# New estimations for the Berezin number inequality

## Abstract

In this paper, by the definition of Berezin number, we present some inequalities involving the operator geometric mean. For instance, it is shown that if $$X, Y, Z\in {\mathcal{L}}(\mathcal{H})$$ such that X and Y are positive operators, then

\begin{aligned} \operatorname{ber}^{r} \bigl( ( X\mathbin{\sharp} Y ) Z \bigr) &\leq \operatorname{ber} \biggl(\frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q}+ \frac{X^{ \frac{rp}{2}}}{p} \biggr) -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2}, \end{aligned}

in which $$X\mathbin{\sharp} Y=X^{\frac{1}{2}} ( X^{-\frac{1}{2}}YX^{- \frac{1}{2}} ) ^{\frac{1}{2}}X^{\frac{1}{2}}$$, $$p\geq q>1$$ such that $$r\geq \frac{2}{q}$$ and $$\frac{1}{p}+\frac{1}{q}=1$$.

## 1 Introduction and preliminaries

We denote the $$C^{*}$$-algebra of all bounded linear operators on a separable complex Hilbert space $${\mathcal{H}}$$ with $$\mathcal{L} ( \mathcal{H} )$$. An operator $$X\in \mathcal{L} ( \mathcal{H} )$$ is called positive if $$\langle Xx,x\rangle \geq 0$$ for every $$x\in {\mathcal{H} }$$, and in this case we write $$X\geq 0$$. The numerical range and numerical radius of $$X\in \mathcal{L} ( \mathcal{H} )$$ are respectively defined by $$W ( X ) := \{ \langle Xf,f \rangle :f \in \mathcal{H}, \Vert f \Vert =1 \}$$ and $$w ( X ) :=\sup \{ \vert f \vert :f \in W ( X ) \}$$. We denote by $$\mathcal{F} ( \varOmega )$$ the set of all complex-valued functions on a nonempty set Ω. Let $$\mathcal{H}=\mathcal{H} ( \varOmega ) \subset \mathcal{F} ( \varOmega )$$ be a Hilbert space. The Riesz representation theorem makes certain that a functional Hilbert space has a reproducing kernel, which is a function $$k_{\lambda }:\varOmega \times \varOmega \rightarrow \mathcal{H}$$, that is called the reproducing kernel enjoying the reproducing property $$k_{\lambda }:=k ( \cdot, \lambda ) \in \mathcal{H}$$ ($$\lambda \in \varOmega$$) such that $$f(\lambda )= \langle f,k_{\lambda } \rangle _{\mathcal{H}}$$, in which $$\lambda \in \varOmega$$ and $$f\in \mathcal{H}$$ (see [18]). For $$\{ \xi _{n} ( z ) \} _{n \geq 0}$$, an orthonormal basis of the space $$\mathcal{H} ( \varOmega )$$, the reproducing kernel can be presented as follows:

$$k_{\lambda } ( z ) =\sum_{n=0}^{\infty } \overline{ \xi _{n} ( \lambda ) }\xi _{n} ( z )$$

(see [2, 18] and the references therein). Throughout the paper, $$\mathcal{H}=\mathcal{H}(\varOmega )$$ for some nonempty set Ω. If $$X\in \mathcal{L}(\mathcal{H})$$, then the Berezin symbol of X is the function with

$$\widetilde{X}(\mu ):= \langle X\widehat{k}_{\mu },\widehat{k} _{\mu } \rangle _{\mathcal{H}} \quad (\mu \in \varOmega ),$$

where $$\widehat{k}_{\lambda }=\frac{k_{\lambda }}{ \Vert k_{ \lambda } \Vert }$$ is the normalized reproducing kernel of $$\mathcal{H}$$ (see [7]). Karaev in [1315] defined the Berezin set and the Berezin number for operator X as follows:

$$\operatorname{Ber} ( X ) := \bigl\{ \widetilde{X} ( \lambda ) :\lambda \in \varOmega \bigr\} \quad \text{and} \quad \operatorname{ber} ( X ) :=\sup \bigl\{ \bigl\vert \widetilde{X} ( \lambda ) \bigr\vert :\lambda \in \varOmega \bigr\} ,$$

respectively. Moreover, the Berezin number of two operators X, Y satisfies the following properties:

1. (i)

$$\operatorname{ber} ( \nu X ) =|\nu | \operatorname{ber} ( X )$$ for all $$\nu \in \mathcal{C}$$;

2. (ii)

$$\operatorname{ber} ( X+Y ) \leq \operatorname{ber} ( X ) +\operatorname{ber} ( X )$$.

Also, we know that

$$\operatorname{ber} ( X ) \leq w ( X ) \leq \Vert X \Vert$$

for all $$X\in \mathcal{L} ( \mathcal{H} )$$. In some recent papers, several Berezin number inequalities have been investigated by authors [36, 9, 10, 12, 21, 22].

