|}^2-{\left|T^{*}\right|}^2\right)\le {\parallel \theta \left|T\right|}^{2r}+(1-\theta ){\left|T^{*}\right|}^{2r}{\parallel }_{\mathit{ber}}.\end{array}$$

(14)

Proof. 

In [23], Kittaneh and Manasrah improved Young’s inequality. Thus

$$a^{\theta }{b}^{1-\theta }+min\{\theta ,1-\theta \}{(\sqrt{a}-\sqrt{b})}^2\le \theta a+(1-\theta )b,$$

(15)

where $a,b\ge 0$ and $\theta \in [0,1].$ Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of space ${\mathcal{H}}_{\mathrm{\Omega }}$. We have the following calculations:

$$\begin{array}{cc}\hfill |⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2& \stackrel{Kato}{\le }⟨{\left|T\right|}^{2\theta }{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨|T^{*}{|}^{2(1-\theta )}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & \stackrel{McCarthy}{\le }{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^{\theta }{⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^{1-\theta }.\hfill \end{array}$$

Since ${(\sqrt{a}-\sqrt{b})}^2\ge {\displaystyle \frac{{(a-b)}^2}{4max\{a,b\}}}$ and using inequality (15) we deduce

$$a^{\theta }{b}^{1-\theta }+min\{\theta ,1-\theta \}{\displaystyle \frac{{(a-b)}^2}{4max\{a,b\}}}\le \theta a+(1-\theta )b,$$

where $a,b>0$ and $\theta \in [0,1].$ Further, let $r\ge 1$. Applying the above inequality for $a=⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩$ and $b=⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩$ we deduce

$$\begin{array}{cc}\hfill |⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2+& \frac{min\{\theta ,1-\theta \}}{4max\{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩,⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\}}{\left(⟨{\left(\right|T|}^2-|T^{*}{|}^2){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)}^2\hfill \\ \hfill & \le \theta ⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+(1-\theta )⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & \stackrel{Power-Mean}{\le }{\left(\theta {⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r+(1-\theta ){⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r\right)}^{\frac{1}{r}}.\hfill \end{array}$$

Therefore, we obtain

$$\begin{array}{cc}\hfill |⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^{2r}+& \frac{{min}^r\{\theta ,1-\theta \}}{4^r{max}^r\{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩,⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\}}{\left(⟨{\left(\right|T|}^2-|T^{*}{|}^2){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)}^{2r}\hfill \\ \hfill & \le \theta {⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r+(1-\theta ){⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r\hfill \\ \hfill & \stackrel{McCarthy}{\le }\theta ⟨{\left|T\right|}^{2r}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+(1-\theta )⟨|T^{*}{|}^{2r}{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & =⟨{\left(\theta \right|T|}^{2r}+(1-\theta )|T^{*}{|}^{2r}){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩.\hfill \end{array}$$

Now, by taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the above inequality and taking into account that the operator ${\theta \left|T\right|}^{2r}+(1-\theta ){\left|T^{*}\right|}^{2r}$ is positive, we deduce the inequality

$$\begin{array}{cc}\hfill {\mathbf{ber}}^{2r}\left(T\right)+& \frac{{min}^r\{\theta ,1-\theta \}}{4^r{max}^r{\{\parallel |T|}^2{\parallel }_{\mathbf{ber}},\parallel |T^{*}{|}^2{\parallel }_{\mathbf{ber}}\}}{c}^{2r}\left({\left|T\right|}^2-{\left|T^{*}\right|}^2\right)\hfill \\ \hfill & \le \mathbf{ber}\left({\theta \left|T\right|}^{2r}+(1-\theta ){\left|T^{*}\right|}^{2r}\right)\hfill \\ \hfill & ={\parallel \theta \left|T\right|}^{2r}+(1-\theta ){\left|T^{*}\right|}^{2r}{\parallel }_{\mathbf{ber}}\hfill \end{array}$$

for all $\theta \in [0,1]$ and $r\ge 1$. Therefore, we obtain the inequality of the statement. □

Remark 6. 

Inequality (14) represents an improvement in the following inequality given by Bhunia et al. in [12]:

$${\mathit{ber}}^{2r}\left(T\right)\le {\parallel \theta \left|T\right|}^{2r}+(1-\theta ){\left|T^{*}\right|}^{2r}{\parallel }_{\mathit{ber}},$$

for all $r\ge 1$, $\theta \in [0,1]$ and $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$.

In our next result, we prove another estimation of ${\mathbf{ber}}^{2r}\left({\sum }_{i=1}^d{T}_i\right)$.

Theorem 9. 

