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: (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 X̃ with where \(\widehat{k}_{\lambda }=\frac{k_{\lambda }}{ \Vert k_{ \lambda } \Vert }\) is the normalized reproducing kernel of \(\mathcal{H}\) (see [7]). Karaev in [13‐15] defined the Berezin set and the Berezin number for operator X as follows: respectively. Moreover, the Berezin number of two operators X, Y satisfies the following properties:
$$ k_{\lambda } ( z ) =\sum_{n=0}^{\infty } \overline{ \xi _{n} ( \lambda ) }\xi _{n} ( z ) $$
$$ \widetilde{X}(\mu ):= \langle X\widehat{k}_{\mu },\widehat{k} _{\mu } \rangle _{\mathcal{H}} \quad (\mu \in \varOmega ), $$
$$ \operatorname{Ber} ( X ) := \bigl\{ \widetilde{X} ( \lambda ) :\lambda \in \varOmega \bigr\} \quad \text{and} \quad \operatorname{ber} ( X ) :=\sup \bigl\{ \bigl\vert \widetilde{X} ( \lambda ) \bigr\vert :\lambda \in \varOmega \bigr\} , $$
(i)
\(\operatorname{ber} ( \nu X ) =|\nu | \operatorname{ber} ( X ) \) for all \(\nu \in \mathcal{C}\);
(ii)
\(\operatorname{ber} ( X+Y ) \leq \operatorname{ber} ( X ) +\operatorname{ber} ( X ) \).
Also, we know that for all \(X\in \mathcal{L} ( \mathcal{H} ) \). In some recent papers, several Berezin number inequalities have been investigated by authors [3‐6, 9, 10, 12, 21, 22].
$$ \operatorname{ber} ( X ) \leq w ( X ) \leq \Vert X \Vert $$
Anzeige
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: If \(p,q>1\) with \(\frac{1}{p}+\frac{1}{q}=1\), then the Young inequality is the inequality where x and y are positive real numbers (see [11]). A refinement of (1) was obtained by Kittaneh and Manasrah [17] where \(r_{0}=\min \{ \frac{1}{p},\frac{1}{q} \} \) or equivalently in which \(\nu \in [0,1]\) and \(r_{0}=\min \{\nu ,1-\nu \}\).
$$ \mathbf{ber}_{p}(X_{1},\ldots ,X_{n}):= \sup_{\lambda \in \varOmega } \Biggl(\sum_{i=1}^{n} \bigl\vert \langle X_{i}\hat{k}_{\lambda },\hat{k} _{\lambda }\rangle \bigr\vert ^{p} \Biggr)^{\frac{1}{p}}. $$
$$ xy\leq \frac{x^{p}}{p}+\frac{y^{q}}{q}, $$
(1)
$$ xy+r_{0} \bigl( x^{\frac{p}{2}}-y^{\frac{q}{2}} \bigr) ^{2}\leq \frac{x ^{p}}{p}+\frac{y^{q}}{q}, $$
(2)
$$ x^{\nu }y^{1-\nu }+r_{0} \bigl(x^{\frac{1}{2}}-y^{\frac{1}{2}}\bigr)^{2}\leq \nu x+(1-\nu )y, $$
(3)
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]).
Anzeige
Lemma 1
Let\(X\in \mathcal{L} ( \mathcal{H} ) \)be a positive operator, and let\(x\in \mathcal{H}\)be any unit vector. If\(r\geq 1\), thenand if\(0\leq r\leq 1\), then
$$ \langle Xx,x \rangle ^{r}\leq \bigl\langle X^{r}x,x \bigr\rangle $$
(4)
$$ \bigl\langle X^{r}x,x \bigr\rangle \leq \langle Xx,x \rangle ^{r}. $$
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\(\frac{1}{p}+ \frac{1}{q}=1\), thenfor all\(r\geq \frac{2}{q}\).