Assume that $$X_{1},\ldots,X_{n}\in \mathcal{L} ( \mathcal{H} )$$ and $$p\geq 1$$. In [3], the generalized Euclidean Berezin number of $$X_{1},\ldots ,X_{n}$$ is defined as follows:

$$\mathbf{ber}_{p}(X_{1},\ldots ,X_{n}):= \sup_{\lambda \in \varOmega } \Biggl(\sum_{i=1}^{n} \bigl\vert \langle X_{i}\hat{k}_{\lambda },\hat{k} _{\lambda }\rangle \bigr\vert ^{p} \Biggr)^{\frac{1}{p}}.$$

If $$p,q>1$$ with $$\frac{1}{p}+\frac{1}{q}=1$$, then the Young inequality is the inequality

$$xy\leq \frac{x^{p}}{p}+\frac{y^{q}}{q},$$
(1)

where x and y are positive real numbers (see [11]). A refinement of (1) was obtained by Kittaneh and Manasrah [17]

$$xy+r_{0} \bigl( x^{\frac{p}{2}}-y^{\frac{q}{2}} \bigr) ^{2}\leq \frac{x ^{p}}{p}+\frac{y^{q}}{q},$$
(2)

where $$r_{0}=\min \{ \frac{1}{p},\frac{1}{q} \}$$ or equivalently

$$x^{\nu }y^{1-\nu }+r_{0} \bigl(x^{\frac{1}{2}}-y^{\frac{1}{2}}\bigr)^{2}\leq \nu x+(1-\nu )y,$$
(3)

in which $$\nu \in [0,1]$$ and $$r_{0}=\min \{\nu ,1-\nu \}$$.

For positive operators $$X,Y\in \mathcal{L} ( \mathcal{H} )$$, the operator geometric mean is the positive operator $$X\mathbin{\sharp} Y=X ^{\frac{1}{2}} ( X^{-\frac{1}{2}}YX^{-\frac{1}{2}} ) ^{ \frac{1}{2}}X^{\frac{1}{2}}$$, where it has the property $$X\mathbin{\sharp} Y=Y \mathbin{\sharp} X$$. A matrix mean inequality was established by Bhatia and Kittaneh in [8], and later this inequality was generalized in [18]. A matrix Young inequality was obtained by Ando in [1]. The matrix mean inequality and the matrix Young inequality were considered with the numerical radius norm by Salemi and Sheikhhosseini in [19, 20].

In this paper, we get some upper bounds for the Berezin number of the $$( X\mathbin{\sharp} Y ) Z$$ on reproducing kernel Hilbert spaces (RKHS), where $$Z\in \mathcal{L} ( \mathcal{H} )$$ is arbitrary, and give some Berezin number inequalities. We also present some inequalities for the generalized Euclidean Berezin number.

## 2 Main results

We need the following lemma to prove our results (see [16]).

### Lemma 1

Let$$X\in \mathcal{L} ( \mathcal{H} )$$be a positive operator, and let$$x\in \mathcal{H}$$be any unit vector. If$$r\geq 1$$, then

$$\langle Xx,x \rangle ^{r}\leq \bigl\langle X^{r}x,x \bigr\rangle$$
(4)

and if$$0\leq r\leq 1$$, then

$$\bigl\langle X^{r}x,x \bigr\rangle \leq \langle Xx,x \rangle ^{r}.$$

Before giving our next result, we set $$\Vert X \Vert _{ \operatorname{ber}}:=\sup \{ \vert \langle X \widehat{k}_{\lambda },\widehat{k}_{\mu } \rangle \vert : \lambda ,\mu \in \varOmega \}$$ and $$m ( X ) := \inf_{\lambda \in \varOmega } \vert \widetilde{X} ( \lambda ) \vert$$.

### Theorem 2

Let$$X,Y,Z\in \mathcal{L} ( \mathcal{H} )$$be operators such thatX, Yare positive. If$$p\geq q>1$$with$$\frac{1}{p}+ \frac{1}{q}=1$$, then

$$\operatorname{ber}^{r} \bigl( ( X\mathbin{\sharp} Y ) Z \bigr) \leq \operatorname{ber} \biggl( \frac{X^{\frac{rp}{2}}}{p}+\frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr) - \frac{1}{p} \inf_{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2}$$

for all$$r\geq \frac{2}{q}$$.

