Some Berezin Number Inequalities for Operator Matrices
By M. Bakherad. 2018
52 citations recorded by the source corpus.
AbstractThe Berezin symbol à of an operator A acting on the reproducing kernel Hilbert space $$\mathcal{H}=\mathcal{H}(\Omega)$$H=H(Ω) over some (nonempty) set is defined by $$\tilde A(\lambda ) = \left\langle {A\hat k_\lambda ,\hat k_\lambda } \right\rangle $$A∼(λ)=⟨Ak^λ,k^λ⟩, λ ∈ Ω, where $$\hat k_\lambda = k_\lambda /\left\| {k_\lambda } \right\|$$k^λ=kλ/∥kλ∥ is the normalized reproducing kernel of $$\mathcal{H}$$H. The Berezin number of the operator A is defined by $$ber(A) = \mathop {\sup }\limits_{\lambda \in \Omega } \left| {\tilde A(\lambda )} \right| = \mathop {\sup }\limits_{\lambda \in \Omega } \left| {\left\langle {A\hat k_\lambda ,\hat k_\lambda } \right\rangle } \right|$$ber(A)=supλ∈Ω|A∼(λ)|=supλ∈Ω|⟨Ak^λ,k^λ⟩|. Moreover, ber(A) ⩽ w(A) (numerical radius). We present some Berezin number inequalities. Among other inequalities, it is shown that if $$T = \left[ {\begin{array}{*{20}c} A & B \\ C & D \\ \end{array} } \right] \in \mathbb{B}(\mathcal{H}(\Omega _1 ) \oplus \mathcal{H}(\Omega _2 ))$$T=[ABCD]∈B(H(Ω1)⊕H(Ω2)), then $$ber(T) \leqslant \frac{1} {2}(ber(A) + ber(D)) + \frac{1} {2}\sqrt {(ber(A) - ber(D))^2 + \left( {\left\| B \right\| + \left\| C \right\|} \right)^2 } .$$ber(T)⩽12(ber(A)+ber(D))+12(ber(A)−ber(D))2+(∥B∥+∥C∥)2.
Subfields: Applied Mathematics, Computational Theory and Mathematics
View Paper: https://www.semanticscholar.org/paper/a00531cb8d050026d0ae48e5e706a84dd400219b
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In Applied Mathematics
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Citations in the corpus
- Improvements of Berezin number inequalities (2018, background): “pplied in the various questions of analysis and uniquely determines the operator (i.e., for all λ ∈ Ω,Ae(λ) = Be(λ) implies A = B). For further information about Berezin symbol we refer the reader to [1, 11, 12, 15] and references therein. In [3], the author showed some Berezin number inequalities a”