$$ \operatorname{ber}^{r} \bigl( ( X\mathbin{\sharp} Y ) Z \bigr) \leq \operatorname{ber} \biggl( \frac{X^{\frac{rp}{2}}}{p}+\frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr) - \frac{1}{p} \inf_{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2} $$
Proof
Using the Cauchy–Schwarz inequality, we get for all \(\lambda \in \varOmega \). By using the Young inequality and (2), we get and it follows from inequality (4) that 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 for all \(\lambda \in \varOmega \). This implies that □
$$\begin{aligned} \bigl\vert \widetilde{ ( X\mathbin{\sharp} Y ) Z } ( \lambda ) \bigr\vert ^{r} & = \bigl\vert \widetilde{ \bigl( X^{\frac{1}{2}} \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z \bigr) } ( \lambda ) \bigr\vert ^{r} \\ & = \bigl\vert \bigl\langle \bigl( X^{\frac{-1}{2}}YX^{ \frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{ \lambda },X^{\frac{1}{2}} \widehat{k}_{\lambda } \bigr\rangle \bigr\vert ^{r} \\ & \leq \bigl\Vert \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{\lambda } \bigr\Vert ^{r}\cdot \bigl\Vert X^{\frac{1}{2}}\widehat{k}_{\lambda } \bigr\Vert ^{r} \\ & = \bigl\langle \bigl( X^{\frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z\widehat{k}_{\lambda }, \bigl( X^{ \frac{-1}{2}}YX^{\frac{-1}{2}} \bigr) ^{\frac{1}{2}}X^{\frac{1}{2}}Z \widehat{k}_{\lambda } \bigr\rangle ^{\frac{r}{2}} \times \bigl\langle X^{\frac{1}{2}}\widehat{k}_{\lambda },X ^{\frac{1}{2}}\widehat{k}_{\lambda } \bigr\rangle ^{\frac{r}{2}} \\ & = \bigl( \widetilde{Z^{\star }YZ} ( \lambda ) \bigr) ^{\frac{r}{2}} \bigl( \widetilde{X} ( \lambda ) \bigr) ^{\frac{r}{2}} \end{aligned}$$
$$\begin{aligned} \bigl( \widetilde{X} ( \lambda ) \bigr) ^{ \frac{r}{2}} \bigl( \widetilde{Z^{\star }YZ} ( \lambda ) \bigr) ^{\frac{r}{2}} & \leq \frac{1}{p} \langle X\widehat{k}_{\lambda },\widehat{k} _{\lambda } \rangle ^{\frac{rp}{2}}+\frac{1}{q} \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}} \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \rangle ^{\frac{rp}{4}}- \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2}, \end{aligned}$$
$$\begin{aligned} & \frac{1}{p} \langle X\widehat{k}_{\lambda }, \widehat{k}_{ \lambda } \rangle ^{\frac{rp}{2}}+\frac{1}{q} \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}}-\frac{1}{p} \bigl( \langle X \widehat{k}_{\lambda },\widehat{k}_{\lambda } \rangle ^{\frac{rp}{4}}- \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \\ &\quad \leq \frac{1}{p} \bigl\langle X^{\frac{rp}{2}} \widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle + \frac{1}{q} \bigl\langle \bigl( Z^{\star }YZ \bigr) ^{\frac{rq}{2}}\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \rangle ^{\frac{rp}{4}}- \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \\ &\quad = \biggl\langle \biggl( \frac{X^{\frac{rp}{2}}}{p}+\frac{ ( Z ^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr) \widehat{k}_{\lambda },\widehat{k}_{\lambda } \biggr\rangle - \frac{1}{p} \bigl( \langle X \widehat{k}_{\lambda }, \widehat{k}_{\lambda } \rangle ^{ \frac{rp}{4}}- \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}$$
$$\begin{aligned} &\sup_{\lambda \in \varOmega } \bigl\vert \widetilde{ ( X\mathbin{\sharp} Y )Z } ( \lambda ) \bigr\vert ^{r} \\ &\quad \leq \sup_{\lambda \in \varOmega }\widetilde{ \biggl( \frac{X^{\frac{rp}{2}}}{p}+ \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr) } ( \lambda ) - \frac{1}{p}\inf _{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2} \end{aligned}$$
$$\begin{aligned} \operatorname{ber}^{r} \bigl( ( X\mathbin{\sharp} Y ) Z \bigr) \leq \operatorname{ber} \biggl( \frac{X ^{\frac{rp}{2}}}{p}+ \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr) -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2}. \end{aligned}$$
(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\(\frac{1}{p}+\frac{1}{q}=1\). Thenfor all\(r\geq \frac{2}{q}\).