### Proof

Using the Cauchy–Schwarz inequality, we get

\begin{aligned} \bigl\vert \widetilde{ ( X\mathbin{\sharp} Y ) Z } ( \lambda ) \bigr\vert ^{r} & = \bigl\vert \widetilde{ \bigl( X^{\frac{1}{2}} \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z \bigr) } ( \lambda ) \bigr\vert ^{r} \\ & = \bigl\vert \bigl\langle \bigl( X^{\frac{-1}{2}}YX^{ \frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{ \lambda },X^{\frac{1}{2}} \widehat{k}_{\lambda } \bigr\rangle \bigr\vert ^{r} \\ & \leq \bigl\Vert \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{\lambda } \bigr\Vert ^{r}\cdot \bigl\Vert X^{\frac{1}{2}}\widehat{k}_{\lambda } \bigr\Vert ^{r} \\ & = \bigl\langle \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{\lambda }, \bigl( X^{ \frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z \widehat{k}_{\lambda } \bigr\rangle ^{\frac{r}{2}} \times \bigl\langle X^{\frac{1}{2}}\widehat{k}_{\lambda },X ^{\frac{1}{2}}\widehat{k}_{\lambda } \bigr\rangle ^{\frac{r}{2}} \\ & = \bigl( \widetilde{Z^{\star }YZ} ( \lambda ) \bigr) ^{\frac{r}{2}} \bigl( \widetilde{X} ( \lambda ) \bigr) ^{\frac{r}{2}} \end{aligned}

for all $$\lambda \in \varOmega$$. By using the Young inequality and (2), we get

\begin{aligned} \bigl( \widetilde{X} ( \lambda ) \bigr) ^{ \frac{r}{2}} \bigl( \widetilde{Z^{\star }YZ} ( \lambda ) \bigr) ^{\frac{r}{2}} & \leq \frac{1}{p} \langle X\widehat{k}_{\lambda },\widehat{k} _{\lambda } \rangle ^{\frac{rp}{2}}+\frac{1}{q} \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}} \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \rangle ^{\frac{rp}{4}}- \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2}, \end{aligned}

and it follows from inequality (4) that

\begin{aligned} & \frac{1}{p} \langle X\widehat{k}_{\lambda }, \widehat{k}_{ \lambda } \rangle ^{\frac{rp}{2}}+\frac{1}{q} \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}}-\frac{1}{p} \bigl( \langle X \widehat{k}_{\lambda },\widehat{k}_{\lambda } \rangle ^{\frac{rp}{4}}- \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \\ &\quad \leq \frac{1}{p} \bigl\langle X^{\frac{rp}{2}} \widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle + \frac{1}{q} \bigl\langle \bigl( Z^{\star }YZ \bigr) ^{\frac{rq}{2}}\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \rangle ^{\frac{rp}{4}}- \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \\ &\quad = \biggl\langle \biggl( \frac{X^{\frac{rp}{2}}}{p}+\frac{ ( Z ^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr) \widehat{k}_{\lambda },\widehat{k}_{\lambda } \biggr\rangle - \frac{1}{p} \bigl( \langle X \widehat{k}_{\lambda }, \widehat{k}_{\lambda } \rangle ^{ \frac{rp}{4}}- \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}

for all $$\lambda \in \varOmega$$. Since $$\widetilde{ ( \frac{X^{\frac{rp}{2}}}{p}+\frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} ) } ( \lambda )$$ is positive, then we have

\begin{aligned} &\sup_{\lambda \in \varOmega } \bigl\vert \widetilde{ ( X\mathbin{\sharp} Y )Z } ( \lambda ) \bigr\vert ^{r} \\ &\quad \leq \sup_{\lambda \in \varOmega }\widetilde{ \biggl( \frac{X^{\frac{rp}{2}}}{p}+ \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr) } ( \lambda ) - \frac{1}{p}\inf _{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2} \end{aligned}

for all $$\lambda \in \varOmega$$. This implies that

\begin{aligned} \operatorname{ber}^{r} \bigl( ( X\mathbin{\sharp} Y ) Z \bigr) \leq \operatorname{ber} \biggl( \frac{X ^{\frac{rp}{2}}}{p}+ \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr) -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2}. \end{aligned}
(5)

□

Taking the $$Z=I$$ in inequality (5), we have the following result.

### Corollary 3

Let$$X,Y\in \mathcal{L} ( \mathcal{H} )$$be positive operators, and let$$p\geq q>1$$with$$\frac{1}{p}+\frac{1}{q}=1$$. Then

$$\operatorname{ber}^{r} ( X\mathbin{\sharp} Y ) \leq \operatorname{ber} \biggl( \frac{X^{\frac{rp}{2}}}{p}+\frac{Y^{ \frac{rq}{2}}}{q} \biggr) -\frac{1}{p}\inf _{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ Y } ( \lambda ) \bigr] ^{\frac{rq}{4}} \bigr) ^{2}$$

for all$$r\geq \frac{2}{q}$$.

