New estimations for the Berezin number inequality

<jats:p>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 <jats:inline-formula><jats:alternatives><jats:tex-math>$X, Y, Z\in {\mathcal{L}}(\mathcal{H})$</jats:tex-math><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"><mml:mi>X</mml:mi><mml:mo>,</mml:mo><mml:mi>Y</mml:mi><mml:mo>,</mml:mo><mml:mi>Z</mml:mi><mml:mo>∈</mml:mo><mml:mi>L</mml:mi><mml:mo>(</mml:mo><mml:mi>H</mml:mi><mml:mo>)</mml:mo></mml:math></jats:alternatives></jats:inline-formula> such that <jats:italic>X</jats:italic> and <jats:italic>Y</jats:italic> are positive operators, then <jats:disp-formula><jats:alternatives><jats:tex-math> $$\begin{aligned} \operatorname{ber}^{r} \bigl( ( X\mathbin{\sharp} Y ) Z \bigr) &\leq \operatorname{ber} \biggl(\frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q}+ \frac{X^{ \frac{rp}{2}}}{p} \biggr) -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2}, \end{aligned}$$ </jats:tex-math><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"><mml:msup><mml:mo>ber</mml:mo><mml:mi>r</mml:mi></mml:msup><mml:mrow><mml:mo>(</mml:mo><mml:mo>(</mml:mo><mml:mi>X</mml:mi><mml:mo>♯</mml:mo><mml:mi>Y</mml:mi><mml:mo>)</mml:mo><mml:mi>Z</mml:mi><mml:mo>)</mml:mo></mml:mrow><mml:mo>≤</mml:mo><mml:mo>ber</mml:mo><mml:mrow><mml:mo>(</mml:mo><mml:mfrac><mml:msup><mml:mrow><mml:mo>(</mml:mo><mml:msup><mml:mi>Z</mml:mi><mml:mo>⋆</mml:mo></mml:msup><mml:mi>Y</mml:mi><mml:mi>Z</mml:mi><mml:mo>)</mml:mo></mml:mrow><mml:mfrac><mml:mrow><mml:mi>r</mml:mi><mml:mi>q</mml:mi></mml:mrow><mml:mn>2</mml:mn></mml:mfrac></mml:msup><mml:mi>q</mml:mi></mml:mfrac><mml:mo>+</mml:mo><mml:mfrac><mml:msup><mml:mi>X</mml:mi><mml:mfrac><mml:mrow><mml:mi>r</mml:mi><mml:mi>p</mml:mi></mml:mrow><mml:mn>2</mml:mn></mml:mfrac></mml:msup><mml:mi>p</mml:mi></mml:mfrac><mml:mo>)</mml:mo></mml:mrow><mml:mo>−</mml:mo><mml:mfrac><mml:mn>1</mml:mn><mml:mi>p</mml:mi></mml:mfrac><mml:munder><mml:mo>inf</mml:mo><mml:mrow><mml:mi>λ</mml:mi><mml:mo>∈</mml:mo><mml:mi>Ω</mml:mi></mml:mrow></mml:munder><mml:msup><mml:mrow><mml:mo>(</mml:mo><mml:msup><mml:mrow><mml:mo>[</mml:mo><mml:mover><mml:mi>X</mml:mi><mml:mo>˜</mml:mo></mml:mover><mml:mo>(</mml:mo><mml:mi>λ</mml:mi><mml:mo>)</mml:mo><mml:mo>]</mml:mo></mml:mrow><mml:mfrac><mml:mrow><mml:mi>r</mml:mi><mml:mi>p</mml:mi></mml:mrow><mml:mn>4</mml:mn></mml:mfrac></mml:msup><mml:mo>−</mml:mo><mml:msup><mml:mrow><mml:mo>[</mml:mo><mml:mover><mml:mrow><mml:mo>(</mml:mo><mml:msup><mml:mi>Z</mml:mi><mml:mo>⋆</mml:mo></mml:msup><mml:mi>Y</mml:mi><mml:mi>Z</mml:mi><mml:mo>)</mml:mo></mml:mrow><mml:mo>˜</mml:mo></mml:mover><mml:mo>(</mml:mo><mml:mi>λ</mml:mi><mml:mo>)</mml:mo><mml:mo>]</mml:mo></mml:mrow><mml:mfrac><mml:mrow><mml:mi>r</mml:mi><mml:mi>q</mml:mi></mml:mrow><mml:mn>4</mml:mn></mml:mfrac></mml:msup><mml:mo>)</mml:mo></mml:mrow><mml:mn>2</mml:mn></mml:msup><mml:mo>,</mml:mo></mml:math></jats:alternatives></jats:disp-formula> in which <jats:inline-formula><jats:alternatives><jats:tex-math>$X\mathbin{\sharp} Y=X^{\frac{1}{2}} ( X^{-\frac{1}{2}}YX^{- \frac{1}{2}} ) ^{\frac{1}{2}}X^{\frac{1}{2}}$</jats:tex-math><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"><mml:mi>X</mml:mi><mml:mo>♯</mml:mo><mml:mi>Y</mml:mi><mml:mo>=</mml:mo><mml:msup><mml:mi>X</mml:mi><mml:mfrac><mml:mrow><mml:mn>1</mml:mn></mml:mrow><mml:mn>2</mml:mn></mml:mfrac></mml:msup><mml:msup><mml:mrow><mml:mo>(</mml:mo><mml:msup><mml:mi>X</mml:mi><mml:mrow><mml:mo>−</mml:mo><mml:mfrac><mml:mrow><mml:mn>1</mml:mn></mml:mrow><mml:mn>2</mml:mn></mml:mfrac></mml:mrow></mml:msup><mml:mi>Y</mml:mi><mml:msup><mml:mi>X</mml:mi><mml:mrow><mml:mo>−</mml:mo><mml:mfrac><mml:mrow><mml:mn>1</mml:mn></mml:mrow><mml:mn>2</mml:mn></mml:mfrac></mml:mrow></mml:msup><mml:mo>)</mml:mo></mml:mrow><mml:mfrac><mml:mrow><mml:mn>1</mml:mn></mml:mrow><mml:mn>2</mml:mn></mml:mfrac></mml:msup><mml:msup><mml:mi>X</mml:mi><mml:mfrac><mml:mrow><mml:mn>1</mml:mn></mml:mrow><mml:mn>2</mml:mn></mml:mfrac></mml:msup></mml:math></jats:alternatives></jats:inline-formula>, <jats:inline-formula><jats:alternatives><jats:tex-math>$p\geq q>1$</jats:tex-math><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"><mml:mi>p</mml:mi><mml:mo>≥</mml:mo><mml:mi>q</mml:mi><mml:mo>></mml:mo><mml:mn>1</mml:mn></mml:math></jats:alternatives></jats:inline-formula> such that <jats:inline-formula><jats:alternatives><jats:tex-math>$r\geq \frac{2}{q}$</jats:tex-math><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"><mml:mi>r</mml:mi><mml:mo>≥</mml:mo><mml:mfrac><mml:mn>2</mml:mn><mml:mi>q</mml:mi></mml:mfrac></mml:math></jats:alternatives></jats:inline-formula> and <jats:inline-formula><jats:alternatives><jats:tex-math>$\frac{1}{p}+\frac{1}{q}=1$</jats:tex-math><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"><mml:mfrac><mml:mn>1</mml:mn><mml:mi>p</mml:mi></mml:mfrac><mml:mo>+</mml:mo><mml:mfrac><mml:mn>1</mml:mn><mml:mi>q</mml:mi></mml:mfrac><mml:mo>=</mml:mo><mml:mn>1</mml:mn></mml:math></jats:alternatives></jats:inline-formula>.</jats:p>