$$ \operatorname{ber}^{r} ( X\mathbin{\sharp} Y ) \leq \operatorname{ber} \biggl( \frac{X^{\frac{rp}{2}}}{p}+\frac{Y^{ \frac{rq}{2}}}{q} \biggr) -\frac{1}{p}\inf _{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ Y } ( \lambda ) \bigr] ^{\frac{rq}{4}} \bigr) ^{2} $$
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 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 □
$$\begin{aligned} \bigl\vert \widetilde{ ( X\mathbin{\sharp} Y ) } ( \lambda ) \bigr\vert & \leq \frac{1}{2} \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{2}+ \bigl[ \widetilde{Y} ( \lambda ) \bigr] ^{2} \bigr) \\ & \leq \frac{1}{2} \bigl( \widetilde{X^{2}} ( \lambda ) + \widetilde{Y^{2}} ( \lambda ) \bigr) =\frac{1}{2} \widetilde{ \bigl( X^{2}+Y^{2} \bigr) } ( \lambda )\quad ( \lambda \in \varOmega ). \end{aligned}$$
$$ \sqrt{2}\operatorname{ber} ( X\mathbin{\sharp} Y ) \leq \operatorname{ber}_{2} ( X,Y ) \leq \operatorname{ber}^{ \frac{1}{2}} \bigl( X^{2}+Y^{2} \bigr). $$
Proposition 5
Let\(X,Y,Z\in \mathcal{L} ( \mathcal{H} ) \)such thatX, Yare positive, and let\(\frac{1}{p}+\frac{1}{q}=1\). Thenfor all\(r\geq \frac{2}{q}\).
$$\begin{aligned} \bigl\Vert ( X\mathbin{\sharp} Y )Z \bigr\Vert _{\mathrm{ber}}^{r} \leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} -\frac{1}{p} \inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}$$
Proof
Indeed, for every \(\lambda ,\mu \in \varOmega \), we have so that if we take the supremum over \(\lambda ,\mu \in \varOmega \) in inequality (6), we get □
$$\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)
$$\begin{aligned} \bigl\Vert ( X\mathbin{\sharp} Y ) Z \bigr\Vert _{\mathrm{ber}}^{r} & \leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} \\ &\quad {} -\frac{1}{p} \inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2}. \end{aligned}$$
Remark 6
It follows from inequality and inequality (6) that
$$\begin{aligned} &\inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X\widehat{k} _{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \\ &\quad = \inf_{\mu ,\lambda \in \varOmega } \bigl( \langle X\widehat{k} _{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{2}}+ \bigl\langle Z ^{\star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}}-2 \langle X\widehat{k}_{\mu }, \widehat{k}_{ \mu } \rangle ^{\frac{rp}{4}} \bigl\langle Z^{\star }YZ \widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{ \frac{rq}{4}} \bigr) \\ &\quad \geq \inf_{\mu \in \varOmega } \langle X\widehat{k}_{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{2}}+ \inf_{\lambda \in \varOmega } \bigl\langle Z^{\star }YZ\widehat{k}_{ \lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}} -2 \sup_{\mu \in \varOmega } \langle X\widehat{k}_{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{4}} \sup_{\lambda \in \varOmega } \bigl\langle Z^{\star }YZ\widehat{k}_{ \lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \\ &\quad = m ( X ) ^{\frac{rp}{2}}+ m \bigl( Z^{\star }YZ \bigr) ^{\frac{rq}{2}}-2 \bigl\Vert Z^{\star }YZ \bigr\Vert _{\mathrm{ber}}^{ \frac{rq}{4}} \Vert X \Vert _{\mathrm{ber}}^{\frac{rp}{4}} \end{aligned}$$
$$\begin{aligned} \bigl\Vert ( X\mathbin{\sharp} Y )Z \bigr\Vert _{\mathrm{ber}}^{r} &\leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} \\ &\quad {} - \bigl(m ( X ) ^{\frac{rp}{2}}+ m \bigl( Z^{\star }YZ \bigr)^{\frac{rq}{2}}-2 \bigl\Vert Z^{\star }YZ \bigr\Vert _{\mathrm{ber}}^{\frac{rq}{4}} \Vert X \Vert _{\mathrm{ber}}^{\frac{rp}{4}} \bigr). \end{aligned}$$
Proposition 7
Let\(X,Y, Z\in \mathcal{L} ( \mathcal{H} ) \)such thatX, Yare positive, and let\(p\geq q>1\), where\(\frac{1}{p}+ \frac{1}{q}=1\). Thenfor all\(r\geq \frac{2}{q}\).