### Corollary 4

Let$$X,Y\in \mathcal{L} ( \mathcal{H} )$$be positive operators. Then

$$\sqrt{2}\operatorname{ber} ( X\mathbin{\sharp} Y ) \leq \operatorname{ber}_{2} ( X,Y ) \leq \operatorname{ber}^{ \frac{1}{2}} \bigl( X^{2}+Y^{2} \bigr).$$

### Proof

As in the same arguments in the proof of Theorem 2, if we put $$r=p=q=2$$, then we get

\begin{aligned} \bigl\vert \widetilde{ ( X\mathbin{\sharp} Y ) } ( \lambda ) \bigr\vert & \leq \frac{1}{2} \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{2}+ \bigl[ \widetilde{Y} ( \lambda ) \bigr] ^{2} \bigr) \\ & \leq \frac{1}{2} \bigl( \widetilde{X^{2}} ( \lambda ) + \widetilde{Y^{2}} ( \lambda ) \bigr) =\frac{1}{2} \widetilde{ \bigl( X^{2}+Y^{2} \bigr) } ( \lambda )\quad ( \lambda \in \varOmega ). \end{aligned}

Since $$[ \widetilde{X} ( \lambda ) ] ^{2} \geq 0$$, $$[ \widetilde{Y} ( \lambda ) ] ^{2} \geq 0$$, and $$\widetilde{ ( X^{2}+Y^{2} ) } ( \lambda ) \geq 0$$, taking the supremum over $$\lambda \in \varOmega$$, we get that

$$\sqrt{2}\operatorname{ber} ( X\mathbin{\sharp} Y ) \leq \operatorname{ber}_{2} ( X,Y ) \leq \operatorname{ber}^{ \frac{1}{2}} \bigl( X^{2}+Y^{2} \bigr).$$

□

### Proposition 5

Let$$X,Y,Z\in \mathcal{L} ( \mathcal{H} )$$such thatX, Yare positive, and let$$\frac{1}{p}+\frac{1}{q}=1$$. Then

\begin{aligned} \bigl\Vert ( X\mathbin{\sharp} Y )Z \bigr\Vert _{\mathrm{ber}}^{r} \leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} -\frac{1}{p} \inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}

for all$$r\geq \frac{2}{q}$$.

### Proof

Indeed, for every $$\lambda ,\mu \in \varOmega$$, we have

\begin{aligned} & \bigl\vert \bigl\langle ( X\mathbin{\sharp} Y ) Z\widehat{k}_{ \lambda }, \widehat{k}_{\mu } \bigr\rangle \bigr\vert ^{r} \\ &\quad = \bigl\vert \bigl\langle X^{\frac{1}{2}} \bigl( X^{\frac{-1}{2}}YX ^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{ \lambda }, \widehat{k}_{\mu } \bigr\rangle \bigr\vert \\ &\quad = \bigl\vert \bigl\langle \bigl( X^{\frac{-1}{2}}YX^{ \frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{ \lambda },X^{\frac{1}{2}} \widehat{k}_{\mu } \bigr\rangle \bigr\vert ^{r} \\ &\quad = \bigl\vert \bigl\langle \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{ \frac{1}{2}}Z\widehat{k}_{\lambda },X^{\frac{1}{2}} \widehat{k}_{ \mu } \bigr\rangle \bigr\vert ^{r} \\ &\quad \leq \bigl\Vert \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{\lambda } \bigr\Vert ^{r}\cdot \bigl\Vert X^{\frac{1}{2}}\widehat{k}_{\mu } \bigr\Vert ^{r} \\ &\quad = \bigl\langle \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{\lambda }, \bigl( X^{ \frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z \widehat{k}_{\lambda } \bigr\rangle ^{\frac{r}{2}}\times \bigl\langle X ^{\frac{1}{2}}\widehat{k}_{\mu },X^{\frac{1}{2}} \widehat{k}_{\mu } \bigr\rangle ^{\frac{r}{2}} \\ &\quad = \langle X\widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{r}{2}} \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{r}{2}} \\ &\quad \leq \frac{1}{p} \langle X\widehat{k}_{\lambda },\widehat{k} _{\lambda } \rangle ^{\frac{rp}{2}}+\frac{1}{q} \bigl\langle Z ^{\star }YZ\widehat{k}_{\mu },\widehat{k}_{\mu } \bigr\rangle ^{ \frac{rq}{2}} \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\lambda }, \widehat{k} _{\lambda } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{\star }YZ \widehat{k}_{\mu },\widehat{k}_{\mu } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \quad (\text{by (1) and (2)}) \\ &\quad \leq \frac{1}{p} \bigl\langle X^{\frac{rp}{2}} \widehat{k}_{\mu }, \widehat{k}_{\mu } \bigr\rangle + \frac{1}{q} \bigl\langle \bigl( Z ^{\star }YZ \bigr) ^{\frac{rq}{2}}\widehat{k}_{\lambda },\widehat{k} _{\lambda } \bigr\rangle \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\mu }, \widehat{k}_{ \mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{\star }YZ \widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{ \frac{rq}{4}} \bigr) ^{2} \quad(\text{by (4)}) \\ &\quad \leq \frac{1}{p} \bigl\langle X^{\frac{rp}{2}} \widehat{k}_{\mu }, \widehat{k}_{\mu } \bigr\rangle + \frac{1}{q} \bigl\langle \bigl( Z ^{\star }YZ \bigr) ^{\frac{rq}{2}}\widehat{k}_{\lambda },\widehat{k} _{\lambda } \bigr\rangle \\ &\qquad {} -\frac{1}{p} \inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}
(6)

