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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 \(\frac{1}{p}+\frac{1}{q}=1\).

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.

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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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

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MSC

  • 47A63
  • 15A60

Keywords

  • Inequalities
  • Berezin number
  • Berezin symbol
  • Geometric mean