$$\begin{aligned} \bigl( \Vert X \Vert _{\mathrm{ber}} \bigl\Vert Z^{\star }YZ \bigr\Vert _{\mathrm{ber}} \bigr) ^{\frac{r}{2}} &\leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} \\ &\quad {} -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}} \bigr] - \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}$$
Proof
By inequality (2), we have for all \(\lambda ,\mu \in \varOmega \) and taking supremum over \(\lambda ,\mu \in \varOmega \) in the above inequality, we get □
$$\begin{aligned} & \langle X\widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{r}{2}} \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{r}{2}} \\ &\quad \leq \frac{1}{p} \langle X\widehat{k}_{\mu }, \widehat{k}_{ \mu } \rangle ^{\frac{rp}{2}}+\frac{1}{q} \bigl\langle Z^{ \star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{2}} \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{ \star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \\ &\quad \leq \frac{1}{p} \bigl\langle X^{\frac{rp}{2}} \widehat{k}_{\mu }, \widehat{k}_{\mu } \bigr\rangle + \frac{1}{q} \bigl\langle \bigl( Z ^{\star }YZ \bigr) ^{\frac{rq}{2}}\widehat{k}_{\lambda },\widehat{k} _{\lambda } \bigr\rangle \\ &\qquad {} -\frac{1}{p} \bigl( \langle X\widehat{k}_{\mu }, \widehat{k}_{\mu } \rangle ^{\frac{rp}{4}}- \bigl\langle Z^{ \star }YZ\widehat{k}_{\lambda },\widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2} \end{aligned}$$
$$\begin{aligned} \bigl( \Vert X \Vert _{\mathrm{ber}} \bigl\Vert Z^{\star }YZ \bigr\Vert _{\mathrm{ber}} \bigr) ^{\frac{r}{2}} & \leq \biggl\Vert \frac{X^{\frac{rp}{2}}}{p} \biggr\Vert _{\mathrm{ber}}+ \biggl\Vert \frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q} \biggr\Vert _{\mathrm{ber}} \\ &\quad {} -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \langle X \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle ^{\frac{rp}{4}} \bigr] - \bigl\langle Z^{\star }YZ\widehat{k}_{\lambda }, \widehat{k}_{\lambda } \bigr\rangle ^{\frac{rq}{4}} \bigr) ^{2}. \end{aligned}$$
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\(f(t)g(t)=t\) (\(t\in [0, \infty )\)), thenwhere\(X \in {\mathcal{L}}({\mathcal{H}})\)and\(x, y\in {\mathcal{H}}\).
$$ \bigl| \langle Xx, y \rangle \bigr| \leq \bigl\| f\bigl(| X | \bigr)x \bigr\| \bigl\| g\bigl(\bigl| X^{*}\bigr| \bigr)x \bigr\| , $$
In the next theorem we show an upper bound for the generalized Euclidean Berezin number.
Theorem 9
Let\(X_{i}, Y_{i}, Z_{i}, \in {\mathcal{L}}({\mathcal{H}})\) (\(1 \leq i \leq n\)). Thenwhere\(f, g: [0, \infty )\longrightarrow \mathcal{R} \)are nonnegative continuous such that\(f(t)g(t)=t\) (\(t\in [0, \infty )\)) and\(p, r\geq 1\).