so that if we take the supremum over $$\lambda ,\mu \in \varOmega$$ in inequality (6), we get

\begin{aligned} \bigl\Vert ( X\mathbin{\sharp} Y ) Z \bigr\Vert _{\mathrm{ber}}^{r} & \leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} \\ &\quad {} -\frac{1}{p} \inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2}. \end{aligned}

□

### Remark 6

It follows from inequality

\begin{aligned} &\inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X\widehat{k} _{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \\ &\quad = \inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X\widehat{k} _{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{2}}+ \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}}-2 \langle X\widehat{k}_{\mu }, \widehat{k}_{ \mu } \rangle ^{\frac{rp}{4}} \bigl\langle Z^{\star }YZ \widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{ \frac{rq}{4}} \bigr) \\ &\quad \geq \inf_{\mu \in \varOmega } \langle X\widehat{k}_{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{2}}+ \inf_{\lambda \in \varOmega } \bigl\langle Z^{\star }YZ\widehat{k}_{ \lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}} -2 \sup_{\mu \in \varOmega } \langle X\widehat{k}_{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{4}} \sup_{\lambda \in \varOmega } \bigl\langle Z^{\star }YZ\widehat{k}_{ \lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \\ &\quad = m ( X ) ^{\frac{rp}{2}}+ m \bigl( Z^{\star }YZ \bigr) ^{\frac{rq}{2}}-2 \bigl\Vert Z^{\star }YZ \bigr\Vert _{\mathrm{ber}}^{ \frac{rq}{4}} \Vert X \Vert _{\mathrm{ber}}^{\frac{rp}{4}} \end{aligned}

and inequality (6) that

\begin{aligned} \bigl\Vert ( X\mathbin{\sharp} Y )Z \bigr\Vert _{\mathrm{ber}}^{r} &\leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} \\ &\quad {} - \bigl(m ( X ) ^{\frac{rp}{2}}+ m \bigl( Z^{\star }YZ \bigr)^{\frac{rq}{2}}-2 \bigl\Vert Z^{\star }YZ \bigr\Vert _{\mathrm{ber}}^{\frac{rq}{4}} \Vert X \Vert _{\mathrm{ber}}^{\frac{rp}{4}} \bigr). \end{aligned}

### Proposition 7

Let$$X,Y, Z\in \mathcal{L} ( \mathcal{H} )$$such thatX, Yare positive, and let$$p\geq q>1$$, where$$\frac{1}{p}+ \frac{1}{q}=1$$. Then

\begin{aligned} \bigl( \Vert X \Vert _{\mathrm{ber}} \bigl\Vert Z^{\star }YZ \bigr\Vert _{\mathrm{ber}} \bigr) ^{\frac{r}{2}} &\leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} \\ &\quad {} -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}} \bigr] - \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}

for all$$r\geq \frac{2}{q}$$.

### Proof

By inequality (2), we have

\begin{aligned} & \langle X\widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{r}{2}} \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{r}{2}} \\ &\quad \leq \frac{1}{p} \langle X\widehat{k}_{\mu }, \widehat{k}_{ \mu } \rangle ^{\frac{rp}{2}}+\frac{1}{q} \bigl\langle Z^{ \star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}} \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{ \star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \\ &\quad \leq \frac{1}{p} \bigl\langle X^{\frac{rp}{2}} \widehat{k}_{\mu }, \widehat{k}_{\mu } \bigr\rangle + \frac{1}{q} \bigl\langle \bigl( Z ^{\star }YZ \bigr) ^{\frac{rq}{2}}\widehat{k}_{\lambda },\widehat{k} _{\lambda } \bigr\rangle \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{ \star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}

for all $$\lambda ,\mu \in \varOmega$$ and taking supremum over $$\lambda ,\mu \in \varOmega$$ in the above inequality, we get

\begin{aligned} \bigl( \Vert X \Vert _{\mathrm{ber}} \bigl\Vert Z^{\star }YZ \bigr\Vert _{\mathrm{ber}} \bigr) ^{\frac{r}{2}} & \leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} \\ &\quad {} -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}} \bigr] - \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2}. \end{aligned}

□

Now, we present the next lemma to obtain our last results.