Introduction and preliminaries
We denote the C * -algebra of all bounded linear operators on a separable complex Hilbert space H with L(H). An operator X ∈ L(H) is called positive if Xx, x ≥ 0 for every x ∈ H, and in this case we write X ≥ 0. The numerical range and numerical radius of X ∈ L(H) are respectively defined by W (X) := { Xf , f : f ∈ H, f = 1} and w(X) := sup{|f | : f ∈ W (X)}. We denote by F(Ω) the set of all complex-valued functions on a nonempty set Ω. Let H = H(Ω) ⊂ F(Ω) be a Hilbert space. The Riesz representation theorem makes certain that a functional Hilbert space has a reproducing kernel, which is a function k λ : Ω × Ω → H, that is called the reproducing kernel enjoying the reproducing property k λ := k(·, λ) ∈ H (λ ∈ Ω) such that f (λ) = f , k λ H , in which λ ∈ Ω and f ∈ H (see [18]). For {ξ n (z)} n≥0 , an orthonormal basis of the space H(Ω), the reproducing kernel can be presented as follows: where k λ = k λ k λ is the normalized reproducing kernel of H (see [7]). Karaev in [13][14][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: Also, we know that for all X ∈ L(H). In some recent papers, several Berezin number inequalities have been investigated by authors [3-6, 9, 10, 12, 21, 22]. Assume that X 1 , . . . , X n ∈ L(H) and p ≥ 1. In [3], the generalized Euclidean Berezin number of X 1 , . . . , X n is defined as follows: If p, q > 1 with 1 p + 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] xy where r 0 = min{ 1 p , 1 q } or equivalently in which ν ∈ [0, 1] and r 0 = min{ν, 1 -ν}. For positive operators X, Y ∈ L(H), the operator geometric mean is the positive opera- where it has the property X Y = Y 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 Y )Z on reproducing kernel Hilbert spaces (RKHS), where Z ∈ L(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]).
Proof Using the Cauchy-Schwarz inequality, we get for all λ ∈ Ω. By using the Young inequality and (2), we get and it follows from inequality (4) that for all λ ∈ Ω. This implies that Taking the Z = I in inequality (5), we have the following result.

Corollary 3 Let X, Y ∈ L(H) be positive operators, and let p
for all r ≥ 2 q .
Proof By inequality (2), we have for all λ, μ ∈ Ω and taking supremum over λ, μ ∈ Ω in the above inequality, we get X ber Z YZ ber In the next theorem we show an upper bound for the generalized Euclidean Berezin number.