Some results related to the Berezin number inequalities

In this paper, we prove reverse inequalities for the so-called Berezin number of some operators. Also, by using the classical Jensen and Young inequalities, we obtain upper bounds for Berezin number of AXB and AαXB1−α for the case when 0 ≤ α ≤ 1 .

In the present paper, by using some ideas from [12,13], we will prove reverse inequalities for the so-called Berezin number of some operators acting in the reproducing kernel Hilbert space. Also, we obtain upper bounds for the Berezin number of A α XB α and A α XB 1−α for the case when 0 ≤ α ≤ 1. The following two lemmas are well known (and very easy to verify).

Relations between numerical radius and Berezin number
Proof Indeed, if f is a multiplier, then we have for all λ ∈ Ω, as desired. 2 Proof In fact, since f ∈ H (Ω) and f g ∈ H (Ω) for all g ∈ H (Ω), by using Lemma 1, we have: Since M f is a closed operator defined in hull space H (Ω), by the closed graph theorem M f is bounded (see, for instance, Aronzajn [2]). The last inequality shows that f is bounded. In this short section, we prove the relations between the numerical radius and Berezin number of reproducing kernel Hilbert space operators, which improves some results in [17,23].

Theorem 1 Let H = H(Ω) be a RKHS of complex-valued functions
on Ω with reproducing kernel k λ such that it has a dense multiplier M (Ω) and k 0 = 1.

Let A : H → H be a bounded linear operator (i.e. A ∈ B (H) ). Then
Proof By assumption M (Ω) is dense in H . Then it is standard to show that According to Lemma 2, M (Ω) consists of bounded functions of the space H(Ω). Then we have: and hence On the other hand, for any V ∈ J and g ∈ H , we have:

Thus, sup
It remains only to combine inequalities (2.1) and (2.2) to get the required result. The theorem is proven.

Reverse inequalities for the Berezin numbers of operators
Our next results in this section are mainly motivated with Dragomir's survey paper [13], where he proved relations only between the norm and numerical radius of operators. Here we investigate similar questions also for Berezin numbers of operators A and |A| 2 := A * A ; here, |A| := (A * A) 1/2 is a so-called module of operator A.
or equivalently, Hence, Taking the supremum over λ ∈ Ω in (3.3), we have the following inequality: By arithmetic-geometric mean inequality, , (3.5) and hence by (3.4) and (3.5) we deduce the desired inequality (3.2), because it is elementary to see that actually . The theorem is proved. 2 Proof Utilizing the fact that in any Hilbert space the following two statements are equivalent, we conclude that (3.6) is equivalent to for any λ ∈ Ω, which in its turn is equivalent with the following inequality: On utilizing (3.4) and (3.8), we deduce inequality (3.9), as desired. ≤ r and |µ| = ber (A) , µ ∈ C , then ber which implies, on dividing with By the arithmetic-geometric mean inequality, and by (3.11) we get √ ber This is equivalent to (3.10), which proves the theorem. 2 is an operator such that either (3.6) or holds true, then and ber ( |A| ) .
Proof If we put µ = ψ + φ 2 and r = 1 2 |ψ − φ| , then |µ| It is easy by applying Theorem 2 to see that under condition (3.12) we have By considering all these and applying Theorem 3 , we obtain the desired results. 2 The next result maybe of interest as well.

Theorem 4 Let A : H (Ω) → H (Ω) be a nonzero bounded linear operator and
Proof From the proof of Theorem 2, we have for all λ ∈ Ω. Now, after dividing (3.14) by |µ| by (3.14), is positive), we obtain for all λ ∈ Ω. Hence, for all λ ∈ Ω. If we subtract in (3.16) the same quantity |µ| from both sides, then we have

by (3.17) we obtain
which implies the inequality for all λ ∈ Ω, which implies that as desired. The proof is complete. 2 is an operator such that either (3.6) or (3.12) holds true, then ber (A) .
which is equivalent to Our next result is based on the following refinement of Schwartz's inequality obtained by Dragomir [11, Inequality (3.19) was proved by a different method earlier by Buzano [7]. Now we are ready to state our result.
Proof Let us choose in (3.19) e = k λ , a = A k λ , and b = A * k λ to get for all λ ∈ Ω. From this we have or all λ ∈ Ω, which obviously implies the desired result.

Other Berezin number inequalities for product of operators
The main goal of this section is to find upper bounds for the Berezin number of A α XB α and A α XB 1−α for the case when 0 ≤ α ≤ 1.
The following lemma is a consequence of the classical Jensen and Young inequalities [11]. By using this lemma, we prove the next results. ) α for all 0 ≤ α ≤ 1.
Proof By using the Cauchy-Schwarz inequality, we have and so ] r/2 for all λ ∈ Ω. From the McCarthy inequality and Lemma 3, we obtain for all λ ∈ Ω. From the concavity of t α , we have for all λ ∈ Ω. Combining (4.1) and (4.2), we get ) α for all λ ∈ Ω. Taking the supremum in the last inequality, we get for all r ≥ 2 and 0 ≤ α ≤ 1 .
Proof By using the Cauchy-Schwarz inequality, as in the proof of Theorem 8, we have and therefore ) r/2 ( A 2α (λ) ) r/2 for all λ ∈ Ω. Then we get from the McCarthy inequality and Lemma 3 that ) α ( B r (λ) ) for all λ ∈ Ω. Taking the supremum in the last inequality, we obtain ber r ( A α XB 1−α ) ≤ ∥X∥ r ber (αA r + (1 − α) B r ) for all positive operators A, B ∈ B (H (Ω)) . This proves the theorem. 2