Let $T_i\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ for all $i\in \{1,2,\dots ,d\}$ with $d\in {\mathbb{N}}^{*}$. Then, for every $r\ge 1$, we have

$${\mathit{ber}}^{2r}\left(\sum _{i=1}^d{T}_i\right)\le \frac{d^{2r-1}}{2}\parallel \sum _{i=1}^d\left(|T_i{|}^{2r}+{\left|T_i^{*}\right|}^{2r}\right){\parallel }_{\mathit{ber}}.$$

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of the space ${\mathcal{H}}_{\mathrm{\Omega }}$. By using Lemmas 2 and 5, and inequality (7), we have

$$\begin{array}{cc}\hfill {\left|⟨\left(\sum _{i=1}^d{T}_i\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|}^{2r}& \le {\left(\sum _{i=1}^d\left|⟨T_i{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\right)}^{2r}\hfill \\ \hfill & \le d^{2r-1}\sum _{i=1}^d|⟨T_i{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^{2r}\hfill \\ \hfill & \le d^{2r-1}\sum _{i=1}^d{⟨|T_i|{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r{⟨|T_i^{*}|{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^r\hfill \\ \hfill & \le d^{2r-1}\sum _{i=1}^d⟨|T_i{|}^r{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨|T_i^{*}{|}^r{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩.\hfill \end{array}$$

Moreover, since $|T_i{|}^r\ge 0$ and $|T_i^{*}{|}^r\ge 0$, then by using the arithmetic-geometric mean inequality together with Lemma 2, we have

$$\begin{array}{cc}\hfill {\left|⟨\left(\sum _{i=1}^d{T}_i\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|}^{2r}& \le \frac{d^{2r-1}}{2}\sum _{i=1}^d{⟨|T_i{|}^r{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2+{⟨|T_i^{*}{|}^r{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2\hfill \\ \hfill & \le \frac{d^{2r-1}}{2}\sum _{i=1}^d⟨\left(|T_i{|}^{2r}+{\left|T_i^{*}\right|}^{2r}\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & =\frac{d^{2r-1}}{2}⟨\sum _{i=1}^d\left(|T_i{|}^{2r}+{\left|T_i^{*}\right|}^{2r}\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & \le \frac{d^{2r-1}}{2}\mathbf{ber}\left(\sum _{i=1}^d\left(|T_i{|}^{2r}+{\left|T_i^{*}\right|}^{2r}\right)\right).\hfill \end{array}$$

Since ${\sum }_{i=1}^d\left(|T_i{|}^{2r}+{\left|T_i^{*}\right|}^{2r}\right)\ge 0$, then by taking (2) into consideration, we deduce that

$$\begin{array}{cc}\hfill {\left|⟨\left(\sum _{i=1}^d{T}_i\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|}^{2r}& \le \frac{d^{2r-1}}{2}\parallel \sum _{i=1}^d\left(|T_i{|}^{2r}+{\left|T_i^{*}\right|}^{2r}\right){\parallel }_{\mathbf{ber}}.\hfill \end{array}$$

Taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the last inequality yields the desired result. □

Remark 7. 

By letting $r=d=1$ in Theorem 9, we reach inequality (3).

Our next result reads as follows.

Theorem 10. 

Let $T,S\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then

$$\begin{array}{cc}\hfill {\mathit{ber}}^4\left(S^{*}T\right)& \le \frac{3}{8}\parallel {\left|T\right|}^8+{\left|S\right|}^8{\parallel }_{\mathit{ber}}+\frac{1}{8}\parallel {\left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathit{ber}}\mathit{ber}\left({\left|S\right|}^2{\left|T\right|}^2\right).\hfill \end{array}$$

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of the space ${\mathcal{H}}_{\mathrm{\Omega }}$. The application of the Cauchy–Schwarz inequality shows that

$$\begin{array}{cc}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4& =|⟨T{\widehat{k}}_{\lambda },S{\widehat{k}}_{\lambda }⟩{|}^4\hfill \\ \hfill & \le \parallel T{\widehat{k}}_{\lambda }{\parallel }^4\parallel S{\widehat{k}}_{\lambda }{\parallel }^4\hfill \\ \hfill & ={⟨T{\widehat{k}}_{\lambda },T{\widehat{k}}_{\lambda }⟩}^2{⟨S{\widehat{k}}_{\lambda },S{\widehat{k}}_{\lambda }⟩}^2\hfill \\ \hfill & ={⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2{⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2={⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2{⟨{\widehat{k}}_{\lambda },{\left|S\right|}^2{\widehat{k}}_{\lambda }⟩}^2.\hfill \end{array}$$