### Lemma 8

([16])

If$$f, g: [0, \infty )\longrightarrow \mathcal{R}$$are nonnegative continuous such that$$f(t)g(t)=t$$ ($$t\in [0, \infty )$$), then

$$\bigl| \langle Xx, y \rangle \bigr| \leq \bigl\| f\bigl(| X | \bigr)x \bigr\| \bigl\| g\bigl(\bigl| X^{*}\bigr| \bigr)x \bigr\| ,$$

where$$X \in {\mathcal{L}}({\mathcal{H}})$$and$$x, y\in {\mathcal{H}}$$.

In the next theorem we show an upper bound for the generalized Euclidean Berezin number.

### Theorem 9

Let$$X_{i}, Y_{i}, Z_{i}, \in {\mathcal{L}}({\mathcal{H}})$$ ($$1 \leq i \leq n$$). Then

\begin{aligned} &\operatorname{ber}_{p}^{p} \bigl(X_{1}^{*}Z_{1}Y_{1}, \ldots,X_{n}^{*}Z_{n}Y_{n} \bigr) \\ &\quad \leq \frac{n^{1-\frac{1}{r}}}{2^{\frac{1}{r}}}\mathrm{ber}^{\frac{1}{r}} \Biggl(\sum _{i=1}^{n}\bigl[Y_{i}^{*} f^{2}\bigl( \vert Z_{i} \vert \bigr)Y_{i} \bigr]^{rp}+\bigl[X_{i}^{*}g^{2} \bigl( \bigl\vert Z_{i} ^{*} \bigr\vert \bigr)X_{i}\bigr]^{rp} \Biggr) \\ &\qquad {} -\frac{1}{2}\inf_{\lambda \in \varOmega } \Biggl( \sum _{i=1}^{n} \bigl(\sqrt{\bigl\langle \bigl(X_{i}^{*}g^{2}\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)X_{i} \bigr)^{p}\hat{k} _{\lambda }, \hat{k}_{\lambda }\bigr\rangle } -\sqrt{ \bigl\langle \bigl(Y_{i}^{*}f ^{2}\bigl( \vert Z_{i} \vert \bigr)Y_{i} \bigr)^{p}\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2} \Biggr), \end{aligned}
(7)

where$$f, g: [0, \infty )\longrightarrow \mathcal{R}$$are nonnegative continuous such that$$f(t)g(t)=t$$ ($$t\in [0, \infty )$$) and$$p, r\geq 1$$.

### Proof

For any $$\hat{k}_{\lambda }\in {\mathcal{H}(\varOmega )}$$, we have

\begin{aligned} &\sum_{i=1}^{n} \bigl\vert \bigl\langle X_{i}^{*}Z_{i}Y_{i} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle \bigr\vert ^{p} \\ &\quad =\sum_{i=1}^{n} \bigl\vert \langle Z_{i}Y_{i}\hat{k}_{\lambda }, X_{i} \hat{k} _{\lambda }\rangle \bigr\vert ^{p} \\ &\quad \leq \sum_{i=1}^{n} \bigl\Vert f \bigl( \vert Z_{i} \vert \bigr)Y_{i} \hat{k}_{\lambda } \bigr\Vert ^{p} \bigl\Vert g\bigl( \bigl\vert Z _{i}^{*} \bigr\vert \bigr)X_{i} \hat{k}_{\lambda } \bigr\Vert ^{p} \quad (\text{by Lemma 8}) \\ &\quad =\sum_{i=1}^{n}\bigl\langle f\bigl( \vert Z_{i} \vert \bigr)Y_{i}\hat{k}_{\lambda }, f\bigl( \vert Z_{i} \vert \bigr)Y _{i} \hat{k}_{\lambda }\bigr\rangle ^{\frac{p}{2}} \bigl\langle g\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)X _{i} \hat{k}_{\lambda }, g\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)X_{i}\hat{k}_{\lambda } \bigr\rangle ^{\frac{p}{2}} \\ &\quad =\sum_{i=1}^{n} \bigl\langle Y_{i}^{*} f^{2}\bigl( \vert Z_{i} \vert \bigr)Y_{i}\hat{k}_{\lambda}, \hat{k}_{\lambda }\bigr\rangle ^{\frac{p}{2}} \bigl\langle X_{i}^{*}g^{2}\bigl( \bigl\vert Z _{i}^{*} \bigr\vert \bigr)Z_{i} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle ^{ \frac{p}{2}} \\ &\quad \leq \sum_{i=1}^{n} \bigl\langle \bigl(Y_{i}^{*} f^{2}\bigl( \vert Z_{i} \vert \bigr)Y_{i}\bigr)^{p} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}} \bigl\langle \bigl(X _{i}^{*}g^{2}\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)Z_{i}\bigr)^{p}\hat{k}_{\lambda }, \hat{k}_{ \lambda }\bigr\rangle ^{\frac{1}{2}} \quad (\text{by (4)}) \\ &\quad \leq \sum_{i=1}^{n} \biggl[ \biggl( \frac{1}{2} \bigl\langle \bigl(Y_{i}^{*} f ^{2}\bigl( \vert Z_{i} \vert \bigr)Y_{i} \bigr)^{pr}\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle + \frac{1}{2} \bigl\langle \bigl(X_{i}^{*}g^{2} \bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)X_{i}\bigr)^{pr}\hat{k} _{\lambda }, \hat{k}_{\lambda }\bigr\rangle \biggr)^{\frac{1}{r}} \biggr] \quad (\text{by (2)}) \\ &\qquad {} -\frac{1}{2}\sum_{i=1}^{n} \bigl(\sqrt{\bigl\langle \bigl(X_{i}^{*}g ^{2}\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)X_{i}\bigr)^{p}\hat{k}_{\lambda }, \hat{k}_{\lambda } \bigr\rangle } -\sqrt{ \bigl\langle \bigl(Y_{i}^{*}f^{2} \bigl( \vert Z_{i} \vert \bigr)Y_{i} \bigr)^{p}, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2} \\ &\quad \leq \frac{n^{1-\frac{1}{r}}}{2^{\frac{1}{r}}} \Biggl\langle \Biggl(\sum _{i=1}^{n} \bigl( \bigl[Y_{i}^{*}f^{2} \bigl( \vert Z_{i} \vert \bigr)Y_{i} \bigr]^{rp}+ \bigl[X_{i}^{*} g^{2}\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)X_{i} \bigr]^{rp} \bigr) \Biggr) \hat{k}_{\lambda }, \hat{k}_{\lambda } \Biggr\rangle ^{\frac{1}{r}} \\ &\qquad {} -\frac{1}{2}\sum_{i=1}^{n} \bigl(\sqrt{\bigl\langle \bigl(X_{i}^{*}g ^{2}\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)X_{i}\bigr)^{p}\hat{k}_{\lambda }, \hat{k}_{\lambda } \bigr\rangle } -\sqrt{ \bigl\langle \bigl(Y_{i}^{*}f^{2} \bigl( \vert Z_{i} \vert \bigr)Y_{i} \bigr)^{p}\hat{k} _{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2}. \end{aligned}