$$\begin{aligned} &\operatorname{ber}_{p}^{p} \bigl(X_{1}^{*}Z_{1}Y_{1}, \ldots,X_{n}^{*}Z_{n}Y_{n} \bigr) \\ &\quad \leq \frac{n^{1-\frac{1}{r}}}{2^{\frac{1}{r}}}\mathrm{ber}^{\frac{1}{r}} \Biggl(\sum _{i=1}^{n}\bigl[Y_{i}^{*} f^{2}\bigl( \vert Z_{i} \vert \bigr)Y_{i} \bigr]^{rp}+\bigl[X_{i}^{*}g^{2} \bigl( \bigl\vert Z_{i} ^{*} \bigr\vert \bigr)X_{i}\bigr]^{rp} \Biggr) \\ &\qquad {} -\frac{1}{2}\inf_{\lambda \in \varOmega } \Biggl( \sum _{i=1}^{n} \bigl(\sqrt{\bigl\langle \bigl(X_{i}^{*}g^{2}\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)X_{i} \bigr)^{p}\hat{k} _{\lambda }, \hat{k}_{\lambda }\bigr\rangle } -\sqrt{ \bigl\langle \bigl(Y_{i}^{*}f ^{2}\bigl( \vert Z_{i} \vert \bigr)Y_{i} \bigr)^{p}\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2} \Biggr), \end{aligned}$$
(7)
Proof
For any \(\hat{k}_{\lambda }\in {\mathcal{H}(\varOmega )}\), we have By taking the supremum on \(\hat{k}_{\lambda }\in {\mathcal{H}}\) with \(\|\hat{k}_{\lambda }\|=1\), we reach the desired inequality. □
$$\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}$$
Corollary 10
Let\(Z_{i}\in {\mathcal{L}}({\mathcal{H}}) \) (\(1\leq i \leq n\)) and\(r, p\geq 1\). Thenwhere\(f, g: [0, \infty )\longrightarrow \mathcal{R} \)are nonnegative continuous such that\(f(t)g(t)=t\) (\(t\in [0, \infty )\)).
$$\begin{aligned} \operatorname{ber}_{p}^{p}(Z_{1},\ldots ,Z_{n})&\leq \frac{n^{1-\frac{1}{r}}}{2^{ \frac{1}{r}}}\mathrm{ber}^{\frac{1}{r}} \Biggl(\sum_{i=1}^{n}[ f^{2} \bigl( \vert Z_{i} \vert \bigr)^{rp}+g ^{2}\bigl( \bigl\vert Z_{i}^{*} \bigr\vert \bigr)^{rp} \Biggr) \\ &\quad {} -\frac{1}{2}\sum_{i=1}^{n} \bigl(\sqrt{\bigl\langle g^{2p}\bigl( \bigl\vert Z_{i} ^{*} \bigr\vert \bigr)\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } -\sqrt{ \bigl\langle f^{2p} \bigl( \vert Z_{i} \vert \bigr)\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2}, \end{aligned}$$
In particular, if\(X, Y\in {\mathcal{L}}({\mathcal{H}})\), then for all\(p\geq 1\)and\(0\leq \nu \leq 1\)where
$$ \operatorname{ber}_{p}^{p}(X, Y)\leq \frac{1}{2} \operatorname{ber} \bigl( \vert X \vert ^{2\nu p}+ \bigl\vert X^{*} \bigr\vert ^{2(1- \nu )p}+ \vert Y \vert ^{2\nu p}+ \bigl\vert Y^{*} \bigr\vert ^{2(1-\nu )p} \bigr)- \inf_{\lambda \in \varOmega }\delta (\hat{k}_{\lambda }), $$
$$\begin{aligned} \delta (\hat{k}_{\lambda })&=\frac{1}{2} \bigl[\bigl( \bigl\langle \vert X \vert ^{2\nu p} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}} - \bigl\langle \bigl\vert X^{*} \bigr\vert ^{2(1-\nu ) p}\hat{k}_{\lambda }, \hat{k}_{\lambda }x \bigr\rangle ^{\frac{1}{2}}\bigr)^{2} \\ &\quad {} + \bigl(\bigl\langle \vert Y \vert ^{2\nu p}\hat{k}_{\lambda }, \hat{k}_{\lambda } \bigr\rangle ^{\frac{1}{2}} - \bigl\langle \bigl\vert Y^{*} \bigr\vert ^{2(1-\nu )p}\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}}\bigr)^{2} \bigr]. \end{aligned}$$
In the last theorem, we show another upper bound for \(\operatorname{ber}_{p}(T_{1},\ldots,T_{n})\).