Moreover, by putting $e={\widehat{k}}_{\lambda }$, $x={\left|T\right|}^2{\widehat{k}}_{\lambda }$ and $y={\left|S\right|}^2{\widehat{k}}_{\lambda }$ in Lemma 6, we obtain

$$\begin{array}{cc}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4& \le \frac{3}{4}{\parallel \left|T\right|}^2{\widehat{k}}_{\lambda }{\parallel }^2{\parallel \left|S\right|}^2{\widehat{k}}_{\lambda }{\parallel }^2+\frac{1}{4}{\parallel \left|T\right|}^2{\widehat{k}}_{\lambda }\parallel \parallel {\left|S\right|}^2{\widehat{k}}_{\lambda }\parallel \left|⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\left|S\right|}^2{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & =\frac{3}{4}⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\left|T\right|}^2{\widehat{k}}_{\lambda }⟩⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\left|S\right|}^2{\widehat{k}}_{\lambda }⟩\hfill \\ \hfill & +\frac{1}{4}\sqrt{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\left|T\right|}^2{\widehat{k}}_{\lambda }⟩⟨{\left|S\right|}^2{\widehat{k}}_{\lambda },{\left|S\right|}^2{\widehat{k}}_{\lambda }⟩}\left|⟨{\left|S\right|}^2{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & =\frac{3}{4}⟨{\left|T\right|}^4{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨{\left|S\right|}^4{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\frac{1}{4}\sqrt{⟨{\left|T\right|}^4{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨{\left|S\right|}^4{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\left|⟨{\left|S\right|}^2{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & \le \frac{3}{8}\left({⟨{\left|T\right|}^4{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2+{⟨{\left|S\right|}^4{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2\right)\hfill \\ \hfill & +\frac{1}{8}\left(⟨{\left|T\right|}^4{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+⟨{\left|S\right|}^4{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)\left|⟨{\left|S\right|}^2{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|,\hfill \end{array}$$

where we have used the arithmetic-geometric mean inequality in the last inequality. Now, by using Lemma 2, we obtain

$$\begin{array}{cc}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4& \le \frac{3}{8}⟨\left({\left|T\right|}^8+{\left|S\right|}^8\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\frac{1}{8}⟨\left({\left|T\right|}^4+{\left|S\right|}^4\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\left|⟨{\left|S\right|}^2{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & \le \frac{3}{8}\mathbf{ber}\left({\left|T\right|}^8+{\left|S\right|}^8\right)+\frac{1}{8}\mathbf{ber}\left({\left|T\right|}^4+{\left|S\right|}^4\right)\mathbf{ber}\left({\left|S\right|}^2{\left|T\right|}^2\right).\hfill \end{array}$$

Since ${\left|T\right|}^8+{\left|S\right|}^8\ge 0$ and ${\left|T\right|}^4+{\left|S\right|}^4\ge 0$, then an application of (2) shows that

$$\begin{array}{cc}\hfill |⟨S^{*}T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4& \le \frac{3}{8}\parallel {\left|T\right|}^8+{\left|S\right|}^8{\parallel }_{\mathbf{ber}}+\frac{1}{4}\parallel {\left|T\right|}^4+{\left|S\right|}^4{\parallel }_{\mathbf{ber}}\mathbf{ber}\left({\left|S\right|}^2{\left|T\right|}^2\right).\hfill \end{array}$$

By taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the last inequality, we deduce the required result. □

Another Berezin number inequality is stated in the following theorem.

Theorem 11. 

Let $T\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$. Then

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

Proof. 

Let $\lambda \in \mathrm{\Omega }$ and ${\widehat{k}}_{\lambda }$ be the normalized reproducing kernel of the space ${\mathcal{H}}_{\mathrm{\Omega }}$. By putting $e={\widehat{k}}_{\lambda }$, $x=T{\widehat{k}}_{\lambda }$ and $y=T^{*}{\widehat{k}}_{\lambda }$ in Lemma 6, we obtain

$$\begin{array}{cc}\hfill |⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4& =\left|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2\right|⟨{\widehat{k}}_{\lambda },T^{*}{\widehat{k}}_{\lambda }⟩{|}^2\hfill \\ \hfill & \le \frac{3}{4}\parallel T{\widehat{k}}_{\lambda }{\parallel }^2\parallel T^{*}{\widehat{k}}_{\lambda }{\parallel }^2+\frac{1}{4}\parallel T{\widehat{k}}_{\lambda }\parallel \parallel T^{*}{\widehat{k}}_{\lambda }\parallel \left|⟨T^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|\hfill \\ \hfill & \le \frac{3}{4}⟨T{\widehat{k}}_{\lambda },T{\widehat{k}}_{\lambda }⟩⟨T^{*}{\widehat{k}}_{\lambda },T^{*}{\widehat{k}}_{\lambda }⟩+\frac{1}{4}\sqrt{⟨T{\widehat{k}}_{\lambda },T{\widehat{k}}_{\lambda }⟩⟨T^{*}{\widehat{k}}_{\lambda },T^{*}{\widehat{k}}_{\lambda }⟩}\mathbf{ber}\left(T^2\right)\hfill \\ \hfill & =\frac{3}{4}⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\frac{1}{4}\sqrt{⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}\mathbf{ber}\left(T^2\right).\hfill \end{array}$$