By taking the supremum on $$\hat{k}_{\lambda }\in {\mathcal{H}}$$ with $$\|\hat{k}_{\lambda }\|=1$$, we reach the desired inequality. □

Selecting $$X_{i}=Y_{i}=I$$ for $$i=1,2,\ldots ,n$$ in Theorem 9, we get the next result.

### Corollary 10

Let$$Z_{i}\in {\mathcal{L}}({\mathcal{H}})$$ ($$1\leq i \leq n$$) and$$r, p\geq 1$$. Then

\begin{aligned} \operatorname{ber}_{p}^{p}(Z_{1},\ldots ,Z_{n})&\leq \frac{n^{1-\frac{1}{r}}}{2^{ \frac{1}{r}}}\mathrm{ber}^{\frac{1}{r}} \Biggl(\sum_{i=1}^{n}[ f^{2} \bigl( \vert Z_{i} \vert \bigr)^{rp}+g ^{2}\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)^{rp} \Biggr) \\ &\quad {} -\frac{1}{2}\sum_{i=1}^{n} \bigl(\sqrt{\bigl\langle g^{2p}\bigl( \bigl\vert Z_{i} ^{*} \bigr\vert \bigr)\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } -\sqrt{ \bigl\langle f^{2p} \bigl( \vert Z_{i} \vert \bigr)\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2}, \end{aligned}

where$$f, g: [0, \infty )\longrightarrow \mathcal{R}$$are nonnegative continuous such that$$f(t)g(t)=t$$ ($$t\in [0, \infty )$$).

In particular, if$$X, Y\in {\mathcal{L}}({\mathcal{H}})$$, then for all$$p\geq 1$$and$$0\leq \nu \leq 1$$

$$\operatorname{ber}_{p}^{p}(X, Y)\leq \frac{1}{2} \operatorname{ber} \bigl( \vert X \vert ^{2\nu p}+ \bigl\vert X^{*} \bigr\vert ^{2(1- \nu )p}+ \vert Y \vert ^{2\nu p}+ \bigl\vert Y^{*} \bigr\vert ^{2(1-\nu )p} \bigr)- \inf_{\lambda \in \varOmega }\delta (\hat{k}_{\lambda }),$$

where

\begin{aligned} \delta (\hat{k}_{\lambda })&=\frac{1}{2} \bigl[\bigl( \bigl\langle \vert X \vert ^{2\nu p} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}} - \bigl\langle \bigl\vert X^{*} \bigr\vert ^{2(1-\nu ) p}\hat{k}_{\lambda }, \hat{k}_{\lambda }x \bigr\rangle ^{\frac{1}{2}}\bigr)^{2} \\ &\quad {} + \bigl(\bigl\langle \vert Y \vert ^{2\nu p}\hat{k}_{\lambda }, \hat{k}_{\lambda } \bigr\rangle ^{\frac{1}{2}} - \bigl\langle \bigl\vert Y^{*} \bigr\vert ^{2(1-\nu )p}\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}}\bigr)^{2} \bigr]. \end{aligned}

In the last theorem, we show another upper bound for $$\operatorname{ber}_{p}(T_{1},\ldots,T_{n})$$.