Theorem 11
Let\(Z_{i}\in {\mathcal{L}}({\mathcal{H}})\) (\(1\leq i \leq n\)). Thenwhere\(p\geq 1\), \(0\leq \nu \leq 1\), and\(\delta (\hat{k}_{\lambda })= (\sqrt{ \langle |Z_{i}^{*}|^{2(1-\nu )}\hat{k}_{\lambda }, \hat{k}_{\lambda }\rangle } - \sqrt{ \langle |Z_{i}|^{2\nu }\hat{k} _{\lambda }, \hat{k}_{\lambda }\rangle } )^{2} \).
$$ \operatorname{ber}_{p}(Z_{1}, \ldots,Z_{n})\leq \frac{1}{2} \Biggl[\sum _{i=}^{n} \Bigl(\operatorname{ber} \bigl( \vert Z_{i} \vert ^{2\nu }+ \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )} \bigr)-2\inf_{ \Vert x \Vert =1} \delta ( \hat{k}_{\lambda }) \Bigr)^{p} \Biggr]^{\frac{1}{p}}, $$
(8)
Proof
Let \(\hat{k}_{\lambda }\in {\mathcal{H}(\varOmega )}\). Then, by using Lemma 8 and inequality (3), we have Thus 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. □
$$\begin{aligned} &\sum_{i=1}^{n} \bigl\vert \langle Z_{i}\hat{k}_{\lambda }, \hat{k}_{\lambda } \rangle \bigr\vert ^{p} \\ &\quad \leq \sum_{i=1}^{n}\bigl(\bigl\langle \vert Z_{i} \vert ^{2\nu }\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}}\bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1- \nu )} \hat{k}_{\lambda },\hat{k}_{\lambda }\bigr\rangle ^{\frac{1}{2}} \bigr)^{p} \quad (\text{by Lemma 8}) \\ &\quad \leq \frac{1}{2^{p}}\sum_{i=1}^{n} [\bigl\langle \vert Z_{i} \vert ^{2\nu } \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle +\bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1- \nu )} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle \\ &\qquad {} - \bigl(\sqrt{ \bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )}\hat{k}_{\lambda }, \hat{k}_{\lambda } \bigr\rangle } -\sqrt{ \bigl\langle \vert Z_{i} \vert ^{2\nu } \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2} ]^{p} \quad (\text{by (3)}) \\ &\quad =\frac{1}{2^{p}}\sum_{i=1}^{n} \bigl[\bigl\langle \vert Z_{i} \vert ^{2\nu }+ \bigl\vert Z_{i} ^{*} \bigr\vert ^{2(1-\nu )} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle - \bigl(\sqrt{ \bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )}\hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } {} -\sqrt{ \bigl\langle \vert Z_{i} \vert ^{2\nu } \hat{k}_{\lambda }, \hat{k} _{\lambda }\bigr\rangle } \bigr)^{2} \bigr]^{p}. \end{aligned}$$
$$\begin{aligned} \Biggl(\sum_{i=1}^{n} \bigl\vert \langle Z_{i}\hat{k}_{\lambda }, \hat{k}_{ \lambda } \rangle \bigr\vert ^{p} \Biggr)^{\frac{1}{p}} &\leq \frac{1}{2} \Biggl[ \sum_{i=1}^{n} \bigl(\bigl\langle \vert Z_{i} \vert ^{2\nu }+ \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle \\ &\quad {} - \bigl(\sqrt{ \bigl\langle \bigl\vert Z_{i}^{*} \bigr\vert ^{2(1-\nu )}\hat{k}_{\lambda }, \hat{k}_{\lambda } \bigr\rangle } - \sqrt{ \bigl\langle \vert Z_{i} \vert ^{2\nu } \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle } \bigr)^{2} \bigr)^{p} \Biggr]^{\frac{1}{p}} \\ &=\frac{1}{2} \Biggl[\sum_{i=1}^{n} \bigl(\bigl\langle \vert Z_{i} \vert ^{2\nu }+ \bigl\vert Z _{i}^{*} \bigr\vert ^{2(1-\nu )} \hat{k}_{\lambda }, \hat{k}_{\lambda }\bigr\rangle -2 \delta ( \hat{k}_{\lambda }) \bigr)^{p} \Biggr]^{\frac{1}{p}}. \end{aligned}$$
Acknowledgements
The first author would like to thank the Tusi Mathematical Research Group (TMRG).
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The authors declare that they have no competing interests.
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