Moreover, by applying the arithmetic-geometric mean inequality together with Lemma 2, we get

$$\begin{array}{cc}\hfill & |⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^4\hfill \\ \hfill & \le \frac{3}{8}\left({⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2+{⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩}^2\right)+\frac{1}{8}\left(⟨{\left|T\right|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+⟨|T^{*}{|}^2{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right)\mathbf{ber}\left(T^2\right)\hfill \\ \hfill & \le \frac{3}{8}⟨\left({\left|T\right|}^4+{\left|T^{*}\right|}^4\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩+\frac{1}{8}⟨\left({\left|T\right|}^2+{\left|T^{*}\right|}^2\right){\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\mathbf{ber}\left(T^2\right)\hfill \\ \hfill & \le \frac{3}{8}\mathbf{ber}\left({\left|T\right|}^4+{\left|T^{*}\right|}^4\right)+\frac{1}{8}\mathbf{ber}\left({\left|T\right|}^2+{\left|T^{*}\right|}^2\right)\mathbf{ber}\left(T^2\right)\hfill \\ \hfill & =\frac{3}{8}\parallel {\left|T\right|}^4+|T^{*}{|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{8}\parallel {\left|T\right|}^2+{|T^{*}|}^2{\parallel }_{\mathbf{ber}}\mathbf{ber}\left(T^2\right),\hfill \end{array}$$

where we have used the last equality in (2) since the operators ${\left|T\right|}^4+{\left|T^{*}\right|}^4$ and ${\left|T\right|}^2+{\left|T^{*}\right|}^2$ are positive. Therefore, we obtain

$$|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩{|}^2\le \frac{3}{8}\parallel {\left|T\right|}^4+|T^{*}{|}^4{\parallel }_{\mathbf{ber}}+\frac{1}{8}\parallel {\left|T\right|}^2+{\left|T^{*}\right|}^2{\parallel }_{\mathbf{ber}}\mathbf{ber}\left(T^2\right).$$

Taking the supremum over all $\lambda \in \mathrm{\Omega }$ in the last inequality yields the desired result. □

4. Conclusions

Reproducing kernel Hilbert spaces (RKHS) arise in many areas, including statistics, approximation theory, group representation theory, etc. Starting from the RKHS, the Berezin set and the Berezin number of an operator are defined as:

$$\begin{array}{c}\hfill \mathbf{Ber}\left(T\right):=\left\{\tilde{T}\left(\lambda \right);\lambda \in \mathrm{\Omega }\right\}\mathrm{and}\mathbf{ber}\left(T\right):=\underset{\lambda \in \mathrm{\Omega }}{sup}\left|\tilde{T}\left(\lambda \right)\right|=\underset{\lambda \in \mathrm{\Omega }}{sup}\left|⟨T{\widehat{k}}_{\lambda },{\widehat{k}}_{\lambda }⟩\right|.\end{array}$$

The Berezin number has been investigated for the Toeplitz and Hankel operators on the Hardy and Bergman spaces.

To characterize the Berezin number and the Berezin norm, we found many inequalities and their properties in some papers.

Therefore, our objective was to study other new upper bounds involving the Berezin number and Berezin norm of bounded linear operators acting on RKHS. This study begins in this paper [24].

This article was structured as follows: In Section 2, a few lemmas teams are required to establish our main results were collected. In Section 3, we presented our main results, which include several Berezin number and norm inequalities of operators. In particular, we establish two estimations of ${\mathbf{ber}}^{2r}\left({\sum }_{i=1}^d{T}_i\right)$ where $T_i\in \mathbb{B}\left({\mathcal{H}}_{\mathrm{\Omega }}\right)$ for all $i\in \{1,\dots ,d\}$ and $r\ge 1$ with $d\in {\mathbb{N}}^{*}$. Here ${\mathbb{N}}^{*}$ denotes the set of all positive integers. Finally, we obtained the upper bounds for ${\mathbf{ber}}^4\left(S^{*}T\right)$ and ${\mathbf{ber}}^4\left(T\right)$.

In this work, the ideas and methodologies presented may serve as a starting point for future investigation in this field. We will look for other connections between the Berezin number and norm inequalities of operators, studying a possible generalization of the Berezin number. We want to find some lower bounds for the Berezin number. Further, some bounds related to the Berezin number and Berezin norm will be studied when an additional semi-inner product structure induced by a positive operator A on ${\mathcal{H}}_{\mathrm{\Omega }}$ is considered [14]. We will also study how our research can contribute to connections with “Fractals and Fractional Calculus” (see [25]).