Berezin number inequalities for operators

Abstract The Berezin transform Ã of an operator A, acting on the reproducing kernel Hilbert space ℋ = ℋ (Ω) over some (non-empty) set Ω, is defined by Ã(λ) = 〉Aǩ λ, ǩ λ〈 (λ ∈ Ω), where k⌢λ=kλ‖ kλ ‖ ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} _\lambda } = {{{k_\lambda }} \over {\left\| {{k_\lambda }} \right\|}}$ is the normalized reproducing kernel of ℋ. The Berezin number of an operator A is defined by ber(A)=supλ∈Ω| A˜(λ) |=supλ∈Ω| 〈 Ak⌢λ,k⌢λ 〉 | ${\bf{ber}}{\rm{(}}A) = \mathop {\sup }\limits_{\lambda \in \Omega } \left| {\tilde A(\lambda )} \right| = \mathop {\sup }\limits_{\lambda \in \Omega } \left| {\left\langle {A{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} }_\lambda },{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} }_\lambda }} \right\rangle } \right|$ . In this paper, we prove some Berezin number inequalities. Among other inequalities, it is shown that if A, B, X are bounded linear operators on a Hilbert space ℋ, then ber(AX±XA)⩽ber12(A*A+AA*)ber12(X*X+XX*) $${\bf{ber}}(AX \pm XA) \leqslant {\bf{be}}{{\bf{r}}^{{1 \over 2}}}\left( {A*A + AA*} \right){\bf{be}}{{\bf{r}}^{{1 \over 2}}}\left( {X*X + XX*} \right)$$ and ber2(A*XB)⩽‖ X ‖2ber(A*A)ber(B*B). $${\bf{be}}{{\bf{r}}^2}({A^*}XB) \leqslant {\left\| X \right\|^2}{\bf{ber}}({A^*}A){\bf{ber}}({B^*}B).$$ We also prove the multiplicative inequality ber(AB)⩽ber(A)ber(B) $${\bf{ber}}(AB){\bf{ber}}(A){\bf{ber}}(B)$$


Introduction
Let H be a complex Hilbert space and B(H ) denote the C * -algebra of all bounded linear operators on H with the identity I. In the case when dimH = n, we identify B(H ) with the matrix algebra Mn of all n × n matrices having entries in the complex eld C. The numerical range and numerical radius of A ∈ B(H ) are de ned by W(A) := Ax, x : x ∈ H , x = and w(A) := sup |λ| : λ ∈ W(A) , respectively. It is well known that w( · ) de nes a norm on B(H ), which is equivalent to the usual operator norm · . In fact, for any A ∈ B(H ), A w(A) A (see [9, p. 9]). A functional Hilbert space is a Hilbert space H = H (Ω) of complex-valued functions on a (non-empty) set Ω, which has the property that point evaluations are continuous, i.e., for each λ ∈ Ω the map f −→ f (λ) is a continuous (linear) functional on H . Then the Riesz representation theorem ensures that for each λ ∈ Ω there is a unique element k λ of H such that f (λ) = f , k λ for all f ∈ H . The collection {k λ : λ ∈ Ω} is called the reproducing kernel of H . If {en} is an orthonormal basis for a functional Hilbert space H , then the reproducing kernel of H is given by k λ (z) = n en(λ)en(z). For λ ∈ Ω, letk λ = k λ k λ be the normalized reproducing kernel of H . For a bounded linear operator A on H , the function A de ned on Ω by A(λ) = Ak λ ,k λ , is the Berezin transform of A, which was introduced by Berezin [4,5]. The Berezin set and the Berezin number of the operator A are de ned by (see [12])  [16] showed that for A = S ⊗ S ∈ B(H ), where S is the shift operator de ned by Sf (z) = zf (z) on the Hardy-Hilbert space H = H (D) over the unit disc D = {z ∈ C : |z| < }, we have A(λ) = |λ| ( − |λ| ), and thus Ber(A) = , [ , ] = W(A) and ber(A) = < = w(A). Moreover, the Berezin number of an operator A satis es the following properties: (iii) ber(A + B) ber(A) + ber(B).
The Berezin transform has been investigated in detail for the Toeplitz and Hankel operators on the Hardy and Bergman spaces. It is widely applied in the various questions of analysis and uniquely determines the operator acting on the reproducing kernel Hilbert space of analytic functions in some suitable set Ω (i.e., For further information about the Berezin transform and Berezin number we refer the reader to [6-8, 13-18, 20-22, 24] and references therein. In this paper, by using some ideas of [1,2,23], we present several Berezin number inequalities. In particular, we show that Theorem: Let A, B, X, Y ∈ B(H ). Then

Some Upper bounds for ber(AX ± XA * )
To prove our rst result, we need the following lemma. where Re(X) = X+X * and Im(X) = X−X * i .
Proof. It follows from (see [23]) If we replace X by iX in the rst equation, we get the second equation.
Remark 2.2. If X = L + iK is the cartesian decomposition of the operator X, i.e., L = X+X * and K = X−X * i , then using this fact and we have ber(L) = ber Re(X) ber(X) ber (L) + ber (K). Now, by applying Lemma 2.1, we show an upper bound for ber(AX ± XA * ).
where X = L + iK is the cartesian decomposition of the operator X.
3 includes a special case as follows. Upper bounds for ber(AX ± XA) and ber(A * XB ± B * YA) The following theorem gives some upper bounds for ber(AX ± XA).

Now, according to the inequality
and a similar argument of the proof of part (i), we get the second inequality.
For the special case A = I, we have the next result.
Also, we have In the special case of Theorem 3.5, for X = I we obtain the next result.       Hence, we get the second inequality. Proof. If we put B = I and X = Y in Theorem 3.8(ii), then we reach the rst inequality and if we take X = Y = I in Theorem 3.8(ii), then we get the second inequality.
Note that Berezin transform is not, in general, multiplicative, i.e., AB = A B (for more information, see Kilič [18]). So, it is natural to ask: when the multiplicative inequality

ber(AB) ≤ ber(A)ber(B)
is held? Our next result proves the inequality ber(AB) ber(A)ber(B) for some operators. Proof. It follows from A * = A that for all λ ∈ Ω we have from which, by using the hypotheses of the theorem, we have that there exists a point ξ ∈ ∂Ω such that lim λ→ξ | AB(λ)| ber(A)ber(B), as desired. The second assertion of the theorem is immediate from the rst one.
Note that all Toeplitz operators Tφ, φ ∈ L ∞ (∂D), on H satisfy the conditions for almost all t ∈ [ , π); see Engliš [6]. Also, the operators of the form Tφ + K, where K is compact, satisfy the condition of Proposition 4.1. Recall that the reproducing kernel Hilbert space H = H (Ω) is called standard [20], ifk λ → weakly whenever λ tends to the boundary of Ω. Then K(λ) vanishes on the boundary for any compact operator K on H . This easily shows that if A or B is a compact operator on the standard reproducing kernel Hilbert space H , then AB(λ) → , whenever λ → ∂Ω, that is A and B satisfy the hypothesis of Proposition 4.2. There are also the following nontrivial examples. Proof. In fact, we have: whence J * T * θ (λ) → , whenever λ → ∂D. Since J = I, J is invertible on H and T * θ is not compact on H , and so J * T * θ is not compact on H .
There is also an unbounded operator satisfying the hypothesis of Proposition 4.2. Namely, let D := d dz denote the di erentiation operator on H , and let J be the same as in Example 4.3. Then we have → as |λ| → − , as desired.