Results 11 to 20 of about 8,614,347 (174)
Some new relations between the Berezin number and the Berezin norm of operators
Rocky Mountain Journal of Mathematics, 2022 For a bounded linear operator \(A\) on a reproducing kernel Hilbert space with reproducing kernel \(K(z,w)\), the Berezin symbol, Berezin norm and Berezin number of \(A\) are defined, respectively, by \[ \tilde A(w):=\langle Ak_w,k_w\rangle, \quad \|A\|_{Ber} := \sup_w \|Ak_w\|, \quad \operatorname{ber}(A) :=\sup_w |\tilde A(w)|, \] where \(k_w(z):=K(z,
Çalisir, Irem +2 more
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Berezin number inequalities via convex functions
Filomat, 2022 The Berezin symbol ?A of an operator A on the reproducing kernel Hilbert space H (?) over some set ? with the reproducing kernel k? is defined by ? (?) = ?A k?/||k?||, k?/||k?||?, ? ? ?. The Berezin number of an operator A is defined by ber(A) := sup ??? |?(?)|.
Başaran, Hamdullah +2 more
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Advanced refinements of Berezin number inequalities
Communications Faculty Of Science University of Ankara Series A1Mathematics and Statistics, 2023 For a bounded linear operator $A$ on a functional Hilbert space $\mathcal{H}\left( \Omega\right) $, with normalized reproducing kernel $\widehat {k}_{\eta}:=\frac{k_{\eta}}{\left\Vert k_{\eta}\right\Vert _{\mathcal{H}}},$ the Berezin symbol and Berezin number are defined respectively by $\widetilde{A}\left( \eta\right) :=\left\langle A\widehat{k}_{\
Mehmet GÜRDAL, Hamdullah BAŞARAN
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Refinements of Kantorovich type, Schwarz and Berezin number inequalities [PDF]
Extracta Mathematicae, 2020 In this article, we use Kantorovich and Kantorovich type inequalities in order to prove some new Berezin number inequalities. Also, by using a refinement of the classical Schwarz inequality, we prove Berezin number inequalities for powers of f (A), where
M. Garayev +3 more
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New estimations for the Berezin number inequality [PDF]
Journal of Inequalities and Applications, 2020 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 X , Y , Z ∈ L ( H ) $X, Y, Z\in {\mathcal{L}}(\mathcal{H})$ such that X and Y are positive operators,
Mojtaba Bakherad, Ulas Yamancı
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Proceedings of the Institute of Mathematics and Mechanics, National Academy of Sciences of Azerbaijan, 2022 Summary: We introduce the notions \((A,r)\)-adjoint of operators and \(A\) Berezin number of operators on the reproducing kernel Hilbert space and prove some inequalities for \(A\)-Berezin number of operators. Some other related questions are also discussed.
GÜRDAL, Mehmet, Basaran, Hamdullah
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FURTHER BEREZIN NUMBER INEQUALITIES OF OPERATOR MATRICES [PDF]
, 2023 In this paper, we have some inequalities for the Berezin number of operator matrices using the convex functions. Also, we obtain some upper bounds for the Berezin number of operator matrices.
Guesba, Messaoud, Yamanci, Ulas
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REVERSE INEQUALITIES FOR THE BEREZIN NUMBER OF OPERATORS
Proceedings of the Institute of Mathematics and Mechanics, National Academy of Sciences of Azerbaijan, 2022 For a bounded linear operator $A$ on a reproducing kernel Hilbert space $\mathcal{H}(Ω)$, with normalized reproducing kernel $\widehat{k}_λ = \frac{k_λ}{\lVert k_λ\lVert}$, the Berezin symbol, Berezin number and Berezin norm are defined respectively by $\widetilde{A}(λ) = \langle A\widehat{k}_λ,\widehat{k}_λ\rangle$, $ber(A) = \sup_{λ\inΩ}\left ...
Garayev, Mubariz +2 more
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More Correct Berezin Symbol Inequalities
Dera Natung Government College Research Journal, 2023 The purpose of this research is to show bounds for some Berezin number inequalities in an innovative approach. Some inequalities have been proven using the improvement of the Hermite-Hadamard inequality.
Hamdullah Başaran, Mehmet Gurdal
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An extension of the Euclidean Berezin number
Filomat, 2023 The Berezin transform ? of an operator A, acting on the reproducing kernel Hilbert space H = H(?) over some (non-empty) set ?, is defined by ?(?) = ?A?k?,?k?? (? ? ?), where ?k? = k?/?k?? is the normalized reproducing kernel of H. The Berezin number of an operator A is defined by ber(A) = sup ??? ??(?)? = sup ??? ??A?k?,?k???.
Nooshin Eslami Mahdiabadi +1 more
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