### Theorem 11

Let$$Z_{i}\in {\mathcal{L}}({\mathcal{H}})$$ ($$1\leq i \leq n$$). Then

$$\operatorname{ber}_{p}(Z_{1}, \ldots,Z_{n})\leq \frac{1}{2} \Biggl[\sum _{i=}^{n} \Bigl(\operatorname{ber} \bigl( \vert Z_{i} \vert ^{2\nu }+ \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )} \bigr)-2\inf_{ \Vert x \Vert =1} \delta ( \hat{k}_{\lambda }) \Bigr)^{p} \Biggr]^{\frac{1}{p}},$$
(8)

where$$p\geq 1$$, $$0\leq \nu \leq 1$$, and$$\delta (\hat{k}_{\lambda })= (\sqrt{ \langle |Z_{i}^{*}|^{2(1-\nu )}\hat{k}_{\lambda }, \hat{k}_{\lambda }\rangle } - \sqrt{ \langle |Z_{i}|^{2\nu }\hat{k} _{\lambda }, \hat{k}_{\lambda }\rangle } )^{2}$$.

### Proof

Let $$\hat{k}_{\lambda }\in {\mathcal{H}(\varOmega )}$$. Then, by using Lemma 8 and inequality (3), we have

\begin{aligned} &\sum_{i=1}^{n} \bigl\vert \langle Z_{i}\hat{k}_{\lambda }, \hat{k}_{\lambda } \rangle \bigr\vert ^{p} \\ &\quad \leq \sum_{i=1}^{n}\bigl(\bigl\langle \vert Z_{i} \vert ^{2\nu }\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}}\bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1- \nu )} \hat{k}_{\lambda },\hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}} \bigr)^{p} \quad (\text{by Lemma 8}) \\ &\quad \leq \frac{1}{2^{p}}\sum_{i=1}^{n} [\bigl\langle \vert Z_{i} \vert ^{2\nu } \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle +\bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1- \nu )} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle \\ &\qquad {} - \bigl(\sqrt{ \bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )}\hat{k}_{\lambda }, \hat{k}_{\lambda } \bigr\rangle } -\sqrt{ \bigl\langle \vert Z_{i} \vert ^{2\nu } \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2} ]^{p} \quad (\text{by (3)}) \\ &\quad =\frac{1}{2^{p}}\sum_{i=1}^{n} \bigl[\bigl\langle \vert Z_{i} \vert ^{2\nu }+ \bigl\vert Z_{i} ^{*} \bigr\vert ^{2(1-\nu )} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle - \bigl(\sqrt{ \bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )}\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } {} -\sqrt{ \bigl\langle \vert Z_{i} \vert ^{2\nu } \hat{k}_{\lambda }, \hat{k} _{\lambda }\bigr\rangle } \bigr)^{2} \bigr]^{p}. \end{aligned}

Thus

\begin{aligned} \Biggl(\sum_{i=1}^{n} \bigl\vert \langle Z_{i}\hat{k}_{\lambda }, \hat{k}_{ \lambda } \rangle \bigr\vert ^{p} \Biggr)^{\frac{1}{p}} &\leq \frac{1}{2} \Biggl[ \sum_{i=1}^{n} \bigl(\bigl\langle \vert Z_{i} \vert ^{2\nu }+ \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle \\ &\quad {} - \bigl(\sqrt{ \bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )}\hat{k}_{\lambda }, \hat{k}_{\lambda } \bigr\rangle } - \sqrt{ \bigl\langle \vert Z_{i} \vert ^{2\nu } \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2} \bigr)^{p} \Biggr]^{\frac{1}{p}} \\ &=\frac{1}{2} \Biggl[\sum_{i=1}^{n} \bigl(\bigl\langle \vert Z_{i} \vert ^{2\nu }+ \bigl\vert Z _{i}^{*} \bigr\vert ^{2(1-\nu )} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle -2 \delta ( \hat{k}_{\lambda }) \bigr)^{p} \Biggr]^{\frac{1}{p}}. \end{aligned}

If we get the supremum over all $$\hat{k}_{\lambda }\in {\mathcal{H}( \varOmega )}$$ with $$\|\hat{k}_{\lambda }\|=1$$, then we reach the desired result. □

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### Acknowledgements

The first author would like to thank the Tusi Mathematical Research Group (TMRG).

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Bakherad, M., Yamancı, U. New estimations for the Berezin number inequality. J Inequal Appl 2020, 40 (2020). https://doi.org/10.1186/s13660-020-2307-0